Reza N. Jazar
Vehicle Dynamics
Theory and Application
2nd Edition
Vehicle Dynamics
Reza N. Jazar
Vehicle Dynamics
Theory and Application
Second Edition
Reza N. Jazar
School of Aerospace, Mechanical and
Manufacturing Engineering
RMIT University
Bundoora, VIC
Australia
ISBN 978-1-4614-8544-5 (eBook)
ISBN 978-1-4614-8543-8
DOI 10.1007/978-1-4614-8544-5
Springer New York Heidelberg Dordrecht London
Library of Congress Control Number: 2013951659
© Springer Science+Business Media New York 2014
This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part
of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations,
recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission
or information storage and retrieval, electronic adaptation, computer software, or by similar or
dissimilar methodology now known or hereafter developed. Exempted from this legal reservation
are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically
for the purpose of being entered and executed on a computer system, for exclusive use by the
purchaser of the work. Duplication of this publication or parts thereof is permitted only under the
provisions of the Copyright Law of the Publisher’s location, in its current version, and permission
for use must always be obtained from Springer. Permissions for use may be obtained through
RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the
respective Copyright Law.
The use of general descriptive names, registered names, trademarks, service marks, etc. in this
publication does not imply, even in the absence of a specific statement, that such names are exempt
from the relevant protective laws and regulations and therefore free for general use.
While the advice and information in this book are believed to be true and accurate at the date of
publication, neither the authors nor the editors nor the publisher can accept any legal responsibility
for any errors or omissions that may be made. The publisher makes no warranty, express or implied,
with respect to the material contained herein.
Printed on acid-free paper
Springer is part of Springer Science+Business Media (www.springer.com)
Contents
Preface
xvii
1 Tire and Rim Fundamentals
1.1 Tires and Sidewall Information . . . . . . . . . . . . . . . .
1.2 Tire Components . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Radial and Non-Radial Tires . . . . . . . . . . . . . . . . .
1.4 Tread . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Tireprint . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Wheel and Rim . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Vehicle Classications . . . . . . . . . . . . . . . . . . . . .
1.7.1 ISO and FHWA Classication . . . . . . . . . . . . .
1.7.2 Passenger Car Classications . . . . . . . . . . . . .
1.7.3 Passenger Car Body Styles . . . . . . . . . . . . . .
1.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
11
15
18
20
21
26
26
29
31
32
34
35
I
37
Vehicle Motion
2 Forward Vehicle Dynamics
2.1 Parked Car on a Level Road . . . . . . . . . . . . . . . . . .
2.2 Parked Car on an Inclined Road . . . . . . . . . . . . . . .
2.3 Accelerating Car on a Level Road . . . . . . . . . . . . . . .
2.4 Accelerating Car on an Inclined Road . . . . . . . . . . . .
2.5 Parked Car on a Banked Road . . . . . . . . . . . . . . . .
2.6 F Optimal Drive and Brake Force Distribution . . . . . . .
2.7 F Vehicles With More Than Two Axles . . . . . . . . . . .
2.8 F Vehicles on a Crest and Dip . . . . . . . . . . . . . . . .
2.8.1 F Vehicles on a Crest . . . . . . . . . . . . . . . . .
2.8.2 F Vehicles on a Dip . . . . . . . . . . . . . . . . . .
2.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.10 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
39
45
50
55
66
70
76
80
81
86
88
90
92
3 Tire Dynamics
99
3.1 Tire Coordinate Frame and Tire Force System . . . . . . . 99
3.2 Tire Stiness . . . . . . . . . . . . . . . . . . . . . . . . . . 103
v
Contents
vi
3.3
3.4
Eective Radius . . . . . . . . . . . . . . . . . . . . . . . .
F Tireprint Forces of a Static Tire . . . . . . . . . . . . . .
3.4.1 F Static Tire, Normal Stress . . . . . . . . . . . . .
3.4.2 F Static Tire, Tangential Stresses . . . . . . . . . .
3.5 Rolling Resistance . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 Eect of Speed on the Rolling Friction Coe!cient .
3.5.2 Eect of In ation Pressure and Load on the Rolling
Friction Coe!cient . . . . . . . . . . . . . . . . . . .
3.5.3 F Eect of Sideslip Angle on Rolling Resistance . .
3.5.4 F Eect of Camber Angle on Rolling Resistance . .
3.6 Longitudinal Force . . . . . . . . . . . . . . . . . . . . . . .
3.7 Lateral Force . . . . . . . . . . . . . . . . . . . . . . . . . .
3.8 Camber Force . . . . . . . . . . . . . . . . . . . . . . . . . .
3.9 Tire Force . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.11 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
107
122
123
126
128
131
135
138
138
139
148
158
164
170
172
174
4 Driveline Dynamics
179
4.1 Engine Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 179
4.2 Driveline and E!ciency . . . . . . . . . . . . . . . . . . . . 186
4.3 Gearbox and Clutch Dynamics . . . . . . . . . . . . . . . . 192
4.4 Gearbox Design . . . . . . . . . . . . . . . . . . . . . . . . . 200
4.4.1 Geometric Ratio Gearbox Design . . . . . . . . . . . 201
4.4.2 F Progressive Ratio Gearbox Design . . . . . . . . . 215
4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
4.6 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 220
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222
II
Vehicle Kinematics
231
5 F Applied Kinematics
233
5.1 Rotation About Global Cartesian Axes . . . . . . . . . . . . 233
5.2 Successive Rotation About Global Cartesian Axes . . . . . 238
5.3 Rotation About Local Cartesian Axes . . . . . . . . . . . . 239
5.4 Successive Rotation About Local Cartesian Axes . . . . . . 243
5.5 General Transformation . . . . . . . . . . . . . . . . . . . . 251
5.6 Local and Global Rotations . . . . . . . . . . . . . . . . . . 258
5.7 Axis-angle Rotation . . . . . . . . . . . . . . . . . . . . . . 259
5.8 Rigid Body Motion . . . . . . . . . . . . . . . . . . . . . . . 264
5.9 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . 267
5.10 F Time Derivative and Coordinate Frames . . . . . . . . . 275
5.11 Rigid Body Velocity . . . . . . . . . . . . . . . . . . . . . . 284
5.12 Angular Acceleration . . . . . . . . . . . . . . . . . . . . . . 288
Contents
5.13 Rigid Body Acceleration . . . . . . . . . . . . . . . . . . . .
5.14 F Screw Motion . . . . . . . . . . . . . . . . . . . . . . . .
5.15 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.16 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vii
293
296
309
312
313
6 Applied Mechanisms
319
6.1 Four-Bar Linkage . . . . . . . . . . . . . . . . . . . . . . . . 319
6.2 Slider-Crank Mechanism . . . . . . . . . . . . . . . . . . . . 339
6.3 Inverted Slider-Crank Mechanism . . . . . . . . . . . . . . . 346
6.4 Instant Center of Rotation . . . . . . . . . . . . . . . . . . . 352
6.5 Coupler Point Curve . . . . . . . . . . . . . . . . . . . . . . 364
6.5.1 Coupler Point Curve for Four-Bar Linkages . . . . . 364
6.5.2 Coupler Point Curve for a Slider-Crank Mechanism . 366
6.5.3 Coupler Point Curve for Inverted Slider-Crank Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . 370
6.6 F Universal Joint . . . . . . . . . . . . . . . . . . . . . . . 371
6.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
6.8 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 381
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
7 Steering Dynamics
387
7.1 Kinematic Steering . . . . . . . . . . . . . . . . . . . . . . . 387
7.2 Vehicles with More Than Two Axles . . . . . . . . . . . . . 404
7.3 F Vehicle with Trailer . . . . . . . . . . . . . . . . . . . . . 407
7.4 Steering Mechanisms . . . . . . . . . . . . . . . . . . . . . . 411
7.5 F Four wheel steering. . . . . . . . . . . . . . . . . . . . . . 417
7.6 F Road Design . . . . . . . . . . . . . . . . . . . . . . . . . 434
7.7 F Steering Mechanism Optimization . . . . . . . . . . . . . 461
7.8 F Trailer-Truck Kinematics . . . . . . . . . . . . . . . . . . 469
7.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
7.10 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 484
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486
8 Suspension Mechanisms
497
8.1 Solid Axle Suspension . . . . . . . . . . . . . . . . . . . . . 497
8.2 Independent Suspension . . . . . . . . . . . . . . . . . . . . 508
8.3 Roll Center and Roll Axis . . . . . . . . . . . . . . . . . . . 513
8.4 F Car Tire Relative Angles . . . . . . . . . . . . . . . . . . 524
8.4.1 F Toe . . . . . . . . . . . . . . . . . . . . . . . . . . 527
8.4.2 F Caster Angle . . . . . . . . . . . . . . . . . . . . . 529
8.4.3 F Camber . . . . . . . . . . . . . . . . . . . . . . . 530
8.4.4 F Thrust Angle . . . . . . . . . . . . . . . . . . . . 532
8.5 F Suspension Requirements and Coordinate Frames . . . . 533
8.5.1 Kinematic Requirements . . . . . . . . . . . . . . . . 533
Contents
viii
8.5.2 Dynamic Requirements . . . . . . . . . . . . . . . .
8.5.3 Wheel, wheel-body, and tire Coordinate Frames . . .
8.6 F Caster Theory . . . . . . . . . . . . . . . . . . . . . . . .
8.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.8 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
III
Vehicle Dynamics
534
534
544
554
557
559
567
9 F Applied Dynamics
569
9.1 Elements of Dynamics . . . . . . . . . . . . . . . . . . . . . 569
9.1.1 Force and Moment . . . . . . . . . . . . . . . . . . . 569
9.1.2 Momentum . . . . . . . . . . . . . . . . . . . . . . . 570
9.1.3 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 571
9.1.4 Equation of Motion . . . . . . . . . . . . . . . . . . 573
9.1.5 Work and Energy . . . . . . . . . . . . . . . . . . . . 573
9.2 Rigid Body Translational Dynamics . . . . . . . . . . . . . 579
9.3 Rigid Body Rotational Dynamics . . . . . . . . . . . . . . . 582
9.4 Mass Moment Matrix . . . . . . . . . . . . . . . . . . . . . 593
9.5 Lagrange’s Form of Newton’s Equations of Motion . . . . . 603
9.6 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . 610
9.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 620
9.8 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 623
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 624
10 Vehicle Planar Dynamics
631
10.1 Vehicle Coordinate Frame . . . . . . . . . . . . . . . . . . . 631
10.2 Rigid Vehicle Newton-Euler Dynamics . . . . . . . . . . . . 637
10.3 Force System Acting on a Rigid Vehicle . . . . . . . . . . . 644
10.3.1 Tire Force and Body Force Systems . . . . . . . . . 644
10.3.2 Tire Lateral Force . . . . . . . . . . . . . . . . . . . 648
10.3.3 Two-wheel Model and Body Force Components . . . 649
10.4 Two-wheel Rigid Vehicle Dynamics . . . . . . . . . . . . . . 659
10.5 Steady-State Turning . . . . . . . . . . . . . . . . . . . . . . 670
10.6 F Linearized Model for a Two-Wheel Vehicle . . . . . . . . 695
10.7 F Transient Response . . . . . . . . . . . . . . . . . . . . . 699
10.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 727
10.9 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 729
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 731
11 F Vehicle Roll Dynamics
741
11.1 F Vehicle Coordinate and DOF . . . . . . . . . . . . . . . . 741
11.2 F Equations of Motion . . . . . . . . . . . . . . . . . . . . 742
11.3 F Vehicle Force System . . . . . . . . . . . . . . . . . . . . 746
Contents
11.3.1 F Tire and Body Force Systems . . . . . . . . . . .
11.3.2 F Tire Lateral Force . . . . . . . . . . . . . . . . . .
11.3.3 F Body Force Components on a Two-wheel Model .
11.4 F Two-wheel Rigid Vehicle Dynamics . . . . . . . . . . . .
11.5 F Steady-State Motion . . . . . . . . . . . . . . . . . . . .
11.6 F Time Response . . . . . . . . . . . . . . . . . . . . . . .
11.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.8 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
IV
Vehicle Vibration
ix
746
749
752
759
763
767
782
783
786
795
12 Applied Vibrations
797
12.1 Mechanical Vibration Elements . . . . . . . . . . . . . . . . 797
12.2 Newton’s Method and Vibrations . . . . . . . . . . . . . . . 805
12.3 Frequency Response of Vibrating Systems . . . . . . . . . . 812
12.3.1 Forced Excitation . . . . . . . . . . . . . . . . . . . 813
12.3.2 Base Excitation . . . . . . . . . . . . . . . . . . . . . 823
12.3.3 Eccentric Excitation . . . . . . . . . . . . . . . . . . 835
12.3.4 F Eccentric Base Excitation . . . . . . . . . . . . . 841
12.3.5 F Classication for the Frequency Responses of OneDOF Forced Vibration Systems . . . . . . . . . . . . 847
12.4 Time Response of Vibrating Systems . . . . . . . . . . . . . 852
12.5 Vibration Application and Measurement . . . . . . . . . . . 864
12.6 F Vibration Optimization Theory . . . . . . . . . . . . . . 869
12.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 880
12.8 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 882
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885
13 Vehicle Vibrations
891
13.1 Lagrange Method and Dissipation Function . . . . . . . . . 891
13.2 F Quadratures . . . . . . . . . . . . . . . . . . . . . . . . . 901
13.3 Natural Frequencies and Mode Shapes . . . . . . . . . . . . 908
13.4 Bicycle Car and Body Pitch Mode . . . . . . . . . . . . . . 915
13.5 Half Car and Body Roll Mode . . . . . . . . . . . . . . . . . 920
13.6 Full Car Vibrating Model . . . . . . . . . . . . . . . . . . . 925
13.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 933
13.8 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . 934
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 936
14 Suspension Optimization
939
14.1 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . 939
14.2 Frequency Response . . . . . . . . . . . . . . . . . . . . . . 945
14.3 RMS Optimization . . . . . . . . . . . . . . . . . . . . . . . 949
Contents
x
14.4 F Time Response Optimization . . . . . . . . . . . . . . . .
14.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14.6 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
972
978
979
981
15 F Quarter Car Model
985
15.1 Mathematical Model . . . . . . . . . . . . . . . . . . . . . . 985
15.2 Frequency Response . . . . . . . . . . . . . . . . . . . . . . 987
15.3 F Natural and Invariant Frequencies . . . . . . . . . . . . . 994
15.4 F RMS Optimization . . . . . . . . . . . . . . . . . . . . 1006
15.5 F Optimization Based on Natural Frequency and Wheel
Travel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1016
15.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022
15.7 Key Symbols . . . . . . . . . . . . . . . . . . . . . . . . . 1023
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1025
References
1027
A Frequency Response Curves
1031
B Trigonometric Formulas
1037
C Unit Conversions
1041
Index
1045
Dedicated to
my son, Kavosh,
my daughter, Vazan,
and my wife, Mojgan.
Nature is not perfect, not even optimum.
Preface to the Second Edition
The second edition of this textbook would not be possible without comments and contribution of my students and colleagues, especially those at
Columbia University in New York and RMIT University in Melbourne. I
am deeply thankful of my friend Mr. Stefan Anthony who shared with me
his 50 years of experiences in automotive industries. Special thanks to Andy
Fu and Hormoz Marzbani who kindly read the book carefully and caught
typos as well as logical mismatches.
New topics introduced in this edition are results of students’ feedback
which assisted me in clarifying and better presenting some aspects of this
book.
The intent of this book is to explain vehicle dynamics in a manner I
would have liked it explained to me as a student. This book can now help
students and engineers by being a great reference that covers all aspects
of vehicle dynamics and providing students with detailed explanations and
information.
The first edition of this book was published in 2008 by Springer. Soon
after its publication, the book become very popular in the automotive field.
It was appreciated by many students and instructors in addition to my own
students and colleagues. Their questions, comments, and suggestions have
helped me in creating the second edition.
xv
Preface
This text is for engineering students. It introduces the fundamental knowledge used in vehicle dynamics. This knowledge can be utilized to develop
computer programs for analyzing the ride, handling, and optimization of
road vehicles.
Vehicle dynamics has been in the engineering curriculum for more than
a hundred years. Books on the subject are available, but most of them
are written for specialists and are not suitable for a classroom application.
A new student, engineer, or researcher would not know where and how
to start learning vehicle dynamics. So, there is a need for a textbook for
beginners. This textbook presents the fundamentals with a perspective on
future trends.
The study of classical vehicle dynamics has its roots in the work of
great scientists of the past four centuries and creative engineers in the
past century who established the methodology of dynamic systems. The
development of vehicle dynamics has moved toward modeling, analysis,
and optimization of multi-body dynamics supported by some compliant
members. Therefore, merging dynamics with optimization theory was an
expected development. The fast-growing capability of accurate positioning,
sensing, and calculations, along with intelligent computer programming are
the other important developments in vehicle dynamics. So, a textbook help
the reader to make a computer model of vehicles, which this book does.
Level of the Book
This book has evolved from nearly a decade of research in nonlinear
dynamic systems and teaching courses in vehicle dynamics. It is addressed
primarily to the last year of undergraduate study and the rst year graduate
student in engineering. Hence, it is an intermediate textbook. It provides
both fundamental and advanced topics. The whole book can be covered
in two successive courses, however, it is possible to jump over some sections and cover the book in one course. Students are required to know the
fundamentals of kinematics and dynamics, as well as a basic knowledge of
numerical methods.
The contents of the book have been kept at a fairly theoretical-practical
level. Many concepts are deeply explained and their application emphasized, and most of the related theories and formal proofs have been explained. The book places a strong emphasis on the physical meaning and
applications of the concepts. Topics that have been selected are of high
xvii
xviii
Preface
interest in the eld. An attempt has been made to expose students to a
broad range of topics and approaches.
There are four special chapters that are indirectly related to vehicle dynamics: Applied Kinematics, Applied Mechanisms, Applied Dynamics, and
Applied Vibrations. These chapters provide the related background to understand vehicle dynamics and its subsystems.
Organization of the Book
The text is organized so it can be used for teaching or for self-study.
Chapter 1 “Fundamentals,” contains general preliminaries about tire and
rim with a brief review of road vehicle classications.
Part L “One Dimensional Vehicle Dynamics,” presents forward vehicle
dynamics, tire dynamics, and driveline dynamics. Forward dynamics refers
to weight transfer, accelerating, braking, engine performance, and gear ratio
design.
Part LL “Vehicle Kinematics,” presents a detailed discussion of vehicle
mechanical subsystems such as steering and suspensions.
Part LLL “Vehicle Dynamics,” employs Newton and Lagrange methods
to develop the maneuvering dynamics of vehicles.
Part LY “Vehicle Vibrations,” presents a detailed discussion of vehicle vibrations. An attempt is made to review the basic approaches and
demonstrate how a vehicle can be modeled as a vibrating multiple degreeof-freedom system. The concepts of the Newton-Euler dynamics and Lagrangian method are used equally for derivation of equations of motion.
The RMS optimization technique for suspension design of vehicles is introduced and applied to vehicle suspensions. The outcome of the optimization
technique is the optimal stiness and damping for a car or suspended equipment.
Method of Presentation
This book uses a "fact-reason-application" structure. The "fact" is the
main subject we introduce in each section. Then the reason is given as a
"proof." The application of the fact is examined in some "examples." The
"examples" are a very important part of the book because they show how
to implement the "facts." They also cover some other facts that are needed
to expand the subject.
Prerequisites
Since the book is written for senior undergraduate and rst-year graduatelevel students of engineering, the assumption is that users are familiar with
matrix algebra as well as basic dynamics. Prerequisites are the fundamentals of kinematics, dynamics, vector analysis, and matrix theory. These
basics are usually taught in the rst three undergraduate years.
Preface
xix
Unit System
The system of units adopted in this book is, unless otherwise stated, the
international system of units (SI). The units of degree (deg) or radian ( rad)
are utilized for variables representing angular quantities.
Symbols
• Lowercase bold letters indicate a vector. Vectors may be expressed in
an  dimensional Euclidian space. Example:
,
,
,
r
p
ω
,
,
,
s
q
α
,
,
,
d
v
²
,
,
,
a
w
θ
b
y
δ
,
,
,
c
z
φ
• Uppercase bold letters indicate a dynamic vector or a dynamic matrix, such as force and moment. Example:
F
M
• Lowercase letters with a hat indicate a unit vector. Unit vectors are
not bold. Example:
ˆ~
Lˆ
,
,
ˆ
Mˆ
,
,
n̂
N̂
,
,
x̂
x̂
,
,
x̂
x̂*
,
,
q̂
x̂#
• Lowercase letters with a tilde indicate a 3 × 3 skew symmetric matrix
associated to a vector. Example:
5
6
6
5
0
d3 d2
d1
0
d1 8
d̃ = 7 d3
a = 7 d2 8
d2 d1
0
d3
• An arrow above two uppercase letters indicates the start and end
points of a position vector. Example:
$
RQ = a position vector from point R to point Q
• The length of a vector is indicated by a non-bold lowercase letter.
Example:
u = |r|
d = |a|
e = |b|
v = |s|
• Capital letter E is utilized to denote a body coordinate frame. Example:
E(r{|})
E(R{|})
E1 (r1 {1 |1 }1 )
xx
Preface
• Capital letter J is utilized to denote a global, inertial, or xed coordinate frame. Example:
J
J([\ ])
J(R[\ ])
• Right subscript on a transformation matrix indicates the departure
frames. Example:
UE = transformation matrix from frame E(r{|})
• Left superscript on a transformation matrix indicates the destination
frame. Example:
J
UE
= transformation matrix from frame E(r{|})
to frame J(R[\ ])
• Capital letter U indicates rotation or a transformation matrix, if it
shows the beginning and destination coordinate frames. Example:
5
6
cos sin 0
J
UE = 7 sin cos 0 8
0
0
1
• Whenever there is no sub or superscript, the matrices are shown in a
bracket. Example:
5
6
cos sin 0
[W ] = 7 sin cos 0 8
0
0
1
• Left superscript on a vector denotes the frame in which the vector
is expressed. That superscript indicates the frame that the vector
belongs to; so the vector is expressed using the unit vectors of that
frame. Example:
J
r = position vector expressed in frame J(R[\ ])
• Right subscript on a vector denotes the tip point that the vector is
referred to. Example:
J
rS
= position vector of point S
expressed in coordinate frame J(R[\ ])
• Right subscript on an angular velocity vector indicates the frame that
the angular vector is referred to. Example:
$ E = angular velocity of the body coordinate frame E(r{|})
Preface
xxi
• Left subscript on an angular velocity vector indicates the frame that
the angular vector is measured with respect to. Example:
J $E
= angular velocity of the body coordinate frame E(r{|})
with respect to the global coordinate frame J(R[\ ])
• Left superscript on an angular velocity vector denotes the frame in
which the angular velocity is expressed. Example:
E2
J $ E1
= angular velocity of the body coordinate frame E1
with respect to the global coordinate frame J>
and expressed in body coordinate frame E2
Whenever the subscript and superscript of an angular velocity are
the same, we usually drop the left superscript. Example:
J
J $E J $E
Also for position, velocity, and acceleration vectors, we drop the left
subscripts if it is the same as the left superscript. Example:
E
E
vS
E vS • Left superscript on derivative operators indicates the frame in which
the derivative of a variable is taken. Example:
J
J
g
{
gw
E
gE
rS
gw
gJ
rS
gw E
If the variable is a vector function, and also the frame in which the
vector is dened is the same frame in which a time derivative is taken,
we may use the following short notation,
J
gJ
rS = J rb S
gw
E
gE
rS = E
bS
r r
gw r
and write equations simpler. Example:
J
J
v=
g J
r(w) = J rb
gw
• If followed by angles, lowercase f and v denote frv and vlq functions
in mathematical equations. Example:
f = cos v* = sin *
xxii
Preface
• Capital bold letter I indicates a unit matrix, which, depending on
the dimension of the matrix equation, could be a 3 × 3 or a 4 × 4
unit matrix. I3 or I4 are also being used to clarify the dimension of
I. Example:
5
6
1 0 0
I = I3 = 7 0 1 0 8
0 0 1
• An asterisk F indicates a more advanced subject or example that is
not designed for undergraduate teaching and can be dropped in the
rst reading.
1
Tire and Rim Fundamentals
Tires are the only component of a vehicle to transfer forces between the
road and the vehicle. Tire parameters such as dimensions, maximum loadcarrying capacity, and maximum speed index are usually indicated on its
sidewall. In this chapter, we review some topics about tires, wheels, roads,
vehicles, and their interactions.
Tireprint width
hT, Section height
Sidewall
Pan width
wT, Section width
FIGURE 1.1. Cross section of a tire on a rim to show tire height and width.
1.1 Tires and Sidewall Information
Pneumatic tires are the only means to transfer forces between the road and
the vehicle. Tires are required to produce the forces necessary to control the
vehicle, and hence, they are an important component of a vehicle. Figure
1.1 illustrates a cross section view of a tire on a rim to show the dimension
parameters that are used to standard tires.
The section height, tire height, or simply height, kW , is a number that
must be added to the rim radius to make the wheel radius. The section
width, or tire width, zW , is the widest dimension of a tire when the tire is
not loaded.
Tires are required to have certain information printed on the tire sidewall.
Figure 1.2 illustrates a side view of a sample tire to show the important
information printed on a tire sidewall.
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_1, © Springer Science+Business Media New York 2014
1
2
1. Tire and Rim Fundamentals
DOT FA
CA
7C
DE
F
6
7
5
PA
1
RE
UR 40 Ps i
ESS
PR
6
2
AX
M
15
0R
4
R
S
TI
TUBELESS
96H
DA
NA
IN
9
L
RADI A
6 0R15 96
H
30
3
E
5/
P21
ED
ORC
INF
RE
125
&S
E 11 0 4
M
MA
D
1
8
P21
5/
0
A+2
FIGURE 1.2. Side view of a tire and the most important information printed on
a tire sidewall.
The codes in Figure 1.2 are:
1
2
3
4
5
6
7
8
Size number.
Maximum allowed in ation pressure.
Type of tire construction.
M&S denotes a tire for mud and snow.
E-Mark is the Europe type approval mark and number.
US Department of Transport (DOT) identication numbers.
Country of manufacture.
Manufacturers, brand name, or commercial name.
The most important information on the sidewall of a tire is the size
number, indicated by 1 . To see the format of the size number, an example
is shown in Figure 1.3 and their denitions are explained as follows.
S Tire type. The rst letter indicates the proper type of car that the
tire is made for. S stands for passenger car. The rst letter can also be
VW for special trailer, W for temporary, and OW for light truck.
215 Tire width. This three-number code is the width of the unloaded
tire from sidewall to sidewall measured in [ mm].
60 Aspect ratio. This two-number code is the ratio of the tire section
height kW to tire width zW , expressed as a percentage. Aspect ratio is shown
1. Tire and Rim Fundamentals
3
P 215 / 60 R 15 96 H
P
Passenger car
215
Tire width [mm]
60
Aspect ratio [%]
R
Radial
15
Rim diameter [in]
96
Load rating
H
Speed rating
FIGURE 1.3. A sample of a tire size number and its meaning.
by vW .
kW
× 100
(1.1)
zW
Generally speaking, tire aspect ratios range from 35, for race car tires, to
75 for tires used on utility vehicles.
U Tire construction type. The letter U indicates that the tire has
a radial construction. It may also be E for bias belt or bias ply, and
G for diagonal.
vW =
15 Rim diameter. This is a number in [ in] to indicate diameter of the
rim that the tire is designed to t on.
96 Load rate or load index. Many tires come with a service description
at the end of the tire size. The service description is made of a two-digit
number (load index) and a letter (speed rating). The load index is a representation of the maximum load each tire is designed to support. Table
1=1 shows some of the most common load indices and their load-carrying
capacities. The load index is generally valid for speeds under 210 km@ h
( 130 mi@ h).
K Speed rate. Speed rate indicates the maximum speed that the tire
can sustain for a ten minute endurance without breaking down.
Table 1=2 shows the most common speed rate indices and their meanings.
The maximum speed of a tire is related to its typical application as is shown
in Table 1=3.
Table 1=2 indicates a confusing point as why we need indices Z and \
while we have ]. It is because when the ] speed rate were rst introduced,
It was assumed to be the highest tire speed rating that would ever be
required. The ] indexed tire is capable of speeds in excess of +240 km@ h +149 mi@ h, however it is not clear how far above 240 km@ h is possible. It
was expected that vehicles will not be driven much higher than 240 km@ h.
4
1. Tire and Rim Fundamentals
Very soon vehicles with higher speeds capability appeared and therefore,
automotive industry add Z and \ indices to identify the tires that meet
the needs of new vehicles that have very high speed capabilities.
Example 1 Weight of a car and load index of its tire.
For a car that weighs 2 wrqv = 2000 kg, we need a tire with a load index
higher than 84. This is because we have about 500 kg per tire and it is in a
load index of 84 as indicated in Table 1=1.
Table 1=1 - Maximum load-carrying capacity tire index.
Index
Maximum load
Index
Maximum load
0
45 kg 99 lbf
···
···
100
800 kg 1764 lbf
71
345 kg 761 lbf
101
825 kg 1819 lbf
72
355 kg 783 lbf
102
850 kg 1874 lbf
73
365 kg 805 lbf
103
875 kg 1929 lbf
74
375 kg 827 lbf
104
900 kg 1984 lbf
75
387 kg 853 lbf
105
925 kg 2039 lbf
76
400 kg 882 lbf
106
950 kg 2094 lbf
77
412 kg 908 lbf
107
975 kg 2149 lbf
78
425 kg 937 lbf
108
1000 kg 2205 lbf
79
437 kg 963 lbf
109
1030 kg 2271 lbf
80
450 kg 992 lbf
110
1060 kg 2337 lbf
81
462 kg 1019 lbf
111
1090 kg 2403 lbf
82
475 kg 1047 lbf
113
1120 kg 2469 lbf
83
487 kg 1074 lbf
113
1150 kg 2581 lbf
84
500 kg 1102 lbf
114
1180 kg 2601 lbf
85
515 kg 1135 lbf
115
1215 kg 2679 lbf
86
530 kg 1163 lbf
116
1250 kg 2806 lbf
87
545 kg 1201 lbf
117
1285 kg 2833 lbf
88
560 kg 1235 lbf
118
1320 kg 2910 lbf
89
580 kg 1279 lbf
119
1360 kg 3074 lbf
90
600 kg 1323 lbf
120
1400 kg 3086 lbf
91
615 kg 1356 lbf
121
1450 kg 3197 lbf
92
630 kg 1389 lbf
122
1500 kg 3368 lbf
93
650 kg 1433 lbf
123
1550 kg 3417 lbf
94
670 kg 1477 lbf
124
1600 kg 3527 lbf
95
690 kg 1521 lbf
125
1650 kg 3690 lbf
96
710 kg 1565 lbf
126
1700 kg 3748 lbf
97
730 kg 1609 lbf
127
1750 kg 3858 lbf
98
750 kg 1653 lbf
128
1800 kg 3968 lbf
99
775 kg 1709 lbf
···
···
199
13600 kg 30000 lbf
1. Tire and Rim Fundamentals
5
Example 2 Height of a tire based on tire numbers.
A tire has the size number S 215@60U15 96K. The aspect ratio 60 means
the height of the tire is equal to 60% of the tire width. To calculate the tire
height in [ mm], we should multiply the rst number (215) by the second
number (60) and divide by 100.
kW = 215 ×
60
= 129 mm
100
(1.2)
This is the tire height from rim to tread.
Index
D1
D2
D3
D4
D5
D6
D7
D8
E
F
G
H
I
J
M
Table 1=2 - Maximum speed tire index.
Maximum speed
Index
Maximum speed
5 km@ h 3 mi@ h
N
110 km@ h 68 mi@ h
10 km@ h 6 mi@ h
O
120 km@ h 75 mi@ h
15 km@ h 9 mi@ h
P
130 km@ h 81 mi@ h
20 km@ h 12 mi@ h
Q
140 km@ h 87 mi@ h
25 km@ h 16 mi@ h
S
150 km@ h 93 mi@ h
30 km@ h 19 mi@ h
T
160 km@ h 100 mi@ h
35 km@ h 22 mi@ h
U
170 km@ h 106 mi@ h
40 km@ h 25 mi@ h
V
180 km@ h 112 mi@ h
50 km@ h 31 mi@ h
W
190 km@ h 118 mi@ h
60 km@ h 37 mi@ h
X
200 km@ h 124 mi@ h
65 km@ h 40 mi@ h
K
210 km@ h 130 mi@ h
70 km@ h 43 mi@ h
Y
240 km@ h 150 mi@ h
80 km@ h 50 mi@ h
Z
270 km@ h 168 mi@ h
90 km@ h 56 mi@ h
\
300 km@ h 188 mi@ h
100 km@ h 62 mi@ h
]
+240 km@ h +149 mi@ h
Table 1=3 - Maximum speed index and vehicle classication.
Index
Vehicle classication
O
O-Road and Light Truck Tires
Q
Temporary Spare Tires
T
O-Road, Studless and Studdable Winter Tires
U
Passenger cars and Light Truck Tires
V
Family Sedans and Vans
W
Family Sedans and Vans
K
Sport Sedans and Coupes
Y
Sport Sedans, Coupes and Sports Cars
Example 3 Alternative tire size indication.
If the load index is not indicated on the tire, then a tire with a size number
such as 255@50U17 Y may also be numbered by 255@50Y U17.
6
1. Tire and Rim Fundamentals
Example 4 Tire and rim widths.
The dimensions of a tire are dependent on the rim on which it is mounted.
For tires with an aspect ratio of 50 and above, the rim width is approximately 70% of the tire’s width, rounded to the nearest 0=5 in. As an example,
a S 255@50U16 tire has a design width of 255 mm = 10=04 in however, 70%
of 10=04 in is 7=028 in, which rounded to the nearest 0=5 in, is 7 in. Therefore,
a S 255@50U16 tire should be mounted on a 7 × 16 rim.
For tires with aspect ratio 45 and below, the rim width is 85% of the tire’s
section width, rounded to the nearest 0=5 in. For example, a S 255@45U17
tire with a section width of 255 mm = 10=04 in, needs an 8=5 in rim because
85% of 10=04 in is 8=534 in 8=5 in. Therefore, a S 255@45U17 tire should
be mounted on an 8 12 × 17 rim.
Example 5 Calculating tire diameter and radius.
To calculate the overall diameter of a tire, we rst nd the tire height by
multiplying the tire width and the aspect ratio. As an example, we use tire
number S 235@75U15.
kW = 235 × 75% = 176=25 mm 6=94 in
(1.3)
Then, we add twice the tire height kW to the rim diameter.
G = 2 × 6=94 + 15 = 28=88 in 733=8 mm
U = G@2 = 366=9 mm
(1.4)
(1.5)
Example 6 Speed rating code.
Two similar tires are coded as S 235@70KU15 and S 235@70U15 100K.
Both tires have speed code of K 210 km@ h. However, the second tire can
sustain the speed only when it is loaded less than the specied load index,
so it states 100K 800 kg 210 km@ h.
Speed ratings generally depend on the type of tire. O road vehicles usually use T-rated tires, passenger cars usually use U-rated tires for typical
street cars or W -rated for performance cars.
Example 7 Tire weight.
The average weight of a tire for passenger cars is 1012 kg 2226 lbf.
The weight of a tire for light trucks is 14 16 kg 30 35 lbf, and the
average weight of commercial truck tires is 135 180 kg 300 400 lbf.
Example 8 Eects of aspect ratio.
A higher aspect ratio provides a softer ride and an increase in de ection
under the load of the vehicle. However, lower aspect ratio tires are normally
used for higher performance vehicles. They have a wider road contact area
and a faster response. This results in less de ection under load, causing a
rougher ride to the vehicle.
Changing to a tire with a dierent aspect ratio will result in a dierent
contact area, therefore changing the load capacity of the tire.
1. Tire and Rim Fundamentals
DOT
DNZE
ABCD
7
2315
FIGURE 1.4. An example of a US DOT tire identication number.
DOT
B3CD
E52X
2112
FIGURE 1.5. An example of a Canadian DOT tire identication number.
Example 9 F BMW tire size code.
BMW, a European car, uses the metric system for sizing its tires. As
an example, W G230@55]U390 is a metric tire size code. W G indicates the
BMW TD model, 230 is the section width in [ mm], 55 is the aspect ratio in
percent, ] is the speed rating, U means radial, and 390 is the rim diameter
in [ mm].
Example 10 F "P V," "P + V," "P@V," and "P &V" signs.
The sign "P V,"and "P + V," and "P@V," and "P &V" indicate that
the tire has some mud and snow capability. Most radial tires have one of
these signs.
Example 11 F U.S. DOT tire identication number.
The US tire identication number is in the format "DOT GQ ]H DEFG
2315." It begins with the letters DOT to indicate that the tire meets US federal standards. DOT stands for Department of Transportation. The next
two characters, GQ , after DOT is the plant code, which refers to the manufacturer and the factory location at which the tire was made.
The next two characters, ]H, are a letter-number combination that refers
to the specic mold used for forming the tire. It is an internal factory code
and is not usually a useful code for customers. The list can be easily found
on internet.
The last four numbers, 2315, represents the week and year the tire was
built. The other numbers, DEFG, are marketing codes used by the manufacturer or at the manufacturer’s instruction. An example is shown in
Figure 1.4.
GQ is the plant code for Goodyear-Dunlop Tire located in Wittlich, Germany. ]H is the tire’s mold size, DEFG is the compound structure code,
23 indicates the 23wk week of the year, and 15 indicates year 2015. So, the
tire is manufactured in the 23wk week of 2015 at Goodyear-Dunlop Tire in
Wittlich, Germany.
Example 12 F Canadian tires identication number.
In Canada, all tires should have a DOT identication number on the
sidewall. An example is shown in Figure 1.5.
This identication number provides the manufacturer, time, and place
8
1. Tire and Rim Fundamentals
that the tire was made. The rst two characters following DOT indicate
the manufacturer and plant code. In this case, E3 indicates Group Michelin
located at Bridgewater, Nova Scotia, Canada. The third and fourth characters, FG, are the tire’s mold size code. The fth, sixth, seventh, and eighth
characters, H52[, are optional and are used by the manufacturer. The nal
four numbers, 2112, indicates the manufacturing date. For example, 2112
indicate the twenty rst week of year 2012. Finally, the maple leaf sign or
the ag sign following the identication number certies that the tire meets
Transport Canada requirements.
Table 1=4 - European county codes for tire manufacturing.
Code Country
Code Country
H1
Germany
H14
Switzerland
H2
France
H15
Norway
H3
Italy
H16
Finland
H4
Netherlands
H17
Denmark
H5
Sweden
H18
Romania
H6
Belgium
H19
Poland
H7
Hungary
H20
Portugal
H8
Czech Republic
H21
Russia
H9
Spain
H22
Greece
H10
Yugoslavia
H23
Ireland
H11
United Kingdom H24
Croatia
H12
Austria
H25
Slovenia
H13
Luxembourg
H26
Slovakia
Example 13 F E-Mark and international codes.
All tires sold in Europe after July 1997 must carry an E-mark. An example is shown by 5 in Figure 1.2. The mark itself is either an upper
or lower case "H" followed by a number in a circle or rectangle, followed
by a further number. An "H" indicates that the tire is certied to comply with the dimensional, performance and marking requirements of HFH
regulation. HFH or X Q HFH stands for the united nations economic commission for Europe. The number in the circle or rectangle is the country
code. Example: 11 is the X N. The rst two digits outside the circle or
rectangle indicate the regulation series under which the tire was approved.
Example: "02" is for HFH regulation 30 governing passenger tires, and
"00" is for HFH regulation 54 governing commercial vehicle tires. The remaining numbers represent the HFH mark type approval numbers. Tires
may have also been tested and met the required noise limits. These tires
may have a second HFH branding followed by an "v" for sound.
Table 1=4 indicates the European country codes for tire manufacturing.
Besides the GRW and HFH codes for US and Europe, we may also see
the other country codes such as: LVR 9001 for international standards organization, F=F=F for China compulsory product certication, MLV G 4230
1. Tire and Rim Fundamentals
9
for Japanese industrial standard and so on.
Example 14 F Light truck tires.
The tire sizes for a light truck may be shown in two formats:
OW 245@70U16
or
32 × 11=50U16OW
In the rst format, OW light truck, 245 tire width in millimeters,
70 aspect ratio in percent, U radial structure, and 16 rim diameter in
inches.
In the second format, 32 tire diameter in inches, 11=50 tire width in
inches, U radial structure, 16 rim diameter in inches, and OW light
truck.
Example 15 F X W TJ ratings.
Tire manufacturers may put some other symbols, numbers, and letters
on their tires supposedly rating their products for wear, wet traction, and
heat resistance. These characters are referred to as X W TJ (Uniform Tire
Quality Grading), although there is no uniformity and standard in how they
appear. There is an index for wear to show the average wearing life time
in mileage. The higher the wear number, the longer the tire lifetime. An
index of 100 is equivalent to approximately 20000 miles or 30000 km. Other
numbers are indicated in Table 1=5.
Table 1=5 - Tread wear rating index.
Index
Life (Dssur{lpdwh)
100
32000 km
20000 mi
150
48000 km
30000 mi
200
64000 km
40000 mi
250
80000 km
50000 mi
300
96000 km
60000 mi
400
129000 km 80000 mi
500
161000 km 100000 mi
The X W TJ also rates tires for wet traction and heat resistance. These
are rated in letters between "D" to "F," where "D" is the best, "E" is
intermediate and "F" is acceptable. An "D" wet traction rating is typically
an indication that the tire has a deep open tread pattern with lots of sipping,
which are the ne lines in the tread blocks.
An "D" heat resistance rating indicates two things: rst, low rolling resistance due to stier tread belts, stier sidewalls, or harder compounds;
second, thinner sidewalls, more stable blocks in the tread pattern. Temperature rating is also indicated by a letter between "D" to "FP " where "D"
is the best, "E" is intermediate, and "F" is acceptable.
10
1. Tire and Rim Fundamentals
There might also be a traction rating to indicate how well a tire grips
the road surface. This is an overall rating for both dry and wet conditions.
Tires are rated as: "DD" for the best, "D" for better, "E" for good, and
"F" for acceptable.
Example 16 F Tire sidewall additional marks.
W O Tubeless
W W Tube type, tire with an inner-tube
Made in Country Name of the manufacturing country
F Commercial tires made for commercial trucks; Example: 185U14F
E Bias ply
VI L Side facing inwards
VI R Side facing outwards
W Z L Tire wear index
It is an indicator in the main tire prole, which shows when the tire is
worn down and needs to be replaced.
VO Standard load; Tire for normal usage and loads
[O Extra load; Tire for heavy loads
UI or ui Reinforced tires
Duurz Direction of rotation
Some tread patterns are designed to perform better when driven in a
specic direction. Such tires will have an arrow showing which way the tire
should rotate when the vehicle is moving forwards.
Example 17 F Plus one (+1) concept.
The plus one (+1) concept describes the sizing up of a rim and matching
it to a proper tire. Generally speaking, each time we add 1 in to the rim
diameter, we should add 20 mm to the tire width and subtract 10% from
the aspect ratio. This compensates the increases in rim width and diameter,
and provides the same overall tire radius. Figure 1.6 illustrates the idea.
By using a tire with a shorter sidewall, we get a quicker steering response
and better lateral stability. However, we will have a stier ride.
Example 18 F Under- and over-in ated tire.
Overheat caused by improper in ation of tires is a common tire failure.
An under-in ated tire will support less of the vehicle weight with the air
pressure in the tire; therefore, more weight will be supported by the tire wall.
This tire load increase causes the tire to have a larger tireprint that creates
more friction and more heat.
In an over-in ated tire, too much of the vehicle weight is supported by the
tire air pressure. The vehicle will be bouncy and hard to steer because the
tireprint is small and only the center portion of the tireprint is contacting
the road surface.
In a properly-in ated tire, approximately 95% of the vehicle weight is
supported by the air pressure in the tire and 5% is supported by the tire
wall.
1. Tire and Rim Fundamentals
15 in
16 in
17 in
205 mm
225 mm
245 mm
205/65 R15
225/55 R16
245/45 R17
11
FIGURE 1.6. The plus one (+1) concept is a rule to nd the tire to a rim with
a 1 inch increase in diameter.
1.2 Tire Components
A tire is an advanced engineering product made of rubber and a series
of synthetic materials cooked together. Fiber, textile, and steel cords are
some of the components that go into the tire’s inner liner, body plies, bead
bundle, belts, sidewalls, and tread. Figure 1.7 illustrates a sample of tire
interior components and their arrangement.
The main components of a tire are:
Bead or bead bundle is a loop of high strength steel cable coated with
rubber. It gives the tire the strength it needs to stay seated on the wheel
rim and to transfer the tire forces to the rim.
Inner layers are made up of dierent fabrics, called plies. The most
common ply fabric is polyester cord. The top layers are also called cap
plies. Cap plies are polyesteric fabric that help hold everything in place.
Cap plies are not found on all tires; they are mostly used on tires with higher
speed ratings to help all the components stay in place at high speeds.
An inner liner is a specially compounded rubber that forms the inside
of a tubeless tire. It inhibits loss of air pressure.
Belts or belt buers are one or more rubber-coated layers of steel, polyester, nylon, Kevlar (para-aramid synthetic ber) or other materials running circumferentially around the tire under the tread. They are designed
to reinforce body plies to hold the tread at on the road and make the
best contact with the road. Belts reduce squirm to improve tread wear and
resist damage from impacts and penetration.
12
1. Tire and Rim Fundamentals
Cap/Base tread
Sidewall
Inner layer
Belt buffer
Body plies/Carcass
Inner liner
Bead bundle
FIGURE 1.7. Illustration of a sample radial tire interior components and arrangement.
The carcass or body plies are the main part in supporting the tension
forces generated by tire air pressure. The carcass is made of rubber-coated
steel or other high strength cords tied to bead bundles. The cords in a
radial tire, as shown in Figure 1.7, run perpendicular to the tread. The
plies are coated with rubber to help them bond with the other components
and to seal in the air.
A tire’s strength is often described by the number of carcass plies. Most
car tires have two carcass plies. By comparison, large commercial jetliners
often have tires with 30 or more carcass plies.
The sidewall provides lateral stability for the tire, protects the body plies,
and helps to keep the air in the tire. It may contain additional components
to help increase the lateral stability.
The tread is the portion of the tire that comes in contact with the road.
Tread designs vary widely depending on the specic purpose of the tire. The
tread is made from a mixture of dierent kinds of natural and synthetic
rubbers. The outer perimeter of a tire is called the crown.
The tread groove is the space or area between two tread rows or blocks.
The tread groove gives the tire traction and is especially useful during rain
or snow.
Example 19 Tire rubber main material.
There are two major ingredients in a rubber compound: the rubber and
the ller. They are combined in such a way to achieve dierent objectives.
The objective may be performance optimization, traction maximization, or
less rolling resistance. The most common llers are dierent types of carbon
1. Tire and Rim Fundamentals
13
black and silica. The other tire ingredients are antioxidants, antiozonant,
and anti-aging agents.
Tires are combined with several components and cooked with a heat treatment. The components must be formed, combined, assembled, and cured together. Tire quality depends on the ability to blend all of the separate components into a cohesive product that satises the driving needs. A modern
tire is basically a mixture of steel, fabric, and rubber. Generally speaking,
the weight percentage of the components of a tire are:
1= Reinforcements: steel, rayon, nylon, 16%
2= Rubber: natural/synthetic, 38%
3= Compounds: carbon, silica, chalk, 30%
4= Softener: oil, resin, 10%
5= Vulcanization: sulfur, zinc oxide, 4%
6= Miscellaneous, 2%
Example 20 Tire cords.
Because tires have to carry heavy loads, steel and fabric cords are used in
their construction to reinforce the rubber compound and provide strength.
The most common materials suitable for the tire application are cotton,
rayon, polyester, steel, berglass, and aramid.
Example 21 Bead components and preparation.
The bead component of tires is a non-extensible composite loop that anchors the carcass and locks the tire into the rim. The tire bead components
include the steel wire loop and apex or bead ller. The bead wire loop is
made from a steel wire covered by rubber and wound around the tire with
several continuous loops. The bead ller is made from a very hard rubber
compound, which is extruded to form a wedge.
Example 22 Tire ply construction.
The number of plies and cords indicates the number of layers of rubbercoated fabric or steel cords in the tire. In general, the greater the number of
plies, the more weight a tire can support. Tire manufacturers also indicate
the number and type of cords used in the tire.
Example 23 F Tire tread extrusion.
Tire tread, or the portion of the tire that comes in contact with the road,
consists of the tread, tread shoulder, and tread base. There are at least
three dierent rubber compounds used in forming the tread prole. The
three rubber compounds are extruded simultaneously into a shared extruder
head.
Example 24 F Dierent rubber types used in tires.
There are ve major rubbers used in tire production: natural rubber,
styrene-butadiene rubber (VEU), polybutadiene rubber (EU), butyl rubber,
and halogenated butyl rubber. The rst three are primarily used for tread
and sidewall compounds, while butyl rubber and halogenated butyl rubber
14
1. Tire and Rim Fundamentals
H
H
C
H
H
H
C
C
H
H
H
C
C
C
H
H
H
H
H
C
S
H
C
C
H
H
S
H
H
C
C
H
C
H
C
H
C
C
H
H
H
(a)
(b)
FIGURE 1.8. Natural rubber chemical compond: (d) the monomer unit, (e) the
vulcanized rubber.
are primarily used for the inner liner and the inside portion that holds the
compressed air inside the tire.
The natural rubber latex comes from the Pará rubber tree. The para rubber tree initially grew in South America, but today Malaysia is the main
producer of this material. When rubber rst arrived in England around
1770, it was observed that a piece of the material was good for rubbing o
pencil marks on paper, and hence the name rubber appeared.
Styrene-butadiene rubber (VEU) provides aging stability, polybutadiene
rubber (EU) provides wear resistance.
Example 25 F History of rubber.
About 2500 years ago, people living in Central and South America used
the sap and latex of a local tree to waterproof their shoes and clothes. This
material was introduced to the rst pilgrim travelers in the 17wk century.
The rst application of this new material was discovered by the English as
an eraser. This application supports the name rubber, because it was used
for rubbing out pencil marks. The rubber pneumatic tires were invented in
1845 and its production began in 1888.
The natural rubber is a mixture of polymers and isomers. The main rubber isomer is shown in Figure 1.8 (d) and is called isoprene. The natural
rubber may be vulcanized to make longer and stronger polyisopren, suitable
for tire production. Vulcanization is usually done by sulfur as cross-links.
Figure 1.8 (e) illustrates a vulcanized rubber polymer.
1. Tire and Rim Fundamentals
15
Example 26 F A world without rubber.
Rubber is the main material used to make a tire compliant. A compliant
tire can stick to the road surface while it goes out of shape and provides
distortion to move in another direction. The elastic characteristic of a tire
allows the tire to be pointed in a direction dierent than the direction the
car is pointed. There is no way for a vehicle to turn without rubber tires,
unless it moves at a very low speed. If vehicles were equipped with only
noncompliant wheels then trains moving on railroads would be the main
travelling vehicles. People could not live too far from the railways and there
would not be much use for bicycles and motorcycles.
Example 27 Tire information tips.
A new front tire with a worn rear tire can cause instability.
Tires stored in direct sunlight for long periods of time will harden and
age more quickly than those kept in a dark area.
Prolonged contact with oil or gasoline causes contamination of the rubber
compound, making the tire life short.
1.3 Radial and Non-Radial Tires
Tires are divided into two classes: radial and non-radial, depending on
the angle between carcass metallic cords and the tire-plane. Each type of
tire construction has its own set of characteristics that are the key to its
performance.
The radial tire is constructed with reinforced steel cable belts that are
assembled in parallel and run side to side, from one bead to another bead at
an angle of 90 deg to the circumferential centerline of the tire. This makes
the tire more exible radially, which reduces rolling resistance and improves
cornering capability. Figure 1.7 shows the interior structure and the carcass
arrangement of a radial tire.
The non-radial tires are also called bias-ply and cross-ply tires. The plies
are layered diagonal from one bead to the other bead at about a 30 deg
angle, although any other angles may also be applied. One ply is set on
a bias in one direction as succeeding plies are set alternately in opposing
directions as they cross each other. The ends of the plies are wrapped
around the bead wires, anchoring them to the rim of the wheel. Figure 1.9
shows the interior structure and the carcass arrangement of a non-radial
tire.
The most important dierence in the dynamics of radial and non-radial
tires is their dierent ground sticking behavior when a lateral force is applied on the wheel. This behavior is shown in Figure 1.10. The radial tire,
shown in Figure 1.10(d), exes mostly in the sidewall and keeps the tread
at on the road. The bias-ply tire, shown in Figure 1.10(e) has less contact
with the road as both tread and sidewalls distort under a lateral load.
16
1. Tire and Rim Fundamentals
Cap/Base tread
Sidewall
Inner layer
Belt buffer
Body plies/Carcass
Inner liner
Bead bundle
FIGURE 1.9. Illustration of a non-radial tire’s interior components and arrangement.
(a) Radial tire
(b) Non-Radial tire
FIGURE 1.10. Ground-sticking behavior of radial and non-radial tires in the
presence of a lateral force.
1. Tire and Rim Fundamentals
17
The radial arrangement of carcass in a radial tire allows the tread and
sidewall act independently. The sidewall exes more easily under the weight
of the vehicle. So, more vertical de ection is achieved with radial tires. As
the sidewall exes under the load, the belts hold the tread rmly and
evenly on the ground which reduces tread scrub. In a cornering maneuver,
the independent action of the tread and sidewalls keeps the tread at on
the road. This allows the tire to hold its path. Radial tires are the preferred
tire in most applications today.
The cross arrangement of carcass in bias-ply tires allows it act as a unit.
When the sidewalls de ect or bend under load, the tread squeezes in and
distorts. This distortion aects the tireprint and decrease traction. Because
of the bias-ply inherent construction, sidewall strength is less than that of
a radial tire’s construction and cornering is less eective.
Example 28 Increasing the strength of tires.
The strength of bias-ply tires increases by increasing the number of plies
and bead wires. However, more plies means more mass, which increases heat
and reduces tire life. To increase a radial tire’s strength, larger diameter
steel cables are used in the tire’s carcass.
Example 29 Tubeless and tube-type tire construction.
A tubeless tire is similar in construction to a tube-type tire, except that a
thin layer of air and moisture-resistant rubber is used on the inside of the
tubeless tire from bead to bead to obtain an internal seal of the casing. This
eliminates the need for a tube and ap. Both tires, in equivalent sizes, can
carry the same load at the same in ation pressure.
Example 30 F New shallow tires.
Low aspect ratio tires are radial tubeless tires that have a section width
wider than their section height. The aspect ratio of these tires is between
50% to 30%. These shallow tires have shorter sidewall heights and wider
tread widths. This feature improves stability and handling from a higher
lateral stiness rates.
Example 31 F Tire function.
A tire is a pneumatic system to support a vehicle’s load. Tires support
a vehicle’s load by using compressed air to create tension in the carcass
plies. Tire carcass are a series of cords that have a high tension strength,
and almost no compression strength. The air pressure creates tension in
the carcass and carries the load. In an in ated and unloaded tire, the cords
pull equally on the bead wire all around the tire. When the tire is loaded,
the tension in the cords between the rim and the ground is relieved while
the tension in other cords is unchanged. Therefore, the cords opposite the
ground pull the bead upwards. This is how pressure is transmitted from the
ground to the rim.
Besides vertical load carrying, a tire must transmit acceleration, braking,
18
1. Tire and Rim Fundamentals
Lugs
Voids
FIGURE 1.11. A sample of tire tread to show lugs and voids.
and cornering forces to the road. These forces are transmitted to the rim
in a similar manner. Acceleration and braking forces also depend on the
friction between the rim and the bead. A tire also acts as a spring between
the rim and the road.
1.4 Tread
The tread pattern is made up of tread lugs and tread voids. The lugs
are the sections of rubber that make contact with the road and voids are
the spaces that are located between the lugs. Lugs are also called slots or
blocks, and voids are also called grooves. The tread pattern of block-groove
congurations aects the tire’s traction and noise level. Wide and straight
grooves running circumferentially have a lower noise level and high lateral
friction. More lateral grooves running from side to side increase traction
and noise levels. A sample of a tire tread is shown in Figure 1.11.
Tires need both circumferential and lateral grooves. The water on the
road is compressed into the grooves by the vehicle’s weight and is evacuated
from the tireprint region, providing better traction at the tireprint contact.
Without such grooves, the water would not be able to escape out to the
sides of the wheel. This would cause a thin layer of water to remain between
the road and the tire, which causes a loss of friction with the road surface.
Therefore, the grooves in the tread provide an escape path for water.
On a dry road, the tire treads reduce grip because they reduce the contact
area between the rubber and the road. This is the reason for using treadless
or slick tires at smooth and dry race tracks.
1. Tire and Rim Fundamentals
19
The mud-terrain tire pattern is characterized by large lugs and large
voids. The large lugs provide large bites in poor traction conditions and
the large voids allow the tire to clean itself by releasing and expelling the
mud and dirt. The all-terrain tire pattern is characterized by smaller voids
and lugs when compared to the mud terrain tire. A denser pattern of lugs
and smaller voids make all-terrain tires quieter on the street. However,
smaller voids cannot clean themselves easily and if the voids ll up with
mud, the tire loses some of it’s traction. The all-terrain tire is good for
highway driving.
Example 32 Asymmetrical and directional tread design.
The design of the tread pattern may be asymmetric and change from one
side to the other. Asymmetric patterns are designed to have two or more
dierent functions and provide a better overall performance.
A directional tire is designed to rotate in only one direction for maximum
performance. Directional tread pattern is especially designed for driving on
wet, snowy, or muddy roads. A non-directional tread pattern is designed to
rotate in either direction without sacricing in performance.
Example 33 Self-cleaning.
Self-cleaning is the ability of a tire’s tread pattern to release mud or
material from the voids of tread. This ability provides good bite on every
rotation of the tire. A better mud tire releases the mud or material easily
from the tread voids.
Example 34 F Hydroplaning
Hydroplaning is sliding of a tire on a lm of water. Hydroplaning can
occur when a car drives through standing water and the water cannot totally
escape out from under the tire. This causes the tire to lift o the ground
and slide on the water. The hydroplaning tire will have little traction and
therefore, the car will not obey the driver’s command.
Deep grooves running from the center front edge of the tireprint to the
corners of the back edges, along with a wide central channel help water to
escape from under the tire. Figure 1.12 illustrates the hydroplaning phenomena when the tire is riding over a water layer.
There are three types of hydroplaning: dynamic, viscous, and rubber hydroplaning. Dynamic hydroplaning occurs when standing water on a wet
road is not displaced from under the tires fast enough to allow the tire to
make pavement contact over the total tireprint. The tire rides on a wedge of
water and loses its contact with the road. The speed at which hydroplaning
happens is called hydroplaning speed.
Viscous hydroplaning occurs when the wet road is covered with a layer
of oil, grease, or dust. Viscous hydroplaning happens with less water depth
and at a lower speed than dynamic hydroplaning.
Rubber hydroplaning is generated by superheated steam at high pressure
in the tireprint, which is caused by the friction-generated heat in a hard
20
1. Tire and Rim Fundamentals
Tire
Z
Water film
Ground plane
FIGURE 1.12. Illustration of hydroplaning phnomena.
braking.
Example 35 F Aeronautic hydroplaning speed.
In aerospace engineering the hydroplaning speed is estimated in [knots]
by
s
(1.6)
yK = 9 s
where, s is tire in ation pressure in [psi].
For main wheels of a E757 aircraft, the hydroplaning speed would be
s
(1.7)
yK = 9 144 = 108 nqrwv 55=5 m@ s
Equation (1.6) for a metric system would be
s
y{ = 5=5753 × 102 s
(1.8)
where y{ is in [ m@ s] and s is in [ Pa]. As an example, the hydroplaning
speed of a car using tires with pressure 28svl 193053 Pa is
s
y{ = 5=5753 × 102 193053 24=5 m@ s
47=6 nqrwv 88=2 km@ h 54=8 mi@ h
(1.9)
1.5 Tireprint
The contact area between a tire and the road is called the tireprint and is
shown by DS . At any point of a tireprint, the normal and friction forces are
transmitted between the road and tire. The eect of the contact forces can
be described by a resulting force system including force and torque vectors
applied at the center of the tireprint.
The tireprint is also called contact patch, contact region, or tire footprint.
A model of tireprint is shown in Figure 1.13.
1. Tire and Rim Fundamentals
21
x
b
a
y
Tireprint
FIGURE 1.13. A tireprint.
The area of the tireprint is inversely proportional to the tire pressure.
Lowering the tire pressure is a technique used for o-road vehicles in sandy,
muddy, or snowy areas, and for drag racing. Decreasing the tire pressure
causes the tire to slump so more of the tire is in contact with the surface,
giving better traction in low friction conditions. It also helps the tire grip
small obstacles as the tire conforms more to the shape of the obstacle, and
makes contact with the object in more places.
Example 36 Uneven wear in front and rear tires.
In most vehicles, the front and rear tires will wear at dierent rates. So,
it is advised to swap the front and rear tires as they wear down to even out
the wear patterns. This is called rotating the tires.
Front tires, especially on front-wheel drive vehicles, wear out more quickly
than rear tires.
1.6 Wheel and Rim
When a tire is installed on a rim and is in ated, it is called a wheel. A wheel
is a combined tire and rim. The rim is the cylindrical part on which the tire
is installed. Most passenger cars are equipped with steel or metallic rims.
The steel rim is made by welding a disk to a shell. However, light alloy rims
made with light metals such as aluminium and magnesium are also popular.
Figure 1.14 illustrates a wheel and the most important dimensional names.
A rim has two main parts: ange and spider. The ange or hub is the ring
or shell on which the tire is mounted. The spider or center section is the
disc section that is attached to the hub. The rim width is also called pan
width and measured from inside to inside of the bead seats of the ange.
Flange provides lateral support to the tire. A ange has two bead seats
22
1. Tire and Rim Fundamentals
Section width
Tread width
Section height
Inside
Outside
Center line
Rim diameter
Offset
Pan width
FIGURE 1.14. Illustration of a wheel and its dimensions.
providing radial support to the tire. The well is the middle part between
the bead seats with su!cient depth and width to enable the tire beads to
be mounted and demounted on the rim. The rim hole or valve aperture is
the hole or slot in the rim that accommodates the valve for tire in ation.
There are two main rim shapes:
1= drop center rim (GF) and,
2= wide drop center rim (Z GF).
The Z GF may also come with a hump. The humped Z GF may be
called Z GFK. Their cross sections are illustrated in Figure 1.15.
Drop center (GF) rims usually are symmetric with a well between the
bead seats. The well is built to make mounting and demounting the tire
easier. The bead seats are around 5 deg tapered. Wide drop center rims
(Z GF) are wider than GF rims and are built for low aspect ratio tires.
The well of Z GF rims are shallower and wider. Today, most passenger
cars are equipped with Z GF rims. The Z GF rims may be manufactured
with a hump behind the bead seat area to prevent the bead from slipping
down.
A sample of rim numbering and its meaning is shown in Figure 1.16.
Rim width, rim diameter, and oset are shown in Figure 1.14. Oset is
the distance between the inner plane and the center plane of the rim. A
rim may be designed with a negative, zero, or positive oset. A rim has a
positive oset if the spider is outward from the center plane.
1. Tire and Rim Fundamentals
Rim width
Rim width
Rim diameter
WDC rim
Rim diameter
DC rim
23
Rim width
Hump
Rim diameter
WDCH rim
FIGURE 1.15. Illustration of GF, Z GF, and Z GFK rims and their geometry.
7 ½ – JJ
7 1/2
JJ
15
55
5
114.3
15
55
5 – 114.3
Rim width [in]
Flange shape code
Rim diameter [in]
Offset [mm]
Number of bolts
Pitch circle diameter
FIGURE 1.16. A sample rim number.
24
1. Tire and Rim Fundamentals
Rim
Spindle
FIGURE 1.17. Illustration of a wheel attched to the spindle axle.
The ange shape code signies the tire-side prole of the rim and can be
E, F, G, H, I , J, M, MM, MN, and N. Usually the prole code follows the
nominal rim width but dierent arrangements are also used. Figure 1.17
illustrates how a wheel is attached to the spindle axle.
Example 37 Wire spoke wheel.
A rim that uses wires to connect the center part to the exterior ange
is called a wire spoke wheel, or simply a wire wheel. The wires are called
spokes. This type of wheel is usually used on classic vehicles. The highpower cars do not use wire wheels because of safety. Figure 1.18 depicts
two examples of wire spoke wheels.
Example 38 Light alloy rim material.
Metal is the main material for manufacturing rims, however, new composite materials are also used for rims. Composite material rims are usually
thermoplastic resin with glass ber reinforcement, developed mainly for low
weight. Their strength and heat resistance still need improvement before
being a proper substitute for metallic rims.
Other than steel and composite materials, light alloys such as aluminum,
magnesium, and titanium are used for manufacturing rims.
Aluminum is very good for its weight, thermal conductivity, corrosion resistance, easy casting, low temperature, easy machine processing, and recycling. Magnesium is about 30% lighter than aluminum, and is excellent for
size stability and impact resistance. However, magnesium is more expensive
and it is used mainly for luxury or racing cars. The corrosion resistance of
magnesium is not as good as aluminum. Titanium is much stronger than
1. Tire and Rim Fundamentals
Center line
25
Center line
FIGURE 1.18. Two samples of wire spoke wheel.
Ground plane Magnesium rim
Aluminum rim
Steel rim
FIGURE 1.19. The dierence between aluminum, magnesium, and steel rims in
regaining road contact after a jump.
aluminum and magnesium with excellent corrosion resistance. However, titanium is expensive and hard to be machine processed.
The dierence between aluminum, magnesium, and steel rims is illustrated in Figure 1.19. Light weight wheels regain contact with the ground
quicker than heavier wheels.
Example 39 Spare tire.
Road vehicles typically carry a spare tire, which is already mounted on a
rim ready to use in the event of at tire. Since 1980, some cars have been
equipped with spare tires that are smaller than normal size. These spare
tires are called doughnuts or space-saver spare tires. Although the doughnut
spare tire is not very useful or popular, it can help to save a little space,
weight, cost, and gas mileage. Doughnut spare tires can not be driven far
or fast.
Example 40 Wheel history.
Stone and wooden wheels were invented and used somewhere in the Mid-
26
1. Tire and Rim Fundamentals
dle East about 5000 years ago. Hard wheels have some ine!cient characteristics namely poor traction, low friction, harsh ride, and poor load carrying
capacity.
Solid rubber tires and air tube tires began to be used in the late nineteen
and early twentieth century.
1.7 Vehicle Classications
Road vehicles are usually classied based on their size and number of axles.
Although there is no standard or universally accepted classication method,
there are a few important and applied vehicle classications.
1.7.1 ISO and FHWA Classication
ISO3833 classies ground vehicles in 7 groups:
1= Motorcycles
2= Passenger cars
3= Busses
4= Trucks
5= Agricultural tractors
6= Passenger cars with trailer
7= Truck trailer/semi trailer road trains
The Federal Highway Administration (I KZ D) classies road vehicles
based on size and application. All road vehicles are classied in 13 classes
as described below:
1= Motorcycles
2= Passenger cars, including cars with a one-axle or two-axle trailer
3= Other two-axle vehicles, including: pickups, and vans, with a one-axle
or two-axle trailer
4= Buses
5= Two axle, six-tire single units
6= Three-axle single units
7= Four or more axle single units
8= Four or fewer axle single trailers
9= Five-axle single trailers
10= Six or more axle single trailers
11= Five or less axle multi-trailers
12= Six-axle multi-trailers
13= Seven or more axle multi-trailers
Figure 1.20 illustrates the I KZ D classication. The denition of I KZ D
classes follow.
1. Tire and Rim Fundamentals
1
2
3
4
5
6
7
8
9
10
11
12
13
FIGURE 1.20. The I KZ D vehicle classication.
27
28
1. Tire and Rim Fundamentals
FIGURE 1.21. A three-wheel motorcycle.
Motorcycles: Any motorvehicle having a seat or saddle and no more
than three wheels that touch the ground is a motorcycle. Motorcycles,
motor scooters, mopeds, motor-powered or motor-assisted bicycles, and
three-wheel motorcycles are in this class. Motorcycles are usually, but not
necessarily, steered by handlebars. Figure 1.21 depicts a three-wheel motorcycle.
Passenger Cars: Street cars, including sedans, coupes, and station wagons manufactured primarily for carrying passengers, are in this class. Passenger cars are also called street cars, automobiles, or autos.
Other Two-Axle, Four-Tire Single-Unit Vehicles: All two-axle, four-tire
vehicles other than passenger cars make up this class. This class includes
pickups, panels, vans, campers, motor homes, ambulances, hearses, carryalls, and minibuses. Other two-axle, four-tire single-unit vehicles pulling
recreational or light trailers are also included in this class. Distinguishing
class 3 from class 2 is not clear, so these two classes may sometimes be
combined into class 2.
Buses: A motor vehicle able to carry more than ten persons is a bus.
Buses are manufactured as traditional passenger-carrying vehicles with two
axles and six tires. However, buses with three or more axles are also manufactured.
Two-Axle, Six-Tire, Single-Unit Trucks: Vehicles on a single frame including trucks, camping and recreational vehicles, motor homes with two
axles, and dual rear wheels.
Three-Axle Single-Unit Trucks: Vehicles having a single frame including
trucks, camping, recreational vehicles, and motor homes with three axles.
Four-or-More-Axle-Single-Unit Trucks: All trucks on a single frame with
four or more axles make up this class.
Four-or-Fewer-Axle Single-Trailer Trucks: Vehicles with four or fewer
axles consisting of two units, one of which is a tractor or straight truck
power unit.
1. Tire and Rim Fundamentals
29
Five-Axle Single-Trailer Trucks: Five-axle vehicles consisting of two units,
one of which is a tractor or straight truck power unit.
Six-or-More-Axle Single-Trailer Trucks: Vehicles with six or more axles
consisting of two units, one of which is a tractor or straight truck power
unit.
Five-or-Fewer-Axle Multi-Trailer Trucks: Vehicles with ve or fewer axles
consisting of three or more units, one of which is a tractor or straight truck
power unit.
Six-Axle Multi-Trailer Trucks: Six-axle vehicles consisting of three or
more units, one of which is a tractor or straight truck power unit.
Seven or More Axle Multi-Trailer Trucks: Vehicles with seven or more
axles consisting of three or more units, one of which is a tractor or straight
truck power unit.
The classes 6 to 13 are also called truck. A truck is a motor vehicle
designed primarily for carrying load or property.
1.7.2 Passenger Car Classications
A passenger car or automobile is a motorvehicle designed for carrying ten
or fewer persons. Automobiles may be classied based on their size and
weight. Size classication is based on wheelbase, the distance between front
and rear axles. Weight classication is based on curb weight, the weight of
an automobile with standard equipment, and a full complement of fuel
and other uids, but with no load, persons, or property. The wheelbase is
rounded to the nearest inch and the curb weight to the nearest 100 lb 50 kg before classication.
For a size classication, passenger car may be classied as a small, midsize, and large car. Small cars have a wheelbase of less than 99 in 2=5 m,
midsize cars have a wheelbase of less than 109 in 2=8 m and greater than
100 in 2=5 m, and large cars have a wheelbase of more than 110 in 2=8 m.
Each class may also be divided further.
For a weight classication, passenger car may be classied as light, midweight, and heavy. Light weight cars have a curb weight of less than 2400 lb 1100 kg, midweight cars have a curb weight of less than 3400 lb 1550 kg
and more than 2500 lb 1150 kg, and heavy cars have a curb weight of
more than 3500 lb 1600 kg. Each class may also be divided in some subdivisions.
Dynamically, passenger cars may be classied by their type of suspension,
engine, driveline arrangement, weight distribution, or any other parameters
that aect the dynamics of a car. However, in the market, passenger cars
are usually divided into the following classes according to the number of
passengers and load capacity.
1= Economy
30
1. Tire and Rim Fundamentals
2= Compact
3= Intermediate
4= Standard Size
5= Full Size
6= Premium Luxury
7= Convertible Premium
8= Convertible
9= Minivan
10= Midsize
11= SUV
In another classication, cars are divided according to size and shape.
However, using size and shape to classify passenger cars is not clear-cut;
many vehicles fall in between classes. Also, not all are sold in all countries,
and sometimes their names dier between countries. Common entries in the
shape classication are the sedan, coupe, convertible, minivan/van, wagon,
and SUV.
A sedan is a car with a four-door body conguration and a conventional
trunk or a sloping back with a hinged rear cargo hatch that opens upward.
A coupe is a two-door passenger car.
A convertible is a car with a removable or retractable top.
A minivan/van is a vehicle usually with a box-shaped body enclosing a
large cargo or passenger area. The identied gross weight of a van is less
than 10000 lb 4500 kg. Vans can be identiable by their enclosed cargo
or passenger area, short hood, and box shape. Vans can be divided into
mini van, small van, midsize van, full-size van, and large van. The van
subdivision has the same specications as SUV subdivisions.
A wagon is a car with an extended body and a roo ine that extends past
the rear doors.
An SUV (Sport Utility Vehicle) is a vehicle with o-road capability. SUV
is designed for carrying ten or fewer persons, and generally considered a
multi-purpose vehicle. Most SUVs are four-wheel-drive with and increased
ground clearance. The SUV is also known as 4 e| 4, 4Z G, 4 × 4 or 4{4.
SUVs can be divided into mini, small, midsize, full-size, and large SUV.
Mini SUVs are those with a wheelbase of less than or equal to 88 in 224 cm. A mini SUV is typically a microcar with a high clearance, and
o-road capability. Small SUVs have a wheelbase of greater than 88 in 224 cm with an overall width of less than 66 in 168 cm. Small SUVs
are short and narrow 4 × 4 multi-purpose vehicles. Midsize SUVs have a
wheelbase of greater than 88 in 224 cm with an overall width greater
than 66 in 168 cm, but less than 75 in 190 cm. Midsize SUVs are 4 × 4
multi-purpose vehicles designed around a shortened pickup truck chassis.
Full-size SUVs are made with a wheelbase greater than 88 in 224 cm and
a width between 75 in 190 cm and 80 in 203 cm. Full-size SUVs are 4×4
1. Tire and Rim Fundamentals
31
multi-purpose vehicles designed around an enlarged pickup truck chassis.
Large SUVs are made with a wheelbase of greater than 88 in 224 cm and
a width more than 80 in 203 cm.
Because of better performance, the vehicle manufacturing companies are
going to make more cars four-wheel-drive. So, four-wheel-drive does not
refer to SUV a specic class of cars anymore.
A truck is a vehicle with two or four doors and an exposed cargo box. A
light truck has a gross weight of less than 10 000 lb 4 500 kg. A medium
truck has a gross weight from 10 000 lb 4 500 kg to 26 000 lb 12 000 kg.
A heavy truck is a truck with a gross weight of more than 26 000 lb 12 000 kg.
Gross weight is the maximum weight of a vehicle as specied by the
manufacturer. Gross weight includes chassis, body, engine, uids, fuel, accessories, driver, passengers and cargo.
1.7.3 Passenger Car Body Styles
Passenger cars are manufactured in so many dierent styles and shapes.
Not all of those classes are made today, and some have new shapes and still
carry the same old names. Some of them are as follows:
Convertible or cabriolet cars are automobiles with removable or retractable
roofs. There are also the subdivisions cabrio coach or semi-convertible with
partially retractable roofs.
Coupé or coupe are two-door automobiles with two or four seats and a
xed roof. In cases where the rear seats are smaller than regular size, it is
called a two-plus-two or 2 + 2. Coupé cars may also be convertible.
Crossover SUV or XUV cars are smaller sport utility vehicles based on
a car platform rather than truck chassis. Crossover cars are a mix of SUV,
minivan, and wagon to encompass some of the advantages of each.
Estate car or just estate is the British/English term for what North
Americans call a station wagon.
Hardtop cars are those having a removable solid roof on a convertible car.
However, today a xed-roof car whose doors have no xed window frame
are also called a hardtops.
Hatchback cars are identied by a rear door, including the back window
that opens to access a storage area that is not separated from the rest of
the passenger compartment. A hatchback car may have two or four doors
and two or four seats. They are also called three-door, or ve-door cars. A
hatchback car is called a liftback when the opening area is very sloped and
is lifted up to open.
A limousine is a chaueur-driven car with a glass-window dividing the
front seats from the rear. Limousines are usually an extended version of a
luxury car.
32
1. Tire and Rim Fundamentals
P 215 / 60 R 15 96 H
FIGURE 1.22. A sample of tire size.
Minivans are boxy wagon cars usually containing three rows of seats,
with a capacity of six or more passengers and extra luggage space.
An MPV (Multi-Purpose Vehicle) is designed as large cars or small buses
having o-road capability and easy loading of goods. However, the idea for a
car with a multi-purpose application can be seen in other classes, especially
SUVs.
Notchback cars are something between the hatchback and sedan. Notchback is a sedan with a separate trunk compartment.
A pickup truck ( or simply pickup) is a small or medium-sized truck with
a separate cabin and rear cargo area. Pickups are made to act as a personal
truck, however they might also be used as light commercial vehicles.
Sedan is the most common body style that are cars with four or more
seats and a xed roof that is full-height up to the rear window. Sedans can
have two or four doors.
Station wagon or wagon is a car with a full-height body all the way to the
rear; the load-carrying space created is accessed via a rear door or doors.
1.8 Summary
A sample of tire size and performance code is shown in Figure 1.22 and
their denitions are explained as follows:
S stands for passenger car. 215 is the unloaded tire width, in [ mm].
kW
60 is the aspect ratio of the tire, vW = z
× 100, which is the section
W
height to tire width, expressed as a percentage. U stands for radial.
15 is the rim diameter that the tire is designed to t in [ in]. 96 is the
load index, and K is the speed rate index.
Road vehicles are usually classied based on their size and number of
axles. There is no universally accepted standard classication, however,
ISO and FHWA present two important classications in North America.
1. Tire and Rim Fundamentals
ISO3833 classies ground vehicles into seven groups:
1=Motorcycles
2=Passenger cars
3=Busses
4=Trucks
5=Agricultural tractors
6=Passenger cars with trailer
7=Truck trailer/semitrailer road trains
FHWA classies all road vehicles into 13 classes:
1=Motorcycles
2=Passenger cars with one or two axles trailer
3=Other two-axle four-wheel single units
4=Buses
5=Two-axle six-wheel single units
6=Three-axle single units
7=Four-or-more-axle single units
8=Four-or-less-axle single trailers
9=Five-axle single trailers
10=Six-or-more-axle single trailers
11=Five-or-less-axle multi-trailers
12=Six-axle-multi-trailers
13=Seven-or-more-axle multi-trailers
33
34
1. Tire and Rim Fundamentals
1.9 Key Symbols
DS
E
G
G
GF
GRW
I KZ D
kW
K
Z GFK
OW
P &V
s
S
U
vW
VW
VX Y
W
yK
y> y{
zW
Z GF
tireprint area
bias ply tire
tire diameter
diagonal
drop center rim
Department of Transportation
Federal Highway Administration
section height
speed rate
humped wide drop center rim
light truck
mud and snow
tire in ation pressure
passenger car
radial tire
aspect ratio
special trailer
sports utility vehicle
temporary tire
hydroplaning speed
forward velocity of vehicle
tire width
wide drop center rim
1. Tire and Rim Fundamentals
35
Exercises
1. Tire size codes.
Explain the meaning of the following tire size codes:
d= 10=00U20 14(J)
e= 18=4U46
f= 480@80U46155D8
g= 18=4 38(10)
h= 76 × 50=00E32 = 1250@45E32
i= OW 255@85E16
j= 33{12=50U15OW
2. Tire height and diameter.
Find the tire height kW and diameter G for the following tires.
d= 480@80U46 155D8
e= S 215@65U15 96K
3. F Plus one.
Increase 1 in to the diameter of the rim of the following tires and nd
a proper tire for the new rim.
d= S 215@65U15 96K
e= S 215@60U15 96K
4. F Problem of tire beads.
Explain what would be the possible problem for a tire that has tight
or loose beads.
5. Tire of Porsche 911 turboW P .
A model of Porsche 911 turboW P uses the following tires.
i urqw : 235@35]U19
uhdu : 305@30]U19
Determine and compare kW , and G for the front and rear tires.
6. Tire of Porsche Cayenne turboW P .
A model of Porsche Cayenne turboW P is an all-wheel-drive that uses
the following tire.
255@55U18
What is the angular velocity of its tires when it is moving at the top
speed y = 171 mi@ h 275 km@ h?
36
1. Tire and Rim Fundamentals
7. Tire of Ferrari P 4@5 by PininfarinaW P .
A model of Ferrari P 4@5 by PininfarinaW P is a rear-wheel-drive sport
car that uses the following tires.
i urqw : 255@35]U20
uhdu : 335@30]U20
What is the angular velocity of its tires when it is moving at the top
speed y = 225 mi@ h 362 km@ h?
8. Tire of Mercedes-Benz SLR 722 EditionW P .
A model of Mercedes-Benz SLR 722 EditionW P uses the following
tires.
i urqw 255@35U19
uhdu 295@30U19
What is the speed of this car if its rear tires are turning at
$ = 2000 usp
At that speed, what would be the angular velocity of the front tires?
9. Tire of Chevrolet Corvette ]06W P .
A model of Chevrolet Corvette ]06W P uses the following tires.
i urqw 275@35]U18
uhdu
325@30]U19
What is the speed of this car if its rear tires are turning at
$ = 2000 usp
At that speed, what would be the angular velocity of the front tires?
10. Tire of Koenigsegg CCXW P .
Koenigsegg CCXW P is a sport car, equipped with the following tires.
i urqw 255@35U19
uhdu
335@30U20
What is the angular speed ratio of the rear tire to the front tire?
11. F Vehicle speed measurement.
Let us assume that the speed indicator of a vehicle is showing the
speed proportional to the angular speed of its wheel. The speed meter
is correct when the tire is new and it is as indicated below:
i urqw 255@35U19
(a) What is the angular speed of the tire when the car is moving at
100 km@ h?
(b) What would be the angular speed of the tire if it loses its tread
such that the radius reduces by 5%?
(c) Let us assume to change the tires to 335@30U20 while using the
same speed meter with the same calibration. What the speed
indicator shows when the vehicle is moving at 100 km@ h?
Part I
Vehicle Motion
2
Forward Vehicle Dynamics
Straight motion of an ideal rigid vehicle is the subject of this chapter.
We ignore air friction and examine the load variation under the tires to
determine the vehicle’s limits of acceleration, road grade, and kinematic
capabilities.
z
a2
a1
x
C
2Fz2
mg
2Fz1
FIGURE 2.1. A parked car on level pavement.
2.1 Parked Car on a Level Road
When a car is parked on level pavement, the normal force, I} , under each
of the front and rear wheels, I}1 , I}2 , are
d2
1
pj
2
o
d1
1
I}2 = pj
2
o
I}1 =
(2.1)
(2.2)
where, d1 is the distance of the car’s mass center, F, from the front axle,
d2 is the distance of F from the rear axle, and o is the wheel base.
o = d1 + d2
(2.3)
Proof. Consider a longitudinally symmetrical car as shown in Figure 2.1.
It can be modeled as a two-axel vehicle which is equivalent to a rigid beam
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_2, © Springer Science+Business Media New York 2014
39
40
2. Forward Vehicle Dynamics
having two supports. The vertical force under the front and rear wheels
can be determined using planar static equilibrium equations.
X
I} = 0
(2.4)
X
P| = 0
(2.5)
Applying the equilibrium equations
2I}1 + 2I}2 pj = 0
2I}1 d1 + 2I}2 d2 = 0
(2.6)
(2.7)
provides the reaction forces under the front and rear tires.
1
1
d2
d2
= pj
pj
2
d1 + d2
2
o
1
1
d1
d1
= pj
I}2 = pj
2
d1 + d2
2
o
I}1 =
(2.8)
(2.9)
Example 41 Reaction forces under wheels.
A model of Lamborghini DiabloW P has 1576 kg mass. Its mass center, F,
is 108=65 cm behind the front wheel axis, and it has a 265 cm wheel base.
d1 = 1=086 m
o = 2=65 m
p = 1576 kg
(2.10)
The force under each front wheel is
I}1 =
1
d2
1
2=65 1=086
pj
= × 1576 × 9=81 ×
= 4560=9 N
2
o
2
2=65
(2.11)
and the force under each rear wheel is
I}2 =
1
d1
1
1=086
pj
= × 1576 × 9=81 ×
= 3169=4 N
2
o
2
2=65
(2.12)
The front/rear load distribution is usually a reported number to be used to
determine the position of the mass center. The front/rear load distribution
of the Lamborghini DiabloW P is 41@59%.
Example 42 Mass center position.
Equations (2.1) and (2.2) can be rearranged to calculate the position of
mass center.
2o
2o
d2 =
(2.13)
d1 =
I}2
I}
pj
pj 1
Reaction forces under the front and rear wheels of a horizontally parked
car, with a wheel base o = 2=65 m, are:
I}1 = 4560=9 N
I}2 = 3169=4 N
(2.14)
2. Forward Vehicle Dynamics
41
l
a1
a2
x
C
2Fz2
mg
2Fz1
FIGURE 2.2. Measuring the force under the front wheels.
Therefore, the longitudinal position of the car’s mass center is at
d1
=
d2
=
2o
2=65
I}2 = 2
× 3169=4 = 1=086 m (2.15)
pj
2 (4560=9 + 3169=4)
2o
2=65
I}1 = 2
× 4560=9 = 1=563 m (2.16)
pj
2 (4560=9 + 3169=4)
Example 43 Longitudinal mass center determination.
The position of mass center F can be determined experimentally. To
determine the longitudinal position of F, we should measure the total weight
of the car as well as the force under the front or the rear wheels. Figure 2.2
illustrates a situation in which we measure the force under the front wheels.
Assuming the force under the front wheels is 2I}1 , the position of the
mass center is calculated by static equilibrium conditions
X
X
I} = 0
P| = 0
(2.17)
Applying the equilibrium equations
2I}1 + 2I}2 pj = 0
2I}1 d1 + 2I}2 d2 = 0
(2.18)
provides the longitudinal position of F and the reaction forces under the
rear wheels.
¶
μ
2o
o
2I}1
I}2 =
(pj 2I}1 ) = o 1 (2.19)
d1 =
pj
pj
pj
1
I}2 = (pj 2I}1 )
(2.20)
2
Example 44 Lateral mass center determination.
Most cars are approximately symmetrical about the longitudinal center
plane and therefore, the lateral position of the mass center F is close to
42
2. Forward Vehicle Dynamics
z
b2
b1
C
y
h
mg
Fz1
Fz2
FIGURE 2.3. A laterally asymmetric car has an o center mass center.
the center plane. However, the lateral position of F may be calculated by
weighing one side of the car. The problem is that although the left and
right sides of cars are almost symmetric, the front and rear of cars are not
symmetric and there is an unequal weight distribution among the front and
rear. Therefore, when we measure the weight of the left and right sides there
would be a signicant dierence between the front and rear, and they must
be analyzed separately.
As an example let us consider the Lamborghini DiabloW P with total mass
of 1576 kg, and front/rear load distribution of 41@59%. Let us assume there
is a 4% dierence between the left and right sides in the front of the car.
The load distribution indicates that the front of the car weighs
2I}i = 0=59 × 1576j = 9121=7 N
(2.21)
As depicted in Figure 2.3, let us name the vertical force under the tire at
left and right sides by I}1 and I}2 , respectively. If the left side of the car
is 4% heavier than the right, then
I}1
I}2
= 0=52 × 2I}i = 0=52 × 9121=7 = 4743=3 N
= 0=48 × 2I}i = 0=48 × 9121=7 = 4378=4 N
(2.22)
(2.23)
The car’s front and rear tracks are reported as
zi = 1540 mm 60=52 in
zu = 1640 mm 64=45 in
(2.24)
therefore
e1
e2
= 0=48 × zi = 0=48 × 1540 = 739=2 mm
= 0=52 × zi = 0=52 × 1540 = 800=8 mm
(2.25)
(2.26)
2. Forward Vehicle Dynamics
z
43
h R sin I
x
a1
a2
h
h R sin I
C
2Fz1
H
h cos I
I
mg
2Fz2
FIGURE 2.4. Measuring the force under the wheels to nd the height of the mass
center.
To determine the exact position of the mass center, we should measure the
vertical force under all tires simultaneously.
Example 45 Height mass center determination.
To determine the height of mass center F, we should measure the force
under the front or rear wheels while the car is on an inclined surface. Experimentally, we use a device such as is shown in Figure 2.4. The car is
parked on a level surface such that the front wheels are on a scale jack. The
front wheels will be locked and anchored to the jack, while the rear wheels
will be left free to turn. The jack lifts the front wheels and the required
vertical force applied by the jacks is measured by a load cell.
Assume that we have the longitudinal position of F and the jack is lifted
such that the car makes an angle ! with the horizontal plane. The slope
angle ! is measurable using level meters. Assuming the force under the
front wheels is measured as 2I}1 , the height of the mass center can be
calculated by static equilibrium conditions
X
X
P| = 0
(2.27)
I] = 0
Applying the equilibrium equations
2I}1 + 2I}2 pj = 0
2I}1 (d1 cos ! + (k U) sin !)
+2I}2 (d2 cos ! (k U) sin !) = 0
(2.28)
(2.29)
provides the vertical position of F and the reaction forces under the rear
44
2. Forward Vehicle Dynamics
wheels.
1
pj I}1
2
I} (U sin ! d1 cos !) + I}2 (U sin ! + d2 cos !)
k= 1
0=5pj sin !
μ
¶
2I}1
d1 I}1 d2 I}2
cot ! = U + d2 o cot !
=U
0=5pj
pj
I}2 =
(2.30)
(2.31)
There are three main assumptions in this calculation:
1= the tires are assumed to be rigid disks with radius U,
2= uid shift, such as fuel, coolant, and oil, are ignored,
3= suspension de ections are assumed to be zero.
Suspension de ection generates the maximum eect on height determination error. To eliminate the suspension de ection, we should lock the
suspension, usually by replacing the shock absorbers with rigid rods to keep
the vehicle at its ride height.
Let us consider a car with the following measured specications
p = 1576 kg
2I}1 = 6300 N
! = 30 deg 0=5236 rad
d1 = 108=6 cm
o = 265 cm
(2.32)
W luh : 245@40]U17
If the tire has a radius of
U 30 cm
(2.33)
then the car has a F at height k.
k = 65=2 cm
(2.34)
Example 46 Dierent front and rear tires.
Depending on the application, it is sometimes necessary to use dierent
type of tires and wheels for front and rear axles. When the longitudinal
position of F for a symmetric vehicle is determined, we can nd the height
of F by measuring the load on only one axle. As an example, consider the
motorcycle in Figure 2.5. It has dierent front and rear tires.
Assume the load under the rear wheel of the motorcycle I} is known.
The height k of F can be found by taking a moment of the forces about the
tireprint of the front tire.
¶
μ
I}
I}
(2.35)
k = Uu + d2 1 o cot ! 1 (Uu Ui )
pj
pj
In case of a four wheel vehicle, the equation will be
¶
μ
2I}1
2I}1
o cot ! (Uu Ui )
k = Uu + d2 pj
pj
(2.36)
2. Forward Vehicle Dynamics
45
h R f sin I
a1
x
a2
h Rr sin I
Rf
C
h
Fz1
mg
Rr
g
H
I
Fz2
FIGURE 2.5. A motorcycle with dierent front and rear tires.
Example 47 Statically indeterminate.
A vehicle with more than three wheels is statically indeterminate. To
determine the vertical force under each tire, we need to know the mechanical
properties and conditions of the tires, such as the value of de ection at the
center of the tire, and its vertical stiness.
2.2 Parked Car on an Inclined Road
Passenger cars are usually equipped with rear wheels hand brakes or park
brakes. When a car is parked on an inclined pavement as shown in Figure
2.6, the normal force, I} , under each of the front and rear wheels, I}1 , I}2 ,
is
1
d2
1
k
pj cos ! pj sin !
2
o
2
o
1
d1
1
k
I}2 = pj cos ! + pj sin !
2
o
2
o
o = d1 + d2
I}1 =
(2.37)
(2.38)
and the brake force, I{2 , is
I{2 =
1
pj sin !
2
(2.39)
46
2. Forward Vehicle Dynamics
a1
z
x
h
a2
C
2F z1
mg
2F x
2
2F z2
g
I
FIGURE 2.6. A parked car on inclined pavement.
where, ! is the angle of the road with the horizon. The horizon is perpendicular to the gravitational acceleration g.
Proof. Consider the car shown in Figure 2.6. Let us assume the parking
brake forces are applied on only the rear tires. It means the front tires are
free to spin. Applying the planar static equilibrium equations
X
(2.40)
I{ = 0
X
I} = 0
(2.41)
X
P| = 0
(2.42)
shows that
2I{2 pj sin ! = 0
2I}1 + 2I}2 pj cos ! = 0
2I}1 d1 + 2I}2 d2 2I{2 k = 0
(2.43)
(2.44)
(2.45)
These equations provide the brake force and reaction forces under the front
and rear tires.
1
d2
1
k
I}1 = pj cos ! pj sin !
(2.46)
2
o
2
o
1
d1
1
k
I}2 = pj cos ! + pj sin !
(2.47)
2
o
2
o
1
I{2 = pj sin !
(2.48)
2
2. Forward Vehicle Dynamics
47
Example 48 Tilting angle.
When ! = 0, Equations (2.37) and (2.38) reduce to (2.1) and (2.2). By
increasing the inclination angle, the normal force under the front tires of
a parked car decreases and the normal force and braking force under the
rear tires increase. The limit for increasing ! is where the weight vector
pg goes through the contact point of the rear tire with the ground. Such an
angle is called a tilting angle, and shown by !W .
At ! = !W , we have I}1 = 0, and therefore,
tan !W =
d2
k
(2.49)
Example 49 Ultimate angle.
The required braking force I{2 increases by the inclination angle. Because
I{2 is equal to the friction force between the tire and pavement, its maximum depends on the tire and pavement conditions. There is an ultimate
angle !P at which the braking force I{2 will saturate and cannot increase
any more. At this maximum angle, the braking force is proportional to the
normal force I}2
I{2 = {2 I}2
(2.50)
where, the coe!cient {2 is the {-direction friction coe!cient for the rear
wheels. At ! = !P , the equilibrium equations (2.43)-(2.45) will reduce to
2{2 I}2 pj sin ! = 0
2I}1 + 2I}2 pj cos ! = 0
2I}1 d1 + 2I}2 d2 2{2 I}2 k = 0
(2.51)
(2.52)
(2.53)
These equations provide
1
1
d2
k
pj cos !P pj sin !P
2
o
2
o
1
1
d1
k
I}2 = pj cos !P + pj sin !P
2
o
2
o
d1 {2
tan !P =
o {2 k
I}1 =
(2.54)
(2.55)
(2.56)
showing that there is a relation between the friction coe!cient {2 , maximum inclination or ultimate angle !P , and the geometrical position of the
mass center F. The angle !P increases by decreasing k.
For a car having the specications
{2 = 1
d1 = 110 cm
o = 230 cm
k = 35 cm
(2.57)
the ultimate angle is
!P 0=514 rad 29=43 deg
(2.58)
48
2. Forward Vehicle Dynamics
Example 50 Front wheel braking.
When the front wheels are the park braking wheels I{2 = 0 and I{1 6= 0.
In this case, the equilibrium equations (2.40)-(2.42) will be
2I{1 pj sin ! = 0
2I}1 + 2I}2 pj cos ! = 0
2I}1 d1 + 2I}2 d2 2I{1 k = 0
(2.59)
(2.60)
(2.61)
These equations provide the brake force I{1 and reaction forces under the
front and rear tires.
1
d2
1
k
pj cos ! pj sin !
2
o
2
o
1
d1
1
k
I}2 = pj cos ! + pj sin !
2
o
2
o
1
I{1 = pj sin !
2
I}1 =
(2.62)
(2.63)
(2.64)
At the ultimate angle ! = !P , we have
I{1 = {1 I}1
(2.65)
2{1 I}1 pj sin !P = 0
2I}1 + 2I}2 pj cos !P = 0
2I}1 d1 2I}2 d2 + 2{1 I}1 k = 0
(2.66)
(2.67)
(2.68)
and therefore,
These equations provide
1
1
d2
k
pj cos !P pj sin !P
2
o
2
o
1
1
d1
k
I}2 = pj cos !P + pj sin !P
2
o
2
o
d2 {1
tan !P =
o {1 k
I}1 =
(2.69)
(2.70)
(2.71)
Let us name the ultimate angle for the front wheel brake in Equation (2.71)
as !Pi , and the ultimate angle for the rear wheel brake in Equation (2.56)
as !Pu . Comparing !Pi and !Pu shows that
!Pi
!Pu
=
¢
¡
d2 {1 o {2 k
¢
¡
d1 {2 o {1 k
(2.72)
We may assume the front and rear tires are the same,
{1 = {2
(2.73)
2. Forward Vehicle Dynamics
z
x
a1
h
a2
C
1
2F x
2F z1
mg
2
2F x
49
2F z2
g
I
FIGURE 2.7. A four wheel brake car, parked uphill.
and therefore,
!Pi
!Pu
=
d2
d1
(2.74)
Hence, if d2 ? d1 then !Pi ? !Pu and therefore, a rear brake is more
eective than a front brake on uphill parking as long as !Pu is less than the
tilting angle, !Wu ? arctan dk1 . At the tilting angle, the weight vector passes
through the contact point of the rear wheel with the ground.
Similarly we may conclude that when parked on a downhill road, the front
brake is more eective than the rear brake.
Example 51 Four-wheel braking.
Consider a four-wheel brake car, parked uphill as shown in Figure 2.7.
In these conditions, there will be two brake forces I{1 on the front wheels
and two brake forces I{2 on the rear wheels.
The equilibrium equations for this car are
2I{1 + 2I{2 pj sin ! = 0
2I}1 + 2I}2 pj cos ! = 0
2I}1 d1 + 2I}2 d2 (2I{1 + 2I{2 ) k = 0
(2.75)
(2.76)
(2.77)
These equations provide the brake force and reaction forces under the front
50
2. Forward Vehicle Dynamics
and rear tires.
d2
1
k
1
pj cos ! pj sin !
2
o
2
o
d1
1
k
1
I}2 = pj cos ! + pj sin !
2
o
2
o
1
I{1 + I{2 = pj sin !
2
I}1 =
(2.78)
(2.79)
(2.80)
At the ultimate angle ! = !P , all wheels will begin to slide simultaneously
and therefore,
I{1 = {1 I}1
I{2 = {2 I}2
(2.81)
The equilibrium equations show that
Assuming
2{1 I}1 + 2{2 I}2 pj sin !P = 0
2I}1 + 2I}2 pj cos !P = 0
¢
¡
2I}1 d1 + 2I}2 d2 2{1 I}1 + 2{2 I}2 k = 0
{1 = {2 = {
(2.82)
(2.83)
(2.84)
(2.85)
will provide
d2
k
1
1
pj cos !P pj sin !P
2
o
2
o
d1
k
1
1
I}2 = pj cos !P + pj sin !P
2
o
2
o
tan !P = {
I}1 =
(2.86)
(2.87)
(2.88)
2.3 Accelerating Car on a Level Road
When a car is speeding with acceleration d on a level road as shown in
Figure 2.8, the vertical forces under the front and rear wheels are
I}1
=
I}2
=
d2 1
k
1
pj pd
2
o
2
o
1
d1 1
k
pj + pd
2
o
2
o
(2.89)
(2.90)
The rst terms, 12 pj do2 and 12 pj do1 , are called static parts, and the second
terms, ± 12 pd ko , are called dynamic parts of the normal forces.
Proof. The vehicle is considered as a rigid body that moves along a horizontal road. The force at the tireprint of each tire may be decomposed
2. Forward Vehicle Dynamics
51
z
a2
a1
g
C
2Fx2
2Fz2
x
a
h
2Fx1
mg
2Fz1
FIGURE 2.8. An accelerating car on a level pavement.
to a normal and a longitudinal force. The equations of motion for the accelerating car come from Newton’s equation in {-direction and two static
equilibrium equations in | and }-direction.
X
I{ = pd
(2.91)
X
I} = 0
(2.92)
X
P| = 0
(2.93)
Expanding the equations produces three equations for four unknowns
I{1 , I{2 , I}1 , I}2 .
2I{1 + 2I{2 = pd
2I}1 + 2I}2 pj = 0
2I}1 d1 + 2I}2 d2 2 (I{1 + I{2 ) k = 0
(2.94)
(2.95)
(2.96)
However, it is possible to eliminate (I{1 + I{2 ) between the rst and third
equations, and solve for the normal forces I}1 , I}2 .
I}1
I}2
1
d2
1
k
pj pd
2
o
2
o
1
d1
1
k
= (I}2 )vw + (I}2 )g|q = pj + pd
2
o
2
o
= (I}1 )vw + (I}1 )g|q =
(2.97)
(2.98)
The static parts
(I}1 )vw
=
(I}2 )vw
=
1
d2
pj
2
o
1
d1
pj
2
o
(2.99)
(2.100)
52
2. Forward Vehicle Dynamics
are weight distribution for a stationary car and depend on the horizontal
position of the mass center. However, the dynamic parts
(I}1 )g|q
(I}2 )g|q
1
k
= pd
2
o
1
k
=
pd
2
o
(2.101)
(2.102)
indicate the weight distribution because of horizontal acceleration, and depend on the vertical position of the mass center.
When accelerating d A 0, the normal forces under the front tires are less
than the static load, and under the rear tires are more than the static load.
Example 52 Front-wheel-drive accelerating on a level road.
When the car is front-wheel-drive, I{2 = 0. Equations (2.94) to (2.96)
will provide the same vertical tireprint forces as (2.89) and (2.90). However,
the required horizontal force, 2I{1 , to achieve the same acceleration, d, must
be provided by solely the front wheels.
2I{1 = pd
(2.103)
Example 53 Rear-wheel drive accelerating on a level road.
If a car is rear-wheel drive then, I{1 = 0 and the required force to achieve
the acceleration, d, must be provided only by the rear wheels.
2I{2 = pd
(2.104)
The vertical force under the wheels will still be the same as (2.89) and
(2.90).
Example 54 Maximum acceleration on a level road.
The maximum acceleration of a car is proportional to the friction under
its tires. We assume the friction coe!cients at the front and rear tires are
equal and all tires reach their maximum tractions at the same time.
I{1 = ±{ I}1
I{2 = ±{ I}2
(2.105)
Newton’s equation (2.94) can now be written as
pd = ±2{ (I}1 + I}2 )
(2.106)
Substituting I}1 and I}2 from (2.97) and (2.98) results in
d = ±{ j
(2.107)
Therefore, the maximum acceleration and deceleration depend directly on
the friction coe!cient.
2. Forward Vehicle Dynamics
53
Example 55 Maximum acceleration for a single-axle drive car.
The maximum acceleration duzg for a rear-wheel-drive car is achieved
when I{2 reaches its maximum frictional force. Substituting I{1 = 0, and
I{2 = { I}2 in Equation (2.94) and using Equation (2.90)
¶
μ
d1 k duzg
+
= pduzg
(2.108)
{ pj
o
o j
yields
d1 {
duzg
=
=
j
o k{
{
d1
(2.109)
k o
1 {
o
If friction is strong enough, the front wheels can leave the ground at I{2 ?
{ I}2 and we have I}1 = 0. Substituting I}1 = 0 in Equation (2.89)
provides the maximum acceleration at which the front wheels are still on
the road.
d2
duzg
=
(2.110)
j
k
Therefore, the maximum attainable acceleration of a rear-wheel-drive car
would be the less value of Equation (2.109) or (2.110).
Similarly, the maximum acceleration di zg for a front-wheel drive car is
achieved when we substitute I{2 = 0, I{1 = { I}1 in Equation (2.94) and
use Equation (2.89).
³
d2 {
di zg
{
d1 ´
=
1
=
(2.111)
k
j
o + k{
o
1 + {
o
To visualize the eect of changing the position of mass center on the
maximum achievable acceleration, we plot Figure 2.9 for a sample car with
{ = 1
k = 0=56 m
o = 2=6 m
(2.112)
Passenger cars are usually in the range 0=4 ? (d1 @o) ? 0=6. In this range,
duzg A di zg and rear-wheel-drive cars can reach higher forward acceleration than front-wheel-drive cars. It is an important applied fact, especially
for race cars.
For the race cars, the maximum acceleration may also be limited by the
lift o condition (2.110).
Example 56 Minimum time for 0 100 km@ h on a level road.
Consider a car with the following characteristics:
ohqjwk = 4245 mm
zlgwk = 1795 mm
khljkw = 1285 mm
k = 220 mm
{ = 1
zkhho edvh = 2272 mm
i urqw wudfn = 1411 mm
uhdu wudfn = 1504 mm
qhw zhljkw = 1500 kg
d1 = d2
(2.113)
54
2. Forward Vehicle Dynamics
1.2
a/g
1
0.8
0.6
0.4
d
a rw
0.2
/g
afw
d/g
0
0.2
0
0.4
a1 / l
0.6
0.8
1
FIGURE 2.9. Eect of mass center position on the maximum achievable acceleration of a front- and a rear-wheel drive car.
Assume the car is rear-wheel-drive and its engine can provide the maximum
traction supported by friction. Equation (2.90) determines the load on the
rear wheels and therefore, the forward equation of motion (2.94) becomes
2I{2
= 2{ I}2 = { pj
d1
k
+ { pd
o
o
= pd
(2.114)
Rearrangement provides the following dierential equation to calculate velocity and displacement:
d1
{ j
o = d1 { j
(2.115)
d={
¨=
k
o k{
1 {
o
Taking an integral between y = 0 and y = 100 km@ h 27=78 m@ s
Z 27=78
Z w
gy =
d gw
(2.116)
0
0
shows that the minimum time for 0 100 km@ h on a level road for the
rear-wheel-drive is
27=78
5=11 s
(2.117)
w=
d1
j{
o k{
If the same car was front-wheel-drive, then the traction force would be
2I{1
= 2{ I}1 = { pj
= pd
d2
k
{ pd
o
o
(2.118)
2. Forward Vehicle Dynamics
55
and the equation of motion would reduce to
d2
d2
o
d={
¨=
= j{
k1
o + k{
1 + { j
oj
{ j
(2.119)
The minimum time for 0 100 km@ h on a level road for the front-wheeldrive car would be
27=78
6= 21 s
(2.120)
w=
d2
j{
o + k{
Now consider the same car to be four-wheel-drive with the assumption
that all wheels reach their maximum possible traction at the same time.
Then, the traction force is
2I{1 + 2I{2
= 2{ (I}1 + I}2 ) =
j
p (d1 + d2 )
o
= pd
(2.121)
and the minimum time for 0100 km@ h on a level road for this four-wheeldrive car can theoretically be reduced to
w=
27=78
2=83 s
j
(2.122)
2.4 Accelerating Car on an Inclined Road
When a car is accelerating on an inclined pavement with angle ! as shown
in Figure 2.10, the normal force under each of the front and rear wheels,
I}1 , I}2 , would be:
μ
¶
d2
k
1
k
1
pj
cos ! sin ! pd
2
o
o
2
o
μ
¶
d1
k
1
k
1
I}2 = pj
cos ! + sin ! + pd
2
o
o
2
o
o = d1 + d2
I}1 =
(2.123)
(2.124)
The dynamic parts, ± 12 pd ko , depend on acceleration d and height k of mass
center F and not on the slope !, while the static parts are in uenced by
the slope angle ! as well as the longitudinal and vertical positions of the
mass center.
Proof. The Newton’s equation in {-direction and two static equilibrium
equations must be examined to nd the equation of motion and ground
56
2. Forward Vehicle Dynamics
x
a1
z
h
a2
a
C
2F x
1
2F z1
mg
2
2F x
2F z2
g
I
FIGURE 2.10. An accelerating car on inclined pavement.
reaction forces.
X
I{ = pd
(2.125)
I} = 0
(2.126)
P| = 0
(2.127)
X
X
Expanding these equations produces three equations for four unknowns
I{1 , I{2 , I}1 , I}2 .
2I{1 + 2I{2 pj sin ! = pd
2I}1 + 2I}2 pj cos ! = 0
2I}1 d1 + 2I}2 d2 2 (I{1 + I{2 ) k = 0
(2.128)
(2.129)
(2.130)
It is possible to eliminate (I{1 + I{2 ) between the rst and third equations,
and solve for the normal forces I}1 , I}2 .
I}1
I}2
= (I}1 )vw + (I}1 )g|q
μ
¶
1
d2
k
1
k
=
pj
cos ! sin ! pd
2
o
o
2
o
(2.131)
= (I}2 )vw + (I}2 )g|q
μ
¶
1
d1
k
1
k
=
pj
cos ! + sin ! + pd
2
o
o
2
o
(2.132)
2. Forward Vehicle Dynamics
57
Example 57 Front-wheel-drive car, accelerating on inclined road.
For a front-wheel-drive car, we should substitute I{1 = 0 in Equations
(2.128) and (2.130) to have the governing equations. However, it does not
aect the expression of the ground reaction forces under the tires (2.131)
and (2.132) as long as the car is driven under its frictional limit conditions.
Example 58 Rear-wheel-drive car, accelerating on inclined road.
Substituting I{2 = 0 in Equations (2.128) and (2.130) and solving for the
normal reaction forces under each tire provides the same results as (2.131)
and (2.132). Hence, the expression of the normal forces applied on the tires
do not make any dierence if the car is front-, rear-, or all-wheel drive. As
long as we drive in a straight inclined path at low acceleration, the drive
wheels can be the front or the rear ones. However, the advantages and disadvantages of front-, rear-, or all-wheel drive cars appear in maneuvering,
slippery roads, or when the maximum acceleration is required.
Example 59 Maximum acceleration on an inclined road.
The maximum acceleration depends on the friction under the tires. Let us
assume the friction coe!cients at the front and rear tires are equal. Then,
the front and rear traction forces are
I{1 { I}1
I{2 { I}2
(2.133)
If we assume the front and rear wheels reach their traction limits at the
same time, then
I{1 = ±{ I}1
I{2 = ±{ I}2
(2.134)
and we may rewrite Newton’s equation (2.125) as
pdP = ±2{ (I}1 + I}2 ) pj sin !
(2.135)
where, dP is the maximum achievable acceleration.
Now substituting I}1 and I}2 from (2.131) and (2.132) results in
dP
= ±{ cos ! sin !
j
(2.136)
Accelerating on an uphill road and braking on a downhill road are the extreme cases in which the car can stall, d = 0. In these cases, the car can
move as long as
{ |tan !|
(2.137)
Following the directions of the body coordinate frame of the vehicle, uphill
road should be assigned by ! ? 0 and downhill by ! A 0, as the slope angle
! should be measured about the |-axis. However, in this section we have
used the absolute value of the slope angle for calculations.
58
2. Forward Vehicle Dynamics
Example 60 F Limits of acceleration and inclination angle.
Considering I}1 A 0 and I}2 A 0, we can write Equations (2.123) and
(2.124) as
d2
d
cos ! sin !
(2.138)
j
k
d1
d
cos ! sin !
(2.139)
j
k
Hence, if there is limit on friction, the maximum achievable acceleration
(d A 0) is limited by d2 , k, !; while the maximum deceleration (d ? 0) is
limited by d1 , k, !. These two equations can be combined to result in
d
d2
d1
cos !
(2.140)
cos ! + sin ! k
j
k
If d $ 0, then the limits of the inclination angle would be
d1
d2
tan ! k
k
(2.141)
This is the maximum and minimum road inclination angles that the car
can stay on without tilting or falling.
Figure 2.11 illustrates the boundaries of maximum acceleration and deceleration at each uphill or downhill angles for a sample car with
d1 = 1=5 m
d1 = 1=7 m
k = 0=7 m
(2.142)
On a at road, ! = 0 and the possible acceleration and deceleration are
limited to
d
d2
d1
(2.143)
k
j
k
d
2=14 2=43
(2.144)
j
By increasing the slope angle of the uphill road, the maximum possible
acceleration drops according to the upper line in Figure 2.11. The friction cannot accelerate the car anymore at ! = arctan dk2 = 67=62 deg. The
car will fall back if the slope is higher. Decreasing the slope of the downhill road, the maximum possible deceleration drops according to the lower
line in the gure. The car cannot decelerate anymore at downhill slope of
! = arctan dk1 = 64=98 deg and the car will fall down if ! is higher. The
limit of Equation (2.141) is illustrated in the magnied gure.
Example 61 Maximum deceleration for a single-axle-brake car.
We can nd the maximum braking deceleration di ze of a front-wheelbrake car on a horizontal road by substituting ! = 0, I{2 = 0, I{1 =
{ I}1 in Equation (2.128) and using Equation (2.123)
¶
μ
d2 k di ze
= di ze
(2.145)
{ j
o
o j
acceleration
2. Forward Vehicle Dynamics
deceleration
a
g
uphill
20
10
30
M > deg @
a1
1.5 > m @
a2
1.7 > m @
h
0.6 > m@
50
59
60
downhill
64.98
M > deg @
67.6
FIGURE 2.11. Limits of acceleration and deceleration versus inclination angle.
which provides us with
di ze
=
j
{
1 {
k
o
³
1
d1 ´
o
(2.146)
Similarly, the maximum braking deceleration duze of a rear-wheel-brake car
can be achieved when we substitute ! = 0, I{1 = 0, I{2 = { I}2 and use
Equation (2.124).
{
d1
duze
=
(2.147)
k
j
o
1 + {
o
The eect of changing the position of the mass center on the maximum
achievable braking deceleration is shown in Figure 2.12 for a sample car
with
{ = 1
k = 0=56 m
o = 2=6 m
(2.148)
60
2. Forward Vehicle Dynamics
0
-0.2
-0.4
afw
-0.6
b/g
-0.8
a/g
-1
a rwb
/g
-1.2
0.2
0
0.4
a1 / l
0.6
0.8
1
FIGURE 2.12. Eect of mass center position on the maximum achievable decceleration of a front-wheel and a rear-wheel-drive car.
Passenger cars are usually in the range 0=4 ? (d1 @o) ? 0=6. In this range,
(di ze @j) ? (duze @j) and therefore, front-wheel-brake cars can reach better
forward deceleration than rear-wheel-brake cars. Hence, front brakes are
much more important than the rear brakes.
Example 62 F A car with a trailer.
Figure 2.13 depicts a car moving on an inclined road while pulling a
trailer. To analyze the car-trailer motion, we need to separate the car and
trailer to see the forces at the hinge, as shown in Figure 2.14. We assume
the mass center of the trailer Fw is at distance e3 in front of the only axle
of the trailer. If Fw is behind the trailer axle, then e3 should be negative in
the following equations.
For an ideal hinge between a car and a trailer moving in a straight path,
there must be a horizontal force I{w and a vertical force I}w , but there would
be no moment. Writing the Newton’s equation in {-direction and two static
equilibrium equations for both the trailer and the vehicle
X
I{ = pw d
X
I} = 0
X
P| = 0
(2.149)
2I{1 + 2I{2 I{w pj sin ! = pd
2I}1 + 2I}2 I}w pj cos ! = 0
2I}1 d1 + 2I}2 d2 2 (I{1 + I{2 ) k
+I{w (k k1 ) I}w (e1 + d2 ) = 0
(2.150)
(2.151)
we nd the following set of equations:
(2.152)
2. Forward Vehicle Dynamics
z
61
a1
x
a2
h
b1
b2
a
C
2F z
1
1
2F x
b3
I
Ct
F x2
2
g
2F z2
mt g
3
2F z
FIGURE 2.13. A car moving on an inclined road and pulling a trailer.
z
a1
x
a2
h
g
b1
C
I mg
F xt
h1
F zt
2
2F x
2F z2
a
1
2F x
2F z1
b2
F zt
b3
h2
h1
F xt
Ct
mt g
I
2F z3
FIGURE 2.14. Free-body-diagram of a car and the trailer when moving on an
uphill road.
62
2. Forward Vehicle Dynamics
I{w pw j sin ! = pw d
2I}3 + I}w pw j cos ! = 0
2I}3 e3 I}w e2 I{w (k2 k1 ) = 0
(2.153)
(2.154)
(2.155)
If the value of traction forces I{1 and I{2 are given, then these are six equations for six unknowns: d, I{w , I}w , I}1 , I}2 , I}3 . Solving these equations
provides the following solutions:
2
(I{1 + I{2 ) j sin !
p + pw
2pw
(I{1 + I{2 )
p + pw
k1 k2 2pw
e3
(I{1 + I{2 ) +
pw j cos !
e2 e3 p + pw
e2 e3
d =
I{w
=
I}w
=
I}1
I}2
I}3
o
=
¶
μ
e3 2d2 e1
d2
pw + p j cos !
2o e2 e3
e3
¸
2d2 e1
I{1 + I{2
+
(k1 k2 ) pw k1 pw kp
e2 e3
o (p + pw )
¶
μ
e3 d1 d2 + e1
d1
=
pw + p j cos !
2o
e2 e3
e3
¸
d1 d2 + e1
I{1 + I{2
+
(k1 k2 ) pw + k1 pw + kp
e2 e3
o (p + pw )
1 e2
k1 k2 pw
=
pw j cos ! +
(I{1 + I{2 )
2 e2 e3
e2 e3 p + pw
= d1 + d2
(2.156)
(2.157)
(2.158)
(2.159)
(2.160)
(2.161)
(2.162)
However, if the value of acceleration d is known, then unknowns are: I{1 +
I{2 , I{w , I}w , I}1 , I}2 , I}3 .
I{1 + I{2
I{w
I}w
I}1
1
(p + pw ) (d + j sin !)
2
= pw (d + j sin !)
k1 k2
e3
=
pw (d + j sin !) +
pw j cos !
e2 e3
e2 e3
=
(2.163)
(2.164)
(2.165)
¶
d2
2d2 e1
=
pw + p j cos !
(2.166)
e2 e3
e3
¸
1 2d2 e1
+
(k1 k2 ) pw k1 pw kp (d + j sin !)
2o e2 e3
e3
2o
μ
2. Forward Vehicle Dynamics
I}2
=
I}3
o
63
μ
¶
d1 d2 + e1
d1
pw + p j cos !
(2.167)
e2 e3
e3
¸
1 d1 d2 + e1
+
(k1 k2 ) pw + k1 pw + kp (d + j sin !)
2o
e2 e3
e3
2o
1 pw
(e2 j cos ! + (k1 k2 ) (d + j sin !))
2 e2 e3
= d1 + d2
=
(2.168)
Example 63 F Maximum inclination angle for a car with a trailer.
For a car and trailer as shown in Figure 2.13, the maximum inclination angle !P is the angle at which the car cannot accelerate the vehicle.
Substituting d = 0 and ! = !P in Equation (2.156) shows that
sin !P =
2
(I{1 + I{2 )
(p + pw ) j
(2.169)
The value of maximum inclination angle !P increases by decreasing the
total weight of the vehicle and trailer (p + pw ) j or increasing the traction
force I{1 + I{2 .
The traction force is limited by the maximum torque on the drive wheels
and the friction under the drive tires. Let us assume the vehicle is fourwheel-drive and friction coe!cients at the front and rear tires are equal.
Then, the front and rear traction forces are
I{1 { I}1
I{2 { I}2
(2.170)
If we assume the front and rear wheels reach their traction limits at the
same time, then
I{1 = { I}1
I{2 = { I}2
(2.171)
and we may rewrite the Equation (2.169) as
sin !P =
2{
(I}1 + I}2 )
(p + pw ) j
(2.172)
Now substituting I}1 and I}2 from (2.159) and (2.160) yields
(pe3 pe2 pw e3 ) { cos !P + (e2 e3 ) (p + pw ) sin !P
pw (k1 k2 )
= 2{
(I{1 + I{2 )
(2.173)
p + pw
If we arrange Equation (2.173) as
D cos !P + E sin !P = F
then
r
F
F2
!P = atan2( s
>± 1 2
) atan2(D> E)
D + E2
D2 + E 2
(2.174)
(2.175)
64
2. Forward Vehicle Dynamics
and
p
F
!P = atan2( s
> ± D2 + E 2 F 2 ) atan2(D> E)
D2 + E 2
(2.176)
where
D = (pe3 pe2 pw e3 ) {
E = (e2 e3 ) (p + pw )
pw (k1 k2 )
(I{1 + I{2 )
F = 2{
p + pw
(2.177)
(2.178)
(2.179)
For a rear-wheel-drive car pulling a trailer with the following characteristics:
o = 2272 mm z = 1457 mm k = 230 mm
d1 = d2
k1 = 310 mm
k2 = 560 mm
e1 = 680 mm e2 = 610 mm
e3 = 120 mm
p = 1500 kg pw = 150 kg
d = 1 m@ s2
{ = 1
! = 10 deg
(2.180)
I}2 = 3877=93 N
I}w = 147=99 N
I{2 = 2230=37 N
(2.181)
we nd
I}1
I}3
I{w
= 3441=78 N
= 798=57 N
= 405=52 N
To check if the required traction force I{2 is applicable, we should compare
it to the maximum available friction force I}2 and it must be
I{2 I}2
(2.182)
Example 64 F Solution of equation d cos + e sin = f=
The rst type of trigonometric equation is
d cos + e sin = f
(2.183)
It can be solved by introducing two new variables u and such that
and therefore,
u=
d = u sin e = u cos (2.184)
p
d2 + e2
= atan2(d> e)
(2.185)
Substituting the new variables shows that
f
sin( + ) =
u
r
cos( + ) = ± 1 f2
u2
(2.186)
2. Forward Vehicle Dynamics
Hence, the solutions of the problem are
r
f2
f
= atan2( > ± 1 2 ) atan2(d> e)
u
u
65
(2.187)
and
f p
= atan2( > ± u2 f2 ) atan2(d> e)
(2.188)
u
Therefore, the equation d cos + e sin = f has two solutions if u2 =
2
d + e2 A f2 , one solution if u2 = f2 , and no solution if u2 ? f2 .
Example 65 F The function arctan2 {| = atan2(|> {).
There are many situations in kinematics in which we need to nd an
angle based on the sin and cos functions of the angle. However, regular
arctan cannot show the eect of the individual sign for the numerator and
denominator. It always represents an angle in the rst or fourth quadrant,
2 2 . To overcome this problem and determine the angle in the
correct quadrant, the atan2 function is
¯ ¯
;
1 ¯ | ¯
A
{ A 0> | 6= 0
sgn
|
tan
¯ ¯
A
A
{
A
A
? sgn |
{ = 0> | 6= 0
atan2(|> {) =
(2.189)
2
¯ | ¯´
³
A
¯
¯
1
A
sgn
|
tan
{
?
0>
|
=
6
0
¯
¯
A
A
A
{
=
sgn {
{ 6= 0> | = 0
where sgn represents the signum function:
;
? 1
0
sgn devroxwh{ydoxh = ri
=
1
{A0
{ = 0 wkh
{?0
(2.190)
In this text, wherever tan1 {| is used, it should be calculated based on
atan2(|> {).
Example 66 Zero vertical force at the hinge.
We can make the vertical force at the hinge equal to zero by examining
Equation (2.158) for the hinge vertical force I}w .
I}w =
k1 k2 2pw
e3
(I{1 + I{2 ) +
pw j cos !
e2 e3 p + pw
e2 e3
(2.191)
To make I}w = 0, it is enough to adjust the position of trailer mass center
Fw exactly on top of the trailer axle and at the same height as the hinge. In
these conditions we have
k1 = k2
e3 = 0
(2.192)
that makes
I}w = 0
(2.193)
66
2. Forward Vehicle Dynamics
z
y
h
C
mg
2
2F y
1
2F y
b1
I
2F z1
b2
2F z2
FIGURE 2.15. Normal force under the uphill and downhill tires of a vehicle,
parked on banked road.
However, to increase safety, the load should also be distributed evenly
throughout the trailer. Heavy items should be loaded as low as possible,
mainly over the axle. Bulkier and lighter items should be distributed to
give a little positive e3 . Such a trailer is called nose weight at the towing
coupling.
2.5 Parked Car on a Banked Road
Figure 2.15 depicts the eect of a bank angle ! on the load distribution
of a vehicle. A bank causes the load on the lower tires to increase and the
load on the upper tires to decrease. The tire reaction forces are:
1 pj
(e2 cos ! k sin !)
2 z
1 pj
(e1 cos ! + k sin !)
I}2 =
2 z
z = e1 + e2
I}1 =
(2.194)
(2.195)
(2.196)
The maximum bank angle is
tan !P = |
(2.197)
at which the car will slides down laterally.
Proof. Starting with equilibrium equations in the body coordinate frame
2. Forward Vehicle Dynamics
67
(F{|})
X
I| = 0
(2.198)
I} = 0
(2.199)
P{ = 0
(2.200)
X
X
we can write
2I|1 + 2I|2 pj sin ! = 0
2I}1 + 2I}2 pj cos ! = 0
2I}1 e1 2I}2 e2 + 2 (I|1 + I|2 ) k = 0
(2.201)
(2.202)
(2.203)
We assumed the force under the lower tires, front and rear, are equal, and
also the forces under the upper tires, front and rear are equal. To calculate
the reaction forces under each tire, we may assume the overall lateral force
I|1 +I|2 as an unknown. The solution of these equations provide the lateral
and reaction forces under the upper and lower tires.
1
e2
1
k
pj cos ! pj sin !
2
z
2
z
1
e1
1
k
I}2 = pj cos ! + pj sin !
2
z
2
z
1
I|1 + I|2 = pj sin !
2
I}1 =
(2.204)
(2.205)
(2.206)
At the ultimate angle ! = !P , all wheels will begin to slide simultaneously and therefore,
I|1
I|2
= |1 I}1
= |2 I}2
(2.207)
(2.208)
The equilibrium equations show that
Assuming
2|1 I}1 + 2|2 I}2 pj sin ! = 0
(2.209)
2I}1 + 2I}2 pj cos ! = 0
¡
¢
2I}1 e1 2I}2 e2 + 2 |1 I}1 + |2 I}2 k = 0
(2.210)
|1 = |2 = |
(2.212)
(2.211)
provides us with
1
1
e2
k
pj cos !P pj sin !P
2
z
2
z
1
1
e1
k
I}2 = pj cos !P + pj sin !P
2
z
2
z
tan !P = |
I}1 =
(2.213)
(2.214)
(2.215)
68
2. Forward Vehicle Dynamics
These calculations are correct as long as
tan !P
|
e2
k
e2
k
(2.216)
(2.217)
If the lateral friction | is higher than e2 @k then the car will roll downhill.
To increase the capability of a car moving on a banked road, the car should
be as wide as possible with a mass center as low as possible.
Example 67 Tire forces of a parked car in a banked road.
A car having
p = 980 kg
k = 0=6 m
z = 1=52 m
e1 = e2
(2.218)
is parked on a banked road with ! = 4 deg. The forces under the lower and
upper tires of the car are:
I}1 = 2265=2 N
I}2 = 2529=9 N
I|1 + I|2 = 335=3 N
(2.219)
The ratio of the uphill force I}1 to downhill force I}2 depends on only
the mass center position.
e2 cos ! k sin !
I}1
=
I}2
e1 cos ! + k sin !
(2.220)
Assuming a symmetric car with e1 = e2 = z@2 simplies the equation to
I}1
z cos ! 2k sin !
=
I}2
z cos ! + 2k sin !
(2.221)
Figure 2.16 illustrates the behavior of force ratio I}1 @I}2 as a function of !
for k = 0=6 m and z = 1=52 m. The rolling down angle !P = tan1 (e2 @k) =
51=71 deg indicates the bank angle at which the force under the uphill wheels
become zero and the car rolls down. The negative part of the curve indicates
the required force to keep the car on the road, which is not applicable in real
situations.
Example 68 Vehicle on a banked round road.
When a vehicle of mass p is moving with speed y on a at round path
of radius U, the direction of the wheels lateral force are inward and provide
the required centripetal acceleration.
2I|1 + 2I|2 = p
y2
U
(2.222)
Knowing that the lateral wheels lateral force are limited by the maximum
friction force between tire and road, we conclude that there is a maximum
2. Forward Vehicle Dynamics
Rolling down angle
69
h=0.6 m
w=1.52 m
b1=b2
Fz1
Fz2
I [rad]
1
I[deg]
FIGURE 2.16. Illustration of the force ratio I}1 @I}2 as a function of road bank
angle !.
speed yP at which the required lateral force will not be produced by tires
and vehicle slides out of the road.
To have a safe road, we have to design round roads such that vehicle
do not need any wheel lateral force to provide the required centripetal force.
Designing roads with a bank angle is a good approximate solution; so a component of weight force provides the required centripetal force. Figure 2.17
illustrates a moving car on a round banked road. Assuming equal vertical
force under each tire, the balance of the applied forces on the vehicle in the
body coordinate frame provides us with
y2
sin ! = 0
U
y2
pj sin ! p cos ! = 0
U
pj cos ! 4I}1 p
(2.223)
(2.224)
The second equation indicates the required bank angle as a function of speed
tan ! =
y2
Uj
(2.225)
The bank angle is independent of the vehicle mass p and is a function
of the road radius of turn U and velocity of the vehicle y. Assuming that
the road radius of turn U is not variable, the bank angle must vary with
the speed of the vehicle. To design the road, we have to decide about the
proper velocity of vehicles on the road and calculate the bank angle based
on (2.225). Because the angle is only a function of the vehicle velocity, the
proper banked road works well for all types of vehicles as long as they keep
their velocity as recommended. Any lower or higher speed would respectively
70
2. Forward Vehicle Dynamics
z
y
h
I
C
a
1
2F z
g
mg
2 F z2
FIGURE 2.17. A moving car on a round banked road.
need some positive or negative lateral force to be generated by tires. The lack
or excessive tire lateral force will be provided by steering and sideslip angles
of vehicles, or by roll and camber angles of bicycles and motorcycles.
As of today, we do not make roads with variable !. However, imagine that
roads on a circular path were smart to adjust their bank angle according to
the speed of vehicles. Then, Equation (2.225) is the function the road must
follow. Figure 2.18 depicts the bank angle as a function of vehicle speed on
a circular path with radius U = 100 m. The angle is an increasing function
of speed with an asymptotic value of ! = 90 deg regardless of U provided
that U 6= 0.
lim ! = 90 deg
y$4
(2.226)
Figure 2.19 illustrates the sensitivity of the bank angle to the radius of
turn at dierent speed.
2.6 F Optimal Drive and Brake Force Distribution
A certain acceleration d can be achieved by adjusting and controlling the
longitudinal forces I{1 and I{2 . The optimal longitudinal forces under
the front and rear tires of a four wheel drive to achieve the maximum
acceleration are
I{1
pj
μ ¶2
1 d2 d
1k d
+
2o j
2 o j
1 k 1 d2
= 2{ + {
2 o
2
o
= (2.227)
2. Forward Vehicle Dynamics
71
90
I[deg]
R=100 m
g=9.81 m/s2
v [m/s]
0
72
144
216
288
0
45
90
134
179
v [km/h]
360
224
v [mi/h]
FIGURE 2.18. The required bank angle ! as a function of vehicle speed on a
circular path with radius U = 100 m.
90
R=10 m
25
I[deg]
50
100
200
500
R=1000 m
g=9.81 m/s2
v [m/s]
0
72
144
216
288
0
45
90
134
179
v [km/h]
360
224
v [mi/h]
FIGURE 2.19. The required road bank angle ! for dierent radius of turn and
dierent speed.
72
2. Forward Vehicle Dynamics
I{2
pj
=
=
μ ¶2
1 d1 d
1k d
+
2o j
2 o j
1 2 k 1 d1
+ {
2 {o
2
o
(2.228)
Proof. The longitudinal equation of motion for a car on a horizontal road
is
2I{1 + 2I{2 = pd
(2.229)
and the maximum traction forces under each tire is a function of normal
force and the friction coe!cient.
I{1
I{2
±{ I}1
±{ I}2
(2.230)
(2.231)
However, the normal forces are a function of the car’s acceleration and
geometry.
I}1
=
I}2
=
1
d2
1
kd
pj pj
2
o
2
oj
1
d1
1
kd
pj + pj
2
o
2
oj
(2.232)
(2.233)
We may generalize the equations by making them dimensionless. Under
the best conditions, we should adjust the traction forces to their maximum
μ
¶
d2
I{1
1
kd
=
(2.234)
pj
2 { o
oj
μ
¶
d1
1
kd
I{2
=
{
+
(2.235)
pj
2
o
oj
and therefore, the longitudinal equation of motion (2.229) becomes
d
= {
j
(2.236)
Substituting this result back into Equations (2.234) and (2.235) shows that
μ ¶2
1 d2 d
1k d
+
2o j
2 o j
μ ¶2
1 d1 d
1k d
+
2o j
2 o j
I{1
pj
= (2.237)
I{2
pj
=
(2.238)
Depending on the geometric parameters of the car (k> d1 > d2 ), and the acceleration d A 0, these two equations determine how much the front and
rear driving forces must be. The same equations are applied for deceleration
2. Forward Vehicle Dynamics
73
Fx2 / mg
Fx / mg
a
1
h
Fx1 / mg
a/g
Fx2 / mg
a2
h
Fx1 / mg
FIGURE 2.20. Optimal driving and braking forces for a sample car.
d ? 0, to determine the value of optimal front and rear braking forces. Figure 2.20 depicts a graphical illustration of the optimal driving and braking
forces for a sample car using the following data:
{ = 1
0=56
k
=
= 0=2154
o
2=6
d1
d2
1
=
=
o
o
2
(2.239)
When accelerating d A 0, the optimal driving force on the rear tire
grows rapidly while the optimal driving force on the front tire drops after
a maximum at
d
j
=
I{1
pj
=
1 d2
2 k
1 d2 d2
8 k o
(2.240)
(2.241)
The value of d@j = d2 @k is the maximum possible acceleration at which
the front tires lose their contact with the ground. The acceleration at which
front (or rear) tires lose their ground contact is called tilting acceleration.
The required force at the rear wheels to achieve d@j = d2 @k would be
I{2 @ (pj) = d2 @ (2k).
The opposite phenomenon happens when decelerating. For d ? 0, the
absolute value of the optimal front brake force increases rapidly and the
rear brake force goes to zero after a maximum at
d
j
I{1
pj
1 d2
2 k
1 d1 d1
= 8 k o
= (2.242)
(2.243)
The deceleration d@j = d1 @k is the maximum possible deceleration at
which the rear tires lose their ground contact.
74
2. Forward Vehicle Dynamics
Fx1 / mg
a2
h
Fx2 / mg
Braking
Driving
a1
h
FIGURE 2.21. Optimal traction and braking force distribution between the front
and rear wheels.
The graphical representation of the optimal driving and braking forces
can be shown better by plotting I{1 @ (pj) versus I{2 @ (pj) using d@j as
a parameter.
d
d2 k
j
(2.244)
I{1 =
d I{2
d1 + k
j
d2 { k
I{1
(2.245)
=
I{2
d1 + { k
Such a plot is shown in Figure 2.21. This is a design curve describing the
relationship between forces under the front and rear wheels to achieve the
maximum acceleration or deceleration.
Adjusting the optimal force distribution is not an automatic procedure
and needs a force distributor control system to measure and adjust the
forces.
Example 69 F Slope at zero.
The initial optimal traction force distribution is the slope of the optimal
curve (I{1 @ (pj) > I{2 @ (pj)) at zero.
μ ¶2
1k d
1 d2 d
I{1
+
g
d2
2o j
2 o j
pj
=
(2.246)
= lim
μ ¶2
I{2
d$0 1 k
d
1
d
1 d1 d
g
+
pj
2o j
2 o j
Therefore, the initial traction force distribution depends on only the position
of mass center F.
2. Forward Vehicle Dynamics
75
Fx1 / mg
Braking
Fx2 / mg
a1
h
FIGURE 2.22. Optimal braking force distribution between the front and rear
wheels, along with a thre-line under estimation.
Example 70 F Brake balance.
When braking, a car is stable if the rear wheels do not lock. Thus, the
rear brake forces must be less than the maximum possible braking force at
all time. This means the brake force distribution should always be in the
shaded area of Figure 2.22, and below the optimal curve. This restricts the
achievable deceleration, especially at low friction values, but increases the
stability of the car.
Whenever it is easier for a force distributor to follow a line, the optimal
brake curve is underestimated using two or three lines, and a control system
adjusts the force ratio I{1 @I{2 . A sample of three-line approximation is
shown in Figure 2.22.
Distribution of the brake force between the front and rear wheels is called
brake balance. Brake balance varies with deceleration. The higher the deceleration, the more load will transfer to the front wheels and the more
braking eort they can support. Meanwhile the rear wheels are unloaded
and they must have less braking force.
Example 71 F Best race car.
Racecars are always supposed to work at their maximum achievable acceleration to nish their race in minimum time. They are usually equipped
with rear-wheel-drive and all-wheel-brake. However, if an all-wheel-drive
race car is reasonable to build, then a force distributor, to follow the curve
shown in Figure 2.23, is what it needs to race better.
Example 72 F Eect of F location on braking.
Load is transferred from the rear wheels to the front when the brakes are
applied. The higher the F, the more load transfer. So, to improve braking,
76
2. Forward Vehicle Dynamics
Fx1 / mg
Driving
0.2
a2
h
0.1
0
0.2
0.4
0.6
0.8
1.0
1.2
Fx2 / mg
FIGURE 2.23. Optimal traction force distribution between the front and rear
wheels.
the mass center F should be as low as possible and as back as possible. This
is not feasible for every vehicle, especially for forward-wheel drive street
cars. However, this fact should be considered when a car is being designed
for better braking performance.
Example 73 F Front and rear wheel locking.
The optimal brake force distribution is according to Equation (2.245) for
an ideal I{1 @I{2 ratio. However, if the brake force distribution is not ideal,
then either the front or the rear wheels will lock up rst. Locking the rear
wheels makes the vehicle unstable, and it loses directional stability. When
the rear wheels lock, they slide on the road and they lose their capacity to
support lateral force. The resultant shear force at the tireprint of the rear
wheels reduces to a dynamic friction force in the opposite direction of the
sliding.
A slight lateral motion of the rear wheels, by any disturbance, develops
a yaw motion because of unbalanced lateral forces on the front and rear
wheels. The yaw moment turns the vehicle about the }-axis until the rear
end leads the front end and the vehicle turns 180 deg. Figure 2.24 illustrates
a 180 deg sliding rotation of a rear-wheel-locked car.
The lock-up of the front tires does not cause a directional instability,
although the car would not be steerable and the driver would lose control.
2.7 F Vehicles With More Than Two Axles
If a vehicle has more than two axles, such as the three-axle car shown
in Figure 2.25, then the vehicle will be statically indeterminate and the
normal forces under the tires cannot be determined by static equilibrium
equations. We need to consider the suspensions’ de ection to determine
their applied forces.
The q normal forces I}l under the tires can be calculated using the
2. Forward Vehicle Dynamics
Fx2
Fx1
77
Fy1
v
Fx2
Fx1
Fy1
FIGURE 2.24. 180 deg sliding rotation of a rear-wheel-locked car.
following q equations.
q
X
I}l pj cos ! = 0
(2.247)
I}l {l + k (d + pj sin !) = 0
(2.248)
2
l=1
2
q
X
l=1
{l {1
I}l
nl
{q {1
μ
I}
I}q
1
nq
n1
¶
I}1
=0
n1
l = 2> 3> · · · > q 1
(2.249)
where I{l and I}l are the longitudinal and normal forces under the tires
attached to the axle number l, and {l is the distance of mass center F
from the axle number l. The distance {l is positive for axles in front of F,
and is negative for the axles in back of F. The parameter nl is the vertical
stiness of the suspension at axle l.
Proof. For a multiple-axle vehicle, the following equations
X
(2.250)
I{ = pd
X
I} = 0
(2.251)
X
P| = 0
(2.252)
provide the same sort of equations as (2.128)-(2.130). However, if the total
number of axles are q, then the individual forces can be substituted by a
summation.
q
X
2
I{l pj sin ! = pd
(2.253)
l=1
q
X
2
l=1
I}l pj cos ! = 0
(2.254)
78
2. Forward Vehicle Dynamics
z
a3
a1
x
a2
h
a
C
2F x1
2F x3
I
2F x2
2 F z3
2F z1
mg
g
2F z2
FIGURE 2.25. A three-axle car moving on an inclined road.
2
q
X
I}l {l + 2k
l=1
q
X
I{l = 0
(2.255)
l=1
Pq
The overall forward force I{ = 2 l=1 I{l can be eliminated between
Equations (2.253) and (2.255) to make Equation (2.248). Then, there remain two equations (2.247) and (2.248) for q unknowns I}l , l = 1> 2> · · · > q.
Hence, we need q 2 extra equations to be able to nd the wheel loads.
The extra equations come from the compatibility among the suspensions’
de ection.
We ignore the tires’ compliance, and use } to indicate the vehicle’s vertical direction as is shown in Figure 2.25. Then, if }l is the suspension
de ection at the center of axle l, and nl is the vertical stiness of the
suspension at axle l, the de ections are
}l =
I}l
nl
(2.256)
For a at road, and a rigid vehicle, we must have
}l }1
}q }1
=
{l {1
{q {1
l = 2> 3> · · · > q 1
(2.257)
which, after substituting with (2.256), reduces to Equation (2.249). The
q 2 equations (2.257) along with the two equations (2.247) and (2.248)
are enough to calculate the normal load under each tire.
The resultant set of equations is linear and may be arranged in a matrix
form
[D] [[] = [E]
(2.258)
2. Forward Vehicle Dynamics
where
[[] =
5
2
9
2{1
9
9 {q {2
9
9
n1 o
9
9
···
[D] = 9 { {
q
l
9
9
n1 o
9
9
···
9
7 {q {q1
n1 o
o = {1 {q
[E] =
£
£
I}1
I}2
2
2{2
1
n2
···
···
···
···
···
···
···
···
···
···
···
···
···
···
···
1
nl
···
···
···
···
···
···
I}3
···
···
···
···
···
···
···
I}q
···
···
···
1
nq1
¤W
79
(2.259)
6
2
:
2{q
:
{2 {1 :
:
:
nq o
:
:
···
(2.260)
{l {1 :
:
:
nq o
:
:
···
:
{q1 {1 8
nq o
(2.261)
pj cos ! k (d + pj sin !) 0 · · ·
0
¤W
(2.262)
Example 74 F Wheel reactions for a three-axle car.
Figure 2.25 illustrates a three-axle car moving on an inclined road. We
start counting the axles of a multiple-axle vehicle from the front axle as
axle-1, and move sequentially to the back as shown in the gure.
The set of equations for the three-axle car is
2I{1 + 2I{2 + 2I{3 pj sin !
2I}1 + 2I}2 + 2I}3 pj cos !
2I}1 {1 + 2I}2 {2 + 2I}3 {3 + 2k (I{1 + I{2 + I{3 )
μ
¶
μ
¶
1
I}2
I}3
I}
I}
1
1 1
{2 {1 n2
n1
{3 {1 n3
n1
= pd
= 0
= 0
(2.263)
(2.264)
(2.265)
= 0
(2.266)
2I}1 + 2I}2 + 2I}3 pj cos ! = 0
2I}1 {1 + 2I}2 {2 + 2I}3 {3 + kp (d + j sin !) = 0
({2 n2 n3 {3 n2 n3 ) I}1 + ({1 n1 n2 {2 n1 n2 ) I}3
({1 n1 n3 {3 n1 n3 ) I}2 = 0
(2.267)
(2.268)
which can be simplied to
(2.269)
The set of equations for wheel loads is linear and may be rearranged in a
matrix form
[D] [[] = [E]
(2.270)
80
2. Forward Vehicle Dynamics
where
5
6
2
2
2
8 (2.271)
2{1
2{2
2{3
[D] = 7
n2 n3 ({2 {3 ) n1 n3 ({3 {1 ) n1 n2 ({1 {2 )
6
5
I}1
(2.272)
[[] = 7 I}2 8
I}3
5
6
pj cos !
[E] = 7 kp (d + j sin !) 8
(2.273)
0
The unknown vector may be found using matrix inversion
[[] = [D]1 [E]
(2.274)
The solution of the equations are
1
]1
I}1 =
n1 p
]0
1
]2
I}2 =
n2 p
]0
1
]3
I}3 =
n2 p
]0
(2.275)
where,
]0 = 4n1 n2 ({1 {2 )2 4n2 n3 ({2 {3 )2 4n1 n3 ({3 {1 )2
]1
]2
]3
= j ({2 n2 {1 n3 {1 n2 + {3 n3 ) k sin !
+d ({2 n2 {1 n3 {1 n2 + {3 n3 ) k
¡
¢
+j n2 {22 {1 n2 {2 + n3 {23 {1 n3 {3 cos !
= j ({1 n1 {2 n1 {2 n3 + {3 n3 ) k sin !
+d ({1 n1 {2 n1 {2 n3 + {3 n3 ) k
¡
¢
+j n1 {21 {2 n1 {1 + n3 {23 {2 n3 {3 cos !
= j ({1 n1 + {2 n2 {3 n1 {3 n2 ) k sin !
+d ({1 n1 + {2 n2 {3 n1 {3 n2 ) k
¡
¢
+j n1 {21 {3 n1 {1 + n2 {22 {3 n2 {2 cos !
{1 = d1
{2 = d2
{3 = d3
(2.276)
(2.277)
(2.278)
(2.279)
(2.280)
2.8 F Vehicles on a Crest and Dip
When a road has an outward or inward curvature, we call the road is a
crest or a dip. The road curvature can decrease or increase the normal
forces under the wheels. In this section we assume the radius of curvature
of the road UK is much bigger than the height of the vehicle mass center,
UK AA k.
2. Forward Vehicle Dynamics
z
a1
v
81
x
a2
T
a
C
I
F x2
2
2F x1 2F z1
mg
RH
z2
2F
FIGURE 2.26. A cresting vehicle at a point where the hill has a radius of curvature
Uk .
2.8.1 F Vehicles on a Crest
Moving on the convex curve of a hill is called cresting. The normal force
under the wheels of a cresting vehicle is less than the force on a at inclined road with the same slope, because of the developed centrifugal force
py 2 @UK in the }-direction.
Figure 2.26 illustrates a cresting vehicle at the point on the hill with a
radius of curvature UK and average slope ! with respect to horizon. The
traction and normal forces under its tires are approximately equal to
I{1 + I{2 I}1
I}2
o
1
p (d + j sin !)
2
(2.281)
μ
¶¸
1
d2
k
1
k 1 y2 d2
pj
cos ! + sin ! pd p
(2.282)
2
o
o
2
o
2 UK o
μ
¶¸
d1
k
1
k 1 y2 d1
1
pj
cos ! sin ! + pd p
(2.283)
2
o
o
2
o
2 UK o
= d1 + d2
(2.284)
Proof. For the cresting car shown in Figure 2.26, the normal and tangential
directions are equivalent to the } and { directions respectively. Hence, the
82
2. Forward Vehicle Dynamics
governing equation of motion for the car is
X
I{ = pd
X
y2
I} = p
UK
X
P| = 0
(2.285)
(2.286)
(2.287)
Expanding these equations yields
2I{1 cos + 2I{2 cos pj sin ! = pd
(2.288)
2
2I}1 cos 2I}2 cos + pj cos ! = p
y
UK
2I}1 d1 cos 2I}2 d2 cos + 2 (I{1 + I{2 ) k cos +2I}1 d1 sin 2I}2 d2 sin 2 (I{1 + I{2 ) k sin = 0
(2.289)
(2.290)
We may eliminate (I{1 + I{2 ) between the rst and third equations, and
solve for the total traction force I{1 + I{2 and wheel normal forces I}1 ,
I}2 .
I{1 + I{2
I}1
I}2
pd + pj sin !
2μ
cos ¶¸
1
d2
k (1 sin 2)
=
pj
cos ! +
sin !
2
o cos o cos cos 2
1
k (1 sin 2) 1 y 2 d2
pd
p
2
o cos cos 2
2 UK o cos μ
¶¸
1
d1
k (1 sin 2)
=
pj
cos ! sin !
2
o cos o cos cos 2
k (1 sin 2) 1 y 2
d1
1
p
+ pd
2
o cos cos 2
2 UK o cos cos =
(2.291)
(2.292)
(2.293)
If the car’s wheel base is much smaller than the radius of curvature, o ¿
UK , then the slope angle is too small, and we may use trigonometric
approximations
cos sin cos 2 1
sin 2 0
(2.294)
(2.295)
Substituting these approximations in Equations (2.291)-(2.293) produces
the following approximate results:
I{1 + I{2
I}1
pj
I}
2 2
pj
2
1
p (d + j sin !)
2
μ
¶
d2
k
d k y2 @UK d2
cos ! + sin ! o
o
j o
j
o
μ
¶
2
d1
k
d k y @UK d1
cos ! sin ! +
o
o
j o
j
o
(2.296)
(2.297)
(2.298)
2. Forward Vehicle Dynamics
83
Example 75 F Wheel loads of a cresting car.
Consider a car with these specications:
o = 2272 mm z = 1457 mm k = 230 mm d1 = d2
p = 1500 kg y = 15 m@ s
d = 1 m@ s2
(2.299)
which is cresting a hill at a point where the road has
UK = 40 m
! = 30 deg
= 2=5 deg
(2.300)
I}2 = 1488=75 N
y2
p
= 8437=5 N
UK
(2.301)
The force information on the car is:
I{1 + I{2 = 4432=97 N I}1 = 666=33 N
I}1 + I}2 = 2155=08 N
pj = 14715 N
If we simplifying the results by assuming small , the approximate values
of the forces would be
I{1 + I{2 = 4428=75 N I}1 628=18 N
I}1 + I}2 2153=03 N
pj = 14715 N
I}2 1524=85 N
y2
p
= 8437=5 N
UK
(2.302)
Example 76 F Losing the road contact on a crest.
When a car moves too fast, it can lose its road contact on a crest. Such
a car is called a ying car. The condition to have a ying car is I}1 = 0
and I}2 = 0.
Assuming a symmetric ying car d1 = d2 = o@2 with no acceleration,
and using the approximate Equations (2.282) and (2.283), we have
μ
¶¸
1
d2
k
1 y2 d2
pj
cos ! + sin ! p
2
o
o
2 UK o
μ
¶¸
d1
k
1 y2 d1
1
pj
cos ! sin ! p
2
o
o
2 UK o
= 0
(2.303)
= 0
(2.304)
and we can nd the critical minimum speed yf to start ying. There are
two critical speeds yf1 and yf2 for losing the contact of the front and rear
wheels, respectively.
s
μ
¶
k
1
yf1 =
2jUK
sin ! + cos !
(2.305)
o
2
s
μ
¶
k
1
2jUK
sin ! cos !
(2.306)
yf2 =
o
2
84
2. Forward Vehicle Dynamics
I
RH
I
FIGURE 2.27. A cresting car over a circular hill.
For any car, the critical speeds yf1 and yf2 are functions of the hill’s
radius of curvature UK and the angular position on the hill, indicated by !.
The angle ! cannot be out of the tilting angles given by Equation (2.141).
d2
d1
tan ! k
k
(2.307)
Figure 2.27 illustrates a cresting car over a circular hill, and Figure 2.28
depicts the critical speeds yf1 and yf2 at a dierent angle ! for 1=371 rad ! 1=371 rad. The specications of the car and the hill are:
o = 2272 mm k = 230 mm d1 = d2
d = 0 m@ s2
UK = 100 m
(2.308)
At the maximum uphill slope ! = 1=371 rad 78=5 deg, the front wheels
can leave the ground at zero speed while the rear wheels are on the ground.
When the car moves over the hill and reaches the maximum downhill slope
! = 1=371 rad 78=5 deg the rear wheels can leave the ground at zero
speed while the front wheels are on the ground. As long as the car is moving
uphill, the front wheels can leave the ground at a lower speed while going
downhill the rear wheels leave the ground at a lower speed. Hence, at each
slope angle ! the lower curve of Figure 2.28 determines the critical speed
yf .
To have a general image of the critical speed, we may plot the lower
values of yf as a function of ! using UK or k@o as a parameter. Figure
2.29 shows the eect of hill radius of curvature UK on the critical speed
yf for a car with k@o = 0=10123 mm@ mm and Figure 2.30 shows the eect
of a car’s height factor k@o on the critical speed yf for a circular hill with
UK = 100 m.
2. Forward Vehicle Dynamics
85
vc [m/s]
30
25
vc1
20
vc2
15
l=2.272 m
h=0.23 m
a1=a2
RH=40 m
10
5
0.5
1.0
60
40
0
20
-0.5
-20
-1.0 I [rad]
-40
-60 I [deg]
FIGURE 2.28. Critical speeds yf1 and yf2 at dierent angle ! for a specic car
and hill.
vc [m/s]
RH=1000 m
500 m
200 m
100 m
h/l=0.10123
a1=a2
100
80
60
40 m
40
20
0.5
1.0
60
40
0
20
-0.5
-20
-1.0 I [rad]
-40
-60 I [deg]
FIGURE 2.29. Eect of hill radius of curvature Uk on the critical speed yf for a
car.
86
2. Forward Vehicle Dynamics
vc [m/s]
h/l=0.1
0.2
0.3
0.4
0.5
RH=100 m
a1=a2
30
25
20
15
10
5
0.5
1.0
60
40
0
20
-0.5
-20
-1.0 I [rad]
-40
-60 I [deg]
FIGURE 2.30. Eect of a car’s height factor k@o on the critical speed yf for a
circular hill.
2.8.2 F Vehicles on a Dip
Moving on the concave curve of a hill is called dipping. The normal force
under the wheels of a dipping vehicle is more than the force on a at
inclined road with the same slope, because of the developed centrifugal
force py 2 @UK in the }-direction.
Figure 2.31 illustrates a dipping vehicle at a point where the hill has a
radius of curvature UK and equivalent slop !. The traction and normal
forces under the tires of the vehicle are approximately equal to
I{1 + I{2 I}1
I}2
o
1
p (d + j sin !)
2
(2.309)
μ
¶¸
1
d2
k
1
k 1 y2 d2
pj
cos ! + sin ! pd + p
(2.310)
2
o
o
2
o
2 UK o
μ
¶¸
d1
k
1
k 1 y2 d1
1
pj
cos ! sin ! + pd + p
(2.311)
2
o
o
2
o
2 UK o
= d1 + d2
(2.312)
Proof. To develop the equations for the traction and normal forces under
the tires of a dipping car, we follow the same procedure as a cresting car.
The normal and tangential directions of a dipping car, shown in Figure 2.31,
are equivalent to the } and { directions respectively. Hence, the governing
2. Forward Vehicle Dynamics
z
a1
a2
RH
2F x1
mg
v
x
T
a
C
2F x 2
87
2F z1
I
2F z2
FIGURE 2.31. A dipping vehicle at a point where the hill has a radius of curvature
Uk .
equations of motion for the car are
X
X
X
I{ = pd
I} = p
(2.313)
y2
UK
(2.314)
P| = 0
(2.315)
Expanding these equations yields
2I{1 cos + 2I{2 cos pj sin ! = pd
2I}1 cos 2I}2 cos + pj cos ! = p
2I}1 d1 cos 2I}2 d2 cos + 2 (I{1 + I{2 ) k cos +2I}1 d1 sin 2I}2 d2 sin 2 (I{1 + I{2 ) k sin = 0
y2
UK
(2.316)
(2.317)
(2.318)
The total traction force (I{1 + I{2 ) may be eliminated between the rst
and third equations. Then, the resultant equations provide us with the total
traction force I{1 + I{2 and wheel normal forces I}1 , I}2 :
I{1 + I{2 =
pd + pj sin !
2 cos (2.319)
88
2. Forward Vehicle Dynamics
μ
¶¸
1
d2
k (1 sin 2)
pj
cos ! +
sin !
2
o cos o cos cos 2
1
k (1 sin 2) 1 y 2 d2
pd
+ p
2
o cos cos 2
2 UK o cos μ
¶¸
d1
k (1 sin 2)
1
=
pj
cos ! sin !
2
o cos o cos cos 2
k (1 sin 2) 1 y 2
d1
1
+ p
+ pd
2
o cos cos 2
2 UK o cos cos I}1
=
I}2
(2.320)
(2.321)
Assuming very small , these forces can be approximated to
I{1 + I{2
I}1
pj
I}
2 2
pj
2
1
p (d + j sin !)
2
μ
¶
d2
k
d k y 2 @UK d2
cos ! + sin ! +
o
o
j o
j
o
μ
¶
2
d1
k
d k y @UK d1
cos ! sin ! +
+
o
o
j o
j
o
(2.322)
(2.323)
(2.324)
Example 77 F Wheel loads of a dipping car.
Consider a car with the following specications:
o = 2272 mm z = 1457 mm k = 230 mm d1 = d2
p = 1500 kg y = 15 m@ s
d = 1 m@ s2
(2.325)
that is dipping on a hill at a point where the road has
UK = 40 m
! = 30 deg
= 2=5 deg
(2.326)
I}2 = 5711=52 N
y2
p
= 8437=5 N
UK
(2.327)
The force information on the car is:
I{1 + I{2 = 4432=97 N
I}1 = 4889=1 N
I}1 + I}2 = 10600=62 N
pj = 14715 N
If we ignore the eect of by assuming a very value, then the approximate
value of the forces are
I{1 + I{2 = 4428=75 N
I}1 4846=93 N
I}1 + I}2 10590=53 N
pj = 14715 N
I}2 5743=6 N
y2
p
= 8437=5 N
UK
(2.328)
2.9 Summary
For straight motion of a symmetric rigid vehicle, we may assume the forces
on the left wheels are equal to the forces on the right wheels, and simplify
the tire force calculation.
2. Forward Vehicle Dynamics
89
When a car is accelerating on an inclined road with slop angle !, the
normal forces under the front and rear wheels, I}1 , I}2 , are
μ
¶
1
d2
k
1
k
I}1 = pj
cos ! sin ! pd
2
o
o
2
o
μ
¶
1
d1
k
1
k
I}2 = pj
cos ! + sin ! + pd
2
o
o
2
o
o = d1 + d2
(2.329)
(2.330)
(2.331)
¡
¢
The rst parenthesis, 12 pj do1 cos ! ± ko sin ! is the static force which depends on slope and mass center position. The term ± 12 pj ko dj is the dynamic
force, because it depends on the acceleration d.
90
2. Forward Vehicle Dynamics
2.10 Key Symbols
d{
¨
di zg
duzg
d1
d2
dl
dP
d> e
D> E> F
e1
e1
e2
e2
e3
F
Fw
g
I
I{
I{1
I{2
I{w
I}
I}1
I}2
I}3
I}w
j> g
k
K
L
nl
o
p
pw
P
U
Ui
Uu
UK
w
y {>
b v
yf
acceleration
front wheel drive acceleration
rear wheel drive acceleration
distance of rst axle from mass center
distance of second axle from mass center
distance of axle number l from mass center
maximum acceleration
arguments for atan2 (d> e)
constant parameters
distance of left wheels from mass center
distance of hinge point from rear axle
distance of right wheels from mass center
distance of hinge point from trailer mass center
distance of trailer axle from trailer mass center
mass center of vehicle
mass center of trailer
height
force
traction or brake force under a wheel
traction or brake force under front wheels
traction or brake force under rear wheels
horizontal force at hinge
normal force under a wheel
normal force under front wheels
normal force under rear wheels
normal force under trailer wheels
normal force at hinge
gravitational acceleration
height of F
height
mass moment of inertia
vertical stiness of suspension at axle number l
wheel base
car mass
trailer mass
moment
tire radius
front tire radius
rear tire radius
radius of curvature
time
velocity
critical velocity
2. Forward Vehicle Dynamics
z
}l
{> |> }
[> \> ]
track
de ection of axil number l
vehicle coordinate axes
global coordinate axes
!
!P
!W
road slope
road angle with horizon
ultimate angle, maximum slope angle
tilting angle
friction coe!cient
Subscriptions
g|q
i
i zg
P
u
uzg
vw
dynamic
front
front-wheel-drive
maximum
rear
rear-wheel-drive
statics
91
92
2. Forward Vehicle Dynamics
Exercises
1. Axle load.
Consider a car is parked on a level road. Find the load on the front
and rear axles if
(a) p = 1765 kg
d1 = 1=22 m
d2 = 1=62 m
(b) p = 1665 kg
d1 = 1=42 m
d2 = 1=42 m
(c) p = 1245 kg
d1 = 1=62 m
d2 = 1=22 m
2. Mass center distance ratio.
Peugeot 907 ConceptW P approximately has
p = 1400 kg
o = 97=5 in
Assume d1 @d2 1=131 and determine the axles load.
3. Parked car on a level road.
Consider a parked car on a level road and show that
d2
I{1
=
I{2
d1
4. Axle load ratio.
(a) Jeep Commander XKW P approximately has
pj = 5091 lb
o = 109=5 in
Assume I}1 @I}2 1=22 and determine the axles load.
(b) BMW [5W P approximately has
p = 2900 kg
o = 2933 mm
if the front/rear load ratio is 1400@1500, determine d1 and d2 .
5. Axle load and mass center distance ratio.
The wheelbase of the 1981 DeLorean SportscarW P is
o = 94=89 in
Find the axles load if we assume:
d1 @d2 0=831
pj = 3000 lb
2. Forward Vehicle Dynamics
93
6. Mass center lateral position.
In Example 44, determine e1 and e2 for the rear of the car. Assume
that the left side of the car is 4% heavier than the right side.
7. F Dierent front and rear tires.
Prove Equations (2.35) and (2.36).
8. Mass center height.
McLaren SLR 722 SportscarW P has the following specications.
p = 1649 kg
front tire : 255@35]U19
o = 2700 mm
rear tire : 295@30]U19
When the front axle is lifted K = 540 mm, assume that
d1 = d2
I}2 = 0=68pj
What is the height k of the mass center?
9. F Mass center height limit.
In the equation of the height of the mass center, k
¶
μ
2I}1
o cot !
k = U + d2 pj
when ! $ 0 then, cot ! $ 4. Explain why it is wrong to say that
k $ 4.
10. F Mass center height in terms of K.
It may be more practical if in the height calculation experiment of
Example 45, we measure g, the height of the cylinder, instead of
measuring the angle !. Use Figure 2.32 and re-calculate k in terms
of g instead of !.
11. A parked car on an uphill road.
Specications of Lamborghini GallardoW P are
p = 1430 kg
o = 2560 mm
d1 = d2
k = 520 mm
(a) Assume
and determine the forces I}1 , I}2 , and I{2 if the car is parked
on an uphill with ! = 30 deg and the hand brake is connected
to the rear wheels.
(b) What would be the maximum road grade !P , that the car can
be parked, if {2 = 1.
94
2. Forward Vehicle Dynamics
h R sin I
x
a1
z
h
h R sin I
a2
C
2Fz1
H
h cos I
I
d
mg
2Fz2
FIGURE 2.32. Determination of the height of mass center using the height of
cylinder g instead of the angle !.
12. Parked on an uphill road.
Rolls-Royce PhantomW P has
p = 2495 kg
o = 3570 mm
Assume the car is parked on an uphill road and
d1 = d2
k = 670 mm
! = 30 deg
Determine the forces under the wheels if the car is
(a) front wheel braking
(b) rear wheel braking
(c) four wheel braking.
13. A parked car on a downhill road.
Solve Exercise 11 if the car is parked on a downhill road.
14. Forces at tilting angle.
Determine the brake force 2I{2 and normal force I}2 when a rear
park brake car is parked uphill.
15. F Angle of an uphill road.
Consider a parked car on an uphill road and show that if the forces
under the front and rear wheels, I}1 and I}2 , are given then the
2. Forward Vehicle Dynamics
95
inclined angle of the road is:
! = arctan
μ
d2 d1 (I}1 @I}2 )
k + k (I}1 @I}2 )
¶
16. Maximum acceleration.
Honda CR-VW P is a midsize SUV car with
p = 1550 kg
o = 2620 mm
Assume
d1 = d2
k = 720 mm
{ = 0=8
and determine the maximum acceleration of the car if
(a) the car is rear-wheel drive
(b) the car is front-wheel drive
(c) the car is four-wheel drive.
17. F Acceleration on a level road.
Consider an accelerating car on a level road. Show that if the forces
under the front and rear wheels, I}1 and I}2 , are given then the
acceleration of the car is
d=
d2 d1 (I}1 @I}2 )
j
k + k (I}1 @I}2 )
18. F Slope of d@j.
Compare the slopes of di zg @j, duzg @j, di ze @j, and duze @j to determine if there is any equation among them.
19. F Acceleration on an inclined road.
Consider an accelerating car on an inclined road. Assume that the
forces under the front and rear wheels, I}1 and I}2 , are given. Determine the acceleration of the car d in terms of I}1 @I}2 .
20. Minimum time for 0 100 km@ h.
RoadRazerW P is a light weight rear-wheel drive sportscar with
p = 300 kg
o = 2286 mm
k = 260 mm
Assume d1 = d2 . If the car can reach the speed 0 100 km@ h in
w = 3=2 s, what would be the minimum friction coe!cient?
96
2. Forward Vehicle Dynamics
21. Axle load of an all-wheel drive car.
Acura CourageW P is an all-wheel drive car with
p = 2058=9 kg
o = 2750=8 mm
Assume d1 = d2 and k = 760 mm. Determine the axles load if the car
is accelerating at d = 1=7 m@ s2 .
22. F A car with a trailer.
Volkswagen TouaregW P is an all-wheel drive car with
p = 2268 kg
o = 2855 mm
Assume d1 = d2 and the car is pulling a trailer with
pw
e1
k1 = k2
e2 = 1350 mm
= 600 kg
= 855 mm
e3 = 150 mm
If the car is accelerating on a level road with acceleration d = 2 m@ s2 ,
what would be the forces at the hinge.
23. A parked car on a banked road.
Cadillac EscaladeW P is a SUV car with
p = 2569=6 kg
zi = 1732=3 mm
o = 2946=4 mm
zu = 1701=8 mm
Assume e1 = e2 , k = 940 mm, and use an average track to determine
the wheels load when the car is parked on a banked road with ! =
12 deg.
24. F A parked car on a banked road with zi 6= zu .
Determine the wheels load of a parked car on a banked road, if the
front and rear tracks of the car are dierent.
25. F Angle of a banked road.
Consider a parked car on a banked road and show that if the forces
under the left and right wheels, I}1 and I}2 , are given then the bank
angle of the road is:
μ
¶
e2 e1 (I}1 @I}2 )
! = arctan
k + k (I}1 @I}2 )
26. Optimal traction force.
Mitsubishi OutlanderW P is an all-wheel drive SUV car with
p = 1599=8 kg
o = 2669=6 mm
z = 1539=3 mm
2. Forward Vehicle Dynamics
97
Assume
d1 = d2
k = 760 mm
{ = 0=75
and nd the optimal traction force ratio I{1 @I{2 to reach the maximum acceleration.
27. F A three-axle car.
Citroën Cruise CrosserW P is a three-axle o-road pick-up car. Assume
p = 1800 kg
d1 = 1100 mm
n1 = 12800 N@ m
d2 = 1240 mm
n2 = 14000 N@ m
d3 = 1500 mm
n3 = 14000 N@ m
and nd the axles load on a level road when the car is moving with
no acceleration.
28. F Car on top of a crest.
Consider a car on top of a crest and show that
¢
¡
d2 y2 @UK j + dk
I}1
I}2
d1 (y2 @UK j) dk
29. F Car at the bottom of a dip.
Consider a car at the bottom of a dip and show that
¢
¡
d2 y2 @UK + j dk
I}1
I}2
d1 (y2 @UK + j) + dk
3
Tire Dynamics
The tire is the main component of a vehicle interacting with the road.
The performance of a vehicle is mainly in uenced by the characteristics
of its tires. Tires aect a vehicle’s handling, traction, ride comfort, and
fuel consumption. To understand its importance, it is enough to remember
that a vehicle can maneuver only by longitudinal, vertical, and lateral force
systems generated under the tires.
Loaded tire
Tire axis
Ground surface
FIGURE 3.1. A vertically loaded stationary tire.
Figure 3.1 illustrates a vertically loaded stationary tire. To model the
tire-road interactions, we determine the tireprint and describe the forces
distributed on the tireprint.
3.1 Tire Coordinate Frame and Tire Force System
To describe the tire-road interaction and its force system, we assume a
at ground and attach a Cartesian coordinate frame at the center of the
tireprint as shown in Figure 3.2. The {-axis is along the intersection line
of the tire-plane and the ground. The tire plane is the plane made by
narrowing the tire to a at disk. The }-axis is perpendicular to the ground
and upward, and the |-axis makes the coordinate system a right-hand triad.
To show the tire orientation, we use two angles: camber angle and
sideslip angle . The camber angle is the angle between the tire-plane and
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_3, © Springer Science+Business Media New York 2014
99
100
3. Tire Dynamics
z
Camber angle J
Tire
e
plan
ane
d pl
n
u
o
Gr
Tire axis
Fz
Mz
Sideslip angle
D M Fy
y
J
Mx
Fx
y
v V
eloc
ity v
ecto
r
x
FIGURE 3.2. Tire coordinate system.
the vertical plane measured about the {-axis. The camber angle can be
recognized better in a front view as shown in Figure 3.3(d). The sideslip
angle , or simply sideslip, is the angle between the velocity vector v and
the {-axis measured about the }-axis. The sideslip can be recognized better
in a top view, as shown in Figure 3.3(e).
The resultant force system that a tire receives from the ground is assumed
to be located at the center of the tireprint and can be decomposed along {,
|, and } axes. Therefore, the interaction of a tire with the road generates
a three dimensional (3D) force system including three forces and three
moments, as shown in Figure 3.2.
1. Longitudinal force I{ . It is the force acting along the {-axis. The
resultant longitudinal force I{ A 0 if the car is accelerating, and
I{ ? 0 if the car is braking. Longitudinal force is also called forward
force.
2. Normal force I} . It is the vertical force, normal to the ground plane.
The resultant normal force I} A 0 if it is upward. The traditional
tires and pavements are unable to provide I} ? 0. Normal force is
also called vertical force or wheel load.
3. Lateral force I| . It is the force, tangent to the ground and orthogonal
to both I{ and I} . The resultant lateral force I| A 0 when it is in
the |-direction.
4. Roll moment P{ . It is the longitudinal moment about the {-axis.
The resultant roll moment P{ A 0 when it tends to turn the tire
3. Tire Dynamics
J
v
z
D
101
x
y
Tire axis
J
y
(a)
(b)
FIGURE 3.3. Camber and sideslip angles illustration. (d) Front view of a tire
and measurment of the camber angle. (e) Top view of a tire and measurment of
the sideslip angle.
about the {-axis. The roll moment is also called the bank moment,
tilting torque, or overturning moment.
5. Pitch moment P| . It is the lateral moment about the |-axis. The
resultant pitch moment P| A 0 when it tends to turn the tire about
the |-axis and move it forward. The pitch moment is also called rolling
resistance torque.
6. Yaw moment P} . It is the upward moment about the }-axis. The
resultant yaw moment P} A 0 when it tends to turn the tire about
the }-axis. The yaw moment is also called the aligning moment, self
aligning moment, or bore torque.
These are the force system that are applied on the tire from the ground.
All other possible forces on a wheel are at the wheel axle. The driving or
braking moment applied to the tire from the vehicle about the tire axis is
called wheel torque W .
Example 78 Origin of tire coordinate frame.
For a cambered tire, it is not always possible to nd or dene a center
point for the tireprint to be used as the origin of the tire coordinate frame.
It is more practical to set the origin of the tire coordinate frame at the
center of the intersection line between the tire-plane and the ground at zero
camber and sideslip angles. So, the origin of the tire coordinate frame is at
102
3. Tire Dynamics
Camber angle J
Tire axis
Tire
e
plan
D
Sideslip angle
e
plan
d
n
u
Gro
y
My
Fy
Fz
Mz
Mx
Fx
v V
eloc
ity v
ecto
r
x
z
FIGURE 3.4. SAE tire coordinate system.
the center of the tireprint when the tire is standing upright and stationary
on a at road.
Example 79 F SAE tire coordinate system.
The tire coordinate system adopted by the Society of Automotive Engineers (SAE) is shown in Figure 3.4. The origin of the coordinate system
is at the center of the tireprint when the tire is standing stationary. The
{-axis is at the intersection of the tire-plane and the ground plane. The
}-axis is downward and perpendicular to the tireprint. The |-axis is on the
ground plane and goes to the right to make the coordinate frame a righthand frame.
Because of this coordinate arrangement, the positive direction of some
angles and forces are in negative direction of the coordinate axes. Besides
some miss-match and confusion, sometimes authors consider forces positive
when they are in the negative direction of the coordinate axes. However, the
signs and positive directions of forces and angles are not commonly accepted
among authors who are using SAE coordinate and they usually refer to right
and left direction instead of coordinate axes. For example, the sideslip angle
is considered positive if the tire is slipping to the right. Furthermore, in
SAE convention, the camber angle for the left and right tires of a vehicle
have opposite signs. So, the camber angle of the left tire is positive when
the tire leans to the right and the camber angle of the right tire is positive
when the tire leans to the left.
The SAE coordinate system is as good as the coordinate system in Figure 3.2 and may be used alternatively. However, having the }-axis directed
downward is sometimes ine!cient and confusing.
3. Tire Dynamics
103
F1 < F2 < F3
F1
F2
F3
F1
F2
z1
z2
F3
z3
FIGURE 3.5. Vertically loaded tire at zero camber.
3.2 Tire Stiness
As an applied approximation, the vertical tire force I} can be calculated
as a linear function of the normal tire de ection 4} measured at the tire
center.
(3.1)
I} = n} 4}
The coe!cient n} is called tire stiness in the }-direction. Similarly, the
reaction of a tire to a lateral and a longitudinal force can be approximated
as
I{
I|
= n{ 4{
= n| 4|
(3.2)
(3.3)
where the coe!cient n{ and n| are called tire stiness in the { and |
directions.
Proof. The deformation behavior of tires to the applied forces in any three
directions {, |, and } are the rst important tire characteristics in tire
dynamics. Calculating the tire stiness is generally based on experiment
and therefore, they are dependent on the tire’s mechanical properties, as
well as environmental characteristics.
Consider a vertically loaded tire on a sti and at ground as shown in
Figure 3.5. The tire will de ect under the load and generate a pressurized
contact area to balance the vertical load.
Figure 3.6 depicts a sample of experimental stiness curve in the (I} > 4})
plane. The curve may be expressed by a curve tted mathematical function
I} = i (4})
(3.4)
104
3. Tire Dynamics
14000
1200
Fz [N]
1000
800
600
400
T
200
0
0
10
20
tan 1 k z
30
40
50
60
Deflection [mm]
FIGURE 3.6. A sample tire vertical stiness curve.
however, we usually use a linear approximation for the range of the usual
application.
Ci
4}
(3.5)
I} =
C (4})
Ci
The coe!cient C(4})
is the slope of the experimental stiness curve at zero
and is shown by the stiness coe!cient n}
n} = tan = lim
Ci
4}$0 C (4})
(3.6)
Therefore, the normal tire de ection 4} remains proportional to the vertical tire force I} by denition.
I} = n} 4}
(3.7)
The stiness curve can be in uenced by many parameters. The most effective one is the tire in ation pressure. The tire can apply only pressure
forces to the road, so normal force is restricted to I} A 0.
Lateral and longitudinal force/de ection behavior is also determined experimentally by applying a force in the appropriate direction. The lateral
and longitudinal forces are limited by the sliding force when the tire is
vertically loaded. Figure 3.7 depicts a sample of longitudinal and lateral
stiness curves compared to a vertical stiness curve.
The practical range of tires’ longitudinal and lateral stiness curves is
the linear part and may be estimated by linear equations.
I{
I|
= n{ 4{
= n| 4|
(3.8)
(3.9)
3. Tire Dynamics
105
7000
Vertical
6000
Fz [N]
5000
Longitudinal
4000
Lateral
3000
2000
1000
0
0
10
20
30
40
50
60
Vertical, longitudinal and lateral deflections [mm]
FIGURE 3.7. Vertical, longitudinal, and lateral stiness curves.
The coe!cients n{ and n| are called the tire stiness in the { and | directions. They are measured by the slope of the experimental stiness curves
in the (I{ > 4{) and (I| > 4|) planes.
n{
n|
Ci
4{$0 C (4{)
Ci
= lim
4|$0 C (4|)
=
lim
(3.10)
(3.11)
When the longitudinal and lateral forces increase, parts of the tireprint
creep and slide on the ground until the whole tireprint starts sliding. At
this point, the applied force saturates and reaches its maximum supportable
value.
When we measure displacements from the equilibrium position at zero
de ection, we may respectively show 4{, 4|, 4} by {, |, }. Generally, a
tire is most sti in the longitudinal direction, {, and least sti in the lateral
direction, |.
(3.12)
n{ A n} A n|
Figure 3.8 illustrates tire deformation under a lateral and a longitudinal
force.
Example 80 F Nonlinear tire stiness.
In a better modeling, the vertical tire force I} is a function of the normal
tire de ection }, and de ection rate }.
b
b = I}v + I}g
I} = I} (}> })
(3.13)
In a rst approximation we may assume I} is a linear combination of a static and a dynamic part. The static part is a nonlinear, usually a polynomial
106
3. Tire Dynamics
Fx
Fy
'x
'y
FIGURE 3.8. Illustration of laterally and longitudinally tire deformation.
function of the vertical tire de ection and the dynamic part is proportional
to the vertical speed of the tire.
I}v
I}g
= n1 } + n2 } 2 + n3 } 3
= n4 }b
(3.14)
(3.15)
The constants n1 , n2 , and n3 are calculated from the rst, second, and third
slopes of the experimental stiness curve in the (I} > }) plane. The constant
b plane, which indicates the
n4 is the rst slope of the curve in the (I} > })
tire damping.
n1
=
n4
=
C I} (0)
C}
C I} (0)
C }b
n2 =
1 C 2 I} (0)
2! C } 2
n2 =
1 C 2 I} (0)
3! C } 3
(3.16)
(3.17)
The value of n1 = 200000 N@ m is a good approximation for a 205@50U15
passenger car tire, and n1 = 1200000 N@ m is a good approximation for a
[31580U22=5 truck tire.
Tires with a larger number of plies have higher damping, because the
damping is a result of the plies’ internal friction. Tire damping decreases
by increasing speeds.
Example 81 F Hysteresis eect.
Rubber is a viscoelastic material, and therefore, their loading and unloading stiness curves are not exactly the same. Figure 3.9 illustrates a sample
of loading and unloading of a rubbery tire. The curves make a loop with the
unloading curve sitting below the loading. The area within the loop is the
amount of dissipated energy during loading and unloading. As a rolling
tire rotates under the weight of a vehicle, it experiences repeated cycles of
deformation and recovery, and it dissipates energy loss as heat. Such a behavior is a common property of hysteretic material and is called hysteresis.
3. Tire Dynamics
107
14000
1200
Fz [N]
1000
g
din
loa
800
600
ing
ad
o
l
un
400
200
0
0
10
20
30
40
50
60
Deflection [mm]
FIGURE 3.9. Histeresis loop in a vertically loading and unloading tire.
So, hysteresis is a characteristic of a deformable material such as rubber,
that the energy of deformation is greater than the energy of recovery. The
amount of dissipated energy depends on the mechanical characteristics of
the tire. Hysteretic energy loss in rubber decreases as temperature increases.
The hysteresis eect causes a loaded rubber not to rebound fully after load
removal. Consider a high hysteresis race car tire turning over road irregularities. The deformed tire recovers slowly, and therefore, it cannot push
the tireprint tail on the road as hard as the tireprint head. The dierence
in head and tail pressures causes a resistance force, which is reason for the
rolling resistance.
Race cars have high hysteresis tires to increase friction and traction.
Street cars have low hysteresis tires to reduce the rolling resistance and low
operating temperature. Hysteresis level of tires inversely aect the stopping
distance. A high hysteresis tire makes the stopping shorter, however, it
wears rapidly and has a shorter life time.
3.3 Eective Radius
Consider a vertically loaded wheel that is turning on a at surface as shown
in Figure 3.10. The eective radius of the wheel Uz , which is also called a
rolling radius, is dened by
y{
(3.18)
Uz =
$
where, y{ is the forward velocity and $ = $ z is the angular velocity of the
wheel. The eective radius Uz is approximately equal to
Uj Uk
(3.19)
Uz Uj 3
108
3. Tire Dynamics
z
Z
v
2M
Rh Rw Rg
x
Ground plane
2a
FIGURE 3.10. Eective radius Uz compared to tire radius Uj and loaded height
Uk .
and is a number between the unloaded or geometric radius Uj and the
loaded height Uk .
Uk ? Uz ? Uj
(3.20)
Proof. An eective radius Uz = y{ @$ z is dened by measuring the tire’s
angular velocity $ = $ z and forward velocity y{ . When the tire rolls, each
part of the circumference is attened as it passes through the contact area.
A practical estimate of the eective radius can be made by substituting the
arc with the straight length of tireprint. The tire vertical de ection is
Uj Uk = Uj (1 cos *)
(3.21)
Uk = Uj cos *
d = Uj sin *
(3.22)
(3.23)
and
If the motion of the tire is compared to the rolling of a rigid disk with
radius Uz , then the tire must move a distance d Uz * for an angular
rotation *.
d = Uj sin * Uz *
(3.24)
Hence,
Uz =
Uj sin *
*
Expanding sin* * in a Taylor series shows that
μ
¶
¡ 4¢
1 2
Uz = Uj 1 * + R *
6
(3.25)
(3.26)
3. Tire Dynamics
109
Using Equation (3.21) we may approximate
1
cos * 1 *2
2
*2
(3.27)
μ
2 (1 cos *) 2 1 Uk
Uj
¶
(3.28)
and therefore,
μ
¶¶
μ
1
1
2
Uk
Uz Uj 1 = Uj + Uk
1
3
Uj
3
3
(3.29)
Because Uk is a function of tire load I} ,
Uk = Uk (I} ) = Uj I}
n}
(3.30)
the eective radius Uz is also a function of the tire load. The angle * is
called tireprint angle or tire contact angle.
The vertical stiness of radial tires is less than non-radial tires under
the same conditions. So, the loaded height of radial tires, Uk , is less than
the non-radials’. However, the eective radius of radial tires Uz , is closer
to their unloaded radius Uj . As a good estimate, for a non-radial tire,
Uz 0=96Uj , and Uk 0=94Uj , while for a radial tire, Uz 0=98Uj , and
Uk 0=92Uj .
Generally speaking, the eective radius Uz depends on the type of tire,
stiness, load conditions, in ation pressure, and tire’s forward velocity.
Example 82 Tire rotation.
The geometric radius of a tire S 235@75U15 is Uj = 366=9 mm, because
kW = 235 × 75% = 176=25 mm 6=94 in
(3.31)
2kW + 15
2 × 6=94 + 15
=
= 14=44 in 366=9 mm
2
2
(3.32)
and therefore,
Uj =
Consider a vehicle with such a tire is traveling at a high speed such as
y = 50 m@ s = 180 km@ h 111=8 mi@ h. The tire is radial, and therefore the
eective tire radius Uz is approximately equal to
Uz 0=98Uj 359=6 mm
(3.33)
After moving a distance g = 100 km, this tire must have been turned q1 =
44259 times because
q1 =
g
100 × 103
= 44259
=
G
2 × 359=6 × 103
(3.34)
110
3. Tire Dynamics
z
Z
Z
Rw
v
Rg
Rw
2M
Rh
x
Effective ground surface
2a
FIGURE 3.11. Equivalent tire is rolling under on the equivalent ground surface
below the ground.
Now assume the vehicle travels the same distance g = 100 km at a low
in ation pressure, such that the eective radius of the tire remained close
to the loaded radius
Uz Uk 0=92Uj = 330=8 mm
(3.35)
This tire must turn q2 = 48112 times to travel g = 100 km, because,
q2 =
100 × 103
g
=
= 48112
G
2 × 330=8 × 103
(3.36)
Example 83 Tire is rolling under the ground.
The equivalent radius of a rolling loaded tire, Uz , is, by denition, the
radius of a solid disc when roles with the same angular velocity of the tire,
$, moves with the same velocity, y{ , as the tire.
y{ = Uz $
(3.37)
However, because Uz A Uk the equivalent disc must be rolling on a solid
surface below the ground by the distance Uz Uk as is shown in Figure
3.11.
Example 84 Compression and expansion of tires in the tireprint zone.
Because of longitudinal deformation, the peripheral velocity of any point
of the tread varies periodically. When it gets close to the starting point of
the tireprint, it slows down and a circumferential compression occurs. The
tire treads are compressed in the rst half of the tireprint and gradually
expanded in the second half. The treads in the tireprint contact zone almost
stick to the ground. Assuming the relative velocity of the tire with respect
3. Tire Dynamics
111
a
Rh
Rg
M
T
r T
Ground plane
x
Rg - Rh
d, displacement of tire
FIGURE 3.12. Radial motion of the tire peripheral points in the contact area.
to the ground is zero at the centerpoint of the tireprint, the relative speed of
the circumferential elements of the tire varies between zero to almost double
of the forward velocity of the tire center y{ at the top.
Example 85 F Radial motion of tire’s peripheral points in tireprint.
The radial displacement of a tire’s peripheral elements during road contact may be modeled by a function
g = g ()
(3.38)
We assume that a peripheral element of the tire moves along only the radial
direction during contact with the ground, as shown in Figure 3.12.
Let us show a radius at an angle , by u = u (). Knowing that
cos =
Uk
u
cos * =
Uk
Uj
(3.39)
where * is the half of the contact angle, we nd
u = Uj
cos *
cos Thus, the displacement function is
³
cos * ´
g = Uj u () = Uj 1 cos (3.40)
*? ?*
(3.41)
Example 86 F Tread travel.
Let us follow an element of tire tread with respect to the center of the tire
when it travels around the spin axis and the vehicle moves at a constant
speed. Although the wheel is turning at constant angular velocity $, the tread
does not travel at constant speed. At the top of the tire, the radius is equal to
112
3. Tire Dynamics
.
r [m/s]
r [m]
T rad
T rad
..
r [m/s2]
Rg=0.5 m
Z 60 r/s
M 15 deg
T rad
FIGURE 3.13. Radial displacement, velocity, and acceleration of tire treads in
the tireprint.
the unloaded geometric radius Uj and the speed of the tread is Uj $ relative
to the wheel center. As the tire turns, the tread approaches the leading edge
of tireprint, and slows down. The tread is compacted radially, and gets
squeezed in the heading part of the tireprint area. Then, it is stretched out
and unpacked in the tail part of the tireprint as it moves to the tail edge.
At the center of the tireprint, the tread velocity is Uk $.
The variable radius of a tire during the motion through the tireprint is
cos *
*? ?*
(3.42)
u = Uj
cos where * is the half of the contact angle, and is the angular rotation of the
tire, as shown in Figure 3.12. The angular velocity of the tire is $ = b
and is assumed to be constant. Then, the radial velocity ub and acceleration
ü of the tread with respect to the wheel center are
sin cos2 ¢
cos * ¡
ü = Uj $ 2 3 2 cos2 cos Figure 3.13 depicts u, u,
b and ü for a sample tire with
ub
= Uj $ cos *
Uj = 0=5 m
* = 15 deg
$ = 60 rad@ s
(3.43)
(3.44)
(3.45)
The minimum u happens at = 0 which is 1=7 cm shorter than Uj .
12
= 0=483 m
Uk = umin = 0=5
cos 0
cos
(3.46)
3. Tire Dynamics
113
Considering the time w = @$ = 12
@60 = 4=363× 103 s that takes for a tire
element to pass the tireprint, the maximum radial speed and acceleration of
the tire tread elements would respectively be 8 m@ s and 2050 m@ s2 .
Example 87 F Impossible kinematics for sticking tire.
Consider a tread element of a rolling tire with a constant angular velocity.
Ideally the tread element sticks to the ground in the tireprint and keeps its
constant velocity with respect to the center of the tire. These con icting
conditions are expressed by
$ = b = fwh
{b = y = Uk $ = fwh
(3.47)
Taking a derivative of the kinematic constraint equations
*? ?*
(3.48)
0 = ub cos + u$ sin (3.49)
y cos = Uk cos = Uj cos * cos $
(3.50)
{ = u sin Uk = u cos provides us with
y = ub sin u$ cos Eliminating ub between them yields
u=
Equation (3.50) is not compatible with (3.42) and (3.47), indicating that
the conditions (3.47) cannot both be satised. So, either $ = b 6= fwh
or {b = y 6= fwh. As, it is possible to keep the angular velocity of the tire
constant, $ = fwh, the tire tread cannot stick to the ground and will slide
on the ground. Therefore,
{b = y 6= fwh
(3.51)
Example 88 F Velocity analysis in tireprint.
Consider an element of a rolling tire tread. The element’s velocity relative
to the center of the tire is a tangential vector with speed y = Uj $ as long as
it is not in the tireprint zone. As soon as the element enters the tireprint, its
sin radius changes with the rate of ub = Uj $ cos * cos
2 . The element’s velocity
has then two components: a radial speed of ub and a tangential speed of y =
u$. As the element must lay on the ground surface, the resultant velocity
vector of the element must be along the surface. Figure 3.14 illustrates the
kinematics of the system.
The speed of the tread element in tireprint is therefore equal to
sμ
¶2 ³
p
sin cos * ´2
2
2
2
$
Uj $ cos * 2
ub + u $ =
+ Uj
y =
cos cos cos *
*? ?*
(3.52)
= Uj $ 2
cos 114
3. Tire Dynamics
a
M
Rh
T
Ground surface
v
Rg
r T
.
r
x
Rg Z
rZw
Rg - Rh
FIGURE 3.14. Velocity analysis of tire tread in tireprint.
There is a constraint equation between the velocity of the ground, yJ , and
the angular velocity $, which allows us to calculate the ground velocity.
yJ
$
=
*
d
d = Uj sin *
(3.53)
Showing the constant speed of the ground by yJ = d$@*, we may determine
the relative velocity yuho of the tire tread and the ground as
¶
μ
sin *
cos *
sin *
cos *
= Uj $
(3.54)
yuho = y yJ = Uj $ 2 Uj $
cos *
cos2 *
Figure 3.15 depicts velocity y for a sample tire at a high speed with
Uj = 0=5 m
* = 15 deg
$ = 60 rad@ s
(3.55)
and Figure 3.16 illustrates the relative velocity yuho for the same tire with
dierent tireprint angle *.Theoretically, as long as the tread is not in
tireprint zone the velocity of tire tread is y = Uj $. As soon as the element
enters the tireprint, its speed jumps to the higher value y = Uj $@ cos *.
Then the speed gradually decreases to the lower value y = Uj $ cos * at the
center of tireprint. The speed will then increase to y = Uj $@ cos * at the
moment it exits the tireprint and jumps back to y = Uj $ after that. Practically, the sharp edges of the velocity prole would be substituted by small
curves.
Example 89 F Displacement analysis in tireprint.
We may express Equation (3.52) as a dierential equation
g{ = Uj $
cos *
cos *
gw = Uj
g
cos2 cos2 (3.56)
3. Tire Dynamics
115
v
Rg=0.5 m
Z 60 rad/s
M 15 deg
Rg Z
M
M
T deg
FIGURE 3.15. Velocity prole of a tire tread element in tireprint zone.
vrel
Rg=0.5 m
Z 60 rad/s
M
M
M 15
M 10
5
20
T deg
FIGURE 3.16. Relative velocity of a tire tread element with respect to the ground
in tireprint zone.
116
3. Tire Dynamics
x m
Rg=0.5 m
M 15 deg
0.1
Rg
T
sin M
M
0.05
M
-10
-15
-5
x
5
x
M
10
15
-0.05
Rg
T
sin M
M
T deg
M
-0.1
FIGURE 3.17. Horizontal displacement of tire tread, {, and the contact-free
displacement, Uj .
and determine the horizontal displacement of the tire tread by integration.
{ = Uj cos * tan *? ?*
(3.57)
Comparing the tread displacement { with respect to the ground displacement
{J
(3.58)
{J = Uj sin *
*
provides us with the relative displacement of tire and ground, {uho .
{uho
= { {J = Uj cos * tan Uj sin *
*
¶
μ
= Uj cos * tan tan *
*
(3.59)
Interestingly, the tire tread displacement in tireprint is independent of the
angular velocity. Therefore, the horizontal displacement of tire tread remains the same at dierent time rate for dierent angular velocity. Figure
3.17 illustrates the tread displacement, {, for a sample tire, and Figure 3.18
depicts the relative displacement of tire tread and the ground, {uho .
When {uho is positive, the tread of the tire is traveling ahead of the road;
and when {uho is negative, the tread is behind the corresponding point on
the road. Figure 3.19 illustrates how the treads of the tire stretch relative to
the ground. The tread stretch of free rolling tire would be symmetric with
respect to the center of the tireprint.
Example 90 F Velocity centers of tireprint.
The relative velocity of tire tread with respect to the ground has two zeros.
The zero relative points indicate the velocity centers of the tireprint at which
3. Tire Dynamics
117
xrel [mm]
M 20
Rg=0.5 m
M 15
T deg
M 5
M 10
FIGURE 3.18. Horizontal displacement of tire tread with respect to the ground.
z
a
Rg
M
Rh
T
r
x
FIGURE 3.19. Tread displacement illustration of a rolling tire.
118
3. Tire Dynamics
T deg
T2
T1
M deg
FIGURE 3.20. The velocity centers 1 and 2 as functions of *.
tire tread is going to start going faster or slower than the road. Solving the
relative velocity Equation (3.54) for the angular position of the velocity
centers yields
r
*
*A0
(3.60)
1 = cos1
tan *
r
*
*?0
(3.61)
2 = cos1
tan *
Figure depicts 1 and 2 as functions of *. Interestingly, the position of
velocity centers are independent from the size and kinematics of the tire as
they are only functions of the tireprint angle *.
The velocity centers 1 and 2 also determine the angular positions of
the maximum and minimum point of the relative displacement of Figure
3.18.
Example 91 F Longitudinal strain and stress.
Having the horizontal stretch of treads of a rolling tire as Equation (3.59),
we are able to determine the longitudinal strain % in tire treads.
¶
μ
Uj cos * tan tan *
tan *
{
*
=
=1
(3.62)
%=
{
Uj cos * tan * tan Assuming that the strain is proportional to the tangential stress, we may
estimate the longitudinal shear stress in tireprint as
¶
μ
tan *
(3.63)
J% = J 1 * tan 3. Tire Dynamics
119
H [m/m]
T deg
Rg=0.5 m
M 15 deg
FIGURE 3.21. Strain distribution in the tireprint.
However, rubbery materials do not behave as linear Hookean material and
their stress-strain relationship does not follow the linear Hooke’s law. The
tangential stress is almost symmetric for a free rolling tire. Applying a
driving moment will shift the symmetric point of the tread displacement of
Figures 3.18 and 3.19 backward. The imbalance tread displacement yields
the resultant forward shear stresses become higher than the resultant backward shear stresses, and providing a forward traction force. The reverse
situation happens in braking. Figure 3.21 illustrates the strain distribution
in the tireprint for a sample tire.
Example 92 F Acceleration analysis in tireprint.
Consider an element of a rolling tire tread. The element’s acceleration
relative to the center of the tire is a combination of tangential and radial
components.
³
³
´
´
2
a = du x̂u + d x̂ = ü ub x̂u + 2ub b + u¨ x̂
¢
¡
= ü u$ 2 x̂u 2u$x̂
b (3.64)
Substituting the values of variables from (3.42)-(3.44) provides us with
sin2 cos3 sin d = 2Uj $ 2 cos * 2
cos q
sin d =
d2u + d2 = 2Uj $ 2 cos * 3
cos du
= 2Uj $ 2 cos *
(3.65)
(3.66)
(3.67)
The absolute acceleration, d, of the element must be on {-axis, as the element only moves along the ground surface. Figure 3.22 depicts du , d , and
120
3. Tire Dynamics
a
a
a
aT
ar
ar
I
I
T rad
Rg=0.5 m
Z 60 rad/s
I 15 deg
aT
FIGURE 3.22. Radial and tangential components of the acceleration of a tire
tread element in tireprint zone.
d for a sample tire with
Uj = 0=5 m
* = 15 deg
$ = 60 rad@ s
(3.68)
Figure 3.23 illustrates the acceleration components of a tread element in
the tireprint.
F Jerk analysis in tireprint.
Having the absolute acceleration of the treads in tireprint as given in
Equation (3.67) is enough to calculate the jerk m of the treads.
gd
Uj
gd
=$
= 2$ 3 4 (cos 2 2) cos *
(3.69)
gw
g
cos ¢
¡
Figure 3.24 illustrates m@ Uj $ 3 as a function of for a tire with * =
15 deg.
The jerk m of the treads is proportional to Uj $ 3 . If we accept that the
noise of a tire is proportional to the treads’ jerk, then to reduce the noise
of a given tire, we should reduce the size of the tire or the angular velocity
of the tire, $. However, these reduction must be along with keeping Uj $
constant, as Uj $ is almost equal to the speed of the car. Therefore, to reduce
the jerk of a tire for a given speed y, we should use a bigger tire to have a
smaller $.
y
(3.70)
$ Uj
y
(3.71)
m = 2$ 2 4 (cos 2 2) cos *
cos m=
By increasing Uj while keeping y constant, we reduce jerk m signicantly.
First because m drops by the order of $ 2 , and second, because * drops by
3. Tire Dynamics
121
ûT
uˆr
a
I
Rh
T
Rg
r
aT
Ground surface
a
x
ar
Rg - Rh
FIGURE 3.23. The acceleration components of a tread element in the tireprint.
j / Rg Z3
M 15 deg
M
M
T rad
FIGURE 3.24. Nondimensional tire tread jerk m@ Uj $3 as a function of for a
tire with * = 15 deg.
122
3. Tire Dynamics
a=0.1 m
j
v / a2
3
T [rad]
Rg [m]
FIGURE 3.25. A plot of the tire tread jerk m@ y 3 @d2 as a function of Uj and for d = 0=1 m.
the order of sin1 U1j as
* = sin1
d
Uj
(3.72)
The half length of the tireprint, d is a function of the total weight on a tire
and is almost constant for a given car using every tire with the same width
z. The jerk as a function of the tire radius Uj is
y3
m = 2 2
Uj
s
1
d2 cos 2 2
Uj2 cos4 (3.73)
¡
¢
A plot of m@ y3 @d2 versus Uj and is given in Figure 3.25 for d = 0=1 m
to show the eect of increasing Uj on the level of tire tread jerk.
3.4 F Tireprint Forces of a Static Tire
The force per unit area applied on a tire in the tireprint can be decomposed
into a component normal to the ground and a tangential component on
the ground. The normal component is the contact pressure } , while the
tangential component can be further decomposed in the { and | directions
to make the longitudinal and lateral shear stresses { and | .
For a stationary tire under normal load, the tireprint is symmetrical. Due
to equilibrium conditions, the overall integral of the normal stress over the
tireprint area DS must be equal to the normal load I} , and the integral of
3. Tire Dynamics
123
z
Stationary
loaded tire
Fz
y
x
Ground surface
V z Normal stress
distribution
FIGURE 3.26. Normal stress } applied on the round because of a stationary
tire under a normal load I} .
shear stresses must be equal to zero.
Z
} ({> |) gD = I}
ZDS
{ ({> |) gD = 0
ZDS
| ({> |) gD = 0
(3.74)
(3.75)
(3.76)
DS
3.4.1 F Static Tire, Normal Stress
Figure 3.26 illustrates a stationary tire under a normal load I} along with
the generated normal stress } applied on the ground. The applied loads
on the tire are illustrated in the side view shown in Figure 3.27. For a
stationary tire, the shape of normal stress } ({> |) over the tireprint area
depends on the tire and load conditions, however its distribution over the
tireprint is generally in the shape shown in Figure 3.28.
The normal stress } ({> |) may be approximated by the function
μ
¶
|6
{6
(3.77)
} ({> |) = }P 1 6 6
d
e
where d and e indicate the dimensions of the tireprint, as shown in Figure
3.29. The tireprints may approximately be modeled by a mathematical
function
{2q | 2q
+ 2q = 1
q5Q
(3.78)
d2q
e
124
3. Tire Dynamics
z
Stationary
loaded tire
Fz
Vz
Ground plane
x
FIGURE 3.27. Side view of a normal force I} and stress } applied on a stationary
tire.
Vz
Ground plane
x
b
a
y
FIGURE 3.28. A model of normal stress } ({> |) in the tireprint area for a stationary tire.
3. Tire Dynamics
125
x
b
a
x6 y6
a 6 b6
y
1
Tireprint
FIGURE 3.29. A mode for tireprint of stationary radial tires under normal load.
For radial tires, q = 3 or q = 2 may be used,
{6 | 6
+ 6 =1
d6
e
(3.79)
while for non-radial tires q = 1 is a better approximation.
|2
{2
= 2 =1
2
d
e
(3.80)
Example 93 Normal stress in tireprint.
A car weighs 800 kg. If d1 = d2 and the tireprint of each radial tire is
DS = 4 × d × e = 4 × 5 cm × 12 cm, then the normal stress distribution
under each radial tire, } , must satisfy the equilibrium equation.
Z
1
} ({> |) gD
I} = 800 × 9=81 =
4
DS
μ
¶
Z 0=05 Z 0=12
{6
|6
g| g{
}P 1 =
0=056 0=126
0=05 0=12
= 1=7143 × 102 }P
(3.81)
Therefore, the maximum normal stress is
}P =
I}
= 1=1445 × 105 Pa
1=7143 × 102
and the stress distribution over the tireprint is
μ
¶
|6
{6
Pa
} ({> |) = 1=1445 × 105 1 0=056
0=126
(3.82)
(3.83)
126
3. Tire Dynamics
Example 94 Normal stress in tireprint for q = 2.
The maximum normal stress }P for an 800 kg car having d1 = d2 and
DS = 4 × d × e = 4 × 5 cm × 12 cm, can be found for non-radial tires with
q = 2 as
Z
1
} ({> |) gD
I} = 800 × 9=81 =
4
DS
μ
¶
Z 0=05 Z 0=12
{4
|4
g| g{
=
}P 1 0=054 0=124
0=05 0=12
= 1=44 × 102 }P
(3.84)
I}
= 1=3625 × 105 Pa
(3.85)
1=44 × 102
Comparing }P = 1=3625 × 105 Pa for q = 2 to }P = 1=1445 × 105 Pa
= 3¢ shows that maximum stress for non-radial tires, q = 2, is
¡for q1=1445
1 1=3625 × 100 = 16% more than the radial tires, q = 3.
}P =
3.4.2 F Static Tire, Tangential Stresses
Because of tire deformation in contact with the ground, a three-dimensional
stress distribution will appear in the tireprint even for a stationary tire.
The tangential stress on the tireprint can be decomposed into { and |
directions. The tangential stress is also called shear stress or friction stress.
For a static tire under a normal load, the tangential stresses are inward
in { direction and outward in | direction. Hence, the tire tries to stretch
the ground in the {-axis and compact the ground on the |-axis. Figure
3.30 depicts the shear stresses on a vertically loaded stationary tire. The
force distribution on the tireprint is not constant and is in uenced by tire
structure, load, in ation pressure, and environmental conditions.
The tangential stress { in the {-direction may be modeled and expressed
by
μ 2q+1 ¶
³| ´
³ ´
{
2 {
cos
q5Q
(3.86)
sin
{ ({> |) = {P
d2q+1
d
2e
{ is negative for { A 0 and is positive for { ? 0, showing an inward longitudinal stress. Figure 3.31 illustrates the absolute value of a { distribution
for q = 1.
The |-direction tangential stress | may be modeled by the equation
μ 2q
¶
³| ´
{
q5Q
(3.87)
1
sin
| ({> |) = |P
d2q
e
where | is positive for | A 0 and negative for | ? 0, showing an outward
lateral stress. Figure 3.32 illustrates the absolute value of a | distribution
for q = 1.
3. Tire Dynamics
z
127
z
x
y
Wy
x
Wx
y
FIGURE 3.30. Direction of tangential stresses on the tireprint of a stationary
vertically loaded tire.
Wx
Stress direction
x
b
a
y
FIGURE 3.31. Absolute value of a { distribution model for q = 1.
128
3. Tire Dynamics
Wy
Stress direction
x
b
a
y
FIGURE 3.32. Absolute value of a | distribution model for q = 1.
3.5 Rolling Resistance
A turning tire on the ground generates a longitudinal force Fu called rolling
resistance. The force is opposite to the direction of motion and is proportional to the normal force on the tireprint.
Fu
Iu
= Iu ˆ~
= u I}
(3.88)
(3.89)
The parameter u is called the rolling friction coe!cient. The value of
u depends on tire speed, in ation pressure, sideslip and camber angles.
It also depends on mechanical properties, speed, wear, temperature, load,
size, driving and braking forces, and road condition.
Proof. When a tire is turning on the road, that portion of the tire’s circumference that passes over the pavement undergoes a de ection. Part of
the energy that is spent in deformation will not be restored in the following relaxation. Hence, a change in the distribution of the contact pressure
makes normal stress } in the heading part of the tireprint higher than the
tailing part. The dissipated energy and stress distortion cause the rolling
resistance.
Figures 3.33 and 3.34 illustrate a model of normal stress distribution
across the tireprint and their resultant force I} for a turning tire.
Because of higher normal stress in the front part of the tireprint, the
resultant normal force moves forward. This forward shift of the normal
force makes a resistance moment in the | direction, opposing the forward
3. Tire Dynamics
129
Vz
Ground plane
x
a
b
y
FIGURE 3.33. A model of normal stress } ({> |) in the tireprint area for a rolling
tire.
z
Z
Rolling
loaded tire
Fz
Vz
x
Ground plane
'x
FIGURE 3.34. Side view of a normal stress } distribution and its resultant force
I} on a rolling tire.
130
3. Tire Dynamics
rotation.
Mu
Pu
= Pu ˆ
= I} {
(3.90)
(3.91)
The rolling resistance moment Mu may be assumed to be the result of a
rolling resistance force Fu parallel to the {-axis.
= Iu ˆ~
(3.92)
1
{
Iu =
Pu =
I}
(3.93)
Uk
Uk
As the rolling resistance force Iu is directly proportional to the normal load,
we may dene the proportionality by using a rolling friction coe!cient u .
Fu
Iu = u I}
(3.94)
Example 95 A model for normal stress of a turning tire.
We may assume that the normal stress of a turning tire is expressed by
¶
μ
{2q | 2q
{
(3.95)
} = }p 1 2q 2q +
d
e
4d
where q = 3 or q = 2 for radial tires and q = 1 for non-radial tires.
We may determine the stress mean value }p by knowing the total load
on the tire. As an example, using q = 3 for an 800 kg car with a tireprint
DS = 4 × d × e = 4 × 5 cm × 12 cm, we have
Z
1
} ({> |) gD
I} = 800 × 9=81 =
4
DS
¶
μ
Z 0=05 Z 0=12
{6
|6
{
g| g{
=
}p 1 +
0=056 0=126 4 × 0=05
0=05 0=12
= 1=7143 × 102 }p
(3.96)
and therefore,
I}
= 1=1445 × 105 Pa
(3.97)
1=7143 × 102
Example 96 Deformation and rolling resistance.
The distortion of stress distribution is proportional to the tire-road deformation that is the reason for shifting the resultant force forward. Hence,
the rolling resistance increases with increasing deformation. A high pressure tire on concrete has lower rolling resistance than a low pressure tire
on soil.
To model the mechanism of dissipation energy for a turning tire, we assume there are many small dampers and springs in the tire structure. Pairs
of parallel dampers and springs are installed radially and circumstantially.
Figures 3.35 illustrates the spring and damping structures of a tire.
}p =
3. Tire Dynamics
Z
131
Z
v
v
Ground plane
2a
2a
FIGURE 3.35. Spring and damping structures of a tire.
3.5.1 Eect of Speed on the Rolling Friction Coe!cient
The rolling friction coe!cient u increases with a second degree of speed.
Experiment shows that we may express u = u (y{ ) by the function
u = 0 + 1 y{2
(3.98)
Proof. The rolling friction coe!cient increases by increasing speed experimentally. We can use a polynomial function
u =
q
X
l y{l
(3.99)
l=0
to t the experimental data. Practically, two or three terms of the polynomial would be enough to express u = u (y{ ). The function
u = 0 + 1 y{2
(3.100)
is simple and good enough for representing experimental data and analytic
calculation. The values of
0
1
= 0=015
= 7 × 106 s2 @ m2
(3.101)
(3.102)
are applicable for most passenger car tires. However, 0 and 1 should be
determined experimentally for any individual tire. Figure 3.36 depicts a
comparison between Equation (3.98) and experimental data for a radial
tire.
Generally speaking, the rolling friction coe!cient of radial tires show to
be less than non-radials. Figure 3.37 illustrates a sample comparison.
132
3. Tire Dynamics
Pr
Rolling friction coefficient
Experiment
Pr
P0 P1v 2
Critical speed
Experiment
vx [m/s]
0
50
100
vx 200
[km/h]
150
FIGURE 3.36. Comparison between the analytic equation and experimental data
for the rolling friction coe!cient of a radial tire.
Pr
Rolling friction coefficient
Non-radial
Radial
vx [m/s]
0
50
100
150
vx 200
[km/h]
FIGURE 3.37. Comparison of the rolling friction coe!cient between radial and
non-radial tires.
3. Tire Dynamics
133
z
Z
v
Ground plane
x
FIGURE 3.38. Illustration of circumferential waves in a rolling tire at its critical
speed.
Equation (3.98) is applied when the speed is below the tire’s critical
speed. The critical speed is the speed at which standing circumferential
waves appear and the rolling friction increases rapidly. The wavelength of
the standing waves are close to the length of the tireprint. Above the critical
speed, overheating happens and tire fails very soon. Figure 3.38 illustrates
the circumferential waves in a rolling tire at its critical speed.
Example 97 Rolling resistance force and vehicle velocity.
For computer simulation purposes, a fourth degree equation is presented
to evaluate the rolling resistance force Iu
Iu = F0 + F1 y{ + F2 y{4
(3.103)
The coe!cients Fl are dependent on the tire characteristics, however, the
following values can be used for a typical radial passenger car tire:
F0
F1
F2
= 9=91 × 103
= 1=95 × 105
= 1=76 × 109
(3.104)
Example 98 F Wave occurrence justication.
The normal stress will move forward when the tire is turning on a road.
By increasing the speed, the normal stress will shift more and concentrate
in the rst half of the tireprint, causing low stress in the second half of the
tireprint. High stress in the rst half along with no stress in the second half
is similar to hammering the tire repeatedly. When the tire is moving faster
than its critical speed, the hammered tire will not be fully recovered before
the next impact with the grounds.
134
3. Tire Dynamics
Example 99 F Race car tires.
Racecars have very smooth tires, known as slicks. Smooth tires reduce the
rolling friction and maximize straight line speed. The slick racing tires are
also pumped up to high pressure. High pressure reduces the tireprint area.
Hence, the normal stress shift reduces and the rolling resistance decreases.
Example 100 Road pavement and rolling resistance.
The eect of the pavement and road condition is introduced by assigning
a value for 0 in equation u = 0 + 1 y{2 . Table 3=1 is a good reference.
Table 3=1 - The value of 0 on dierent pavements.
Road and pavement condition
Very good concrete
Very good tarmac
Average concrete
Very good pavement
Very good macadam
Average tarmac
Concrete in poor condition
Good block paving
Average macadam
Tarmac in poor condition
Dusty macadam
Good stone paving
Good natural paving
Stone pavement in poor condition
Snow shallow (5 cm)
Snow thick (10 cm)
Unmaintained natural road
Sand
0
0=008 0=01
0=01 0=0125
0=01 0=015
0=015
0=013 0=016
0=018
0=02
0=02
0=018 0=023
0=023
0=023 0=028
0=033 0=055
0=045
0=085
0=025
0=037
0=08 0=16
0=15 0=3
Example 101 F Eect of tire structure, size, wear, and temperature on
u .
The tire material and the arrangement of tire plies aect the rolling friction coe!cient and the critical speed. Radial tires have around 20% lower
u , and 20% higher critical speed than a similar non-radial tires.
Tire radius Uj and aspect ratio kW @zW are the two size parameters that
aect the rolling resistance coe!cient. A tire with larger Uj and smaller
kW @zW has lower rolling resistance and higher critical speed.
Generally speaking, the rolling friction coe!cient decreases with wear in
both radial and non-radial tires, and increases by increasing temperature.
3. Tire Dynamics
135
3.5.2 Eect of In ation Pressure and Load on the Rolling
Friction Coe!cient
The rolling friction coe!cient u decreases by increasing the in ation pressure s. The eect of increasing pressure is equivalent to decreasing normal
load I} .
The following empirical equation has been suggested to show the eects
of both pressure s and load I} on the rolling friction coe!cient.
N
u =
1000
μ
5=5 × 105 + 90I}
1100 + 0=0388I} 2
5=1 +
+
y{
s
s
¶
(3.105)
The parameter N is equal to 0=8 for radial tires, and is equal to 1=0 for nonradial tires. The value of I} , s, and y{ must be in [ N], [ Pa], and [ m@ s]
respectively. Low tire pressure increases roll resistance, fuel consumption,
tire wear, and tire temperature.
Example 102 Motorcycle rolling friction coe!cient.
The following equations are suggested for calculating rolling friction coe!cient u of motorcycles. They can be only used as a rough estimate for
passenger cars. The equations consider the in ation pressure and forward
velocity of the motorcycle, but not considering the load on tire I} .
u
=
;
1800 2=0606 2
A
A
? 0=0085 + s + s y{
A
1800 3=7714 2
A
=
+
y{
s
s
46 m@ s 165 km@ h
y{ 46 m@ s
(3.106)
y{ A 46 m@ s
The speed y{ must be expressed in [ m@ s] and the pressure s must be in
[ Pa]. Figure 3.39 illustrates this equation for y{ 46 m@ s ( 165 km@ h).
Increasing the in ation pressure s decreases the rolling friction coe!cient
u .
Example 103 Dissipated power because of rolling friction.
Rolling friction reduces the vehicle’s power. The dissipated power because
of rolling friction is equal to the rolling friction force Iu times the forward
velocity y{ . Using Equation (3.105), the rolling resistance power is
S
= Iu y{ = u y{ I}
μ
¶
5=5 × 105 + 90I}
1100 + 0=0388I} 2
N y{
5=1 +
+
y{ I} (3.107)
=
1000
s
s
The resultant power S is in [ W] when the normal force I} is expressed in
[ N], velocity y{ in [ m@ s], and pressure s in [ Pa].
136
Rolling friction coefficient
Pr
3. Tire Dynamics
p=100 kPa
p=200 kPa
p=300 kPa
p=400 kPa
vx [m/s]
vx [km/h]
FIGURE 3.39. Motorcycle rolling friction coe!cient.
The rolling resistance dissipated power for motorcycles can be found based
on Equation (3.106).
; μ
¶
1800 2=0606 2
A
A
A
? 0=0085 + s + s y{ y{ I} y{ 46 m@ s
S =
(3.108)
μ
¶
A
A
A 1800 + 3=7714 y{2 y{ I}
y{ A 46 m@ s
=
s
s
46 m@ s 165 km@ h
Example 104 Rolling resistance dissipated power.
If a vehicle is moving at 100 km@ h 27=78 m@ s 62 mi@ h and each
radial tire of the vehicle is pressurized up to 220 kPa 32 psi and loaded
by 220 kg, then the dissipated power, because of rolling resistance, is
¶
μ
N y{
5=5 × 105 + 90I}
1100 + 0=0388I} 2
S = 4×
5=1 +
+
y{ I}
1000
s
s
= 2424=1 W 2=4 kW
(3.109)
To compare the given equations, assume the vehicle has motorcycle tires
with power loss given by Equation (3.108).
μ
¶
1800 2=0606 2
S = 4 × 0=0085 +
+
y{ y{ I}
s
s
= 5734=1 W 5=7 kW
(3.110)
It shows that if the vehicle uses motorcycle tires, it dissipates more power.
3. Tire Dynamics
Proper Inflation
Over Inflation
137
Under Inflation
FIGURE 3.40. Tire-road contact of an over- and under-in ated tire compared to
a properly in ated tire.
Example 105 Eects of improper in ation pressure.
High in ation pressure increases stiness of tires, which reduces ride
comfort and generates vibration. Tireprint area and traction force are reduced when tires are over in ated. Over-in ation causes the tire to transmit
shock loads to the suspension, and reduces the tire’s ability to support the
required load for cornerability, braking, and acceleration.
Under-in ation results in higher internal shear stress, cracking and tire
component separation. It also increases sidewall exing and rolling resistance that cause heat and mechanical failure. A tire’s load capacity is largely
determined by its in ation pressure. Therefore, under-in ation results in
an overloaded tire that operates at high de ection with low fuel economy
and low handling.
Figure 3.40 illustrates the eect of over and under in ation on tire-road
contact compared to a proper in ated tire. Proper in ation pressure is necessary for optimum tire performance, safety, and fuel economy. Correct
in ation is especially signicant to the endurance and performance of radial tires because it may not be possible to nd a 5 psi 35 kPa underin ation in a radial tire just by visual observation. However, under-in ation
of 5 psi 35 kPa can reduce up to 25% of the tire performance and life.
A tire may lose 1 to 2 psi ( 7 to 14 kPa) every month. The in ation
pressure can also change by 1 psi 7 kPa for every 10 F 5 C of temperature change. As an example, if a tire is in ated to 35 psi 240 kPa
on an 80 F 26 C summer day, it could have an in ation pressure of
23 psi 160 kPa on a 20 F 6 C day in winter. This represents a normal loss of 6 psi 40 kPa over the six months and an additional loss of
6 psi 40 kPa due to the 60 F 30 C change. At 23 psi 160 kPa, this
tire is functioning under-in ated.
Example 106 Small / large and soft / hard tires.
If the driving tires are small, the vehicle becomes twitchy with low traction
and low top speed. On the other hand, when the driving tires are big, then
the vehicle has slow steering response and high tire distortion in turns,
decreasing the stability.
Softer front tires show more steerability, less stability, and more wear
while hard front tires show the opposite. Soft rear tires have more rear
traction, but they make the vehicle less steerable, more bouncy, and less
138
3. Tire Dynamics
stable. Hard rear tires have less rear traction, but they make the vehicle
more steerable, less bouncy, and more stable.
3.5.3 F Eect of Sideslip Angle on Rolling Resistance
When a tire is rolling on the road with a sideslip angle , a signicant
increase in rolling resistance occurs. The rolling resistance force Iu would
then be
(3.111)
Iu = I{ cos + I| sin I{ F 2
where, I{ is the longitudinal force opposing the motion, and I| is the
lateral force, both in tire coordinate frame.
Proof. Figure 3.41 illustrates the top view of a rolling tire on the ground
under a sideslip angle . The rolling resistance force is dened as the force
opposite to the velocity vector of the tire, which has angle with the {axis. Assume a longitudinal force I{ in {-direction is applied on the tire.
Sideslip increases I{ and generates a lateral force I| in |-direction.
The sum of the components of the longitudinal force I{ and the lateral
force I| makes the rolling resistance force Iu .
Iu = I{ cos + I| sin (3.112)
For small values of the sideslip angle , the lateral force is proportional to
and therefore,
(3.113)
Iu I{ F 2
3.5.4 F Eect of Camber Angle on Rolling Resistance
When a tire travels with a camber angle , the component of rolling moment Mu on rolling resistance Fu will be reduced, however, a component
of aligning moment P} on rolling resistance will appear.
Fu
Iu
= Iu ˆ~
1
1
=
Pu cos +
P} sin Uk
Uk
(3.114)
(3.115)
Proof. Rolling moment Pu appears when a tire is rolling and the normal
force I} shifts forward. However, only the component Pu cos is perpendicular to the tire-plane and prevents the tire’s spin. Furthermore, when a
moment in }-direction is applied on the tire, only the component P} sin will prevent the tire’s spin. Therefore, the camber angle will aect the
rolling resistance according to
Fu
Iu
= Iu ˆ~
1
1
=
Pu cos +
P} sin Uk
Uk
(3.116)
3. Tire Dynamics
v
D
139
x
y
Fy
Fx
Fr
FIGURE 3.41. Eect of sideslip angle on rolling resistance force Iu .
where Uk is the height of tire center from the road surface as is shown
in Figure 3.10. The rolling moment Pu may be substituted by Equation
(3.90) to show the eect of normal force I} .
Iu =
{
1
I} cos +
P} sin Uk
Uk
(3.117)
3.6 Longitudinal Force
The longitudinal slip ratio of a tire is dened by
v=
Uj $ z
1
y{
(3.118)
where, Uj is the tire’s geometric and unloaded radius, $ z is the tire’s
angular velocity, and y{ is the tire’s forward velocity. Slip ratio is positive
for driving and is negative for braking.
To accelerate or brake a vehicle, longitudinal forces must develop between
the tire and the ground. When a moment is applied to the spin axis of
the tire, slip ratio occurs and a longitudinal force I{ is generated at the
tireprint. The force I{ is proportional to the normal force,
F{
I{
= I{ ˆ~
= { (v) I}
(3.119)
(3.120)
140
3. Tire Dynamics
Px
P dp
P ds
Sliding
Braking
-0.5
-0.3
Sliding
Driving
-0.1
0.1
0.3
0.5
s
P bs
P bp
FIGURE 3.42. Longitudinal friction coe!cient as a function of slip ratio v, in
driving and braking.
where the coe!cient { (v) is called the longitudinal friction coe!cient and
is a function of slip ratio v as shown in Figure 3.42. The friction coe!cient
reaches a driving peak value gs at v 0=1, before dropping to an almost
steady-state driving slide value gv . The friction coe!cient { (v) may be
assumed proportional to v when v is very small
{ (v) = Fv v
v ?? 1
(3.121)
where Fv is called the longitudinal slip coe!cient.
The tire will spin when v & 0=1 and the friction coe!cient remains almost
constant. The same phenomena happen in braking at the values es and
ev .
Proof. Slip ratio, or simply slip, is dened as the dierence between the
actual speed of the tire y{ and the equivalent tire speed Uz $ z . Figure
3.43 illustrates a rolling tire on the ground. The ideal distance that the tire
would freely travel with no slip is denoted by gI , while the actual distance
the tire travels is denoted by gD . Thus, for a spinning tire, gI A gD , and
for a slipping tire, gI ? gD .
The dierence gI gD is the tire slip and therefore, the slip ratio of the
tire is
gI gD
(3.122)
v=
gD
To have the instant value of v, we must measure the travel distances in an
innitesimal time period, and therefore,
v
gbI gbD
gbD
(3.123)
3. Tire Dynamics
141
Z
v
Ground plane
x
dA
dF
FIGURE 3.43. A turning tire on the ground to show the no slip travel distance
gI , and the actual travel distance gD .
If the angular velocity of the tire is $ z then, gbI = Uj $ z and gbD = Uz $ z
where, Uj is the geometric tire radius and Uz is the eective radius. Therefore, the slip ratio v can be dened based on the actual speed y{ = Uz $ z ,
and the freely rolling speed Uj $ z
v=
Uj $ z Uz $ z
Uj $ z
=
1
Uz $ z
y{
(3.124)
A tire can exert longitudinal force only if a longitudinal slip is present.
During acceleration, the actual velocity y{ is less than the free velocity
Uj $ z , and therefore, v A 0. However, during braking, the actual velocity
y{ is higher than the free velocity Uj $ z and therefore, v ? 0.
The frictional force I{ between a tire and the road surface is a function
of normal load I} , vehicle speed y{ , and wheel angular speed $ z . In addition to these variables there are a number of parameters that aect I{ ,
such as tire pressure, tread design, wear, and road surface conditions. It
has been determined empirically that a contact friction force of the form
I{ = { ($ z > y{ )I} can model experimental measurements obtained with
constant y{ , $ z .
Longitudinal slip is also called circumferential or tangential slip.
Example 107 Slip ratio is 0 ? v ? 4 in driving.
When we drive, a driving moment is applied to the tire axis. The tread
of the tire will be compressed circumstantially in the tireprint zone. Hence,
the tire is moving slower than a free tire
Uz $ z ? Uj $ z
(3.125)
and therefore v A 0. The equivalent radius for a driving tire is less than the
geometric radius
Uz ? Uj
(3.126)
142
3. Tire Dynamics
Equivalently, we may express the condition using the equivalent angular
velocity $ ht and deduce that a driving tire turns faster than a free tire
Uj $ ht ? Uj $ z
(3.127)
The driving moment can be high enough to overcome the friction and turn
the tire on pavement while the car is not moving. In this case y{ = 0
and therefore, v = 4. It shows that the longitudinal slip would be between
0 ? v ? 4 when accelerating.
0 ? v ? 4 for d A 0
(3.128)
The tire speed Uz $ z equals vehicle speed y{ only if acceleration is zero.
In this case, the normal force acting on the tire and the size of the tireprint
are constant in time.
Applying a driving moment generates a positive v A 0 and shifts the symmetric point of the tread displacement of Figures 3.18 and 3.19 backward.
This results more forward shear stresses which provides a forward traction
force.
Example 108 Samples for longitudinal friction coe!cients gs and gv .
Table 3=2 shows the average values of longitudinal friction coe!cients
gs and gv for a passenger car tire 215@65U15. It is practical to assume
gs = es , and gv = ev .
Table 3=2 - Average of longitudinal friction coe!cients.
Road surface Peak value, gs Sliding value, gv
Asphalt, dry
0=8 0=9
0=75
Concrete, dry
0=8 0=9
0=76
Asphalt, wet
0=5 0=7
0=45 0=6
Concrete, wet
0=8
0=7
Gravel
0=6
0=55
Snow, packed
0=2
0=15
Ice
0=1
0=07
Example 109 Slip ratio is 1 ? v ? 0 in braking.
When we brake, a braking moment is applied to the wheel axis. The tread
of the tire will be stretched circumstantially in the tireprint zone. Hence,
the tire is moving faster than a free tire
Uz $ z A Uj $ z
(3.129)
and therefore, v ? 0. The equivalent radius for a braking tire is more than
the free radius
Uz A Uj
(3.130)
3. Tire Dynamics
143
Equivalently, we may express the condition using the equivalent angular
velocity $ h and deduce that a braking tire turns slower than a free tire
Uj $ ht A Uj $ z
(3.131)
The brake moment can be high enough to lock the tire. In this case $ z = 0
and therefore, v = 1. It shows that the longitudinal slip would be between
1 ? v ? 0 when braking.
1 ? v ? 0 for d ? 0
(3.132)
Example 110 Slip ratio based on equivalent angular velocity $ ht .
It is possible to dene an eective angular velocity $ ht as an equivalent
angular velocity for a tire with radius Uj to proceed with the actual speed
y{ = Uj $ ht . Using $ ht we have
y{ = Uj $ ht = Uz $ z
(3.133)
Uj $ z Uj $ ht
$z
=
1
Uj $ ht
$ ht
(3.134)
and therefore,
v=
Example 111 Power and maximum velocity.
Consider a moving car with power S = 100 kW 134 hp that can attain
279 km@ h 77=5 m@ s 173=3 mi@ h. The total driving force must be
I{ =
S
100 × 103
=
= 1290=3 N
y{
77=5
(3.135)
If we assume that the car is rear-wheel-drive and the rear wheels are driving
at the maximum traction under the load 1600 N, then the longitudinal friction coe!cient { is
{ =
I{
1290=3
=
0=806
I}
1600
(3.136)
Example 112 Slip of hard tire on hard road.
A tire with no slip cannot create any tangential force. Assume a toy
car equipped with steel tires is moving on a glass table. Such a car cannot
accelerate or steer easily. If the car can accelerate at very low rate, it is
because there is su!cient microscopic slip to generate forces to drive. The
glass table and the small contact area of the small metallic tires deform and
stretch each other, although such a deformation is in microscopic scale. If
there is any friction between the tire and the surface, there must be slip to
maneuver.
144
3. Tire Dynamics
Z
Fz
FIGURE 3.44. The molecular binding between the tire and road surfaces.
Example 113 F Friction mechanisms.
Rubber tires generate friction in three mechanisms: 1= adhesion, 2= deformation, and 3= wear.
I{ = Idg + Igh + Izh
(3.137)
Adhesion friction is equivalent to sticking. The rubber resists sliding
on the road because adhesion causes it stick to the road surface. Adhesion
occurs as a result of molecular binding between the rubber and surfaces.
Because the real contact area is much less than the observed contact area,
high local pressure make molecular binding, as shown in Figure 3.44. Bound
occurs at the points of contact and welds the surfaces together. The adhesion
friction is equal to the required force to break these molecular bounds and
separate the surfaces. The adhesion is also called cold welding and is
attributed to pressure rather than heat. Higher load increases the contact
area, makes more bounds, and increases the friction force. So the adhesion
friction conrms the traditional friction equation
I{ = { (v) I}
(3.138)
The main contribution to tire traction force on a dry road is the adhesion
friction. The adhesion friction decreases considerably on a road covered by
water, ice, dust, or lubricant. Water on a wet road prevents direct contact
between the tire and road and reduces the formation of adhesion friction.
The main contribution to tire friction when it slides on a road surface is the
viscoelastic energy dissipation in the tireprint area. This dissipative energy
is velocity and is time-history dependent.
Deformation friction is the result of deforming rubber and lling the
microscopic irregularities on the road. The surface of the road has many
peaks and valleys called asperities. Movement of a tire on a rough surface
results in the deformation of the rubber by peaks and high points on the
surface. A load on the tire causes the peaks of irregularities to penetrate
the tire and the tire drapes over the peaks. The deformation friction force,
3. Tire Dynamics
145
needed to move the irregularities in the rubber, comes from the local high
pressure across the irregularities. Higher load increases the penetration of
the irregularities in the tire and therefore increases the friction force. So
the deformation friction also conrms the friction equation (3.138).
The main contribution to the tire traction force on a wet road is the
deformation friction. The adhesion friction decreases considerably on a road
covered by water, ice, dust, or lubricant.
Deformation friction exists in relative movement between any contacted
surfaces. No matter how much care is taken to form a smooth surface, the
surfaces are irregular with microscopic peaks and valleys. Opposite peaks
interact with each other and cause damage to both surfaces.
Wear friction is the result of excessive local stress over the tensile
strength of the rubber. High local stresses deform the structure of the tire
surface past the elastic point. The polymer bonds break, and the tire surface
tears in microscopic scale. This tearing makes the wear friction mechanism. Wear results in separation of material. Higher load eases the tire
wear and therefore increases the wear friction force. So the wear friction
also conrms the friction equation (3.138).
Example 114 tanh model of slip friction coe!cient { .
We introduce the tanh describing function to provide a mathematical
equation for analytical calculations.
μ ¶
v
(3.139)
{ = v tanh
vf
The equation needs two numbers v and vf which should be measured experimentally. Assuming a symmetric { in driving and braking, v is the
sliding value of { , and vf is the critical value of v at which { = 0=76v .
Figure 3.45 illustrates the mathematical model of slip friction coe!cient
{ .
In case the v and vf are dierent in driving and braking, the function
¶
μ
μ ¶
v
v
+ ev K (v) tanh
(3.140)
{ = gv K (v) tanh
vgf
vef
models the friction coe!cient { , where K (v) is the Heaviside function,
vgf and vef are respectively the critical values of v in driving and braking.
Example 115 F Empirical slip models.
Based on experimental data and curve tting methods, some mathematical equations were presented to simulate the longitudinal tire force as a
function of longitudinal slip v. Most of these models are too complicated
to be useful in vehicle dynamics. However, a few of them are simple and
accurate enough to be applied.
The Pacejka model, which was presented in 1991, has the form
¡
¡
¡
¢¢¢
(3.141)
I{ (v) = f1 sin f2 tan1 f3 v f4 f3 v tan1 (f3 v)
146
3. Tire Dynamics
Px
Ps
Sliding
0.76 P s
Braking
-sc
Driving
sc
Sliding
s
0.76 P s
Ps
FIGURE 3.45. The mathematical model of slip friction coe!cient { .
where f1 , f2 , f3 , and f4 are four constants based on the tire experimental
data.
The 1987 Burckhardt model is a simpler equation that also needs three
numbers.
¡
¢
(3.142)
I{ (v) = f1 1 hf2 v f3 v
There is another Burckhardt model that includes the velocity dependency.
¢
¡ ¡
¢
I{ (v) = f1 1 hf2 v f3 v hf4 y
(3.143)
This model needs four numbers to be measured from experiment.
By expanding and approximating the 1987 Burckhardt model, the simpler
model by Kiencke and Daviss was suggested in 1994. This model is
v
I{ (v) = nv
(3.144)
1 + f1 v + f2 v2
where nv is the slope of I{ (v) versus v at v = 0
nv = lim
v$0
4Iv
4v
and f1 , f2 are two experimental numbers.
Another simple model is the 2002 De-Wit model
s
I{ (v) = f1 v f2 v
(3.145)
(3.146)
that is based on two numbers f1 , f2 .
In either case, we need at least one experimental curve such as shown
in Figure 3.42 to nd the constant numbers fl . The constants fl are the
numbers that best t the associated equation with the experimental curve.
The 1997 Burckhardt model (3.143) needs at least two similar tests at two
dierent speeds.
3. Tire Dynamics
147
Z
Fz
v
T1
T T2
h
Sand
W
V
FIGURE 3.46. A tire turning on sand.
Example 116 F Alternative slip ratio denitions.
An alternative method to dene the slip ratio is
;
y{
A
A
? 1 Uj $ z Uj $ z A y{ driving
v=
A
U $
A
= j z 1 Uj $ z ? y{ braking
y{
(3.147)
where y{ is the speed of the wheel center, $ z is the angular velocity of the
wheel, and Uj is the tire radius.
In another alternative denition, the following equation is used for longitudinal slip:
¶q
½
μ
Uj $ z
+1 Uj $ z y{
where q =
(3.148)
v = 1
1 Uj $ z A y{
y{
v 5 [0> 1]
In this denition v is always between zero and one. When v = 1, then the
tire is either locked while the car is sliding, or the tire is spinning while the
car is not moving.
Example 117 F Tire on soft sand.
Figure 3.46 illustrates a tire rolling on sand. The sand will be packed
when the tire passes. The applied stresses from the sand on the tire are
developed during the angle 1 ? ? 2 measured counterclockwise from
vertical direction.
It is possible to dene a relationship between the normal stress and
tangential stress under the tire
´
³
Uj
(3.149)
= (f + tan ) 1 h n [1 +(1v)(sin )sin 1 ]
where v is the slip ratio dened in Equation (3.147), and
P = f + tan (3.150)
148
3. Tire Dynamics
is the maximum shear stress in the sand applied on the tire. In this equation,
f is the cohesion stress of the sand, and n is a constant.
Example 118 F Lateral slip ratio.
Analytical expressions can be established for the force contributions in {
and | directions using adhesive and sliding concept by dening longitudinal
and lateral slip ratios v{ and v|
v{ =
Uj $ z
1
y{
v| =
Uj $ z
y|
(3.151)
where y{ is the longitudinal speed of the wheel and y| is the lateral speed of
the wheel. The unloaded geometric radius of the tire is denoted by Uj and
$ z is the angular velocity of the wheel.
At very low slips, the resulting tire forces are proportional to the slip
I{ = Fv{ v{
I| = Fv| v|
(3.152)
where Fv{ is the longitudinal slip coe!cient and Fv| is the lateral slip
coe!cient.
3.7 Lateral Force
When a rolling tire is under a vertical force I} and a lateral force I| at
the tire axis, its path of motion on the road makes an angle with respect
to the tire-plane. The angle is called sideslip angle and is proportional to
the lateral force
F|
I|
= I| ˆ
= F where F is called the cornering stiness of the tire.
¯
¯
C (I| ) ¯¯
CI| ¯¯
F = lim
= ¯ lim
$0
$0 C ¯
C
(3.153)
(3.154)
(3.155)
The lateral force F| at tireprint is at a distance d{ behind the centerline
of the tireprint and makes a moment M} called aligning moment.
M}
P}
= P} n̂
= I| d{
(3.156)
(3.157)
For small , the aligning moment M} tends to turn the tire about the
}-axis and make the {-axis align with the velocity vector v. The aligning
moment always tends to reduce .
3. Tire Dynamics
Fz /2
149
Fz /2
Fy
y
'y
FIGURE 3.47. Front view of a laterally de ected tire.
Tireprint
head
Symmetric line
Tireprint
tail
Laterally deflected tire
Tireprint
FIGURE 3.48. Bottom view of a laterally de ected tire.
Proof. When a wheel is under a constant load I} and then a lateral force
is applied on the rim, the tire will de ect laterally as is shown in Figure
3.47. The tire acts as a linear spring under small lateral forces I| with a
lateral stiness n| .
(3.158)
I| = n| |
The wheel will start sliding laterally when the lateral force reaches a
maximum value I|P . At this point, the lateral force approximately remains
constant and is proportional to the vertical load
I|P = | I}
(3.159)
where, | is the tire friction coe!cient in the |-direction. A bottom view
of the tireprint of a laterally de ected tire is shown in Figure 3.48.
If the laterally de ected tire is rolling forward on the road, the tireprint
will also ex longitudinally. A bottom view of the tireprint for such a laterally de ected and rolling tire is shown in Figure 3.49. Although the tireplane of such a tire remains perpendicular to the road, the path of the
150
3. Tire Dynamics
Sticking
region
Sliding line
Sliding
region
Laterally deflected
rolling tire
Tireprint
FIGURE 3.49. Bottom view of a laterally de ected and turning tire.
wheel makes an angle with tire-plane. As the wheel rolls forward, unde ected treads enter the tireprint region and de ect laterally as well as
longitudinally. When a tread moves toward the end of the tireprint, its lateral de ection increases until it approaches the tailing edge of the tireprint.
The normal load decreases at the tail of the tireprint, so the friction force is
lessened and the tread can slide back to its original position when leaving
the tireprint region. The point where the laterally de ected tread slides
back is called sliding line.
A rolling tire under lateral force and the associated sideslip angle are
shown in Figure 3.50. Lateral distortion of the tire treads is a result of
a tangential stress distribution | over the tireprint. Assuming that the
tangential stress | is proportional to the distortion, the resultant lateral
force I|
Z
I| =
| gDs
(3.160)
DS
is at a distance d{ behind the center line.
Z
1
d{ =
{ | gDs
I| DS
(3.161)
The distance d{ is called the pneumatic trail, and the resultant moment
M} is called the aligning moment.
M}
P}
= P} n̂
= I| d{
(3.162)
(3.163)
The aligning moment tends to turn the tire about the }-axis and make it
align with the direction of tire velocity vector v. A stress distribution | ,
the resultant lateral force I| , and the pneumatic trail d{ are illustrated
in Figure 3.50.
3. Tire Dynamics
x
th
x
v
Wh
eel
pa
Tread path
v
D
151
x
D
Wy
y
Fy
y
y
a xD
Pneumatic trail
Bottom view
Top view
FIGURE 3.50. The stress distribution | , the resultant lateral force I| , and the
pneumatic trail d| for a turning tire going on a positive slip angle .
There is also a lateral shift in the tire vertical force I} because of slip
angle , which generates a slip moment P{ about the forward {-axis.
M{
P{
= P{ ˆ~
= I} d|
(3.164)
(3.165)
The slip angle always increases by increasing the lateral force I| .
However, the sliding line moves toward the tail at rst and then moves
forward by increasing the lateral force I| . Slip angle and lateral force I|
work as action and reaction. A lateral force generates a slip angle, and a
slip angle generates a lateral force. Hence, we can steer the tires of a car
to make a slip angle and produce a lateral force to turn the car. Steering
causes a slip angle in the tires and creates a lateral force. The slip angle
A 0 if the tire should be turned about the }-axis to be aligned with the
velocity vector v. A positive slip angle generates a negative lateral force
I| . Hence, steering to the right about the }-axis makes a positive slip
angle and produces a negative lateral force to move the tire to the right.
Using the velocity vector of the tire, and its components, v =y{ˆ~ + y| ˆ, we
may also dene the sideslip angle as
= arctan
y|
y{
(3.166)
A sample of measured lateral force I| as a function of slip angle for
a constant vertical load is plotted in Figure 3.51. The lateral force I| is
linear for small slip angles, however the rate of increasing I| decreases
for higher . The lateral force remains constant or drops slightly when 152
3. Tire Dynamics
7000
6000
-Fy [N]
5000
4000
Radial
Non-Radial
3000
2000
tan 1 CD
1000
0
0
2
4
6
8
10
12
D >deg @
FIGURE 3.51. Lateral force I| as a function of slip angle for a constant vertical
load.
reaches a critical value at which the tire slides on the road. Therefore, we
may assume the lateral force I| is proportional to the slip angle for low
values of .
I|
= F F
= lim
$0
(3.167)
CI|
C
(3.168)
The cornering stiness F of radial tires are higher than F for non-radial
tires. This is because radial tires need a smaller slip angle to produce the
same amount of lateral force I| .
Examples of aligning moments for radial and non-radial tires are illustrated in Figure 3.52. The pneumatic trail d{ increases for small slip angles
up to a maximum value, and decreases to zero and even negative values for
high slip angles. Therefore, the behavior of aligning moment M} is similar
to what is shown in Figure 3.52.
The lateral force I| = F can be decomposed to I| cos , parallel to
the path of motion y, and I| sin , perpendicular to y as shown in Figure
3.53. The component I| cos , normal to the path of motion, is called
cornering force, and the component I| sin , along the path of motion, is
called drag force.
The lateral force I| is also called side force or grip. We may combine
the lateral forces of all tires of a vehicle and have them acting at the car’s
mass center F.
Example 119 Eect of tire load on lateral force curve.
When the wheel load I} increases, the tire treads can stick to the road
better. Hence, the lateral force increases at a constant slip angle , and the
3. Tire Dynamics
153
70
Radial
Non-Radial
60
Mz [Nm]
50
40
30
20
10
0
0
2
4
6
8
10
12
D > deg @
FIGURE 3.52. Aligning moment P} as a function of slip angle for a constant
vertical load.
v
D x
y
Fy cos D
Fy sin D
Fy
FIGURE 3.53. The cornering and drag components of a lateral force I| .
154
3. Tire Dynamics
7000
Fz =7000 N
6000
Fz =6000 N
-Fy [N]
5000
Fz =5000 N
4000
Fz =4000 N
3000
Line of maxima
2000
1000
0
0
2
4
6
8
10
12
D > deg @
FIGURE 3.54. Lateral force behavior of a sample tire for dierent normal loads
as a function of slip angle .
slippage occurs at the higher slip angles. Figure 3.54 illustrates the lateral
force behavior of a sample tire for dierent normal loads.
Increasing the load not only increases the maximum attainable lateral
force, it also pushes the maximum of the lateral force to higher slip angles.
Sometimes the eect of load on lateral force is presented in a dimensionless variable to make it more practical. Figure 3.55 depicts a sample.
Example 120 F Gough diagram.
The slip angle is the main aective parameter on the lateral force I|
and aligning moment P} = I| d{ . However, I} and P} depend on many
other parameters such as speed y, pressure s, temperature, humidity, and
road conditions. A better method to show I} and P} is to plot them versus
each other for a set of parameters. Such a graph is called a Gough diagram. Figure 3.56 depicts a sample Gough diagram for a radial passenger
car tire. Every tire has its own Gough diagram, although we may use an
average diagram for radial or non-radial tires.
Example 121 Eect of velocity.
The curve of lateral force as a function of the slip angle I| () decreases
as velocity increases. Hence, we need to increase the sideslip angle at higher
velocities to generate the same lateral force. Sideslip angle increases by
increasing the steer angle. Figure 3.57 illustrates the eect of velocity on
I| for a radial passenger tire. Because of this behavior, the curvature of
trajectory of a one-wheel-car at a xed steer angle increases by increasing
the driving speed.
3. Tire Dynamics
Fz =4000 N
Fz =5000 N
Fz =6000 N
Fz =7000 N
1.05
0.90
0.75
-Fy / Fz
155
Line of maxima
0.60
0.45
0.30
0.15
0
0
2
4
6
8
10
12
D >deg @
m
FIGURE 3.55. Eect of load on lateral force as a function of slip angle presented
in a dimensionless fashion.
5k
N
-Fy [N]
5000
m
m
mm
30
40
N
ay
6000
0
6 k2
N
7k
=1
0m
7000
9q
4000 Fz =
4 kN
3000
m
55 m
6q 80 mm
4q 100 mm
2000
1000
0
mm
D
0
20
40
60
80
2q
100
Mz [Nm]
FIGURE 3.56. Gough diagram for a radial passenger car tire.
120
156
3. Tire Dynamics
7000
v =10 m/s
v =15 m/s
-Fy [N]
6000
5000
v =20 m/s
4000
v =30 m/s
3000
2000
1000
0
0
2
4
6
8
10
12
D > deg @
FIGURE 3.57. Eect of velocity on I| and P} for a radial tire.
Example 122 F Cubic model for lateral force.
When the sideslip angle is not small, the linear approximation (3.154)
cannot show the tire behavior. Based on a parabolic normal stress distribution on the tireprint, the following third-degree function was presented in
the 1950v to calculate the lateral force at high sideslips
Ã
¯
¯
μ
¶2 !
1 ¯¯ F ¯¯
1 F (3.169)
+
I| = F 1 ¯
3 I|P ¯ 27 I|P
where I|P is the maximum lateral force that the tire can support. I|P is
set by the tire load and the lateral friction coe!cient | . Let us show the
sideslip angle at which the lateral force I| reaches its maximum value I|P
by P . Equation (3.169) shows that
P =
3I|P
F
(3.170)
and therefore,
I|
I|
I|P
Ã
μ
¶2 !
1
= F 1 +
P
3 P
Ã
μ
¶2 !
3
1
=
+
1
P
P
3 P
(3.171)
(3.172)
Figure 3.58 shows the cubic curve model for lateral force as a function
of sideslip angle. The Equation is applicable only for 0 P .
3. Tire Dynamics
157
1
0.8
Fy 0.6
Fy M 0.4
0.2
0
0
0.2
0.4
D
DM
0.6
0.8
1
FIGURE 3.58. A cubic curve model for lateral force as a function of the sideslip
angle.
Example 123 F A model for lateral stress.
Consider a tire rolling on a dry road at a low sideslip angle . Assume
the developed lateral stress on tireprint can be expressed by the equation
¶
μ
³
³| ´
{´
{3
| ({> |) = f |P 1 1 3 cos2
(3.173)
d
d
2e
The coe!cient f is proportional to the tire load I} sideslip , and longitudinal slip v. If the tireprint area is DS = 4 × d × e = 4 × 5 cm × 12 cm, then
the lateral force under the tire, I| , for f = 1 is
Z
I| =
| ({> |) gD
DS
=
¶
μ
³
³ | ´
{ ´
{3
2
|P 1 cos
1
g| g{
0=05
0=053
0=24
0=12
Z 0=05 Z 0=12
0=05
(3.174)
= 0=0144 |P
If we calculate the lateral force I| = 1000 N by measuring the lateral acceleration, then the maximum lateral stress is
|P =
I}
= 69444 Pa
0=014 4
and the lateral stress distribution over the tireprint is
¶
μ
³
³ | ´
{ ´
{3
cos2
1
Pa
| ({> |) = 69444 1 3
0=05
0=05
0=24
(3.175)
(3.176)
158
3. Tire Dynamics
y
x
y
x
FIGURE 3.59. The unbalanced longitudinal stress distribution on a free rolling
tire in a turn.
Example 124 Tire on a circle.
Ideally a tire must be a disc to be able to turn slip free on a circle. A
free rolling wide tire on a turn will have an unbalanced longitudinal stress
distribution as is shown in Figure 3.59. The inner part of the tireprint
contracts, while the outer part of the tireprint extracts. The amount of
contraction and extraction increases by increasing the distance from the tire
plane indicated by the {-axis. Furthermore, the leading part of the tireprint
is shorter than the tail part; therefore, the |-axis of the tireprint coordinate
frame is not a symmetric line. As a result, there is an asymmetric stress
distribution in the tireprint which generates an aligning moment, an antispin moment, and a rolling resistance force.
3.8 Camber Force
Camber angle is the tilting angle of tire about the longitudinal {-axis.
Camber angle generates a lateral force I| called camber thrust or camber
force. Figure 3.60 illustrates a front view of a cambered tire and the generated camber force I| . Camber angle is considered positive A 0, if it is in
the positive direction of the {-axis, measured from the }-axis to the tireplane. A positive camber angle generates a camber force along the |-axis.
3. Tire Dynamics
J
159
z
Tire axis
Fy
y
FIGURE 3.60. A front view of a cambered tire and the generated camber force.
The camber force is proportional to at low camber angles, and depends
directly on the wheel load I} . Therefore,
F|
I|
= I| ˆ
= F (3.177)
(3.178)
where F is called the camber stiness of tire.
F = lim
$0
C (I| )
C
(3.179)
In presence of both, camber and sideslip , the overall lateral force I|
on a tire is a superposition of the corner force and camber thrust.
I| = F F (3.180)
Proof. When a wheel is under a constant load and then a camber angle is
applied on the rim, the tire will de ect laterally such that the tireprint area
is longer in the cambered side and shorter in the other side. Figure 3.61
compares the tireprint of a straight and a cambered tire, turning slowly
on a at road. As the wheel rolls forward, unde ected treads enter the
tireprint region and de ect laterally as well as longitudinally. However, because of the shape of the tireprint, the treads entering the tireprint closer
to the cambered side have more time to be stretched laterally. Because the
developed lateral stress is proportional to the lateral stretch, the nonuniform tread stretching generates an asymmetric stress distribution and more
lateral stress will be developed on the cambered side. The result of the
160
3. Tire Dynamics
nonuniform lateral stress distribution over the tireprint of a cambered tire
produces the camber thrust I| in the cambered direction.
= I| ˆ
Z
=
| gD
F|
I|
(3.181)
(3.182)
DS
The camber thrust is proportional to the camber angle for small angles.
I| = F (3.183)
When the tire is rolling, the resultant camber thrust I| shifts forward
by a distance d{ . The resultant moment in the }-direction is called camber
torque, and the distance d{ is called camber trail.
M}
P}
= P} n̂
= I| d{
(3.184)
(3.185)
Camber trail is usually very small and hence, the camber torque can be
ignored in linear analysis of vehicle dynamics.
Because the tireprint of a cambered tire deforms to be longer in the
cambered side, the resultant vertical force I} that supports the wheel load,
Z
} gD
(3.186)
I} =
DS
shifts laterally by a distance d| from the center of the tireprint.
Z
1
d| =
| } gDs
I} DS
(3.187)
The distance d| is called the camber arm, and the resultant moment M{
is called the camber moment.
M{
P{
= P{ n̂
= I} d|
(3.188)
(3.189)
The camber moment tends to turn the tire about the {-axis and make the
tire-plane align with the }-axis. The camber arm d| is proportional to the
camber angle for small angles.
d| = F| (3.190)
Figure 3.62 shows the camber force I| for dierent camber angle at a
constant tire load I} = 4500 N. Radial tires generate lower camber force
due to their higher exibility.
3. Tire Dynamics
z
J
z
Fy
y
Fz
161
y
ay
Fz
Tireprint
Wy
Straight tire
Wy
Lateral stress
Cambered tire
FIGURE 3.61. The tireprint of a straight and a cambered tire, turning slowly on
a at road.
It is better to illustrate the eect of I} graphically to visualize the camber
force. Figure 3.63 depicts the variation of camber force I| as a function of
normal load I} at dierent camber angles for a sample radial tire.
If we apply a slip angle to a rolling cambered tire, the tireprint will
distort similar to the shape in Figure 3.64 and the path of treads become
more complicated. The resultant lateral force would be at a distance d{
and d| from the center of the tireprint. Both distances d{ and d| are
functions of angles and . Camber force due to , along with the corner
force due to , give the total lateral force applied on a tire. Therefore, the
lateral force can be calculated as
I| = F F (3.191)
that is acceptable for . 10 deg and . 5 deg. Presence of both camber
angle and slip angle makes the situation interesting because the total
lateral force can be positive or negative according to the directions of 162
3. Tire Dynamics
J
1400
z
1200
-Fy [N]
1000
800
y
Fy
600
400
Radial
Non-Radial
200
0
0
2
4
6
8
10
12
J > deg @
FIGURE 3.62. The camber force I| for dierent camber angle at a constant
tire load.
J
1400
z
J
1200
-Fy [N]
1000
800
10q
J
8q
J
6q
J
4q
J
2q
y
Fy
600
400
200
0
0
2000
4000
6000
8000
10000
12000
Fz [N]
FIGURE 3.63. The variation of camber force I| as a function of normal load I}
at dierent camber angles for a sample radial tire.
3. Tire Dynamics
163
x
y
FIGURE 3.64. Tireprint of a cambered tire under a sideslip.
J 10q
J 5q
J 0q
J 5q
J 10q
5000
4000
-Fy [N]
3000
2000
Fz =4000 N
1000
0
-1000
-2000
0
2
4
6
8
10
12
D >deg @
FIGURE 3.65. An example for lateral force as a function of and at a constant
load I} = 4000 N.
and . Figure 3.65 illustrates an example of lateral force as a function of and at a constant load I} = 4000 N. Similar to lateral force, the aligning
moment P} can be approximated as a combination of the slip and camber
angle eects
(3.192)
P} = FP + FP For a radial tire, FP 0=013 N m@ deg and FP 0=0003 N m@ deg, while
for a non-radial tire, FP 0=01 N m@ deg and FP 0=001 N m@ deg.
Example 125 F Banked road.
Consider a vehicle moving on a road with a transversal slope , while
its tires remain vertical. There is a downhill component of weight, I1 =
pj sin , that pulls the vehicle down. There is also an uphill camber force
due to camber of tires with respect to the road I2 = F . The
164
3. Tire Dynamics
resultant lateral force I| = F pj sin depends on camber stiness
F and determines if the vehicle goes uphill or downhill. Since the camber
stiness F is higher for non-radial tires, it is more possible for a non-radial
than a radial tire to go uphill.
The eects of cambering are particularly important for motorcycles that
produce a large part of the lateral force by cambering. For cars and trucks,
the camber angles are much smaller and in many applications their effect can be negligible. However, some suspensions are designed to make the
wheels cambered when the axle load varies, or when they steered.
Example 126 Camber importance and tireprint model.
Cambering of a tire creates a lateral force, even though there is no sideslip.
The eects of cambering are particularly important for motorcycles that
produce a large part of the lateral force by camber. The following equations
are presented to model the lateral deviation of a cambered tireprint from the
straight tireprint, and express the lateral stress | due to camber
|
|
q
³q
´
= sin Uj2 {2 Uj2 d2
¡
¢
= n d2 {2
(3.193)
(3.194)
where n is chosen such that the average camber defection is correct in the
tireprint
Z d
Z d
| g{ =
| g{
(3.195)
d
d
Therefore,
n
=
¶
μ q
3 sin d
2 d2 + U2 sin1
U
d
j
j
4d3 Uj
q
2
2
3 Uj Uj d
4
d2
and
3 Uj
| = 4
q
Uj2 d2 ¡
¢
d2 {2
2
d
(3.196)
(3.197)
(3.198)
3.9 Tire Force
Tires may be considered as a force generator with two major outputs: forward force I{ , lateral force I| , and three minor outputs: aligning moment
P} , roll moment P{ , and pitch moment P| . The input of the force generator is the tire load I} , sideslip , longitudinal slip v, and the camber
3. Tire Dynamics
165
angle .
I{
I|
P{
P|
P}
=
=
=
=
=
I{ (I} > > v> )
I| (I} > > v> )
P{ (I} > > v> )
P| (I} > > v> )
P} (I} > > v> )
(3.199)
(3.200)
(3.201)
(3.202)
(3.203)
Ignoring the rolling resistance and aerodynamic force, the major output
forces can be approximated by a set of linear equations for a given load I}
I{
I|
= { (v) I}
{ (v) = Fv v
= F F (3.204)
where, Fv is the longitudinal slip coe!cient, F is the lateral stiness, and
F is the camber stiness.
When the tire has a combination of tire inputs, > v> , the tire forces
are called tire combined force. The most important tire combined force
is the shear force because of longitudinal slip and sideslips. However, as
long as the angles and slips are within the linear range of tire behavior, a
superposition can be utilized to estimate the output forces.
Driving and braking forces change the lateral force I| generated at any
sideslip angle . This is because the longitudinal force pulls the tireprint
in the direction of the driving or braking force and hence, the length of
lateral displacement of the tireprint will also change.
Figure 3.66 illustrates how a sideslip aects the longitudinal force
ratio I{ @I} as a function of slip ratio v. Figure 3.67 illustrates the eect
of sideslip on the lateral force ratio I| @I} as a function of slip ratio v.
Figure 3.68 and 3.69 illustrate the same force ratios as Figures 3.66 and
3.67 when the slip ratio v is a parameter.
Proof. Consider a turning tire under a sideslip angle . The tire develops a lateral force I| = F . Applying a driving or braking force on
this tire will reduce the lateral force while developing a longitudinal force
I{ = { (v) I} . Experimental data shows that the reduction in lateral
force in presence of a slip ratio v is similar to Figure 3.67. Now assume
the sideslip is reduced to zero. Reduction will increase the longitudinal
force while decreasing the lateral force. Increasing the longitudinal force is
experimentally similar to Figure 3.68.
A turning tire under a slip ratio v develops a longitudinal force I{ =
{ (v) I} . Applying a sideslip angle will reduce the longitudinal force
while developing a lateral force. Experimental data shows that the reduction in longitudinal force in presence of a sideslip is similar to Figure 3.66.
Now assume the slip ratio v and hence, the driving or breaking force is reduced to zero. Reduction in v will increase the lateral force while decreasing
166
3. Tire Dynamics
Px
Fx / Fz
D 1
D 2
Sliding
1.0
0.75
D 4
D 8
D 12
0.5
Braking
0.25
-0.3
-0.1
-0.5
-0.25
0.3
0.5
D
20
Driving
0.7
s
-0.5
-0.75
-1.0
Sliding
FIGURE 3.66. Longitudinal force ratio I{ @I} as a function of slip ratio v for
dierent sideslip .
Fy / Fz
1.0
0.75
0.5
-0.5
-0.3
Braking
0.25
D 20
D 12
D 8
D 4
D 1D 2
-0.1
0.1
0.3
s
0.5
Driving
FIGURE 3.67. Lateral force ratio I| @I} as a function of slip ratio v for dierent
sideslip .
3. Tire Dynamics
167
Fx / Fz
1.0
s=0.5
0.75
s=0.2
0.5
15
deg
9.0
3.0
Braking
s=0.3
s=0.1
s=0.5
0.25
D
s=0.8
3.0
9.0
15
Driving
deg
21
D
FIGURE 3.68. Longitudinal force ratio I{ @I} as a function of sideslip for
dierent slip ratio v.
Px
Fy / Fz
s=0
s=0.1
s=0.2
s=0.3
s=0.4
s=0.6
1.0
0.75
0.5
0.25
D
15
deg
9.0
3.0
Braking
3.0
Driving
9.0
15
deg
21
D
FIGURE 3.69. Lateral force ratio I| @I} as a function of sideslip for dierent
slip ratio v.
168
3. Tire Dynamics
v
A
x
D
B
A
C
Fshear
Fx
y
O
E
Fy
D
G
FIGURE 3.70. Friction ellipse.
the longitudinal force. Increasing the lateral force is similar to Figure 3.67.
Example 127 Friction ellipse.
When the tire is under both longitudinal slip and sideslips, the tire is
under combined slip. The shear force on the tireprint of a tire under a
combined slip can approximately be found using a friction ellipse model.
¶2 μ
¶2
μ
I{
I|
+
=1
(3.205)
I|P
I{P
A friction ellipse is shown in Figure 3.70.
Proof. The shear force Fvkhdu , applied on the tire at tireprint, parallel to
the ground surface, has two components: the longitudinal force I{ and the
lateral force I| .
Fvkhdu
I{
I|
= I{ ˆ~ + I| ˆ
= Fv v I}
= F (3.206)
(3.207)
(3.208)
These forces cannot exceed their maximum values I|P and I{P .
I|P
I{P
= | I}
= { I}
(3.209)
(3.210)
3. Tire Dynamics
169
The tire shown in Figure 3.70 is moving along the velocity vector y at
a sideslip angle . The {-axis indicates the tire-plane. When there is no
$
sideslip, the maximum longitudinal force is I{P = { I} = RD. Now, if
$
a sideslip angle is applied, a lateral force I| = RH is generated and the
$
longitudinal force reduces to I{ = RE. The maximum lateral force would
$
be I|P = | I} = RG where there is no longitudinal slip.
In presence of the longitudinal and lateral forces, we may assume that
the tip point of the maximum shear force vector is on a friction ellipse:
¶2 μ
¶2
μ
I{
I|
+
=1
(3.211)
I|P
I{P
When { = | = , the friction ellipse would be a circle and
Ivkhdu = I}
(3.212)
Example 128 Wide tires.
A wide tire has a shorter tireprint than a narrow tire. Assuming the
same vehicle and same tire pressure, the area of tireprint would be equal in
both tires. The shorter tireprint at the same sideslip has more of its length
stuck to the road than longer tireprint. So, a wider tireprint generates more
lateral force than a narrower tireprint for the same tire load and sideslip.
Generally speaking, tire performance and maximum force capability decrease with increasing speed in both wide and narrow tires.
Example 129 Sinus tire forces model.
A few decades ago, a series of applied sine functions were developed based
on experimental data to model tire forces. The sine functions may eectively
be used to model tire forces, especially for computer purposes.
The sine function model for the lateral force of a tire is
¡
¢
(3.213)
I| = D sin E tan1 (F )
= (1 H) ( + ) I}
(3.214)
¶
μ
F
I
}
F = F1 sin 2 tan1
(3.215)
F =
DE
F2
D> E = Shape factors
(3.216)
F1 = Maximum cornering stiness
(3.217)
F2 = Tire load at maximum cornering stiness
(3.218)
Example 130 F Empirical tire force model.
Based on experimental data and curve tting, the Pacejka model was
presented in 1991, to model nonlinear tire forces.
¡
¡
¢¢
(3.219)
I (x) = f1 sin f2 tan1 f3 x (1 f4 ) + f4 tan1 (f3 v)
170
3. Tire Dynamics
where f1 , f2 , f3 , and f4 are three constants based on the tire experimental
data. When the variable x is the longitudinal slip v, then I (v) = I{ represents the longitudinal traction force. When the variable x is the sideslip
angle , then I () = I| represents the lateral force. The coe!cients f1 ,
f2 , f3 , and f4 are referred to as the stiness, shape, peak, and curvature
coe!cients respectively.
When the tire is under both longitudinal and lateral forces, then I{ and
I| get coe!cients to satisfy the friction ellipse (3.211).
3.10 Summary
We attach a coordinate frame (r{|}) to the tire at the center of the
tireprint, called the tire frame. The {-axis is along the intersection line of
the tire-plane and the ground. The }-axis is perpendicular to the ground,
and the |-axis makes the coordinate system right-hand. We show the tire
orientation using two angles: camber angle and sideslip angle . The
camber angle is the angle between the tire-plane and the vertical plane
measured about the {-axis, and the sideslip angle is the angle between
the velocity vector v and the {-axis measured about the }-axis.
A vertically loaded wheel rolling on a at surface has an eective radius
Uz , called rolling radius
y{
Uz =
(3.220)
$z
where y{ is the forward velocity, and $ z is the angular velocity of the
wheel. The eective radius Uz is approximately equal to
Uj Uk
(3.221)
3
and is a number between the unloaded or geometric radius Uj and the
loaded height Uk .
Uk ? Uz ? Uj
(3.222)
Uz Uj The tire force in the {-direction is a combination of the longitudinal force
I{ and the rolling resistance Iu . The longitudinal force is
I{ = { (v) I}
(3.223)
where v is the longitudinal slip ratio of the tire
Uj $ z
1
y{
v ?? 1
{ (v) = Fv v
v =
(3.224)
(3.225)
The wheel force in the tire |-direction, I| , is a combination of the cornering and camber thrust forces.
I| = F F (3.226)
3. Tire Dynamics
171
The rst term, F , is the camber thrust and is proportional to the
camber angle of the tire. The second term, F , is called the cornering
force and is proportional to the sideslip angle of the tire. The sideslip
coe!cient F and the camber coe!cient F are proportional to the normal
load on the tire and are measured experimentally.
A rolling tire on the ground also generates a longitudinal force called
rolling resistance. The force is opposite to the direction of motion and is
proportional to the normal force on the tireprint.
Iu = u I}
(3.227)
The parameter u is called the rolling friction coe!cient and is a function
of tire mechanical properties, speed, wear, temperature, load, size, driving
and braking forces, and road condition. Because Iu is in y-direction, it
aects both I{ and I| when 6= 0.
172
3. Tire Dynamics
3.11 Key Symbols
d{
¨
d> e
d{
d{
d|
DS
f1 > f2 > f3 > f4
F0 > F1 > F2
Fv
Fv{ > Fv|
F
F
g
gI
gD
G
H
i
in
Iu Fu
I{
I|
I|P
I}
jg
m
n
n1 > n2 > n3 > n4
nht
nv
n{
n|
n}
N
p
Pu Mu
P{ > M{
P|
P}
q
q1
s
acceleration
semiaxes of DS
aligning arm
camber trail
camber arm
tireprint area
coe!cients of the function I{ = I{ (v)
coe!cients of the polynomial function Iu = Iu (y{ )
longitudinal slip coe!cient
longitudinal and lateral slip coe!cients
sideslip coe!cient, sideslip stiness
camber coe!cient, camber stiness
radial displacement of tire elements
no slip tire travel
actual tire travel
tire diameter
Young modulus
function
spring force
rolling resistance force
longitudinal force, forward force
lateral force
maximum lateral force
normal force, vertical force, wheel load
gravitational acceleration
jerk of tire tread in tireprint
stiness
nonlinear tire stiness coe!cients
equivalent stiness
slope of I{ (v) versus v at v = 0
tire stiness in the {-direction
tire stiness in the |-direction
tire stiness in the }-direction
radial and non-radial tires parameter in u = u (s> y{ )
mass
rolling resistance moment
roll moment, bank moment, tilting torque,
pitch moment, rolling resistance torque
yaw moment, aligning moment, self aligning moment
bore torque
exponent for shape and stress distribution of DS
number of tire rotations
tire in ation pressure
3. Tire Dynamics
173
S
u
Uj
Uk
Uz
v
vf
v|
W
y {>
b v
yj
yuho
{> |> }> x
{> |> }
{j
{uho
4{
4|
4}
}b
rolling resistance power
radial position of tire periphery
geometric radius
loaded height
rolling radius, eective radius
longitudinal slip
critical longitudinal slip
lateral slip
wheel torque
velocity, tread velocity in tireprint
velocity of the ground
relative velocity of tire tread with respect to ground
displacement
coordinate axes
displacement of the ground
relative displacement of tread with respect to ground
tire de ection in the {-direction, rolling resistance arm
tire de ection in the |-direction
tire de ection in the }-direction
tire de ection rate in the }-direction
P
%
4{
sideslip angle
maximum sideslip angle
transversal slope
camber angle
de ection
strain
tire de ection in the {-direction,
rolling resistance arm
tire de ection in the |-direction
tire de ection in the }-direction
tire angular rotation
nonlinear rolling friction coe!cient
rolling friction coe!cient
longitudinal friction coe!cient
friction coe!cient driving peak value
friction coe!cient steady-state value
maximum normal stress
normal stress over the tireprint
normal stress mean value
shear stress
shear stresses over the tireprint
maximum shear stresses
contact angle, angular length of DS
equivalent tire angular velocity
angular velocity of a wheel, actual tire angular velocity
4|
4}
0 > 1
u
{ (v)
gs
gv
}P
} ({> |)
}p
{> |
{P > |P
*
$ ht
$> $ z
174
3. Tire Dynamics
Exercises
1. Tireprint size and average normal stress.
The curb weight of a model of Land Rover OU3W P is
p = 2461 kg 5426 lb
while the gross vehicle weight might be
p = 3230 kg 7121 lb
Assume a front to rear load ratio
I}i
1450 kg
=
I}u
1875 kg
and use the following data
o = 2885 mm 113=6 in
W luhv = 255@55U19
to determine the the size parameters of the tireprints d and e, for the
front and rear tires as a function of in ation pressure s. Assume a
uniform normal stress on tireprints.
2. Tireprint size, radial tire.
Holden TK BarinaW P is a hatchback car with the following characteristics.
p = 860 kg
o = 2480 mm
W luhv = 185@55U15 82Y
Assume
d1
= 1=1
d2
and determine the size of its tireprints for q = 3 and assuming a
maximum pressure in the tireprint equal to }P [ Pa].
3. F Equivalent viscous damping.
The loading and unloading of rubbery materials make a loop in the
force-displacement, (I> {), plane.
(a) Show that the area of the loop determines the wasted energy in
one cycle of loading and unloading.
(b) Explain why the loading curve must be higher than the unloading curve.
(c) Assuming the material is elastic, determine the equivalent viscous damping fht such that the amount of the wasted energy in
a cycle remains the same.
3. Tire Dynamics
175
4. Tireprint size and load on tire.
Assuming a uniform upward pressure s under the tire that is carrying
a load I} ,
sDS = s (2d × 2e) = I}
(a) determine the half length of the tireprint, d, as a function of
pressure s.
(b) determine the loaded height Uk as a function of d, and eventually
as a function of I} .
(c) determine the vertical stiness of a tire based on the curve of
I} versus Uk .
(d) determine the eective rolling radius Uz as a function of tire
load I} .
5. Average strain in tireprint.
Consider a loaded tire with a tireprint area of DS = 2d × 2e. The arc
Uj * of the tire becomes 2d under the load. Determine the contact
strain % = o@o of the tire in the tireprint.
6. Radial jerk.
...
Having the function u = u (), determine and plot the radial jerck u
in tireprint.
7. F Tread velocity in tireprint.
The speed of tread of a rolling tire in tireprint zone is
y = Uj $
cos *
cos2 *? ?*
(a) Plot y@Uj $ as a function of for * = 15 deg.
(b) Show that y is equal to Uj $ at two symmetric points and determine their associated .
8. Maximum yuho in tireprint.
The extremes of yuho in tireprint occurs at the two ends of the tireprint
and at the middle point = 0. Determine and plot yuho at = 0 as a
function of *.
9. Rolling resistance coe!cient.
Alfa Romeo SpiderW P has the following characteristics.
p = 1690 kg 3725=8 lb
W luhv = S 225@50U17
o = 2530 mm 99=6 in
176
3. Tire Dynamics
Determine the rolling resistance coe!cient u for the front and rear
tires of the car at the top speed yP .
yP = 235=0 km@ h 146=0 mi@ h
Assume d1 @d2 = 1=2 and use s = 27 svl.
10. Rolling resistance power.
A model of Mitsubishi GalantW P has the following specications.
p = 1700 kg
o = 2750 mm
W luhv = S 235@45U18
yP 190 km@ h
Assume d1 @d2 = 1=2 and s = 27 svl to nd the rolling resistance
power at the maximum speed.
11. Rolling resistance force.
Is the direction of the rolling resistance force Fu , in v or ˆ~? Discuss
the conditions and assumptions.
12. Longitudinal slip.
(a) Determine the longitudinal slip v for the tire S 225@50U17 if
Uz = 0=98Uj .
(b) If the speed of the wheel is y{ = 100 km@ h, what would be the
wheel angular velocity $ z and equivalent angular velocity $ ht
of the tire.
13. Cornering and drag force on a tire.
Consider the tire for which we have estimated the lateral force behavior shown in Figure 3.54. If the sideslip angle is 4 deg and
I} = 5000 N, calculate the cornering and drag force on the tire.
14. Required camber angle.
Consider the tire for which we have estimated the behavior shown
in Figure 3.65. Assume I} = 4000 N and we need a lateral force
I| = 3000 N. If = 4 deg, what would be the required camber
angle ? Estimate the coe!cients F and F .
15. High camber angle.
Consider a tire with F = 300 N@ deg and F = 700 N@ deg. If the
camber angle is = 18 deg how much lateral force will develop for a
zero sideslip angle? How much sideslip angle is needed to reduce the
value of the lateral force to I| = 3000 N?
3. Tire Dynamics
177
16. Sideslip and longitudinal slip.
Consider the tire for which we have estimated the behavior shown in
Figure 3.67. Assume a vehicle with that tire is turning with a constant
speed on a circle such that = 4 deg. What should be the sideslip
angle if we accelerate the vehicle such that v = 0=05, or decelerate
the vehicle such that v = 0=05, to provide the same lateral force?
17. F Motion of the air in tire.
What do you think about the motion of the pressurized air within
the tires, when the vehicle moves with constant velocity or constant
acceleration?
4
Driveline Dynamics
The maximum achievable acceleration of a vehicle is limited by two factors: maximum torque at driving wheels, and maximum traction force at
tireprints. The rst one depends on engine and transmission performance,
and the second one depends on tire-road friction. In this chapter, we examine engine and transmission performance.
4.1 Engine Dynamics
The maximum attainable power Sh of an internal combustion engine is
a function of the engine angular velocity $ h . This function must be determined experimentally, however, the power performance function, Sh =
Sh ($ h ), can be estimated by a third-order polynomial.
180
60
50
120
T
30
90
20
60
10
30
0
0
1000
2000
3000
0
100
200
300
4000
400
5000
500
6000
T [Nm]
40
P [kW]
150
P
0
7000
Z [rpm]
600
700
Z [rad/s]
FIGURE 4.1. A sample of power and torque performances for a spark ignition
engine.
Sh =
3
X
Sl $ lh = S1 $ h + S2 $ 2h + S3 $ 3h
(4.1)
l=1
If we indicate the maximum power of an engine by SP , measured in
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_4, © Springer Science+Business Media New York 2014
179
180
4. Driveline Dynamics
[ W = N m@ s], which occurs at the angular velocity $ P , measured in
[ rad@ s] then for spark ignition engines we use
S1 =
SP
$P
S2 =
SP
$ 2P
S3 = SP
$ 3P
(4.2)
Figure 4.1 illustrates a sample for power performance of a spark ignition
engine that provides SP = 50 kW at $ P = 586 rad@ s 5600 rpm. The
curve begins at an angular velocity at which the engine idles smoothly.
For indirect injection Diesel engines we use
S1 = 0=6
SP
$P
S2 = 1=4
SP
$ 2P
S3 = SP
$ 3P
(4.3)
SP
$ 3P
(4.4)
and for direct injection Diesel engines we use
S1 = 0=87
SP
$P
S2 = 1=13
SP
$ 2P
S3 = The driving torque Wh of the engine is the torque that provides Sh
Wh =
Sh
= S1 + S2 $ h + S3 $ 2h
$h
(4.5)
Example 131 Porsche 911W P and Corvette ]06W P engines.
A model of Porsche 911 turbo has a at-6 cylinder twin-turbo engine with
3596 cm3 220 in3 total displacement. The engine provides a maximum
power SP = 353 kW 480 hp at $ P = 6000 rpm 628 rad@ s, and a maximum torque WP = 620 N m 457 lb ft at $ h = 5000 rpm 523 rad@ s. The
car weighs around 1585 kg 3494 lb and can move from 0 to 96 km@ h 60 mi@ h in 3=7 s. Porsche 911 has a top speed of 310 km@ h 193 mi@ h.
The power performance equation for the Porsche 911 engine has the coe!cients
S1
S2
S3
SP
353000
= 562=1 W s
=
$P
628
SP
353000
=
=
= 0=89507 W s2
$ 2P
6282
SP
353000
= 3 =
= 1=4253 × 103 W s3
$P
6283
=
(4.6)
(4.7)
(4.8)
and, its power performance function is
Sh = 562=1 $ h + 0=89507 $ 2h 1=4253 × 103 $ 3h
(4.9)
A model of Corvette ]06 uses a Y 8 engine with 6997 cm3 427 in3 total
displacement. The engine provides a maximum power SP = 377 kW 512 hp at $ P = 6300 rpm 660 rad@ s and a maximum torque WP =
4. Driveline Dynamics
400
181
Pcorvette-Z06
P [kW]
300
PPorsche-911
200
100
0
0
1000
2000
3000
0
100
200
300
4000
400
5000
500
6000
7000
Z [rpm]
600
700
Z [rad/s]
FIGURE 4.2. Power performance curves for the Porsche 911 and Corvette ]06.
637 N m 470 lb ft at $ h = 4800 rpm 502 rad@ s. The Corvette weighs
around 1418 kg 3126 lb and can move from 0 to 100 km@ h 62 mi@ h in
3=9 s in rst gear. Its top speed is 320 km@ h 198 mi@ h.
The power performance equation for the engine of Corvette ]06 has the
coe!cients
S1
S2
S3
SP
377000
=
= 571=2 W s
$P
660
SP
377000
=
=
= 0=86547 W s2
2
$P
6602
SP
377000
= 3 =
= 1=3113 × 103 W s3
$P
6603
=
(4.10)
(4.11)
(4.12)
and therefore, its power performance function is
Sh = 571=2 $ h + 0=86547 $ 2h 1=3113 × 103 $ 3h
(4.13)
The power performance curves for the Porsche 911 and Corvette ]06 are
plotted in Figure 4.2 for comparison.
Although there is almost no limit for developing a powerful engine, any
engine with power around 100 hp would have enough for street cars with
usual urban applications. It seems that engines with 1000 hp pass the limits
of application for street cars. However, race cars may have higher power
depending on the race regulations. As an example, formula-1 regulations
dictate the type of engine permitted. It must be a four-stroke engine, less
than 3000 cm3 swept volume, no more than ten cylinders, and no more than
ve valves per cylinder, but there is no limit on power.
182
4. Driveline Dynamics
Example 132 Below the curves Sh = Sh ($ h ) and Wh = Wh ($ h ).
An engine can theoretically work at any point under the performance
curve Sh = Sh ($ h ). Therefore, the power performance curve Sh = Sh ($ h )
indicates the limit of power capacity of an engine. Assume the angular
velocity of an engine is kept constant by applying a braking force. Then,
by opening the throttle, we produce more power until the throttle is widely
open, and the maximum power at that angular velocity is gained.
Power performance increases by increasing $ h and continues to climb
up to the maximum power SP at $ P . The power performance then starts
decreasing. The torque Wh = Sh @$ h also increases with $ h but reaches a
maximum at a lower angular velocity before the maximum power. Hence,
the torque starts decreasing sooner than the power when $ h is increasing.
When the power starts decreasing, the torque is very far from its peak value.
Drivers usually do not feel the engine power, however they may feel the
engine torque when accelerating.
Example 133 Engine e!ciency curves.
Engines are supposed to convert the chemical energy, embedded in the
fuel, into mechanical energy at the engine output shaft. Depending on the
working conditions, this conversion occurs at a specic e!ciency. The constant e!ciency contours can be added to the performance map of the engine
to show the e!ciency at an operating condition. Hence, every point under
the curve Sh = Sh ($ h ) can be an operating condition at a specic e!ciency.
The maximum e!ciency usually happens around the angular velocity corresponding to the maximum torque when the throttle is almost wide open.
A sample of power performance of a spark ignition engine with constant
e!ciency contours is shown in Figure 4.3.
Example 134 Power units.
There are some dierent units for expressing power. The metric unit for
power is Watt [ W].
1Nm
1J
=
(4.14)
1W =
1s
1s
Horsepower [ hp] is also commonly used in vehicle dynamics and vehicle
industries,
1 W = 0=001341 hp
1 hp = 745=699872 W
(4.15)
However, there are four denitions for horsepower: international, metric,
water, and electric. They slightly dier.
1 hp(lqwhuqdwlrqdo)
1 hp(hohfwulfdo)
1 hp(zdwhu)
1 hp(phwulf)
=
=
=
=
745=699872 W
746 W
746=043 W
735=4988 W
(4.16)
(4.17)
(4.18)
(4.19)
4. Driveline Dynamics
183
60
50
P [kW]
40
Pe
30
0.27
0.3
0.26
0.25
0.24
20
0.22
10
0.20
0
-10
0
1000
2000
3000
0
100
200
300
4000
400
0.15
K 0 .1
5000
500
6000
7000
Z [rpm]
600
700
Z [rad/s]
FIGURE 4.3. An example of power performance in a spark ignition engine with
constant e!ciency contours.
Depending on the application, other units may also be used.
1W
1W
1W
= 0=239006 cal@ s
= 0=000948 Btu@ s
= 0=737561 ft lb@ s
(4.20)
(4.21)
(4.22)
James Watt (1736 1819) experimented and concluded that a horse can
lift a weight of 550 lb for one foot in one second. It means the horse performs work at the rate of 550 ft lb@ s 745=701 W, or 33000 ft lb@ min. Watt
then stated that 33000 ft lb@ min of work was equivalent to the power of one
horse, or, one horsepower. The following formulas apply for calculating
horsepower from a torque measurement in the English unit system:
S [ hp] =
W [ ft lb] $[rpm]
5252
S [ hp] =
I [ lb] y{ [ mi@ h]
374
(4.23)
Example 135 Fuel consumption at constant speed.
Consider a vehicle moving straight at a constant speed y{ . The energy
required to travel can be calculated by multiplying the power at the drive
wheels by time
g
H = Sw = S
(4.24)
y{
where g is the distance traveled and H is the energy needed to turn the
wheels. To nd the actual energy needed to run the whole vehicle, we should
include the coe!cients of e!ciencies. We use h to indicate engine e!ciency, K to indicate thermal value of fuel, and i to indicate density of
184
4. Driveline Dynamics
the fuel. When the vehicle moves at constant speed, the traction force I{
is equal to the resistance forces. Therefore, the fuel consumption per unit
distance, t, is
I{
(4.25)
t=
h w i K
¤
£
The dimension of t in SI is m3 @ m , however, liter per 100 km is more
common. In the United States, the fuel consumption of vehicles is called
[ mi@ gal].
Example 136 F Changing the curve Sh = Sh ($ h ).
The whole power performance curve moves up when the engine’s compression ratio increases. The angular velocity associated to the engine’s
peak torque can be moved by changing the cam, header lengths, and intake
manifold runner lengths.
The wheel power curve, or the power delivered to the ground, may have
a dierent shape and a dierent peak $ h , because of transmission losses.
The best result is obtained from a power curve measured by a chassis dynamometer.
Example 137 F Power peak versus torque peak.
When the engine is operating at its torque peak (say Sh = 173=4 kW 232=5 hp at $ h = 3600 rpm) in a gear, it is generating some level of torque
(say WP = 460 N m 340 ft lb times the overall gearing ratio) at the drive
wheels. This is the best performance in that gear. By changing the gear
and making the engine to operate at the power peak (say Sh = 209 kW 280 hp at $h = 5000 rpm), it delivers less torque Wh = 400 N m 295 ft lbf.
However, it will deliver more torque to the drive wheels, at the same car
speed. This is because we gear it up by nearly 39%( [5000 3600] @3600),
while the engine torque is dropped by 13%( [460 400] @460). Hence, we
gain 26% in drive wheel torque at the power peak versus the torque peak, at
a given car speed.
As long as the performance curves of engines are similar to those in
Figure 4.1, any engine speed, other than the power peak speed $ P , at a
given car speed will provide a lower torque value at the drive wheels. Therefore, theoretically the best top speed will always occur when the vehicle is
operating at its power peak.
A car running at its power peak can accelerate no faster at the same
vehicle speed. There is no better gear to choose, even if another gear would
place the engine closer to its torque peak. A car running at peak power at a
given vehicle speed is delivering the maximum possible torque to the tires,
although the engine will not be running at its torque peak. The transmission
amplies the torque coming from the engine by a factor equal to the gear
ratio.
4. Driveline Dynamics
480
60
P
50
400
30
240
T
20
160
10
80
0
0
1000
2000
3000
0
100
200
300
4000
400
5000
500
6000
T [Nm]
320
40
P [kW]
185
0
7000
Z [rpm]
600
700
Z [rad/s]
FIGURE 4.4. Power and torque performance curves for an ideal engine.
Example 138 F Ideal engine performance.
It is said that an ideal engine is one that produces a constant power
regardless of speed. For such an ideal engine we have
Sh = S0
Wh =
S0
$h
(4.26)
Figure 4.4 depicts a sample of the power and torque performance curves
for an ideal engine having S0 = 50 kW.
In vehicle dynamics, we introduce a gearbox to keep the engine running
at the maximum power or in a working range around the maximum power.
So, practically we keep the power of the engine, and therefore, the power
at wheels constant at the maximum value. Hence, the torque at the wheels
should be similar to the torque of an ideal engine.
A constant power performance is an applied approximation for electrical
motors.
Another idea of an ideal engine is the one that provides a linear torquespeed relationship. For such an ideal engine we have
Wh = Fh $ h
Sh = Fh $ 2h
(4.27)
However, internal combustion engines do not work like this engine. Figure
4.5 illustrates such an ideal performance for Fh = 0=14539.
Example 139 F Maximum power and torque at the same $ P .
Ideal performance for an engine would be having maximum power and
maximum torque at the same angular velocity $ P . However, it is impossible
to have such an engine because the maximum torque WP of a spark ignition
4. Driveline Dynamics
60
120
50
100
40
80
60
30
T
20
40
P
20
10
0
0
1000
2000
3000
0
100
200
300
T [Nm]
P [kW]
186
4000
400
5000
500
6000
0
7000
Z [rpm]
600
700
Z [rad/s]
FIGURE 4.5. Performance curves of an ideal engine having a linear torque-speed
relationship Wh = 0=14539 $h .
engine occurs at
gWh
g$ h
= S2 + 2S3 $ h = 0
$h
=
S2
SP @$ 2P
1
=
= $P
2S3
2SP @$ 3P
2
(4.28)
(4.29)
that is half of the speed at which the power is maximum.
When the torque is maximum, the power is at
Sh = S1
³ $ ´2
³ $ ´3 5
$P
P
P
+ S3
= SP
+ S2
2
2
2
8
(4.30)
1
SP
$P
(4.31)
However, when the power is maximum at $ h = $ P , the torque is
Wh =
4.2 Driveline and E!ciency
We use the word driveline, equivalent to transmission, to call the systems
and devices that transfer torque and power from the engine to the drive
wheels of a vehicle. Most vehicles use one of two common transmission
types: manual gear transmission, and automatic transmission with torque
convertor. A driveline includes the engine, clutch, gearbox, propeller shaft,
4. Driveline Dynamics
Wheel
187
Engine
Clutch
Propeller
Shaft
Drive
shaft
Differential
Gearbox
FIGURE 4.6. Driveline components of a rear wheel drive vehicle.
Ze
Fuel
Engine
Ze
Zd
Td
Zd
Zw
Tw
Tw
Te
Propeller
Td
Zw
Differential
Ze
Clutch
Te
Zd
Zd
Td
Te
Gearbox
Te
Ze
Td
v
Wheel
Fx
FIGURE 4.7. The input and output torque and angular velocity of each driveline
component.
188
4. Driveline Dynamics
dierential, drive shafts, and drive wheels. Figure 4.6 illustrates how the
driveline for a rear-wheel-drive vehicle is assembled.
The engine is the power source in the driveline. The output from the
engine is an engine torque, Wh , at an associated engine speed $ h .
The clutch connects and disconnects the engine to the rest of the driveline
when the vehicle is equipped with a manual gearbox.
The gearbox is used to change the transmission ratio between the engine
and the drive wheels.
The propeller shaft connects the gearbox to the dierential. The propeller
shaft does not exist in front-engined front-wheel-drive and rear-engined
rear-wheel-drive vehicles. In those vehicles, the dierential is integrated
with the gearbox in a unit that is called the transaxle.
The dierential is a constant transmission ratio gearbox that allows the
drive wheels to have dierent speeds. So, they can handle the car in a curve.
The drive shafts connect the dierential to the drive wheels.
The drive wheels transform the engine torque to a traction force on the
road.
The input and output torque and angular velocity for each device in a
driveline are indicated in Figure 4.7.
The available power Sz at the drive wheels is
Sz = Sh
(4.32)
where ? 1 indicates the overall e!ciency between the engine and the
drive wheels
(4.33)
= f w
f ? 1 is the convertor e!ciency and w ? 1 is the transmission e!ciency.
The relationship between the angular velocity of the engine and the velocity of the vehicle is
Uz $ h
(4.34)
y{ =
qj qg
where qj is the transmission ratio of the gearbox, qg is the transmission
ratio of the dierential, $ h is the engine angular velocity, and Uz is the
eective tire radius.
Transmission ratio or gear reduction ratio of a gearing device, q, is the
ratio of the input velocity to the output velocity
q=
$ lq
$ rxw
(4.35)
while the speed ratio $u is the ratio of the output velocity to the input
velocity.
$ rxw
(4.36)
$u =
$ lq
4. Driveline Dynamics
189
Proof. The engine is connected to the drive wheels through a driveline.
Because of friction in the driveline, especially in the gearbox and torque
convertor, the power at the drive wheels is always less than the power at
the engine output shaft. The ratio of output power to input power is a
number called e!ciency
Srxw
(4.37)
=
Slq
If we show the e!ciency of transmission by w and the e!ciency of torque
convertor by f , then the overall e!ciency of the driveline is = f w . The
power at the wheel is the output power of driveline Srxw = Sz and the
engine power is the input power to the driveline Slq = Sh . Therefore,
Sz = Sh
(4.38)
Figure 4.8 illustrates a driving wheel with radius Uz that is rolling with
angular velocity $ z on the ground and moving with velocity y{ .
y{ = Uz $ z
(4.39)
There are two gearing devices between the engine and the drive wheel:
gearbox and dierential. Assigning qj for the transmission ratio of the
gearbox and qg for the transmission ratio of the dierential, the overall
transmission ratio of the driveline is
q = qj qg
(4.40)
Therefore, the angular velocity of the engine $ h is q times of the angular
velocity of the drive wheel $ z .
$ h = q $ z = qj qg $ z
(4.41)
These yields
y{ =
Uz $ h
qj qg
(4.42)
Example 140 Front and rear-engined, front and rear drive.
The engine may be installed in the front or back of a car. They are called
front-engined and rear-engined vehicle respectively. The driving wheels may
also be the front, the rear, or all wheels. Therefore, there are six possible
combinations. Out of those six combinations, the front-engined front-wheeldrive I Z G, front-engined rear-wheel-drive UZ G, and front-engined allwheel-drive DZ G vehicles are the most common. There are only a few
manufacturers that make cars with rear-engined rear-wheel-drive. However,
there is no rear-engined front-wheel-drive vehicle.
190
4. Driveline Dynamics
Zw
Tw
Rw
v
FIGURE 4.8. A tire with radius Uz rolling on the ground and moving with
velocity y and angular velocity $z .
Example 141 Torque at the wheel.
The power at the wheel is Sz = Sh , and the angular velocity at the
wheel is $ z = $ h @ (qj qg ). Knowing S = W $, we nd out that the available
torque at the wheel, Wz , is
Wz =
Sz
Sh
= qj qg
= qj qg Wh
$z
$h
(4.43)
Example 142 Power law.
For any mechanical device in the driveline of a car, there is a simple law
to remember.
S rzhu rxw = S rzhu lq plqxv orvvhv
Srxw = Slq Sorvv
(4.44)
S rzhu
S
(4.45)
Also, because of
= W rutxh × dqjxodu yhorflw|
= W$
any gearing device in the driveline of a car can reduce or increase the input
torque by increasing or decreasing the angular velocity.
Example 143 F Volumetric, thermal, and mechanical e!ciencies.
There is an e!ciency between the attainable power in fuel and the power
available at the engine’s output shaft.
0 = Y W P
(4.46)
Y is the engine volumetric e!ciency, W is the thermal e!ciency,
and P is the mechanical e!ciency.
Volumetric e!ciency Y identies how much fueled air gets into the
cylinder.
4. Driveline Dynamics
191
The fueled air mixture that lls the cylinder volume in the intake stroke
is what will be used to create the power. Volumetric e!ciency Y indicates
the amount of fueled air in the cylinder relative to atmospheric air. If the
cylinder is lled with fueled air at atmospheric pressure, then the engine
has 100% volumetric e!ciency. Super and turbo chargers increase the pressure entering the cylinder, giving the engine a volumetric e!ciency greater
than 100%. However, if the cylinder is lled with less than the atmospheric
pressure, then the engine has less than 100% volumetric e!ciency. Engines
typically run between 80% and 100% of Y .
Volumetric e!ciency Y can be changed by any occurrence that aects
the fueled air ow into the cylinder. The power of an engine is proportionally dependent on the mass ratio of fuel/air that gets into the cylinders of
the engine.
Thermal e!ciency W identies how much of the fuel is converted to
usable power.
Although having more fueled air into the cylinder means more fuel energy is available to make power, not all of the available energy converts
to mechanical energy. The best engines can convert only about 1@3 of the
chemical energy to mechanical energy.
Thermal e!ciency is changed by the compression ratio, ignition timing,
plug location, and chamber design. Low compression engines may have W 0=26. A high compression racing engine may have W 0=34. Therefore,
racing engines may produce about 30% more power because of their higher
W .
Any improvement in the thermal e!ciency W signicantly improves the
nal power that the engine produces. Therefore, a huge investment is expended in research to improve W .
Mechanical e!ciency P identies how much power is consumed by
the engine to run itself.
Some of the produced power is consumed by the engine’s moving parts.
It takes power to overcome the friction between parts and to run engine accessories. So, depending on how much fuel goes into the cylinder and how
much converts to power, some of this power is used by the engine to run itself. The leftover power is what we can measure on an engine dynamometer.
The dierence between the engine output power and the generated power in
the cylinders provides the mechanical e!ciency P .
Mechanical e!ciency is aected by mechanical components of the engine
or the devices attached to the engine. It also depends on the engine speed.
The greater the speed, the more power it takes to turn the engine. This
means that P drops with speed. The mechanical e!ciency P is also
called friction power because it indicates how much power is needed to
overcome the engine friction.
The engine power performance curve supplied by a car manufacturer is
usually the gross engine performance and does not include the mechanical
e!ciency. Therefore, the eective engine power available at the transmis-
192
4. Driveline Dynamics
sion input shaft is reduced by the power needed for accessories such as the
fan, electric alternator, power steering pump, water pump, braking system,
and air conditioning compressor.
4.3 Gearbox and Clutch Dynamics
The internal combustion engine cannot operate below a minimum engine
speed $ plq . Consequently, the vehicle cannot move slower than a minimum
speed yplq while the engine is engaged to the drive wheels.
yplq =
Uz $ plq
qj qg
(4.47)
At starting and stopping stages of motion, the vehicle needs to have speeds
less than yplq . A clutch or a torque converter must be used for starting,
stopping, and gear shifting.
Consider a vehicle with only one drive wheel. The forward velocity y{ of
the vehicle is proportional to the angular velocity of the engine $ h , and the
tire traction force I{ is proportional to the engine torque Wh
$h
=
Wh
=
ql qg
y{
Uz
1 Uz
I{
ql qg
(4.48)
(4.49)
where Uz is the eective tire radius, qg is the dierential transmission
ratio, ql is the gearbox transmission ratio in gear number l, and is the
overall driveline e!ciency. Equation (4.48) is called the speed equation, and
Equation (4.49) is called the traction equation. The speed equation will be
used to design the gear ratios of the gearbox of the vehicle.
Proof. The froward velocity y{ of a driving wheel with radius Uz is
y{ = Uz $ z
(4.50)
and the traction force I{ on the driving wheel is
I{ =
Wz
Uz
(4.51)
Wz is the applied spin torque on the wheel, and $ z is the wheel angular
velocity.
The wheel inputs Wz and $ z are the output torque and angular velocity
of dierential. The dierential input torque Wg and angular velocity $ g are
Wg
$g
1
Wz
g qg
= qg $ z
=
(4.52)
(4.53)
4. Driveline Dynamics
193
where qg is the dierential transmission ratio and g is the dierential
e!ciency.
The dierential inputs Wg and $ g are the output torque and angular
velocity to the vehicle’s gearbox. The engine’s torque Wh and angular velocity $ h are the inputs of the gearbox. The input-output relationships for
a gearbox depend on the engaged gear ratio ql .
1
Wg
j ql
Wh
=
$h
= ql $ g
(4.54)
(4.55)
j is the gearbox e!ciency, and ql is the gear reduction ratio in the gear
number l. Therefore, the forward velocity of a driving wheel y{ , is proportional to the engine angular velocity $ h , and the tire traction force I{ is
proportional to the engine torque Wh , when the driveline is engaged to the
engine.
ql qg
y{
(4.56)
$h =
Uz
1
1 Uz
1 Uz
1
Wh =
Wz =
I{ =
I{
(4.57)
j g ql qg
j g ql qg
ql qg
Having the torque performance function Wh = Wh ($ h ) enables us to determine the wheel torque Wz as a function of vehicle speed y{ at each gear
ratio ql .
Wz = ql qg Wh ($ h )
(4.58)
Using the approximate equation (4.5) for Wh provides
Ã
μ
¶
μ
¶2 !
ql qg
ql qg
y{ + S3
y{
Wz = ql qg S1 + S2
Uz
Uz
= S1 qg ql + S2 2 2
S3
q q y{ + 2 q3g q3l y{2
Uz g l
Uz
(4.59)
Example 144 A six-gear gearbox.
Consider an ine!cient passenger car with the following specications:
= 0=24
p = 1550 kg
Uz = 0=326 m
torque = 392 N m at 4400 rpm 460=7 rad@ s
power = 206000 W at 6800 rpm 712=1 rad@ s
1st gear ratio = q1 = 3=827
3rd gear ratio = q3 = 1=685
5th gear ratio = q5 = 1
reverse gear ratio = qu = 3=28
(4.60)
2nd gear ratio = q2 = 2=36
4th gear ratio = q4 = 1=312
6th gear ratio = q6 = 0=793
nal drive ratio = qg = 3=5451
(4.61)
194
4. Driveline Dynamics
800
PM
n1
n2
600
0
10
20
0
30
500
n5
n4
n3
700
n6
TM
400
300
200
100
0
60
30
90
120
40
50
60
150
180
210
70
v [m/s]
240
v [km/h]
FIGURE 4.9. A sample of a gear-speed plot for a gearbox.
Based on the speed equation (4.48),
$h =
ql qg
3=5451ql
y{ = 10=875ql y{
y{ =
Uz
0=326
(4.62)
we nd the gear-speed relationships at dierent gears and plot them as is
shown in Figure 4.9.
The angular velocities associated to maximum power and maximum torque
are indicated by thin lines. The power and torque performance equations for
the engine can be approximated by
Sh
Wh
= 289=29 $ h + 0=40624 $ 2h 5=7049 × 104 $ 3h
= 289=29 + 0=406 24$ h 5=704 9 × 104 $ 2h
S1
=
(4.63)
(4.64)
because
S2
S3
SP
206000
= 289=29 W@ s
=
$P
712=1
SP
206000
=
=
= 0=40624 W@ s2
$ 2P
712=12
SP
206000
= 3 =
= 5=7049 × 104 W@ s3
$P
712=13
(4.65)
(4.66)
(4.67)
Using the torque equation (4.64) and the traction equation (4.58), we plot
the wheel torque as a function of vehicle speed at dierent gears.
Wz
= ql qg Wh
¡
¢
= ql qg 289=29 + 0=406 24$ h 5=704 9 × 104 $ 2h
= 5=7405 × 102 q3l y{2 + 3=7588q2l y{ + 246=13ql
(4.68)
4. Driveline Dynamics
1200
n1
1000
Envelope
800
Tw [Nm]
195
n2
600
n3
n4
n5
400
200
0
n6
0
0
20
50
40
100
150
60
200
80
250
300
100
vx [m/s]
350
vx [km/h]
FIGURE 4.10. Wheel torque-speed Equation (4.68) at each gear ql of a gearbox,
and the envelope curve simulating an ideal engine behavior.
Figure 4.10 shows the wheel torque-speed Equation (4.68) at each gear ql .
The envelope curve for the series of torque-speed equations is similar to the
torque curve of a constant power ideal engine.
Example 145 F Envelope curve for torque-speed family.
The torque-speed equation of a car is similar to Equation (4.68) that is
a second degree of speed having the gear ratio q = ql as a parameter.
W = dq3 y2 + eq2 y + fq
(4.69)
A variation of the parameter generates a series of curves called family. An
envelope is a curve tangent to all members of the family. To nd the envelope of a family, we should eliminate the parameter between the equation of
the family and its derivative with respect to the parameter. The derivative
of the family (4.69) with respect to the parameter q
CW
= 3dq2 y 2 + 2eqy + f = 0
Cq
yields
(4.70)
s
e2 3df
q=
(4.71)
3dy
Substituting the positive parameter back into the equation of the family
provides the equation of the envelop.
s
s
¡
¢¡
¢
e + e2 3df e2 + e e2 3df 6df
(4.72)
W =
27d2 y
e ±
196
4. Driveline Dynamics
Therefore, the equation of envelope for the wheel torque-speed family at
dierent gears is equivalent to
W F
y
(4.73)
where F is a constant.
¡
¢3@2
2e3 9def + 2 e2 3df
(4.74)
F=
27d2
Such a torque equation belongs to an ideal constant power device introduced
in Example 138.
Example 146 Mechanical and hydraulic clutches.
Mechanical clutches are widely used in passenger cars and are normally
in the form of a dry single-disk clutch. The adhesion between input and
output shafts is produced by circular disks that rub against each other.
Engagement begins with the engine running at $ h = $ plq . The clutch
then being released gradually from time w = 0 to w = w1 . Therefore, the
transmitted torque Wf from the engine to the gearbox increases almost linearly in time from Wf = 0 to the maximum value of Wf = Wf1 that can be
handled in slipping mode. The transmitted torque remains constant until
the input and output disks stick together and a speed equality is achieved.
At this time, the clutch is fully engaged and Wf = Wh .
The transmitted torque Wf should overcome the resistance force and the
vehicle should accelerate sometime in 0 ? w w1 . The magnitude of the
transferable torque depends on the applied force between the disks, the frictional coe!cient between clutch disks, the eective frictional area, and the
number of frictional pairs. The axial force is generally produced by a preloaded spring. The driver can control the spring force by using the clutch
pedal, and adjust the transferred torque.
The hydraulic clutch consists of a pump wheel connected to the engine
and a clutch-ended turbine that is equipped with radial vanes. A torque
is transferred between the pump wheel and the turbine over uid, which is
accelerated by the pump and decelerated in the turbine. The hydraulic clutch
is also called Foettinger clutch.
The transferred torque can be calculated according to the Foettinger’s law
Wf = Ff $ 2s G2
(4.75)
where Ff is the slip factor, is the oil density, $ s is the pump angular
velocity, and G is the clutch diameter.
Example 147 Acceleration capacity at dierent speed.
Assume an engine is working at speed $ P associated to the maximum
power SP .
1
(4.76)
SP = Wh $ P = I{ y{
4. Driveline Dynamics
197
ax [m/s2]
m 860 kg
K 0.25
PM=200 kW
150
100
50
vx [m/s]
0
30
60
90
120
150
180
210
vx [km/h]
FIGURE 4.11. An example for the acceleration capacity d{ as a fucntion of
forward speed y{ .
Substituting
I{ = pd{
indicates that
SP =
(4.77)
p
d{ y{
(4.78)
1
p y{
(4.79)
and therefore,
d{ = SP
Equation (4.79) is called acceleration capacity and expresses the achievable acceleration of a vehicle at speed y{ . The acceleration capacity decreases
by increasing velocity and increases by the maximum power. As an example,
Figure 4.11 depicts the acceleration capacity d{ as a function of the forward
speed y{ for a vehicle with mass p = 860 kg, and a low e!ciency = 0=25
for the maximum powers of SP = 50 kW 67 hp, SP = 100 kW 134 hp,
SP = 150 kW 201 hp, and SP = 200 kW 268 hp.
Example 148 Power-limited and traction-limited accelerations.
Acceleration capacity is power-limited and is based on the assumption
that the driving force does not reach the tire traction limit. Therefore, the
vehicle reaches its peak acceleration because the engine cannot deliver any
more power.
The traction-limited acceleration happens when the engine delivers more
power, but vehicle acceleration is limited because the tires cannot transmit
any more driving force to the ground. Equation I{ = { I} gives the maximum transmittable force. If more driving torque is applied to the wheel,
198
4. Driveline Dynamics
the tire slips and enters the dynamic friction regime where the coe!cient
of friction, and hence the traction force, are less.
Example 149 F Gearbox stability condition.
Consider a vehicle moving at speed y{ when the gearbox is engaged in
gear number l with transmission ratio ql . To be safe, we have to select the
transmission ratios such that when the engine reaches the maximum torque
it can shift to a lower gear ql1 without reaching the maximum permissible
engine speed. The maximum permissible engine speed is usually indicated
by a red line or red region on engine rpm indicator.
Let us show the engine speed for the maximum torque WP by $ h = $ W .
The speed of the vehicle at $ h = $ W is
y{ =
Uz
$W
ql qg
(4.80)
When we shift the gear to ql1 the engine speed $ h jumps to a higher speed
$ h = $ l1 A $ W at the same vehicle speed
$ l1 =
ql1 qg
y{
Uz
(4.81)
The stability condition requires that $ l1 be less than the maximum permissible engine speed $ Pd{
$ l1 $ Pd{
(4.82)
Using Equations (4.80) and (4.81), we may dene the following condition
between transmission ratios at two successive gears and the engine speed:
$ Pd{
ql1
$ l1
=
=
$l
$W
ql
(4.83)
A constant relative gear ratio, at a constant vehicle speed, is a simple
rule for a stable gearbox design
ql1
= fj
ql
(4.84)
Example 150 Transmission ratios and stability condition.
Consider a passenger car with the following gearbox transmission ratios:
1st gear ratio = q1 = 3=827
2nd gear ratio = q2 = 2=36
3rd gear ratio = q3 = 1=685
4th gear ratio = q4 = 1=312
5th gear ratio = q5 = 1
6th gear ratio = q6 = 0=793
nal drive ratio = qg = 3=5451
(4.85)
4. Driveline Dynamics
199
The stability condition requires that ql1 @ql = fwh. Examination of the gear
ratios indicates that the relative gear ratios are not constant.
q5
q6
q3
q4
q1
q2
1
= 1=261
0=793
1=685
= 1=284 3
1=312
3=827
= 1=621 6
2=36
=
=
=
1=312
q4
=
= 1=312
q5
1
2=36
q2
=
= 1=4
q3
1=685
(4.86)
We may change the gear ratios to have ql1 @ql = fwh. Let us start from
the highest gears and nd the lower gears using fj = q6 @q5 = 1=261.
q5
q6
q4
q3
q2
q1
=
=
=
=
=
=
1
0=793
fj q5 = 1=261
fj q4 = 1=261 × 1=261 = 1=59
fj q3 = 1=261 × 1=59 = 2
fj q2 = 1=261 × 2 = 2=522
(4.87)
We may also start from the rst two gears and nd the higher gears using
fj = q1 @q2 = 3=827@2=36 = 1=6216.
q1
q2
q3
q4
q5
q6
= 3=827
= 2=36
q2
2=36
=
=
= 1=455
fj
1=6216
q3
1=455
=
=
= 0=897
fj
1=6216
q4
0=897
=
=
= 0=553
fj
1=6216
q5
0=553
=
=
= 0=341
fj
1=6216
(4.88)
None of these two sets shows a practical design. The best way to apply
a constant relative ratio is to use the rst and nal gears and t four
intermittent gears such that ql1 @ql = fwh. Using q1 and q6 we have,
3=827
q1 q2 q3 q4 q5
q1
=
= f5j
=
q6
0=793
q2 q3 q4 q5 q6
(4.89)
fj = 1=37
(4.90)
and therefore,
4. Driveline Dynamics
180
60
50
P [kW]
40
150
P
120
T
Working
Range
30
20
60
30
10
0
0
90
T [Nm]
200
0
1000
2000
3000
4000
Z [rpm]
5000
Z1
Z6000
Z2 7000
M
FIGURE 4.12. The angular velocity range ($ 1 > $ 2 ) around $P , and engine’s
working range.
Now we are able to nd the gear ratios to meet the rst and sixth gear ratios
requirements.
q1
q2
q3
q4
q5
q6
= 3=827
q1
3=827
= 2=793
=
=
fj
1=37
q2
2=793
= 2=039
=
=
fj
1=37
q3
2=039
= 1=488
=
=
fj
1=37
q4
1=488
= 1=086
=
=
fj
1=37
= 0=793
(4.91)
4.4 Gearbox Design
The speed and traction equations (4.48) and (4.49) can be used to calculate
the gear ratios of a gearbox and the vehicle performance. Theoretically the
engine should work at its maximum power to have the best performance.
However, to control the speed of the vehicle, we need to vary the engine’s
angular velocity. Hence, we pick an angular velocity range ($ 1 > $ 2 ) around
$ P , which is associated to the maximum power SP , and sweep the range
($ 1 > $ 2 ) repeatedly at dierent gears. The range ($1 > $ 2 ) is called the engine’s working range as is shown in Figure 4.12.
As a general guideline, we may use the following recommendations to
4. Driveline Dynamics
201
design the transmission ratios of a vehicle gearbox:
1. We may design the dierential transmission ratio qg and the nal gear
qq such that the nal gear qq is a direct gear, qq = 1, when the vehicle is moving at the moderate highway speed. Using qq = 1 implies
that the input and output of the gearbox are directly connected with
each other. Direct engagement maximizes the mechanical e!ciency
of the gearbox.
2. We may design the dierential transmission ratio qg and the nal
gear qq such that the nal gear qq is a direct gear, qq = 1, when the
vehicle is moving at the maximum attainable speed.
3. The rst gear q1 may be designed by the maximum desired torque
at driving wheels. Maximum torque is determined by the slope of a
desired climbing road.
4. We can nd the intermediate gears using the gear stability condition.
Stability condition provides that the engine speed must not exceed
the maximum permissible speed if we gear down from ql to ql1 ,
when the engine is working at the maximum torque in ql .
5. The value of fj for relative gear ratios
ql1
= fj
ql
(4.92)
1 fj 2
(4.93)
can be chosen in the range.
To determine the middle gear ratios, there are two recommended methods:
1= Geometric ratios
2= Progressive ratios
4.4.1 Geometric Ratio Gearbox Design
When the jump of engine speed in any two successive gears is constant at
a vehicle speed, we call the gearbox geometric. The design condition for a
geometric gearbox is
ql1
(4.94)
ql =
fj
where fj is the constant relative gear ratio and is called step jump.
Proof. A geometric gearbox has constant engine speed jump in any gear
shift. So, a geometric gearbox must have a gear-speed plot such as that
shown in Figure 4.13.
202
4. Driveline Dynamics
Ze
'v2
'v3
'v4
n4
n3
n2
n1
Z2
Z1
v1
v2
v3
v4
v
FIGURE 4.13. A gear-speed plot for a geometric gearbox design.
The engine working range is dened by two speeds ($ 1 > $ 2 )
{($ 1 > $ 2 ) > $ 1 ? $ P ? $ 2 }
(4.95)
When the engine reaches the maximum speed $ 2 in the gear number l with
ratio ql , we gear up to ql+1 to jump the engine speed down to $ 1 . The
engine’s speed jump is kept constant for any gear change from ql to ql+1 .
Employing the speed equation (4.48), we have
ql1 qg
ql qg
4$ = $ 2 $ 1 =
y{ y{
Uz
Uz
qg
= (ql1 ql )
y{
(4.96)
Uz
and therefore,
$2 $1
ql1 ql
=
$1
ql
$2
ql1
1 =
1
$1
ql
$2
ql1
=
= fj
(4.97)
$1
ql
Let us indicate the maximum vehicle speed in gear ql , when $ h = $ 2 ,
by yl and in gear ql1 by yl1 , then,
ql qg
ql1 qg
yl =
yl1
(4.98)
$2 =
Uz
Uz
and therefore, the maximum speed in gear l to the maximum speed in gear
l 1 is equal to the inverse of the gear ratios
ql1
yl
fj =
=
(4.99)
ql
yl1
4. Driveline Dynamics
203
The change in vehicle speed between gear ql1 and ql is indicated by
4yl = yl yl1
(4.100)
and is called speed span.
Having the step jump fj and knowing the maximum speed yl of the
vehicle in gear ql , are enough to nd the maximum velocity of the car in
the other gears
yl
yl1
yl+1
= fj yl1
1
=
yl
fj
= fj yl
(4.101)
(4.102)
(4.103)
The speed span of vehicle with a geometric gearbox increases by gearing
up.
Example 151 A gearbox with three gears.
Consider an p = 860 kg car having an engine with = g j = 0=84 and
the power-speed relation of
Sh = 100 100
2
($ h 398) kW
3982
(4.104)
where $ h is in [ rad@ s]. We dene the working range for the engine as
272 rad@ s ( 2600 rpm) $ h 524 rad@ s ( 5000 rpm)
(4.105)
when the power is 100 kW Sh 90 kW. The power performance curve
(4.104) is illustrated in Figure 4.14 and the working range is shaded.
The dierential of the vehicle uses qg = 4, and the eective tire radius
is Uz = 0=326 m. We would like to design a three-gear geometric gearbox
to have the minimum time required to reach the speed y{ = 100 km@ h 27=78 m@ s 62 mi@ h. We assume that the total resistance force is constant,
and the engine cannot accelerate the car at y{ = 180 km@ h = 50 m@ s 112 mi@ h anymore. Assume that every gear change takes 0=47 s and we need
w0 = 2=58 s to adjust the engine speed to the car speed in the rst gear.
Using the speed equation (4.48), the relationship between vehicle and engine speeds is
Uz
0=326
$h =
$h
(4.106)
y{ =
qg ql
4 ql
At the maximum speed y{ = 50 m@ s, the engine is rotating at the upper
limit of the working range $ h = 524 rad@ s and the gearbox is operating in
the third gear. Therefore, Equation (4.106) provides us with
q3 =
0=326 $ h
0=326 524
=
= 0=85412
4 y{
4 50
(4.107)
4. Driveline Dynamics
Pe [kW]
204
Working
Range
Z [rad/s]
0
1000
2000
3000
4000
5000
6000
Z [rpm]
FIGURE 4.14. The power performance curve (4.104) and its working range.
The speed equation
0=326
(4.108)
$h
4 × 0=85412
is applied as long as the gearbox is operating in the third gear ql = q3 ,
and $ h is in the working range. By decreasing $ h and sweeping down over
the working range, the speed of the car will reduce. At the lower range
$ h = 272 rad@ s, the vehicle’s speed would be
y{ =
y{
0=326
× 272 = 25=95 m@ s
4 × 0=85412
93=43 km@ h 58 mi@ h
=
(4.109)
At this speed we should gear down to q2 , so the engine jumps back to the
higher range $ h = 524 rad@ s. This provides us with
q2 =
0=326 $ h
0=326 524
=
= 1=6457
4 y{
4 25=95
(4.110)
Therefore, the engine and vehicle speed relationship in the second gear would
be
0=326
(4.111)
$h
y{ =
4 × 1=6457
that is applicable as long as ql = q2 , and $ h is in the working range.
Sweeping down the engine’s angular velocity reduces the vehicle speed to
y{
0=326
× 272 = 13=47 m@ s
4 × 1=6457
48=49 km@ h 30=1 mi@ h
=
(4.112)
205
600
500
400
n3
n2
n1
[rad/s]
4. Driveline Dynamics
Ze 300
0
0
10
0
30
20
60
180 km/h
100
100 km/h
25.2 km/h
200
30
90
120
40
50
60
150
180
210
70
v [m/s]
240
v [km/h]
FIGURE 4.15. The gear-speed plot for a three-gear gearbox.
At this speed we should gear down to q1 and jump the engine again to the
higher range $h = 524 rad@ s. This provides us
q1 =
0=326 $ h
0=326 524
= 3=1705
=
4 y{
4 13=47
(4.113)
and therefore, the speed equation for the rst gear is
y{ =
0=326
$h
4 × 3=1705
(4.114)
At the lower range of the engine’s speed in the rst gear ql = q1 , the speed
of the vehicle is
y{ =
0=326
× 272 = 7 m@ s 25=2 km@ h 15=6 mi@ h
4 × 3=1705
(4.115)
Therefore, the three-gear gearbox uses the following gear ratios:
q1 = 3=1705
q2 = 1=6457
q3 = 0=85412
(4.116)
The speed equations for the three gears are shown in the gear-speed plot
of Figure 4.15. The gure also shows the gear switching points and how the
vehicle speed is reducing from y{ = 50 m@ s to y{ = 7 m@ s.
To evaluate the required time to reach the desired speed, we need to nd
the traction force I{ from the traction equation and integrate.
μ
¶
100
ql qg Sh
ql qg
2
100 I{ = =
($ h 398)
Uz $ h
$ h Uz
3982
25 q q (796Uz qg ql y{ ) kN
(4.117)
=
2 g l
39 601 Uz
206
4. Driveline Dynamics
At the maximum speed, the gearbox is in the third gear and the traction
force I{ is equal to the total resistance force IU , which is assumed to be
constant.
Sh
0=84 × 90
=
I{ = IU =
= 1=512 kN
(4.118)
y{
50
Therefore, the traction force in the rst gear is
I{
25 q q (796Uz qg q1 y{ )
2 g 1
39601 Uz
25 0=84
=
× 4 × 3=1705 (796 × 0=326 4 × 3=1705y{ )
39601 0=3262
= 16=421 0=80252y{ kN
(4.119)
=
Based on Newton’s equation of motion
I{ IU = p
gy{
gw
(4.120)
we can evaluate the required time to sweep the velocity from zero to y{ =
13=47 m@ s
Z 13=47
1
w1 = p
gy{
I
IU
{
0
Z 13=47
103
= 860
gy{ = 1=3837 s (4.121)
16=421 0=80252y{ 1=512
0
In second gear, we have
I{
25 q q (796Uz qg q2 y{ )
2 g 2
39601 Uz
25 0=84
=
× 4 × 1=6457 (796 × 0=326 4 × 1=6457y{ )
39601 0=3262
= 8=5235 0=21622y{ kN
(4.122)
=
and therefore, the sweep time in the second gear is
Z 25=95
1
w2 = p
gy{
13=47 I{ IU
Z 25=95
103
= 860
gy{ = 4=2712 s (4.123)
13=47 8=5235 0=21622y{ 1=512
Finally, the traction equation in the third gear is
I{
25 q q (796Uz qg q3 y{ )
2 g 3
39601 Uz
25 0=84
=
× 4 × 0=85412 (796 × 0=326 4 × 0=85412y{ )
39601 0=3262
= 4=4237 5=8242 × 102 y{ kN
(4.124)
=
4. Driveline Dynamics
207
and the sweep time to y{ = 27=78 m@ s is
Z 27=78
1
gy{
w3 = p
25=95 I{ IU
Z 27=78
103
gy{ = 1=169 s(4.125)
= 860
2 y 1=512
{
25=95 4=4237 5=8242 × 10
Therefore, the total time to reach the speed y{ = 100 km@ h 27=78 m@ s
is then equal to
w = w0 + w1 + w2 + w3 + 3 × 0=47
= 2=58 + 1=3837 + 4=2712 + 1=169 + 3 × 0=47 = 10=814 s
(4.126)
Example 152 Better performance with a four-speed gearbox.
A car equipped with a small engine has
p = 860 kg
Uz = 0=326 m
= 0=84
qg = 4
(4.127)
and the engine operates based on the performance equation of
Sh = 100 100
($ h 398)2 kW
3982
(4.128)
where $ h is in [ rad@ s]. We chose the engine working range of
272 rad@ s ( 2600 rpm) $ h 524 rad@ s ( 5000 rpm)
(4.129)
when the power is 100 kW Sh 90 kW. We would like to design a gearbox
to minimize the time to reach y{ = 100 km@ h 27=78 m@ s 62 mi@ h.
The power performance equation (4.128) is illustrated in Figure 4.14 and
the working range is shaded. To make this example comparable to Example
151 we assume that the total resistance force is constant and the engine
cannot accelerate the car at y{ = 180 km@ h. Furthermore, we assume that
every gear change takes 0=47 s and a time w0 = 2=58 s is needed to completely
engage the engine in the rst gear.
Let us design a four-gear gearbox and set the third gear such that we
reach the desired speed y{ = 27=78 m@ s at the higher limit of working range
$ h = 524 rad@ s. The gear-speed plot for such a design is plotted in Figure
4.16.
Using the speed equation (4.48), the relationship between vehicle and engine speeds is
Uz
0=326
$h =
$h
(4.130)
y{ =
qg ql
4 ql
At the speed y{ = 100 km@ h 27=78 m@ s, the engine’s speed is at the upper
limit of the working range $ h = 524 rad@ s and the gearbox is operating in
third gear ql = q3 . Therefore,
q3 =
0=326 $ h
0=326 524
=
= 1=5373
4 y{
4 27=78
(4.131)
4. Driveline Dynamics
600
500
n2
n3
n1
[rad/s]
208
0
10
20
0
30
n4
400
300
25.2 km/h
100
0
60
180 km/h
200
100 km/h
Ze
30
90
120
40
50
60
150
180
210
70
v [m/s]
240
v [km/h]
FIGURE 4.16. The gear-speed plot for Example 152.
and the speed equation in the third gear ql = q3 is
y{ =
0=326
$h
4 × 1=5373
(4.132)
while $ h is in the working range. By sweeping down to the lower limit of
the working range, $ h = 272 rad@ s, the speed of the car will reduce to
y{
0=326
× 272 = 14=42 m@ s
4 × 1=5373
51=91 km@ h 32=25 mi@ h
=
(4.133)
At this speed we should gear down to q2 and jump to the higher angular
speed $ h = 524 rad@ s. This provides us with
q2 =
0=326 $ h
0=326 524
=
= 2=9616
4 y{
4 14=42
(4.134)
Therefore, the gear-speed relationship in second gear ql = q2 is
y{ =
0=326
$h
4 × 2=9616
(4.135)
Sweeping down the engine’s angular velocity to $ h = 272 rad@ s, reduces the
vehicle speed to
y{
0=326
× 272 = 7=48 m@ s
4 × 2=9616
26=9 km@ h 16=7 mi@ h
=
(4.136)
4. Driveline Dynamics
209
At this speed, we gear down to q1 and jump again to the higher angular
speed $ h = 524 rad@ s. This provides us with
q1 =
0=326 $ h
0=326 524
=
= 5=7055
4 y{
4 7=48
(4.137)
and therefore, the speed equation for the rst gear would be
y{ =
0=326
$h
4 × 5=7055
(4.138)
In the rst gear, ql = q1 , and the vehicle’s speed at the lower range of the
engine’s speed is
y{
0=326
× 272 = 3=88 m@ s
4 × 5=7055
14 km@ h 8=7 mi@ h
=
(4.139)
To calculate the fourth gear ql = q4 we may use the gear-speed equation
and set the engine speed to the lower limit $ h = 272 rad@ s while the car is
moving at the maximum speed in the third gear.
q4 =
0=326 $ h
0=326 272
=
= 0=79798
4 y{
4 27=78
(4.140)
Therefore, the four-gear gearbox uses the following ratios:
q1
q3
= 5=7055
= 1=5373
q2 = 2=9616
q4 = 0=79798
(4.141)
To calculate the required time to reach the desired speed y{ = 100 km@ h 27=78 m@ s, we need to use the traction equations and nd the traction force
I{
μ
¶
ql qg Sh
ql qg
100
2
I{ = =
($
398)
100 h
Uz $ h
$ h Uz
3982
25 q q (796Uz qg ql y{ ) kN
(4.142)
=
2 g l
39 601 Uz
At the maximum speed, the gearbox is in the fourth gear and the traction
force I{ is equal to the total resistance force IU , which is assumed to be
constant.
Sh
0=84 × 90
I{ = IU =
=
= 1=512 kN
(4.143)
y{
50
Therefore, the traction force in the rst gear is
I{
25 q q (796Uz qg q1 y{ )
2 g 1
39601 Uz
25 0=84
=
× 4 × 5=7055 (796 × 0=326 4 × 5=7055y{ )
39601 0=3262
(4.144)
= 29=55 2=5989y{ kN
=
210
4. Driveline Dynamics
Using Newton’s equation of motion
I{ IU = p
gy{
gw
(4.145)
we can evaluate the required time to reach the velocity y{ = 7=48 m@ s
Z 7=48
1
w1 = p
gy{
I
IU
{
0
Z 7=48
103
= 860
gy{ = 0=39114 s (4.146)
29=55 2=5989y{ 1=512
0
In second gear we have
I{
25 q q (796Uz qg q2 y{ )
2 g 2
39601 Uz
25 0=84
=
× 4 × 2=9616 (796 × 0=326 4 × 2=9616y{ )
39601 0=3262
(4.147)
= 15=339 0=70025y{ kN
=
and therefore, the sweep time in second gear is
Z 14=42
1
gy{
w2 = p
I
IU
{
7=48
Z 14=42
103
= 860
gy{ = 1=0246 s (4.148)
15=339 0=70025y{ 1=512
7=48
The traction equation in the third gear is
I{
25 q q (796Uz qg q3 y{ )
2 g 3
39601 Uz
25 0=84
=
× 4 × 1=5373 (796 × 0=326 4 × 1=5373y{ )
39601 0=3262
= 7=9621 0=18868y{ kN
(4.149)
=
and the sweep time is
Z 27=78
w3 = p
= 860
1
gy{
I{ IU
14=42
Z 27=78
14=42
103
gy{ = 5=1359 s (4.150)
7=9621 0=18868y{ 1=512
The total time to reach the speed y{ = 100 km@ h 27=78 m@ s is then
equal to
w = w0 + w1 + w2 + w3 + 3 × 0=07
= 2=58 + 0=39114 + 1=0246 + 5=1359 + 3 × 0=47 = 10=542 s (4.151)
4. Driveline Dynamics
211
Example 153 Changing the working range.
Consider a car that is equipped with a small engine having the power
performance equation of
Sh = 100 100
($ h 398)2 kW
3982
(4.152)
where $ h is in [ rad@ s]. The characteristics of the car are
p = 860 kg
Uz = 0=326 m
= 0=84
qg = 4
(4.153)
The engine provides a maximum power SP = 100 kW at $ P = 400 rad@ s.
The total resistance force is assumed to be constant, and the maximum
attainable speed is assumed to be y{ = 180 km@ h. Furthermore, we assume
that every gear change takes 0=07 s and a minimum time w0 = 0=18 s is
needed to adjust the engine speed to the car speed in the rst gear. We
would like to design a four-gear gearbox to minimize the time to reach
y{ = 100 km@ h 27=78 m@ s.
To nd the best working range for the engine, we set the third gear to
reach the desired speed y{ = 100 km@ h at the upper limit of the working
range. Therefore, when we gear up, the fourth gear starts with the lower
limit of the working range. If the fourth gear is set such that the car reaches
the maximum speed y{ = 180 km@ h 50 m@ s at the upper limit of the
working range, then the gear-speed equation
$h =
ql qg
y{
Uz
(4.154)
provides us with
$ Pd{ =
4q4
× 50
0=326
$ plq =
4q4
× 27=78
0=326
(4.155)
By setting $ plq and $ P d{ to an equal distance from $ P = 400 rad@ s,
$ Pd{ + $ plq
= 400
2
(4.156)
we nd
q4 = 0=83826
$ plq = 285=73 rad@ s
$ Pd{ = 514=27 rad@ s (4.157)
We are designing a gearbox such that the ratio $ h @y{ is kept constant in
each gear. The engine speed jumps up from $ plq to $ Pd{ when we gear
down from q4 to q3 at $ plq , hence,
$ Pd{
q3
4q3
× 27=78 = 514=27
0=326
= 1=5087
=
(4.158)
(4.159)
212
4. Driveline Dynamics
Therefore, the speed of the car in the third gear at the lower limit of the
engine speed is
y{ = 27=78
$ plq
27=78
= 27=78 ×
= 15=435 m@ s
$ Pd{
50
(4.160)
The engine’s speed jumps again to $ Pd{ when we gear it down from q3 to
q2 , hence,
$ Pd{
q2
4q2
× 15=435 = 514=27
0=326
= 2=715 5
=
(4.161)
(4.162)
Finally the speed of the car in second gear at the lower limit of the engine
speed is
y{ = 15=435
$ plq
27=78
= 8=5757 m@ s
= 15=435 ×
$ Pd{
50
(4.163)
that provides us with the following gear ratio in the rst gear
$ Pd{
q1
4q1
× 8=5757 = 514=27
0=326
= 4=8874
=
(4.164)
(4.165)
The speed of the car in the rst gear at the lower limit of the engine speed
is then equal to
y{ = 8=5757
$ plq
27=78
= 4=7647 m@ s
= 8=5757 ×
$ Pd{
50
(4.166)
Therefore, the four gears of the gearbox have the following ratios:
q1
q3
= 4=8874
= 1=5087
q2 = 2=7155
q4 = 0=83826
(4.167)
and the working range for the engine is
285=73 rad@ s ( 2730 rpm) $ h 514=27 rad@ s ( 4911 rpm)
(4.168)
The power performance curve (4.152) is illustrated in Figure 4.17 and the
working range is shaded. The gear-speed plot of this design is also plotted
in Figure 4.18.
Balance of the traction force I{ and the total resistance force IU at the
maximum speed provides us with
I{ = IU =
Sh
0=84 × 90
=
= 1=512 kN
y{
50
(4.169)
Pe [kW]
4. Driveline Dynamics
213
Working
Range
Z [rad/s]
0
1000
2000
3000
4000
5000
6000
Z [rpm]
600
n2
n3
500
0
10
20
0
30
n1
[rad/s]
FIGURE 4.17. The power performance curve (4.152) and its working range.
n4
400
Ze 300
100
0
60
180 km/h
100 km/h
200
30
90
120
40
50
60
150
180
210
70
v [m/s]
240
v [km/h]
FIGURE 4.18. The gear-speed plot for Example 153.
214
4. Driveline Dynamics
The traction force in the rst gear is
I{
25 q q (796Uz qg q1 y{ )
2 g 1
39601 Uz
25 0=84
=
× 4 × 4=8874 (796 × 0=326 4 × 4=8874y{ )
39601 0=3262
(4.170)
= 25=313 1=907y{ kN
=
The time in the rst gear q1 can be calculated by integrating Newton’s
equation of motion
gy{
I{ IU = p
(4.171)
gw
and sweep the velocity from y{ = 0 to y{ = 8=5757 m@ s
Z 8=5757
1
gy{
w1 = p
I
IU
{
0
Z 8=5757
103
= 860
gy{ = 0=52398 s (4.172)
25=313 1=907y{ 1=512
0
In the second gear, the traction force is
I{
25 q q (796Uz qg q2 y{ )
2 g 2
39601 Uz
25 0=84
=
× 4 × 2=7155 (796 × 0=326 4 × 2=7155y{ )
39601 0=3262
(4.173)
= 14=064 0=5887y{ kN
=
and therefore, the sweep time in the second gear is
Z 15=435
1
gy{
w2 = p
8=5757 I{ IU
Z 15=435
103
= 860
gy{ = 1=1286 s (4.174)
8=5757 14=064 0=5887y{ 1=512
In the third gear, the traction force is
I{
25 q q (796Uz qg q3 y{ )
2 g 3
39601 Uz
25 0=84
=
× 4 × 1=5087 (796 × 0=326 4 × 1=5087y{ )
39601 0=3262
= 7=814 0=18172y{ kN
(4.175)
=
and the third sweep time is
Z 27=78
1
gy{
w3 = p
I
IU
15=435 {
Z 27=78
103
= 860
gy{ = 4=8544 s
15=435 7=814 0=18172y{ 1=512
(4.176)
Ze
'v2
'v3
n2
4. Driveline Dynamics
n3
215
'v4
n1
Z2
v1
v2
n4
v3
v
v4
FIGURE 4.19. A gear-speed plot for a progressive gearbox design.
The total time to reach the speed y{ = 100 km@ h 27=78 m@ s is then
equal to
w = w0 + w1 + w2 + w3 + 3 × 0=07
= 2=58 + 0=52398 + 1=1286 + 4=8544 + 3 × 0=47 = 10=497 s (4.177)
4.4.2 F Progressive Ratio Gearbox Design
When the speed span of a vehicle in any two successive gears is kept constant, we call the gearbox progressive. The design condition for a progressive
gearbox is
ql ql1
(4.178)
ql+1 =
2ql1 ql
where ql1 , ql , and ql+1 are the transmission ratios of three successive
gears.
Proof. A progressive gearbox has constant vehicle speed span in any gear.
Therefore, a progressive gearbox must have a gear-speed plot such as that
shown in Figure 4.19.
Indicating the maximum vehicle speed in gear ql by yl , in gear ql1 by
yl1 , and in gear ql+1 by yl+1 , we have
$2 =
ql qg
ql1 qg
ql+1 qg
yl =
yl1 =
yl+1
Uz
Uz
Uz
(4.179)
The dierence in vehicle speed at maximum engine speed is
4y = yl yl1 = yl+1 yl
(4.180)
216
4. Driveline Dynamics
and therefore,
yl+1 + yl1
yl1
yl+1
+
yl
yl
ql
ql
+
ql+1 ql1
ql+1
= 2yl
(4.181)
= 2
(4.182)
= 2
(4.183)
=
ql ql1
2ql1 ql
(4.184)
The step jump of a progressive gearbox decreases in higher gears. If the
step jump fjl between ql and ql+1 is
ql
= fjl
ql+1
(4.185)
then,
fjl = 2 1
fjl1
(4.186)
Example 154 A progressive gearbox with three gears.
The power performance of the engine of a car with = g j = 0=84 is
Sh = 100 100
2
($ h 398) kW
3982
(4.187)
where $ h is in [ rad@ s] and
SP = 100 kW
$ P = 400 rad@ s
(4.188)
The dierential ratio and the equivalent tire radius are
qg = 4
Uz = 0=326 m
(4.189)
The total resistance force is assumed to be constant and the engine cannot
accelerate the car at maximum speed y{ = 180 km@ h = 50 m@ s 112 mi@ h
anymore.
Using the speed equation (4.48), the relationship between vehicle and engine speed is
Uz
0=326
y{ =
$h =
$h
(4.190)
qg ql
4 ql
We would like to design a four-gear progressive gearbox with equal speed
spans, such that the car achieves the maximum speed in the fourth gear when
the engine is at the maximum permissible speed $ 2 = $ h = 524 rad@ s 5000 rpm.
4. Driveline Dynamics
217
Dividing the maximum velocity of the car into four equal segments determines the speed span at each gear
4y =
50
yP
=
= 12=5 m@ s
4
4
(4.191)
Therefore, the maximum car speed for the four gears are
y1
y2
y3
y4
=
=
=
=
12=5
y1 + 4y = 12=5 + 12=5 = 25 m@ s
y2 + 4y = 25 + 12=5 = 37=5 m@ s
y3 + 4y = 37=5 + 12=5 = 50 m@ s
(4.192)
(4.193)
(4.194)
(4.195)
Considering that the maximum car speeds are all achieved at the maximum permissible engine speed, $ h = 524 rad@ s, the speed equation (4.190)
determines the gear ratios.
q4
=
q3
=
q2
=
q1
=
0=326
0=326
524 = 0=85412
$h =
4y4
4 × 50
0=326
0=326
524 = 1=1388
$h =
4y3
4 × 37=5
0=326
0=326
524 = 1=7082
$h =
4y2
4 × 25
0=326
0=326
524 = 3=4165
$h =
4y1
4 × 12=5
(4.196)
(4.197)
(4.198)
(4.199)
The minimum engine speed, $ 1 , at which the gear should be changed is
not a constant. Assume the gearbox is in q1 and the engine speed reaches
the maximum value of $ h = 524 rad@ s. At this time we need to gear up to q2
and drop the engine speed. Substituting y1 in Equation (4.198) determines
the engine speed at the rst gear up
$h =
4 × 12=5 × 1=7082
4y1 q2
=
= 261=99 rad@ s
0=326
0=326
(4.200)
Substituting y2 in Equation (4.197) determines the engine speed at the second gear up
$h =
4 × 25 × 1=1388
4y2 q3
=
= 349=33 rad@ s
0=326
0=326
(4.201)
Substituting y3 in Equation (4.196) determines the engine speed at the third
gear up
$h =
4 × 37=5 × 0=85412
4y3 q4
=
= 393 rad@ s
0=326
0=326
(4.202)
Figure 4.20 illustrates the speed equations of the gearbox to show the relations of the engine speed and car velocity in dierent gears.
4. Driveline Dynamics
n4
n3
n2
n1
218
v [m/s]
FIGURE 4.20. The relations between the engine speed and car velocity in dierent
gears.
4.5 Summary
The maximum attainable power Sh of an internal combustion engine is
a function of the engine angular velocity $ h . This function must be determined by experiment however, the power performance function Sh =
Sh ($ h ), can be estimated by a mathematical function such as
Sh = S1 $ h + S2 $ 2h + S3 $ 3h
(4.203)
where,
S1 =
SP
$P
S2 =
SP
$ 2P
S3 = SP
$ 3P
(4.204)
$ P is the angular velocity, measured in [ rad@ s], at which the engine
power reaches the maximum value SP , measured in [ W = N m@ s].
The engine torque Wh is that provides Sh is
Wh =
Sh
= S1 + S2 $ h + S3 $ 2h
$h
(4.205)
An ideal engine is the one that produces a constant power regardless of
speed. For the ideal engine, we have
Sh = S0
Wh =
S0
$h
(4.206)
We use a gearbox to make the engine approximately work at a constant
power close to the SP . To design a gearbox we use two equations: the speed
equation
ql qg
$h =
y{
(4.207)
Uz
4. Driveline Dynamics
219
and the traction equation
Wh =
1 Uz
I{
ql qg
(4.208)
These equations state that the forward velocity y{ of a vehicle is proportional to the angular velocity of the engine $ h , and the tire traction force
I{ is proportional to the engine torque Wh , where, Uz is the eective tire radius, qg is the dierential transmission ratio, ql is the gearbox transmission
ratio in gear number l, and is the overall driveline e!ciency.
220
4. Driveline Dynamics
4.6 Key Symbols
d{
¨
dl > l = 0> · · · > 6
d{
DZ G
fj
Ff
g
G
H
I{
IZG
K
p
q = $ lq @$ rxw
ql
qg
qj
S
S0
S1 > S2 > S3
Sh
Sh = Sh ($ h )
SP
t
u = $@$ q
UZ G
Wg
Wh
WP
Wz
y {>
b v
yplq
4y
{> |> }> x
acceleration
coe!cients of function Wh = Wh ($ h )
acceleration capacity
all-wheel-drive
constant relative gear ratio
slip factor
distance traveled
clutch diameter
energy
traction force
front-wheel-drive
thermal value of fuel
vehicle mass
gear reduction ratio
gearbox transmission ratio in gear number l
transmission ratio
overall transmission ratio
power
ideal engine constant power
coe!cients of the power performance function
maximum attainable power of an engine
power performance function
maximum power
fuel consumption per unit distance
frequency ratio
rear-wheel-drive
dierential input torque
engine torque
maximum torque
wheel torque
velocity
minimum vehicle speed corresponding to $ plq
dierence in maximum vehicle speed at two dierent gears
displacement
f
h
P
w
w
W
overall e!ciency
convertor e!ciency
engine e!ciency
mechanical e!ciency
transmission e!ciency
thermal e!ciency
thermal e!ciency
4. Driveline Dynamics
Y
{
i
!
$g
$h
$ plq
$P
$ Pd{
$s
$ u = $ rxw @$ lq
volumetric e!ciency
traction coe!cient
oil density
fuel density
slope of the road
dierential input angular velocity
engine angular velocity
minimum engine speed
engine angular velocity at maximum power
maximum engine speed
pump angular velocity
speed ratio
221
222
4. Driveline Dynamics
Exercises
1. Power performance.
Audi U8W P with p = 1558 kg, has a Y 8 engine with
SP = 313 kW 420 hp dw $ P = 7800 usp
and Audi W W CoupeW P with p = 1430 kg, has a Y 6 engine with
SP = 184 kW 250 hp dw $ P = 6300 usp
Determine the power performance equations of their engines and compare the power mass ratio, SP @p of the cars.
2. Power and torque performance.
A model of Nissan Q LVP R 350] with p = 1522 kg, has a Y 6 engine
with
SP
WP
= 228 kW 306 hp dw $ P = 6800 rpm
= 363 N m 268 lb ft dw $ = 4800 rpm
Determine the power and torque performance equations, and compare
WP from the torque equation with the above reported number.
3. Power performance and modeling.
BMW X3W P is oered in four models with four dierent engines:
xDrive20d, xDrive25i, xDrive30i, and xDrive30d. The maximum power
and torque of the models and their associated engine speed are
xDrive20d
xDrive25i
xDrive30i
xDrive30d
SP kW
130
160
200
160
$ P rpm WP N m $ rpm p kg
4000
350
1750 1750
6500
250
2750 1755
6650
315
2750 1765
4000
500
1750 1810
(a) Employ Equation (4.1) to express the power performance of
these vehicles.
(b) Determine the maximum torque of each vehicle based on the
analytic power performance equation. Compare the calculated
maximum torques with the actual values and determine which
vehicle can better be modeled by analytic equation.
4. Fuel consumption conversion.
4. Driveline Dynamics
223
A model of Subaru Impreza Z U[ STIW P with p = 1521 kg, has a
turbocharged at-4 engine with
SP = 219 kW 293 hp dw $ P = 6000 rpm
Fuel consumption of the car is 19 mi@ gal in city and 25 mi@ gal in
highway. Determine the fuel consumption in liter per 100 km.
5. Traction force.
A model of Mercedes-Benz VOU 722 EditionW P with p = 1724 kg,
has a supercharged Y 8 engine with
SP = 485 kW 650 hp dw $ P = 6500 rpm
The maximum speed of the car is
yP = 337 km@ h 209 mi@ h
Assume the maximum speed happens at the maximum power and use
an overall e!ciency = 0=75 to determine the traction force at the
maximum speed.
6. Gearbox speed equation.
BMW X3W P has four models with four dierent gearboxes: xDrive20d,
xDrive25i, xDrive30i, and xDrive30d. The mass of the vehicles and
the gear ratio of their gearbox are given as
xDrive20d
xDrive25i
xDrive30i
xDrive30d
q1
4=17
4=07
4=07
4=17
q2
2=34
2=37
2=37
2=34
q3
1=52
1=55
1=55
1=52
q4
1=14
1=16
1=16
1=14
q5
0=87
0=85
0=85
0=87
q6
qu
qg
0=69 3=4 3=73
0=67 3=20 4=44
0=67 3=20 4=44
0=69 3=40 3=46
All models are using the same tire and rim: 235@55 U 17, 8M{17,
and some other specications of the vehicles are given in Exercise 3
if needed.
(a) Employ the speed equation and plot the them on four gures
for the vehicles.
(b) Assuming = 1, plot the acceleration capacity of the vehicles
on the same gure to compare.
(c) Assume $ 2 is 1=2$ P and make a table of dierent $ 1 at each
gear change for all vehicles.
(d) Examine the gearbox stability condition for all gearboxes.
(e) F Determine the best constant $ 1 and $ 2 for each gearbox such
that the gearbox is as close as possible to a geometric design.
224
4. Driveline Dynamics
7. Car speed and engine speed.
A model of Toyota CamryW P has a 3=5-liter, 6-cylinder engine with
SP = 268 hp dw $ P = 6200 rpm
The car uses transaxle/front-wheel drive and is equipped with a sixspeed ECT-i automatic transmission.
1vw gear ratio = q1 = 3=300
2qg gear ratio = q2 = 1=900
3ug gear ratio = q3 = 1=420
4wk gear ratio = q4 = 1=000
5wk gear ratio = q5 = 0=713
6wk gear ratio = q6 = 0=609
reverse gear ratio = qu = 4=148 nal drive ratio = qg = 3=685
Determine the speed of the car at each gear, when the engine is
running at $ P , and it is equipped with
(a) S 215@55U17 tires
(b) S 215@60U16 tires.
8. Gear-speed equations.
A model of Ford MondeoW P is equipped with a 2=0-liter engine, which
has
WP = 185 N m dw $ h = 4500 rpm
It has a manual ve-speed gearbox.
1vw gear ratio = q1 = 3=42
2qg gear ratio = q2 = 2=14
3ug gear ratio = q3 = 1=45
4wk gear ratio = q4 = 1=03
5wk gear ratio = q5 = 0=81
reverse gear ratio = qu = 3=46 nal drive ratio = qg = 4=06
If the tires of the car are 205@55U16, determine the gear-speed equations for each gear.
9. Final drive and gear ratios.
A model of Renault/Dacia LoganW P with p = 1115 kg, has a fourcylinder engine with
SP
WP
yP
W luhv
=
=
=
=
77 kW 105 hp dw $ P = 5750 rpm
148 N m dw $ h = 3750 rpm
183 km@ h
185@65U15
4. Driveline Dynamics
225
It has a ve-speed gearbox. When the engine is running at 1000 rpm
the speed of the car at each gear is
speed in the 1vw gear
speed in the 2qg gear
speed in the 3ug gear
speed in the 4wk gear
speed in the 5wk gear
=
=
=
=
=
yP1 = 7=25 km@ h
yP2 = 13=18 km@ h
yP3 = 19=37 km@ h
yP4 = 26=21 km@ h
yP5 = 33=94 km@ h
Evaluate the gear ratios ql > l = 1> 2> · · · 5 for qg = 4.
10. Traction equation.
A model of Jeep WranglerW P is equipped with a Y 6 engine and has
the following specications.
SP
WP
= 153 kW 205 hp dw $ P = 5200 rpm
= 325 N m 240 lb ft dw $ h = 4000 rpm
A model of the car may have a six-speed manual transmission with
the following gear ratios
1vw gear ratio = q1 = 4=46
2qg gear ratio = q2 = 2=61
3ug gear ratio = q3 = 1=72
4wk gear ratio = q4 = 1=25
5wk gear ratio = q5 = 1=00
6wk gear ratio = q6 = 0=84
reverse gear ratio = qu = 4=06 nal drive ratio = qg = 3=21
or a four-speed automatic transmission with the following gear ratios.
2qg gear ratio = q2 = 1=57
1vw gear ratio = q1 = 2=84
3ug gear ratio = q3 = 1=0
4wk gear ratio = q4 = 0=69
reverse gear ratio = qu = 2=21 nal drive ratio = qg = 4=10
Assume
= 0=8
W luhv = 245@75U16
and determine the traction equation for the two models.
11. Acceleration capacity.
Lamborghini MurcielagoW P is equipped with a 6=2-liter Y 12 engine
and has the following specications.
SP
WP
= 631 hp dw $ P = 8000 rpm
= 487 lb ft dw $ h = 6000 rpm
p = 3638 lb
I urqw wluh = S 245@35]U18
Uhdu wluh = S 335@30]U18
226
4. Driveline Dynamics
The gearbox of the car uses ratios close to the following values.
2qg gear ratio = q2 = 2=056
1vw gear ratio = q1 = 2=94
3ug gear ratio = q3 = 1=520
4wk gear ratio = q4 = 1=179
5wk gear ratio = q5 = 1=030
6wk gear ratio = q6 = 0=914
reverse gear ratio = qu = 2=529 nal drive ratio = qg = 3=42
If = 0=8, then
(a) determine the wheel torque function at each gear
(b) determine the acceleration capacity of the car.
12. F Gearbox stability.
A model of Jaguar [M W P is a rear-wheel drive car with a 4=2-liter
Y 8 engine. Some of the car’s specications are close to the following
values.
p
I urqw wluh
Uhdu wluh
SP
=
=
=
=
3638 lb
o = 119=4 in
S 235@50U18
S 235@50U18
300 hp dw $ P = 6000 rpm
If gear ratios of the car’s gearbox are
1vw gear ratio = q1 = 4=17
2qg gear ratio = q2 = 2=34
3ug gear ratio = q3 = 1=52
4wk gear ratio = q4 = 1=14
5wk gear ratio = q5 = 0=87
6wk gear ratio = q6 = 0=69
reverse gear ratio = qu = 3=4 nal drive ratio = qg = 2=87
Check the gearbox stability condition. In case the relative gear ratio
is not constant, determine the new gear ratios using the relative ratio
of the rst two gears.
13. F Geometric gearbox design.
Lamborghini DiabloW P is a rear-wheel drive car that was built in
years 1990 2000. The car is equipped with a 5=7-liter Y 12 engine.
Some of the car’s specications are given.
SP
WP
yP
= 492 hp at $ P = 7000 rpm
= 580 N m 428 lb ft dw $ h = 5200 rpm
= 328 km@ h 203 mi@ h
p = 1576 kg 3474 lb
o = 2650 mm 104 in
zi = 1540 mm 60=6 in
zu = 1640 mm 64=6 in
Front tire = 245@40]U17
Rear tire = 335@35]U17
4. Driveline Dynamics
227
The gear ratios of the car’s gearbox are close to the following values.
1vw gear ratio
2qg gear ratio
3ug gear ratio
4wk gear ratio
5wk gear ratio
reverse gear ratio
nal drive ratio
=
=
=
=
=
=
=
q1 = 2=31
q2 = 1=52
q3 = 1=12
q4 = 0=88
q5 = 0=68
qu = 2=12
qg = 2=41
yP = 97=3 km@ h 60=5 mi@ h
yP = 147=7 km@ h 91=8 mi@ h
yP = 200=2 km@ h 124 mi@ h
yP = 254=8 km@ h 158=4 mi@ h
yP = 325 km@ h 202 mi@ h
yP = 105=7 km@ h 65=7 mi@ h
Assume = 0=9 and
(a) Determine the step jump fj for each gear change.
(b) Determine the speed span for each gear change.
(c) Determine the engine speed at the maximum car speed for each
gear.
(d) Determine the power performance equation and nd the engine
power at the maximum car speed for each gear.
(e) There is a dierence between the car’s top speed and the maximum speed in the 5wk gear. Find the engine power at the car’s
top speed. Based on the top speed, determine the overall resistance forces.
14. Manual and auto transmission comparison.
A model of Nissan X 12 PintaraW P may come with manual or auto
transmission. A model with a manual transmission has gear ratios
and characteristics close to the following values
1vw gear ratio = q1 = 3=285
2qg gear ratio = q2 = 1=850
3ug gear ratio = q3 = 1=272
4wk gear ratio = q4 = 0=954
5wk gear ratio = q5 = 0=740
reverse gear ratio = qu = 3=428 nal drive ratio = qg = 3=895
and the model with an auto transmission has gear ratios close to the
following values.
1vw gear ratio = q1 = 2=785
2qg gear ratio = q2 = 1=545
3ug gear ratio = q3 = 1=000
4wk gear ratio = q4 = 0=694
reverse gear ratio = qu = 2=272 nal drive ratio = qg = 3=876
Compare the transmissions according to geometric design condition
and determine which one has the maximum deviation.
228
4. Driveline Dynamics
15. F Progressive and geometric gearbox design.
An all wheel drive model of Hyundai Santa FeW P has specications
close to the following numbers.
SP
WP
p
o
W luhv
=
=
=
=
=
242 hp dw $ P = 6000 rpm
226 lb ft dw $ h = 4500 rpm
1724 kg 4022 lb
2700 mm 106=3 in
S 235@70U16
1vw gear ratio = q1 = 3=79
2qg gear ratio = q2 = 2=06
3ug gear ratio = q3 = 1=42
4wk gear ratio = q4 = 1=03
5wk gear ratio = q5 = 0=73
reverse gear ratio = qu = 3=81 nal drive ratio = qg = 3=68
The car can reach a speed y = 200=2 km@ h 124 mi@ h at the maximum power SP in the nal gear q5 = 0=73. Accept q1 and q5 and
redesign the gear ratios based on a progressive and a geometric gearbox.
16. F Engine performance estimation.
Consider a UZ G vehicle with the following specications.
p
o
I}1 @I}2
W luhv
=
=
=
=
6300 lb
153 in
4410@6000
245@75U16
If an experiment shows that
yP = 62=6 mi@ h at 3% slope
yP = 52=1 mi@ h at 6% slope
yP = 0
at 33=2% slope
estimate the maximum torque and maximum power of the vehicle.
Assume = 0=85.
Hint: assume that when the vehicle is stuck on a road with the maximum slope, the engine is working at the maximum torque. However,
when the vehicle is moving on a slope at the maximum speed, the
engine is working at the maximum power. Slope 3% means the angle
of the road with horizon is
! = tan1
3
100
4. Driveline Dynamics
229
17. F Gearbox design.
Consider a UZ G vehicle with the following specications.
SP
WP
yP
=
=
=
=
141 kW 189 hp dw $ P = 7800 rpm
181 N m 133 lb ft dw $ h = 6800 rpm
237 km@ h 147 mi@ h
0=90
p = 875 kg
I urqw wluhvw = 195@50U16
o = 2300 mm
Uhdu wluhvw = 225@45U17
1vw gear ratio = q1 = 3=116
2qg gear ratio = q2 = 2=050
3ug gear ratio = q3 = 1=481
4wk gear ratio = q4 = 1=166
5wk gear ratio = q5 = 0=916
6wk gear ratio = q6 = 0=815
reverse gear ratio = qu = 3=250 nal drive ratio = qg = 4=529
Based on the maximum velocity at the 6wk gear q6 , redesign the gear
ratios. Use ±20% around the maximum power for the working range.
Part II
Vehicle Kinematics
5
F Applied Kinematics
Position, velocity, and acceleration are called kinematics information. Rotational position analysis is the key to calculate kinematics of relatively
moving rigid bodies. In this chapter, we review kinematics and show applied
methods to calculate the relative kinematic information of rigid bodies. A
vehicle has many moving sub-systems such as suspensions, and the vehicle
can be treated as a moving rigid body in an inertia coordinate frame.
5.1 Rotation About Global Cartesian Axes
Consider an orthogonal Cartesian body coordinate frame E (R{|}) xed
to a rigid body E that is attached to the ground J at the origin point R.
The orientation of the rigid body E with respect to the global coordinate
Z
ZP
B
G
z
P
zP
r
yP
y
O
XP
xP
YP
Y
X
x
FIGURE 5.1. Rotation of a body coordinate frame E about point R in the global
coordinate frame J.
frame J (R[\ ]) xed to the ground is known when the orientation of
R{|} with respect to R[\ ] is determined. Figure 5.1 illustrates a body
coordinate E rotating about point R in the global coordinate frame J.
If the rigid body, E rotates degrees about the ]-axis of the global
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_5, © Springer Science+Business Media New York 2014
233
234
5. F Applied Kinematics
coordinate frame, then coordinates of any point S of the rigid body in the
local and global coordinate frames are related by the equation
r = U]> E r
(5.1)
6
cos sin 0
U]> = 7 sin cos 0 8
0
0
1
(5.2)
J
where
and
5
5
6
[
J
r=7 \ 8
]
5
6
{
E
r =7 | 8
}
(5.3)
Similarly, rotation degrees about the \ -axis, and degrees about the
[-axis of the global frame relate the local and global coordinates of point
S by the following equations:
J
J
r = U\> E r
r = U[> E r
(5.4)
(5.5)
where
5
U\>
U[>
6
cos 0 sin 0
1
0 8
= 7
sin 0 cos 5
6
1
0
0
= 7 0 cos sin 8
0 sin cos (5.6)
(5.7)
ˆ M>
ˆ N̂) be the unit vectors along the coordinate
Proof. Let (ˆ~> ˆ> n̂) and (L>
axes of R{|} and R[\ ] respectively. The rigid body has a space xed
point at R, which is the common origin of R{|} and R[\ ]. Consider a
body xed point S of a rigid body that turns rad about the }-axis. Figure
5.2 illustrate the top view of the coordinate frames after the rotation while
the dashed lines indicate the point at initial position ([1 > \1 > ]1 ).
The position vector r of S can be expressed in body and global coordinate
frames by
E
J
r = {ˆ~ + |ˆ
 + } n̂
r = [ Lˆ + \ Mˆ + ] N̂
(5.8)
(5.9)
where E r refers to the position vector r expressed in the body coordinate
frame E, and J r refers to the position vector r expressed in the global
coordinate frame J.
5. F Applied Kinematics
235
Y
P
Y
y
y
G
r
x
x
B
D
X
X
FIGURE 5.2. Position vectors of point S before and after the rotation of the
local frame about the ]-axis of the global frame.
Using Equation (5.8) and the denition of the inner product, we may
write
[ = Lˆ · r = Lˆ · {ˆ~ + Lˆ · |ˆ
 + Lˆ · } n̂
(5.10)
= Mˆ · r = Mˆ · {ˆ~ + Mˆ · |ˆ
 + Mˆ · } n̂
\
]
(5.11)
= N̂ · r = N̂ · {ˆ~ + N̂ · |ˆ
 + N̂ · } n̂
(5.12)
6 5 ˆ
65
6
L · ˆ~ Lˆ · ˆ Lˆ · n̂
[
{
7 \ 8 = 7 Mˆ · ˆ~ Mˆ · ˆ Mˆ · n̂ 8 7 | 8
]
}
N̂ · ˆ~ N̂ · ˆ N̂ · n̂
(5.13)
or equivalently
5
Analyzing Figure 5.2 indicates that
Lˆ · ˆ~ = cos Lˆ · ˆ = sin Lˆ · n̂ = 0
Mˆ · ˆ~ = sin Mˆ · ˆ = cos Mˆ · n̂ = 0
N̂ · ˆ~ = 0
N̂ · ˆ = 0
N̂ · n̂ = 1
(5.14)
Combining Equations (5.13) and (5.14) shows that
J
r = U]> E r
6
5
65
6
[
cos sin 0
{
7 \ 8 = 7 sin cos 0 8 7 | 8
]
0
0
1
}
5
(5.15)
(5.16)
The elements of the ]-rotation matrix, U]> , are called the direction cosines
of E r with respect to R[\ ].
5
6
cos sin 0
U]> = 7 sin cos 0 8
(5.17)
0
0
1
236
5. F Applied Kinematics
Equation (5.15) states that the vector r at the second position in the
global coordinate frame is equal to U]> times the position vector in the
local coordinate frame. Hence, we are able to nd the global coordinates of
a point of a rigid body after rotation about the ]-axis, if we have its local
coordinates.
Similarly, rotation about the \ -axis and rotation about the [-axis
are expressed by the \ -rotation matrix U\> and the [-rotation matrix
U[> respectively.
5
6
cos 0 sin 0
1
0 8
(5.18)
U\> = 7
sin 0 cos 5
1
0
U[> = 7 0 cos 0 sin 6
0
sin 8
cos (5.19)
The rotation matrices U]> , U\> , and U[> are called basic global rotation matrices. We usually refer to the rst, second, and third rotations
about the axes of the global coordinate frame by , , and respectively.
Example 155 Successive rotation about global axes.
The nal position of the point S (1> 2> 3) after a 30 deg rotation about
the ]-axis, followed by 30 deg about the [-axis, and then 90 deg about the
\ -axis can be found by rst multiplying U]>30 by [1> 2> 3]W to get the new
global position after the rst rotation
5
6 5
65 6 5
6
[2
cos 6 sin 6 0
1
0=134
7 \2 8 = 7 sin cos 6
0 8 7 2 8 = 7 2=23 8
(5.20)
6
0
0
1
3
3
]2
and then multiplying U[>30 by [0=134> 2=23> 3]W to calculate the position of
S after the second rotation
5
6 5
65
6 5
6
[3
1
0
0
0=134
0=134
7 \3 8 = 7 0 cos sin 8 7 2=23 8 = 7 0=433 8
6
6
3
3=714
]3
0 sin 6
cos 6
(5.21)
and nally multiplying U\>90 by [0=134> 0=433> 3=714]W to nd the nal position of S after the third rotation.
5
6 5
65
6 5
6
[4
cos 2
0=134
3=714
0 sin 2
7 \4 8 = 7
0
1
0 8 7 0=433 8 = 7 0=433 8
(5.22)
3=714
0=134
sin 2 0 cos 2
]4
5. F Applied Kinematics
237
Example 156 Global rotation, local position.
If a point S is moved to J r = [2> 3> 2]W after a 60 deg rotation about the
]-axis, its position in the local coordinate is
1 J
r = U]>60
r
5
6
5
{
cos 3 sin 3
7 | 8 = 7 sin cos 3
3
0
0
}
E
6
61 5 6 5
0
2
3=6
0 8 7 3 8 = 7 0=23 8
2
2
1
(5.23)
Example 157 Time-dependent global rotation.
Assume a rigid body E is continuously turning about the ]-axis at a rate
of 0=1 rad@ s. The transformation matrix of the body is
5
6
cos 0=1w sin 0=1w 0
J
UE = 7 sin 0=1w cos 0=1w 0 8
(5.24)
0
0
1
The rotation
5
6
5
65
6
[
cos 0=1w sin 0=1w 0
{
7 \ 8 = 7 sin 0=1w cos 0=1w 0 8 7 | 8
]
0
0
1
}
5
6
{ cos 0=1w | sin 0=1w
= 7 | cos 0=1w + { sin 0=1w 8
}
will move any point of E on a circle with radius U =
to the ([> \ )-plane:
[2 + \ 2
[ 2 + \ 2 parallel
2
2
= ({ cos 0=1w | sin 0=1w) + (| cos 0=1w + { sin 0=1w)
= {2 + | 2 = U2
(5.26)
Consider a point S at E r =
5
s
£
1 0 0
¤W
. The point will be seen at
6 5
65 6 5
6
[
cos 0=1 sin 0=1 0
1
0=995
7 \ 8 = 7 sin 0=1 cos 0=1 0 8 7 0 8 = 7 0=0998 8
]
0
0
1
0
0
after w = 1 s, and at
5
6 5
65 6 5
6
[
cos 0=6 sin 0=6 0
1
0=825
7 \ 8 = 7 sin 0=6 cos 0=6 0 8 7 0 8 = 7 0=564 8
]
0
0
1
0
0
after w = 6 s,
(5.25)
(5.27)
(5.28)
5. F Applied Kinematics
238
5.2 Successive Rotation About Global Cartesian
Axes
The nal global position of a point S of a rigid body E with position vector
r, after a sequence of rotations U1 , U2 , U3 , ===, Uq about the global axes
can be found by
J
r = J UE E r
(5.29)
where,
J
J
E
UE = Uq · · · U3 U2 U1
(5.30)
and r and r indicate the position vector r expressed in the global and
local coordinate frames, respectively. The transformation matrix J UE is
called the global rotation matrix which maps the local coordinates to their
corresponding global coordinates.
Example 158 Successive global rotations, global position.
The point S of a rigid body that is attached to the global frame at R is
located at
5
6 5
6
[
0=0
7 \ 8 = 7 0=26 8
(5.31)
]
0=97
The rotation matrix to nd the new position of the point after a 29 deg
rotation about the [-axis, followed by 30 deg about the ]-axis, and again
132 deg about the [-axis is
5
6
0=87 0=44 0=24
J
UE = U[>132 U]>30 U[>29 = 7 0=33 0=15 0=93 8
(5.32)
0=37
0=89 0=27
Therefore, its new position is at
6
5
6 5
65
6 5
[
0=87 0=44 0=24
0=0
0=35
7 \ 8 = 7 0=33 0=15 0=93 8 7 0=26 8 = 7 0=94 8
0=031
]
0=37
0=89 0=27
0=97
(5.33)
Example 159 Order of rotation, and order of matrix multiplication.
Changing the order of global rotation matrices is equivalent to changing
the order of rotations.
£
¤W
.
The position of a point S of a rigid body E is located at E r = 1 2 3
Its global position after rotation 30 deg about the [-axis and then 45 deg
about the \ -axis is at
J
r1
= U\>45 U[>30 E r
5
65 6 5
6
0=53 0=84 0=13
1
0=76
0=15
0=99 8 7 2 8 = 7 3=27 8
= 7 0=0
0=85 0=52 0=081
3
1=64
(5.34)
5. F Applied Kinematics
239
and if we change the order of rotations, then its position would be at
J
r2
= U[>30 U\>45 E r
5
65 6 5
6
0=53
0=0
0=85
1
3=08
0=52 8 7 2 8 = 7 1=02 8
= 7 0=84 0=15
0=13 0=99 0=081
3
1=86
(5.35)
¯
¯
These two nal positions of S are g = ¯J r1 J r2 ¯ = 4=456 apart.
Example 160 Global roll-pitch-yaw angles.
The rotation about the [-axis of the global coordinate frame is called
roll, the rotation about the \ -axis of the global coordinate frame is called
pitch, and the rotation about the ]-axis of the global coordinate frame is
called yaw. The global roll-pitch-yaw rotation matrix is
J
UE
= U]> U\> U[>
5
ff fv + fvv
= 7 fv ff + vvv
v
fv
6
vv + ffv
fv + fvv 8
ff
(5.36)
Given the roll, pitch, and yaw angles, we can compute the overall rotation
matrix using Equation (5.36). Also, we are able to compute the equivalent
roll, pitch, and yaw angles when a rotation matrix is given. Suppose that ulm
indicates the element of row l and column m of the roll-pitch-yaw rotation
matrix (5.36), then the roll angle is
μ
¶
u32
1
= tan
(5.37)
u33
and the pitch angle is
= sin1 (u31 )
and the yaw angle is
= tan
1
μ
u21
u11
¶
(5.38)
(5.39)
provided that cos 6= 0.
5.3 Rotation About Local Cartesian Axes
Consider a rigid body E with a space-xed point at point R. The local
body coordinate frame E(R{|}) is coincident with the global coordinate
frame J(R[\ ]), where the origin of both frames are on the xed point R.
If the body undergoes a rotation * about the }-axis of its local coordinate
240
5. F Applied Kinematics
Y
B
P
Y
y
G
y
r
x
x
M
X
X
FIGURE 5.3. Position vectors of point S before and after rotation of the local
frame about the }-axis of the local frame.
frame, then coordinates of any point of the rigid body in the local and
global coordinate frames are related by the equation
E
r = U}>* J r
(5.40)
Figure 5.3 depicts the top view of the system. The vectors J r and E r are
the position vectors of the point in the global and local frames respectively
5
6
5
6
[
{
J
E
r=7 \ 8
r =7 | 8
(5.41)
]
}
and U}>* is the }-rotation matrix
5
6
cos * sin * 0
U}>* = 7 sin * cos * 0 8
0
0
1
(5.42)
Similarly, rotation about the |-axis and rotation # about the {-axis
are expressed by the |-rotation matrix U|> and the {-rotation matrix U{>#
respectively.
5
U|>
U{>#
6
cos 0 sin 8
1
0
= 7 0
sin 0 cos 5
6
1
0
0
= 7 0 cos # sin # 8
0 sin # cos #
(5.43)
(5.44)
5. F Applied Kinematics
241
Proof. Vector r indicates the position of a point S of the rigid body E
after the rotation. Using the unit vectors (ˆ~> ˆ> n̂) along the axes of the
ˆ M>
ˆ N̂) along the axes of the global
local coordinate frame E(R{|}), and (L>
coordinate frame J(R[\ ]), the expression of the position vectors r in
both coordinate frames would be
E
J
r = {ˆ~ + |ˆ
 + } n̂
r = [ Lˆ + \ Mˆ + ] N̂
(5.45)
(5.46)
The components of E r can be found if we have the components of J r. Using
Equation (5.45) and the denition of the inner product, we may write
{ = ˆ~ · r = ˆ~ · [ Lˆ + ˆ~ · \ Mˆ + ˆ~ · ] N̂
| = ˆ · r = ˆ · [ Lˆ + ˆ · \ Mˆ + ˆ · ] N̂
} = n̂ · r = n̂ · [ Lˆ + n̂ · \ Mˆ + n̂ · ] N̂
(5.47)
(5.48)
(5.49)
or equivalently
65
6
6 5
ˆ~ · Lˆ ˆ~ · Mˆ ˆ~ · N̂
[
{
7 | 8 = 7 ˆ · Lˆ ˆ · Mˆ ˆ · N̂ 8 7 \ 8
]
}
n̂ · Lˆ n̂ · Mˆ n̂ · N̂
5
(5.50)
The elements of the }-rotation matrix U}>* are the direction cosines of
r2 with respect to R{|}. So, the elements of the matrix in Equation (5.50)
are
ˆ~ · Lˆ = cos *
ˆ~ · Mˆ = sin * ˆ~ · N̂ = 0
(5.51)
ˆ · Lˆ = sin * ˆ · Mˆ = cos * ˆ · N̂ = 0
ˆ
ˆ
n̂ · L = 0
n̂ · M = 0
n̂ · N̂ = 1
J
E
Combining Equations (5.50) and (5.51), we can nd the components of
r by multiplying }-rotation matrix U}>* and vector J r.
5
6 5
65
6
{
cos * sin * 0
[
7 | 8 = 7 sin * cos * 0 8 7 \ 8
(5.52)
}
0
0
1
]
It can also be shown in a short form:
E
where
r = U}>* J r
(5.53)
6
cos * sin * 0
U}>* = 7 sin * cos * 0 8
0
0
1
(5.54)
5
Equation (5.53) says that after rotation about the }-axis of the local
coordinate frame, the position vector in the local frame is equal to U}>*
242
5. F Applied Kinematics
times the position vector in the global frame. Hence, after rotation about
the }-axis, we are able to nd the coordinates of any point of a rigid body
in a local coordinate frame, if we have its coordinates in the global frame.
Similarly, rotation about the |-axis and rotation # about the {-axis
are described by the |-rotation matrix U|> and the {-rotation matrix U{>#
respectively.
5
6
cos 0 sin 8
1
0
U|> = 7 0
(5.55)
sin 0 cos 5
6
1
0
0
(5.56)
U{># = 7 0 cos # sin # 8
0 sin # cos #
We usually indicate the rst, second, and third rotations about the local
axes by *, , and # respectively.
Example 161 Local rotation, local position.
If a local coordinate frame E (R{|}) has been rotated 60 deg about the }axis and a point S in the global coordinate frame J (R[\ ]) is at (4> 3> 2),
its coordinates in the local coordinate frame E are
5
6 5
65 6 5
6
{
cos 3
sin 3 0
4
4=60
7 | 8 = 7 sin cos 0 8 7 3 8 = 7 1=97 8
(5.57)
3
3
}
0
0
1
2
2=0
Example 162 Local rotation, global position.
If a local coordinate frame E (R{|}) has been rotated 60 deg about the
}-axis and a point S in the local frame E is at (4> 3> 2), its position in the
global coordinate frame J (R[\ ]) is at
6 5
5
6
61 5 6 5
[
cos 3
4
0=60
sin 3 0
7 \ 8 = 7 sin cos 0 8 7 3 8 = 7 4=96 8
(5.58)
3
3
]
2
2=0
0
0
1
Example 163 Successive local rotation, global position.
Let us turn a rigid body 90 deg about the |-axis and then 90 deg about
£
¤W
, then its
the {-axis. If a body point S is at E r = 9=5 10=1 10=1
position in the global coordinate frame is at
J
r2
1 E
1
1 E
r = U|>90
U{>90
r
5
6
10=1
W
W
E
= U|>90
U{>90
rS = 7 10=1 8
9=5
= [U{>90 U|>90 ]
(5.59)
The inverse and transpose of the rotation transformation matrices are equal
U1 = UW
such matrices are called orthogonal matrices.
(5.60)
5. F Applied Kinematics
243
Example 164 Global position and postmultiplication of rotation matrix.
£
¤W
. If
The local position of a point S after rotation is E r = 1 2 3
the local rotation matrix to transform J r to E r is given as
5
6 5
6
cos * sin * 0
cos 6
sin 6 0
(5.61)
U}>* = 7 sin * cos * 0 8 = 7 sin 6 cos 6 0 8
0
0
1
0
0
1
then we may nd the global position vector J r by postmultiplication E U}>*
by the local position vector E rW ,
5
6
cos 6
sin 6 0
£
¤
J W
r
= E rW U}>* = 1 2 3 7 sin 6 cos 6 0 8
0
0
1
£
¤
0=13 2=23 3=0
=
(5.62)
1
by E r.
instead of premultiplication of U}>*
J
1 E
U}>*
r
5
cos 6 sin 6
7
sin 6
cos 6
=
0
0
r =
65 6 5
6
0
1
0=13
0 8 7 2 8 = 7 2=23 8
1
3
3
(5.63)
5.4 Successive Rotation About Local Cartesian
Axes
The nal global position of a point S in a rigid body E with position
vector r, after some rotations U1 , U2 , U3 , ===, Uq about the local axes, can
be found by
E
r = E UJ J r
(5.64)
where
E
UJ = Uq · · · U3 U2 U1
(5.65)
E
The transformation matrix UJ is called the local rotation matrix and it
maps the global coordinates to their corresponding local coordinates.
Example 165 Successive local rotation, local position.
A local coordinate frame E(R{|}) that initially is coincident with a global
coordinate frame J(R[\ ]) undergoes a rotation * = 30 deg about the }axis, then = 30 deg about the {-axis, and then # = 30 deg about the |-axis.
The local coordinates of a point S located at [ = 5, \ = 30, ] = 10 can
be found by
5
6
5
6
{
5
7 | 8 = E UJ 7 30 8
(5.66)
}
10
5. F Applied Kinematics
244
z
yaw
B
\
M
pitch
roll
T
x
y
FIGURE 5.4. Local roll-pitch-yaw angles.
The local rotation matrix is
5
0=63
0=65
E
0=75
UJ = U|>30 U{>30 U}>30 = 7 0=43
0=65 0=125
6
0=43
0=50 8
0=75
and coordinates of S in the local frame are
5
6 5
65
6 5
6
{
0=63
0=65
0=43
5
18=35
7 | 8 = 7 0=43
0=75
0=50 8 7 30 8 = 7 25=35 8
}
0=65 0=125
0=75
10
7=0
(5.67)
(5.68)
Example 166 Successive local rotation.
The rotation matrix for a body point S ({> |> }) after rotation U}>* followed by U{> and U|># is
E
UJ
= U|># U{> U}>*
5
f*f# vv*v# f#v* + f*vv#
fv*
ff*
= 7
f*v# + vf#v* v*v# f*vf#
6
fv#
v 8 (5.69)
ff#
Example 167 Local roll-pitch-yaw angles
Rotation about the {-axis of the local frame is called roll or bank, rotation about |-axis of the local frame is called pitch or attitude, and rotation
about the }-axis of the local frame is called yaw, spin, or heading. The
local roll-pitch-yaw angles of a car are shown in Figure 5.4.
The local roll-pitch-yaw rotation matrix is
E
UJ
= U}># U|> U{>*
5
6
ff# f*v# + vf#v* v*v# f*vf#
= 7 fv# f*f# vv*v# f#v* + f*vv# 8 (5.70)
v
fv*
ff*
5. F Applied Kinematics
245
z'
z' Z
y"
z"
T
y'
y'
M
X
Y
x'
x'
x"
(a)
(b)
Z
z
y"
z"
z
y \
e\
T
eM
y
M
\
x
x
x"
X
Y
eT
(c)
(d)
FIGURE 5.5. (a) First Euler angle. (b) Second Euler angle. (c) Third Euler angle.
Example 168 F Euler Angles
The rotation about the ]-axis of the global coordinate is called precession, the rotation about the {-axis of the local coordinate is called nutation, and the rotation about the }-axis of the local coordinate is called
spin. The precession-nutation-spin rotation angles are also called Euler angles. Rotation matrix based on Euler angles has application in rigid body
kinematics. To nd the Euler angles rotation matrix to go from the global
frame J(R[\ ]) to the nal body frame E(R{|}), we employ a body frame
E 0 (R{0 | 0 } 0 ) as shown in Figure 5.5(d) that before the rst rotation coincides with the global frame. Let there be a rotation * about the } 0 -axis.
Because ]-axis and } 0 -axis are coincident, we have
E0
E0
r =
UJ
E0
UJ J r
5
6
cos * sin * 0
= U}>* = 7 sin * cos * 0 8
0
0
1
(5.71)
(5.72)
Next we consider the E 0 (R{0 | 0 } 0 ) frame as a new xed global frame and introduce a new body frame E 00 (R{00 | 00 } 00 ). Before the second rotation, the two
frames coincide. Then, we execute a rotation about {00 -axis as shown in
5. F Applied Kinematics
246
Figure 5.5(e). The transformation between E 0 (R{0 | 0 } 0 ) and E 00 (R{00 | 00 } 00 )
is
E 00
E 00
UE 0
E 00
0
UE 0 E r
5
1
0
= U{> = 7 0 cos 0 sin r =
6
0
sin 8
cos (5.73)
(5.74)
Finally, we consider the E 00 (R{00 | 00 } 00 ) frame as a new xed global frame
and consider the nal body frame E(R{|}) to coincide with E 00 before the
third rotation. We now execute a # rotation about the } 00 -axis as shown in
Figure 5.5(f). The transformation between E 00 (R{00 | 00 } 00 ) and E(R{|}) is
E
E
r =
UE 00
E
00
UE 00 E r
5
cos #
= U}># = 7 sin #
0
sin #
cos #
0
6
0
0 8
1
(5.75)
(5.76)
By the rule of composition of rotations, the transformation from J(R[\ ])
to E(R{|}) is
E
r = E UJ J r
(5.77)
where
E
UJ
= U}># U{> U}>*
5
f*f# fv*v#
f#v* + ff*v#
= 7 f*v# ff#v* v*v# + ff*f#
vv*
f*v
6
vv#
vf# 8
f
(5.78)
6
f*v# ff#v* vv*
v*v# + ff*f# f*v 8
vf#
f
(5.79)
and therefore,
E
1
UJ
=
E
5
W
UJ
= J UE = [U}># U{> U}>* ]W
f*f# fv*v#
= 7 f#v* + ff*v#
vv#
Given the angles of precession *, nutation , and spin #, we can compute
the overall rotation matrix using Equation (5.78). Also we are able to compute the equivalent precession, nutation, and spin angles when a rotation
matrix is given.
If ulm indicates the element of row l and column m of the precessionnutation-spin rotation matrix, then,
= cos1 (u33 )
μ
¶
u31
1
* = tan
u32
μ
¶
u13
1
# = tan
u23
(5.80)
(5.81)
(5.82)
5. F Applied Kinematics
247
provided that sin 6= 0.
Example 169 F Euler angles of a local rotation matrix.
The local rotation matrix after a rotation 30 deg about the }-axis, 30 deg
about the {-axis, and 30 deg about the |-axis is
E
UJ
= U|>30 U{>30 U}>30
5
0=63
0=65
0=75
= 7 0=43
0=65 0=125
6
0=43
0=50 8
0=75
(5.83)
and therefore, the local coordinates of a sample point at [ = 5, \ = 30,
and ] = 10 are
5
6 5
65
6 5
6
{
0=63
0=65
0=43
5
18=35
7 | 8 = 7 0=43
0=75
0=50 8 7 30 8 = 7 25=35 8
(5.84)
}
0=65 0=125
0=75
10
7=0
The Euler angles of the corresponding precession-nutation-spin rotation
matrix are
= cos1 (0=75) = 41=41 deg
μ
¶
0=65
* = tan1
= 79=15 deg
0=125
μ
¶
0=43
# = tan1
= 40=7 deg
0=50
(5.85)
(5.86)
(5.87)
We can examine that the precession-nutation-spin rotation matrix for * =
79=15 deg, = 41=41 deg, and # = 40=7 deg is also
E
UJ
= U}>40=7 U{>41=41 U}>79=15
5
6
0=63
0=65
0=43
0=75
0=50 8
= 7 0=43
0=65 0=125
0=75
(5.88)
Therefore, U|>30 U{>30 U}>30 = U}># U{> U}>* when * = 79=15 deg, =
41=41 deg, and # = 40=7 deg. In other words, the rigid body attached to the
local frame moves to the nal conguration by undergoing either three consecutive rotations * = 79=15 deg, = 41=41 deg, and # = 40=7 deg about
the }, {, and } axes respectively, or three consecutive rotations 30 deg,
30 deg, and 30 deg about the }, {, and | axes. Hence the triple rotations
between the initial and nal conguration of a rigid body is not unique.
Example 170 F Relative rotation matrix of two bodies.
Consider a rigid body E1 with an orientation matrix 1 UJ made by Euler
angles * = 30 deg, = 45 deg, # = 60 deg, and another rigid body E2
248
5. F Applied Kinematics
having * = 10 deg, = 25 deg, # = 15 deg, with respect to the global
frame. To nd the relative rotation matrix 1 U2 to map the coordinates of
the second body frame E2 to the rst body frame E1 , we need to nd the
individual rotation matrices rst.
1
2
UJ
UJ
= U}>60 U{>45 U}>30
5
6
0=127
0=78
0=612
= 7 0=927 0=127 0=354 8
0=354 0=612
0=707
= U}>10 U{>25 U}>15
5
0=992
6=33 × 102
7
0=103
0=907
=
7=34 × 102
0=416
6
0=109
0=408 8
0=906
(5.89)
(5.90)
The desired rotation matrix 1 U2 may be found by
1
U2 = 1 UJ J U2
(5.91)
which is equal to
1
U2
1
W
UJ 2 UJ
5
0=992
= 7 6=33 × 102
0=109
=
6
0=103 7=34 × 102
8
0=907
0=416
0=408
0=906
(5.92)
Example 171 F Euler angles rotation matrix for small angles.
The Euler rotation matrix E UJ = U}># U{> U}>* for very small Euler
angles *> > and # is approximated by
5
6
1
0
E
UJ = 7 1 8
(5.93)
0 1
where
=*+#
(5.94)
Therefore, when the angles of rotation are small, the angles * and # are
indistinguishable.
If $ 0 then the Euler rotation matrix E UJ = U}># U{> U}>* approaches
to
5
6
f*f# v*v#
f#v* + f*v# 0
E
UJ = 7 f*v# f#v* v*v# + f*f# 0 8
0
0
1
5
6
cos (* + #) sin (* + #) 0
= 7 sin (* + #) cos (* + #) 0 8
(5.95)
0
0
1
5. F Applied Kinematics
249
and therefore, the angles * and # are indistinguishable even if the value of
* and # are nite. Hence, the Euler set of angles in rotation matrix (5.78)
is not unique when = 0.
Example 172 F Angular velocity vector in terms of Euler frequencies.
A Eulerian local frame H (R> ĥ* > ĥ > ĥ# ) can be introduced by dening unit
vectors ĥ* , ĥ , and ĥ# as shown in Figure 5.5(g). Although the Eulerian
frame is not necessarily orthogonal, it is very useful in rigid body kinematic
analysis.
The angular velocity vector J $ E of the body frame E(R{|}) with respect
to the global frame J(R[\ ]) can be written in Euler angles frame H as
the sum of three Euler angle rate vectors:
H
b + #ĥ
b #
b * + ĥ
J $ E = *ĥ
(5.96)
b and #b are called Euler frequencies.
where the rate of Euler angles, *,
b ,
To nd J $ E in the body frame we must express the unit vectors ĥ* , ĥ ,
£
¤W
and ĥ# in the body frame. The unit vector ĥ* = 0 0 1
= N̂ is in the
global frame and can be transformed to the body frame after three rotations
5
6
sin sin #
E
ĥ* = E UJ N̂ = U}># U{> U}>* N̂ = 7 sin cos # 8
cos (5.97)
£
¤W
= ˆ~0 is in the intermediate frame
The unit vector ĥ = 1 0 0
0 0 0
R{ | } and needs to get two rotations U{> and U}># to be transformed
to the body frame
5
6
cos #
E
ĥ = E UR{0 |0 } 0 ˆ~0 = U}># U{> ˆ~0 = 7 sin # 8
0
(5.98)
£
¤W
= n̂.
The unit vector ĥ# is already in the body frame, ĥ# = 0 0 1
Therefore, J $ E is expressed in the body coordinate frame as
5
E
J $E
6
5
6
5 6
sin sin #
cos #
0
= *b 7 sin cos # 8 + b 7 sin # 8 + #b 7 0 8
cos 0
1
³
´
³
´
=
*b sin sin # + b cos # ˆ~ + *b sin cos # b sin # ˆ
³
´
+ *b cos + #b n̂
(5.99)
and therefore, components of J $ E in body frame E (R{|}) are related to
250
5. F Applied Kinematics
the Euler angle frame H (R*#) by the following relationship:
5
E
J $E
6
=
E
5
UH H
J $E
sin sin #
${
7 $ | 8 = 7 sin cos #
cos $}
cos #
sin #
0
65
6
*b
0
0 8 7 b 8
1
#b
(5.100)
(5.101)
Then, J $ E can be expressed in the global frame using an inverse transformation of Euler rotation matrix (5.78)
6
5
*b sin sin # + b cos #
1 E
J
E 1 7
(5.102)
= E UJ
UJ
*b sin cos # b sin # 8
J $E =
J $E
*b cos + #b
³
´
³
´
=
b cos * + #b sin sin * Lˆ + b sin * #b cos * sin Mˆ
³
´
+ *b + #b cos N̂
(5.103)
and hence, components of J $ E in global coordinate frame J(R[\ ]) are
related to the Euler angle coordinate frame H (R*#) by the following relationship:
5
J
J $E
6
=
J
5
UH H
J $E
65
6
*b
0 cos *
sin sin *
$[
7 $ \ 8 = 7 0 sin * cos * sin 8 7 b 8
1
0
cos $]
#b
(5.104)
(5.105)
Example 173 F Euler frequencies based on a Cartesian angular velocity
vector.
The vector E
J $ E , that indicates the angular velocity of a rigid body E
with respect to the global frame J written in frame E, is related to the
Euler frequencies by
E
J $E
E
J $E
UH H
J $E
6 5
sin sin #
${
= 7 $ | 8 = 7 sin cos #
cos $}
=
E
5
cos #
sin #
0
65
6
*b
0
0 8 7 b 8 (5.106)
1
#b
The matrix of coe!cients is not an orthogonal matrix because,
E
E
E
W
UH
6=
W
UH
= 7
1
UH
=
E
5
1
UH
(5.107)
6
sin sin # sin cos # cos cos #
sin #
0 8
0
0
1
5
sin #
cos #
1 7
sin cos #
sin sin #
sin cos sin # cos cos #
(5.108)
6
0
0 8
1
(5.109)
5. F Applied Kinematics
251
It is because the Euler angles coordinate frame H (R*#) is not an orthogonal frame. For the same reason, the matrix of coe!cients that relates the
Euler frequencies and the components of J
J $E
J
J $E
J
J $E
J
UH H
J $E
6 5
65
6
*b
0 cos *
sin sin *
$[
= 7 $ \ 8 = 7 0 sin * cos * sin 8 7 b 8
1
0
cos $]
#b
=
(5.110)
5
(5.111)
is not an orthogonal matrix. Therefore, the Euler frequencies based on local
and global decomposition of the angular velocity vector J $ E must solely be
found by the inverse of coe!cient matrices
5
H
J $E
6
=
*b
7 b 8 =
#b
1 E
UH
J $E
5
sin #
1 7
sin cos #
sin cos sin #
E
cos #
sin sin #
cos cos #
65
6
(5.112)
0
${
0 8 7 $ | 8 (5.113)
$}
1
and
5
H
J $E
6
=
*b
7 b 8 =
#b
1 J
UH
(5.114)
J $E
6
5
65
cos sin * cos cos * 1
$[
1 7
sin cos *
sin sin * 0 8 7 $ \ 8 (5.115)
sin sin *
cos *
0
$]
J
Using (5.112) and (5.114), it can be veried that the transformation ma1
would be the same as Euler transformation matrix
trix E UJ = E UH J UH
(5.78).
The angular velocity vector can thus be expressed as
6
5
6
5
£
¤ $[
£
¤ ${
=
ˆ~ ˆ n̂ 7 $ | 8 = Lˆ Mˆ N̂ 7 $ \ 8
J$E
$}
$]
5
6
£
¤ *b
=
(5.116)
N̂ ĥ n̂ 7 b 8
#b
5.5 General Transformation
Consider a general situation in which two coordinate frames, J(R[\ ])
and E(R{|}) with a common origin R, are employed to express the components of a vector r. There is always a transformation matrix J UE to
map the components of r from the coordinate frame E(R{|}) to the other
coordinate frame J(R[\ ]).
5. F Applied Kinematics
252
J
r = J UE E r
(5.117)
1 J
In addition, the inverse map, E r = J UE
r, can be done by E UJ
E
r = E UJ J r
(5.118)
When the coordinate frames E and J are orthogonal, the rotation matrix
UE is called an orthogonal matrix. The transpose UW and inverse U1 of
an orthogonal matrix [U] are equal:
J
1
E
W
U
= J UE
= J UE
¯J J¯
¯E
¯
¯ UE ¯ = ¯ UJ ¯ = 1
(5.119)
(5.120)
Lˆ = (Lˆ · ˆ~)ˆ~ + (Lˆ · ˆ)ˆ
 + (Lˆ · n̂)n̂
Mˆ = (Mˆ · ˆ~)ˆ~ + (Mˆ · ˆ)ˆ
 + (Mˆ · n̂)n̂
(5.121)
Because of the matrix orthogonality condition, only three of the nine elements of J UE are independent.
Proof. Decomposition of the unit vectors of J(R[\ ]) along the axes of
E(R{|})
N̂
= (N̂ · ˆ~)ˆ~ + (N̂ · ˆ)ˆ
 + (N̂ · n̂)n̂
(5.122)
(5.123)
introduces the transformation matrix J UE to map the local frame to the
global frame
5 ˆ 6 5 ˆ
L · ˆ~
L
7 Mˆ 8 = 7 Mˆ · ˆ~
65
6
5
6
Lˆ · ˆ Lˆ · n̂
ˆ~
ˆ~
Mˆ · ˆ Mˆ · n̂ 8 7 ˆ 8 = J UE 7 ˆ 8
n̂
n̂
N̂ · ˆ~ N̂ · ˆ N̂ · n̂
N̂
(5.124)
where
J
UE
5 ˆ
6
L · ˆ~ Lˆ · ˆ Lˆ · n̂
= 7 Mˆ · ˆ~ Mˆ · ˆ Mˆ · n̂ 8
N̂ · ˆ~ N̂ · ˆ N̂ · n̂
5
ˆ ~) cos(L>
ˆ ˆ) cos(L>
ˆ n̂) 6
cos(L>ˆ
ˆ ~) cos(M>
ˆ ˆ) cos(M>
ˆ n̂) 8
= 7 cos(M>ˆ
(5.125)
cos(N̂> ˆ~) cos(N̂> ˆ) cos(N̂> n̂)
The elements of J UE are direction cosines of the axes of J(R[\ ]) in
frame E(R{|}). This set of nine direction cosines then completely species
the orientation of the frame E(R{|}) in the frame J(R[\ ]), and can be
used to map the local coordinates of any point ({> |> }) to its corresponding
global coordinates ([> \> ]).
5. F Applied Kinematics
253
Alternatively, using the method of unit vector decomposition to develop
the matrix E UJ leads to
E
E
1 J
UJ J r = J UE
r
5
6
ˆ~ · Lˆ ˆ~ · Mˆ ˆ~ · N̂
= 7 ˆ · Lˆ ˆ · Mˆ ˆ · N̂ 8
n̂ · Lˆ n̂ · Mˆ n̂ · N̂
5
6
ˆ
ˆ
cos(ˆ~> L)
cos(ˆ~> M)
cos(ˆ~> N̂)
ˆ cos(ˆ
ˆ cos(ˆ
= 7 cos(ˆ
> L)
> M)
> N̂) 8
ˆ
ˆ
cos(n̂> L) cos(n̂> M) cos(n̂> N̂)
r =
UJ
E
(5.126)
(5.127)
and shows that the inverse of a transformation matrix is equal to the transpose of the transformation matrix,
J
1
W
UE
= J UE
(5.128)
A matrix with condition (5.128) is called orthogonal. Orthogonality of
U comes from the fact that it maps an orthogonal coordinate frame to
another orthogonal coordinate frame.
The transformation matrix U has only three independent elements. The
constraint equations among the elements of U will be found by applying
the orthogonality condition (5.128).
J
5
u11
7 u21
u31
u12
u22
u32
65
u11
u13
u23 8 7 u12
u33
u13
W
UE · J UE
= L
(5.129)
6
5
6
1 0 0
u21 u31
u22 u32 8 = 7 0 1 0 8 (5.130)
0 0 1
u23 u33
Therefore, the dot product of any two dierent rows of J UE is zero, and
the dot product of any row of J UE with the same row is one.
2
2
2
+ u12
+ u13
u11
2
2
2
u21
+ u22
+ u23
2
2
2
u31
+ u32
+ u33
u11 u21 + u12 u22 + u13 u23
u11 u31 + u12 u32 + u13 u33
u21 u31 + u22 u32 + u23 u33
=
=
=
=
=
=
1
1
1
0
0
0
(5.131)
These relations are also true for columns of J UE , and evidently for rows
and columns of E UJ . The orthogonality condition can be summarized in
the following equation:
r̂Kl · r̂Km = r̂WKl r̂Km =
3
X
l=1
ulm uln = mn
(m> n = 1> 2> 3)
(5.132)
254
5. F Applied Kinematics
where ulm is the element of row l and column m of the transformation matrix
U, and mn is the Kronecker’s delta
½
1 m=n
mn =
(5.133)
0 m 6= n
Equation (5.132) gives six independent relations satised by nine direction
cosines. As a result, there are only three independent direction cosines.
The independent elements of the matrix U cannot be in the same row or
column, or any diagonal.
The determinant of a transformation matrix is equal to one,
¯J ¯
¯ UE ¯ = 1
(5.134)
It is because of Equation (5.129), and noting that
¯J
¯ ¯J ¯ ¯J W ¯ ¯J ¯ ¯J ¯ ¯J ¯2
W¯
¯ UE · J UE
= ¯ UE ¯ · ¯ UE ¯ = ¯ UE ¯ · ¯ UE ¯ = ¯ UE ¯ = 1 (5.135)
Using linear algebra and row vectors r̂K1 > r̂K2 > and r̂K3 of J UE , we know
that
¯J ¯
¯ UE ¯ = r̂WK · (r̂K2 × r̂K3 )
(5.136)
1
and¯ because
¯ the coordinate system is right handed, we have r̂K2 ×r̂K3 = r̂K1
so ¯J UE ¯ = r̂WK1 · r̂K1 = 1.
Example 174 Elements of transformation matrix.
The position vector r of a point S may be expressed in terms of its
components with respect to either J (R[\ ]) or E (R{|}) frames. If J r =
100Lˆ50Mˆ+150N̂, and we are looking for components of r in the E (R{|})
frame, then we have to nd the proper rotation matrix E UJ rst. Assume
that the angle between the { and [ axes is 40 deg, and the angle between
the | and \ axes is 60 deg.
The row elements of E UJ are the direction cosines of the E (R{|}) axes
in the J (R[\ ]) coordinate frame. The {-axis lies in the [] plane at
40 deg from the [-axis, and the angle between | and \ is 60 deg. Therefore,
5
6 5
6
cos 40
0
sin 40
ˆ~ · Lˆ ˆ~ · Mˆ ˆ~ · N̂
E
UJ = 7 ˆ · Lˆ ˆ · Mˆ ˆ · N̂ 8 = 7 ˆ · Lˆ cos 60 ˆ · N̂ 8
n̂ · Lˆ
n̂ · Mˆ n̂ · N̂
n̂ · Lˆ n̂ · Mˆ n̂ · N̂
5
6
0=766
0
0=643
ˆ
7
ˆ · L
0=5 ˆ · N̂ 8
=
(5.137)
ˆ
n̂ · L n̂ · Mˆ n̂ · N̂
W
=L
and by using E UJ J UE = E UJ E UJ
65
6 5
6
5
0=766 u21 u31
1 0 0
0=766 0 0=643
7 u21
0=5 u23 8 7 0
0=5 u32 8 = 7 0 1 0 8
0 0 1
u31
u32
u33
0=643 u23 u33
(5.138)
5. F Applied Kinematics
255
we obtain a set of six equations to nd the missing elements.
0=766 u21 + 0=643 u23
0=766 u31 + 0=643 u33
2
2
u21
+ u23
+ 0=25
u21 u31 + 0=5u32 + u23 u33
2
2
2
u31
+ u32
+ u33
=
=
=
=
=
0
0
1
0
1
(5.139)
Solving these equations provides the following transformation matrix:
5
6
0=766
0
0=643
E
0=5 0=663 8
UJ = 7 0=557
(5.140)
0=322 0=866 0=383
and then we can nd the components of E r.
E
UJ J r
65
6 5
6
0=766
0
0=643
100
173=05
0=5 0=663 8 7 50 8 = 7 68=75 8 (5.141)
= 7 0=557
0=322 0=866 0=383
150
18=05
r =
E
5
Example 175 Global position, using E r and E UJ .
The position vector r of a point S may be expressed in either J (R[\ ])
or E (R{|}) frames. If E r = 100ˆ~ 50ˆ
 + 150n̂, and E UJ is the transforJ
E
mation matrix to map r to r
5
6
0=766
0
0=643
E
0=5 0=663 8 J r
r = E UJ J r = 7 0=557
(5.142)
0=322 0=866 0=383
then the components of J r in J (R[\ ]) would be
J
W E
UE E r = E UJ
r
65
6 5
6
0=766 0=557 0=322
100
0=45
0=5
0=866 8 7 50 8 = 7 104=9 8 (5.143)
= 7 0
0=643 0=663 0=383
150
154=9
r =
J
5
Example 176 Two points transformation matrix.
The global position vector of two points, S1 and S2 , of a rigid body E are
5
6
5
6
1=077
0=473
J
J
rS1 = 7 1=365 8
rS2 = 7 2=239 8
(5.144)
2=666
0=959
The origin of the body E (R{|}) is xed on the origin of J (R[\ ]), and
the points S1 and S2 are lying on the local {-axis and |-axis respectively.
256
5. F Applied Kinematics
To nd J UE , we use the local unit vectors Jˆ~ and J ˆ
5
6
5
6
0=338
0=191
J
J
rS
rS
J
J
ˆ~ = J 1 = 7 0=429 8
ˆ = J 2 = 7 0=902 8
| rS1 |
|
rS2 |
0=838
0=387
(5.145)
to obtain J n̂
J
n̂
= ˆ~ × ˆ = ˜~ ˆ
5
65
6
0
0=838 0=429
0=191
0
0=338 8 7 0=902 8
= 7 0=838
0=429 0=338
0
0=387
5
6
0=922
= 7 0=029 8
0=387
(5.146)
where ˜~ is the skew-symmetric matrix corresponding to ˆ~, and ˜~ ˆ is an
alternative for ˆ~ × ˆ.
Hence, the transformation matrix using the coordinates of two points
J
rS1 and J rS2 would be
£ J J J ¤
J
UE =
ˆ~
ˆ
n̂
5
6
0=338 0=191 0=922
0=029 8
= 7 0=429 0=902
(5.147)
0=838 0=387
0=387
Example 177 Length invariant of a position vector.
Describing a vector in dierent frames utilizing rotation matrices does
not aect the length and direction properties of the vector. Therefore, the
length of a vector is an invariant
¯ ¯ ¯ ¯
|r| = ¯J r¯ = ¯E r¯
(5.148)
The length invariant property can be shown by
£
¤W
2
|r| = J rW J r = J UE E r J UE E r
=
E W J
r
W J
UE
UE E r = E rW E r
(5.149)
Example 178 F Skew symmetric matrices for ˆ~, ˆ, and n̂.
The denition of skew symmetric matrix d̃ corresponding to a vector a
is dened by
5
6
0
d3 d2
0
d1 8
d̃ = 7 d3
(5.150)
d2 d1
0
Hence,
5
6
5
6
5
6
0 0 1
0 1 0
0 0 0
˜~ = 7 0 0 1 8 m̃ = 7 0 0 0 8 ñ = 7 1 0 0 8 (5.151)
1 0 0
0 0 0
0 1 0
5. F Applied Kinematics
257
Example 179 Inverse of Euler angles rotation matrix.
Precession-nutation-spin or Euler angle rotation matrix (5.78)
E
UJ
= U}># U{> U}>*
5
f*f# fv*v#
f#v* + ff*v#
= 7 f*v# ff#v* v*v# + ff*f#
vv*
f*v
6
vv#
vf# 8
f
(5.152)
must be inverted to be a transformation matrix to map body coordinates to
global coordinates.
J
UE
=
E
5
1
W
W
W
UJ
= U}>*
U{>
U}>#
f*f# fv*v#
= 7 f#v* + ff*v#
vv#
6
f*v# ff#v* vv*
v*v# + ff*f# f*v 8
vf#
f
(5.153)
The transformation matrix (5.152) is called a local Euler rotation matrix,
and (5.153) is called a global Euler rotation matrix.
Example 180 F Alternative proof for transformation matrix.
Starting with an identity
5
6
£
¤ ˆ~
ˆ~ ˆ n̂ 7 ˆ 8 = 1
n̂
we may write
6
5
5 ˆ 6 5 ˆ 6
L
L
¤ ˆ~
£
7 Mˆ 8 = 7 Mˆ 8 ˆ~ ˆ n̂ 7 ˆ 8
n̂
N̂
N̂
(5.154)
(5.155)
Because the matrix multiplication can be performed in any order, we nd
5 ˆ 6 5 ˆ
65
6
5
6
L · ˆ~ Lˆ · ˆ Lˆ · n̂
ˆ~
ˆ~
L
7 Mˆ 8 = 7 Mˆ · ˆ~ Mˆ · ˆ Mˆ · n̂ 8 7 ˆ 8 = J UE 7 ˆ 8
(5.156)
n̂
n̂
N̂
N̂ · ˆ~ N̂ · ˆ N̂ · n̂
where,
5 ˆ 6
L
£
¤
J
UE = 7 Mˆ 8 ˆ~ ˆ n̂
N̂
Following the same method we can show that
5
6
ˆ~ £
¤
E
UJ = 7 ˆ 8 Lˆ Mˆ N̂
n̂
(5.157)
(5.158)
258
5. F Applied Kinematics
5.6 Local and Global Rotations
The global rotation matrix J UE is the inverse of the local rotation matrix
E
UJ and vice versa,
J
1
UE = E UJ
E
1
UJ = J UE
(5.159)
The premultiplication of the global rotation matrix is equal to postmultiplication of the local rotation matrix.
Proof. Let us show rotation matrices about the global axes by [T] and
rotation matrices about the local axes by [D]. Consider a sequence of global
rotations and their resultant global rotation matrix J TE to transform a
position vector E r to J r:
J
r = J TE E r
(5.160)
The global position vector J r can also be transformed to E r using a local
rotation matrix E DJ :
E
r = E DJ J r
(5.161)
Combining Equations (5.160) and (5.161) yields
TE E DJ J r
E
DJ J TE E r
J
J
r =
E
r =
(5.162)
(5.163)
and hence,
J
TE E DJ = E DJ J TE = I
(5.164)
Therefore, the global and local rotation matrices are inverse of each other:
J
J
TE
T1
E
=
=
E
E
D1
J
DJ
(5.165)
(5.166)
Assume that
J
E
TE
DJ
= Tq · · · T3 T2 T1
= Dq · · · D3 D2 D1
(5.167)
(5.168)
then
J
E
TE
DJ
=
=
E
J
1 1 1
1
D1
J = D1 D2 D3 · · · Dq
1
1 1 1
TE = T1 T2 T3 · · · T1
q
(5.169)
(5.170)
and Equation (5.164) becomes
Tq · · · T2 T1 Dq · · · D2 D1 = Dq · · · D2 D1 Tq · · · T2 T1 = I
(5.171)
Therefore,
Tq · · · T3 T2 T1
Dq · · · D3 D2 D1
1 1
1
= D1
1 D2 D3 · · · Dq
1 1
1
= T1
1 T2 T3 · · · Tq
(5.172)
(5.173)
5. F Applied Kinematics
259
or
1 1
1
T1
1 T2 T3 · · · Tq Tq · · · T3 T2 T1
1 1
1
D1
1 D2 D3 · · · Dq Dq · · · D3 D2 D1
= I
= I
(5.174)
(5.175)
Hence, the eect of order of rotations about the global coordinate axes
is equivalent to the eect of the same rotations about the local coordinate
axes performed in reverse order:
J
E
TE
DJ
1 1
1
= D1
1 D2 D3 · · · Dq
1 1 1
= T1 T2 T3 · · · T1
q
(5.176)
(5.177)
Therefore the global rotation matrix J UE is the inverse of the local rotation
matrix E UJ and vice versa,
J
1
UE = E UJ
E
1
UJ = J UE
(5.178)
Example 181 Postmultiplication of rotation matrix.
Assume that the local position of a point S after a rotation is at E r =
£
¤W
1 2 3
. If the local rotation matrix to transform J r to E r is given
as
5
6 5
6
cos * sin * 0
cos 30 sin 30 0
E
U}>* = 7 sin * cos * 0 8 = 7 sin 30 cos 30 0 8
(5.179)
0
0
1
0
0
1
then we may nd the global position vector J r by postmultiplying E U}>*
and the local position vector E rW ,
5
6
cos 30 sin 30 0
£
¤
J W
r
= E rW E U}>* = 1 2 3 7 sin 30 cos 30 0 8
0
0
1
£
¤
0=13 2=23 3
=
(5.180)
1
instead of premultiplying of E U}>*
by E r:
J
r =
E
5
1 E
U}>*
r
65 6 5
6
cos 30 sin 30 0
1
0=13
= 7 sin 30 cos 30 0 8 7 2 8 = 7 2=23 8
0
0
1
3
3
(5.181)
5.7 Axis-angle Rotation
The nial orientation of a rigid body after a nite number of rotations is
equivalent to a unique rotation about a unique axis. Determination of the
260
5. F Applied Kinematics
Z
z
I
T
B
G
u
y
O
M
Y
X
x
FIGURE 5.6. Axis-angle rotation when the rotation axix x̂ is coincident with the
local }-axis.
angle and axis is called the orientation kinematics of rigid bodies. Assume
the body frame E(R{|}) rotate ! about a xed line in the global frame
J(R[\ ]). The line is indicated by a unit vector x̂ with directional cosines
x1 > x2 > x3 ,
x̂ = x1 Lˆ + x2 Mˆ + x3 N̂
q
x21 + x22 + x23 = 1
(5.182)
(5.183)
Two parameters are needed to dene the axis of rotation that goes through
R and one parameter is needed to dene the amount of rotation about the
axis. This is called the axis-angle representation of a rotation. An axis-angle
rotation needs three independent parameters to be dened.
The axis-angle transformation matrix J UE that transforms the coordinates of the body frame E(R{|}) to the associated coordinates in the
global frame J(R[\ ]),
J
r = J UE E r
(5.184)
is
J
UE = Ux̂>! = I cos ! + x̂x̂W vers ! + x̃ sin !
(5.185)
6
x21 vers ! + f!
x1 x2 vers ! x3 v! x1 x3 vers ! + x2 v!
J
x22 vers ! + f!
x2 x3 vers ! x1 v! 8
UE = 7 x1 x2 vers ! + x3 v!
x1 x3 vers ! x2 v! x2 x3 vers ! + x1 v!
x23 vers ! + f!
(5.186)
where
!
vers ! = yhuvlqh ! = 1 cos ! = 2 sin2
(5.187)
2
5
5. F Applied Kinematics
261
and x̃ is the skew-symmetric matrix associated with the vector x̂,
6
5
0
x3 x2
0
x1 8
(5.188)
x̃ = 7 x3
x2 x1
0
A matrix x̃ is skew-symmetric if
x̃W = x̃
(5.189)
For any transformation matrix J UE , we may determine the axis x̂ and
angle ! to provide the same matrix by
¢
1 ¡J
W
(5.190)
UE J UE
x̃ =
2 sin !
¢
1 ¡ ¡J ¢
(5.191)
tr UE 1
cos ! =
2
Equation (5.186) is called the angle-axis or axis-angle rotation matrix
and is the most general transformation matrix for rotation of a body frame
E in a global frame J. If the axis of rotation (5.182) coincides with a global
coordinate axis ], \ , or [, then Equation (5.186) reduces to the principal
local rotation matrices.
Proof. The rotation ! about an axis x̂ is equivalent to a sequence of rotations about the axes of a body frame E such that the local frame is rst
rotated to bring one of its axes, say the }-axis, into coincidence with the
rotation axis x̂ followed by a rotation ! about that local axis, then the
reverse of the rst sequence of rotations.
Figure 5.6 illustrates an axis of rotation x̂ = x1 Lˆ + x2 Mˆ + x3 N̂ , the
global frame J (R[\ ]), and the rotated body frame E (R{|}) when the
local }-axis is coincident with x̂. Assume that the body and global frames
were coincident initially. When we apply a sequence of rotations * about
the }-axis and about the |-axis on the body frame E (R{|}), the local
}-axis will become coincident with the rotation axis x̂. Now, we apply the
rotation ! about x̂ and then perform the sequence of rotations * and backward. Following Equation (5.169), the rotation matrix J UE to map
the coordinates of a point in the body frame to its coordinates in the
global frame after rotation ! about x̂ is
J
UE
=
E
1
W
UJ
= E UJ
= Ux̂>!
= [U}>* U|> U}>! U|> U}>* ]W
W
W
W
W
W
= U}>*
U|>
U}>!
U|>
U}>*
(5.192)
Substituting equations
x2
sin * = p 2
x + x22
p 1
sin = x21 + x22
sin sin * = x2
x1
cos * = p 2
x1 + x22
cos = x3
sin cos * = x1
(5.193)
5. F Applied Kinematics
262
in J UE will provide us with the axis-angle rotation matrix
J
UE = Ux̂>! =
(5.194)
5
6
2
x1 vers ! + f!
x1 x2 vers ! x3 v! x1 x3 vers ! + x2 v!
7 x1 x2 vers ! + x3 v!
x22 vers ! + f!
x2 x3 vers ! x1 v! 8
x1 x3 vers ! x2 v! x2 x3 vers ! + x1 v!
x23 vers ! + f!
The matrix (5.194) can be decomposed to
5
6
6
5
1 0 0
x1 £
Ux̂>! = cos ! 7 0 1 0 8 + (1 cos !) 7 x2 8 x1
0 0 1
x3
5
6
0
x3 x2
0
x1 8
+ sin ! 7 x3
x2 x1
0
x2
x3
¤
(5.195)
which can be summarized as (5.185). Showing the rotation matrix by its
elements J UE = Ux̂>! = [ulm ], we have
ulm = lm cos ! + xl xm (1 cos !) lmn xn sin !
(5.196)
The axis-angle rotation equation (5.185) is also called the Rodriguez rotation formula. We can show that the Rodriguez rotation formula may also
be expressed by any of the following equivalent forms:
Ux̂>!
Ux̂>!
Ux̂>!
Ux̂>!
Ux̂>!
!
= I + x̃ sin ! + 2x̃2 sin2
2
μ
¶
!
!
!
= I + 2x̃ sin
I cos + x̃ sin
2
2
2
= I + x̃ sin ! + x̃2 vers !
£
¤
= Ix̂x̂W cos ! + x̃ sin ! + x̂x̂W
= I + x̃2 + x̃ sin ! x̃2 cos !
(5.197)
(5.198)
(5.199)
(5.200)
(5.201)
The inverse of an angle-axis rotation is
E
W
UJ = J UE
= Ux̂>! = I cos ! + x̂x̂W vers ! x̃ sin !
(5.202)
It means that the orientation of E in J when E is rotated ! about x̂ is
the same as the orientation of J in E when E is rotated ! about x̂. The
rotation Ux̂>! is also called the reverse rotation.
We may examine Equations (5.190) and (5.191) by direct substitution
5
6
0
2x3 sin ! 2x2 sin !
J
W
0
2x1 sin ! 8
UE J UE
= 7 2x3 sin !
2x2 sin ! 2x1 sin !
0
5
6
0
x3 x2
0
x1 8 = 2x̃ sin ! (5.203)
= 2 sin ! 7 x3
x2 x1
0
5. F Applied Kinematics
and
tr
¡J
UE
¢
263
= u11 + u22 + u33
= 3 cos ! + x21 (1 cos !) + x22 (1 cos !) + x23 (1 cos !)
¡
¢
= 3 cos ! + x21 + x22 + x23 x21 + x22 + x23 cos !
= 2 cos ! + 1
(5.204)
The axis of rotation x̂ is also called the Euler axis or the eigenaxis of
rotation.
Example 182 F Axis-angle rotation when x̂ = N̂.
If the local frame E (R{|}) rotates about the ]-axis, then
x̂ = N̂
(5.205)
and the transformation matrix (5.186) reduces to
5
6
0 vers ! + cos ! 0 vers ! 1 sin ! 0 vers ! + 0 sin !
J
UE = 7 0 vers ! + 1 sin ! 0 vers ! + cos ! 0 vers ! 0 sin ! 8
0 vers ! 0 sin ! 0 vers ! + 0 sin ! 1 vers ! + cos !
5
6
cos ! sin ! 0
= 7 sin ! cos ! 0 8
(5.206)
0
0
1
which is equivalent to the rotation matrix about the ]-axis of the global
frame.
Example 183 F Rotation about a rotated local axis.
If the body coordinate frame E (R{|}) rotates * deg about the global ]axis, then the {-axis would be along
x̂{
=
J
5
U]>* ˆ~
65 6 5
6
cos * sin * 0
1
cos *
= 7 sin * cos * 0 8 7 0 8 = 7 sin * 8
0
0
1
0
0
(5.207)
Rotation about x̂{ = (cos *) Lˆ + (sin *) Mˆ is dened by Rodriguez’s
formula (5.186)
6
5
cos2 * vers + cos cos * sin * vers sin * sin J
Ux̂{ > = 7 cos * sin * vers sin2 * vers + cos cos * sin 8
sin * sin cos * sin cos (5.208)
Now, rotation * about the global ]-axis followed by rotation about the
local {-axis is transformed by
J
UE
=
J
5
Ux̂{ > J U]>*
6
cos * cos sin *
sin sin *
= 7 sin * cos cos * cos * sin 8
0
sin cos (5.209)
264
5. F Applied Kinematics
W
W
that must be equal to [U{> U}>* ]1 = U}>*
U{>
.
Example 184 F Axis and angle of a rotation matrix.
A body coordinate frame, E, undergoes three Euler rotations (*> > #) =
(30> 45> 60) deg with respect to a global frame J. The rotation matrix to
transform coordinates of E to J is
J
UE
W
W
W
W
W
UJ
= [U}># U{> U}>* ] = U}>*
U{>
U}>#
6
0=126 83 0=926 78 0=353 55
= 7 0=780 33 0=126 83 0=612 37 8
0=612 37 0=353 55
0=707 11
=
E
5
(5.210)
The unique angle-axis of rotation for this rotation matrix can then be found
by Equations (5.190) and (5.191).
¶
μ
¢
1 ¡ ¡J ¢
tr UE 1 = cos1 (0=146 45) = 98 deg (5.211)
! = cos1
2
¢
¡J
1
W
UE J UE
2 sin !
5
6
0=0
0=862 85 0=130 82
0=0
0=488 22 8
= 7 0=862 85
0=130 82 0=488 22
0=0
5
6
0=488 22
x̂ = 7 0=130 82 8
0=862 85
x̃ =
(5.212)
(5.213)
As a double check, we may verify the angle-axis rotation formula and derive
the same rotation matrix.
J
UE
= Ux̂>! = I cos ! + x̂x̂W vers ! + x̃ sin !
5
6
0=126 82 0=926 77 0=353 54
= 7 0=780 32 0=126 83 0=612 37 8
0=612 36 0=353 55
0=707 09
(5.214)
5.8 Rigid Body Motion
Consider a rigid body with a local coordinate frame E (r{|}) that is moving
in a global coordinate frame J(R[\ ]). The rigid body can rotate in J,
while point r of E can translate relative to the origin R of J, as is shown
in Figure 5.7. The vector J d indicates the position of the moving origin r
relative to the xed origin R. The coordinates of a body point S in local
and global frames are related by
J
rS = J UE E rS + J d
(5.215)
5. F Applied Kinematics
Z
265
z
z
\
T
G
y
y d
o
B
O
M
X
Y
x
x
FIGURE 5.7. Rotation and translation of a body frame.
where
5
6
[S
J
r = 7 \S 8
]S
5
6
{S
E
r = 7 |S 8
}S
5
6
[r
J
d = 7 \r 8
]r
(5.216)
The vector J d is called the displacement or translation of E with respect to
J, and J UE is the rotation matrix that transforms E r to J r when J d = 0.
Such a combination of a rotation and translation in Equation (5.215) is
called the rigid body motion in which the conguration of the body can be
expressed by the position of the origin r of E and the orientation of E.
Decomposition of a rigid motion into a rotation and a translation is
the practical method for expressing spatial displacement of rigid bodies.
We show the translation by a vector and the rotation by the orthogonal
Cartesian transformation matrix.
Proof. Consider a body frame E that is initially coincident with a globally
xed frame J. Figure 5.7 illustrates a translated and rotated frame E in J.
The most general rotation is represented by the Rodriguez rotation formula,
(5.185)
J
UE = Ux̂>! = I cos ! + x̂x̂W vers ! + x̃ sin !
(5.217)
All points of the body have the same displacement for the translation J d.
Therefore, translation of a rigid body is independent of the local position
vector E r. Because of that, we can represent the most general displacement
of a rigid body by a rotation, and a translation:
¡
¢
J
r = I cos ! + x̂x̂W vers ! + x̃ sin ! E r + J d = J UE E r + J d (5.218)
Equation (5.218) indicates that the most general displacement of a rigid
body is a rotation about an axis and a translation along an axis. The choice
5. F Applied Kinematics
266
of the point of reference r is arbitrary; however, when this point is chosen
and the body coordinate frame is set up, the rotation and translation are
determined.
Based on translation and rotation, the conguration of a body can be
uniquely determined by six independent parameters: three translation components [r , \r , ]r and three rotational components. If a body moves in
such a way that its rotational components remain constant, the motion is
called a pure translation; if it moves in such a way that [r , \r , and ]r
remain constant, the motion is called a pure rotation. Therefore, a rigid
body has three translational and three rotational degrees of freedom.
Example 185 Rotation and translation of a body coordinate frame.
A body coordinate frame E(r{|}) that is originally coincident with global
coordinate frame J(R[\ ]) rotates 60 deg about the [-axis and translates
£
¤W
£
¤W
to 3 4 5
. The global position of a point at E r = { | }
is
J
UE E r + J d
6
1
0
0
5
6 5 6
: {
3
9
sin : 7
9 0 cos
= 9
3
3 : | 8+7 4 8
7
8 }
5
0 sin
cos
3Ã
3
!
s
¶
μ
s
1
3
1
ˆ
= ({ + 3) L +
|
} + 4 Mˆ +
} + 3| + 5 N̂ (5.219)
2
2
2
r =
J
5
Example 186 Composition of rigid body motion.
Consider a rigid motion of body E1 with respect to body E2 , and then, a
rigid motion of body E2 with respect to frame J such that
2
r =
J
r =
2
U1 1 r + 2 d1
J
U2 2 r + J d2
(5.220)
(5.221)
These two motions may be combined to determine a rigid motion that transforms 1 r to J r:
¡
¢
J
r = J U2 2 U1 1 r + 2 d1 + J d2
=
=
J
U2 2 U1 1 r+ J U2 2 d1 + J d2
J
U1 1 r + J d1
(5.222)
Therefore,
J
U1
J
d1
=
=
J
U2 2 U1
J
U2 2 d1 + J d2
(5.223)
(5.224)
which shows that the transformation from frame E1 to frame J can be done
by rotation J U1 and translation J d1 .
5. F Applied Kinematics
Example 187 Rotation of a translated rigid body.
Point S of a rigid body E has the initial position vector E rS .
5 6
1
E
rS = 7 2 8
3
267
(5.225)
If the body rotates 45 deg about the {-axis, and then translates to J d =
£
¤W
3 2 1
, the nal position of S would be
J
r =
E
5
W
E
U{>45
rS + J d
(5.226)
6W 5
6 5 6 5
6
0
1
3
4
:
sin 4 8 7 2 8 + 7 2 8 = 7 1=29 8
3
1
4=53
cos 4
1
0
9 0 cos = 7
4
0 sin 4
5.9 Angular Velocity
Consider a rotating rigid body E(R{|}) with a xed point R in a reference
frame J(R[\ ]). The motion of the body can be expressed by a time
varying rotation transformation matrix between the global and body frames
to map the instantaneous coordinates of any xed point in body frame E
into their coordinates in the global frame J
J
r(w) = J UE (w) E r
(5.227)
The velocity of a body point in the global frame is
J
rb (w) = J v(w) = J Ub E (w) E r = J $̃ E J r(w) = J $ E × J r(w)
(5.228)
where J $ E is the angular velocity vector of E with respect to J. It is equal
to a rotation with angular rate !b about an instantaneous axis of rotation
x̂.
5
6
$1
$ = 7 $ 2 8 = !b x̂
(5.229)
$3
The angular velocity vector is associated with a skew symmetric matrix
J $̃ E called the angular velocity matrix
5
0
$̃ = 7 $ 3
$ 2
$ 3
0
$1
6
$2
$ 1 8
0
(5.230)
where
J $̃ E =
W
UE J UE
= !b x̃
J b
(5.231)
268
5. F Applied Kinematics
Proof. Consider a rigid body with a xed point R and an attached frame
E(R{|}). The body frame E is initially coincident with the global frame
J. Therefore, the position vector of a body point S is
J
r(w0 ) = E r
(5.232)
The global time derivative of J r is
J
v
=
=
J
¤
g J
g £J
r(w) =
UE (w) E r
gw
gw
J
¤
g £J
UE (w) J r(w0 ) = J Ub E (w) E r
gw
J
J
rb =
(5.233)
Eliminating E r between (5.227) and (5.233) determines the velocity of the
point in global frame
J
W
v = J Ub E (w) J UE
(w) J r(w)
(5.234)
We denote the coe!cient of J r(w) by $̃
J b
W
UE J UE
(5.235)
v = J $̃ E J r(w)
(5.236)
v = J $ E × J r(w)
(5.237)
J $̃ E =
and write Equation (5.234) as
J
or as
J
W
= I, inThe time derivative of the orthogonality condition, J UE J UE
troduces an important identity
J b
W
W
UE J UE
+ J UE J Ub E
=0
(5.238)
W
which can be utilized to show that J $̃ E = [J Ub E J UE
] is a skew-symmetric
matrix because
h
iW
J
W
W
UE J Ub E
= J Ub E J UE
(5.239)
The vector J
J $ E is called the instantaneous angular velocity of the body E
relative to the global frame J as seen from the J frame.
Since a vectorial equation can be expressed in any coordinate frame, we
may use either of the following expressions for the velocity of a body point
in body or global frames
J
J vS
E
J vS
=
=
J
J
J $ E × rS
E
E
rS
J $E ×
(5.240)
(5.241)
5. F Applied Kinematics
269
where J
J vS is the global velocity of point S expressed in the global frame
and E
v
J S is the global velocity of point S expressed in the body frame.
J
J
J
UE E
UE
J vS =
J vS =
¡E
J $E ×
E
rS
¢
(5.242)
J
E
J vS and J vS can be converted to each other using a rotation matrix
E
J vS
=
=
W J
J W
UE
UE J $̃ E J
J rS
J vS =
J W J
J W J b
J W J b
UE UE UE J rS = UE UE E
J rS
J
(5.243)
showing that
E
J W J b
UE UE
J $̃ E =
(5.244)
which is called the instantaneous angular velocity of E relative to the global
frame J as seen from the E frame. From the denitions of J $̃ E and E
J $̃ E
we are able to transform the two angular velocity matrices by
J $̃ E
E
J $̃ E
=
=
J W
UE E
UE
J $̃ E
J
J W J
UE J $̃ E UE
(5.245)
(5.246)
J
UE
J $̃ E
E
J
UE J $̃ E
J
UE E
J $̃ E
(5.247)
(5.248)
(5.249)
J
or equivalently
J b
UE
J b
UE
J
$̃
U
J E
E
=
=
=
The angular velocity of E in J is negative of the angular velocity of J
in E if both are expressed in the same coordinate frame.
J
J $̃ E
E
J $̃ E
= J
E $̃ J
= E
E $̃ J
(5.250)
(5.251)
J $ E can always be expressed in the form
J $ E = $x̂
(5.252)
where x̂ is a unit vector parallel to J $ E and indicates the instantaneous
axis of rotation.
Example 188 Rotation of a body point about a global axis.
Consider a rigid body that is turning about the ]-axis with b = 10 deg @v.
The global velocity of a point S (5> 30> 10), when the body is turned =
5. F Applied Kinematics
270
30 deg, is
J
J b
(5.253)
UE (w) E rS
35
64 5
6
cos sin 0
5
J
g C7
sin cos 0 8D 7 30 8
=
gw
0
0
1
10
5
65
6
sin cos 0
5
= b 7 cos sin 0 8 7 30 8
0
0
0
10
65
6 5
6
5
5
4=97
sin 6 cos 6 0
10 7
cos 6
sin 6 0 8 7 30 8 = 7 1=86 8
=
180
0
0
0
10
0
=
vS
at this moment, point S is at
J
rS
J
UE E rS
cos 6 sin 6
cos 6
= 7 sin 6
0
0
=
5
65
6 5
6
0
5
10=67
0 8 7 30 8 = 7 28=48 8
1
10
10
(5.254)
Example 189 Rotation of a global point about a global axis.
£
¤W
. When it is turned
A point S of a rigid body is at E rS = 5 30 10
= 30 deg about the ]-axis, the global position of S is
J
rS
UE E rS
cos 6 sin 6
7
sin 6
cos 6
=
0
0
=
J
5
65
6
5
6
(5.255)
0
5
10=67
0 8 7 30 8 = 7 28=48 8
1
10
10
If the body is turning with b = 10 deg @ s, the global velocity of the point S
will be
J
vS
J b
W J
rS
UE J UE
5
f 6
v
10 7 6
f6
v 6
=
180
0
0
5
6
4=97
= 7 1=86 8
0
=
0
0
0
65
f 6
8 7 v
6
0
v 6
f 6
0
(5.256)
6
10=67
0
0 8 7 28=48 8
10
1
6W 5
5. F Applied Kinematics
271
Example 190 F Principal angular velocities.
The principal rotational matrices about the axes [, \ , and ] are
5
6
5
6
1
0
0
cos 0 sin 0
1
0 8
U\> = 7
U[> = 7 0 cos sin 8
0 sin cos sin 0 cos 5
6
cos sin 0
U]> = 7 sin cos 0 8
(5.257)
0
0
1
and hence, their time derivatives are
5
6
0
0
0
Ub [> = b 7 0 v f 8
0 f v
5
6
v f 0
Ub ]> = b 7 f v 0 8
0
0
0
5
v
Ub \> = b 7 0
f
6
0 f
0
0 8
0 v
(5.258)
Therefore, the principal angular velocity matrices about axes [, \ , and ]
are
5
6
0 0 0
W
b
b 7 0 0 1 8
(5.259)
J $̃ [ = U[> U[> = 0 1 0
5
6
0 0 1
W
b7 0 0 0 8
b
(5.260)
J $̃ \ = U\> U\> = 1 0 0
5
6
0 1 0
W
b
b7 1 0 0 8
(5.261)
J $̃ ] = U]> U]> = 0 0 0
which are equivalent to
˜
bL
J $̃ [ = J $̃ \
= b M˜
b Ñ
J $̃ ] = (5.262)
and therefore, the principal angular velocity vectors are
J $[
J$\
J$]
= $ [ Lˆ = b Lˆ
= $ \ Mˆ = b Mˆ
= $ ] N̂ = b N̂
(5.263)
(5.264)
(5.265)
Utilizing the same technique, we can nd the following principal angular
velocity matrices about the local axes:
5
6
0 0 0
E
W
b 7 0 0 1 8 = #b ˜~
b
(5.266)
J $̃ { = U{># U{># = #
0 1 0
272
5. F Applied Kinematics
5
6
1
0 8 = b ˜
0
6
0
0 8 = *b ñ
0
0 0
E
W b
b7 0 0
J $̃ | = U|> U|> = 1 0
5
0 1
E
W
b
7
1
0
$̃
=
U
=
*
b
U
J }
}>* }>*
0 0
(5.267)
(5.268)
Example 191 Decomposition of an angular velocity vector.
Every angular velocity vector can be decomposed to three principal angular velocity vectors.
³
´
³
´
³
´
ˆ ˆ
ˆ ˆ
=
(5.269)
J$E
J $ E · L L + J $ E · M M + J $ E · N̂ N̂
= $ [ Lˆ + $ \ Mˆ + $ ] N̂ = b Lˆ + b Mˆ + b N̂
= $[ + $\ + $]
Example 192 Combination of angular velocities.
Starting from a combination of rotations
0
U2 = 0 U1 1 U2
(5.270)
and taking a time derivative, we nd
0 b
U2 = 0 Ub 1 1 U2 + 0 U1 1 Ub 2
(5.271)
Now, substituting the derivative of rotation matrices with
0 b
0 b
U2 = 0 $̃ 2 0 U2
U1 = 0 $̃ 1 0 U1
1 b
U2 = 1 $̃ 2 1 U2
(5.272)
U1 1 U2 + 0 U1 1 $̃ 2 1 U2
0
0 W 0
1
0 $̃ 1 U2 + U1 1 $̃ 2 U1 U1 U2
0
0
0
0 $̃ 1 U2 + 1 $̃ 2 U2
(5.273)
U1 1 $̃ 2 0 U1W = 01 $̃ 2
(5.274)
0
0 $̃ 2 = 0 $̃ 1 + 1 $̃ 2
(5.275)
results in
0 $̃ 2
0
U2
0 $̃ 1
=
=
=
0
0
where
0
Therefore, we nd
which indicates that the angular velocities may be added relatively:
0
0 $2 = 0 $1 + 1 $2
(5.276)
This result also holds for any number of angular velocities
0
0
0
0 $ q = 0 $ 1 + 1 $ 2 + 2 $ 3 + · · · + q1 $ q =
q
X
l=1
0
l1 $ l
(5.277)
5. F Applied Kinematics
273
Example 193 F Angular velocity in terms of Euler frequencies.
The angular velocity vector can be expressed by Euler frequencies. Therefore,
b + #ĥ
b #
= $ {ˆ~ + $ | ˆ + $ } n̂ = *ĥ
b * + ĥ
5
6
5
6
5 6
sin sin #
cos #
0
= *b 7 sin cos # 8 + b 7 sin # 8 + #b 7 0 8
cos 0
1
5
65
6
*b
sin sin # cos # 0
7
8
7
sin cos # sin # 0
b 8
=
(5.278)
cos 0
1
#b
E
J $E
and
6
*b sin sin # + b cos #
1 E
E 1 7
= E UJ
UJ
*b sin cos # b sin # 8
J $E =
*b cos + #b
5
6
65
*b
0 cos *
sin sin *
= 7 0 sin * cos * sin 8 7 b 8
(5.279)
1
0
cos #b
5
J
J $E
where the inverse of the Euler transformation matrix is
5
f*f# fv*v#
E 1
UJ = 7 f#v* + ff*v#
vv#
6
f*v# ff#v* vv*
v*v# + ff*f# f*v 8
vf#
f
(5.280)
Example 194 F Angular velocity in terms of rotation frequencies.
Consider the Euler angles transformation matrix:
E
UJ = U}># U{> U}>*
(5.281)
The angular velocity matrix is then equal to
E $̃ J
E b E W
UJ UJ
μ
¶
gU}>#
gU}>* b
gU{>
=
*b U}># U{>
+ U}>#
U}>* + #b
U{> U}>*
gw
gw
gw
=
W
× (U}># U{> U}>* )
gU}>* W
gU{> W W
W
W
U}>* U{>
U{> U}>#
= *b U}># U{>
U}>#
+ b U}>#
gw
gw
gU}># W
+#b
(5.282)
U}>#
gw
274
5. F Applied Kinematics
which, in matrix form, is
5
6
0
cos sin cos #
0
sin sin # 8
b 7 cos E $̃ J = *
sin cos # sin sin #
0
5
6
5
6
0
0
sin #
0 1 0
0
0
cos # 8 + #b 7 1 0 0 8 (5.283)
+b 7
sin # cos #
0
0 0 0
or
5
#b + *f
b
0
b *vv#
f#
b
0
b
7
$̃
=
# *f
b
E J
b + *vf#
v#
b
The corresponding angular velocity vector is
6
5 b
f# + *vv#
b
b + *vf#
8
7 v#
b
E $J = #b + *f
b
5
sin sin # cos #
= 7 sin cos # sin #
cos 0
6
b *vf#
v#
b
b + *vv#
8
f#
b
0
(5.284)
65
6
*b
0
0 8 7 b 8
1
#b
(5.285)
However,
E
E $̃ J
E
E $J
= E
J $̃ E
= E
J $E
(5.286)
(5.287)
and therefore,
5
sin sin #
E
7
sin
cos #
$
=
J E
cos cos #
sin #
0
65
6
*b
0
0 8 7 b 8
1
#b
(5.288)
Example 195 F Coordinate transformation of angular velocity.
Angular velocity 1 $ 2 of coordinate frame E2 with respect to E1 and expressed in E1 can be expressed in base coordinate frame E0 according to
0
U1 1 $̃ 2 0 U1W = 01 $̃ 2
(5.289)
To show this equation, it is enough to apply both sides on an arbitrary
vector 0 r. Therefore, the left-hand side would be
0
U1 1 $̃ 2 0 U1W 0 r =
=
=
U1 1 $̃ 2 1 U0 0 r = 0 U1 1 $̃ 2 1 r
¢
¡
0
U1 1 $ 2 × 1 r = 0 U1 1 $ 2 × 0 U1 1 r
0
0
0
1 $2 × r
(5.290)
5. F Applied Kinematics
275
which is equal to the right-hand side after applying on the vector 0 r
0
0
0
0
1 $̃ 2 r = 1 $ 2 × r
(5.291)
Example 196 F Time derivative of unit vectors.
Using Equation (5.241) we can dene the time derivative of unit vectors
of a body coordinate frame E(ˆ~> ˆ> n̂), rotating in the global coordinate frame
ˆ M>
ˆ N̂)
J(L>
J
gˆ~
~
= E
J $E × ˆ
gw
J
gˆ

ˆ
= E
J $E × 
gw
J
gn̂
= E
J $ E × n̂
gw
(5.292)
5.10 F Time Derivative and Coordinate Frames
The time derivative of a vector depends on the coordinate frame in which
we are taking the derivative. The time derivative of a vector r in the global
frame is called the J-derivative and is denoted by
J
g
r
gw
while the time derivative of the vector in the body frame is called the
E-derivative and is denoted by
E
g
r
gw
The left superscript on the derivative symbol indicates the frame in which
the derivative is taken, and hence, its unit vectors are considered constant.
Time derivative is called simple if the vector is expressed in the same
coordinate frame that we are taking the derivative, because the unit vectors
are constant and scalar coe!cients are the only time variables. The simple
derivatives of E rS in E and J rS in J are
E
gE
rS
gw
J
gJ
rS
gw
=
E
rb S = E vS = {ˆ
b ~ + |b ˆ + }b n̂
(5.293)
=
J
rb S = J vS = [b Lˆ + \b Mˆ + ]b N̂
(5.294)
It is also possible to nd the J-derivative of E rS and the E-derivative
of J rS . We dene the J-derivative of a body vector E rS by
E
J vS =
J
gE
rS
gw
(5.295)
and similarly, a E-derivative of a global vector J rS by
J
E vS =
E
gJ
rS
gw
(5.296)
5. F Applied Kinematics
276
When point S is moving in frame E while E is rotating in J, the Jderivative of E rS (w) is dened by
J
gE
E
rS (w) = E rb S + E
rS = E
bS
J $E ×
Jr
gw
(5.297)
and the E-derivative of J rS is dened by
E
gJ
rS (w) = J rb S J $ E × J rS = J
bS
Er
gw
(5.298)
ˆ M,
ˆ and N̂ be the global coProof. Let J(R[\ ]) with unit vectors L,
ordinate frame, and let E(R{|}) with unit vectors ˆ~, ˆ, and n̂ be a body
coordinate frame. The position vector of a moving point S , can be expressed
in the body and global frames
E
J
rS (w) = { (w) ˆ~ + | (w) ˆ + } (w) n̂
rS (w) = [ (w) Lˆ + \ (w) Mˆ + ] (w) N̂
(5.299)
(5.300)
The time derivative of E rS in E and J rS in J are
E
gE
rS
gw
J
gJ
rS
gw
=
E
rb S = E vS = {ˆ
b ~ + |b ˆ + }b n̂
(5.301)
=
J
rb S = J vS = [b Lˆ + \b Mˆ + ]b N̂
(5.302)
because the unit vectors of E in Equation (5.299) and the unit vectors of
J in Equation (5.300) are considered constant.
Using Equation (5.241) for the global velocity of a body xed point S ,
expressed in body frame
E
E
E
rS =
J vS = J $ E ×
J
gE
rS
gw
(5.303)
and denition (5.295), we can nd the J-derivative of the position vector
rS as
E
J
gE
rS
gw
´
g³
{ˆ~ + |ˆ
 + } n̂
gw
J
=
J
J
J
gˆ~
gˆ

gn̂
+|
+}
gw
gw
gw
E
rb S + { E
~+|E
ˆ+ } E
J $E × ˆ
J $E × 
J $ E × n̂
³
´
E
rb S + E
~ + |ˆ
 + } n̂
J $ E × {ˆ
= {ˆ
b ~ + |b ˆ + }b n̂ + {
=
=
=
E
E
rb S + E
rS =
J $E ×
E
gE
E
rS + E
rS
J $E ×
gw
(5.304)
5. F Applied Kinematics
277
We achieved this result because the {, |, and } components of E rS are
scalar. Scalars are invariant with respect to frame transformations. Therefore, if { is a scalar then,
J
E
g
g
{=
{ = {b
gw
gw
(5.305)
The E-derivative of J rS can be found similarly
E
gJ
rS
gw
´
g³ ˆ
[ L + \ Mˆ + ] N̂
gw
E
=
E ˆ
E ˆ
E
gL
gM
gN̂
= [b Lˆ + \b Mˆ + ]b N̂ + [
+\
+]
gw
gw
gw
J
= J rb S + J
rS
(5.306)
E $J ×
and therefore,
E
gJ
rS = J rb S J $ E × J rS
(5.307)
gw
The angular velocity of E relative to J is a vector quantity and can be
expressed in either frames.
J
J $E
E
J $E
= $ [ Lˆ + $ \ Mˆ + $ ] N̂
(5.308)
= $ {ˆ~ + $ | ˆ + $ } n̂
(5.309)
Example 197 F Time derivative of a moving point in E.
Consider a local frame E that is rotating in J by b about the ]-axis,
and a moving point at E rS (w) = wˆ~ . Therefore,
J
rS
UE E rS = U]> (w) E rS
65 6
cos sin 0
w
= 7 sin cos 0 8 7 0 8
0
0
1
0
= w cos Lˆ + w sin Mˆ
=
J
5
(5.310)
The angular velocity matrix is
J $̃ E =
J b
W
= b Ñ
UE J UE
(5.311)
that gives
b N̂
J $E = (5.312)
E
J W J
UE J $̃ E J UE = b ñ
J $̃ E =
(5.313)
It can also be veried that
5. F Applied Kinematics
278
and therefore,
E
b n̂
J $E = (5.314)
Now we can nd the following derivatives:
E
gE
rS = E rb S = ˆ~
gw
(5.315)
J
gJ
rS = J rb S = (cos wb sin ) Lˆ + (sin + wb cos ) Mˆ
gw
For the mixed derivatives, we start with
J
(5.316)
E
gE
rS
gw
gE
E
rS + E
rS
J $E ×
gw
5 6
5 6 5 6 5
6
1
0
w
1
= 7 0 8 + b 7 0 8 × 7 0 8 = 7 wb 8
0
1
0
0
=
= ˆ~ + wˆ
b = E
bS
Jr
(5.317)
which is the global velocity of S expressed in E. We may, however, transform E
b S to the global frame and nd the global velocity expressed in J.
Jr
J
rb S
UE E
bS
Jr
65
6 5
6
cos sin 0
1
cos wb sin = 7 sin cos 0 8 7 wb 8 = 7 sin + wb cos 8
0
0
1
0
0
= (cos wb sin ) Lˆ + (sin + wb cos ) Mˆ
(5.318)
=
J
5
The next derivative is
E
gJ
rS
gw
rb S J $ E × J rS
5
6
5 6 5
6
cos wb sin 0
w cos = 7 sin + wb cos 8 b 7 0 8 × 7 w sin 8
0
1
0
5
6
cos = 7 sin 8 = (cos ) Lˆ + (sin ) Mˆ = J
bS
(5.319)
Er
0
=
J
which is the velocity of S relative to E and expressed in J. To express this
velocity in E we apply a frame transformation
E
rb S
=
J
5
W J
UE
bS
Er
6W 5
6 5 6
cos sin 0
cos 1
= 7 sin cos 0 8 7 sin 8 = 7 0 8 = ˆ~
0
0
1
0
0
(5.320)
5. F Applied Kinematics
279
Sometimes it is more applied if we transform the vector to the same frame in
which we are taking the derivative and then apply the dierential operator.
Therefore,
¢
g ¡J
UE E rS
gw 5
6 5
6
w cos cos wb sin J
g 7
w sin 8 = 7 sin + wb cos 8
gw
0
0
J
J
gE
rS
gw
=
=
and
5 6 5 6
w
1
E ¡
gJ
g J W J ¢ Eg 7 8 7 8
0 = 0
rS =
UE rS =
gw
gw
gw
0
0
(5.321)
E
(5.322)
Example 198 F Orthogonality of position and velocity vectors.
If the position vector of a body point in global frame is denoted by r then
gr
·r=0
gw
(5.323)
To show this property we may take a derivative from
r · r = u2
(5.324)
and nd
g
gr
gr
gr
(r · r) =
·r+r·
=2
·r=0
(5.325)
gw
gw
gw
gw
Equation (5.323) is correct in every coordinate frame and for every constant length vector, as long as the vector and the derivative are expressed
in the same coordinate frame.
Example 199 F Derivative transformation formula.
The global velocity of a xed point in the body coordinate frame E (R{|})
can be found by Equation (5.228). Now consider a point S that can move
in E (R{|}). In this case, the body position vector E rS is not constant, and
therefore, the global velocity of such a point expressed in E is
J
E
gE
gE
E
rS =
rS + E
rS = E
bS
J $E ×
Jr
gw
gw
(5.326)
Sometimes the result of Equation (5.326) is utilized to dene transformation of the dierential operator from a body to a global coordinate frame
J
E
g
g
E
b
¤= E
¤=
¤+ E
J $E ×
J¤
gw
gw
(5.327)
b
The nal result is E
J ¤ showing the global (J) time derivative expressed in
body frame (E). The vector ¤ might be any vector such as position, velocity,
280
5. F Applied Kinematics
angular velocity, momentum, angular velocity, or even a time-varying force
vector.
Equation (5.327) is called the derivative transformation formula
and relates the time derivative of a vector as it would be seen from frame
J to its derivative as seen in frame E. The derivative transformation formula (5.327) is general and can be applied to every vector for derivative
transformation between every two relatively moving coordinate frames.
Example 200 F Dierential equation for rotation matrix.
Equation (5.231) for dening the angular velocity matrix may be written
as a rst-order dierential equation
g J
UE J UE J $̃ E = 0
gw
(5.328)
The solution of the equation conrms the exponential denition of the rotation matrix as
J
UE = h$̃w
(5.329)
or
¢
¡
$̃w = !b x̃ = ln J UE
(5.330)
Example 201 F Acceleration of a body point in the global frame.
The angular acceleration vector of a rigid body E(R{|}) in the global
frame J(R[\ ]) is denoted by J E and is dened as the global time derivative of J $ E .
J
g
(5.331)
J E =
J $E
gw
Using this denition, the acceleration of a xed body point in the global
frame is
J
¢
g ¡
J
J
J
J $ E × rS = J E × rS + J $ E × ( J $ E × rS ) (5.332)
gw
J
aS =
Example 202 F Alternative denition of angular velocity vector.
The angular velocity vector of a rigid body E(ˆ~> ˆ> n̂) in global frame
ˆ M>
ˆ N̂) can also be dened by
J(L>
E
~(
J $E = ˆ
J
J
J
gˆ

gn̂
gˆ~
· n̂) + ˆ(
· ˆ~) + n̂(
· ˆ)
gw
gw
gw
(5.333)
Proof. Consider a body coordinate frame E moving with a xed point in
the global coordinate frame J. The xed point of the body is taken as the
origin of both coordinate frames, as shown in Figure 5.8. To describe the
motion of the body, it is su!cient to describe the motion of the local unit
vectors ˆ~, ˆ, n̂ . Let rS be the position vector of a body point S . Then, E rS
is a vector with constant components.
E
rS = {ˆ~ + |ˆ
 + } n̂
(5.334)
5. F Applied Kinematics
Z
281
B
z
k
i
G
P
rP
j
x
X
y
Y
FIGURE 5.8. A body coordinate frame moving with a xed point in the global
coordinate frame.
When the body moves, it is only the unit vectors ˆ~, ˆ, and n̂ that vary
relative to the global coordinate frame. Therefore, the vector of dierential
displacement is
grS = { gˆ~ + | gˆ
 + } gn̂
(5.335)
which can also be expressed by
³
´
grS = (grS · ˆ~) ˆ~ + (grS · ˆ) ˆ + grS · n̂ n̂
Substituting (5.335) in the right-hand side of (5.336) yields
³
´
grS =
{ˆ~ · gˆ~ + |ˆ~ · gˆ
 + }ˆ~ · gn̂ ˆ~
³
´
+ {ˆ
 · gˆ~ + |ˆ
 · gˆ
 + }ˆ
 · gn̂ ˆ
³
´
+ {n̂ · gˆ~ + | n̂ · gˆ
 + } n̂ · gn̂ n̂
(5.336)
(5.337)
Utilizing the unit vectors’ relationships
the grS reduces to
grS
=
ˆ · gˆ~ = ˆ~ · gˆ

(5.338)
ˆ~ · gn̂
= n̂ · gˆ~
(5.340)
ˆ~ · ˆ = ˆ · n̂ = n̂ · ˆ~ = 0
(5.342)
n̂ · gˆ
 = ˆ
 · gn̂
(5.339)
ˆ~ · gˆ~ = ˆ · gˆ
 = n̂ · gn̂ = 0
(5.341)
ˆ~ · ˆ~ = ˆ · ˆ = n̂ · n̂ = 1
(5.343)
³
´
³
´
}ˆ~ · gn̂ |ˆ
 · gˆ~ ˆ~ + {ˆ
 · gˆ~ } n̂ · gˆ
 ˆ
³
´
+ | n̂ · gˆ
 {ˆ~ · gn̂ n̂
(5.344)
282
5. F Applied Kinematics
This equation can be rearranged to be expressed as a vector product
³
´ ³
´
grS = (n̂ · gˆ
)ˆ~ + (ˆ~ · gn̂)ˆ
 + (ˆ
 · gˆ~)n̂ × {ˆ~ + |ˆ
 + } n̂
(5.345)
or
E
bS =
Jr
Ã
J
J
J
gˆ

gn̂
gˆ~
(n̂ ·
)ˆ~ + (ˆ~ ·
)ˆ
 + (ˆ
·
)n̂
gw
gw
gw
!
³
´
× {ˆ~ + |ˆ
 + } n̂
(5.346)
Comparing this result with
rb S = J $ E × rS
shows that
E
~
J $E = ˆ
Ã
!
¶
μJ
¶
J
gˆ

gn̂
gˆ~
· n̂ + ˆ
· ˆ~ + n̂
· ˆ
gw
gw
gw
μJ
(5.347)
(5.348)
Example 203 F Alternative proof for angular velocity denition (5.333).
The angular velocity denition presented in Equation (5.333) can also be
shown by direct substitution for J UE in the angular velocity matrix E
J $̃ E
J W J b
E
UE UE
J $̃ E =
(5.349)
Therefore,
5
E
J $̃ E
6
5 ˆ
6
ˆ~ · Lˆ ˆ~ · Mˆ ˆ~ · N̂
L · ˆ~ Lˆ · ˆ Lˆ · n̂
J
g
7 Mˆ · ˆ~ Mˆ · ˆ Mˆ · n̂ 8
= 7 ˆ · Lˆ ˆ · Mˆ ˆ · N̂ 8 ·
gw
ˆ
ˆ
n̂ · L n̂ · M n̂ · N̂
N̂ · ˆ~ N̂ · ˆ N̂ · n̂
Ã
! 6
5 μ
¶
μ
¶
J
J
J
gˆ~
gˆ

gn̂
ˆ~ ·
ˆ~ ·
:
9 ˆ~ ·
:
9
gw
gw
gw
:
9
:
9 μ
Ã
!
¶
μ
¶
:
9
J
J
J
gˆ~
gˆ

gn̂
:
9
(5.350)
= 9 ˆ ·
ˆ ·
ˆ ·
:
:
9
gw
gw
gw
:
9
9 μ
¶ μ J ¶ Ã J ! :
9
J
gˆ~
gˆ

gn̂ :
8
7
n̂ ·
n̂ ·
n̂ ·
gw
gw
gw
which shows that
5 μJ
¶ 6
gˆ

· n̂ :
9
gw
9
:
9 Ã
! :
9 J gn̂
:
9
:
E
$
=
·
ˆ
~
9
:
E
J
9
:
gw
9
:
9 μJ
¶ :
7
8
gˆ~
· ˆ
gw
(5.351)
5. F Applied Kinematics
283
Example 204 F Second derivative.
In general, J g r@gw is a variable vector in J(R[\ ]) and in any other
coordinate frame such as E (r{|}). Therefore, it can be dierentiated in
either coordinate frames J or E. However, the order of dierentiating is
important. In general,
E J
J E
g gr
g gr
6=
(5.352)
gw gw
gw gw
As an example, consider a rotating body coordinate frame about the ]-axis,
and a variable vector as
J
r = wLˆ
(5.353)
Therefore,
J
gr
= J rb = Lˆ
gw
(5.354)
and hence,
E
μJ
gr
gw
¶
5
65 6
cos * sin * 0
1
h i
E
W
b = U]>*
Lˆ = 7 sin * cos * 0 8 7 0 8
Jr
0
0
1
0
=
= cos *ˆ~ sin *ˆ

(5.355)
which provides
E
g J gr
= *b sin *ˆ~ *b cos *ˆ

gw gw
μE J ¶
g gr
J
= *b Mˆ
gw gw
and
Now
E
that provides
h i
W
r = U]>*
wLˆ = w cos *ˆ~ w sin *ˆ

(5.356)
(5.357)
(5.358)
E
gr
= (w*b sin * + cos *) ˆ~ (sin * + w*b cos *) ˆ
gw
(5.359)
and
J
μE
gr
gw
¶
which shows
=
J
b = U]>* ((w*b sin * + cos *) ˆ~ (sin * + w*b cos *) ˆ)
Er
5
65
6
cos * sin * 0
w*b sin * + cos *
= 7 sin * cos * 0 8 7 sin * w*b cos * 8
0
0
1
0
ˆ
ˆ
= L w*b M
J
E J
g E gr
g gr
= (*b + w¨
*) Mˆ 6=
gw gw
gw gw
(5.360)
284
5. F Applied Kinematics
z
B
Z
B
rP
P
x
G
dB
G
X
G
rP
y
Y
FIGURE 5.9. A rigid body with an attached coordinate frame E (r{|}) moving
freely in a global coordinate frame J(R[\ ]).
5.11 Rigid Body Velocity
Consider a rigid body with an attached local coordinate frame E (r{|})
moving freely in a xed global coordinate frame J(R[\ ]), as shown in
Figure 5.9. The rigid body can rotate in the global frame, while the origin of
the body frame E can translate relative to the origin of J. The coordinates
of a body point S in local and global frames are related by the following
equation:
J
rS = J UE E rS + J dE
(5.361)
where J dE indicates the position of the moving origin r relative to the
xed origin R.
The velocity of the point S in J is
J
vS
Jb
rb S = J Ub E E rS + J db E = J $̃ E J
dE
E rS +
¡J
¢
J
Jb
rS dE + dE
J $̃ E
¢
¡J
rS J dE + J db E
(5.362)
J$E ×
J
=
=
=
Proof. Direct dierentiating shows
J
J
¢
g J
g ¡J
rS = J rb S =
UE E rS + J dE
gw
gw
J b E
UE rS + J db E
J
vS
=
=
(5.363)
The local position vector E rS can be substituted from (5.361) to obtain
¡
¢
J
W J
vS = J Ub E J UE
rS J dE + J db E
¡
¢
= J $̃ E J rS J dE + J db E
¢
¡
= J $ E × J rS J dE + J db E
(5.364)
5. F Applied Kinematics
285
It may also be written using relative position vector
J
Jb
vS = J $ E × J
E rS + dE
(5.365)
Example 205 Geometric interpretation of rigid body velocity.
Consider a body point S of a moving rigid body. The global velocity of
point S
J
Jb
vS = J $ E × J
E rS + dE
is a vector addition of rotational and translational velocities, both expressed
in the global frame. At the moment, the body frame is assumed to be coincident with the global frame, and the body frame has a velocity J db E with
respect to the global frame. The translational velocity J db E is a common
property for every point of the body, but the rotational velocity J $ E × J
E rS
diers for dierent points of the body.
Example 206 Velocity of a moving point in a moving body frame.
Assume that point S in Figure 5.9 is moving in frame E, indicated by
time varying position vector E rS (w). The global velocity of S is a composition of the velocity of S in E, rotation of E relative to J, and velocity of
E relative to J.
J
J ¡
¢
g J
g J
rS =
dE + J UE E rS
gw
gw
J
J ¡
¢
gJ
g J
=
dE +
UE E rS
gw
gw
J
= J db E + J
r
b
(5.366)
E S + J $ E × E rS
Example 207 Velocity of a body point in multiple coordinate frames.
Consider three frames, E0 , E1 and E2 , as shown in Figure 5.10. Let us
analyze the velocity of point S . If the point is stationary in a frame, say
E2 , then the time derivative of 2 rS in E2 is zero. If frame E2 is moving
relative to frame E1 , then, the time derivative of 1 rS is a combination
of the rotational component due to rotation of E2 relative to E1 and the
velocity of E2 relative to E1 . In forward velocity kinematics, the velocities
must be measured in the base frame E0 . Therefore, the velocity of point S
in the base frame is a combination of the velocity of E2 relative to E1 and
the velocity of E1 relative to E0 .
The global coordinate of the body point S is
0
rS = 0 d1 + 01 d2 + 02 rS = 0 d1 + 0 U1 1 d2 + 0 U2 2 rS
(5.367)
Therefore, the velocity of point S can be found by combining the relative
velocities
0
rb S
=
=
d1 + ( 0 Ub 1 1 d2 + 0 U1 1 db 2 ) + 0 Ub 2 2 rS
0b
d1 + 00 $ 1 × 01 d2 + 0 U1 1 db 2 + 00 $ 2 × 02 rS
0b
(5.368)
5. F Applied Kinematics
286
z
B0
B1
B2
Z
z
2
1
rP
P
x
d2
y
0
rP
y
0
d1
x
Y
X
FIGURE 5.10. A rigid body coordinate frame E2 is moving in a frame E1 that
is moving in the base coordinate frame E0 .
It is usually easier to use the relative velocity method and write
0
0
0
0
0 vS = 0 v1 + 1 v2 + 2 vS
(5.369)
because
0
0b
0 v1 = 0 d1
0
0
0
0
1b
1 v2 = 0 $ 1 × 1 d2 + U1 d2
0
0
0
2 vS = 0 $ 2 × 2 rS
(5.370)
and therefore,
0
vS = 0 db 1 + 00 $ 1 × 01 d2 + 0 U1 1 db 2 + 00 $ 2 × 02 rS
(5.371)
Example 208 Velocity vectors in dierent coordinate frames.
To express the velocity vectors in dierent coordinate frames we need
only to premultiply them by a rotation matrix. Hence, considering nm vl as
the velocity of the origin of the El coordinate frame with respect to the
origin of frame Em expressed in frame En , we can write
n
n
m vl = l vm
(5.372)
p
n
n
m vl = Up m vl
(5.373)
gl
rS = l vS = lm vS + ll $ m × lm rS
gw l
(5.374)
and
and therefore,
l
5. F Applied Kinematics
287
Example 209 F Zero velocity points.
To answer whether there is always a point with zero velocity, we may
utilize Equation (5.364) and write
¡J
¢
r0 J dE + J db E = 0
(5.375)
J $̃ E
to search for J r0 which refers to a point with zero velocity
J
Jb
dE
r0 = J dE J $̃ 1
E
(5.376)
however, the skew symmetric matrix J $̃ E is singular and has no inverse.
In other words, there is no general solution for Equation (5.375).
If we restrict ourselves to planar motions, say [\ -plane, then J $ E =
$ N̂ and J $̃ 1
E = 1@$. Hence, in 2D space there is a point at any time with
zero velocity at position J r0 given by
J
r0 (w) = J dE (w) 1 Jb
dE (w)
$
(5.377)
The zero velocity point is called the pole or instantaneous center of
rotation. The position of the pole is generally a function of time and the
path of its motion is called a centroid.
Example 210 F Eulerian and Lagrangian view points.
When a variable quantity is measured within the stationary global coordinate frame, it is called absolute or the Lagrangian viewpoint. When the
variable is measured within a moving body coordinate frame, it is called
relative or the Eulerian viewpoint.
In 2D planar motion of a rigid body, there is always a pole of zero velocity
at
1 Jb
J
r0 = J dE (5.378)
dE
$
The position of the pole in the body coordinate frame can be found by substituting for J r from (5.361)
J
Jb
UE E r0 + J dE = J dE J $̃ 1
dE
E
(5.379)
and solving for the position of the zero velocity point in the body coordinate
frame E r0 .
h
i1
1 J b
Jb
E
W
W
J b J W
r 0 = J UE
dE = J UE
UE UE
dE
J $̃ E
h
i
1 J b
1 J b
W
J
= J UE
UE J Ub E
(5.380)
dE = J Ub E
dE
Therefore, J r0 indicates the path of motion of the pole in the global frame,
while E r0 indicates the same path in the body frame. The J r0 refers to the
Lagrangian centroid and E r0 refers to the Eulerian centroid.
5. F Applied Kinematics
288
5.12 Angular Acceleration
Consider a rotating rigid body E(R{|}) with a xed point R in a reference
frame J(R[\ ]). Equation (5.228), for the velocity vector of a point in a
xed origin body frame,
J
rb (w) = J v(w) = J $̃ E J r(w) = J $ E × J r(w)
(5.381)
can be utilized to nd the acceleration vector of the body point
J
¡
¢
gJ
rb (w) = J E × J r + J $ E × J $ E × J r (5.382)
³gw
´
¡
¢
¨ + !b ub × J r + !b 2 x̂ × x̂ × J r
(5.383)
=
!x̂
J
r̈ =
J E is the angular acceleration vector of the body with respect to the J
frame.
J
g
J $E
gw
Proof. Dierentiating Equation (5.381) gives
J E =
J
a =
=
and because
vb = J r̈ = J $
b E × J r + J $ E × J rb
¡
¢
J
r + J $E × J $E × J r
J E ×
(5.384)
J
b
$ = !x̂
¨ + !b ub
= !x̂
(5.385)
(5.386)
(5.387)
we derive Equation (5.383). Therefore, the position, velocity, and acceleration vectors of a body point are
E
J
J
=
aS
=
rS = {ˆ~ + |ˆ
 + } n̂
vS = J rb S =
(5.388)
J
gE
rS = J $ E × J r
gw
(5.389)
J 2
g E
rS = J E × J r + J $ E × J rb
gw2
J
J
r)
(5.390)
J E × r + J $ E × ( J $ E ×
J
vb S = J r̈S =
The angular acceleration expressed in the body frame is the body derivative
of the angular velocity vector. To show this, we use the derivative transport
formula (5.327)
E
J E =
J
E
E
gE
gE
gE
E
E
$E = E
b E (5.391)
J $E =
J $E + J $E × J $E =
J$
gw
gw
gw J
5. F Applied Kinematics
Y
289
y
B
X
l
G
g
I
m
x
FIGURE 5.11. A simple pendulum.
The angular acceleration of E in J can always be expressed in the form
J E = J E x̂
(5.392)
where x̂ is a unit vector parallel to J E . The angular velocity and angular
acceleration vectors are not parallel in general, and therefore,
x̂
J E
6= x̂$
6
=
bE
J$
(5.393)
(5.394)
However, the only special case is when the axis of rotation is xed in both
J and E frames. In this case
¨
b x̂ = ! x̂
J E = x̂ = $
(5.395)
Example 211 Velocity and acceleration of a simple pendulum.
A point mass attached to a massless rod and hanging from a revolute joint
is called a simple pendulum. Figure 5.11 illustrates a simple pendulum. A
local coordinate frame E is attached to the pendulum that rotates in a global
frame J. The position vector of the bob and the angular velocity vector J $ E
are
5
6
o sin !
E
J
r = oˆ~
r = J UE E r = 7 o cos ! 8
(5.396)
0
E
J W E
$ E = !b n̂
U
$ E = !b N̂
(5.397)
J$E =
J
E J
5. F Applied Kinematics
290
z
Z
Z
Z
Z
G
x
x
E
z
r
30 deg
T y
y
Y
x
X
z
(a)
Y
B
(b)
FIGURE 5.12. The motion of a vehicle at 30 deg latitude and heading north on
the Earth.
¡
¢
¡
¢
6
cos ¡ 32 + !¢ sin¡ 32 + !¢ 0
= 7 sin 32 + !
cos 32 + !
0 8
0
0
1
5
6
sin !
cos ! 0
= 7 cos ! sin ! 0 8
0
0
1
5
J
UE
(5.398)
Its velocity is therefore given by
E
Jv
=
J
=
v
E
b
b
rb + E
~ = o !ˆ
J $ E × J r = 0 + !n̂ × oˆ
5
6
o !b cos !
E
J
7
UE v = o !b sin ! 8
0
E
(5.399)
(5.400)
The acceleration of the bob is then equal to
E
Ja
J
=
a =
E
E
¨  + !b n̂ × o !ˆ
b  = o !ˆ
¨  o !b 2ˆ~
b+E
Jv
J $ E × J v = o !ˆ
6
5
2
(5.401)
J
(5.402)
UE
E
¨ cos ! o !b sin !
o!
:
9
a = 7 o!
¨ sin ! + o !b 2 cos ! 8
0
Example 212 Motion of a vehicle on the Earth.
Consider the motion of a vehicle on the Earth at latitude 30 deg and
heading north, as shown in Figure 5.12. The vehicle has the velocity y =
5. F Applied Kinematics
291
E
2
b = 80 km@ h = 22=22 m@ s and acceleration d = E
Hr
H r̈ = 0=1 m@ s , both with
respect to the road. The radius of the Earth is U, and hence, the vehicle’s
kinematics are
E
Hr
b
= Un̂ m
y
=
rad@ s
U
E
b = 22=22ˆ~ m@ s
Hr
¨ = d rad@ s2
U
E
~ m@ s2
H r̈=0=1ˆ
(5.403)
(5.404)
There are three coordinate frames involved. A body coordinate frame E is
attached to the vehicle. A global coordinate J is set up at the center of the
Earth. Another local coordinate frame H is rigidly attached to the Earth
and turns with the Earth. The frames H and J are assumed coincident at
the moment. The angular velocity of E is
´
³
E
J
E
b Lˆ
N̂
+
$
=
$
+
$
=
U
$
E
J
H
E
J
H
J
H
b
= ($ H cos ) ˆ~ + ($ H sin ) n̂ + ˆ
y
= ($ H cos ) ˆ~ + ($ H sin ) n̂ + ˆ
U
(5.405)
Therefore, the velocity and acceleration of the vehicle are
E
Jv
E
E
= E rb + E
J $ E × J r = 0 + J $ E × Un̂
= yˆ~ (U$ H cos ) ˆ
(5.406)
E
E
E
E
b + J $E × J v
Ja =
Jv
6 5
5
6
$ H cos y
³
´
y
:
9
= dˆ~ + U$ H b sin ˆ + 7
8 × 7 U$ H cos 8
U
0
$ H sin 5
6
2
d + U$ H cos sin 8
= 7
(5.407)
2U$ H b sin 1 2
y U$ 2H cos2 U
 is the
The term dˆ~ is the acceleration relative to Earth, (2U$ H b sin )ˆ
2
Coriolis acceleration, yU n̂ is the centrifugal acceleration due to traveling,
and (U$ 2H cos2 ) is the centrifugal acceleration due to Earth’s rotation.
Substituting the numerical values and accepting U = 6=3677 × 106 m provides
μ
¶
2
366=25
E
6
~ 6=3677 × 10
cos ˆ
J v = 22=22ˆ
24 × 3600 365=25
6
= 22=22ˆ~ 402=13ˆ
 m@ s
(5.408)
E
Ja
= 1=5662 × 102ˆ~ + 1=6203 × 103 ˆ 2=5473 × 102 n̂ m@ s2 (5.409)
292
5. F Applied Kinematics
Example 213 F Combination of angular accelerations.
It is shown that the angular velocity of several bodies rotating relative to
each other can be related according to (5.277)
0
0
0
0 $ q = 0 $ 1 + 1 $ 2 + 2 $ 3 + · · · + q1 $ q
(5.410)
The angular accelerations of several relatively rotating rigid bodies follow
the same rule:
0
0
0
0 q = 0 1 + 1 2 + 2 3 + · · · + q1 q
(5.411)
Let us consider a pair of relatively rotating rigid links in a base coordinate
frame E0 with a xed point at R. The angular velocities of the links are
related by
0
0 $2 = 0 $1 + 1 $2
(5.412)
So, their angular accelerations are
0
0 1
=
0 2
=
g
0 $1
gw
0
g
0
0 $ 2 = 0 1 + 1 2
gw
(5.413)
(5.414)
Example 214 F Angular acceleration and Euler angles.
The angular velocity E
J $ E in terms of Euler angles is
5
J
J $E
6 5
65
6
*b
$[
0 cos *
sin sin *
= 7 $ \ 8 = 7 0 sin * cos * sin 8 7 b 8
1
0
cos $]
#b
6
5 b
cos * + #b sin sin *
7
(5.415)
=
b sin * #b cos * sin 8
*b + #b cos The angular acceleration is then equal to
J
J E
J
gJ
(5.416)
J $E
³
´
³
´ 6
5gw
cos * ¨ + *b #b sin + sin * #̈ sin + b #b cos b *b
³
´
³
´ :
9
:
¨
b
b
b
b
= 9
sin
*
+
*
b
#
sin
+
cos
*
*
b
#̈
sin
#
cos
8
7
*
¨ + #̈ cos b #b sin =
The angular acceleration vector in the body coordinate frame is then equal
5. F Applied Kinematics
293
to
E
J E
W J
UE
(5.417)
J E
6
f*f# fv*v#
f#v* + ff*v# vv#
= 7 f*v# ff#v* v*v# + ff*f# vf# 8 J
J E
vv*
f*v
f
³
´
³
´ 6
5
cos # ¨ + *b #b sin + sin # *
¨ sin + b *b cos b #b
9
³
´
³
´ :
:
= 9
¨ sin + b *b cos b #b sin # ¨ + *b #b sin 8
7 cos # *
*
¨ cos #̈ b *b sin =
J
5
5.13 Rigid Body Acceleration
Consider a rigid body with an attached local coordinate frame E (r{|})
moving freely in a xed global coordinate frame J(R[\ ]). The rigid
body can rotate in the global frame, while the origin of the body frame E
can translate relative to the origin of J. The coordinates of a body point
S in local and global frames, as shown in Figure 5.13, are related by the
equation
J
rS = J UE E rS + J dE
(5.418)
where J dE indicates the position of the moving origin r relative to the
xed origin R.
The acceleration of point S in J is
J
aS
=
¢
¡
vb S = J r̈S = J E × J rS J dE
¢¢
¡
¡
+ J $ E × J $ E × J rS J dE + J d̈E
J
(5.419)
Proof. The acceleration of point S is a consequence of dierentiating the
velocity equation (5.364) or (5.365).
J
J
aS
g J
J
vS = J E × J
b S + J d̈E
E rS + J $ E × E r
gw
¢ J
¡
J
= J E × J
E rS + J $ E × J $ E × E rS + d̈E
¢
¡
= J E × J rS J dE
¢¢
¡
¡
+ J $ E × J $ E × J rS J dE + J d̈E
(5.420)
=
¢
¡
is called centripetal acceleration and is
The term J $ E × J $ E × J
E rS
independent of the angular acceleration. The term J E × J
E rS is called
r
.
tangential acceleration and is perpendicular to J
E S
5. F Applied Kinematics
Z (G
rP - Gd
B ))
294
z
B
D (G
rP - G
dB )
B
rP
Z
Z
P
x
G
dB
y
G
rP
G
X
Y
FIGURE 5.13. A rigid body with coordinate frame E (r{|}) moving freely in a
xed global coordinate frame J(R[\ ]).
Example 215 Acceleration of joint 2 of a 2R planar manipulator.
A 2R planar manipulator is illustrated in Figure 5.14. The elbow joint
has a circular motion about the base joint. Knowing that
b
0 $1 = 1
0
n̂0
(5.421)
we can write
¡
0 1
b 1 × 0 r1
0$
0 $1 × 0 $1 ×
0
r1
¢
=
=
=
b 1 = ¨1 0 n̂0
0$
¨1 0 n̂0 × 0 r1 = ¨1 U]>+90 0 r1
2
b 1 0 r1
(5.422)
(5.423)
(5.424)
and calculate the acceleration of the elbow joint
0
2
r̈1 = ¨1 U]>+90 0 r1 b 1 0 r1
(5.425)
Example 216 Acceleration of a moving point in a moving body frame.
Assume the point S in Figure 5.13 is indicated by a time varying local
position vector E rS (w). Then, the velocity and acceleration of S can be
found by applying the derivative transformation formula (5.327).
J
vS
=
=
Jb
E
dE + E rb S + E
rS
J $E ×
Jb
E
E
E
dE + vS + J $ E × rS
(5.426)
5. F Applied Kinematics
y2
295
x2
y1
l2
y0
T2
x1
T1
x0
l1
FIGURE 5.14. A 2R planar manipulator.
J
aS
=
=
J
E
d̈E + E r̈S + E
rb S + E
b E × E rS
J $E ×
J$
¡
¢
E
E
+E
rb S + E
rS
J $E ×
J $E ×
J
E
d̈E + E aS + 2 E
vS + E
b E × E rS
J $E ×
J$
¡
¢
E
E
+E
rS
J $E × J $E ×
(5.427)
It is also possible to take derivative from Equation (5.361) with the assumption E rb S 6= 0 and nd the acceleration of S .
J
J
J
r̈S
=
=
rb S
=
=
rS = J UE E rS + J dE
(5.428)
J b
UE E rS + J UE E rb S + J db E
E
E
J
J
rb S + J db E
J $ E × UE rS + UE
(5.429)
b E × J UE E rS + J $ E × J Ub E E rS + J $ E × J UE E rb S
J$
J b
+ UE E rb S + J UE E r̈S + J d̈E
¡
¢
J
J
bE × J
bS
J$
E rS + J $ E × J $ E × rS + 2 J $ E × E r
J
J
+ E r̈S + d̈E
(5.430)
The third term on the right-hand side is called the Coriolis acceleration.
The Coriolis acceleration is perpendicular to both J $ E and E rb S .
Example 217 F Acceleration of a body point.
Consider a rigid body is moving and rotating in a global frame. The
acceleration of a body point can be found by taking twice the time derivative
296
5. F Applied Kinematics
of its position vector
J
=
=
=
=
rS
J
rb S
J
r̈S
J
UE E rS + J dE
J b
UE E rS + J db E
J
ÜE E rS + J d̈E
¡
¢
J
W J
rS J dE + J d̈E
ÜE J UE
(5.431)
(5.432)
(5.433)
Dierentiating the angular velocity matrix
J b
W
UE J UE
(5.434)
g
J
W
W
+ J Ub E J Ub E
ÜE J UE
J $̃ E =
gw
J
W
+ J $̃ E J $̃ WE
ÜE J UE
(5.435)
J $̃ E =
yields
J
·
J $̃ E
=
=
and therefore,
J
·
W
= J $̃ E J $̃ E J $̃ WE
ÜE J UE
The acceleration vector of the body point becomes
¶
μ
·
¢
¡
J
r̈S = J $̃ E J $̃ E J $̃ WE J rS J dE + J d̈E
where
5
0
7
$
b
$̃
=
˜
=
3
J E
J E
$b 2
·
and
5
W
7
J $̃ E J $̃ E =
$ 22 + $ 23
$ 1 $ 2
$ 1 $ 3
$b 3
0
$b 1
$ 1 $ 2
$ 21 + $ 23
$ 2 $ 3
(5.436)
(5.437)
6
$b 2
$b 1 8
0
(5.438)
6
$ 1 $ 3
$ 2 $ 3 8
$ 21 + $ 22
(5.439)
5.14 F Screw Motion
Based on the Chasles theorem, any rigid body motion can be expressed by a
single translation along an axis combined with a unique rotation about the
axis. Such a motion is called screw. Consider the screw motion illustrated
in Figure 5.15. Point S rotates about the screw axis indicated by x̂ and
simultaneously translates along the same axis. Hence, any point on the
screw axis moves along the axis, while any point o the axis moves along
a helix.
5. F Applied Kinematics
297
u
p
P
FIGURE 5.15. A screw motion is the translation along a line combined with a
rotation about the line.
The angular rotation of the rigid body about the screw is called twist.
The Pitch of a screw, s, is the ratio of translation, k, to rotation, !.
s=
k
!
(5.440)
So, pitch is the rectilinear distance through which the rigid body translates
parallel to the axis of screw for a unit rotation. If s A 0, then the screw is
right-handed, and if s ? 0, it is left-handed.
A screw is shown by v(k> !> x̂> s) and is indicated by a unit vector x̂, a
location vector s, a twist angle !, and a translation k (or pitch s). The
location vector s indicates the global position of a point on the screw axis.
The twist angle !, the twist axis x̂, and the translation k (or pitch s) are
called screw parameters.
The screw is another transformation method to express the motion of a
rigid body. A linear displacement along an axis combined with an angular
displacement about the same axis arises in steering kinematics of vehicles.
If E rS indicates the position vector of a body point, its position vector in
the global frame after a screw motion is
J
rS = v(k> !> x̂> s) E rS
(5.441)
that is equivalent to a translation J dE along with a rotation J UE .
J
rS = J UE E rS + J dE
(5.442)
We may introduce a 4×4 matrix [W ], that is called the homogeneous matrix,
J
¸
UE J d
J
WE =
(5.443)
0
1
5. F Applied Kinematics
298
and combine the translation and rotation to express the motion with only
a matrix multiplication
J
rS = J WE E rS
(5.444)
where, J rS and E rS are expanded with an extra zero element to be consistent with the 4 × 4 matrix [W ].
5
6
5
6
[
{
9 \ :
9 | :
J
E
:
:
rS = 9
rS = 9
(5.445)
7 ] 8
7 } 8
0
0
Homogeneous matrix representation can be used for screw transformations
to combine the screw rotation and screw translation about the screw axis.
If x̂ passes through the origin of the coordinate frame, then s = 0 and
the screw motion is called central screw v(k> !> x̂). For a central screw we
have
J
vE (k> !> x̂) = Gx̂>k Ux̂>!
(5.446)
where,
5
1
9 0
Gx̂>k = 9
7 0
0
0
1
0
0
6
0 kx1
0 kx2 :
:
1 kx3 8
0
1
Ux̂>! =
5
x21 vers ! + f!
x1 x2 vers ! x3 v! x1 x3 vers ! + x2 v!
9 x1 x2 vers ! + x3 v!
x22 vers ! + f!
x2 x3 vers ! x1 v!
9
7 x1 x3 vers ! x2 v! x2 x3 vers ! + x1 v!
x23 vers ! + f!
0
0
0
and hence,
(5.447)
6
0
0 :
:
0 8
1
(5.448)
¸
d
=
1
6
5
x21 vers ! + f!
x1 x2 vers ! x3 v! x1 x3 vers ! + x2 v! kx1
9 x1 x2 vers ! + x3 v!
x22 vers ! + f!
x2 x3 vers ! x1 v! kx2 :
:
9
7 x1 x3 vers ! x2 v! x2 x3 vers ! + x1 v!
x23 vers ! + f!
kx3 8
0
0
0
1
(5.449)
As a result, a central screw transformation matrix includes the pure or fundamental translations and rotations as special cases because a pure translation corresponds to ! = 0, and a pure rotation corresponds to k = 0 (or
s = 4).
When the screw is not central and x̂ is not passing through the origin, a
screw motion to move p to p00 is denoted by
J
J
vE (k> !> x̂) =
p00
UE
0
J
= (p s) cos ! + (1 cos !) (x̂ · (p s)) x̂
+ (x̂ × (p s)) sin ! + s + kx̂
(5.450)
5. F Applied Kinematics
299
P''
h
I
z
B
Z
P'
P
p'
r
x
p
r'
y
u
G
p''
s
O
X
Y
FIGURE 5.16. Screw motion of a rigid body.
or
p00 = J UE (p s) + s + kx̂ = J UE p + s J UE s + kx̂
(5.451)
and therefore,
p00 = v(k> !> x̂> s)p = [W ] p
where
J
[W ] =
UE
0
J
s J UE J s + kx̂
1
¸
J
=
UE
0
(5.452)
J
d
1
¸
(5.453)
The vector J s, called location vector, is the global position of the body
frame before screw motion. The vectors p00 and p are global positions of a
point S after and before screw, as shown in Figure 5.16.
The screw axis is indicated by the unit vector x̂. Now a body point S
moves from its rst position to its second position S 0 by a rotation about
x̂. Then it moves to S 00 by a translation k parallel to x̂. The initial position
of S is pointed by p and its nal position is pointed by p00 .
A screw motion is a four variable function v(k> !> x̂> s). A screw has a line
of action x̂ at J s, a twist !, and a translation k.
The instantaneous screw axis was rst used by Mozzi (1730 1813) in
1763 although Chasles (1793 1880) is credited with this discovery.
Proof. The angle-axis rotation formula (5.185) relates r0 and r, which are
position vectors of S after and before rotation ! about x̂ when s = 0,
k = 0.
r0 = r cos ! + (1 cos !) (x̂ · r) x̂ + (x̂ × r) sin !
(5.454)
300
5. F Applied Kinematics
However, when the screw axis does not pass through the origin of J(R[\ ]),
then r0 and r must accordingly be substituted with:
r = ps
r0 = p00 s kx̂
(5.455)
(5.456)
where r0 is a vector after rotation and hence in J coordinate frame, and r
is a vector before rotation and hence in E coordinate frame.
Therefore, the relationship between the new and old positions of the
body point S after a screw motion is
p00
= (p s) cos ! + (1 cos !) (x̂ · (p s)) x̂
+ (x̂ × (p s)) sin ! + (s + kx̂)
(5.457)
Equation (5.457) is the Rodriguez formula for the most general rigid body
motion. Dening new notations J p = p00 and E p = p and also noting that
s indicates a point on the rotation axis and therefore rotation does not
aect s, we may factor out E p and write the Rodriguez formula in the
following form
¡
¢
J
p = I cos ! + x̂x̂W (1 cos !) + x̃ sin ! E p
¡
¢
I cos ! + x̂x̂W (1 cos !) + x̃ sin ! J s + J s + kx̂ (5.458)
which can be rearranged to show that a screw can be represented by a
homogeneous transformation
J
J
WE
p =
=
UE E p +J s J UE J s + kx̂
J
UE E p + J d = J WE E p
J
J
vE (k> !> x̂> s)
J
¸ J
UE J s J UE J s + kx̂
UE
=
=
0
1
0
(5.459)
=
J
d
1
¸
(5.460)
where,
J
UE = I cos ! + x̂x̂W (1 cos !) + x̃ sin !
¡¡
¢
¢
J
d =
I x̂x̂W (1 cos !) x̃ sin ! J s + kx̂
(5.461)
(5.462)
Direct substitution shows that:
5
6
x21 vers ! + f!
x1 x2 vers ! x3 v! x1 x3 vers ! + x2 v!
J
x22 vers ! + f!
x2 x3 vers ! x1 v! 8
UE = 7 x1 x2 vers ! + x3 v!
x1 x3 vers ! x2 v! x2 x3 vers ! + x1 v!
x23 vers ! + f!
(5.463)
5. F Applied Kinematics
301
¡¡
¢
¢
6
kx1 + ¡¡1 x21 ¢ v1 x1 (v2 x2 + v3 x3 )¢ vers ! + (v2 x3 v3 x2 ) v!
J
d = 7 kx2 + ¡¡1 x22 ¢ v2 x2 (v3 x3 + v1 x1 )¢ vers ! + (v3 x1 v1 x3 ) v! 8
kx3 + 1 x23 v3 x3 (v1 x1 + v2 x2 ) vers ! + (v1 x2 v2 x1 ) v!
(5.464)
This representation of a rigid motion requires six independent parameters, namely one for rotation angle !, one for translation k, two for screw
axis x̂, and two for location vector J s. It is because three components of x̂
are related to each other according to
5
x̂W x̂ = 1
(5.465)
and the location vector J s can locate any arbitrary point on the screw axis.
It is convenient to choose the point where it has the minimum distance from
R to make J s perpendicular to x̂. Let us indicate the shortest location vector
by J s0 , then there is a constraint among the components of the location
vector
J W
s0 x̂ = 0
(5.466)
If s = 0 then the screw axis passes through the origin of J and (5.460)
reduces to (5.449).
The screw parameters ! and k, together with the screw axis x̂ and location vector J s, completely dene a rigid motion of E(r{|}) in J(R[\ ]).
Having the screw parameters and screw axis, we can nd the elements of
the transformation matrix by Equations (5.463) and (5.464). So, given the
transformation matrix J WE , we can nd the screw angle and axis by
¢ 1 ¡ ¡J ¢
¢
1 ¡ ¡J ¢
tr UE 1 =
tr WE 2
2
2
1
=
(u11 + u22 + u33 1)
2
¢
1 ¡J
W
UE J UE
x̃ =
2 sin !
cos ! =
hence,
6
u32 u23
1 7
u13 u31 8
x̂ =
2 sin !
u21 u12
(5.467)
(5.468)
5
(5.469)
To nd all the required screw parameters, we must also nd k and coordinates of one point on the screw axis. Because the points on the screw
axis are invariant under the rotation, we must have
5
65
6 5
65
6
u11 u12 u13 u14
[
1 0 0 kx1
[
9 u21 u22 u23 u24 : 9 \ : 9 0 1 0 kx2 : 9 \ :
9
:9
: 9
:9
:
7 u31 u32 u33 u34 8 7 ] 8 = 7 0 0 1 kx3 8 7 ] 8 (5.470)
0
0
0
1
1
0 0 0
1
1
where ([> \> ]) are coordinates of points on the screw axis.
302
5. F Applied Kinematics
We may nd the intersection point of the screw line with \ ]-plane, as
£
¤W
.
a sample point, by setting [v = 0 and searching for s = 0 \v ]v
Therefore,
5
65
6 5 6
u11 1
u12
u13
u14 kx1
0
0
9 u21
: 9 \v : 9 0 :
u
1
u
u
kx
22
23
24
2
9
:9
:=9 :
(5.471)
7 u31
u32
u33 1 u34 kx3 8 7 ]v 8 7 0 8
0
0
0
0
1
0
which generates three equations to be solved for \v , ]v , and k.
5
6 5
k
x1
7 \v 8 = 7 x2
]v
x3
u12
1 u22
u32
61 5
6
u14
u13
u23 8 7 u24 8
1 u33
u34
(5.472)
Now we can nd the shortest location vector J s0 by
J
s0 = s (s · x̂)x̂
(5.473)
Example 218 F Central screw transformation of a base unit vector.
Consider two initially coincident frames J(R[\ ]) and E(r{|}). The
body performs a screw motion along the \ -axis for k = 2 and ! = 90 deg.
£
¤W
can be found by applying
The position of a body point at 1 0 0 1
the central screw transformation.
ˆ
ˆ U(M>
ˆ )
v(k> !> x̂) = v(2> > M)
= G(2M)
2
5
65
6 25
1 0 0 0
0 0 1 0
0
9 0 1 0 2 :9 0 1 0 0 : 9 0
:9
: 9
= 9
7 0 0 1 0 8 7 1 0 0 0 8 = 7 1
0 0 0 1
0 0 0 1
0
0
1
0
0
1
0
0
0
(5.474)
6
0
2 :
:
0 8
1
Therefore,
ˆ E
ˆ~ = v(2> > M)
ˆ~
5 2
0 0 1
9 0 1 0
= 9
7 1 0 0
0 0 0
J
(5.475)
65
6
5
6
0
1
0
9 0 : 9 2 :
2 :
:9 : = 9
:
0 8 7 0 8 7 1 8
1
1
1
The pitch of this screw is
s=
k
4
2
= = = 1=2732
!
2
unit/rad
(5.476)
5. F Applied Kinematics
303
Example 219 F Screw transformation of a point.
Consider two initially parallel coordinate frames J(R[\ ]) and E(r{|}).
The body performs a screw motion along [ = 2 and parallel to the \ axis for k = 2 and ! = 90 deg. Therefore, the body coordinate frame
£
¤W
is at location s = 2 0 0
. The position of a body point at E r =
£
¤W
3 0 0 1
can be found by applying the screw transformation, which
is
5
6
0 0 1 2
J
¸
9 0 1 0 2 :
UE s J UE s + kx̂
J
:
WE =
=9
(5.477)
7 1 0 0 2 8
0
1
0 0 0 1
because,
5
6
0 0 1
J
UE = 7 0 1 0 8
1 0 0
5
6
2
s=7 0 8
0
5
6
0
x̂ = 7 1 8
0
(5.478)
Therefore, the position vector of J r would then be
J
J
WE E r
0 0
9 0 1
= 9
7 1 0
0 0
r =
5
1
0
0
0
65 6 5
6
2
3
2
9 : 9
:
2 :
:9 0 : = 9 2 :
2 8 7 0 8 7 1 8
1
1
1
(5.479)
Example 220 F Rotation of a vector.
Transformation equation J r = J UE E r and Rodriguez rotation formula
(5.185) describe the rotation of any vector xed in a rigid body. However,
the vector can conveniently be described in terms of two points xed in the
body to derive the screw equation.
A reference point S1 with position vector r1 at the tail, and a point S2
with position vector r2 at the head, dene a vector in the rigid body. Then
the transformation equation between body and global frames can be written
as
J
(r2 r1 ) = J UE E (r2 r1 )
(5.480)
Assume the original and nal positions of the reference point S1 are along
the rotation axis. Equation (5.480) can then be rearranged in a form suitable
for calculating coordinates of the new position of point S2 in a transformation matrix form
J
r2
=
=
UE E (r2 r1 ) + J r1 = J UE E r2 + J r1 J UE E r1
J
WE E r2
(5.481)
J
304
5. F Applied Kinematics
Y
P0
Q2(X,Y)
58.0°
P2(4,1.5)
Q1(3,1)
P1(1,1)
X
FIGURE 5.17. Motion in a plane.
where
J
J
WE =
UE
0
J
r1 J UE E r1
1
¸
(5.482)
It is compatible with screw motion (5.460) for k = 0.
Example 221 F Special cases for screw determination.
There are two special cases for screws. The rst one occurs when u11 =
u22 = u33 = 1, then, ! = 0 and the motion is a pure translation k parallel
to x̂, where,
u14 v1 ˆ u24 v2 ˆ u34 v3
x̂ =
L+
M+
N̂
(5.483)
k
k
k
Since there is no unique screw axis in this case, we cannot locate any specic
point on the screw axis.
The second special case occurs when ! = 180 deg. In this case
5 q
1
(u + 1)
9 q 2 11
6
:
:
9
1
x̂ = 9
:
(u
+
1)
22
8
7 q2
1
2 (u33 + 1)
(5.484)
however, k and ([> \> ]) can again be calculated from (5.472).
Example 222 F Rotation and translation in a plane.
Assume that a plane is displaced from position 1 to position 2 according
to Figure 5.17. New coordinates of T2 are
5. F Applied Kinematics
rT2
305
2
U1 (rT1 rS1 ) + rS2
(5.485)
5
6 35 6 5 64 5
6
cos 58 sin 58 0
3
1
4
= 7 sin 58 cos 58 0 8 C7 1 8 7 1 8D + 7 1=5 8
0
0
1
0
0
0
5
6 5
6 5
6
1=06
4
5=06
= 7 1=696 8 + 7 1=5 8 = 7 3=196 8
0
0
0=0
=
or equivalently
rT2
¸
rS2 2 U1 rS1
rT1
=
W1 rT1 =
(5.486)
1
5
65 6 5
6
cos 58 sin 58 0 4=318
3
5=06
9 sin 58 cos 58 0 0=122 : 9 1 : 9 3=196 :
:9 : = 9
:
= 9
7 0
0
1
0 87 0 8 7 0 8
0
0
0
1
1
1
2
2
U1
0
Example 223 F Pole of planar motion.
In the planar motion of a rigid body, going from position 1 to position
2, there is always one point in the plane of motion that does not change
its position. Hence, the body can be considered as having rotated about this
point, which is known as the nite rotation pole. The transformation matrix
can be used to locate the pole. Figure 5.17 depicts a planar motion of a
triangle. To locate the pole of motion S0 ([0 > \0 ) we need the transformation
of the motion. Using the data given in Figure 5.17 we have
¸
2
U1 rS2 2 U1 rS1
2
(5.487)
W1 =
0
1
5
6
f v 0 f + v + 4
9 v f 0 f v + 3=5 :
:
= 9
7 0
8
0
1
0
0
0
0
1
The pole would be conserved under the transformation. Therefore,
rS0 = 2 W1 rS0
6
5
[0
cos sin 9 \0 :
9 sin cos 9
:
9
7 0 8 = 7 0
0
1
0
0
5
0
0
1
0
(5.488)
6
cos + sin + 4
[0
:
9
cos sin + 1=5 :
: 9 \0 :
8
7
0 8
0
1
1
65
which for = 58 deg provides
[0
\0
= 1=5 sin + 1 4 cos = 2=049
= 4 sin + 1 1=5 cos = 3=956
(5.489)
(5.490)
306
5. F Applied Kinematics
Example 224 F Determination of screw parameters.
We are able to determine screw parameters when we have the original
and nal position of three non-colinear points of a rigid body. Assume p0 ,
q0 , and r0 denote the position of points S , T, and U before the screw
motion, and p1 , q1 , and r1 denote their positions after the screw motion.
To determine screw parameters, !, x̂, k, and s, we should solve the following three simultaneous Rodriguez equations:
p1 p0
q1 q0
r1 r0
!
= tan x̂ × (p1 + p0 2s) + kx̂
2
!
= tan x̂ × (q1 + q0 2s) + kx̂
2
!
= tan x̂ × (r1 + r0 2s) + kx̂
2
(5.491)
(5.492)
(5.493)
We start with subtracting Equation (5.493) from (5.491) and (5.492).
!
(p1 p0 ) (r1 r0 ) = tan x̂ × [(p1 + p0 ) (r1 r0 )]
2
!
(q1 q0 ) (r1 r0 ) = tan x̂ × [(q1 + q0 ) (r1 r0 )]
2
(5.494)
(5.495)
Now multiplying both sides of (5.494) by [(q1 q0 ) (r1 r0 )] which is
perpendicular to x̂
[(q1 q0 ) (r1 r0 )] × [(p1 p0 ) (r1 r0 )]
!
= tan [(q1 q0 ) (r1 r0 )] × {x̂ × [(p1 + p0 ) (r1 r0 )]}
2
(5.496)
gives us
[(q1 q0 ) (r1 r0 )] × [(p1 + p0 ) (r1 r0 )]
!
= tan {[(q1 q0 ) (r1 r0 )] · [(p1 + p0 ) (r1 r0 )]} x̂
2
(5.497)
and therefore, the rotation angle can be found by equating tan !2 with the
norm of the right-hand side of the following equation:
!
[(q1 q0 ) (r1 r0 )] × (p1 + p0 ) (r1 r0 )
tan x̂ =
2
[(q1 q0 ) (r1 r0 )] · (p1 + p0 ) (r1 r0 )
(5.498)
To nd s, we may start with the cross product of x̂ with Equation (5.491).
¸
!
(5.499)
x̂ × (p1 p0 ) = x̂ × tan x̂ × (p1 + p0 2s) + kx̂
2
!
= tan {[x̂ · (p1 + p0 )] x̂ (p1 + p0 ) + 2 [s (x̂ · s) x̂]}
2
Note that s (x̂ · s) x̂ is the component of s perpendicular to x̂, where s
is a vector from the origin of the global frame J(R[\ ]) to an arbitrary
5. F Applied Kinematics
307
point on the screw axis. This perpendicular component indicates a vector
with the shortest distance between R and x̂. Let’s assume s0 is the name of
the shortest s. Therefore,
s0
= s (x̂ · s) x̂
#
"
1 x̂ × p1 p0
=
[x̂ · (p1 + p0 )] x̂ + p1 + p0
2
tan !2
(5.500)
The last parameter of the screw is the pitch k, which can be found from any
one of the Equations (5.491), (5.492), or (5.493).
k = x̂ · (p1 p0 ) = x̂ · (q1 q0 ) = x̂ · (r1 r0 )
(5.501)
Example 225 F Alternative derivation of screw transformation.
Assume the screw axis does not pass through the origin of J. If J s is the
position vector of some point on the axis x̂, then we can derive the matrix
representation of screw v(k> !> x̂> s) by translating the screw axis back to the
origin, performing the central screw motion, and translating the line back
to its original position.
v(k> !> x̂> s) = G(J s) v(k> !> x̂) G( J s)
= G(J s) G(kx̂) U(x̂> !) G( J s)
¸ J
¸
¸
I Js
UE kx̂
I Js
=
0 1
0
1
0
1
J
¸
UE J s J UE J s + kx̂
=
(5.502)
0
1
Example 226 F Rotation about an o-center axis.
Rotation of a rigid body about an axis indicated by x̂ and passing through
a point at J s, where J s×x̂ 6= 0 is a rotation about an o-center axis. The
transformation matrix associated with an o-center rotation can be obtained
from the screw transformation by setting k = 0. Therefore, an o-center
rotation transformation is
J
¸
UE J s J UE J s
J
WE =
(5.503)
0
1
Example 227 F Principal central screw.
There are three principal central screws, namely the {-screw, |-screw,
and }-screw, which are
5
6
cos sin 0
0
9 sin cos 0
0 :
:
v(k] > > N̂) = 9
(5.504)
7 0
0
1 s] 8
0
0
0
1
308
5. F Applied Kinematics
6
cos 0 sin 0
9
0
1
0
s\ :
ˆ =9
:
v(k\ > > M)
7 sin 0 cos 0 8
0
0
0
1
6
5
1
0
0
s[ 9
0 :
:
ˆ = 9 0 cos sin v(k[ > > L)
7 0 sin cos 0 8
0
0
0
1
5
(5.505)
(5.506)
Example 228 F Proof of Chasles theorem.
Let [W ] be an arbitrary spatial displacement, and decompose it into a
rotation U about x̂ and a translation G.
[W ] = [G][U]
(5.507)
We may also decompose the translation [G] into two components [Gk ] and
[GB ], parallel and perpendicular to x̂, respectively.
[W ] = [Gk ][GB ][U]
(5.508)
Now [GB ][U] is a planar motion, and is therefore equivalent to some rotation [U0 ] = [GB ][U] about an axis parallel to the rotation axis x̂. This yields
the decomposition [W ] = [Gk ][U0 ]. This decomposition completes the proof,
since the axis of [Gk ] can be taken equal to x̂.
Example 229 F Every rigid motion is a screw.
To show that any proper rigid motion can be considered as a screw motion, we must show that a homogeneous transformation matrix
J
¸
UE J d
J
WE =
(5.509)
0
1
can be written in the form
J
UE
J
WE =
0
(I J UE ) s + kx̂
1
¸
(5.510)
This problem is then equivalent to the following equation to nd k and x̂.
J
d = (I J UE ) s + kx̂
(5.511)
The matrix [I J UE ] is singular because J UE always has 1 as an eigenvalue. This eigenvalue corresponds to x̂ as an eigenvector. Therefore,
W
]x̂ = 0
[I J UE ]x̂ = [I J UE
(5.512)
and an inner product shows that
£
¤
x̂ · J d = x̂ · I J UE s + x̂ · kx̂
£
¤
= I J UE x̂ · s + x̂ · kx̂
(5.513)
5. F Applied Kinematics
309
which leads to
k = x̂ · J d
(5.514)
Now we may use k to nd s
£
¤1 J
s = I J UE
( d kx̂)=
(5.515)
5.15 Summary
To analyze the relative motion of rigid bodies, we instal a body coordinate
frame at the mass center of each body. The relative motion of the bodies
can be expressed by the relative motion of the frames.
Coordinates of a point in two Cartesian coordinate frames with a common origin are convertible based on nine directional cosines of the three
axes of a frame in the other. The conversion of coordinates in the two
frames can be cast in matrix transformation
J
r = J UE E r
6
5 ˆ
65
6
5
L · ˆ~ Lˆ · ˆ Lˆ · n̂
{2
[2
7 \2 8 = 7 Mˆ · ˆ~ Mˆ · ˆ Mˆ · n̂ 8 7 |2 8
]2
}2
N̂ · ˆ~ N̂ · ˆ N̂ · n̂
where,
ˆ ~) cos(L>
ˆ ˆ) cos(L>
ˆ n̂) 6
cos(L>ˆ
J
ˆ ~) cos(M>
ˆ ˆ) cos(M>
ˆ n̂) 8
UE = 7 cos(M>ˆ
cos(N̂> ˆ~) cos(N̂> ˆ) cos(N̂> n̂)
(5.516)
(5.517)
5
(5.518)
The transformation matrix J UE is orthogonal and therefore its inverse is
equal to its transpose.
J 1
W
UE = J UE
(5.519)
When a body coordinate frame E and a global frame J have a common origin and frame E rotates continuously with respect to frame J, the
rotation matrix J UE is time dependent.
J
r(w) = J UE (w) E r
(5.520)
Then, the global velocity of a point in E is
J
rb (w) = J v(w) = J Ub E (w) E r = J $̃ E J r(w)
(5.521)
where J $̃ E is the skew symmetric angular velocity matrix
J $̃ E
J $̃ E
J b
W
UE J UE
0
$ 3
0
= 7 $3
$ 2 $ 1
=
5
6
$2
$ 1 8
0
(5.522)
(5.523)
310
5. F Applied Kinematics
The matrix J $̃ E is associated with the angular velocity vector J $ E = !b x̂,
which is equal to an angular rate !b about the instantaneous axis of rotation
x̂.
Angular velocities of connected rigid bodies may be added relatively to
nd the angular velocity of the qth body in the base coordinate frame
0
0
0
0 $ q = 0 $ 1 + 1 $ 2 + 2 $ 3 + · · · + q1 $ q =
q
X
0
l1 $ l
(5.524)
l=1
Relative time derivatives between the global and a coordinate frames
attached to a moving rigid body must be taken according to the following
rules.
E
gE
rS
gw
J
gJ
rS
gw
=
E
rb S = E vS = {ˆ
b ~ + |b ˆ + }b n̂
(5.525)
=
J
rb S = J vS = [b Lˆ + \b Mˆ + ]b N̂
(5.526)
E
E
rb S + E
rS = E
bS
J $E ×
Jr
(5.527)
J
rb S J $ E × J rS = J
bS
Er
(5.528)
J
gE
rS (w) =
gw
E
gJ
rS (w) =
gw
The global velocity of a point S in a moving frame E at
J
rS = J UE E rS + J dE
(5.529)
¡
¢
rb S = J $̃ E J rS J dE + J db E
¡J
¢
rS J dE + J db E
J$E ×
(5.530)
is
J
vS
=
=
J
When a body coordinate frame E and a global frame J have a common
origin, the global acceleration of a point S in frame E is
J
¡
¢
gJ
vS = J E × J r + J $ E × J $ E × J r
gw
J
r̈ =
(5.531)
where, J E is the angular acceleration of E with respect to J
J
J E =
g
J $E
gw
(5.532)
However, when the body coordinate frame E has a rigid motion with respect to J, then
J
¡
¢
g J
vS = J E × J rS J dE
gw
¡
¡
¢¢
+ J $ E × J $ E × J rS J dE + J d̈E
J
aS
=
(5.533)
5. F Applied Kinematics
311
where J dE indicates the position of the origin of E with respect to the
origin of J.
Angular accelerations of two connected rigid bodies are related according
to
0
0
(5.534)
0 2 = 0 1 + 1 2 + 0 $ 1 × 1 $ 2
312
5. F Applied Kinematics
5.16 Key Symbols
E> R{|}
J
dE
0
d
1 2
ĥ* > ĥ > ĥ#
J> R[\ ]
ˆ~> ˆ> n̂
˜~> m̃> ñ
ˆ M>
ˆ N̂
L>
s = k@!
S
J
r
E
r
r̂K1 > r̂K2 > r̂K3
U]
U\
U[
Ub
J
UE
UW
U1
U]
U\
U[
E1
UE2
E
UJ
v(k> !> x̂> s)
w
x̂> !
x̂
x̂$
{> |> }
[> \> ]
{> |> }> x
body Cartesian coordinate frame
position vector of body coordinate frame E in J
position of frame E2 respect to E1 expressed in E0
Euler angle coordinate frame unit vectors
global Cartesian coordinate frame
body coordinate frame unit vectors
skew symmetric matrix associated to ˆ~> ˆ> n̂
global coordinate frame unit vectors
pitch of screw
point
position vector in global coordinate frame
position vector in body coordinate frame
row vectors of a rotation matrix
rotation matrix about the global ]-axis
rotation matrix about the global \ -axis
rotation matrix about the global [-axis
time derivative of a rotation matrix U
rotation matrix from local frame to global frame
transpose of a rotation matrix
inverse of a rotation matrix
rotation matrix about the body }-axis
rotation matrix about the body |-axis
rotation matrix about the body {-axis
rotation matrix from coordinate frame E1 to E2
rotation matrix from global to local coordinate frame
screw motion
time
axis and angle of rotation
instant angular acceleration axis
instant angular velocity axis
body coordinates of a point
body coordinates of a point
displacement
J E
angular acceleration of body E expressed in J
Kronecker’s delta
permutation symbol
Euler frequencies
angular velocity of rigid body E expressed in J
skew symmetric matrix associated to $
mn
lmn
b #b
*>
b >
J $E
$̃
5. F Applied Kinematics
313
Exercises
1. Body point and global rotations.
The point S is at rS = (1> 2> 1) in a body coordinate E(R{|}). Find
the nal global position of S after a rotation of 30 deg about the
[-axis, followed by a 45 deg rotation about the ]-axis.
2. Body point after global rotation.
Find the position of a point S in the local coordinate frame if it is
moved to J rS = [1> 3> 2]W after
(a) a 60 deg rotation about ]-axis.
(b) a 60 deg rotation about [-axis.
(c) a 60 deg rotation about ]-axis followed by a 60 deg rotation
about [-axis.
(d) F Is it possible to combine the rotations in part (f) and do only
one rotation about the bisector of [ and ]-axes?
3. Invariant of a vector.
A point was at E rS = [1> 2> }]W . After a rotation of 60 deg about the
[-axis, followed by a 30 deg rotation about the ]-axis, it is at
5
6
[
J
rS = 7 \ 8
2=933
Find }, [, and \ .
4. F Constant length vector.
Show that the length of a vector will not change by rotation.
¯J ¯ ¯ J
¯
¯ r¯ = ¯ UE E r¯
Show that the distance between two body points will not change by
rotation.
¯E
¯ ¯
¯
¯ p1 E p2 ¯ = ¯ J UE E p1 J UE E p2 ¯
5. Global roll-pitch-yaw rotation angles.
Calculate the role, pitch, and yaw angles for the following rotation
matrix:
5
6
0=53 0=84 0=13
E
0=15
0=99 8
UJ = 7 0=0
0=85 0=52 0=081
314
5. F Applied Kinematics
6. Body point, local rotation.
What is the global coordinates of a body point at E rS = [2> 2> 3]W ,
after a rotation of 60 deg about the {-axis?
7. Two local rotations.
Find the global coordinates of a body point at E rS = [2> 2> 3]W after
a rotation of 60 deg about the {-axis followed by 60 deg about the
}-axis.
8. Combination of local and global rotations.
Find the nal global position of a body point at E rS = [10> 10> 10]W
after a rotation of 45 deg about the {-axis followed by 60 deg about
the ]-axis.
9. Combination of global and local rotations.
Find the nal global position of a body point at E rS = [10> 10> 10]W
after a rotation of 45 deg about the [-axis followed by 60 deg about
the }-axis.
10. F Euler angles from rotation matrix.
Find the Euler angles for the following rotation matrix:
5
6
0=53 0=84 0=13
E
0=15
0=99 8
UJ = 7 0=0
0=85 0=52 0=081
11. F Equivalent Euler angles to two rotations.
Find the Euler angles corresponding to the rotation matrix E UJ =
D|>45 D{>30 .
12. F Equivalent Euler angles to three rotations.
Find the Euler angles corresponding to the rotation matrix E UJ =
D}>60 D|>45 D{>30 .
13. F Local and global positions, Euler angles.
Find the conditions between the Euler angles to transform J rS =
[1> 1> 0]W to E rS = [0> 1> 1]W .
14. Elements of rotation matrix.
The elements of rotation matrix J UE are
5
ˆ ˆ~) cos(L>
ˆ ˆ) cos(L>
ˆ n̂) 6
cos(L>
J
ˆ ˆ~) cos(M>
ˆ ˆ) cos(M>
ˆ n̂) 8
UE = 7 cos(M>
cos(N̂>ˆ~) cos(N̂> ˆ) cos(N̂> n̂)
Find J UE if J rS1 = (0=7071> 1=2247> 1=4142) is a point on the {-axis,
and J rS2 = (2=7803> 0=38049> 1=0607) is a point on the |-axis.
5. F Applied Kinematics
315
15. Local position, global velocity.
A body is turning about the ]-axis at a constant angular rate b =
2 rad@ sec. Find the global velocity of a point at
5
6
5
E
r = 7 30 8
10
16. Global position, constant angular velocity.
A body is turning about the ]-axis at a constant angular rate b =
2 rad@ s. Find the global position of a point at
5
6
5
E
r = 7 30 8
10
after w = 3 sec if the body and global coordinate frames were coincident at w = 0 sec.
17. Turning about {-axis.
Find the angular velocity matrix when the body coordinate frame is
turning 35 deg @ sec at 45 deg about the {-axis.
18. Combined rotation and angular velocity.
Find the rotation matrix for a body frame after 30 deg rotation about
the ]-axis, followed by 30 deg about the [-axis, and then 90 deg
about the \ -axis. Then calculate the angular velocity of the body
if it is turning with b = 20 deg @ sec, b = 40 deg @ sec, and b =
55 deg @ sec about the ], \ , and [ axes respectively.
19. Angular velocity, expressed in body frame.
The point S is at rS = (1> 2> 1) in a body coordinate E(R{|}). Find
when the body frame is turned 30 deg about the [-axis at a
rate b = 75 deg @ sec, followed by 45 deg about the ]-axis at a rate
b = 25 deg @ sec.
E
J $̃ E
20. Global roll-pitch-yaw angular velocity.
Calculate the angular velocity for a global roll-pitch-yaw rotation
of = 30 deg, = 30 deg, and = 30 deg with b = 20 deg @ sec,
b = 20 deg @ sec, and b = 20 deg @ sec.
21. Roll-pitch-yaw angular velocity.
Find E
b =
J $̃ E and J $̃ E for the role, pitch, and yaw rates equal to b
20 deg @ sec, = 20 deg @ sec, and b = 20 deg @ sec respectively, and
316
5. F Applied Kinematics
having the following rotation matrix:
5
6
0=53 0=84 0=13
E
0=15
0=99 8
UJ = 7 0=0
0=85 0=52 0=081
22. F Dierentiating in local and global frames.
Consider a local point at E rS = wˆ~ + ˆ. The local frame E is rotating
E
J
E
in J by b about the ]-axis. Calculate gwg E rS , gwg J rS , gwg J rS , and
J
g E
rS .
gw
23. F Transformation of angular velocity exponents.
Show that
E q
J W
UE J $̃ qE J UE
J $̃ E =
24. Local position, global acceleration.
A body is turning about the ]-axis at a constant angular acceleration ¨ = 2 rad@ sec2 . Find the global velocity of a point, when
b = 2 rad@ sec, = @3 rad and
5
6
5
E
r = 7 30 8
10
25. Global position, constant angular acceleration.
A body is turning about the ]-axis at a constant angular acceleration
¨ = 2 rad@ sec2 . Find the global position of a point at
5
6
5
E
r = 7 30 8
10
after w = 3 sec if the body and global coordinate frames were coincident at w = 0 sec.
26. Turning about {-axis.
Find the angular acceleration matrix when the body coordinate frame
is turning 5 deg @ sec2 , 35 deg @ sec at 45 deg about the {-axis.
27. Angular acceleration and Euler angles.
Calculate the angular velocity and acceleration vectors in body and
global coordinate frames if the Euler angles and their derivatives are:
* = =25 rad
= =25 rad
# = =5 rad
*b = 2=5 rad@ sec
b = 4=5 rad@ sec
#b = 3 rad@ sec
*
¨ = 25 rad@ sec2
¨ = 35 rad@ sec2
#̈ = 25 rad@ sec2
5. F Applied Kinematics
317
28. Combined rotation and angular acceleration.
Find the rotation matrix for a body frame after 30 deg rotation about
the ]-axis, followed by 30 deg about the [-axis, and then 90 deg
about the \ -axis. Then calculate the angular velocity of the body
if it is turning with b = 20 deg @ sec, b = 40 deg @ sec, and b =
55 deg @ sec about the ], \ , and [ axes respectively. Finally, calculate the angular acceleration of the body if it is turning with
¨ = 4 deg @ sec2 , and ¨ = 6 deg @ sec2 about the
¨ = 2 deg @ sec2 , ], \ , and [ axes.
6
Applied Mechanisms
The mechanisms that are used in vehicle subsystems are mostly made of
four-bar linkages. Double D-arm for independent suspension, and trapezoidal steering are two subsystems examples in vehicle . In this chapter, we
review the analysis and design methods for such mechanisms.
6.1 Four-Bar Linkage
An individual rigid member that can have relative motion with respect to
all other members is called a link. A link may also be called a bar, body,
arm, or a member. Any two or more links connected together, such that no
relative motion can occur among them, are considered a single link.
Axis of joint
Axis of joint
Revolute
Prismatic
FIGURE 6.1. A revolute and a prismatic joint.
Two links are connected by a joint where their relative motion can be
expressed by a single coordinate. Joints are typically revolute (rotary) or
prismatic (translatory). Figure 6.1 illustrates the revolute and prismatic
joints. A revolute joint (R), is like a hinge that allows relative rotation
between the two connected links. A prismatic joint (P), allows a relative
translation between the two connected links.
Relative rotation or translation, between two connected links by a revolute or prismatic joint, occurs about a line called joint axis. The value of
the single variable describing the relative position of two connected links
at a joint is called the joint coordinate or joint variable. It is an angle for
a revolute joint, or a distance for a prismatic joint.
A set of connected links is called a mechanism. A linkage is made by
attaching, and xing, one link of a mechanism to the ground. The xed
link is called the ground link. There are two types of linkages, closed loop
or parallel, and open loop or serial. In vehicle subsystems we usually use
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_6, © Springer Science+Business Media New York 2014
319
320
6. Applied Mechanisms
B
3
A
T3
4
2
T2
T4
T1
1
M
N
FIGURE 6.2. A four-bar linkage.
closed-loop linkages. Open-loop linkages are used in robotic systems where
an actuator controls the joint variable at each joint.
A four-bar linkage is shown in Figure 6.2. Link number 1 is the ground
link P Q . The ground link is the base and used as a reference link. We
measure all the variables with respect to the ground link. Link number
2 P D is usually the input link which is controlled by the input angle 2 .
Link number 4 Q E is usually the output link with angular position 4 ,
and link number 3 DE is the coupler link with angular position 3 that
connects the input and output links.
The angular position of the output and coupler links, 4 and 3 , are
functions of the links’ length and the value of the input variable 2 . The
angles 4 and 3 can be calculated by the functions
4
3
Ã
!
s
E 2 4DF
= 2 tan
2D
Ã
!
s
2 4GI
H
H
±
= 2 tan1
2G
1
E ±
(6.1)
(6.2)
where,
D
E
F
G
H
I
=
=
=
=
=
=
M3 M1 + (1 M2 ) cos 2
2 sin 2
M1 + M3 (1 + M2 ) cos 2
M5 M1 + (1 + M4 ) cos 2
2 sin 2
M5 + M1 (1 M4 ) cos 2
(6.3)
(6.4)
(6.5)
(6.6)
(6.7)
(6.8)
6. Applied Mechanisms
y
321
B
3
A
2
r3
T3
4
r4
r2
T2
r1
G
T1
1
M
T4
x
N
FIGURE 6.3. Expressing a four-bar linkage with a vector loop.
and
M1
=
M2
=
M3
=
M4
=
M5
=
g
d
g
f
d2 e2 + f2 + g2
2df
g
e
f2 g2 d2 e2
2de
(6.9)
(6.10)
(6.11)
(6.12)
(6.13)
Proof. We may show a four-bar linkage by a vector loop as is shown in
Figure 6.3. The direction of each vector is arbitrary. However, the angle of
each vector should be measured with respect to the positive direction of
the {-axis. The vector expression of each link is shown in Table 6=1.
Table 6=1 - Vector representation of the four-bar linkage
shown in Figure 6.3.
Link
Name
Vector Length
Angle
Variable
1
Ground
r1
g
1 = 180 deg
2
Input
r2
d
2
2
3
Coupler
r3
e
3
3
4
Output
r4
f
4
4
The vector loop in the global coordinate frame J is
J
r4 + J r1 + J r2 J r3 = 0
(6.14)
322
6. Applied Mechanisms
where,
J
r1
r2
J
r3
J
r4
J
=
=
=
=
g ˆ~
d (cos 2 ˆ~ + sin 2 ˆ)
e (cos 3 ˆ~ + sin 3 ˆ)
f (cos 4 ˆ~ + sin 4 ˆ)
(6.15)
(6.16)
(6.17)
(6.18)
and the left superscript J reminds that the vectors are expressed in the
global coordinate frame attached to the ground link. Substituting the Cartesian expressions for the planar vectors in Equation (6.14) yields
gˆ~ + d (cos 2 ˆ~ + sin 2 ˆ) + e (cos 3 ˆ~ + sin 3 ˆ)
f (cos 4 ˆ~ + sin 4 ˆ) = 0
(6.19)
We may decompose Equation (6.19) into sin and cos components.
d sin 2 + e sin 3 f sin 4
g + d cos 2 + e cos 3 f cos 4
= 0
= 0
(6.20)
(6.21)
To derive the relationship between the input angle 2 and the output
angle 4 , the coupler angle 3 must be eliminated between Equations (6.20)
and (6.21). Transferring the terms not containing 3 to the other side of
the equations, and squaring both sides, provides the following equations:
(e sin 3 )
2
(e cos 3 )
2
2
= (d sin 2 + f sin 4 )
(6.22)
2
= (d cos 2 + f cos 4 + g)
(6.23)
By adding Equations (6.22) and (6.23), and simplifying, we derive the following equation:
M1 cos 4 M2 cos 2 + M3 = cos (4 2 )
(6.24)
where
g
g
d2 e2 + f2 + g2
M2 =
M3 =
(6.25)
d
f
2df
Equation (6.24) is called the Freudenstein’s equation. The Freudenstein’s
equation may be expanded by using trigonometry
M1 =
4
2
sin 4 =
2 4
1 + tan
2
2 tan
4
2
cos 4 =
2 4
1 + tan
2
1 tan2
(6.26)
to provide a more practical equation
D tan2
4
4
+ E tan
+F =0
2
2
(6.27)
6. Applied Mechanisms
323
where D, E, and F are functions of the input variable.
D = M3 M1 + (1 M2 ) cos 2
E = 2 sin 2
F = M1 + M3 (1 + M2 ) cos 2
(6.28)
(6.29)
(6.30)
Equation (6.27) is quadratic in tan (4 @2) and can be used to nd the
output angle 4 as a function of the input angle 2 .
Ã
!
s
E ± E 2 4DF
1
4 = 2 tan
(6.31)
2D
To nd the relationship between the input angle 2 and the coupler angle
3 , the output angle 4 must be eliminated between Equations (6.20) and
(6.21). Transferring the terms not containing 4 to the right-hand side of
the equations, and squaring both sides, provides
(f sin 4 )2
2
(f cos 4 )
= (d sin 2 + e sin 3 )2
= (d cos 2 + e cos 3 g)
(6.32)
2
(6.33)
By adding Equations (6.32) and (6.33), and simplifying, we derive the equation:
M1 cos 3 + M4 cos 2 + M5 = cos (3 2 )
(6.34)
where
g
f2 g2 d2 e2
M5 =
e
2de
Equation (6.34) may be expanded and transformed to
M4 =
G tan2
3
3
+ H tan
+I =0
2
2
(6.35)
(6.36)
where G, H, and I are functions of the input variable.
G
H
I
= M5 M1 + (1 + M4 ) cos 2
= 2 sin 2
= M5 + M1 (1 M4 ) cos 2
(6.37)
(6.38)
(6.39)
Equation (6.36) is quadratic in tan (3 @2) and can be solved to nd the
coupler angle 3 .
Ã
!
s
H ± H 2 4GI
1
3 = 2 tan
(6.40)
2G
Equations (6.31) and (6.40) calculate the output and coupler angles 4
and 3 as two functions of the input angle 2 , provided that the lengths d,
e, f, and g are given.
324
6. Applied Mechanisms
Example 230 Two possible congurations for a four-bar linkage.
At any angle 2 , and for suitable values of d, e, f, and g, Equations (6.1)
and (6.2) provide us with two values for the output and coupler angles, 4
and 3 . Both solutions are possible and provide two dierent congurations
for each input angle 2 .
A suitable set of (d, e, f, g) make the four-bar linkage a closed loop.
A suitable set of numbers makes the radicals in Equations (6.1) and (6.2)
real.
As an example, consider a linkage with
d=1
e=2
f = 2=5
g=3
(6.41)
The Ml > l = 1> 2> 3> 4> 5 are functions of the links’ length and are equal to
M1
=
M2
=
M3
=
M4
=
M5
=
g
=3
d
3
g
=
= 1=2
f
2=5
d2 e2 + f2 + g2
= 2=45
2df
g
= 1=5
e
2
f g2 d2 e2
= 1=9375
2de
(6.42)
The coe!cients of the quadratic equations are then calculated.
D = 0=6914213562 E = 1=414213562
F = 3=894365082
G = 3=169733048
H = 1=414213562 I = 1=416053390
(6.43)
Using the minus sign of (6.1) and (6.2), the output and coupler angles at
2 = @4 rad = 45 deg are
4
3
2 rad 114=73 deg
0=897 rad 51=42 deg
(6.44)
and using the plus sign, they are
4
3
2=6 rad 149 deg
1=495 rad 85=7 deg
(6.45)
Figure 6.4 depicts the two possible congurations of the linkage for 2 =
45 deg. The conguration in Figure 6.4(d) is called convex, non-crossed,
or elbow-up, and the conguration in Figure 6.4(e) is called concave,
crossed, or elbow-down.
6. Applied Mechanisms
325
B
51.4q
A
A
114.7q
45q
M
N
45q
85.7q
M
N
B
(a)
( b)
149q
FIGURE 6.4. Two possible conguration of a four-bar linkage having the same
input angle 2 .
Example 231 Velocity analysis of a four-bar linkage.
The velocity analysis of a four-bar linkage is possible by taking a time
derivative of Equations (6.20) and (6.21),
g
(d sin 2 + e sin 3 f sin 4 )
gw
= d $ 2 cos 2 + e $ 3 cos 3 f $ 4 cos 4 = 0
(6.46)
g
(g + d cos 2 + e cos 3 f cos 4 )
gw
= d $ 2 sin 2 e $ 3 sin 3 + f $ 4 sin 4 = 0
(6.47)
where
$ 2 = b 2
$ 3 = b 3
$ 4 = b 4
(6.48)
Assuming 2 and $ 2 are given, and 3 , 4 are known from Equations
(6.1) and (6.2), we solve Equations (6.46) and (6.47), for $ 3 and $ 4 .
$4
=
$3
=
d sin (2 3 )
$2
f sin (4 3 )
d sin (2 4 )
$2
e sin (4 3 )
(6.49)
(6.50)
Example 232 Velocity of moving joints for a four-bar linkage.
Having the coordinates 2 , 3 , 4 and velocities $ 2 , $ 3 , $ 4 enables us to
calculate the absolute and relative velocities of points D and E shown in
Figure 6.3. The absolute velocity of a point refers to the its velocity relative
to the ground link, and relative velocity of a point refers to its velocity
relative of another moving point.
326
6. Applied Mechanisms
The absolute velocity of points D and E are
J
vD
J
vE
$ 2 × J r2
6 5
6 5
6
d cos 2
0
d$ 2 sin 2
= 7 0 8 × 7 d sin 2 8 = 7 d$ 2 cos 2 8
$2
0
0
=
J
5
$ 4 × J r4
6 5
6 5
6
0
f cos 4
f$ 4 sin 4
= 7 0 8 × 7 f sin 4 8 = 7 f$ 4 cos 4 8
$4
0
0
=
(6.51)
J
5
(6.52)
and the velocity of point E with respect to point D is
5
6 5
6
f$ 4 sin 4
d$ 2 sin 2
J
vE@D = J vE J vD = 7 f$ 4 cos 4 8 7 d$ 2 cos 2 8
0
0
5
6
d$ 2 sin 2 f$ 4 sin 4
= 7 f$ 4 cos 4 d$ 2 cos 2 8
(6.53)
0
The velocity of point E with respect to D can also be found as
¡
¢
J
vE@D = J U2 2 vE = J U2 2 vE = J U2 2 $ 3 × 2 r3 = J $ 3 × J r3
5
6 5
6 5
6
0
e cos 3
e$ 3 sin 3
= 7 0 8 × 7 e sin 3 8 = 7 e$ 3 cos 3 8
(6.54)
$3
0
0
Equations (6.53) and (6.54) are both correct and convertible to each other.
Example 233 Acceleration analysis of a four-bar linkage.
The acceleration analysis of a four-bar linkage is possible by taking a time
derivative from Equations (6.46) and (6.47),
g
(d $ 2 cos 2 + e $ 3 cos 3 f $ 4 cos 4 )
gw
= d2 cos 2 + e3 cos 3 f4 cos 4
d$ 22 sin 2 e$ 23 sin 3 + f$ 24 sin 4 = 0
(6.55)
g
(d $ 2 sin 2 e $ 3 sin 3 + f $ 4 sin 4 )
gw
= d2 sin 2 e3 sin 3 + f4 sin 4
d$ 22 cos 2 e$ 23 cos 3 + f$ 24 cos 4 = 0
(6.56)
where
2 = $b 2
3 = $b 3
4 = $b 4
(6.57)
6. Applied Mechanisms
327
Assuming 2 , $ 2 , and 2 are given as the kinematics of the input link,
3 , 4 are known from Equations (6.1) and (6.2), and $ 3 , $ 4 are known
from Equations (6.49) and (6.50), we solve Equations (6.55) and (6.56),
for 3 and 4 .
4
=
3
=
F3 F4 F1 F6
F1 F5 F2 F4
F3 F5 F2 F6
F1 F5 F2 F4
(6.58)
(6.59)
where
F1
F3
F4
F6
f sin 4
F2 = e sin 3
2
d2 sin 2 + d$ 2 cos 2 + e$ 23 cos 3 f$ 24 cos 4
f cos 4
F5 = e cos 3
2
d2 cos 2 d$ 2 sin 2 e$ 23 sin 3 + f$ 24 sin 4
=
=
=
=
(6.60)
Example 234 Acceleration of moving joints for a four-bar linkage.
Having the angular kinematics of a four-bar linkage 2 , 3 , 4 , $ 2 , $ 3 ,
$ 4 , 2 , 3 , and 4 is necessary and enough to calculate the absolute and
relative accelerations of points D and E shown in Figure 6.3.
The absolute acceleration of points D and E are
¢
¡
J
aD = J 2 × J r2 + J $ 2 × J $ 2 × J r2
6
5
d2 sin 2 d$ 22 cos 2
(6.61)
= 7 d2 cos 2 d$ 22 sin 2 8
0
J
where
aE
=
J 4 ×
J
r4 + J $ 4 ×
¡
J $4 ×
6
f4 sin 4 f$ 24 cos 4
= 7 f4 cos 4 f$ 24 sin 4 8
0
5
6
d cos 2
J
r2 = 7 d sin 2 8
0
6
5
0
7 0 8
J$2 =
$2
5
6
0
7 0 8
J 2 =
2
5
J
r4
6
f cos 4
J
r4 = 7 f sin 4 8
0
6
5
0
7 0 8
J $4 =
$4
5
6
0
7 0 8
J 4 =
4
¢
(6.62)
5
(6.63)
(6.64)
(6.65)
328
6. Applied Mechanisms
The acceleration of point E with respect to point D is
¢
¡
J
aE@D = J 3 × J r3 + J $ 3 × J $ 3 × J r3
6
5
e3 sin 3 e$ 23 cos 3
= 7 e3 cos 3 e$ 23 sin 3 8
0
where
6
e cos 3
J
r3 = 7 e sin 3 8
0
5
5
6
0
7 0 8
J$3 =
$3
6
0
7 0 8
J 3 =
3
(6.66)
5
(6.67)
Example 235 Grasho criterion.
The ability of a four-bar linkage to have a rotary link is determined by
the Grasho criterion. Assume the four links have the lengths v, o, s, and
t, where
o = orqjhvw olqn
v = vkruwhvw olqn
s> t = wkh rwkhu wzr olqnv
then, the Grasho criterion states that the linkage has a rotary link if
o+v?s+t
(6.68)
Dierent types of a Grasho mechanism are:
1= Shortest link is the input link, then the mechanism is a crank-rocker.
2= Shortest link is the ground link, and the mechanism is a crank-crank.
3= At all other conditions, the mechanism is a rocker-rocker.
A crank-crank mechanism is also called a drag-link.
Example 236 Limit positions for a four-bar linkage.
When the output link of a four-bar linkage stops while the input link
can turn, we call the linkage is at a limit position. It occurs when the
angle between the input and coupler links is either 180 deg or 360 deg. Limit
positions of a four-bar linkage, if there are any, must be determined by the
designer to make sure the linkage is designed properly. A sample of limit
positions for a four-bar linkage are shown in Figure 6.5.
We show the limit angles of the output link by 4O1 , 4O2 , and the corresponding input angles by 2O1 , 2O2 . They can be calculated as
"
#
2
2
2
1 (d + e) + g f
2O1 = cos
(6.69)
2g (d + e)
"
#
2
2
2
1 (d + e) g f
4O1 = cos
(6.70)
2fg
6. Applied Mechanisms
329
B
A
B
180q
M
N
M
A
360q
(a)
N
(b)
FIGURE 6.5. Limit position for a four-bar linkage.
2O2
4O2
= cos
1
= cos
1
"
"
(e d)2 + g2 f2
2g (e d)
(e d)2 g2 f2
2fg
#
#
(6.71)
(6.72)
The sweep angle of the output link would be
! = 4O2 4O1
(6.73)
Example 237 Dead positions for a four-bar linkage.
When the input link of a four-bar linkage locks, we say the linkage is at a
dead position. It happens when the angle between the output and coupler
links is either 180 deg or 360 deg. Dead positions of a four-bar linkage, if
there are any, must be determined by the designer to make sure the linkage
is never stuck in a dead position. A dead position for a four-bar linkage is
shown in Figure 6.6.
We show the dead angle of the output link by 4G1 , 4G2 , and the corresponding input angles by 2G1 , 2G2 . They can be calculated as
"
#
2
2
2
+
g
(e
+
f)
d
2G1 = cos1
(6.74)
2dg
"
#
2
2
2
1 d g (e + f)
4G1 = cos
(6.75)
2 (e + f) g
2G2
4G2
= cos
1
= cos1
"
"
d2 + g2 (e f)
2dg
2
d2 g2 (e f)2
2dg
#
#
(6.76)
(6.77)
330
6. Applied Mechanisms
A
180q
B
B
M
N
(a)
M
N
(b)
A
360q
FIGURE 6.6. Dead position for a four-bar linkage.
Example 238 F Designing a four-bar linkage using Freudenstein’s equation.
Designing a mechanism can be thought of as determining the required
lengths of the links to accomplish a specic task.
Freudenstein’s equation (6.24)
M1 cos 4 M2 cos 2 + M3 = cos (4 2 )
(6.78)
g
g
d2 e2 + f2 + g2
M2 =
M3 =
(6.79)
d
f
2df
determines the input-output relationship of a four-bar linkage. This equation can be utilized to design a four-bar linkage for three associated inputoutput angles.
Figure 6.7 illustrates the four popular windshield wiper systems. Doublearm parallel method is the most popular wiping system that serves most
passenger cars. The double-arm opposing method has been using since last
century, however, it was never very popular. The single-arm simple method
is not very e!cient, so the controlled single-arm is designed to maximize
the wiped area.
Wipers are used on windshields, and headlights. Figure 6.8 illustrates
a sample of double-arm parallel windshield wiper mechanism. A four-bar
linkage makes the main mechanism match the angular positions of the left
and right wipers. A dyad or a two-link connects the driving motor to the
main four-bar linkage and converts the rotational output of the motor into
the back-and-forth motion of the wipers.
The input and output links of the main four-bar linkage at three dierent
positions are shown in Figure 6.9. We show the beginning and the end angles
for the input link by 21 and 23 , and for the output link by 41 and 43
respectively. To design the mechanism we must match the angular positions
of the left and right blades at the beginning and at the end positions. Let
us add another match point approximately in the middle of the total sweep
M1 =
6. Applied Mechanisms
Double arm parallel
Double arm opposed
Single arm, simple
Single arm, cam controlled
331
FIGURE 6.7. Four popular windshield wiper systems.
4
2
x
T4
P
M
A
Z
1
3
N
T2
B
y
C
FIGURE 6.8. A sample of double-arm parallel windshield wiper mechanism.
332
6. Applied Mechanisms
4
2
x
26.8°
T43
M
97.5°
T42
113.1°
157.2°
T41
N
157.6°
69.5° T
22
T21
T23
y
FIGURE 6.9. The input and output links of the main four-bar linkage of a windshield wiper at three dierent positions.
angles and design a four-bar linkage to match the angles indicated in Table
6=2.
Table 6=2 - Matching angles for a four-bar linkage of the double-arm
parallel mechanism shown in Figure 6.9.
Matching
Input angle
Output angle
1
21 = 157=6 deg 2=75 rad 41 = 157=2 deg 2=74 rad
2
22 = 113=1 deg 1=97 rad 42 = 97=5 deg 1=7 rad
3
23 = 69=5 deg 1=213 rad 43 = 26=8 deg 0=468 rad
Substituting the input and output angles in Freudenstein’s equation (6.24)
M1 cos 41 M2 cos 21 + M3
M1 cos 42 M2 cos 22 + M3
M1 cos 43 M2 cos 23 + M3
= cos (41 21 )
= cos (42 22 )
= cos (43 23 )
(6.80)
= cos (2=74 2=75)
= cos (1=7 1=97)
= cos (0=468 1=213)
(6.81)
provides us with equations
M1 cos 2=74 M2 cos 2=75 + M3
M1 cos 1=7 M2 cos 1=97 + M3
M1 cos 0=468 M2 cos 1=213 + M3
The set of equations (6.81) is linear for the unknowns M1 , M2 , and M3
5
65
6 5
6
0=92044
0=9243
1
M1
0=99995
7 0=12884 0=38868 1 8 7 M2 8 = 7 0=96377 8
(6.82)
M3
0=89247 0=35021 1
0=73509
6. Applied Mechanisms
333
75.00
29.6
0
19.7
0
84.28
FIGURE 6.10. The main four-bar linkage of the windshield wiper at the initial
position measured in [ cm].
with the following solution:
5
6 5
6
2=5284
M1
7 M2 8 = 7 3=8043 8
M3
0=18911
(6.83)
The three factors M1 , M2 , M3 should be used to nd the four links’ length.
M1 =
g
d
M2 =
g
f
M3 =
d2 e2 + f2 + g2
2df
(6.84)
So, we may set the length of one of the links, based on the physical situation
and solve the equations for the other three lengths. Traditionally, we use
d = 1 and nd the remaining lengths. Then, the designed mechanism can
be magnied or shrunk to t the required geometry. In this example, we nd
d=1
e = 2=8436
f = 0=66462
g = 2=5284
(6.85)
Assuming a distance g = 75 cm 29=5 in for a real passenger car, between
the left and right xed joints P and Q , we nd the following dimensions:
d = 296 mm
f = 197 mm
e = 843 mm
g = 750 mm
(6.86)
Such a mechanism is shown in Figure 6.10 at the initial position.
Example 239 F Equal sweep angles for input and output links.
Let us place the second matching point of the windshield wiper mechanism
in Example 238 exactly in the middle of the total sweep angles
22
=
42
=
157=6 + 69=5
= 113=55 deg 1=982 rad
2
157=2 + 26=8
= 92 deg 1=605 rad
2
(6.87)
334
6. Applied Mechanisms
The rst and second sweep angles for such matching points would be equal.
Having equal sweep angles makes the motion of the wipers more uniform,
although it cannot guarantee that the angular speed ratio of the left and
right blades remains constant.
The matching points for the main four-bar linkage of the windshield wiper
with equal sweep angles are indicated in Table 6=3.
Table 6=3 - Equal sweep angle matching points for the four-bar linkage of
the double-arm parallel mechanism shown in Figure 6.9.
Matching
Input angle
Output angle
1
21 = 157=6 deg 2=75 rad
41 = 157=2 deg 2=74 rad
2
22 = 113=55 deg 1=982 rad 42 = 92 deg 1=605 rad
3
23 = 69=5 deg 1=213 rad
43 = 26=8 deg 0=468 rad
Substituting the angles in Freudenstein’s equation (6.24) provides us with
three equations:
M1 cos 2=74 M2 cos 2=75 + M3
M1 cos 1=605 M2 cos 1=982 + M3
M1 cos 0=468 M2 cos 1=213 + M3
= cos (2=74 2=75)
= cos (1=605 1=982)
= cos (0=468 1=213)
(6.88)
The set of equations can be written in a matrix form for the three unknowns
M1 , M2 , and M3
5
65
6 5
6
0=920 44
0=924 3
1
M1
0=99995
7 =0332
=3993
1 8 7 M2 8 = 7 =929589 8
(6.89)
M3
0=892 47 0=350 21 1
0=73509
with the solution.
5
6 5
6
M1
0=276
7 M2 8 = 7 0=6 8
M3
0=699
(6.90)
Using d = 1 and the three factors M1 , M2 , and M3
M1 =
g
d
M2 =
g
f
M3 =
d2 e2 + f2 + g2
2df
(6.91)
we can nd the links’ length.
d=1
e = 0=803
f = 0=46
g = 0=276
(6.92)
Assuming a distance g = 75 cm 29=5 in between the left and right xed
joint P and Q , we nd the following dimensions for a real passenger car:
d = 2717 mm
f = 1250 mm
e = 2182 mm
g = 750 mm
(6.93)
6. Applied Mechanisms
335
These dimensions do not show a practical design because the links may
be longer than the width of the vehicle.
It shows that the designed mechanism is highly dependent on the second
match point. So, it might be possible to design a desirable mechanism by
choosing a suitable second match point.
Example 240 F Second match point and link’s length.
To examine how the design of the windshield wiper mechanism in Example 238 is dependent on the second match point, let us set
22 =
157=6 + 69=5
= 113=55 deg 1=982 rad
2
(6.94)
and make 42 a variable. The three matching points for the main four-bar
linkage of the windshield wiper are indicated in Table 6=4.
Table 6=4 - Variable second match point for the four-bar linkage of
the double-arm parallel mechanism shown in Figure 6.9.
Matching
Input angle
Output angle
1
21 = 157=6 deg 2=75 rad
41 = 157=2 deg 2=74 rad
2
22 = 113=55 deg 1=982 rad
42
3
23 = 69=5 deg 1=213 rad
43 = 26=8 deg 0=468 rad
The Freudenstein’s equation (6.24) provides us with
M1 cos 2=74 M2 cos 2=75 + M3
M1 cos 42 M2 cos 1=982 + M3
M1 cos 0=468 M2 cos 1=213 + M3
= cos (2=74 2=75)
= cos (42 1=982)
= cos (0=468 1=213)
(6.95)
The set of equations gives the following solutions:
M1
=
M2
=
M3
=
79=657 cos(42 1=9815) 70=96
79=657 cos 42 + 13=828
93=642 cos(42 1=981) + 13=681 cos 42 81=045
65=832 cos 42 + 11=428
32=357 25=959 cos(42 1=981) + 53=184 cos(42 )
11=428 + 65=83 cos(42 )
(6.96)
Having g = 75 cm 29=5 in between the left and right xed joint P and Q
as the ground link, and using the factors M1 , M2 , and M3
M1 =
g
d
M2 =
g
f
M3 =
d2 e2 + f2 + g2
2df
(6.97)
we can nd the length of the other links d, e, and f as functions of 42 .
Figure 6.11 illustrates how the angle 42 aects the lengths of the links.
To t the mechanism under the hood in a small space, we need to have
the lengths d and f much shorter than the ground g. Based on Figure 6.11,
336
6. Applied Mechanisms
Length [cm]
c
a
b
T42 [rad]
26.8
45
64
82
101
120
138
157.2
T42 [deg]
FIGURE 6.11. The length of links d, e, and f as functions of 42 .
Length [cm]
a
b
c
T42 [rad]
91.7
94
98
101
104
108
111
114.6
T42 [deg]
FIGURE 6.12. Maginication of the plot for the length of links d, e, and f as
functions of 42 , around the optimal design.
6. Applied Mechanisms
75.00
337
7.70
5.33
77.21
FIGURE 6.13. The nalized main four-bar linkage of the windshield wiper at the
initial position measured in [ cm].
a possible solution would be around 42 = 100 deg. Figure 6.12 illustrates a
magnied view around 42 = 100 deg.
To have the length of d and f less than 100 mm 3=94 in we pick 42 =
99=52 deg 1=737 rad. Then the factors M1 , M2 , and M3 are
6 5
6
9=740208376
M1
7 M2 8 = 7 14=06262379 8
M3
3=032892944
5
(6.98)
and the links for g = 75 cm 29=5 in are
d = 77 mm
f = 53=3 mm
e = 772 mm
g = 750 mm
(6.99)
These numbers show a compact and reasonable mechanism. Figure 6.13
illustrates the nalized four-bar linkage of the windshield wiper at the initial
position.
Example 241 F Designing a dyad to attach a motor.
The main four-bar linkage of a windshield wiper is a rocker-rocker mechanism because both the input and the output links must oscillate between
two specic limits. To run the wipers and lock them at the limits, a two-link
dyad can be designed. First we set the point of installing a rotary motor
according to the physical conditions. Let point S , as shown in Figure 6.14,
be the point at which we install the electric motor to run the mechanism.
The next step would be to select a point on the input link to attach the second link of the dyad. Although joint E is usually the best choice, we select
a point on the extension of the input link, indicated by G. There must be a
dyad between joints G and S with lengths s and t. When the mechanism
is at the initial position, joint G is at the longest distance form the motor
S , and when it is at the nal position, joint G is at the shortest distance
form the motor S . Let us show the longest distance between S and G by o
338
6. Applied Mechanisms
75.00
M
7.70
N
A
P
5.33
B
D
C
Z
69.5°
7.08
38.29
45.38
157.6°
77.21
38.29
FIGURE 6.14. The nal design of the windshield mechanism at the initial and
nal positions.
and the shortest distance by v.
o = orqjhvw glvwdqfh ehwzhhq S dqg G
v = vkruwhvw glvwdqfh ehwzhhq S dqg G
s> t = g|dg ohqjwkv ehwzhhq S dqg G
When S and G are at the maximum distance, the two-link dyad must
be along each other, and when S and G are at the minimum distance, the
two-link dyad must be on top of each other. Therefore,
o =t+s
v=ts
(6.100)
where s is the shortest link, and t is the longest link of the dyad. Solving
Equations (6.100) for s and t provides
s=
ov
2
o+v
2
(6.101)
v = 312=1 mm
(6.102)
t = 382=9 mm
(6.103)
t=
In this example we measure
o = 453=8 mm
and calculate for s and t
s = 70=8 mm
The nal design of the windshield mechanism and the running motor is
shown in Figure 6.14 at the initial and nal positions. The shorter link of
the running dyad, s, must be attached to the motor at S , and the larger
6. Applied Mechanisms
B
339
N
C
A
M
FIGURE 6.15. Double A arm suspension in a four-bar linkage mechanism.
link, t, connects joint G to the shorter link to F. The motor will turn the
shorter link, S F continuously at an angular speed $, while the longer link,
FG, will run the mechanism and protect the wiper links from going beyond
the initial and nal angles.
Example 242 Application of four-bar linkage in a vehicle.
The double A-arm suspension is a very popular mechanism for independent suspension of road vehicles. Figure 6.15 illustrates a double A-arm
suspension and its equivalent kinematic model. We attach the wheel to a
coupler point at F. The double A-arm is also called double wishbone
suspension.
6.2 Slider-Crank Mechanism
A slider-crank mechanism is shown in Figure 6.16. A slider-crank mechanism is a four-bar linkage. Link number 1 is the ground, which is the base
and reference link. Link number 2 P D is usually the input link, which
is controlled by the input angle 2 . Link number 4 is the slider link that is
usually considered as the output link. The output variable is the horizontal
distance v between the slider and a xed point on the ground, which is
usually the revolute joint at P . If the slider slides on a at surface, we
dene the horizon by a straight line parallel to the at surface and passing
through P . The link number 3 DE is the coupler link with angular position 3 , which connects the input link to the output slider. This mechanism
is called the slider-crank because in most applications, the input link is a
crank link that rotates 360 deg, and the output is a slider.
The position of the output slider, v, and the angular position of the
coupler link, 3 , are functions of the link’s length and the value of the
input variable 2 . The functions are
s
J ± J2 4K
v =
(6.104)
¶
μ 2
h d sin 2
(6.105)
3 = sin1
e
340
6. Applied Mechanisms
A
3
B
4
2
T2
s
M
1
FIGURE 6.16. A slider-crank mechanism.
y
A
2
3
r2
T2
M
r3
r1
T3
B
G
4
r4
T4
T1
1
x
FIGURE 6.17. Expressing a slider-crank mechanism by a vector loop.
where
J = 2d cos 2
K = d2 + h2 e2 2dh sin 2
(6.106)
(6.107)
Proof. We show the slider-crank mechanism by a vector loop, as shown in
Figure 6.17. The direction of each vector is arbitrary, however the angles
should be associated to the vector’s direction and be measured from the
positive direction of the {-axis. The links and their expression vectors are
shown in Table 6=5.
Table 6=5 - Vector representation of the slider-crank mechanism
shown in Figure 6.17.
Link Vector Length
Angle
Variable
J
1
r1
v
1 = 180 deg
v
J
2
r2
d
2
2
J
3
r3
e
3
3
J
4
r4
h
4 = 90 deg
6. Applied Mechanisms
341
The equation of the vector loop is
J
r1 + J r2 J r3 J r4 = 0
(6.108)
We may decompose the vector equation (6.108) into sin and cos components
to generate two independent equations.
d sin 2 e sin 3 h = 0
d cos 2 e cos 3 v = 0
(6.109)
(6.110)
To derive the relationship between the input angle 2 and the output
position v, the coupler angle 3 must be eliminated between Equations
(6.109) and (6.110). Transferring the terms containing 3 to the other side
of the equations, and squaring both sides, we get
(e sin 3 )2
= (d sin 2 h)2
(6.111)
2
2
(6.112)
(e cos 3 )
= (d cos 2 v)
By adding Equations (6.111) and (6.112), we derive the following equation:
v2 2dv cos 2 + d2 + h2 e2 2dh sin 2 = 0
(6.113)
v2 + Jv + K = 0
(6.114)
J = 2d cos 2
K = d2 + h2 e2 2dh sin 2
(6.115)
(6.116)
or
where
Equation (6.114) is quadratic in v and provides two solutions:
s
J ± J2 4K
v=
2
(6.117)
To nd the relationship between the input angle 2 and the coupler angle
3 , we can use Equations (6.109) or (6.110) to solve for 3 .
μ
¶
h d sin 2
1
3 = sin
(6.118)
e
μ
¶
v d cos 2
3 = cos1
(6.119)
e
Equations (6.117) and (6.118) calculate the output and coupler variables
v and 3 as functions of the input angle 2 , provided that the lengths d, e,
and h are given.
342
6. Applied Mechanisms
174q
A
M
A
M
0.5 0.5
45q
2.69
1.28
(a)
(b)
5.9q
45q
FIGURE 6.18. Two possible congurations of a slider-crank mechanism having
the same input angle 2 .
Example 243 Two possible congurations for a slider-crank mechanism.
At any angle 2 , and for suitable values of d, e, and h, Equation (6.117)
provides two values for the output variable v. Both solutions are possible
and provide two dierent congurations for the mechanism. A suitable set
of (d, e, h) is the numbers that make the radical in Equations (6.117) real.
As an example, consider a slider-crank mechanism at 2 = @4 rad =
45 deg with the lengths
d=1
e=2
h = 0=5
(6.120)
To solve the possible congurations, we start by calculating the coe!cients
of the quadratic equation (6.114)
J = 2d cos 2 = 1=4142
K = d2 + h2 e2 2dh sin 2 = 3=4571
Employing Equation (6.117) provides
s
½
J ± J2 4K
2=696
=
v=
1=282
2
(6.121)
(6.122)
(6.123)
The corresponding coupler angle 3 can be calculated form either Equation
(6.118) or (6.119).
¶
¶
μ
μ
h d sin 2
v d cos 2
1
1
= cos
3 = sin
e
e
½
3=037 rad 174 deg
(6.124)
0=103 rad 5=9 deg
Figure 6.18 depicts the two possible congurations of the mechanism for
2 = 45 deg.
6. Applied Mechanisms
343
Example 244 Velocity analysis of a slider-crank mechanism.
The velocity analysis of a slider-crank mechanism is possible by taking a
time derivative of Equations (6.109) and (6.110),
g
(d sin 2 e sin 3 h) = d $ 2 cos 2 e $ 3 cos 3 = 0
gw
(6.125)
g
(d cos 2 e cos 3 v) = d $ 2 sin 2 + e $ 3 sin 3 vb = 0
gw
(6.126)
where
$ 2 = b 2
$ 3 = b 3
(6.127)
Assuming 2 and $ 2 are given values, and v, 3 are known from Equations (6.104) and (6.105), we may solve Equations (6.125) and (6.126) for
vb and $ 3 .
vb =
$3
=
sin (3 2 )
d$ 2
cos 3
cos 2 d
$2
cos 3 e
(6.128)
(6.129)
Example 245 Velocity of moving joints for a slider-crank mechanism.
Having the coordinates 2 , 3 , v and velocities $ 2 , $ 3 , vb enables us to
calculate the absolute and relative velocities of points D and E shown in
Figure 6.17. The absolute velocities of points D and E are
J
vD
$ 2 × J r2
6 5
6
6 5
d$ 2 sin 2
d cos 2
0
= 7 0 8 × 7 d sin 2 8 = 7 d$ 2 cos 2 8
0
0
$2
=
J
5
(6.130)
5
6
sin (3 2 )
d$ 2 :
9
cos 3
J
:
vE = vb ˆ~ = 9
7
8
0
0
(6.131)
and the velocity of point E with respect to point D is
J
vE@D
=
J
5
vE J vD
6
6
5
sin (3 2 )
d$2 sin 2
d$ 2 :
9
cos
3
: 7 d$ 2 cos 2 8
= 9
7
8
0
0
0
5
6
$2
d$ 2 sin 2 + d cos
3 sin ( 3 2 )
8
= 7
d$2 cos 2
0
(6.132)
344
6. Applied Mechanisms
The velocity of point E with respect to D can also be found as
¡
¢
J
vE@D = J U2 2 vE = J U2 2 vE = J U2 2 $ 3 × 2 r3 = J $ 3 × J r3
6 5
6
5
6 5
e$ 3 sin 3
0
e cos 3
(6.133)
= 7 0 8 × 7 e sin 3 8 = 7 e$ 3 cos 3 8
0
0
$3
Equations (6.132) and (6.133) are both correct and convertible to each
other.
Example 246 Acceleration analysis of a slider-crank mechanism.
The acceleration analysis of a slider-crank mechanism is possible by taking another time derivative from Equations (6.125) and (6.126),
g
(d $ 2 cos 2 e $ 3 cos 3 )
gw
= d2 cos 2 e3 cos 3 d$ 22 sin 2 + e$ 23 sin 3 = 0
(6.134)
g
(d $ 2 sin 2 + e $ 3 sin 3 v)
b
gw
= d2 sin 2 e3 sin 3 + d$ 22 cos 2 + e$ 23 cos 3 v̈ = 0
(6.135)
where,
2 = $b 2
3 = $b 3
(6.136)
Assuming 2 , $ 2 , and 2 are given values as the kinematics of the input
b $ 3 are
link, v, 3 , are known from Equations (6.104) and (6.105), and v,
known from Equations (6.128) and (6.129), we may solve Equations (6.134)
and (6.135) for v̈ and 3 .
v̈ =
3
=
d2 sin (2 + 3 ) + e$ 23 cos 23 + d$ 22 cos (2 3 )
cos 3
d2 cos 2 d$ 22 sin 2 + e$ 23 sin 3
e cos 3
(6.137)
(6.138)
Example 247 Acceleration of moving joints of a slider-crank mechanism.
Having the angular kinematics of a slider-crank mechanism 2 , 3 , v, $ 2 ,
b 2 , 3 , and v̈ are necessary and enough to calculate the absolute and
$ 3 , v,
relative accelerations of points D and E, shown in Figure 6.17.
The absolute acceleration of points D and E are
¢
¡
J
aD = J 2 × J r2 + J $ 2 × J $ 2 × J r2
6
5
d2 sin 2 d$ 22 cos 2
(6.139)
= 7 d2 cos 2 d$ 22 sin 2 8
0
6. Applied Mechanisms
B
345
B
A
M
180q
M
A
(a)
360q
(b)
FIGURE 6.19. Limit position for a slider-crank mechanism.
J
aE
= v̈ ˆ~
5
6
d2 sin (2 + 3 ) + e$ 23 cos 23 + d$ 22 cos (2 3 )
9
:
cos 3
:(6.140)
= 9
7
8
0
0
The acceleration of point E with respect to point D is
¢
¡
J
aE@D = J 3 × J r3 + J $ 3 × J $ 3 × J r3
6
5
e3 sin 3 e$ 23 cos 3
= 7 e3 cos 3 e$ 23 sin 3 8
0
(6.141)
Example 248 Limit positions for a slider-crank mechanism.
When the output slider of a slider-crank mechanism stops while the input
link can turn, we say the slider is at a limit position. It occurs when the
angle between the input and coupler links is either 180 deg or 360 deg. Limit
positions of a slider-crank mechanism are usually dictated by the design
requirements. A limit position for a slider-crank mechanism is shown in
Figure 6.19.
We show the limit angle of the input link by 2O1 , 2O2 , and the corresponding horizontal distance of the slider by vPd{ , vplq . They can be
calculated by the equations:
¸
h
1
(6.142)
2O1 = sin
e+d
q
2
(e + d) h2
(6.143)
vPd{ =
2O2
vplq
1
h
ed
= + sin
q
=
(e d)2 h2
¸
(6.144)
(6.145)
346
6. Applied Mechanisms
The length of stroke that the slider travels repeatedly would be
q
q
v = vPd{ vplq = (e + d)2 h2 (e d)2 h2
(6.146)
Example 249 F Quick return slider-crank mechanism.
Consider a slider-crank with a rotating input link at a constant angular
velocity $2 . The required time for the slider to move from vplq to vPd{ is
¸
¸¶
μ
h
h
2O2 2O1
1
1
1
sin
(6.147)
w1 =
+ sin
=
$2
$2
ed
e+d
and the required time for returning from vPd{ to vplq is
¸
¸¶
μ
h
h
2 2O2
1
sin1
w2 = O1
sin1
=
$2
$2
e+d
ed
(6.148)
If h = 0, then
2O1 = 0
2O2 = 180 deg
and therefore,
w1 = w2 =
$2
(6.149)
(6.150)
However, when h ? 0 then,
w2 ? w1
(6.151)
and the slider returns to vplq faster. Such a mechanism is called a quick
return mechanism.
6.3 Inverted Slider-Crank Mechanism
An inverted slider-crank mechanism is shown in Figure 6.20. It is a fourlink mechanism. Link number 1 is the ground link, which is the base and
reference link. Link number 2 P D is usually the input link, which is
controlled by the input angle 2 . Link number 4 is the slider link and is
usually considered as the output link. The slider link has a revolute joint
with the ground and a prismatic joint with the coupler link 3 DE. The
output variable can be the angle of the slider with the horizon, or the length
DE. The link number 3 DE is the coupler link with angular position 3 .
If we attach the coupler link of a slider-crank mechanism to the ground,
an inverted slider-crank mechanism is made. Changing the grounded link
produces a new mechanism that is called an inversion of the previous mechanism. Hence, the inverted slider-crank is an inversion of a slider-crank
mechanism.
6. Applied Mechanisms
347
T3
3
A
4
B
2
T2
T1
1
M
T4
N
FIGURE 6.20. An inverted slider-crank mechanism.
The angular position of the output slider 4 and the length of the coupler
link e are functions of the length of the links and the value of the input
variable 2 . These variables are:
p
(6.152)
e = ± d2 + g2 h2 2dg cos 2
Ã
!
s
2
K ± K 4JL
4 = 3 + = 2 tan1
(6.153)
2
2J
where
J = g h d cos 2
K = 2d sin 2
L = d cos 2 g h
(6.154)
(6.155)
(6.156)
Proof. We show the inverted slider-crank mechanism by a vector loop as
shown in Figure 6.21. The direction of each vector is arbitrary, however,
the angles should be associated with the vector’s direction and be measured
from positive direction of the {-axis. The links and their expression vectors
are shown in Table 6=6.
Table 6=6 - Vector representation of the inverted slider-crank
mechanism shown in Figure 6.21.
Link Vector Length
Angle
Variable
J
1
r1
g
1 = 180 deg
g
J
2
r2
d
2
2
J
3
r3
e
3
3 or 4
J
4
r4
h
4 = 3 + 90 deg
The vector loop is
J
r1 + J r2 J r3 J r4 = 0
(6.157)
348
6. Applied Mechanisms
y
T3
A
3
r2
2
T2
B
4
r3
r4
T1
r1
1
M
T4
x
N
FIGURE 6.21. Kinematic model of an inverted slider-crank mechanism.
which can be decomposed into sin and cos components.
³
´
d sin 2 e sin 4 h sin 4 = 0
2
³
´
g + d cos 2 e sin 4 h cos 4 = 0
2
(6.158)
(6.159)
To derive the relationship between the input angle 2 and the output 4 ,
we eliminate e between Equations (6.158) and (6.159) and nd
(d cos 2 g) cos 4 + d sin 2 sin 4 h = 0
(6.160)
To have a better expression suitable for computer programming, we may
use trigonometric equations
4
2
sin 4 =
2 4
1 + tan
2
2 tan
4
2
cos 4 =
2 4
1 + tan
2
1 tan2
(6.161)
to transform Equation (6.160) to a more useful equation
J tan2
4
4
+ K tan
+L =0
2
2
(6.162)
where, J, K, and L are functions of the input variable.
J = g h d cos 2
K = 2d sin 2
L = d cos 2 g h
(6.163)
(6.164)
(6.165)
Equation (6.162) is quadratic in tan (4 @2) and can be used to nd the
output angle 4 .
Ã
!
s
K ± K 2 4JL
1
4 = 2 tan
(6.166)
2J
6. Applied Mechanisms
349
To nd the relationship between the input angle 2 and the coupler length
e, we may solve Equations (6.158) and (6.159) for sin 4 and cos 4
sin 4
cos 4
de cos 2 dh sin 2 + eg
e2 + h2
de sin 2 dh cos 2 + hg
= e2 + h2
=
(6.167)
(6.168)
or substitute (6.166) in (6.160) and solve for e. By squaring and adding
Equations (6.167) and (6.168), we nd the following equation:
d2 e2 + g2 h2 2dg cos 2 = 0
(6.169)
which must be solved for e.
p
e = ± d2 + g2 h2 2dg cos 2
(6.170)
Example 250 Two possible congurations for an inverted slider-crank mechanism.
At any angle 2 , and for suitable values of d, g, and h, Equations (6.152)
and (6.153) provide two values for the output e and coupler angles 4 . Both
solutions are possible and provide two dierent congurations for the mechanism. A suitable set of (d, g, h) are the numbers that make the radicals in
Equations (6.152) and (6.153) real.
Consider an inverted slider-crank mechanism at 2 = @4 rad = 45 deg
with the lengths
d=1
h = 0=5
g=3
(6.171)
The parameters of Equation (6.152) would be
J = g h d cos 2 = 1=792 9
K = 2d sin 2 = 1=414 2
L = d cos 2 g h = 2=792 9
Now, Equation (6.162) gives two real values for 4
½
1=48 rad 84=8 deg
4 2=08 rad 120 deg
(6.172)
(6.173)
Using 4 = 1=48 rad, Equations (6.167) provides us with
e 2=33
(6.174)
and when 4 = 1=732 rad we get
e = 2=28
(6.175)
Figure 6.22 depicts the two congurations of the mechanism for 2 =
45 deg.
350
6. Applied Mechanisms
2.33
84.8q
A
M
B
45q
N
3
2.2
8
A
45q
0.5
0.5
M
(a)
N
B
120q
(b)
FIGURE 6.22. Two congurations of an inverted slider crank mechanism for
2 = 45 deg.
Example 251 Velocity analysis of an inverted slider-crank mechanism.
The velocity analysis of a slider-crank mechanism can be found by taking
a time derivative of Equations (6.158) and (6.159),
g
(d sin 2 + e cos 4 h sin 4 )
gw
= d $ 2 cos 2 e $ 4 sin 4 + eb cos 4 h $ 4 cos 4 = 0
(6.176)
g
(d cos 2 + e cos 4 h cos 4 g)
gw
= d $2 sin 2 e $ 4 sin 4 + eb cos 4 + h $ 4 sin 4 = 0
(6.177)
where
$ 2 = b 2
$ 4 = $ 3 = b 4
(6.178)
Assuming 2 and $ 2 are given, and e, 4 are known from Equations
(6.152) and (6.153), we can solve Equations (6.176) and (6.177) for eb and
$4.
d
$ 2 [e cos (4 2 ) h sin (4 2 )]
e
d
= $ 3 = $ 2 sin (2 4 )
e
eb =
(6.179)
$4
(6.180)
Example 252 Velocity of moving joints of inverted slider-crank mechanism.
Having the coordinates 2 , 4 , e and velocities $ 2 , $ 4 , eb enables us to
calculate the absolute and relative velocities of points D and E shown in
Figure 6.21. The absolute and relative velocities of points D and E are
J
vD
$ 2 × J r2
6 5
6
6 5
d$ 2 sin 2
0
d cos 2
= 7 0 8 × 7 d sin 2 8 = 7 d$ 2 cos 2 8
0
0
$2
=
5J
(6.181)
6. Applied Mechanisms
J
J
J $4 ×
5
J
6
r4
5
6 5
6
0
h cos 4
h$ 4 sin 4
= 7 0 8 × 7 h sin 4 8 = 7 h$ 4 cos 4 8
$4
0
0
¡ J ¢
r3
5
6 5
6 5
6
0
e cos 4
e$ 4 sin 4
= 7 0 8 × 7 e sin 4 8 = 7 e$ 4 cos 4 8
$4
0
0
=
vE3 @D
J
J
=
vE4
vE3 @E4
vE3
=
351
(6.182)
J $3 ×
vE3 @D + J vD
6 5
6
e$ 4 sin 4
d$ 2 sin 2
= 7 e$ 4 cos 4 8 + 7 d$ 2 cos 2 8
0
0
5
6
e$ 4 sin 4 d$ 2 sin 2
= 7 d$ 2 cos 2 e$ 4 cos 4 8
0
=
J
5
(6.183)
J
5
(6.184)
vE3 J vE4
6 5
6
e$ 4 sin 4 d$ 2 sin 2
h$ 4 sin 4
= 7 d$ 2 cos 2 e$ 4 cos 4 8 7 h$ 4 cos 4 8
0
0
5
6
$ 4 h sin 4 d$ 2 sin 2 + e$ 4 sin 4
= 7 d$ 2 cos 2 $ 4 h cos 4 e$ 4 cos 4 8
(6.185)
0
Example 253 Acceleration analysis of inverted slider-crank mechanism.
The acceleration analysis of an inverted slider-crank mechanism can be
found by taking another time derivative from Equations (6.176) and (6.177),
´
g ³
d $ 2 cos 2 e $ 4 sin 4 + eb cos 4 h $ 4 cos 4
gw
= d2 cos 2 d$ 22 sin 2 e4 sin 4 + e$ 24 cos 4
b 4 sin 4 h4 cos 4 + h$ 2 sin 4 = 0
+ë cos 4 e$
(6.186)
4
´
g ³
d $ 2 sin 2 e $ 4 sin 4 + eb cos 4 + h $ 4 sin 4
gw
= d2 sin 2 + d$ 22 cos 2 e4 sin 4 e$ 24 cos 4
b 4 sin 4 + h4 sin 4 + h$ 2 cos 4 = 0
+ë cos 4 e$
4
(6.187)
where
2 = $b 2
4 = 3 = $b 4 = $b 3
(6.188)
352
6. Applied Mechanisms
Assuming 2 , $ 2 , and 4 are given values as the kinematics of the input
b $4
link, e, 4 , are known from Equations (6.152) and (6.153), and also e,
are known from Equations (6.179) and (6.180), we may solve Equations
(6.186) and (6.187), for ë and 4
ë =
4
=
F7 F12 F9 F10
F7 F11 F8 F10
F9 F11 F8 F12
F7 F11 F8 F10
(6.189)
(6.190)
where,
F7
F8
F9
F10
F11
F12
= sin 4
= e cos 4 + h sin 4
b 4 cos 4
= d2 sin 2 + d$ 22 cos 2 2e$
2
2
+e$ 4 sin 4 h$ 4 cos 4
= cos 4
= e sin 4 + h cos 4
b 4 sin 4
= d2 cos 2 d$ 22 sin 2 + 2e$
2
2
+e$ 4 cos 4 + h$ 4 sin 4
(6.191)
Example 254 Application of inverted slider mechanism in vehicles.
The McPherson strut suspension is a very popular mechanism for independent front suspension of street cars. Figure 6.23 illustrates a McPherson
strut suspension and its equivalent kinematic model. We attach the wheel
to a coupler point at F.
The piston rod of the shock absorber serves as a kingpin axis at the top
of the strut. At the bottom, the shock absorber pivots on a ball joint on
a single lower arm. The McPherson strut, also called the Chapman strut,
was invented by Earl McPherson in the 1940v. It was rst introduced on
the 1949 Ford Vedette, and also adopted in the 1951 Ford Consul, and then
become one of the dominating suspension systems because of its compactness
and low cost.
6.4 Instant Center of Rotation
In the general planar motion of a rigid body, at a given instant, the velocity
of any point of the body can be expressed as a rotation about an axis
perpendicular to the plane. The axis intersects the plane at a point called
the instantaneous center of rotation of the body with respect to the ground.
The instantaneous center of rotation is also called instant center, centro,
and pole.
6. Applied Mechanisms
353
N
B
C
M
A
FIGURE 6.23. The McPherson strut suspension is an inverted slider mechanism.
If the directions of the velocities of two dierent body points D and E
are known, the instant center of rotation L is at the intersection of the lines
perpendicular to the velocity vectors vD and vE . Such a situation is shown
in Figure 6.24(d).
In the exceptional case where the points D and E are such that the
velocity vectors vD and vE are perpendicular to the line DE then, the
instantaneous center of rotation L is at the intersection of DE with the line
joining the extremities of the velocity vectors. Such a situation is shown in
Figure 6.24(e).
There is an instant center of rotation between every two relatively moving
links. The instant center is a point common to both bodies that has the
same velocity in each body coordinate frame.
The three instant centers , L12 , L23 , and L13 between three links numbered
1, 2, and 3 lie on a straight line. This statement is called the Kennedy
theorem for three instant centers.
Proof. Consider the two ground connected and relatively moving bodies
in Figure 6.25. The ground is link number 1. The links number 2 and 3 are
pivoted to the ground at points P and Q , and are rotating with angular
velocities $ 2 and $ 3 with respect to the ground. The two links are contacted
at point F.
The revolute joint at P is the instant center L12 between link 2 and 1,
and the revolute joint at Q is the instant center L13 between links 3 and 1.
The velocity of any point of the link 2 with respect to the ground is perpendicular to the connection line of the point to P . Similarly, The velocity
of any point of the link 3 with respect to the ground is perpendicular to
the connection line of the point to Q . Therefore, the velocity of the contact
354
6. Applied Mechanisms
vB
B
I
vA
A
(a)
vA
vB
A
B
I
(b)
FIGURE 6.24. Determination of the instantaneous center of rotation L for a
moving rigid body.
nn
vC3
vC2
3
B
2
C
A
Z2
I12
M
v2
l2
t-t
v3
I23
1
l3
d
Z3
I13
N
FIGURE 6.25. A 3-link mechanism with the ground as link number 1, and two
moving links, numbers 2 and 3.
6. Applied Mechanisms
355
point F as a point of link 2 is vF2 , perpendicular to the radius P F, and
the velocity of F as a point of link 3 is vF3 , perpendicular to the radius
Q F.
The instant center of rotation L23 must be a point with the same ground
velocity in both bodies. Let us draw the normal line q q, and tangential
line w w to the curves of links 2 and 3, at the contact point F. The normal
components of vF2 and vF3 must be equal to keep the bodies in touch. So
is the velocity of any other point on q q. The only dierence between
the velocities vF2 and vF3 is in their tangent components. Therefore, the
instant center of rotation L23 must be on the normal line q q at a position
where the velocity of the point on both bodies are equal on the line w w.
The intersection of the normal line qq with the center line P Q is the only
possible point for the instant center of rotation L13 at which the velocity on
both bodies coincide. As their components on q q are equal, they must
be equal.
Let us dene
L12 L23
L13 L23
= o2
= o3
(6.192)
(6.193)
then, because the velocities of the two bodies must be equal at the common
instant center of rotation, we have
o2 $ 2 = o3 $ 3
or
o3
$2
=
=
$3
o2
(6.194)
1
1+
g
o2
(6.195)
where, g is the length of the ground link P Q .
Example 255 Number of instant centers.
There is one instant center between every two relatively moving bodies.
So, there are three instant centers between three bodies. The number Q of
instant centers between q relatively moving bodies is
Q=
q (q 1)
2
(6.196)
Thus, a four-bar linkage has six instant centers, L12 , L13 , L14 , L23 , L24 , L34 .
The symbol Llm indicates the instant center of rotation between links l and
m. Because two links have only one instant center, we have
Llm = Lml
(6.197)
356
6. Applied Mechanisms
I13
I14
I13
I34
3
4
I23
I24
I12
2
I14
I24
I23
I12
2
1
3
I34 4
1
(a)
I34
(b)
I13
2
I12
I23
4
3
I24
1
I14
(c)
FIGURE 6.26. The instant centers of rotation for a four-bar linkage, a slider-crank, and an inverted slider-crank mechanism.
The four instant centers of rotation for a four-bar linkage, a slider-crank,
and an inverted slider-crank mechanisms are shown in Figure 6.26. The
instant center of rotation for two links that slide on each other is at innity,
on a line normal to the common tangent. So, L14 in Figure 6.26(e) is on a
line perpendicular to the ground at innity, and L34 in Figure 6.26(f) is on
the perpendicular to the link 3 at innity.
Figure 6.27 depicts the 15 instant centers for a six-link mechanism.
Example 256 Application of instant center of rotation in vehicles.
Figure 6.28 illustrates a double D-arm suspension and its equivalent fourbar linkage kinematic model. The wheel will be fastened to the coupler link
DE, witch connects the upper D-arm EQ to the lower D-arm DQ . The Darms are connected to the body with two revolute joints at Q and P . The
body of the vehicle acts as the ground link for the suspension mechanism.
Points Q and P are, respectively, the instant centers of rotation for the
upper and lower arms with respect to the body. The intersection point of the
extension line for the upper and lower D-arms indicates the instant center
of rotation for the coupler with respect to the body. When the suspension
6. Applied Mechanisms
357
I13
I36
I35
I34
I26
I25
I45
3
I23
4
2
I24
I12
I46
5
I14
I56 6
1
I16
I15
FIGURE 6.27. Fifteen instant center of rotations for a 6-link mechanism.
moves, the wheel will rotate about point L with respect to the body. Point L
is called the roll center of the wheel and body.
Example 257 The instant centers of rotation may not be stationary.
When a mechanism moves, the instant centers of rotation may move,
if they are not at a xed joint with the ground. Figure 6.29 illustrates a
four-bar linkage at a few dierent positions and shows the instant centers
of rotation for the coupler with respect to the ground L13 . Point L13 will
move when the linkage moves, and traces a path shown in the gure.
For a given set of the lengths d> e> f> g, the coordinates of points P> D> E> Q
are as given in Table 6=7.
Table 6=7 - Coordinates of the joints of a four-bar linkage.
point {-coordinate |-coordinate
P
0
0
D
d cos 2
d sin 2
E
g + f cos 4
f sin 4
Q
g
0
358
6. Applied Mechanisms
I
N
B
I
M
A
FIGURE 6.28. The roll center of a double A-arm suspension and its equivalent
kinematic model.
I13
y
B
A
I12
M
N
I14
x
FIGURE 6.29. Path of motion for the instant center of rotation L13 .
6. Applied Mechanisms
y
x
x
y
I13 coordinate
x
359
T2 [rad]
y
FIGURE 6.30. The coordinates ({> |) of the instant center of rotation L13 as
functions of 2 .
The equation of the line P D is
|D
{ = (tan 2 ) {
{D
(6.198)
|E
({ g) = (tan 4 ) ({ g)
{E g
(6.199)
|=
and the equation of Q E is
|=
The instant center of rotation L23 of the coupler link DE with respect to
the ground P Q is at the intersection of the lines P D and Q E.
tan 4
tan 4 tan 2
tan 4 tan 2
= g
tan 4 tan 2
{ = g
(6.200)
|
(6.201)
Figure 6.30 depicts the coordinates ({> |) of the instant center of rotation
L13 for
d=1
e=2
f = 2=5
g=3
(6.202)
The position of the center of rotation L23 goes to innity whenever 4 =
2 and the input and output links become parallel. The Freudenstein’s equation (6.24)
(6.203)
M1 cos 4 M2 cos 2 + M3 = cos (4 2 )
under the parallel condition, yields
2
cos 2 =
1 M3
(d f) e2 + g2
=
M1 M2
2g (d f)
(6.204)
360
6. Applied Mechanisms
f
4
y
a=1
b=2
c = 2.5
d=3
2.1598 T2 <5.3014
f
2
1
-0.9818 T2 2.1598
x
5
3
FIGURE 6.31. The path of motion of the instant center of rotation L23 when
0 $ 2 $ 2.
The values of 2 = 4 for the sample data (6.202) are:
½
2
2
2
2=1598 rad
1 (d f) e + g
=
2 = cos
5=3014 rad
2g (d f)
(6.205)
Figure 6.31 shows the path of motion of the instant center of rotation L23
when 0 2 2. Starting from 2 = 0, L23 moves from (3> 0), indicated
by 1, to point 2 and goes to innity when 2 $ 2=1598. When 2 goes
beyond 2=1598, L23 appears at point 3 and moves to point 4 and innity
when 2 $ 5=3014. The instant center L23 again appears at point 5 and
moves to (3> 0) when 5=3014 ? 2 2.
Example 258 Sliding a slender on the wall.
Figure 6.32 illustrates a slender bar DE sliding at points D and F. We
have the velocity axes of the points D and F, and therefore, we can nd the
instant center of rotation L.
The coordinates of point L are a function of the parameter :
{L
|L
= k cot ¡
¢
= k + {L cot = k 1 + cot2 (6.206)
(6.207)
Eliminating between { and |, generates the path of motion for L.
¶
μ
{2
(6.208)
|L = k 1 + L2
k
Example 259 F Plane motion of a rigid body.
The plane motion of a rigid body is such that all points of the body move
only in parallel planes. So, to study the motion of the body, it is enough to
examine the motion of points in one plane.
6. Applied Mechanisms
y
B
361
G
I
C
G
vC
H
T
O
G
vA
x
A
FIGURE 6.32. A slender bar DE sliding at points D and F.
Y
y
G
B
P
G
x
rP
B
dB
o
G
rP
X
O
FIGURE 6.33. A rigid body in a planar motion.
362
6. Applied Mechanisms
Y
G
B
G
y
vP/Q
P
B
rP
G
o
G
dB
vo/Q
G
x
rP/Q
G
ro/Q
G
rQ
Q
X
O
FIGURE 6.34. Instant center of rotation T, for planar motion of a rigid body.
Figure 6.33 illustrates a rigid body in a planar motion and the corresponding coordinate frames. The position and velocity of a body point S
are:
J
rS = J dE + J UE E rS = J dE + J
(6.209)
E rS
¢
¡
J
vS = J db E + J $ E × J rS J dE = J db E + J $ E × J
(6.210)
E rS
where, J dE indicates the position of the moving origin r relative to the
xed origin R. The term J db E is the velocity of point r and, J $ E × J
E rS
is the velocity of point S relative to r, both in J.
J
vS@r = J $ E × J
E rS
(6.211)
Although it is not a correct view, it might sometimes help if we interpret
dE as the translational velocity and J $ E × J
E rS as the rotational velocity
components of J vS . Then, the velocity of any point S of the rigid body is
a superposition of the velocity J db E of an arbitrary point r and the angular
velocity J $ E × J
E rS of the points S about r.
The relative velocity vector J vS@r is perpendicular to the relative position
vector J
E rS . Employing the same concept we interpret that the velocity of
points S and r with respect to another point T are perpendicular to J
E rS@T
and J
E rr@T respectively. We may search for a point T, as the instantaneous
center of rotation, at which the velocity is zero. Points r, S , and T are
shown in Figure 6.34.
Assuming a position vector J rr@T for the instant center T, we dene
Jb
J
rr@T = dT J db E + eT J $ E × J db E
(6.212)
6. Applied Mechanisms
363
then, following (6.210), the velocity of point T can be expressed by
J
vT
Jb
dE + J $ E × J rT@r = J db E J $ E × J rr@T
³
´
Jb
dE J $ E × dT J db E + eT J $ E × J db E
³
´
Jb
dE dT J $ E × J db E eT J $ E × J $ E × J db E
=
=
=
= 0
(6.213)
Now, using the equations
J $E ×
³
we nd
J $E
J $E ×
Jb
J $E ·
dE
Jb
´
dE
= $ N̂
³
´
Jb
dE J $ E $ 2 J db E
=
J $E ·
= 0
¢
¡
1 + eT $ 2 J db E dT J $ E × J db E = 0
(6.214)
(6.215)
(6.216)
(6.217)
Because J db E and J $ E × J db E must be perpendicular, Equation (6.217)
provides
1 + eT $ 2 = 0
dT = 0
(6.218)
and therefore,
J
rT@r =
´
1 ³
Jb
$
×
d
J
E
E
$2
(6.219)
Example 260 F Instantaneous center of acceleration.
For planar motions of rigid bodies, it is possible to nd a body point with
zero acceleration. Such a point may be called the instantaneous center of
acceleration. When a rigid body is in a planar motion, we can express the
acceleration of a body point S , such as shown in Figure 6.34, as
¢
¡
J
aS = J d̈E + J E × J rS J dE
¢¢
¡
¡
+ J $ E × J $ E × J rS J dE
¢
¡
J
(6.220)
= J d̈E + J E × J
E rS + J $ E × J $ E × E rS
¢
¡
The term J $ E × J $ E × J
E rS is the centripetal acceleration, and the term
J
J E × E rS is the tangential acceleration. Because the motion is planar,
the angular velocity vector is always parallel to n̂ and N̂ unit vectors.
J $ E = $ N̂
J E = N̂
(6.221)
Therefore, the velocity J vS and acceleration J aS can be simplied to
J
2J
aS = J d̈E + J E × J
E rS $ E rS
(6.222)
364
6. Applied Mechanisms
We now look for a zero acceleration point V and express its position vector
by
J
rV@r = dV J d̈E + eV J E × J d̈E
(6.223)
and based on (6.222) we have,
J
aV
2J
d̈E + J E × J
E rV $ E rV
J
d̈E + J E × J rV@r $ 2 J rV@r
³
´
= J d̈E + J E × dV J d̈E + eV J E × J d̈E
³
´
$ 2 dV J d̈E + eV J E × J d̈E
³
´
= J d̈E + dV J E × J d̈E + eV J E × J E × J d̈E
=
=
J
dV $ 2 J d̈E eV $ 2 J E × J d̈E = 0
(6.224)
Simplifying the above equation yields
¢
¢
¡
¡
1 dV $ 2 eV 2 J d̈E + dV eV $ 2 J E × J d̈E = 0
(6.225)
and because J d̈E and J E × J d̈E are perpendicular, we must have
1 dV $ 2 eV 2 = 0
dV eV $ 2 = 0
(6.226)
and hence,
$2
1
eV = 2
(6.227)
2
2
$ +
$ + 2
The position vector of the instant center of acceleration is then equal to
³
´
1
J
2J
J
rV@r = 2
+
×
$
(6.228)
d̈
d̈
E
J
E
E
$ + 2
dV =
6.5 Coupler Point Curve
To provide exibility between wheels and the vehicle chassis, wheels are
attached to chassis by a mechanism. The most common suspension mechanisms are double D-arm and inverted slider-crank. The wheel of the vehicle
will then be attached to a point of the coupler link of the mechanism. The
relative motion of the wheel and chassis will be calculated by analysis of
the displacement of the coupler point.
6.5.1 Coupler Point Curve for Four-Bar Linkages
Figure 6.35 illustrates a four-bar linkage P Q DE and a coupler point at F.
When the mechanism moves, the coupler point F will move on a path.
6. Applied Mechanisms
365
y
C(xC, yC)
e
B
\
b
D
A
G
T3
J
l
E
c
a
T2
d
J
T4
T1
M
x
N
FIGURE 6.35. A four-bar linkage PQDE and a coupler point at F.
The path of the coupler point is called the coupler point curve. Considering 2 as the input variable of the mechanism, the parametric coordinates
of the coupler point curve ({F > |F ) are
{F
|F
= d cos 2 + h cos ( + )
= d sin 2 + h sin ( + )
(6.229)
(6.230)
where,
i
d sin 2
g d cos 2
q
4e2 i 2 (e2 + i 2 f2 )2
= tan1
e2 + i 2 f2
p
=
d2 + g2 2dg cos 2
= tan1
(6.231)
(6.232)
(6.233)
Proof. The position of the coupler point F in Figure 6.35 is dened by
the polar coordinates length h and angle with respect to the coupler
link DE, and by ({F > |F ) in the Cartesian coordinate frame J attached to
the ground. The length of the links are indicated by P D = d, DE = e,
Q E = f, and P Q = g. We show the angle _DQ P by and _EDQ by .
Let us draw a line o through D and parallel to the ground link P Q , then,
_Q Do
_FDo
#
= _DQ P = = #
= +
(6.234)
(6.235)
(6.236)
The global coordinates of point F are then
{F
|F
= d cos 2 + h cos #
= d sin 2 + h sin #
(6.237)
(6.238)
366
6. Applied Mechanisms
where, # comes from Equation (6.236). The angle can be calculated from
the cosine law in 4EDQ ,
cos =
e2 + i 2 f2
2ei
(6.239)
where, i = DQ . Applying the cosine law in 4DP Q shows that i is equal
to
p
i = d2 + g2 2dg cos 2
(6.240)
given by Equation (6.233).
For computer calculation ease, it is better to nd from the trigonometric equation
(6.241)
tan2 = sec2 1
and substitute sec from Equation (6.239).
q
2
4e2 i 2 (e2 + i 2 f2 )
= tan1
e2 + i 2 f2
(6.242)
The angle can be found from a tan equation based on the vertical
distance of point D from the ground link P Q .
= tan1
d sin 2
g d cos 2
(6.243)
Therefore, the coordinates {F and |F can be calculated as two parametric
functions of 2 for a given set of d, e, f, g, h, and .
Example 261 A poorly designed double D-arm suspension mechanism.
Figure 6.36 illustrates a double D-arm suspension mechanism and its
equivalent four-bar linkage kinematic model. Points P and Q are xed
joints on the body, and points D and E are moving joints attached to the
wheel supporting coupler link. Point F is on the spindle and supposed to
be the wheel center. When the wheel moves up and down, the wheel center
moves on a the couple point curve shown in the gure. The wheel center of
proper suspension mechanism is supposed to move vertically for the working
range suspension travel. However, the wheel center of the suspension of
Figure 6.36 moves on a high curvature path and generates an undesired
camber.
A small motion of the kinematic model of suspension is shown in Figure
6.37, and the actual suspension and wheel congurations are shown in the
gure.
6.5.2 Coupler Point Curve for a Slider-Crank Mechanism
Figure 6.38 illustrates a slider-crank mechanism and a coupler point at F.
When the mechanism moves, coupler point F will move on a coupler point
6. Applied Mechanisms
367
N
B
C
A
M
FIGURE 6.36. A double A arm suspension mechanism and its equivalent four-bar
linkage kinematic model.
curve with the following parametric equation:
{F
|F
= d cos 2 + f cos ( )
= d sin 2 + f sin ( )
(6.244)
(6.245)
The angle 2 is the input angle and acts as the parameter of the equations.
The angle can be calculated from
= sin1
d sin 2 h
e
(6.246)
Proof. We attach a planar Cartesian coordinate frame to the ground link
at P . The {-axis is parallel to the ground indicated by the sliding surface,
as shown in Figure 6.38. Drawing a line o through D and parallel to the
ground shows that
=
(6.247)
where is the angle between the coupler link and the ground.
The coordinates ({F > |F ) for the coupler point F are
{F
|F
= d cos 2 + f cos = d sin 2 + f sin (6.248)
(6.249)
To calculate the angle , we examine 4DHE and nd
sin =
d sin 2 h
DH
=
DE
e
(6.250)
that nalizes the proof of Equation (6.246). Therefore, the coordinates {F
and |F can be calculated as two parametric functions of 2 for a given set
of d, e, f, h, and .
368
6. Applied Mechanisms
B
N
N
B
C
C
A
M
A
M
N
N
B
B
A
A
M
M
FIGURE 6.37. A small motion of the kinematic model and the actual suspension
congurations.
Example 262 A centric and symmetric slider-crank mechanism.
Point F ({F > |F ) is the coupler point of a centric and symmetric slidercrank mechanism shown in Figure 6.39. It is centric because h = 0, and is
symmetric because d = e, and therefore, 2 = 4 . Point F is on the coupler
link DE and is at a distance ne from D, where 0 ? n ? 1.
The coordinates of point F are
{F
|F
= d cos 2 + nd cos 2 = d (1 + n) cos 2
= d sin 2 nd sin 2 = d (1 n) sin 2
(6.251)
(6.252)
and therefore,
cos 2 =
{F
d (1 + n)
sin 2 =
|F
d (1 n)
(6.253)
Using cos2 2 + sin2 2 = 1, we can show that the coupler point F will
move on an ellipse.
{2F
2
|F
+
2
2 =1
d2 (1 + n)
d2 (1 n)
(6.254)
6. Applied Mechanisms
y
C(xC, yC)
c
E
D
A
l
J
G
369
a
b
B
J
E
T2
e
x
s
M
FIGURE 6.38. A slider-crank mechanism and a coupler point at F.
y
A
b
a
yC
M
kb
C(xC, yC)
T4
T2
xC
B
FIGURE 6.39. A centric and symmetric slider-crank mechanism.
x
370
6. Applied Mechanisms
y
C(xC, yC)
c
J
D
A
E
a
E3
E
T2
E1
d
F
M
B
E2
T4 e
x
N
FIGURE 6.40. An inverted slider-crank mechanism and a coupler point at F.
6.5.3 Coupler Point Curve for Inverted Slider-Crank
Mechanism
Figure 6.40 illustrates an inverted slider-crank mechanism and a coupler
point at F. When the mechanism moves, the coupler point F will move on
a curve with the following parametric equation:
{F
|F
= d cos 2 + f cos ( 4 )
= d sin 2 + f sin ( 4 )
(6.255)
(6.256)
The angle 2 is the input angle and acts as a parameter, and 4 is the angle
of the output link, given by Equation (6.153).
Ã
!
s
K ± K 2 4JL
1
4 = 2 tan
(6.257)
2J
J = g h d cos 2
K = 2d sin 2
L = d cos 2 g h
(6.258)
(6.259)
(6.260)
Proof. We attach a planar Cartesian coordinate frame to the ground link
at P Q . Drawing a vertical line through F denes the variable angle =
_DFI as shown in Figure 6.40. We also dene three angles 1 = _DQ P ,
2 = _DQ E, and 3 = _EHI to simplify the calculations. From the
triangle 4DFH, we nd
(6.261)
= 3 and from quadrilateral ¤HI Q E, we nd
3 + _HI Q + 2 + 1 + _HEQ = 2
(6.262)
6. Applied Mechanisms
371
However,
_HEQ
=
_HI Q
=
2
2
(6.263)
(6.264)
and therefore,
3 + 2 + 1 = (6.265)
4 = ( 2 + 1 )
(6.266)
4 = 3
(6.267)
The output angle 4 is
and thus,
Now the angle may be written as
= 4 (6.268)
where 4 is the output angle, found in Equation (6.153).
Therefore, the coordinates {F and |F can be calculated as two parametric
functions of 2 for a given set of d, g, f, h, and .
6.6 F Universal Joint
The universal joint shown in Figure 6.41 is a mechanism to connect rotating
shafts that intersect at an angle *. The universal joint is also known as
Hook’s coupling, Hook joint, Cardan joint, or Yoke joint.
Figure 6.42 illustrates a practical universal joint. There are four links in
a universal joint: link number 1 is the ground, which has a revolute joint
with the input link 2 and the output link 4. The input and the output links
are connected with a cross-link 3. The universal joint is a three-dimensional
four-bar linkage for which the cross-link acts as a coupler link.
The driver and driven shafts make a complete revolution at the same
time, but the velocity ratio is not constant throughout the revolution. The
angular velocity of the output shaft 4 relative to the input shaft 2 is called
speed ratio and is a function of the angular position of the input shaft ,
and the angle between the shafts *.
=
cos *
$4
=
$2
1 sin2 * cos2 (6.269)
Proof. A universal joint may appear in many shapes, however, regardless
of how it is constructed, it has essentially the form shown in Figure 6.42.
Each connecting shaft ends in a X -shaped yoke. The yokes are connected
by a rigid cross-link. The ends of the cross-link are set in bearings in the
372
6. Applied Mechanisms
M
FIGURE 6.41. A universal joint.
1
C
Z2
4 Driven
B
1
Driver 2
M
Z4
3
A
Cross-link
D
FIGURE 6.42. A universal joint with four links: link 1 is the ground, link 2 is the
input, link 4 is the output, and the cross-link 3 is a coupler link.
6. Applied Mechanisms
373
y3
z1
x3
y1
y4
x1
Z2
z2
C
Driven
y2
Driver
B
A
Cross-link
x4
M
x2
D
Z4
x2
FIGURE 6.43. A separate illustration of the input, output, and the cross links
for a universal joint.
yokes. When the driver yoke turns, the cross-link rotates relative to the
yoke about its axis DE. The cross-link also rotates relative to the driven
yoke about the axis FG.
Although the driver and driven shafts make a complete revolution together, their velocity ratio is not constant throughout the revolution. A
disassembled illustration of the input, output, and the cross links is shown
in Figure 6.43.
The angular velocity of the cross-link may be shown by
1
1
1 $3 = 1 $2 + 2 $3 = 1 $4 + 4 $3
(6.270)
where, 1 $ 2 is the angular velocity of the driver yoke about the {2 -axis
and 12 $ 3 is the angular velocity of the cross-link about the axis DE relative
to the driver yoke expressed in the ground coordinate frame.
Figure 6.44 shows that the unit vectors ˆ2 and ˆ3 are along the arms of
the cross link, and the unit vectors ˆ~2 and ˆ~4 are along the shafts. Having
the angular velocity vectors,
6 5
5
6
$ 21 ˆ~1
$ 21
7
ˆ1 8 = 7 0 8
(6.271)
1 $2 =
0
n̂1
6 5
5
6
$ 41 ˆ~4
$ 41
7
ˆ4 8 = 7 0 8
(6.272)
1 $4 =
0
n̂4
374
6. Applied Mechanisms
y3 y4
Driver
T34
x2
x3
C
T2
y2
T32
B
x4
M
T4
A
Driven
D
FIGURE 6.44. A kinematic model for a universal joint.
5
6
ˆ~2
7 $ 32 ˆ2 8
2 $3 =
n̂2
5
6
ˆ~4
7 $ 34 ˆ4 8
4 $3 =
n̂4
5
6
$ 32 ˆ~3
3
7
ˆ3 8
2 $3 =
n̂3
5
6
ˆ~3
3
7 $ 34 ˆ3 8
4 $3 =
n̂3
(6.273)
(6.274)
we can simplify Equation (6.270) to
$ 32 ˆ~3 + $ 21 ˆ~2 = $ 41 ˆ~4 + $ 34 ˆ3
(6.275)
However, because the cross-link coordinate frame is right-handed, we have
ˆ~3 × ˆ3 = n̂3
(6.276)
($ 32 ˆ~3 + $ 21 ˆ~2 ) · n̂3 = ($ 41 ˆ~4 + $ 34 ˆ3 ) · n̂3
(6.277)
$ 21 ˆ~2 · n̂3 = $ 41 ˆ~4 · n̂3
(6.278)
and therefore,
that yields
Now the required equation for the speed ratio
=
= $ 41 @$ 21 is
ˆ~3 × ˆ3 · ˆ~2
$ 41
=
$ 21
ˆ~3 × ˆ3 · ˆ~4
(6.279)
The unit vector ˆ3 is perpendicular to ˆ~3 and ˆ~4 , we may write
ˆ3 = dˆ~4 × ˆ~3
(6.280)
ˆ~3 × ˆ3 = ˆ~3 × (dˆ~4 × ˆ~3 ) = d [ˆ~4 (ˆ~3 · ˆ~4 ) ˆ~3 ]
(6.281)
where d is a coe!cient. Now
6. Applied Mechanisms
375
and because
ˆ~3 · ˆ~2 = 0
(6.282)
we nd
$ 41
ˆ~2 · d [ˆ~4 (ˆ~3 · ˆ~4 ) ˆ~3 ]
=
$ 21
ˆ~3 · d [ˆ~4 (ˆ~3 · ˆ~4 ) ˆ~3 ]
cos *
ˆ~2 · ˆ~4
2 =
2
1 (ˆ~3 · ˆ~4 )
1 (ˆ~3 · ˆ~4 )
=
=
(6.283)
If we show the angular position of the input yoke by , then
= cos ˆ1 + sin n̂1
= cos * ˆ~1 sin * ˆ1
ˆ~3
ˆ~4
(6.284)
(6.285)
and the nal equation for the speed ratio is found as
=
cos *
$41
=
$ 21
1 sin2 * cos2 (6.286)
This formula shows that although both shafts complete one revolution
at the same time, the ratio of their angular speeds varies with the angle of
rotation (w) of the driver and is also a function of the shaft angle *. Thus,
even if the angular speed $ 21 of the drive shaft is constant, the angular
speed $ 41 of a driven shaft will not be uniform.
Example 263 F Graphical illustration of the universal joint speed ratio
.
Figure 6.45 depicts a three-dimensional plot for as a function of and
*. The -surface is plotted for one revolution of the drive shaft and every
possible angle between the two shafts.
? ? ?*?
(6.287)
A two-dimensional view for
is depicted in Figure 6.46. When * /
10 deg there is not much uctuation in speed ratio, however, when the angle
between the two shafts is more than 10 deg then the speed ratio cannot be
assumed constant. The universal joint get stuck when * = 90 deg, because
theoretically
= lqghi lqlwh
(6.288)
lim
*$90
The behavior of
as a function of and * can be better viewed in a
polar coordinate, as shown in Figure 6.47.
376
6. Applied Mechanisms
:
M
T
FIGURE 6.45. A three-dimensional plot for the speed ratio of a universal joint,
l as a function of the input angle and the angle between input and output
shafts *.
50
40
30
20
:
M 10q
T> rad @
FIGURE 6.46. A two-dimensional view of l as a function of the input angle and the angle between input and output shafts *.
6. Applied Mechanisms
10
M 0q
:
377
50
40
30
20
:
FIGURE 6.47. The behavior of speed ratio l as a function of and * in a polar
coordinate.
Example 264 F Maximum and minimum of $ 41 in one revolution.
The maximum value of is
P =
1
cos *
(6.289)
that occurs at
and the minimum value of
= 0> (6.290)
p = cos *
(6.291)
3
>
2 2
(6.292)
is
that occurs at
=
Example 265 F Double universal joint.
To eliminate the non-uniform speed ratio between the input and output
shafts, connected by a universal joint, we can connect a second joint to
make the intermediate shaft have a variable speed ratio with respect to both
the input and the output shafts in such a way that the overall speed ratio
between the input and output shafts remains equal to one. Figure 6.48 depicts the two possible arrangements to cancel the alternative speed ratio of
the universal joints between two shafts in an angle or oset.
Example 266 F Alternative proof for universal joint equation.
Consider a universal joint of Figure 6.42. Looking along the axis of the
input shaft, we see points D and E moving on a circle and points F and G
moving on an ellipse as shown in Figure 6.49(d). This is because D and E
378
6. Applied Mechanisms
M
M
M
M
FIGURE 6.48. Two possible arrangements to cancel the alternative speed ratio
of the universal joints between two shafts in an angle or oset.
trace a circle in a normal plane, and F and G trace a circle in a rotated
plane by the angle *. Assume the universal joint starts rotating when the
axis FG of the cross link is at the intersection of the planes of motion FG
and DE, as shown in Figure 6.49(d). If the axis DE turns an angle , then
the projection of the axis FG will turn on the ellipse, as can be seen in
Figure 6.49(e). However, the angle of rotation FG is 4 dierent than when we look at the axis FG along the output shaft.
Looking along the input shaft, the axis DE starts from D1 E1 and moves
to D2 E2 after rotation . From the same viewpoint, the axis FG starts
from F1 G1 and moves to F2 G2 , however, FG would be at F20 G20 , if it
were looking along the output shaft. The geometric relationship between the
angles are
F20 U
= tan 4
RU
F2 U
= tan RU
F2 U
= cos *
F20 U
(6.293)
Therefore,
tan = tan 4 cos *
which after dierentiation becomes
$2
$4
cos *
=
csc2 csc2 4
(6.294)
(6.295)
Eliminating 4 between (6.294) and (6.295), we nd the relationship between the input and output shafts’ angular velocities.
cos *
(6.296)
$2
$4 =
sin2 + cos2 cos2 *
6. Applied Mechanisms
C
C1
379
T4
B2
R
B
A
B1
C2
A1
O
D2' D2
C2'
T
A2
D
D1
(a)
(b)
FIGURE 6.49. Rotation of the cross link from a viewpoint along the input shaft.
The speed ratio would then be the same as (6.269).
=
cos *
cos *
$4
=
=
$2
sin2 + cos2 cos2 *
1 sin2 * cos2 (6.297)
Example 267 F History of the universal joint.
The need to transmit a rotary motion from one shaft to another, which
are intersecting at an angle, was a problem for installing clocktowers in
the 1300v. The transmission of the rotation to the hands should be displaced because of tower construction. Cardano (15011576) in 1550, Hooke
(1635 1703) in 1663, and Schott (1608 1666) in 1664 used the joint for
transferring rotary motion. Hooke was the rst who said that the rotary motion between the input and output shafts is not uniform. However, Monge
(17461818) was the rst person who published the mathematical principles
of the joint in 1794, and later Poncelet (1788 1867) in 1822.
As an alternative proof for the universal joint speed ratio equation, Poncelet (17881867) in 1824 used spherical trigonometry to nd the universal
joint speed ratio equation, and Jazar in 2011 employed the coordinate transformation method to derive the equation.
The universal joint can be used to transfer torque at larger angles than
exible couplings. One universal joint may be used to transmit power up to
* = 15 deg depending on the application. Universal joints are available in
a wide variety of torque capacities.
380
6. Applied Mechanisms
6.7 Summary
Every movable component of a vehicle, such as the doors, hoods, windshield wipers, axles, wheels, and suspensions, are connected to the vehicle
body using some mechanisms. The four-bar linkage and inverted slidercrank mechanism are the two common mechanisms that we use to connect
wheels of independent suspensions to the vehicle chassis. There are analytic equations for determining conguration of the all links of a mechanism
with respect to one of the links that is assumed to be xed and called the
ground link.
The wheels are installed on a spindle, which is rigidly attached to the
coupler link of the mechanisms. The center of the wheel will move on a
coupler point curve. Analytic expressions determine the path of motion of
the center as well as the angle of the wheel with respect to the body.
6. Applied Mechanisms
381
6.8 Key Symbols
d{
¨
d> e
d> e> f> · · ·
a
D> E> · · ·
e
eb
ë
F1 > F2 > · · ·
d
i = 1@W
j
ˆ~> ˆ> n̂
L
M1 > M2 > · · ·
n
o
q
Q
s> t
T
r
v
v
V
vb
v̈
w
W
{> |> }> x
{> |> }
{F > |F
v
l
l
*
$
$l
acceleration
coe!cients of equation
length of the links of a linkage
acceleration vector
coe!cients of quadratic equations
relative position of an inverted slider
relative speed of a slider
relative acceleration of a slider
link acceleration parameters of linkages
position vector of a moving frame
cyclic frequency [ Hz]
gravitational acceleration
unit vectors of Cartesian coordinate frames
instant center of rotation
link position parameters of linkages
a parameter between zero and one, 0 ? n ? 1
length, length of the longest link
number of links
number of instant center of rotations
length of the middle links
instantaneous center of velocity
joint relative position vector
displacement position of an slider
length of the shortest link
instantaneous center of acceleration
speed of a slider
acceleration of a slider
time
period
displacement
Cartesian coordinates
coupler point coordinates
velocity vector
angular acceleration vector
angular acceleration of link number l
angular position of link number l
angular position of input and output axles of a universal joint
angle between the input and output axles of a universal joint
angular velocity vector
angular velocity of link number l
angular velocity ratio
382
6. Applied Mechanisms
Exercises
1. Two possible congurations for a four-bar linkage.
Consider a four-bar linkage with the following links.
d = 10 cm
e = 25 cm
f = 30 cm
g = 25 cm
If 2 = 30 deg what would be the angles 3 and 4 for a convex
conguration?
2. Angular velocity of a four-bar linkage output link.
Consider a four-bar linkage with the following links.
d = 10 cm
e = 25 cm
f = 30 cm
g = 25 cm
Determine the angular velocity of the output link $ 4 at 2 = 30 deg
if $ 2 = 2 rad@ s.
3. Angular acceleration of a four-bar linkage output link.
Consider a four-bar linkage with the following links.
d = 10 cm
e = 25 cm
f = 30 cm
g = 25 cm
Determine the angular acceleration of the output link 4 at 2 =
30 deg if 2 = 0=2 rad@ s2 and $ 2 = 2 rad@ s.
4. Grasho criterion.
Consider a four-bar linkage with the following links.
d = 10 cm
e = 25 cm
f = 30 cm
Determine the limit values of the length g to satisfy the Grasho
criterion.
5. Limit and dead positions.
Consider a four-bar linkage with the following links.
d = 10 cm
e = 25 cm
f = 30 cm
g = 25 cm
Determine if there is any limit or dead positions for the linkage.
6. F Limit position determination.
Explain how we may be able to determine the limit positions of a
four-bar linkage by the following condition.
g
=0
gw
6. Applied Mechanisms
x
T6
383
N
M
T4
P
A
Z
B
C
FIGURE 6.50. A six-link windshield wiper mechanism.
7. F Design a windshield wiper mechanism.
Figure 6.50 illustrates a six-link windshield wiper mechanism by combining two four-bar linkages. The link S F is attached to a driver motor. Assume S F = 1 and P S = Q S . Design the mechanism such
that the links P D and Q E are matching the required angles given
in the following table.
Matching
1
3
Q E angle
41 = 157=6 deg 2=75 rad
43 = 69=5 deg 1=213 rad
P D angle
61 = 157=2 deg 2=74 rad
63 = 26=8 deg 0=468 rad
8. Two possible congurations for a slider-crank mechanism.
Consider a slider-crank mechanism with the following links.
d = 10 cm
e = 45 cm
h=0
If 2 = 30 deg what would be angle 3 and position of the slides v for
a convex conguration?
9. Angular velocity and acceleration of the slider of a slider-crank mechanism.
Consider a slider-crank mechanism with the following links.
d = 10 cm
e = 45 cm
h=0
Determine the angular velocity and acceleration of the slider at 2 =
30 deg if $ 2 = 2 rad@ s and 2 = 0=2 rad@ s2 .
384
6. Applied Mechanisms
5
3
6
4
2
1
FIGURE 6.51. A 6-bar linkage.
1
M
N
6
P
A
B
2
4
C
5
3
FIGURE 6.52. A 6-bar linkage.
10. Quick return time.
Consider a slider-crank mechanism with the following links.
d = 10 cm
e = 45 cm
h = 3 cm
Determine the dierence time between go and return half cycle of the
slider motion if $ 2 = 2 rad@ s.
11. Two possible congurations for an inverted slider-crank mechanism.
Consider an inverted slider-crank mechanism with the following links.
d = 10 cm
g = 45 cm
h = 5 cm
If 2 = 30 deg, what would be the angle 3 and position of the slides
e?
12. Instant center of rotation.
Find the instant center of rotations for the 6-bar linkage shown in
(a) Figure 6.51.
(b) Figure 6.52.
6. Applied Mechanisms
385
13. A coupler point of a four-bar linkage.
Consider a four-bar linkage with the following links
d = 10 cm
e = 25 cm
f = 30 cm
g = 25 cm
and a coupler point with the following parameters.
h = 10 cm
= 30 deg
Determine the coordinates of the coupler point if 2 = 30 deg.
14. A coupler point of a slider-crank mechanism.
Consider a slider-crank mechanism with the following parameters.
d = 10 cm
f = 10 cm
e = 45 cm
= 30 deg
h = 3 cm
Determine the coordinates of the coupler point.
15. A coupler point of an inverted slider-crank mechanism if 2 = 30 deg.
Consider an inverted slider-crank mechanism with the following parameters.
d = 10 cm
f = 10 cm
g = 45 cm
= 30 deg
h = 5 cm
Determine the coordinates of the coupler point if 2 = 30 deg.
7
Steering Dynamics
To maneuver a vehicle we need a steering mechanism to turn steerable
wheels. Steering dynamics which we review in this chapter, introduces the
requirements and challenges to have a steering system to guide a vehicle
on non-straight paths.
7.1 Kinematic Steering
Consider a front-wheel-steering (I Z V) vehicle that is turning to the left,
as is shown in Figure 7.1. When the vehicle is moving very slowly, there is
a kinematic condition between the inner and outer wheels that allows them
to turn slip-free. The kinematic condition is called the Ackerman condition
and is expressed by
z
(7.1)
cot r cot l =
o
where, l is the steer angle of the inner wheel, and r is the steer angle
of the outer wheel. The inner and outer wheels are dened based on the
turning center R.
Go
Gi
w
l
Center of
rotation
O
R1
FIGURE 7.1. A front-wheel-steering vehicle and the Ackerman condition.
The distance between the steer axes of the steerable wheels is called the
track and is shown by z. The distance between the front and real axles
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_7, © Springer Science+Business Media New York 2014
387
388
7. Steering Dynamics
is called the wheelbase and is shown by o. Track z and wheelbase o are
considered as kinematic width and length of the vehicle.
We may also use the following more general equation for the kinematic
condition between the steer angles of a I Z V vehicle
cot i u cot i o =
zi
o
(7.2)
where, i o and i u are the steer angles of the front left and front right
wheels. In this equation the steer angle is measured from the {-axis and is
positive if it is about positive }-axis.
The mass center of a steered vehicle will turn on a circle with radius U,
q
(7.3)
U = d22 + o2 cot2 where is the cot-average of the inner and outer steer angles.
cot =
cot r + cot l
2
(7.4)
The angle is the equivalent steer angle of a bicycle having the same
wheelbase o and radius of rotation U.
Proof. To have all wheels turning freely on a curved road, all the tire axes
must intersect at a common point. This criteria is the Ackerman condition.
The tire axis is the normal line to the tire-plane at the center of the tire.
Figure 7.2 illustrates a vehicle turning left about the turning center R.
Therefore, the inner wheels are the left wheels that are closer to the center
of rotation. The inner and outer steer angles l and r may be calculated
from the triangles 4RDG and 4REF as:
tan l
=
tan r
=
o
U1 o
U1 +
z
2
(7.5)
z
2
(7.6)
Eliminating U1
U1 =
o
o
1
1
z+
= z+
2
tan l
2
tan r
(7.7)
provides us with the Ackerman condition (7.1), which is a direct relationship between l and r .
cot r cot l =
z
o
(7.8)
To nd the vehicle’s turning radius U, we dene an equivalent bicycle
model, as shown in Figure 7.3. The radius of rotation U is perpendicular
7. Steering Dynamics
389
w
Go
Gi
Outer
wheel
Inner
wheel
A
B
a1
Center of
rotation
Go
l
C
R
a2
Gi
O
D
C
R1
FIGURE 7.2. A front-wheel-steering vehicle and steer angles of the inner and
outer wheels.
to the vehicle’s velocity vector v at the mass center F. Using the geometry
shown in the bicycle model, we have
U2
cot and therefore,
= d22 + U12
U1
1
=
= (cot l + cot r )
o
2
q
U = d22 + o2 cot2 (7.9)
(7.10)
(7.11)
The Ackerman condition is needed when the speed of the vehicle is very
small, and slip angles are very close to zero. In these conditions, there
would be no lateral force and no centrifugal force to balance each other. The
Ackerman steering condition is also called the kinematic steering condition,
because it is only a static condition at zero velocity.
A device that provides steering according to the Ackerman condition
(7.1) is called Ackerman steering, Ackerman mechanism, or Ackerman geometry. There is no practical four-bar linkage steering mechanism that can
provide the Ackerman condition perfectly. However, we may design a multibar linkages to work close to the condition and be exact at a few angles.
390
7. Steering Dynamics
G
v
R
Center of
rotation
a1
l
C
a2
G
O
R1
FIGURE 7.3. Equivalent bicycle model for a front-wheel-steering vehicle.
Figure 7.4 illustrates the Ackerman condition for dierent values of z@o.
The dierence between the inner and outer steer angles decreases by decreasing z@o.
Example 268 Steering angles and radii.
Consider a vehicle with the following dimensions and inner steer angle:
o
d2
= 103=1 in 2=619 m
= 60 in 1=524 m
z = 61=6 in 1=565 m
l = 12 deg 0=209 rad
The kinematic steering characteristics of the vehicle would be
³z
´
+ cot l = 0=186 rad 10=661 deg
r = cot1
¶
μo
cot r + cot l
1
= 0=19684 rad 11=278 deg
= cot
2
1
= o cot l + z = 516=9 in 13=129 m
2
q
2
2
U =
d2 + o cot2 = 520=46 in 13=219 m
U1
(7.12)
(7.13)
(7.14)
(7.15)
(7.16)
7. Steering Dynamics
w/l=0.2
0.4
0.6
Go [deg]
391
0.8
1.0
1.2
1.4
1.6
2.0
w/l=3.0
Gi [deg]
FIGURE 7.4. Eect of z@o on the Ackerman condition for front-wheel-steering
vehicles.
Example 269 z is the front track.
Most cars have dierent tracks in front and rear. The track z in the
kinematic condition (7.1) refers to the front track zi . The rear track has
no eect on the kinematic condition of a front-wheel-steering vehicle. The
rear track zu of a I Z V vehicle can be zero with the same kinematic steering
condition (7.1).
Example 270 Width requirement.
The kinematic steering condition can be used to calculate the space requirement of a vehicle during a turn. Consider the front wheels of a two-axle
vehicle, steered according to the Ackerman geometry as shown in Figure 7.5.
The outer point of the front of the vehicle will run on the maximum radius
UPd{ , whereas a point on the inner side of the vehicle at the location of the
rear axle will run on the minimum radius Uplq . The front outer point has
an overhang distance j from the front axle. The maximum radius UPd{ is
q
2
2
UPd{ = (Uplq + z) + (o + j)
(7.17)
Therefore, the required space for turning is a ring with a width 4U
q
2
2
(7.18)
4U = UPd{ Uplq = (Uplq + z) + (o + j) Uplq
The required space 4U can be calculated based on the steer angle by
substituting Uplq
o
1
o
Uplq = U1 z =
=
z
2
tan l
tan r
(7.19)
392
7. Steering Dynamics
w
RMax
Gi
Go
Rmin
g
l
O
R1
FIGURE 7.5. The required space for a turning two-axle vehicle.
and getting
sμ
¶2
o
o
2
+ 2z + (o + j) tan l
tan l
sμ
¶2
o
o
=
+ z + (o + j)2 +z
tan r
tan r
4U =
(7.20)
(7.21)
In this example the width of the car zy and the track z are assumed to
be equal. The width of vehicles are always greater than their track.
zy A z
(7.22)
The requirement for more space becomes more critical when the vehicle is
longer and z@o increases. Figure 7.6 compares the required space 4U for a
bus and a long truck.
Example 271 4U for a semitrailer.
Figure 7.7 illustrates a semitrailer in a turn. Analysis of the tractor provides us with the radius U1 as a function of the steer angle and front width
of the tractor.
1
1
d1
d1
= z+
(7.23)
U1 = z +
2
tan l
2
tan r
Therefore, the maximum radius, which belongs to the tractor, is
r³
z ´2
+ d23
(7.24)
U1 +
UPd{ =
2
60
30
°
30
°
7. Steering Dynamics
°
60 °
90°
12
393
90°
0°
12
0°
15
0
15
0°
°
180°
180°
FIGURE 7.6. The required space 4U for a bus and a long truck.
The minimum radius Uplq belongs to the trailer and is
q
1
(7.25)
Uplq = U1 d22 zu
2
Therefore, the required space for turning is a ring with a width 4U.
r³
q
z ´2
zu
+ d23 U1 d22 U1 +
4U = UPd{ Uplq =
(7.26)
2
2
Example 272 Trapezoidal steering mechanism.
Figure 7.8 illustrates a symmetric four-bar linkage called trapezoidal
steering mechanism, used for more than 100 years as a steering connection. The mechanism has two characteristic parameters: angle and oset
arm length g. A steered position of the trapezoidal mechanism is shown in
Figure 7.9 to illustrate the inner and outer steer angles l and r .
The relationship between the inner and outer steer angles of a trapezoidal
steering mechanism is given by an implicit function:
sin ( + l ) + sin ( r )
r³
´2
z
z
2
= 2 sin (cos ( r ) cos ( + l )) (7.27)
g
g
To prove this equation, we examine Figure 7.10. In the triangle 4DEF
we can write
(z 2g sin )2
= (z g sin ( r ) g sin ( + l ))2
+ (g cos ( r ) g cos ( + l ))2
(7.28)
7. Steering Dynamics
a4
a2
a3
a1
wr
R1
w
394
Go
Rmin
Gi
RMax
FIGURE 7.7. A semitrailer in a turn.
R
d
E
w
E
FIGURE 7.8. A trapezoidal steering mechanism.
d
R
Inner
wheel
Outer
wheel
E
Gi
w
E
Go
FIGURE 7.9. Steered conguration of a trapezoidal steering mechanism.
7. Steering Dynamics
395
d
A
Inner
wheel
B
Gi
C
E
Outer
wheel
E
w
Go
FIGURE 7.10. Trapezoidal steering triangle DEF.
Go [deg]
40
34
E
l=2.93 m
w=1.66 m
d=0.4 m
28
G [rad]
6 d eg o
10
0.6
14
18
22
0.4
23
Ackerman
18
0.2
12
6
Gi [deg]
FIGURE 7.11. Behavior of a trapezoidal steering mechanism, compared to the
associated Ackerman mechanism g = 0=4 m.
and derive Equation (7.27) with some manipulation.
The functionality of a trapezoidal steering mechanism, compared to the
associated Ackerman condition, is shown in Figure 7.11 and 7.12 for o =
2=93 m 9=61 ft, z = 1=66 m 5=45 ft and respectively for g = 0=4 m 1=3 ft and g = 0=2 m 0=65 ft. The horizontal axis shows the inner steer
angle and the vertical axis shows the outer steer angle. It shows that for
given o and z, a mechanism with 18 deg / / 22 deg is the best simulator
of the Ackerman mechanism if l ? 50 deg.
To examine the trapezoidal steering mechanism and compare it with the
Ackerman condition, we dene an error parameter h = Gr Dr . The error
h is the dierence between the outer steer angles calculated by the trapezoidal
mechanism and the Ackerman condition at the same inner steer angle l .
h = r = Gr Dr
(7.29)
Figure 7.13 and 7.14 depicts the error h for the same sample steering mechanisms using the angle as a parameter.
396
7. Steering Dynamics
Go [deg]
40
34
6 d eg
10
14
22
E
l=2.93 m
w=1.66 m
d=0.2 m
18
28
Go [rad]
0.6
0.4
23
Ackerman
18
0.2
12
6
Gi [deg]
FIGURE 7.12. Behavior of a trapezoidal steering mechanism, compared to the
associated Ackerman mechanism g = 0=2 m.
'Go [deg]
6
3
l=2.93 m
w=1.66 m
d=0.4 m
E
10
14
0
-3
6 d eg
Gi [deg]
18
22
-6
-9
FIGURE 7.13. The error parameter h = Gr 3 Dr for a sample trapezoidal
steering mechanism g = 0=4 m.
7. Steering Dynamics
'Go [deg]
6
3
l=2.93 m
w=1.66 m
d=0.2 m
E
6 d eg
10
14
18
0
-3
397
Gi [deg]
22
-6
-9
FIGURE 7.14. The error parameter h = Gr 3 Dr for a sample trapezoidal
steering mechanism g = 0=2 m.
Example 273 F Locked rear axle.
In a design of simple vehicles, we sometimes eliminate the dierential
and use a locked rear axle in which no relative rotation between the left and
right wheels is possible. Such a simple design is usually used in toy cars, or
small o-road vehicles such as a mini Baja.
Consider the vehicle shown in Figure 7.2. In a slow left turn, the speed
of the inner rear wheel must be
³
z´
(7.30)
u = Uz $ ul
yul = U1 2
and the speed of the outer rear wheel must be
³
z´
u = Uz $ ur
yur = U1 +
2
(7.31)
where, u is the yaw velocity of the vehicle, Uz is rear wheels radius, and
$ ul , $ ur should respectively be the angular velocities of the rear inner and
outer wheels about their common axle. If the rear axle is locked, we have
$ ul = $ ur = $
however,
(7.32)
³
z´ ³
z´
U1 6= U1 +
(7.33)
2
2
which shows it is impossible to have a locked axle for a nonzero z.
Turning reduces the load on the inner wheels and makes the rear inner
wheel of a locked axle overcome the friction force and spin. Hence, the
traction of the inner wheel drops to the maximum friction force under a
reduced load. However, the load on the outer wheels increases and hence,
the higher friction limit of the outer wheel helps to have higher traction
force on the outer rear wheel.
398
7. Steering Dynamics
R1
O
Center of
rotation
l
w
Gi
Go
FIGURE 7.15. A rear-wheel-steering vehicle.
Eliminating the dierential and using a locked drive axle is an impractical
design for street cars. However, it can be an acceptable design for small
and light cars moving on dirt or slippery surfaces. It reduces the cost and
simplies the design signicantly.
In a conventional two-wheel-drive motor vehicle, the rear wheels are
driven using a dierential, and the vehicle is steered by changing the direction of the front wheels. With an ideal dierential, equal torque is delivered to each drive wheel. The rotational speed of the drive wheels are
determined by the dierential and the tire-road characteristics. However,
a vehicle using a dierential has disadvantages when one wheel has lower
traction. Dierences in traction characteristics of each of the drive wheels
may come from dierent tire-road characteristics or weight distribution.
Because a dierential delivers equal torque, the wheel with greater tractive
ability can deliver only the same amount of torque as the wheel with the
lower traction. The steering behavior of a vehicle with a dierential is relatively stable under changing tire-road conditions. However, the total thrust
may be reduced when the traction conditions are dierent for drive wheel.
Example 274 F Rear-wheel-steering.
Rear-wheel-steering is used where high maneuverability is a necessity on
a low-speed vehicle, such as forklifts. Rear-wheel-steering is not used on
street vehicles because it is unstable at high speeds. The center of rotation
for a rear-wheel-steering vehicle is always a point on the extension of the
front axle.
Figure 7.15 illustrates a rear-wheel-steering vehicle. The kinematic steer-
7. Steering Dynamics
399
w
Go
Gi
Outer
wheel
Inner
wheel
a1
Center of
rotation
Go
Gi
l
C
R
w
Zi
a2
Zo
O
R1
FIGURE 7.16. Kinematic condition of a I Z V vehicle using the angular velocity
of the inner and outer wheels.
ing condition (7.1) remains the same for a rear-wheel steering vehicle.
cot r cot l =
z
o
(7.34)
Example 275 F Alternative kinematic steer angles equation.
Consider a rear-wheel-drive vehicle with front steerable wheels as shown
in Figure 7.16. Assume that the front and rear tracks of the vehicle are
equal and the drive wheels are turning without slip. If we show the angular
velocities of the inner and outer drive wheels by $ l and $ r , respectively,
the kinematic steer angles of the front wheels can be expressed by
μ μ
¶¶
o $r
1
1
(7.35)
l = tan
z $l
μ μ
¶¶
$l
o
r = tan1
1
(7.36)
z
$r
To prove these equations, we may start from the non-slipping condition
for the drive wheels:
Uz $ l
Uz $ r
(7.37)
z =
z
U1 +
U1 2
2
400
7. Steering Dynamics
Equation (7.37) can be rearranged to
z
U1 +
$r
2
=
z
$l
U1 2
(7.38)
and substituted in Equations (7.35) and (7.36) to reduce them to Equations
(7.5) and (7.6).
The equality (7.37) is the yaw rate of the vehicle about the }-axis, which
is also the vehicle’s angular velocity about the center of rotation.
u=
Uz $ l
Uz $ r
z =
z
U1 +
U1 2
2
(7.39)
Example 276 F Unequal front and rear tracks.
We usually design vehicles with dierent tracks in the front and rear. It
is a common design for race cars, which are usually equipped with wider
and larger rear tires to increase traction and stability. Such a vehicle is
illustrated in Figure 7.17. For street cars we usually use the same tires in
the front and rear, however, it is common to have a few centimeters of
larger rear track.
The yaw rate or angular velocity of the vehicle is
u=
Uz $ l
Uz $ r
zu =
zu
U1 +
U1 2
2
(7.40)
and the kinematic steer angles of the front wheels are
l
r
2o ($ r + $ l )
zi ($ r $ l ) + zu ($ r + $ l )
2o ($ r $ l )
= tan1
zi ($ r $ l ) + zu ($ r + $ l )
= tan1
(7.41)
(7.42)
To show these equations, we should nd U1 from Equation (7.40)
U1 =
zu $ r + $ l
2 $r $l
(7.43)
and substitute it in the following geometric equations.
tan l
=
tan r
=
o
U1 o
U1 +
zi
2
(7.44)
zi
2
(7.45)
In the above equations, zi is the front track, zu is the rear track, and Uz
is the wheel eective radius.
7. Steering Dynamics
401
wf
Go
Gi
Outer
wheel
Inner
wheel
a1
Center of
rotation
Go
Gi
l
C
R
wr
Zi
a2
Zo
O
R1
FIGURE 7.17. Kinematic steering condition for a vehicle with dierent tracks in
the front and in the back.
Example 277 F Independent rear-wheel-drive.
For some special-purpose vehicles, such as moon rovers and autonomous
mobile robots, we may attach each drive wheel to an independently controlled motor to apply any desired angular velocity. The steerable wheels of
such vehicles are usually designed to be able to turn more than 90 deg to
the left and right. Such a vehicle is highly maneuverable at low speeds.
Figure 7.18 illustrates the advantages of such a high steerable vehicle
and its possible turnings. Figures 7.18(d)-(f) illustrate forward maneuvering. The arrows next to the rear wheels, illustrate the magnitude of the
angular velocity of the wheel, and the arrows on the front wheels illustrate
the direction of their motion. The maneuvering in backward motion is illustrated in Figures 7.18(g)-(i ). Having such a vehicle allows us to turn
the vehicle about any point on the rear axle including the inner points. In
Figure 7.18(j) the vehicle is turning about the center of the rear right wheel,
and in Figure 7.18(k) about the center of the rear left wheel. Figure 7.18(l)
illustrates a rotation about the center point of the rear axle.
In any of the above scenarios, the steer angle of the front wheels should
be determined using a proper equation, such as (7.44) and (7.45). The ratio
of the outer to inner angular velocities of the drive wheels $ r @$ l may be
402
7. Steering Dynamics
(b)
(a)
(c)
(g)
(f)
(d)
(h)
(e)
(i)
FIGURE 7.18. A highly steerable vehicle.
determined using either the outer or inner steer angles.
$r
$l
$r
$l
=
=
r (zi zu ) 2o
r (zi + zu ) 2o
l (zi + zu ) + 2o
l (zi zu ) + 2o
(7.46)
(7.47)
Example 278 F Race car steering.
The Ackerman or kinematic steering is a correct condition when the turning speed of the vehicle is very slow. When the vehicle turns very fast, signicant lateral acceleration is needed and therefore, the wheels operate at
high slip angles. Furthermore, the loads on the inner wheels will be much
lower than the outer wheels. Tire performance curves show that by increasing the wheel load, less slip angle is required to reach the peak of the lateral
force. Under these conditions the inner front wheel of a kinematic steering
vehicle might be at a higher slip angle than required for maximum lateral
force. Therefore, the inner wheel of a vehicle in a high speed turn must operate at a lower steer angle than kinematic steering and the outer wheel must
operate at a higher steer angle than kinematic steering. These requirements
reduces the dierence between steer angles of the inner and outer wheels.
For race cars, it is common to use parallel or reverse steering. Ackerman,
parallel, and reverse Ackerman steering are illustrated in Figure 7.19.
The correct steer angle is a function of the instant wheel load, road condition, speed, and tire characteristics. Furthermore, the vehicle must also be
able to turn at a low speed under an Ackerman steering condition. Hence,
there is no ideal steering mechanism unless we control the steer angle of
7. Steering Dynamics
Ackerman
Parallel
403
Reverse
FIGURE 7.19. By increasing the speed at a turn, parallel or reverse steering is
needed instead of Ackerman steering.
each steerable wheel independently using a steer by wire and smart system.
Example 279 F Speed dependent steering system.
There is a speed adjustment idea that says it is better to have a harder
steering system at high speeds that needs higher torques. This idea can be
applied in power steering systems to make them speed dependent, such that
the steering be heavily assisted at low speeds and lightly assisted at high
speeds. The idea is supported by this fact that the drivers might need large
steering for parking, and small steering when traveling at high speeds.
Example 280 F Ackerman condition history.
Correct steering geometry was a major problem in the early days of carriages, horse-drawn vehicles, and cars. Four- or six-wheel cars and carriages always left rubber marks behind. This is why there were so many
three-wheeled cars and carriages in the past. The problem was making a
mechanism to give the inner wheel a smaller turning radius than the outside wheel when the vehicle was driven in a circle.
The required geometric condition for a front-wheel-steering four-wheelcarriage was introduced in 1816 by George Langensperger in Munich, Germany. Langensperger’s mechanism is illustrated in Figure 7.20.
Rudolf Ackerman met Langensperger and saw his invention and acted
as Langensperger’s patent agent in London to introduce the invention to
British carriage builders. Car manufacturers have been adopting and improving the Ackerman geometry for their steering mechanisms since 1881.
The basic design of vehicle steering systems has changed little since the
invention of the steering mechanism. The driver’s steering input is transmitted by a shaft through some type of gear reduction mechanism to generate
404
7. Steering Dynamics
FIGURE 7.20. Langensperger invention for the steering geometry condition.
steering motion at the front wheels.
7.2 Vehicles with More Than Two Axles
If a vehicle has more than two axles, all the axles, except one, must be
steerable to provide slip-free turning at zero velocity. When an q-axle vehicle has only one non-steerable axle, there are q 1 geometric steering
conditions. A three-axle vehicle with two steerable axles is shown in Figure
7.22.
To indicate the geometry of a multi-axle vehicle, we start from the front
axle and measure the longitudinal distance dl between axle l and the mass
center F. Hence, d1 is the longitudinal distance between the front axle
and F, and d2 is the distance between the second axle and F, and so on.
Furthermore, we number the wheels in a clockwise rotation starting from
the front left wheel as number 1. In the United States, Europe, and other
left-hand drive (OKG) system countries, the front left wheel is the driver’s
wheel, while in Britain and other right-hand drive (UKG) system countries
the front left wheel is the passenger wheel
For the three-axle vehicle shown in Figure 7.22, there are two independent Ackerman conditions:
cot 2 cot 1
=
cot 3 cot 4
=
z
d1 + d3
z
d2 + d3
(7.48)
(7.49)
7. Steering Dynamics
405
w
Inner
wheels
G2
G1
Outer
wheels
G4
a1
a2
R
Center of
rotation
C
G3
a3
O
R1
FIGURE 7.21. Steering of a three-axle vehicle.
Example 281 A six-wheel vehicle with one steerable axle.
When a multi-axle vehicle has only one steerable axle, slip-free rotation
is impossible for the non-steering wheels. The kinematic length or wheelbase of the vehicle is not clearly dened, and it is not possible to dene a
kinematic steering condition. Strong wear occurs for the tires, especially at
low speeds and large steer angles. Hence, such a combination is not recommended. However, in case of a long three-axle vehicle with two non-steerable
axles close to each other, an approximated analysis is possible for low-speed
steering.
Figure 7.22 illustrates a six-wheel vehicle with only one steerable axle in
front. We design the steering mechanism such that the center of rotation R
is on a lateral line, called the midline, between the couple rear axles. The
kinematic length of the vehicle, o, is the distance between the front axle and
the midline. For this design we have
z
(7.50)
cot r cot l =
o
and
d3 d2
o = d1 + d2 +
(7.51)
2
z
z
(7.52)
U1 = o cot r = o cot l +
2
2
406
7. Steering Dynamics
w
Go
Gi
Inner
wheel
l
Outer
wheel
Rf
a1
Center of
rotation
Gi
C
R
Go
a2
a3
O
R1
FIGURE 7.22. A six-wheel vehicle with one steerable axle in front.
The center of the front axle and the mass center of the vehicle are turning
about R by radii Ui and U.
Ui =
U
¶
μ 1
1 o
cos tan
U1
U=
U1
¶
μ
1 d3 d2
cos tan
2U1
(7.53)
If the radius of rotation is large compared to the wheelbase, we may approximate Equations (7.52) and (7.53) to:
U1
=
Ui
o
(cot r + cot l )
2
U
U
¶
μ1 ¶
μ 1
U
o
d3 d2
cos
cos
U1
2U1
(7.54)
(7.55)
To avoid strong wear of tires of heavy trucks, it is possible to lift an axle
when the vehicle is not carrying heavy loads. For such a vehicle, we may
design the steering mechanism to follow an Ackerman condition based on
a wheelbase for the non-lifted axle. However, when this vehicle is carrying
a heavy load and using all the axles, the liftable axle encounters huge wear
at large steer angles.
7. Steering Dynamics
407
FIGURE 7.23. A self-steering axle mechanism for locomotive wagons.
Another option for multi-axle vehicles is to use self-steering wheels
that can adjust themselves to minimize sideslip. Such wheels cannot provide
lateral force, and hence, cannot help in maneuvers very much. Self-steering
wheels may be installed on buggies and trailers. Such a self-steering axle
mechanism for locomotive wagons is shown in Figure 7.23.
7.3 F Vehicle with Trailer
If a four-wheel vehicle has a trailer with one axle, it is possible to derive a
kinematic condition for slip-free steering. Figure 7.24 illustrates a vehicle
with a one-axle trailer. The mass center of the vehicle is turning on a circle
with radius U, while the trailer is turning on a circle with radius Uw .
sμ
¶2
1
o cot l + z + e21 e22
Uw =
(7.56)
2
sμ
¶2
1
o cot r z + e21 e22
Uw =
(7.57)
2
At a steady-state condition, the angle between the trailer and the vehicle
is
=
;
³
1
A
1
A
Uw
? 2 tan
e e
1
2
A
A
= 2 tan1 1 (e1 + e2 )
2Uw
´¸
p
2
2
2
Uw e1 + e2
e1 e2 6= 0
(7.58)
e1 e2 = 0
Proof. Using the right triangle 4RDE in Figure 7.24, we may write the
trailer’s radius of rotation as
q
(7.59)
Uw = U12 + e21 e22
408
7. Steering Dynamics
w
Inner
wheel
Outer
wheel
a1
Gi
Center of
rotation
Go
C
R
a2
Gi
Go
T
O
l
b1
R1
B
Rt
T
A
b2
FIGURE 7.24. A vehicle with a one-axle trailer.
because the length RE is
2
RE = Uw2 + e22 = U12 + e21
(7.60)
Substituting U1 from Equation (7.7) shows that the trailer’s radius of rotation is related to the vehicle’s geometry and steer angles by
sμ
¶2
1
o cot l + z + e21 e22
(7.61)
Uw =
2
sμ
¶2
1
o cot r z + e21 e22
Uw =
(7.62)
2
q
Uw =
U2 d22 + e21 e22
(7.63)
Using the equation
Uw sin = e1 + e2 cos (7.64)
and employing trigonometry, we calculate the angle between the trailer
and the vehicle as (7.58).
The minus sign in (7.58), in case e1 e2 6= 0, is the usual case in forward
motion, and the plus sign is a solution associated with a backward motion.
Both possible for a set of conguration (Uw > e1 > e2 ) are shown in Figure
7. Steering Dynamics
409
w
Inner
wheel
Outer
wheel
a1
Gi
Center of
rotation
Go
C
O
l
a2
b1
R1
Rt
b2
FIGURE 7.25. Two possible angle for a set of (Uw > e1 > e2 ).
7.25. The unstable and undesirable 2 is called a jackkning conguration.
The second conguration usually interferes with physical boundaries of
vehicles and is not achievable without crash.
Example 282 F Two possible trailer-vehicle angles.
Consider a four-wheel vehicle that is pulling a one-axle trailer with the
following dimensions:
o
e1
d2
= 104 in 2=641 m
= 24 in 0=61 m
= 50 in 1=27 m
z = 64 in 1=626 m
e2 = 90 in 2=286 m
l = 12 deg 0=209 rad
The kinematic steering characteristics of the vehicle would be
³z
´
+ cot l 0=185 rad 10=626 deg
r = cot1
o
sμ
¶2
1
o cot l + z + e21 e22 = 515=09 in 13=083 m
Uw =
2
1
U1 = o cot l + z = 522=34 in 13=268 m
2
(7.65)
(7.66)
(7.67)
(7.68)
410
7. Steering Dynamics
wv
Go
Gi
g
Center of
rotation
l
O
Rt
B
RM
ax
w
in
t
T
b2
Rm
b1
R1
A
FIGURE 7.26. A two-axle vehicle with a trailer is steered according to the Ackerman condition.
¶
μ
cot r + cot l
= 0=20274 rad 11=616 deg (7.69)
= cot1
2
q
U =
d22 + o2 cot2 = 508=39 in 12=913 m
(7.70)
μ
¶¸
q
1
= 2 tan1
Uw
Uw2 e21 + e22
e1 e2
½
1 = 0=21890 rad 12=542 deg
(7.71)
=
2 = 3=0145 rad 172=72 deg
Example 283 F Space requirement.
The kinematic steering condition can be used to calculate the space requirement of a vehicle with a trailer during a turn. Consider that the front
wheels of a two-axle vehicle with a trailer are steered according to the Ackerman geometry, as shown in Figure 7.26.
The outer point of the front of the vehicle will run on the maximum radius
UPd{ , whereas a point on the inner side of the wheel at the trailer’s rear
7. Steering Dynamics
411
axle will run on the minimum radius Uplq . The maximum radius UPd{ is
r³
zy ´2
2
+ (o + j)
(7.72)
UPd{ =
U1 +
2
where
r³
zw ´2
+ e22 e21
U1 =
Uplq +
2
(7.73)
4U = UPd{ Uplq
(7.74)
and the width of the vehicle is shown by zy .
The required space for turning the vehicle and trailer is a ring with a
width 4U, which is a function of the vehicle and trailer geometry.
The required space 4U can be calculated based on the steer angle by
substituting Uplq
sμ
¶2
1
1
1
o cot l + z + e21 e22 zw
Uplq = Uw zw =
2
2
2
sμ
¶2
1
1
o cot r z + e21 e22 zw
=
2
2
q
1
=
U2 d22 + e21 e22 zw
(7.75)
2
7.4 Steering Mechanisms
A steering system begins with the steering wheel or steering handle. The
driver’s steering input is transmitted by a shaft through a gear reduction
system, usually rack-and-pinion or recirculating ball bearings. The steering
gear output goes to steerable wheels to generate motion through a steering
mechanism. The lever, which transmits motion from the steering gear to
the steering linkage, is called Pitman arm.
The direction of each wheel is controlled by one steering arm. The steering arm is attached to the steerable wheel hub by a keyway, locking taper,
and a hub. In some vehicles, it is an integral part of a one-piece hub and
steering knuckle.
To achieve good maneuverability, a minimum steering angle of approximately 35 deg must be provided at the front wheels of passenger cars.
A sample parallelogram steering mechanism and its components are
shown in Figure 7.27. The parallelogram steering linkage is common on independent front-wheel vehicles. There are many varieties of steering mechanisms each with some advantages and disadvantages.
412
7. Steering Dynamics
Tie rod
Intermediate rod
Pitman arm
Tie rod
Idler arm
FIGURE 7.27. A sample parallelogram steering linkage and its components.
Rack
GS
uR
Steering box
Drag link
FIGURE 7.28. A rack-and-pinion steering system.
Example 284 Steering ratio.
The steering ratio is the rotation angle of a steering wheel divided by
the steer angle of the steerable wheels. The steering ratio of street cars is
around 10 : 1 and the steering ratio of race cars is about 5 : 1. Generally
speaking, the steering ratio of ground vehicles varies between 1 : 1 to 20 : 1.
The steering ratio of Ackerman steering is dierent for inner and outer
wheels. Furthermore, it has a nonlinear behavior and is a function of the
wheel angle.
Example 285 Rack-and-pinion steering.
Rack-and-pinion is the most common steering system of passenger cars.
Figure 7.28 illustrates a sample rack-and-pinion steering system. The rack
is either in front or behind the steering axle. The driver’s rotary steering
command V is transformed by a steering box to translation xU = xU ( V )
of the racks, and then by the drag links to the wheel steering l = l (xU ),
r = r (xU ). The overall steering ratio depends on the ratio of the steering
box and on the kinematics of the steering linkage. The drag link is also
called the tie rod.
Example 286 Lever arm steering system.
Figure 7.29 illustrates a steering linkage that sometimes is called a lever
arm steering system. Using a lever arm steering system, large steering
angles at the wheels are possible. This steering system is used on trucks
7. Steering Dynamics
413
GS
Drag link
FIGURE 7.29. A lever arm steering system.
GS
Drag link
FIGURE 7.30. A drag link steering system.
with large wheel bases and independent wheel suspensions at the front axle.
The steering box and triangle can also be placed outside of the axle’s center.
Example 287 Drag link steering system.
It is sometimes better to send the steering command to only one wheel
and connect the other one to the rst wheel by a drag link, as shown in
Figure 7.30. Such steering linkages are usually used for trucks and busses
with a front solid axle. The rotations of the steering wheel are transformed
by a steering box to the motion of the steering arm and then to the rotation
of the left wheel. A drag link transmits the rotation of the left wheel to the
right wheel.
Figure 7.31 shows a sample for connecting a steering mechanism to the
Pitman arm of the left wheel and using a trapezoidal linkage to connect the
right wheel to the left wheel.
Example 288 Multi-link steering mechanism.
In busses and big trucks, the driver may sit more than 2 m 7 ft in
front of the front axle. These vehicles need large steering angles at the front
wheels to achieve good maneuverability. So a more sophisticated multi-link
steering mechanism is needed. A sample multi-link steering mechanism is
shown in Figure 7.32.
414
7. Steering Dynamics
d
GS
Drag link
E
FIGURE 7.31. Connection of the Pitman arm to a trapezoidal steering mechanism.
GS
Drag link
Drag link
FIGURE 7.32. A multi-link steering mechanism.
7. Steering Dynamics
415
The rotations of the steering wheel are transformed by the steering box to
a steering lever arm. The lever arm is connected to a distributing linkage,
which turns the left and right wheels by a long tire rod.
Example 289 F Reverse e!ciency.
The ability of the steering mechanism to feedback the road inputs to the
driver is called reverse e!ciency. Feeling the applied steering torque or
aligning moment helps the driver to make smoother turn.
Rack-and-pinion and recirculating ball steering gears have a feedback of
the wheels steering torque to the driver. However, worm and sector steering
gears have very weak feedback. Low feedback may be desirable for o-road
vehicles, to reduce the driver’s fatigue.
Because of safety, the steering torque feedback should be proportional to
the speed of the vehicle. In this way, the required torque to steer the vehicle
is higher at higher speeds. Such steering prevents a sharp and high steer
angle. A steering damper with a damping coe!cient increasing with speed
is the mechanism that provides such behavior. A steering damper can also
reduce shimmy vibrations.
Example 290 F Power steering.
Power steering has been developed in the 1950s when a hydraulic power
steering assist was rst introduced. Since then, power assist has gradually
become a standard component in automotive steering systems. Using hydraulic pressure, supplied by an engine-driven pump, amplies the driverapplied torque at the steering wheel. As a result, the steering eort is reduced.
In recent years, electric torque ampliers were introduced in automotive
steering systems as a substitute for hydraulic ampliers. Electrical steering eliminates the need for the hydraulic pump. Electric power steering is
more e!cient than conventional power steering, because the electric power
steering motor needs to provide assistance only when the steering wheel is
turned, whereas the hydraulic pump runs constantly. The assist level is also
tunable by vehicle type, road speed, and driver preference.
Example 291 Bump steering.
The steer angle caused by the vertical motion of the wheel with respect to
the body is called bump steering. Bump steering is usually an undesirable
phenomenon and is a function of the suspension and steering mechanisms.
If the vehicle has a bump steering character, then the wheel steers when
it runs over a bump or when the vehicle rolls in a turn. As a result, the
vehicle will travel in a path not selected by the driver.
Bump steering occurs when the end of the tie rod is not at the instant
center of rotation of the suspension mechanism. Hence, in a suspension deection, the suspension and steering mechanisms will rotate about dierent
centers.
416
7. Steering Dynamics
w
e
d
R
E
E
FIGURE 7.33. An oset design for wheel attachment to an steering mechanism.
Go
Gi
w
R
Path of motion for the center
of the tireprint
e
FIGURE 7.34. Oset attachment of steerable wheels to a trapezoidal steering
mechanism.
Example 292 F Oset steering axis.
Theoretically, the steering axis of each steerable wheel must vertically go
through the center of the wheel at the tire-plane to minimize the required
steering torque. Figure 7.31 is an example of matching the center of a wheel
with the steering axis. However, it is possible to attach the wheels to the
steering mechanism, using an oset design, as shown in Figure 7.33.
Figure 7.34 depicts a steered trapezoidal mechanism with oset wheels
attachment. The path of motion for the center of the tireprint for an oset
design is a circle with oset arm radius h. Such a design is not recommended for street vehicles, especially because of the huge steering torque of
stationary vehicles. However, the steering torque reduces dramatically to an
acceptable value when the vehicle is moving. Furthermore, an oset design
sometimes makes more room to attach other devices, and simplies manufacturing. So, it may be used for small o-road vehicles, such as a mini
Baja, and toy vehicles.
7. Steering Dynamics
417
Gio
Gif
wf
R1
a1
l
C
R
Center of
rotation
O
Gir
wr
a2
Gor
FIGURE 7.35. A positive four-wheel steering vehicle.
7.5 F Four wheel steering.
At very low speeds, the kinematic steering condition that the perpendicular
lines to each tire meet at one point, must be applied. The intersection point
is the turning center of the vehicle.
Figure 7.35 illustrates a positive four-wheel steering vehicle, and Figure
7.36 illustrates a negative 4Z V vehicle. In a positive 4Z V conguration
the front and rear wheels steer in the same direction, and in a negative
4Z V conguration the front and rear wheels steer opposite to each other.
The kinematic condition between the steer angles of a 4Z V vehicle is
cot ri cot li =
zu cot ri cot li
zi
o
o cot ru cot lu
(7.76)
where, zi and zu are the front and rear tracks, li and ri are the steer
angles of the front inner and outer wheels, lu and ru are the steer angles
of the rear inner and outer wheels, and o is the wheelbase of the vehicle.
We may also use the following more general equation for the kinematic
condition between the steer angles of a 4Z V vehicle
cot i u cot i o =
zu cot i u cot i o
zi
o
o cot uu cot uo
(7.77)
where, i o and i u are the steer angles of the front left and front right
wheels, and uo and uu are the steer angles of the rear left and rear right
wheels.
If we dene the steer angles according to the sign convention shown in
Figure 7.37 then, Equation (7.77) expresses the kinematic condition for
418
7. Steering Dynamics
Gif
wf
R1
Gio
a1
R
l
C
O
Center of
rotation
Gir
wr
a2
Gor
FIGURE 7.36. A negative four-wheel steering vehicle.
both, positive and negative 4Z V systems. Employing the wheel coordinate
frame ({z > |z > }z ), we dene the steer angle as the angle between the vehicle
{-axis and the wheel {z -axis, measured about the }-axis. Therefore, a steer
angle is positive when the wheel is turned to the left, and it is negative when
the wheel is turned to the right.
Proof. The slip-free condition for wheels of a 4Z V in a turn requires that
the normal lines to the center of each tire-plane intersect at a common
point. This is the kinematic steering condition.
Figure 7.38 illustrates a positive 4Z V vehicle in a left turn. The inner
wheels are the left wheels that are closer to the turning center R. The
longitudinal distance between point R and the axles of the car are indicated
by f1 , and f2 measured in the body coordinate frame.
The front inner and outer steer angles li , ri may be calculated from
the triangles 4RDH and 4REI , while the rear inner and outer steer
angles lu , ru may be calculated from the triangles 4RGJ and 4RFK
as follows.
tan li
=
tan ri
=
f1
U1 f1
U1 +
zi
2
(7.78)
zi
2
(7.79)
7. Steering Dynamics
x
xw
x
G
G
419
xw
yw
y
y
yw
Positive steer angle
Negative steer angle
FIGURE 7.37. Sign convention for steer angles.
tan lu
=
tan ru
=
f2
U1 f2
U1 +
zu
2
(7.80)
zu
2
(7.81)
Eliminating U1
U1 =
1
f1
1
f1
= zi +
zi +
2
tan li
2
tan ri
(7.82)
between (7.78) and (7.79) provides the kinematic condition between the
front steering angles li and ri .
cot ri cot li =
zi
f1
(7.83)
Similarly, we may eliminate U1
U1 =
1
f2
1
f2
= zu +
zu +
2
tan lu
2
tan ru
(7.84)
between (7.80) and (7.81) to provide the kinematic condition between the
rear steering angles lu and ru .
cot ru cot lu =
zu
f2
(7.85)
420
7. Steering Dynamics
wf
Gof
Gif
Inner
wheel
Outer
wheel
A
B
a1
Gor
Gir
R
c1
a2
D
Gof
C
c2
Gif
O
l
C
Gor
Gir
R1
wr
E
G
F
H
FIGURE 7.38. Illustration of a positive four-wheel steering vehicle in a left turn.
Using the constraint
f1 f2 = o
(7.86)
we may combine Equations (7.83) and (7.85)
zi
zu
=o
cot ri cot li
cot ru cot lu
(7.87)
to nd the kinematic condition (7.76) between the steer angles of the front
and rear wheels for a positive 4Z V vehicle.
Figure 7.39 illustrates a negative 4Z V vehicle in a left turn. The inner
wheels are the left wheels that are closer to the turning center R. The front
inner and outer steer angles li , ri can be calculated from the triangles
4RDH and 4REI , while the rear inner and outer steer angles lu , ru
may be calculated from the triangles 4RGJ and 4RFK as follows.
tan li
=
tan ri
=
f1
U1 f1
U1 +
zi
2
(7.88)
zi
2
(7.89)
7. Steering Dynamics
wf
Gif
Gof
Inner
wheel
Outer
wheel
A
B
Gor
c1
Gir
Gif
R
Gir
Gor
Gof
O
c2
421
G
a1
l
C
E
F
D
H
a2
C
R1
wr
FIGURE 7.39. Illustration of a negative four-wheel steering vehicle in a left turn.
tan lu
=
tan ru
=
f2
zu
U1 2
f2
zu
U1 +
2
(7.90)
(7.91)
Eliminating U1
U1 =
1
f1
1
f1
= zi +
zi +
2
tan li
2
tan ri
(7.92)
between (7.88) and (7.89) provides us with the kinematic condition between
the front steering angles li and ri .
zi
cot ri cot li =
(7.93)
f1
Similarly, we may eliminate U1
U1 =
1
f2
1
f2
= zu +
zu +
2
tan lu
2
tan ru
(7.94)
between (7.90) and (7.91) to provide the kinematic condition between the
rear steering angles lu and ru .
zu
cot ru cot lu =
(7.95)
f2
422
7. Steering Dynamics
Using the constraint
f1 f2 = o
(7.96)
we may combine Equations (7.93) and (7.95)
zu
zi
=o
cot ri cot li
cot ru cot lu
(7.97)
to nd the kinematic condition (7.76) between the steer angles of the front
and rear wheels for a negative 4Z V vehicle.
Using the sign convention of Figure 7.37, we may re-examine Figures
7.38 and 7.39. When the steer angle of the front wheels are positive then,
the steer angle of the rear wheels are negative in a negative 4Z V system,
and are positive in a positive 4Z V system. Therefore, Equation (7.77)
cot i u cot i o =
zu cot i u cot i o
zi
o
o cot uu cot uo
(7.98)
expresses the kinematic condition for both, positive and negative 4Z V
systems. Similarly, the following equations can uniquely determine f1 and
f2 regardless of the positive or negative 4Z V system.
f1
=
f2
=
zi
cot i u cot i o
zu
cot uu cot uo
(7.99)
(7.100)
Four-wheel steering or all wheel steering DZ V may be applied on vehicles to improve steering response, increase the stability at high speed
maneuvering, or decrease turning radius at low speeds. A negative 4Z V
has shorter turning radius U than a front-wheel steering (I Z V) vehicle.
For a I Z V vehicle, the perpendicular to the front wheels meet at a
point on the extension of the rear axle. However, for a 4Z V vehicle, the
intersection point can be any point in the {| plane. The point is the turning
center of the car and its position depends on the steer angles of the wheels.
Positive steering is also called same steer, and a negative steering is also
called counter steer.
Example 293 F Steering angles relationship.
Consider a car with the following dimensions.
o = 2=8 m
zi = 1=35 m
zu = 1=4 m
(7.101)
The set of equations (7.78)-(7.81) which are the same as (7.88)-(7.91) must
be used to nd the kinematic steer angles of the wheels. Assume one of the
angles, such as
(7.102)
li = 15 deg
is a known as an input steer angle. To nd the other steer angles, we need
to know the position of the turning center R. The position of the turning
7. Steering Dynamics
423
center can be determined if we have one of the three parameters f1 , f2 , U1 .
To clarify this fact, let us assume that the car is turning left and we know
the value of li . Therefore, the perpendicular line to the front left wheel is
known. The turning center can be any point on this line. When we pick a
point, the other wheels can be adjusted accordingly.
The steer angles for a 4Z V system is a set of four equations, each with
two variables.
li
lu
= li (f1 > U1 )
= lu (f2 > U1 )
ri = ri (f1 > U1 )
ru = ru (f2 > U1 )
(7.103)
(7.104)
If f1 and U1 are known, we will be able to determine the steer angles li ,
ri , lu , and ru uniquely. However, a practical situation is when we have
one of the steer angles, such as li , and we need to determine the required
steer angle of the other wheels, ri , lu , ru . It can be done if we know f1
or U1 .
The turning center is the curvature center of the path of motion. If the
path of motion is known, then at any point of the road, the turning center
can be found in the vehicle coordinate frame.
In this example, let us assume
U1 = 50 m
therefore, from Equation (7.78), we have
μ
¶
³
zi ´
1=35
f1 = U1 tan li = 50 tan
= 13=217 m
2
2
12
(7.105)
(7.106)
Because f1 A o and li A 0 the vehicle is in a positive 4Z V conguration
and the turning center is behind the rear axle of the car.
f2 = f1 o = 13=217 2=8 = 10=417 m
(7.107)
Now, employing Equations (7.79)-(7.81) provides us the other steer angles.
ri
lu
f1
13=217
1
zi = tan
1=35
U1 +
50 +
2
2
= 0=25513 rad 14=618 deg
= tan1
f2
1 10=417
zu = tan
1=4
U1 50 2
2
= 0=20824 rad 11=931 deg
= tan1
(7.108)
(7.109)
424
7. Steering Dynamics
f2
1 10=417
zu = tan
1=4
U1 +
50 +
2
2
= 0=20264 rad 11=61 deg
= tan1
ru
(7.110)
Example 294 F Position of the turning center.
The turning center of a vehicle, in the vehicle body coordinate frame, is
at a point with coordinates ({R > |R ). The coordinates of the turning center
are
zu
{R = d2 f2 = d2 (7.111)
cot ru cot lu
1
o + (zi tan li zu tan lu )
2
(7.112)
|R = U1 =
tan li tan lu
Equation (7.112) is found by substituting f1 and f2 from (7.92) and (7.94)
in (7.96), and dening |R in terms of li and lu . It is also possible to
dene |R in terms of ri and ru .
Equations (7.111) and (7.112) can be used to dene the coordinates of
the turning center for both positive and negative 4Z V systems.
As an example, let us examine a car with
o = 2=8 m
zi = 1=35 m
li
ri
lu
ru
=
=
=
=
zu = 1=4 m
d1 = d2
0=26180 rad 15 deg
0=25513 rad 14=618 deg
0=20824 rad 11=931 deg
0=20264 rad 11=61 deg
(7.113)
(7.114)
and nd the position of the turning center.
{R
|R
zu
cot ru cot lu
2=8
1=4
= = 11=802 m
(7.115)
2
cot 0=20264 cot 0=20824
1
o + (zi tan li zu tan lu )
2
=
tan li tan lu
1
2=8 + (1=35 tan 0=26180 1=4 tan 0=20824)
2
=
= 50=011 m (7.116)
tan 0=26180 tan 0=20824
= d2 The position of turning center for a I Z V vehicle is at
{R = d2
|R =
1
o
zi +
2
tan li
(7.117)
7. Steering Dynamics
425
and for a UZ V vehicle is at
{R = d1
|R =
1
o
zu +
2
tan lu
(7.118)
Example 295 F Curvature radius.
Consider a road as a path of motion that is expressed mathematically by
a function \ = i ([), in a global coordinate frame. The radius of curvature
U of such a road at point [ is
¡
¢3@2
1 + \ 02
U =
\ 00
(7.119)
where
\0 =
\ 00 =
g2 \
g[ 2
[
100
\ 00 =
g\
g[
(7.120)
Consider a road with a given equation
\ =
[2
200
\0 =
1
100
(7.121)
where both [ and \ are measured in meter [ m]. The curvature radius of
the road is.
¡
¢3@2
μ
¶3@2
1 + \ 02
1
2
[
=
100
+
1
U =
\ 00
10000
(7.122)
At [ = 30 m, we have
\ =
9
m
2
\0 =
3
10
\ 00 =
1
m1
100
(7.123)
and therefore,
U = 113=80 m
(7.124)
Example 296 F Symmetric four-wheel steering system.
Figure 7.40 illustrates a symmetric 4Z V vehicle that the front and rear
wheels steer opposite to each other equally. The kinematic steering condition
for a symmetric steering is simplied to
cot r cot l =
zu
zi
+
o
o
(7.125)
and f1 and f2 are reduced to
1
f1 = f2 = o
2
(7.126)
426
7. Steering Dynamics
Gi
wf
a1
R
O
Center of
rotation
Go
l
C
Gi
wr
a2
Go
FIGURE 7.40. A symmetric four-wheel steering vehicle.
Example 297 F f2 @f1 ratio.
Longitudinal distance of the turning center of a vehicle from the front
axle is f1 and from the rear axle is f2 . We show the ratio of these distances
by fv and call it the 4Z V factor.
fv =
f2
zu cot i u cot i o
=
f1
zi cot uu cot uo
(7.127)
fv is negative for a negative 4Z V vehicle and is positive for a positive
4Z V vehicle. When fv = 0, the car is I Z V, and when fv = 4, the car
is UZ V. A symmetric 4Z V system has fv = 12 .
Example 298 F Steering length ov .
For a 4Z V vehicle, we may dene a steering length ov as
ov
=
=
f1 + f2
o
+ 2fv
=
o
f1
¶
μ
zi
1
zu
+
o cot i u cot i o
cot uu cot uo
(7.128)
Steering length ov is 1 for a I Z V car, zero for a symmetric steering car,
and 1 for a UZ V car. When a car has a negative 4Z V system then,
1 ? ov ? 1, and when the car has a positive 4Z V system then, ov A 1 or
ov ? 1. The case ov A 1 occurs when the turning center is behind the car,
and the case ov ? 1 occurs when the turning center is ahead of the car.
A comparison of I Z V and 4Z V is that the turning center of a I Z V
car is always on the extension of the rear axel, and its steering length ov is
always equal to 1. However, the turning center of a 4Z V car can be:
7. Steering Dynamics
427
O
O
(a)
(b)
O
O
(d)
(c)
FIGURE 7.41. A comparison among the dierent steering lengths.
1= ahead of the front axle, if ov ? 1
2= between the front and rear axles, if 1 ? ov ? 1
3= behind the rear axle, if ov A 1
Figure 7.41 illustrates the comparison among the dierent steering lengths.
A I Z V car is shown in Figure 7.41(d), while the 4Z V systems with
ov ? 1, 1 ? ov ? 1, and ov ? 1 are shown in Figures 7.41(e)-(g) respectively.
Example 299 F I Z V and Ackerman condition.
By substituting
uu $ 0
uo $ 0
in the 4Z V kinematic condition (7.77) the equation reduces to the Ackerman condition (7.1) for the I Z V vehicles.
cot i u cot i o =
z
o
(7.129)
Writing the Ackerman condition as this equation frees us from checking the
inner and outer wheels.
Example 300 F Turning radius.
To nd the vehicle’s turning radius U, we may dene equivalent bicycle
models as shown in Figure 7.42 and 7.43 for positive and negative 4Z V
vehicles. The radius of rotation U is perpendicular to the vehicle’s velocity
vector v at the mass center F.
428
7. Steering Dynamics
Gf
a1
v
c1
l
C
R
Center of
rotation
O
a2
Gf
c2
Gr
Gr
R1
FIGURE 7.42. Bicycle model for a positive 4Z V vehicle.
Gf
Center of
rotation
c2
a1
v
c1
Gf
O
R
l
C
a2
Gr
R1
Gr
FIGURE 7.43. Bicycle model for a negative 4Z V vehicle.
7. Steering Dynamics
429
Let us examine the positive 4Z V conguration of Figure 7.42. Using the
geometry shown in the bicycle model, we have
U2
cot i
= (d2 + f2 )2 + U12
U1
1
=
= (cot li + cot ri )
f1
2
(7.130)
(7.131)
and therefore,
U=
q
2
(d2 + f2 ) + f21 cot2 i
(7.132)
Examining Figure 7.43 shows that the turning radius of a negative 4Z V
vehicle can be determined from the same equation (7.132).
Example 301 F Passive and active four-wheel steering.
The negative 4Z V is not recommended at high speeds because of high yaw
rates, and the positive steering is not recommended at low speeds because
of increasing radius of turning. Therefore, to maximize the advantages of
a 4Z V system, we need a smart system to allow the vehicle to change
the mode of steering according to the speed of the vehicle and adjust the
steer angles for dierent purposes. A smart steering is also called active
steering system.
An active system may provide a negative steering at low speeds and a
positive steering at high speeds. In a negative steering, the rear wheels are
steered in the opposite directions as the front wheels to turn in a signicantly
smaller radius, while in positive steering, the rear wheels are steered in the
same direction as the front wheels to increase the lateral force.
When the 4Z V system is passive, there is a constant proportional ratio
between the front and rear steer angles which is equivalent to have a constant fv . A passive steering may be applied in vehicles to compensate some
undesirable vehicle tendencies. As an example, in a I Z V system, the rear
wheels tend to steer slightly to the outside of a turn. Such tendency can
reduce stability.
Example 302 F Autodriver.
Consider a car at the global position ([> \ ) that is moving on a road, as
shown in Figure 7.44. Point F indicates the center of curvature of the road
at the car’s position. The center of curvature of the road is supposed to be
the turning center of the car at the instant of consideration.
There is a global coordinate frame J attached to the ground, and a vehicle
coordinate frame E attached to the car at its mass center F. The } and ]
axes are parallel and the angle # indicates the angle between the [ and {
axes. If ([F > \F ) are the coordinates of F in the global coordinate frame J
430
7. Steering Dynamics
Y
G
Y
C
y
f (X )
B
v
a
x
E
\
d
X
FIGURE 7.44. Illustration of a car that is moving on a road at the point that R
is the center of curvature.
then, the coordinates of F in E would be
¢
¡
E
rF = U}># J rF J d
(7.133)
5
6
5
6 35
6 5
64
{F
cos # sin # 0
[F
[
7 |F 8 = 7 sin # cos # 0 8 C7 \F 8 7 \ 8D
0
0
0
1
0
0
5
6
([F [) cos # + (\F \ ) sin #
= 7 (\F \ ) cos # ([F [) sin # 8
(7.134)
0
Having coordinates of F in the vehicle coordinate frame is enough to determine U1 , f1 , and f2 .
U1
f2
f1
= |F = (\F \ ) cos # ([F [) sin #
(7.135)
= d2 {F = ([F [) cos # (\F \ ) sin # d2 (7.136)
= f2 + o = ([F [) cos # (\F \ ) sin # + d1
(7.137)
Then, the required steer angles of the wheels can be uniquely determined by
Equations (7.78)-(7.81).
It is possible to dene a road by a mathematical function \ = i ([)
in a global coordinate frame. At any point [ of the road, the position of
the vehicle and the position of the turning center in the vehicle coordinate
frame can be determined. The required steer angles can accordingly be set
to keep the vehicle on the road and run the vehicle in the correct direction.
This principle may be used to design an autodriver.
7. Steering Dynamics
431
As an example, let us consider a car that is moving tangent to a road
with a given equation
[2
\ =
(7.138)
200
where both [ and \ are measured in meter [ m]. At [ = 30 m, we have
\ = 4=5 m and \ 0 = 0=3, \ 00 = 0=01, and therefore
# = arctan
g\
= arctan 0=3 = 0=29146 rad 16=7 deg
g[
(7.139)
The curvature radius at (30> 4=5), from Example 295, is
U = 113=80 m
(7.140)
The tangent line to the road is
\ 4=5 = 0=3 ([ 30)
(7.141)
and therefore the perpendicular line to the road is
10
([ 30)
3
(7.142)
¢3@2
¡
1 + 0=32
= 113=8 m
0=01
(7.143)
\ 4=5 = Having U = 113=80 m,
U =
we have
¡
1 + \ 02
\ 00
¢3@2
=
2
2
([ [F ) + (\ \F ) = U2
(7.144)
and we can nd the global coordinates of the curvature center ([F > \F ) at
the proper intersection of the line (7.142) and circle (7.144).
[F = 2=7002 m
\F = 113=5 m
(7.145)
The coordinates of the turning center in the body frame would then be
5
6 5
6 35
6 5
64 5
6
{F
cos # sin # 0
[F
[
0
7 |F 8 = 7 sin # cos # 0 8 C7 \F 8 7 \ 8D = 7 113=8 8
0
0
0
1
0
0
0
(7.146)
Example 303 F Curvature equation.
Consider a vehicle that is moving on a path \ = i ([) with velocity v
and acceleration a. The curvature = 1@U of the path that the vehicle is
moving on is
1
dq
=
(7.147)
= 2
U
y
432
7. Steering Dynamics
where, dq is the normal component of the acceleration a. The normal component dq is toward the rotation center and is equal to
¯v
¯ 1
¯
¯
dq = ¯ × a¯ = |v × a|
y
y
¨ \b
\¨ [b [
1
(7.148)
(d\ y[ d[ y\ ) = p
=
y
[b 2 + \b 2
and therefore,
¨ \b
¨ \b
1
\¨ [b [
\¨ [b [
= ³
´3@2 =
Ã
!3@2
3
b
[
\b 2
[b 2 + \b 2
1+
[b 2
(7.149)
However,
\0
\ 00
=
=
g\
\b
=
g[
[b
g2 \
g
=
g[ 2
g{
(7.150)
Ã
\b
[b
!
=
g
gw
Ã
\b
[b
!
¨ \b
1
\¨ [b [
=
[b
[b 3
(7.151)
and we nd the equation for the curvature of the path and radius of the
curvature based on the equation of the path.
=
U
=
\ 00
(7.152)
(1 + \ 02 )3@2
´3@2
³
¢
¡
b 2 + \b 2
02 3@2
[
1
+
\
1
=
=
¨ \b
\ 00
\¨ [b [
(7.153)
As an example, let us consider a road with a given equation
\ =
[2
200
(7.154)
At a point with [ = 30 m, we have
\ =
9
m
2
\0 =
3
10
1
m1
100
(7.155)
U = 113=80 m
(7.156)
\ 00 =
and therefore,
= 8=7874 × 103 m1
Example 304 F Center of curvature in global coordinate frame.
Assume a planar road being expressed by a parametric equation in a global
frame as
[ = [ (w)
\ = \ (w)
(7.157)
7. Steering Dynamics
433
The perpendicular line to the road at a point ([0 > \0 ) is:
\
tan = \0 ([ [0 ) @ tan g\
g\ @gw
\b
=
=
=
g[
g[@gw
[b
(7.158)
(7.159)
The curvature center ([F > \F ) is on the perpendicular line and is at a
distance U from the point ([> \ ) = ([0 > \0 ). Therefore,
´
´
³
³
\b [b 2 + \b 2
[b [b 2 + \b 2
\F = \ +
(7.160)
[F = [ ¨ \b
¨ \b
\¨ [b [
\¨ [b [
Employing Equation (7.146), the curvature center in the body coordinate
frame of a moving vehicle that its {-axis makes the angle # with the global
[-axis is
{F
|F
= ([F [) cos # + (\F \ ) sin #
³
´ [b 2 + \b 2
=
[b sin # \b cos #
¨ \b
\¨ [b [
(7.161)
= (\F \ ) cos # ([F [) sin #
³
´ [b 2 + \b 2
=
[b cos # + \b sin #
¨ \b
\¨ [b [
(7.162)
As and example, the global coordinate frame of a parabolic path
[ =w
\ =
w2
200
(7.163)
would be
[F
\F
[b 2 + \b 2 b
\ = 2=5244 × 1029 w 0=0001w3
¨ \b
\¨ [b [
[b 2 + \b 2 b
[ = 0=015 w2 + 100
= \ +
¨ \b
\¨ [b [
= [
(7.164)
(7.165)
Example 305 F An elliptic path and curvature center.
Consider an elliptic path with equations
[ = d cos w
d = 100 m
\ = e sin w
e = 65 m
(7.166)
(7.167)
The curvature center of the road in the global coordinate frame is at
´
³
\b [b 2 + \b 2
d2 e2
231
[F = [ =
(7.168)
cos3 w =
cos3 w
¨
b
¨
b
d
4
\ [ [\
´
³
[b [b 2 + \b 2
d2 e2
1155
=
\F = \ +
sin3 w =
sin3 w (7.169)
¨
b
¨
b
e
13
\ [ [\
434
7. Steering Dynamics
Y
X
FIGURE 7.45. An elliptic path and its curvature center.
Therefore, the curvature center in the vehicle coordinate frame would be
{F
|F
³
´ [b 2 + \b 2
[b sin # \b cos #
¨ \b
\¨ [b [
¶
μ 2
2
d e
=
cos3 w d cos w cos #
d
¶
μ 2
d e2
3
sin w + e sin w sin #
e
=
³
´ [b 2 + \b 2
[b cos # + \b sin #
¨ \b
\¨ [b [
¶
μ 2
d e2
sin3 w + e sin w cos #
= d
¶
μ 2
d e2
cos3 w d cos w sin #
d
(7.170)
=
(7.171)
Figure 7.45 illustrates the elliptic path and its curvature center.
7.6 F Road Design
Roads are made by continuously connecting straight and circular paths by
proper transition turning sections. Having a continues and well behaved
curvature is a necessary criterion in road design. The clothoid spiral is the
best smooth transition connecting curve in road design which is expressed
7. Steering Dynamics
435
y
a
1
S d t d S
x
t
§S 2·
t
§S 2·
x
³0 cos ¨© 2 u ¸¹du
y
³0 sin ¨© 2 u ¸¹du
FIGURE 7.46. The clothoide curve for d = 1 and 3 $ w $ .
by parametric equations called Fresnel Integrals:
Z w
³ ´
x2 gx
cos
[ (w) = d
2
0
Z w
³ ´
x2 gx
\ (w) = d
sin
2
0
(7.172)
(7.173)
The curvature of the clothoid curve varies linearly with arc length and this
linearity makes clothoid the smoothest driving transition curve. Figure 7.46
illustrates the clothoid curve for the scaling parameter d = 1 and variable
w . The scaling parameter d is only a magnication factor that
shrinks or magnies the curve. The range of w determines the variation of
curvature within the clothoid, as well as the initial and nal tangent angles
of the clothoid curve.
The arc length, v, of a clothoid for a given value of w is
v = dw
(7.174)
If the variable w indicates time then, d would be the speed of motion along
the path. The curvature and radius of curvature U of a clothoid at a
given w is
=
U
=
w
d
1
d
=
w
(7.175)
(7.176)
The tangent angle of a clothoid for a given value w is
=
2
w
2
(7.177)
436
7. Steering Dynamics
Having a road with linearly increasing curvature is equivalent to entering
the path with a steering wheel at the neutral position and turning the
steering wheel with a constant angular velocity. This is a desirable and
natural driving action.
Proof. Arc length v of a parametric planar curve [ = [ (w), \ = \ (w)
between w1 and w2 is calculated by
s
¶2 μ
¶2
Z w2 μ
g\
g[
+
gw
(7.178)
v=
gw
gw
w1
Substituting the clothoid spiral parametric expression (7.172)-(7.173), we
have
s
¶2 μ
¶2
Z w2 μ
g\
g[
v =
+
gw
gx
gx
w1
Z wr
Z w
³ ´
³ ´
= d
cos2
gw = dw
(7.179)
x2 + sin2
x2 gw = d
2
2
0
0
Curvature of a planar curve [ = [ (v), \ = \ (v) that is expressed
parametrically by its arc length v is
s
μ 2 ¶2 μ 2 ¶2
g \
g [
=
+
(7.180)
gv2
gv2
Using the result of (7.179), we can replace the variable w with arc length v
w=
v
d
(7.181)
and dened the clothoid spiral parametric equations (7.172)-(7.173) as
Z v@d
³ ´
[ (v) = d
cos
(7.182)
x2 gx
2
0
Z v@d
³ ´
\ (v) = d
sin
(7.183)
x2 gx
2
0
Therefore, the curvature of clothoid spiral is
s
μ 2¶
μ 2¶
v
v
v
v
w
2
= 2 cos2
+
sin
= 2 =
d
2 d2
2 d2
d
d
The slope tan of the tangent to clothoid spiral at a point w is
³ ´
³ ´
d
sin
w2
g\ @gw
g\
2
´
³
=
=
w2
=
tan
tan =
2
g[
g[@gw
2
d cos
w
2
(7.184)
(7.185)
7. Steering Dynamics
437
and therefore, the slope angle of the tangent line is
=
2 v2
w =
2
2 d2
The clothoid curve approaches the point (d@2> d@2) at innity
μ Z w
³ ´ ¶ d
[ (w) = lim d
x2 gx =
cos
w$4
2
2
0
μ Z w
³ ´ ¶ d
\ (w) = lim d
x2 gx =
sin
w$4
2
2
0
(7.186)
(7.187)
(7.188)
Combining the equations for arc length and curvature, we can express
the curvature of a clothoid curve such that it varies linearly with its
arc-length v. The curvature of such a curve is
(v) =
v = nv
d2
(7.189)
where, v is the arc length, n is the sharpness or the rate of change of
curvature.
Figure 7.47 illustrates a design graph of the relationship between the
clothoid and parameters of magnication factor d, curvature , and slope
. The higher magnication factor d the larger the clothoid. The clothoid
curves of dierent d are intersected by the constant slope lines of . The
curves for constant curvature n intersect both, the constant d and constant
curves.
The clothoid transition equation is a proper solution for any required
change in any parameter of a road. As an example the change of the bank
angle from a at straight road to a tilted road on a circular path needs a
transition bank angle.
Example 306 History of clothoid.
The clothoid spiral is also called Cornu spiral, referring to Alfred Cornu
(1841 1902), a French physicist who rediscovered the clothoid spiral. It
may also be called Euler spiral, as Leonard Euler (1707 1783) was the the
rst codiscoverer of the curve with Jacques Bernoulli (16541705) who formulated the clothoid spiral on deformations of elastic members. It is also
called Fresnel spiral credited to Augustin-Jean Fresnel spiral (1788 1827)
who independently rediscovered the curve in his work on the fringes of diffraction of light through a slot. In vehicle dynamics and road design industry
it may also be called the transition spiral to refer to the road connections
corners.
The clothoid spiral is also called Cornu spiral, referring to Alfred Cornu
(18411902), a French physicist who rediscovered the clothoid spiral. It may
also be called Euler spiral, as Leonard Euler (1707 1783) was the the rst
codiscoverer of the curve Jacques Bernoulli (1654 1705) who formulated
0.0
1
180
0. 0
0. 0
140
0 .0
80
0. 0
47
55
0
13
0
12
10
31 180
35
0.0 150
39
11
0
0
90
80
01
1
100
0. 0
27
a=210
0
0.
120
0 .0
23
14
0.01
3
160
70
120
0.0 0. 90
60 79 067
0. 0
99 70
55 40
40
30 20
20
a=10
0
9
a=250
0.015
y
15
0
7. Steering Dynamics
16
0
438
60
50
40
30
20
10
40
80
120
160
200
x
FIGURE 7.47. A design graph of the relationship between the clothoid and parameters of maginication factor d, curvature , and slope .
the clothoid spiral on deformations of elastic members. It is also called
Augustin-Jean Fresnel spiral (1788 1827) who independently rediscovered
the curve in his work on the fringes of diraction of light through a slot.
In vehicle dynamics and road design industry it may also be called the
transition spiral to refer to the road connections corners.
In the 19th century it became clear that we need a track shape with gradually varying curvature. Although circles were being used for most of the
path, a correct transition curve was needed to gradually change the curvature from one path to the other. Arthur Talbot (18571942) in 1880 derived
the same integrals as Bernoulli and Fresnel and introduced the railway transition spirals. Because of this the clothoid spiral is also called Talbot curve.
Talbot curve has bee used in railways and road construction.
It is said that Clotho was one of the three Fates who spun the thread of
human life, by winding it around the spindle. At the beginning of the 20th
century, the Italian mathematician Ernesto Cesàro (18591906), from this
poetic reference gave the name "clothoid" to the curve with a double spiral
shape.
7. Steering Dynamics
439
Example 307 F Derivative of an denite integral.
Dierentiation of an denite integral is based on the Leibniz formula
g
gw
Z e(w)
i (x> w) gx =
d(w)
Z e(w)
gi
ge
gd
gx + i (e (w) > w)
i (d (w) > w)
(7.190)
gw
gw
gw
d(w)
Taking derivative of the clothoid spiral parametric expression in calculating
the arc length v of (7.178) is based on the Leibniz formula.
g[
gw
g\
gw
Z
³ ´
³ ´
g w
cos
= d
x2 gx = d cos
w2
gw 0
2
2
Z
³ ´
³ ´
g w
sin
= d
x2 gx = d sin
w2
gw 0
2
2
(7.191)
(7.192)
The calculation of the curvature of (7.180) is also based on the Leibniz
formula.
μ 2¶
Z
³ ´
v
g v@d
g[
2
cos
(7.193)
= d
x gx = cos
gv
gw 0
2
2 d2
μ 2¶
Z
³ ´
v
g v@d
g\
sin
(7.194)
= d
x2 gx = sin
gv
gw 0
2
2 d2
g2 [
gv2
g2 \
gv2
=
=
μ 2¶
μ 2¶
v
g
v
v
= 2 sin
cos
gv
2 d2
d
2 d2
μ 2¶
μ 2¶
v
v
g
v
= 2 cos
sin
gv
2 d2
d
2 d2
(7.195)
(7.196)
Example 308 A connecting road with given d and .
Let us set d = 200 and plot a clothoid road starting from (0> 0) and end
up at a point with a given curvature of = 0=01. Using
w=
d
(7.197)
we can dene the parametric equations of the transition road (7.172)-(7.173)
as
Z d@
³ ´
cos
(7.198)
[ () = d
x2 gx
2
0
Z d@
³ ´
\ () = d
(7.199)
sin
x2 gx
2
0
The coordinates of the clothoid road at = 0=01 and d = 200 are
[0 = 122=2596310
\0 = 26=24682756
(7.200)
440
7. Steering Dynamics
y
C
36.5°
x
FIGURE 7.48. The tangent line, normal line, and the tangent circle to the clothoid
at the point where, = 0=01 for a given d = 200. The clothoid is plotted up to
= 0=025.
The slope of the road at the point is
=
1 2 2
d = 0=6366197722 rad = 36=475 deg
2
(7.201)
and therefore the tangent line to the road is
\ = 64=14007833 + 0=7393029502[
(7.202)
and the normal line to the road is
\ = 191=6183183 1=352625469[
(7.203)
Having the radius of the tangent curvature circle, U = 1@ = 100, we are
able to nd the coordinates of the center F of the tangent circle on the line
(7.203).
[F = 62=81155414
\F = 106=6578104
(7.204)
Figure 7.48 illustrates the tangent line, normal line, and the tangent circle
to the clothoid at the point where, = 0=01. The clothoid is plotted up to
= 0=025.
Example 309 F Connecting a straight road to a circle.
Assume that we need to dene a clothoid road to begin with zero curvature
and meet a given circular curve. Let us consider the road to be on the [-axis
and the circle of U = 100 m at center F (62=811> 106=658).
([ 62=811)2 + (\ 106=658)2 = 1002
(7.205)
7. Steering Dynamics
441
Therefore, the transition road must begin with = 0 on the [-axis and
touch the circle at a point when its curvature is = 1@100. Because of
=
v
w
=
d2
d
(7.206)
we can dene the parametric equations of the transition road (7.172)-(7.173)
by [ () = d
Z d@
cos
0
\ () = d
Z d@
sin
0
³
2
³
2
´
x2 gx
´
x2 gx
(7.207)
Having = 0=01 at the destination point, we nd the coordinates of the end
of the clothoid as functions of d.
[ (d) = d
Z 0=01d@
cos
0
\ (d) = d
Z 0=01d@
0
sin
³
2
³
2
´
x2 gx
´
x2 gx
(7.208)
We need to nd the magnifying factor, d, such that the clothoid (7.207)
touches the circle (7.205) with the same slope. The slope of the circle at
([> \ ) is
\ 0 = tan = ([ 62=811)
(\ 106=658)
(7.209)
and the slope angle of the clothoid is
=
2
1 2 2
w =
d = 1=5915 × 105 d2
2
2
(7.210)
To make the clothoid have the same slope, we derive an equation that relates
the magnication factor, d, to the components of the nal point of the
clothoid.
arctan
([ 62=811)
= 1=5915 × 105 d2
\ 106=658
(7.211)
Equations (7.211), (7.208) and (7.205) provide us with an equation to nd
442
7. Steering Dynamics
e
a
FIGURE 7.49. The error h = arctan
function of d.
3 ([ 3 62=811)
3 1=5915 × 1035 d2 as a
\ 3 106=658
d. Let us dene and plot an error equation versus d in Figure 7.49
([ 62=811)
1=5915 × 105 d2
\ 106=658
([ 62=811)
= arctan q
1=5915 × 105 d2
2
2
100 ([ 62=811)
³ R
³ ´
´
0=01d@
x2 gx 62=811
d 0
cos
2
= arctan r
³ R
³ ´
´2
0=01d@
1002 d 0
cos
x2 gx 62=811
2
1=5915 × 105 d2
(7.212)
h = arctan
and solve the equation for d.
d = 200
(7.213)
The clothoid equation are
[ () = d
Z d@
cos
0
\ () = d
Z d@
0
³
2
³
´
x2 gx
´
x2 gx
(7.214)
\0 = 26=24682756
(7.215)
sin
2
where at = 0=01 reaches to
[0 = 122=2596310
The slope of the road at the point, the tangent line to the road and the
7. Steering Dynamics
443
normal line to the road are
\
\
1 2 2
d = 0=6366197722 rad = 36=475 deg
2
= 64=14007833 + 0=7393029502[
= 191=6183183 1=352625469[
=
(7.216)
(7.217)
(7.218)
Figure 7.48 illustrates the clothoid, tangent line, normal line, and the tangent circle to the clothoid at the point where, = 0=01. The clothoid is
plotted up to = 0=025.
Example 310 F Connecting a straight road to another circle.
Assume that we need to determine a clothoid road to begin with zero
curvature on the [-axis and meet a given circular curve of U = 80 m at
center F (100> 100).
2
2
([ 100) + (\ 100) = 802
(7.219)
The expression of the transition road (7.172)-(7.173) by is:
[ () = d
Z d@
cos
0
\ () = d
Z d@
sin
0
³
2
³
2
´
x2 gx
´
x2 gx
(7.220)
Using = 1@U = 0=0125 at the destination point determines the coordinates
of the end of the clothoid as functions of d.
[ (d) = d
Z 0=0125d@
cos
0
\ (d) = d
Z 0=0125d@
0
sin
³
2
³
2
´
x2 gx
´
x2 gx
(7.221)
The slope of the tangent to the circle (7.219) at a point ([> \ ) is
\ 0 = tan = ([ 100)
\ 100
(7.222)
and the slope angle of the clothoid as a function of d is
=
1 2 2
2
w =
d = 2=4868 × 105 d2
2
2
(7.223)
The clothoid should have the same slope, therefore,
arctan
([ 100)
= 2=4868 × 105 d2
\ 100
(7.224)
444
7. Steering Dynamics
N 0.0125
e
a
e
arctan
X 100
Y 100
FIGURE 7.50. Plot of h = arctan
d.
a 2 N2
2S
3 ([ 3 100)
t
3 2=4868 × 1035 d2 versus
2
2
± 80 3 ([ 3 100)
N 0.0125
e
a
e
FIGURE 7.51. Plot of h =
d.
X 100
Y 100
tan
a 2 N2
2S
3 ([ 3 100)
t
3 tan 2=4868 × 1035 d2 versus
± 802 3 ([ 3 100)2
7. Steering Dynamics
445
y
75.8°
x
FIGURE 7.52. The transition road starting on the [-axis and goes to a circle of
radius U = 80 m at center F (100 m> 100 m).
Equations (7.224) and (7.219) along with (7.221) provide us with an
equation to nd d. However, substituting \ = \ ([) and replacing tan and
arctan generate four equations to be solved for possible d. To visualize the
possible solutions, let us dene two error equations (7.225)-(7.226).
h = arctan
h =
([ 100)
q
2=4868 × 105 d2
2
2
± 80 ([ 100)
¢
¡
([ 100)
q
tan 2=4868 × 105 d2
2
± 802 ([ 100)
(7.225)
(7.226)
Figure 7.50 depicts Equation (7.225) and Figure 7.51 shows Equation (7.226).
Equation (7.225) provides the solutions of
d = 230=7098693
d = 130=8889343
(7.227)
and Equation (7.226) provide the solutions of
d = 230=7098693
d = 394=0940573
d = 130=8889343
d = 463=5589702
(7.228)
The correct answer is d = 230=7098693 and Figure 7.52 depicts the circle
and the proper clothoid. Using d, we dene the clothoid equation
Z d@
³ ´
[ () = d
x2 gx
cos
2
0
Z d@
³ ´
x2 gx
sin
\ () = d
(7.229)
2
0
446
7. Steering Dynamics
which at = 0=0125 reaches
\0 = 82=38074640
(7.230)
1 2 2
d = 1=3236 rad 75=84 deg
2
(7.231)
[0 = 177=5691613
at angle
=
Example 311 Using the design chart.
Assume we are asked to nd a clothoid transition road to connect a
straight road to a circle of radius U = 58=824 m. Having U is equivalent
to have the destination curvature = 1@U = 0=017. The desired circle
must be tangent to a clothoid with a given d at the point that the clothoid
is intersecting the curve of = 0=017.
The clothoid for d = 250 m hits the curve of = 0=017 at a point for
which we have
[
= 147=3884878 m
= 164=7102491 deg
\ = 176=4421850 m
v = 338=2042540 m
(7.232)
(7.233)
The clothoid for d = 210 m hits the curve of = 0=017 at a point for which
we have
[
= 157=4739501 m
= 116=2195518 deg
\ = 119=7133227 m
v = 238=6369216 m
(7.234)
(7.235)
The clothoid for d = 180 m hits the curve of = 0=017 at a point for which
we have
[
= 140=1918463 m
= 85=38579313 deg
\ = 74=21681673 m
v = 175=3250853 m
(7.236)
(7.237)
The clothoid for d = 150 m hits the curve of = 0=017 at a point for which
we have
[
= 109=3442240 m
= 59=29568967 deg
\ = 38=89541829 m
v = 121=7535314 m
(7.238)
(7.239)
The clothoid for d = 120 m hits the curve of = 0=017 at a point for which
we have
[
= 74=57259185 m
= 37=94924139 deg
\ = 16=67204291 m
v = 77=92226012 m
(7.240)
(7.241)
Figure 7.53 illustrates these solutions. The number of solutions is practically innite and the best solution depends on safety, cost, and physical
constraints of the eld.
7. Steering Dynamics
y
N 0.017
180
a=250
447
T 164.71$
160
2
6.2
11
a=210
140
180
120
100
80
386
85.
150
120
95
59.2
60
37.95
40
20
0
40
80
120
160
200
x
FIGURE 7.53. A few clothoid transition road to connect a straight road to a
circle of radius U = 58=824 m.
Example 312 F Clothoid spiral as an optimal curve problem.
Clothoid spiral is the shortest curve connecting two given points with
given initial and nal tangent angles and curvatures. The angle and curvature are varying. The problem can be formulated as follows:
Given two points ([1 > \1 ) and ([2 > \2 ) and two angles 1 and 2 , nd a
curve (clothoid segment) which satises:
[ (0) = [1
\ (0) = \1
[ (v) = [2
\ (v) = \2
g\ (0)
g\ (0) @gw
=
(7.242)
g[ (0)
g[ (0) @gw
g\ (v)
g\ (v) @gw
tan 2 =
=
(7.243)
g[ (v)
g[ (v) @gw
tan 1 =
with minimal arc length v.
Example 313 F Clothoid shift to meet a given circle.
It is not generally possible to design a clothoid starting at the origin and
meet a given circle at an arbitrary center and radius. However, it is possible
to start the clothoid from other points on the {-axis to meet the given circle.
Assume we need to design a clothoid starting on the {-axis to meet a
448
7. Steering Dynamics
given circle
({ {F )2 + (| |F )2 = U2
(7.244)
where ({F > |F ) indicates the coordinates of the center of the circle, and U
is the radius of the circle. Substituting w in terms of d and U
w=
d
d
=
U
(7.245)
we dene the equation of clothoid to have the same radius of curvature as
the circle at the end point.
{ () = d
Z d@(U)
cos
0
| () = d
Z d@(U)
0
sin
³
2
³
2
´
x2 gx
´
x2 gx
(7.246)
Let us assume there is a | at which the slope of the clothoid
μ 2 ¶
μ 2¶
w
d
= tan
tan = tan
2
2U2
and the circle
tan = are equal.
tan
μ
d2
2U2
¶
{ {F
| |F
=
(7.248)
{ {F
| |F
Searching for a match point in the right half circle
q
| |F = U2 ({ {F )2
(7.247)
(7.249)
(7.250)
makes the slope equation to be a function of d
μ 2 ¶q
d
{ {F
tan
U2 ({ {F )2 +
=0
2
2U
| |F
(7.251)
or
tan
μ
v
!2
à Z
¶u
d@(U)
³ ´
u
d
tU2 d
cos
x2 gx {F
2U2
2
0
Z d@(U)
³ ´
+d
cos
x2 gx {F
2
0
2
= 0 (7.252)
Solution of this equation provides us with an d for which the clothoid ends
at a point with the same curvature as the circle. At the same | of the end
7. Steering Dynamics
449
y
R
xC , yC
y = 38.2
x
FIGURE 7.54. A clothoid starting at origin ends at a point with the same slope
and curvature as the given circle, at the same |.
point, the slope of the clothoid is also equal to the slope of the circle. A
proper shift of the clothoid on the {-axis will match the clothoid and the
circle.
As an example, let us assume that the circle is
({ 60)2 + (| 60)2 = 502
(7.253)
and therefore, the slope equation will be
v
à Z
!2
μ 2 ¶u
d@(50)
³ ´
u
d
t
x2 gx 60
tan
502 d
cos
2502
2
0
Z d@(50)
³ ´
+d
cos
x2 gx 60 = 0 (7.254)
2
0
Numerical solution of the equation is
d = 132=6477323
(7.255)
The plot of the clothoid and the circle at this moment are shown in Figure
7.54.
For the calculated d, the value of w at the end point of the clothoid is
d
2=652954646
=
= 0=84446
U
and therefore, the coordinates of the end point are
Z 0=84
³ ´
x2 gx = 98=75389126
{ () = 132=6
cos
2
0
Z 0=84
³ ´
x2 gx = 38=22304651
| () = 132=6
sin
2
0
w=
(7.256)
(7.257)
450
7. Steering Dynamics
y
R
xC , yC
y = 38.2
xdis
x
FIGURE 7.55. A shifted clothoid starting on a point on the {-axis ends at a point
on the given circle with the same slope and curvature.
At the point, the radius of curvature of the clothoid is
d
132=6477323
1
=
=
= 50
w
0=84446
(7.258)
2 w = 0=844462 = 1=1202 rad
2
2
(7.259)
U=
and the slope is
=
The {-coordinate of the circle at the same | = 38=22304651,
q
2
38=22304651 60 = 502 ({ 60)
(7.260)
is
{flufoh = 105=0084914
(7.261)
If we shift the clothoid by the dierence between {flufoh and {forwkrlg
{glv
= {flufoh {forwkrlg
= 105=0084914 98=75389126 = 6=2546
(7.262)
then the clothoid and circle meet at a point on the circle with all requirements to have a smooth transition. Figure 7.55 illustrates the result.
Example 314 F Complex expression and proof of curvature.
Let us dene a curve in complex plane as
F (v) = d
Z v@d
0
2
hlx @2 gx
(7.263)
7. Steering Dynamics
451
The derivative of the curve is an equation with absolute value of d.
2
2
gF (v)
= hlv @(2d )
gv
(7.264)
The curvature of the curve is
¯
¯ 2
¯ g F (v) ¯ ¯¯ v lv2 @(2d2 ) ¯¯ v
¯ = ¯l h
¯
=¯
¯= 2
gv2 ¯
d2
d
(7.265)
Example 315 F Parametric form of a straight road.
The equation of a straight road that connects two points S1 ({1 > |1 > }1 )
and S2 ({2 > |2 > }3 ) is
| |1
} }1
{ {1
=
=
{2 {1
|2 |1
}2 }1
(7.266)
This line may also be expressed by the following parametric equations.
{ = {1 + ({2 {1 ) w
| = |1 + (|2 |1 ) w
} = }1 + (}2 }1 ) w
(7.267)
Example 316 F Arc length of a planar road.
A planar road in the ({> |)-plane
| = i ({)
(7.268)
r = {ˆ~ + | ({) ˆ
(7.269)
can be expressed vectorially by
The displacement element on the curve
gr
g|
= ˆ~ +
ˆ
g{
g{
provides us with
μ
gv
g{
¶2
gr gr
=
·
=1+
g{ g{
(7.270)
μ
g|
g{
¶2
(7.271)
Therefore, the arc length of the curve between { = {1 and { = {2 is
s
μ ¶2
Z {2
g|
1+
g{
(7.272)
v=
g{
{1
In case the curve is given parametrically,
{ = {(w)
| = |(w)
(7.273)
452
7. Steering Dynamics
we have
μ
and hence,
gv
gw
¶2
gr gr
=
·
=
gw gw
μ
g{
gw
¶2
+
μ
g|
gw
¶2
s
Z w2 ¯ ¯ Z w2 μ ¶2 μ ¶2
¯ gr ¯
g|
g{
¯ ¯=
+
gw
v=
¯ gw ¯
gw
gw
w1
(7.274)
(7.275)
w1
Let us show a circle with radius U by its polar expression using the angle
as a parameter, as an example:
{ = U cos | = U sin (7.276)
The arc length between = 0 and = @2 would then be one-fourth the
perimeter of the circle. Therefore, the equation for calculating the perimeter
of a circle with radius U is
s
Z @2 μ ¶2 μ ¶2
Z @2 p
g|
g{
v = 4
+
g = U
sin2 + cos2 g
g
g
0
0
Z @2
= 4U
g = 2U
(7.277)
0
Using Equation (7.272), we can dene the equation of the road as
s
Z {2 μ ¶2
gv
1 g{
(7.278)
|=
g{
{1
Example 317 A gure 8 as an approximately correct road.
Sometimes, matching slopes, instead of matching curvatures, can be used
to design an approximately correct road. Let us make a closed road in the
shape of a symmetric gure eight with two 180 deg circular paths. Assuming
d = 200
(7.279)
the equations of the clothoid road starting from the origin are:
[ (w) = 200
Z w
cos
³
´
x2 gx
2
Z w
³ ´
sin
\ (w) = 200
x2 gx
2
0
0
(7.280)
(7.281)
The slope (7.177) of the curve would be parallel to the symmetric line \ =
[ when
r
s
2
2
w=
=
= w0
(7.282)
=
4
2
R
tl
XC , Y C
Ta
ng
en
Y
453
in
e
7. Steering Dynamics
X0 , Y0
e
rp
Pe
X
e
lin
ar
ul
ic
nd
FIGURE 7.56. A clothoid between two point at which the tangent lines are
parallel to \ = [.
At w = w0 the clothoid is at
[0
= 200
Z s2@2
cos
0
\0
= 200
Z s2@2
0
sin
³
2
³
2
´
x2 gx = 132=943
´
x2 gx = 35=424
(7.283)
(7.284)
where the tangent and perpendicular lines respectively are
\
\
= 35=424 + ([ 132=943) tan = 97=519 + [
= 35=424 ([ 132=943) @ tan = 168=37 [
(7.285)
(7.286)
as are shown in Figure 7.56 for the clothoid from w = w0 to w = w0 .
The perpendicular line hits the symmetric line \ = [ at
[F = \F = 84=184
(7.287)
which would be the center of a circular path with
q
2
2
U = ([0 [F ) + (\0 \F ) = 68=956
(7.288)
to connect ([0 > \0 ) to its mirror point with respect to \ = [ at
[1 = 35=424
\1 = 132=943
(7.289)
The mirror clothoid
[
= 200
\
= 200
Z s2@2
s
2@2
s
Z 2@2
s
2@2
sin
cos
³
2
³
2
´
x2 gx
´
x2 gx
(7.290)
(7.291)
454
7. Steering Dynamics
Y
C R
X
FIGURE 7.57. A slope match symmetric gure 8 road based on two clothoid and
two circular parts.
will complete the gure 8 road as is shown in Figure 7.57.
Therefore, the parametric equations of the road, beginning from the origin and moving in the [-direction are as follows. The parameter w is not
continuous and not with the same dimension in all equations.
s
Z w
³ ´
2
2
cos
0w
x gx
(7.292)
[ (w) = 200
2
2
0
s
Z w
³ ´
2
sin
0w
\ (w) = 200
x2 gx
(7.293)
2
2
0
w 2=3562
4
w 2=3562
\ (w) = \F + U sin w
4
s
s
Z w
³ ´
2
2
2
x gx
w
sin
[ (w) = 200
2
2
2
0
s
s
Z w
³ ´
2
2
\ (w) = 200
x2 gx
w
cos
2
2
2
0
[ (w) = [F + U cos w
[ (w) = [F + U cos w
\ (w) = \F + U sin w
[ (w) = 200
Z w
cos
0
\ (w) = 200
Z w
0
sin
³
2
³
2
2=3562 w 5=4978
2=3562 w 5=4978
´
x2 gx
´
x2 gx
s
2
w0
2
s
2
w0
2
(7.294)
(7.295)
(7.296)
(7.297)
(7.298)
(7.299)
(7.300)
(7.301)
7. Steering Dynamics
455
Using v as the road length, we may dene the equations with a smooth and
continuos parameter.
Z v@200
³ ´
x2 gx
cos
(7.302)
[ (w) = 200
2
0
Z v@200
³ ´
sin
\ (w) = 200
(7.303)
x2 gx
2
0
s
(7.304)
0 v 100 2
Ã
!
s
v 100 2 [ (w) = [F + U cos
U
4
Ã
!
s
v 100 2 \ (w) = \F + U sin
U
4
s
100 2 v 358=05
[ (w) = 200
Z (499=47v)@200
sin
0
\ (w) = 200
Z (499=47v)@200
0
cos
³
2
³
´
x2 gx
´
x2 gx
2
358=05 v 640=89
¶
640=89 v [ (w) = [F + U cos
U
4
¶
μ
640=89 v \ (w) = \F + U sin
U
4
640=89 v 857=52
μ
[ (w) = 200
Z (v857=52100s2)@200
cos
2
0
\ (w) = 200
Z (v857=52100s2)@200
0
sin
³
³
857=52 v 998=94
2
´
x2 gx
´
x2 gx
(7.305)
(7.306)
(7.307)
(7.308)
(7.309)
(7.310)
(7.311)
(7.312)
(7.313)
(7.314)
(7.315)
(7.316)
The variation of [ and \ for 0 v 998=94 are depicted in Figure 7.58.
Example 318 A gure 8 correct road.
Let us design a closed road in the shape of a symmetric gure eight with a
curvature transition between the clothoids and the circular paths. Assuming
d = 200
(7.317)
456
7. Steering Dynamics
Y
X
X
Y
s
FIGURE 7.58. The variation of [ and \ for 0 $ v $ 998=94 of the gure 8 road.
the equations of the clothoid road starting from the origin are:
Z w
³ ´
cos
[ (w) = 200
x2 gx
2
0
Z w
³ ´
\ (w) = 200
sin
x2 gx
2
0
(7.318)
(7.319)
The slope (7.177) at w of the curve is
=
2
w
2
(7.320)
where the tangent and perpendicular lines respectively are:
\
\
= \ (w) + ([ [ (w)) tan = \ (w) ([ [ (w)) @ tan (7.321)
(7.322)
The radius of curvature of the clothoid at w is
U=
1
d
=
w
(7.323)
If a clothoid point ([> \ ) exist at which the radius of curvature is equal to
the distance of the point from the line \ = [ on the perpendicular line,
then we can have a circular path starting at the point. The intersection
of the clothoid point ([> \ ) and the symmetric line \ = [ is the point
\F = [F where
[ (w) + \ (w) tan
\F = [F =
tan
w2
+1
2
w2
2
(7.324)
7. Steering Dynamics
457
d-R
t
d
FIGURE 7.59. Plot of the equation g 3 U = g 3
= 0 to nd a w at which the
w
curvature of the clothoid matches with a symmetric circle.
The distance of the clothoid and the point ([F > \F ) on the perpendicular
line is
q
2
2
(\ (w) \F ) + ([ (w) [F )
g =
v
u3
42 3
42
u
w2
u
([ (w) \ (w)) tan
uE \ (w) [ (w) F
E
2 F
= u
(7.325)
D +C
D
2
tC
w
w2
+1
+1
tan
tan
2
2
Equating g and U provides us with an equation to be solved for the w at
which the clothoid terminates and a circle with the same curvature starts.
d
=0
(7.326)
g
w
As is shown in Figure 7.59, the equation has multiple solutions, and the
rst solution is at
(7.327)
w = w0 = 0=9371211755
At w = w0 the clothoid and its kinematics are
Z 0=9371211755
³ ´
x2 gx = 154=77
cos
[0 = 200
2
0
Z 0=9371211755
³ ´
x2 gx = 75=154
\0 = 200
sin
2
0
1
U =
= 67=93355959
= 1=379467204 rad = 79=038 deg
(7.328)
(7.329)
(7.330)
(7.331)
458
7. Steering Dynamics
The intersection of the perpendicular to the clothoid at ([0 > \0 ) with \ = [
is at
[F
\F
= 88=0724138 ? [0
= 88=0724138 A \0
(7.332)
(7.333)
We can check the distance of ([0 > \0 ) and ([F > \F ) to be equal to U.
Therefore, using the parametric v as the road length, the equations of
the road, beginning from the origin and moving in the [-direction can be
expressed as
[ (w) = 200
Z v@d
cos
0
\ (w) = 200
Z v@d
sin
0
v0
= dw0 = 187=42
³
2
³
2
´
x2 gx
´
x2 gx
0 v v0
(7.334)
0 v v0
(7.335)
(7.336)
μ
¶
v v0
[ (w) = [F + U cos
v0 v v1
0
U
¶
μ
v v0
\ (w) = \F + U sin
v0 v v1
0
U
\F \0
0 = arctan
= 0=19132 rad = 10=962 deg
[0 [F
³
´
+ 20 = 320=12
v1 = v0 + U
2
[ (w) = 200
Z (v1 +v0 v)@d
sin
0
\ (w) = 200
Z (v1 +v0 v)@d
cos
0
v2
= v1 + 2v0 = 694=96
³
2
³
2
´
x2 gx
´
x2 gx
(7.337)
(7.338)
(7.339)
(7.340)
v1 v v2 (7.341)
v1 v v2 (7.342)
(7.343)
μ
´¶
v2 v ³ v2 v v3 (7.344)
[ (w) = [F + U cos
0
U
2
μ
´¶
v2 v ³ \ (w) = \F + U sin
v2 v v3 (7.345)
0
U
2
´
³
v3 = v2 + U
(7.346)
+ 20 = 827=66
2
7. Steering Dynamics
459
Y
X
FIGURE 7.60. A curvature match symmetric gure 8 road based on two clothoid
and two circular parts.
[ (w) = 200
Z (vv3 v0 )@200
cos
0
\ (w) = 200
Z (vv3 v0 )@200
0
v4
= v3 + v0 = 1015=1
sin
³
2
³
2
´
x2 gx
´
x2 gx
v3 v v4 (7.347)
v3 v v4 (7.348)
(7.349)
The road is shown in Figure 7.60.
Example 319 F Spatial road.
If the position vector J rS of a moving car is such that each component
is a function of a variable w,
J
r = J r (w) = { (w) ˆ~ + | (w) ˆ + } (w) n̂
(7.350)
then the end point of the position vector indicates a curve F in J. The
curve J r = J r (w) reduces to a point on F if we x the parameter w. The
functions
{ = { (w)
| = | (w)
} = } (w)
(7.351)
are the parametric equations of the curve. When the parameter w is the arc
length v, the innitesimal arc distance gv on the curve is
gv2 = gr · gr
(7.352)
The arc length v of a curve is dened as the limit of the diagonal of a
rectangular box as the length of the sides uniformly approaches zero.
When the space curve is a straight line that passes through point S ({0 > |0 > }0 )
460
7. Steering Dynamics
where {0 = {(w0 ), |0 = |(w0 ), }0 = }(w0 ), its equation can be shown by
| |0
} }0
{ {0
=
=
(7.353)
(7.354)
2 + 2 + 2 = 1
where , , and are the directional cosines of the line. The equation of
the tangent line to the space curve (7.351) at a point S ({0 > |0 > }0 ) is
| |0
} }0
{ {0
=
=
(7.355)
g{@gt
g|@gt
g}@gt
μ ¶2 μ ¶2 μ ¶2
g|
g}
g{
+
+
= 1
(7.356)
gt
gt
gt
To show this, let us consider a position vector J r = J r (v) that describes
a space curve using the length parameter v:
J
r = J r (v) = { (v) ˆ~ + | (v) ˆ + } (v) n̂
(7.357)
The arc length v is measured from a xed point on the curve. By a very
small change gv, the position vector will move to a very close point such
that the increment in the position vector would be
gr = g{ (v) ˆ~ + g| (v) ˆ + g} (v) n̂
(7.358)
The lengths of gr and gv are equal for innitesimal displacement:
p
(7.359)
gv = g{2 + g| 2 + g} 2
The arc length has a better expression in the square form:
gv2 = g{2 + g| 2 + g} 2 = gr · gr
(7.360)
If the parameter of the space curve is t instead of v, the increment arc
length would be
μ ¶2
gv
gr gr
=
·
(7.361)
gw
gw gw
Therefore, the arc length between two points on the curve can be found by
integration:
s
Z w2 r
Z w2 μ ¶2 μ ¶2 μ ¶2
g|
g}
g{
gr gr
v=
+
+
gw
(7.362)
· gw =
gw
gw
gw
gw
gw
w1
w1
Let us expand the parametric equations of the curve (7.351) at a point
S ({0 > |0 > }0 ),
g{
1 g2 { 2
4w + · · ·
4w +
gw
2 gw2
g|
1 g2 | 2
4w + · · ·
= |0 + 4w +
gw
2 gw2
g}
1 g2 } 2
4w + · · ·
= }0 + 4w +
gw
2 gw2
{ = {0 +
|
}
(7.363)
7. Steering Dynamics
461
and ignore the nonlinear terms to nd the tangent line to the curve at the
point:
{ {0
| |0
} }0
=
=
= 4w
(7.364)
g{@gw
g|@gw
g}@gw
Example 320 F Length of a spatial road.
Consider a spatial closed road with the following parametric equations:
{ = (d + e sin ) cos | = (d + e sin ) sin } = e + e cos (7.365)
The total length of the road can be found by the integral of gv for from 0
to 2:
s
Z 2 μ ¶2 μ ¶2 μ ¶2
Z 2 r
C{
C{
C{
gr gr
+
+
g
· g =
v =
g
g
C
C
C
1
1
Z 2 s p
2
2d2 + 3e2 e2 cos 2 + 4de sin g
(7.366)
=
2
0
7.7 F Steering Mechanism Optimization
Optimization of a steering mechanism means to design a system that works
as closely as possible to a desired function. Assume the Ackerman kinematic condition is the desired function for a steering system. Comparing
the function of the designed steering mechanism to the Ackerman condition, we dene an error function h to compare the two systems. An example
for the h function can be the dierence between the outer steer angles of the
designed mechanism Gr and the Ackerman Dr for the same inner angle
l.
The error function may be the absolute value of the maximum dierence,
h = max | Gr Dr |
(7.367)
or the root mean square (UP V) of the dierence between the two functions
s
Z 2
1
2
( Gr Dr ) g l
(7.368)
h=
2 1 1
or simply
h=
sZ
2
1
2
( Gr Dr ) g l
in a specic range of the inner steer angle 1 ? l ? 2 .
(7.369)
462
7. Steering Dynamics
The error h, would be a function of a set of parameters. Minimization
of the error function for a parameter, over the working range of the steer
angle l , generates the optimized value of the parameter.
The UP V function (7.368) is dened for continuous variables Gr and
Dr . However, depending on the designed mechanism, it is not always possible to nd a closed-form equation for h. In such a case, the error function
cannot be dened explicitly, and hence, the error function should be evaluated for q dierent values of the inner steer angle l numerically. The UP V
error function for a set of discrete values of h is dene by
v
u q
u1 X
2
h=t
( Gr Dr )
(7.370)
q l=1
where q is the number of subdivision of l in the working range 1 l 2 . The error function (7.368) or (7.370) must be evaluated for dierent
values of a parameter. Then a plot for h = h(sdudphwhu) can show the
trend of variation of h as a function of the parameter. If there is a minimum
for h, then the optimal value for the parameter can be found. Otherwise,
the trend of the h function can show the direction for minimum searching.
Example 321 F Optimized trapezoidal steering mechanism.
The inner-outer angles relationship for a trapezoidal steering mechanism,
shown in Figure 7.8 is
sin ( + l ) + sin ( r )
r³
´2
z
z
2
= 2 sin (cos ( r ) cos ( + l ))
g
g
(7.371)
Comparing Equation (7.371) with the Ackerman condition,
cot r cot l =
z
o
(7.372)
we dene an error function
v
u q
u1 X
h=t
( Gr Dr )2
q l=1
(7.373)
and search for its minimum to optimize the trapezoidal steering mechanism.
Consider a vehicle with the dimensions
z = 1=66 m
o = 2=93 m
(7.374)
Let’s optimize a trapezoidal steering mechanism for
g = 0=4 m
(7.375)
7. Steering Dynamics
463
l=2.93 m
w=1.66 m
d=0.4 m
e
E [rad]
16
17.2
18.3
19.5
20.6
21.8
22.9
24.1
E [deg]
FIGURE 7.61. Error function h = h() for a specic trapezoidal steering mechanism, with a minimum at E 19=5 deg.
and use as a parameter.
A plot of comparison between such a mechanism and the Ackerman condition, for a set of dierent , is shown in Figure 7.11, and their dierence
r = Gr Dr is shown in Figure 7.13.
We may set a value for , say = 6 deg, and evaluate Gr and Dr at
q = 100 dierent values of l for a working range such as 40 deg l 40 deg. Then, we calculate the associated error function h
v
u
100
u 1 X
( Gr Dr )2
(7.376)
h=t
100 l=1
for the specic . Now we conduct our calculation again for new values of
, such as = 10 deg> 11 deg> · · · . Figure 7.61 depicts the function h = h()
with a minimum at 19=5 deg.
The geometry of the optimal trapezoidal steering mechanism is shown in
Figure 7.62(d). The two side arms intersect at point J on their extensions.
For an optimal mechanism, the intersection of point J may be in front or
behind of the rear axle. However, it is recommended to put the intersection
point at the center of the rear axle and design a near optimal trapezoidal
steering mechanism. Using the recommendation, it is possible to eliminate
the optimization process and get a good enough design. Such an estimated
design is shown in Figure 7.62(e). The angle for the optimal design is
= 19=5 deg, and for the estimated design is 15=8 deg.
arctan
1=66@2
z@2
= arctan
= 0=27604 rad = 15=816 deg
o
2=93
(7.377)
464
7. Steering Dynamics
d
19.5o
E
E
15.8o
w
(a)
(b)
w
G
G
FIGURE 7.62. The geometry of the optimal trapezoidal steering mechanism and
the estimated design.
Example 322 F No exact Ackerman mechanism.
It is not possible to make a simple steering linkage to work exactly based
on the Ackerman steering condition. However, it is possible to optimize
various steering linkages, for a working range, to function close to the Ackerman condition, and be exact at a few points. An isosceles trapezoidal
linkage is not as exact as the Ackerman steering at every arbitrary turning
radius, however, it is simple enough to be mass produced, and exact enough
to be installed in street cars.
The trapezoidal linkage cannot act as Ackerman mechanism because the
Ackerman condition is a function of the wheel base o, while the trapezoidal
linkage is not a function of o.
Example 323 F Optimization of a multi-link steering mechanism.
Assume that we wish to design a multi-link steering mechanism for a
vehicle with the following dimensions.
z = 2=4 m
o = 4=8 m
d2 = 0=45o = 2=16 m
(7.378)
Due to space constraints, the positions of some joints of the mechanism
are xed as shown in Figure 7.63. However, we may vary the length { to
7. Steering Dynamics
465
54.6q
x
b
R
A
M
a
B
c
b
C
GS
0.22 m
D
N
P
d=1.2 m
FIGURE 7.63. A multi-link steering mechanism that must be optimized by varying {.
design the best mechanism to function as close as possible to the Ackerman
condition.
z
1
cot 2 cot 1 =
=
(7.379)
o
2
The steering wheel input v turns the triangle S EF which turns both the
left and the right wheels.
The vehicle must be able to turn in a minimum circle with radius Up .
Up = 10 m
(7.380)
The minimum turning radius determines the maximum steer angle P
q
Up =
d22 + o2 cot2 P
q
10 =
(0=45 × 4=8)2 + 4=82 cot2 P
P
= 0=4569 rad 26=178 deg
(7.381)
where P is the cot-average of the inner and outer steer angles. Having U
and P is enough to determine r and l .
U1
l
r
= o cot P = 4=8 cot 0=4569 = 9=7642 m
o
= tan1
z = 0=51085 rad 29=270 deg
U1 2
o
1
= tan
z = 0=41265 rad 23=643 deg
U1 +
2
(7.382)
(7.383)
(7.384)
Because the mechanism is symmetric, each wheel of the steering mechanism
in Figure 7.63 must be able to turn at least 29=27 deg. To be safe, we try
to optimize the mechanism for a wider range, say = ±35 deg, however,
466
7. Steering Dynamics
54.6q
x/2
b1
A
x/2
B
M
D
C
0.22 m
c1 G S
a1
b2
P
d1=1.2 m
N
a2
P
c2
d2=1.2 m
FIGURE 7.64. The multi-link steering is a 6-link mechanism that may be treeted
as two combined 4-bar linkages.
54.6q
x/2
b1
a1
2
T2
3
c1
4
x/2
T4
4
b2
M 2 0.22 m
d=1.2 m
5
M4
6
d=1.2 m
FIGURE 7.65. The input and output angles of the two 4-bar linkages.
vehicles in normal driving conditions barely receive steer angles beyond =
±15 deg. Therefore, we try to optimize the steering mechanism for this
practical range.
The multi-link steering mechanism is a six-link Watt linkage. We divide
the mechanism into two four-bar linkages. The linkage 1 is on the left and
the linkage 2 is on the right, as shown in Figure 7.64. We may assume
that P D is the input link of the left linkage and S E is its output link.
Link S E is rigidly attached to S F, which is the input of the right linkage.
The output of the right linkage is Q G. To nd the inner-outer steer angles
relationship, we need to nd the angle of Q G as a function of the angle of
P D. In the following analysis, d1 and d2 are the length of the input links
of the mechanism 1 and 2.
Figure 7.65 illustrates the link numbers, and the input-output angles of
the four-bar linkages. The steer angles can be calculated based on the angle
of these two links.
1
2
= 2 (90 54=6) deg
= *4 (90 + 54=6) deg
(7.385)
(7.386)
The lengths of the links for the mechanisms are collected in Table 7=1.
7. Steering Dynamics
467
Table 7=1 - Link numbers, and the input-output angles
for the multi-link steering mechanism
Left linkage
Link
Length
angle
1
g1 = 1=2
180
2
d1 = 0=22@ cos 54=6 = 0=37978
2
{
3
e1 = 1=2 0=22 tan 54=6 2
3
{
=
0=89043
2
p
4
f1 = 0=222 + {2 @4
4
Right linkage
Length
angle
g
=
1=2
180
p2
{
2
2
d2 = 0=22 + { @4
*2 = 4 2 tan1 0=44
{
e2 = 0=89043 2
*3
f2 = 0=22@ cos 54=6 = 0=37978
*4
Link
1
4
5
6
Equation (6.1) that is repeated below, provides the angle 4 as a function
of 2 .
Ã
!
p
2 4D F
±
E
E
1
1
1
1
4 = 2 tan1
(7.387)
2D1
D1
F1
= M3 M1 + (1 M2 ) cos 2
= M1 + M3 (1 + M2 ) cos 2
M11
=
M14
=
g1
d1
g1
e1
E1 = 2 sin 2
(7.388)
g1
d2 e21 + f21 + g21
M13 = 1
f1
2d1 f1
2
2
2
2
f g1 d1 e1
M15 = 1
2d1 e1
M12 =
(7.389)
The same equation (7.387) can be used to connect the input-output angles
of the right four-bar linkage.
Ã
!
p
2 4D F
±
E
E
2
2
2
2
(7.390)
*4 = 2 tan1
2D2
D2
F2
= M23 M21 + (1 M22 ) cos *2
= M21 + M23 (1 + M22 ) cos *2
M21
=
M24
=
g2
d2
g2
e2
E2 = 2 sin *2
g2
d2 e22 + f22 + g22
M23 = 2
f2
2d2 f2
f22 g22 d22 e22
M25 =
2d2 e2
(7.391)
M22 =
(7.392)
468
7. Steering Dynamics
G Ac [deg]
w/l=0.5
x=0.0
Muti-link
G2 [deg]
Ackerman
-15
-10
-5
0
G1
5
10
15
G fl [deg]
FIGURE 7.66. Steer angles 2 and Df versus 1 .
Starting with a guess value for {, we are able to calculate the length of the
links. Using Equations (7.387) and (7.390), along with (7.385) and (7.386),
we calculate 2 for a given value of 1 .
Let us start with { = 0, then
d1
d2
= 0=37978 m
e1 = =89043 m
= 0=22 m
e2 = 0=89043 m
f1 = 0=22 m
f2 = 0=37978 m
(7.393)
(7.394)
Using Equations (7.385) and (7.387), we may calculate the output of the
rst four-bar linkage, 4 , for a range of the left steer angle 15 deg ? 1 ?
15 deg. The following constraint, provides the numerical values for *2 to be
used as the input of the right four-bar linkage.
*2 = 4 2 tan1
{
0=44
(7.395)
Then, using Equations (7.390) and (7.387), we can calculate the steer angle
2 for the right wheel.
Figure 7.66 depicts the numerical values of the steer angles 2 and Df
versus 1 . The angle Df is the steer angle of the right wheel based on the
Ackerman equation (7.379).
Having 2 and Df , we calculate the dierence = 2 Df
(7.396)
for q dierent values of 1 in the working range angle 15 deg ? 1 ?
15 deg. Based on the q numbers for , we may nd the error h.
r
2
(7.397)
h=
q
7. Steering Dynamics
469
e [deg]
0.5
w/l=0.5
Muti-link
0.4
0.3
0.2
Ackerman
0.1
-0.9 -0.84 -0.86 -0.84 -0.82 -0.8 -0.78 -0.76 -0.74 -0.72 -0.7
x [m]
FIGURE 7.67. Illustration of the error function h = h({).
Changing the value of { and recalculating h evaluates the function h = h({).
Figure 7.67 illustrates the result of the calculation. It shows that the error
is minimum for { 0=824 m, which is the best length for the base of the
triangle S EF.
The behavior of the multi-link steering mechanism for dierent values
of {, is shown in Figure 7.68. The Ackerman condition is also plotted to
compare with the optimal multi-link mechanism. The optimality of { =
0=824 m may be clearer in Figure 7.69 that shows the dierence =
2 Df for dierent values of {. However, because the calculation has been
done for the range 15 deg ? 1 ? 15 deg, there is no guarantee that the
mechanism exists or is optimal out of this range.
The optimal multi-link steering mechanism along with the length of its
links is shown in Figure 7.70. The mechanism and the meaning of negative
value for { are shown in Figure 7.71 where the mechanism is in a positive
turning position.
7.8 F Trailer-Truck Kinematics
Consider a car pulling a one-axle trailer, as shown in Figure 7.72. We
may normalize the dimensions such that the length of the trailer is 1. The
positions of the car at the hinge point and the trailer at the center of its
axle are shown by vectors r and s.
Assuming r is a given dierentiable function of time w, we would like to
examine the steering behavior of the trailer by calculating s, and predict
jackkning. When the car is moving forward, we say the car and trailer are
470
7. Steering Dynamics
G Ac [deg]
x=-0.5
x=-0.2
w/l=0.5
20
10
Ackerman
G 2 deg] 0
x=-1.4
x=-1.1
-10
x=-0.824
Optimal multi-link
-20
-30
-15
-10
-5
0
G1
5
10
15
G fl [deg]
FIGURE 7.68. The behavior of the multi-link steering mechanism for dierent
values of {.
' G2 G Ac
10
x=-0.824
Optimal multi-link
x=-0.2
x=0.1
x=0.4
5
x=-0.5
0
5
-10
w/l=0.5
-15
-15
-10
-5
0
5
10
15
G1 G fl [deg]
FIGURE 7.69. Illustration of the dierence { = 2 3 Df for dierent values of
{.
7. Steering Dynamics
471
54.6q
0.824
1.302
1.302
0.22 m
0.3
79
0.4
67
d=1.2 m
FIGURE 7.70. The optimal multi-link steering mechanism along with the length
of its links.
FIGURE 7.71. The optimal multi-link steering mechanism in a positive turning
position.
.
r
y
1
T
s
r
x
FIGURE 7.72. A car pulling a one-axle trailer.
472
7. Steering Dynamics
y
.
r
z
s
r
x
FIGURE 7.73. A jackknifed conguration of a car pulling a one-axle trailer.
jackknifed if
rb · z ? 0
(7.398)
z=rs
(7.399)
where
A jackknifed conguration is shown in Figure 7.73, while Figure 7.72 is
showing an unjackknifed conguration.
Mathematically, we want to know if the truck-trailer will jackknife for a
given path of motion r = r(w) and what conditions we must impose on r(w)
to prevent jackkning.
The velocity of the trailer can be expressed by
sb = f (r s)
(7.400)
f = rb · z
(7.401)
where
and the unjackkning condition is
fA0
(7.402)
Assume the twice continuously dierentiable function r is the path of the
car motion. If |z| = 1, and r has a radius of curvature U(w) A 1, and
rb (0) · z(0) A 0
(7.403)
rb (w) · z(w) A 0
(7.404)
then
for all w A 0.
7. Steering Dynamics
473
Therefore, if the car is moving forward and the car-trailer combination
is not originally jackknifed, then it will remain unjackknifed.
Proof. Let us set the normalized constant trailer length as 1 to have z as
a unit vector
|z| = |r s| = 1
(7.405)
and
(r s) · (r s) = 1
(7.406)
The nonslip condition of the wheels of the trailer constrains the trailer
velocity vector sb to be directed along the trailer axis z and therefore,
sb = f (r s) = fz
(7.407)
Dierentiating (7.406) yields
2 (br sb ) · (r s) = 0
rb · (r s) = sb · (r s)
(7.408)
(7.409)
rb · (r s) = f (r s) · (r s)
f = rb · (r s) = rb · z
(7.410)
(7.411)
and therefore,
Having f enables us to express the trailer velocity vector sb of Equation
(7.407) as
sb = [br · (r s)] (r s) = (br · z) z
(7.412)
There are three situations
1. When f A 0, the velocity vector of the trailer sb is along the trailer
axis, z. The trailer follows the car and the system is stable.
2. When f = 0, the velocity vector of the trailer sb is zero. In this case,
the trailer spins about the center of its axle and the system is neutralstable.
3. When f ? 0, the velocity vector of the trailer sb is along the negative
trailer axis, z. The trailer does not follow the car and the system is
unstable.
Using a Cartesian coordinate expression, we may show the car and trailer
position vectors by
¸
{f
r =
(7.413)
|f
¸
{w
s =
(7.414)
|w
474
7. Steering Dynamics
and therefore,
sb =
=
{b w
|bw
¸
= [br · (r s)] (r s)
{b f ({f {w )2 + ({f {w ) (|f |w ) |b f
{b f ({f {w ) (|f |w ) + (|f |w )2 |bf
f = ({f {w ) {b f + (|f |w ) |b f
= {b f {f + |bf |f ({b f {w + |b f |w )
¸
(7.415)
(7.416)
Let us dene a function i (w) = rb · z and assume that conclusion (7.404)
is wrong while assumption (7.403) is correct. Then there exists a time
w1 A 0 such that i (w1 ) = 0 and i 0 (w1 ) 0. Using |z| = 1 and rb 6= 0, we
have rb (w1 ) · z(w1 ) = 0 and therefore, rb (w1 ) is perpendicular to z(w1 ). The
derivative i 0 (w) would be
i 0 (w) = r̈ · z + rb · zb = r̈ · z + rb · (br sb )
= r̈ · z + |br|2 rb · sb = r̈ · z + |br|2 rb · ((br · z) z)
= r̈ · z + |br|2 (br · z)2 = r̈ · z + |br|2 i 2 (w)
and therefore,
2
i 0 (w1 ) = r̈ · z + |br|
(7.417)
(7.418)
The acceleration r̈ in a normal-tangential coordinate frame (x̂q > x̂w ) is
r̈ =
=
g |br|
2
x̂w + |br| x̂q
gw
1
U
(7.419)
(7.420)
where x̂q and x̂w are the unit normal and tangential vectors. x̂w is parallel
to rb (w1 ), and x̂q is parallel to z(w1 ). Hence,
r̈ · z = ±(w1 ) |br(w1 )|2
(7.421)
and
i 0 (w1 ) = |br(w1 )|2 ± (w1 ) |br(w1 )|2 = [1 ± (w1 )] |br(w1 )|2
(7.422)
where is the curvature of the path. Because (w1 ) = 1@U(w) A 0, we
conclude that i 0 (w1 ) A 0, and it is not possible to have i 0 (w1 ) 0.
Example 324 F Straight motion of the car with constant velocity.
Consider a car moving forward in a straight line with a constant velocity.
We may use a normalization and set the speed of the car as 1 moving in
positive { direction starting from { = 0. Using a two-dimensional vector
expression we have
¸ ¸
{f
w
r=
=
(7.423)
0
|f
7. Steering Dynamics
475
y
z
.
r
T
x
FIGURE 7.74. The initial position of a one-axle trailer pulled by a car moving
forward in a straight line with a constant velocity is a circle about the hinge
point.
Because of (7.405), we have
z(0) = r(0) s(0) = s(0)
(7.424)
and therefore, the initial position of the trailer must lie on a unit circle as
shown in Figure 7.74.
Using two dimensional vectors, we may express z(0) as a function of ¸ ¸
cos {w (0)
=
(7.425)
z(0) = s(0) =
sin |w (0)
and simplify Equation (7.415) as
¸
¸ {b w
(w {w )2
=
sb =
|b w
|w (w {w )
(7.426)
Equation (7.426) is a set of two coupled rst-order ordinary dierential
equations with the solution
5
6
h2w F1
¸
w + 2w
9
h
+ F1 :
{w
:
=9
s=
(7.427)
7
8
|w
F2 hw
h2w + F1
Applying the initial conditions (7.425) yields
F1 =
cos 1
cos + 1
F2 =
2 sin cos + 1
(7.428)
476
7. Steering Dynamics
If 6= n, then the solution depends on time, and when time goes to
innity, the solution leads to the following limits asymptotically:
lim {w = w 1
w$4
lim |w = 0
(7.429)
w$4
When the car is moving with a constant velocity, this solution shows that
the trailer will approach the position of straight forward moving, following
the car.
We may also consider that the car is reversing, the solution shows that,
except for the unstable initial condition = , all solutions eventually
approach the jackknifed conguration.
If = 0, then
F1 = 0
F2 = 0
{w = w + 1
|w = 0
(7.430)
and the trailer moves in an unstable conguration. Any deviation from
= 0 leads to the stable limit solution (7.429).
If = , then
F1 = 4
F2 = 4
{w = w 1
|w = 0
(7.431)
and the trailer follows the car in an stable conguration. Any deviation
from = 0 will disappear after a while.
Example 325 F Straight car motion with dierent initial .
Consider a car moving on the {-axis with constant speed. The car is
pulling a trailer, which is initially at 0 as shown in Figure 7.74. Using a
normalized length, we assume the distance between the center of the trailer
axle and the hinge is the length of trailer, and is equal to 1.
¤W
£
and
If we show the global position of the car at hinge by r = {f |f
¤W
£
, then the position of
the global position of the trailer by s = {w |w
the trailer is a function of the car’s motion. When the position of the car
is given by a time-dependent vector function
¸
{f (w)
(7.432)
r=
|f (w)
the trailer position can be found by solving two coupled dierential equations.
{b w
2
= ({f {w ) {b f + ({f {w ) (|f |w ) |b f
(7.433)
2
= ({f {w ) (|f |w ) {b f + (|f |w ) |b f
(7.434)
£
¤W
, Equations (7.433) and
For a constantly uniform car motion r = w 0
(7.434) reduce to
|bw
{b w
|b w
= (w {w )2
= |w (w {w )
(7.435)
(7.436)
7. Steering Dynamics
Normalized displacement
2.0
477
45 deg
T0
1.5
xt
xc
1.0
0.5
yt
0.0
-0.5
-1.0
0
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.6
3.0
time
FIGURE 7.75. Trailer kinematics for a = 45 deg starting position.
The rst equation (7.435) is independent of the second equation (7.436)
and can be solved independently.
{w =
h2w F1
F1 h2w (w 1) w 1
= w + 2w
2w
F1 h 1
h
+ F1
(7.437)
Substituting Equation (7.437) in (7.436) generates an independent dierential equation for |w :
h2w F1
|b w = 2w
|w
(7.438)
h
+ F1
with the solution
|w =
F2 hw
2w
h
+ F1
(7.439)
£
¤W
, the constants of integral
When the trailer starts from s = cos sin would be equal to Equations (7.428) and therefore,
h2w (cos + 1) cos + 1
h2w (cos + 1) + cos 1
2hw sin h2w (cos + 1) + cos 1
{w
= w+
(7.440)
|w
=
(7.441)
Figures 7.75, 7.76, and 7.77 illustrate the behavior of the trailer starting
from = 45 deg, = 90 deg, = 135 deg.
Example 326 F Circular motion of the car with constant velocity.
Consider a car pulling a trailer as Figure 7.72 shows. The car is traveling
along a circle of radius U A 1, based on a normalized length in which the
length of the trailer is 1. In a circular motion with a normalized angular
478
7. Steering Dynamics
Normalized displacement
2.0
T0
1.5
90 deg
xc
1.0
xt
0.5
yt
0.0
-0.5
-1.0
0
0.3
0.6
0.9
1.2
1.5
time
1.8
2.1
2.4
2.6
3.0
FIGURE 7.76. Trailer kinematics for a = 90 deg starting position.
Normalized displacement
2.0
T0
1.5
135 deg
xc
1.0
xt
0.5
yt
0.0
-0.5
-1.0
0
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.6
3.0
time
FIGURE 7.77. Trailer kinematics for a = 135 deg starting position.
7. Steering Dynamics
479
.
r
y
R
r
s
T
x
1
FIGURE 7.78. Steady state conguration of a car-trailer combination.
velocity $ = 1 and period W = 2, the position of the car is given by a
time-dependent vector function:
¸ ¸
U cos(w)
{f (w)
=
(7.442)
r=
U sin(w)
|f (w)
The initial position of the trailer must lie on a unit circle with a center at
£
¤W
r(0) = {f (0) |f (0)
.
¸ ¸ ¸
{f (0)
cos {w (0)
=
+
(7.443)
s(0) =
sin |w (0)
|f (0)
The car-trailer combination approaches a steady-state conguration as shown
in Figure 7.78.
Substituting
¸
U sin(w)
rb =
(7.444)
U cos(w)
and the initial conditions (7.443) in (7.415) will generate two dierential
equations for the trailer position.
{b w
|b w
= U (U cos w {w ) ({w sin w |w cos w)
= U (U sin w |w ) ({w sin w |w cos w)
(7.445)
(7.446)
480
7. Steering Dynamics
Assuming r(0) =
are
£
0 0
¤W
{w
|w
the steady-state solutions of these equations
= Uf cos(w )
= Uf sin(w )
(7.447)
(7.448)
where Uf is the trailer’s radius of rotation, and is the angular position
of the trailer behind the car.
p
1
Uf
Uf = U2 1
sin =
cos =
(7.449)
U
U
We may check the solution by employing two new variables, x and y,
such that
x = {w sin w |w cos w
y = {w cos w + |w sin w
(7.450)
(7.451)
xb = y
(7.452)
and
Using the new variables we nd
{w
|w
{b w
|bw
= x sin w + y cos w
= x cos w + y sin w
(7.453)
(7.454)
= Ux(U cos w x sin w y cos w)
= Ux(U sin w + x cos(w) y sin w)
(7.455)
(7.456)
Direct dierentiating from (7.450), (7.451), (7.453), and (7.454) shows
that
xb = {w cos w + |w sin w + {b w (sin w) |b w (cos w)
yb = {w sin w + |w cos w + {b w cos w + |bw sin w
{b w
|bw
= xb sin w + yb cos w + x cos w y sin w
= xb cos w yb sin w x sin w y cos w
(7.457)
(7.458)
(7.459)
(7.460)
and therefore, the problem can be expressed in a new set of equations.
xb = y Ux2
¡
¢
yb = x U2 Uy 1
(7.461)
(7.462)
y Ux2 = 0
¡ 2
¢
x U Uy 1 = 0
(7.463)
(7.464)
In the steady-state condition the time dierentials must be zero, and
therefore, the steady-state solutions would be the answers to the following
algebraic equations:
7. Steering Dynamics
481
There are three sets of solutions.
{x = 0> y = 0}
Uf
U2
>y = f}
{x =
U
U
U2
Uf
{x = > y = f }
U
U
(7.465)
(7.466)
(7.467)
The rst solution is associated with s = 0,
|w = 0
{w = 0
(7.468)
which shows that the center of the trailer’s axle remains at the origin while
the car is turning on a circle U = 1. This is a stable motion.
The second solution is associated with
{w =
Uf
U2
sin w + f cos w
U
U
|w = Uf
U2
cos w + f sin w
U
U
(7.469)
which are equivalent to (7.447) and (7.448).
To examine the stability of the second solution, we may substitute a perturbed solution
Uf
U2
x=
+s
y = f +t
(7.470)
U
U
in the linearized equations of motion (7.461) and (7.462) at the second set
of solutions
xb = y 2Uf x
yb = Uf y
(7.471)
Therefore, we derive two equations for the perturbed functions s and t.
sb = t 2Uf s
tb = Uf t
The set of linear perturbed equations can be set in a matrix form
¸ ¸ ¸
sb
2Uf
s
1
=
tb
t
0
Uf
(7.472)
(7.473)
The stability of Equation (7.473) is determined by the eigenvalues l of the
coe!cient matrix, which are
1 = Uf
2 = 2Uf
(7.474)
Because both eigenvalues 1 and 2 are negative, the solution of the perturbed equations symptomatically goes to zero. Therefore, the second set of
solutions (7.466) is stable and it absorbs any near path that starts close to
it.
The third solution is associated with
{w = Uf
U2
sin w + f cos w
U
U
|w =
Uf
U2
cos w + f sin w
U
U
(7.475)
482
7. Steering Dynamics
The linearized equations of motion at the third set of solutions (7.467) are
xb = y + 2Uf x
yb = Uf y
(7.476)
The perturbed equations would then be
¸ ¸
¸ s
sb
2Uf 1
=
t
0
Uf
tb
(7.477)
with two positive eigenvalues
1 = Uf
2 = 2Uf
(7.478)
Positive eigenvalues show that the solution of the perturbed equations diverges and goes to innity. Therefore, the third set of solutions (7.467) is
unstable and repels any near path that starts close to it.
Example 327 F Slalom maneuver.
To analyze the behavior of a trailer, we need to have the motion of the
car as a given function of time
¸
{f (w)
r=
(7.479)
|f (w)
and substitute the function into equations
{b w
|bw
2
= ({f {w ) {b f + ({f {w ) (|f |w ) |b f
2
= ({f {w ) (|f |w ) {b f + (|f |w ) |b f
and solve the equations to determine the the trailer position
¸
{w
s=
|w
(7.480)
(7.481)
(7.482)
Slalom is a usual test to examine the handling behavior of vehicles. Let us
assume that a car with a trailer is moving on a steady state sinusoidal path
about the {-axis
¸
yw
r=
(7.483)
sin w
The behavior of the car should be found by solving two coupled dierential
equation
{b w
|b w
2
= (yw {w ) y + (yw {w ) (sin w |w ) cos w
= (yw {w ) ((yw {w ) y + (sin w |w ) cos w)
(7.484)
2
= (yw {w ) (sin w |w ) y + (sin w |w ) cos w
= (sin w |w ) ((yw {w ) y + (sin w |w ) cos w)
(7.485)
7. Steering Dynamics
483
In steady state condition, we have
{b w = 0
|bw = 0
(7.486)
and therefore the possible solutions for the trailer motion is a similar motion
to the car:
¸
yw
s=
(7.487)
sin w
7.9 Summary
Steering is required to guide a vehicle to a desired direction. When a vehicle turns, the wheels closer to the center of rotation are called the inner
wheels, and the wheels further from the center of rotation are called the
outer wheels. If the speed of a vehicle is very slow, there is a kinematic condition between the inner and outer steerable wheels, called the Ackerman
condition.
Street cars are four-wheel vehicles and usually have front-wheel-steering.
The kinematic condition between the inner and outer steered wheels is
cot r cot l =
z
o
(7.488)
where l is the steer angle of the inner wheel, r is the steer angle of the
outer wheel, z is the track, and o is the wheelbase of the vehicle. Track
z and wheel base o are considered the kinematic width and length of the
vehicle.
The mass center of a steered vehicle will turn on a circle with radius U,
q
U = d22 + o2 cot2 (7.489)
where is the cot-average of the inner and outer steer angles.
cot =
cot r + cot l
2
(7.490)
The angle is the equivalent steer angle of a bicycle having the same
wheelbase o and radius of rotation U.
484
7. Steering Dynamics
7.10 Key Symbols
4Z V
d> e> f> g
dl
D> E> F
DZ V
e1
e2
f
f1
f2
fv
F
F1 > F2 > · · ·
g
h
h
IZV
j
M
o
ov
q
R
s
t
u
r
U
U1
U1
Uf
Uw
Uz
U
UZ V
s
w
x
xU
x̂
y {>
b v
four-wheel-steering
lengths of the links of a four-bar linkage
distance of the axle number l from the mass center
input angle parameters of a four-bar linkage
all-wheel-steering
distance of the hinge point from rear axle
distance of trailer axle from the hinge point
stability index of a trailer motion
longitudinal distance of turn center and front axle of a 4ZV car
longitudinal distance of turn center and rear axle of a 4ZV car
4Z V factor
mass center, curvature center
constants of integration
arm length in trapezoidal steering mechanism
error
length of the oset arm
front-wheel-steering
overhang distance
link parameters of a four-bar linkage
wheelbase
steering length
number of increments
center of rotation in a turn, curvature center
perturbation in x
perturbation in y
yaw velocity of a turning vehicle
position vector of a car at the hinge
radius of rotation at mass center
radius of rotation at the center of the rear axle for I Z V
horizontal distance of R and the center of axles
trailer’s radius of rotation
radius of rotation at the center of the trailer axle
radius of the rear wheel
curvature radius
rear wheel steering
position vector of a trailer at the axle center
time
temporary variable in car-trailer analysis
steering rack translation
unit vector
vehicle velocity, temporary variable in car-trailer analysis
7. Steering Dynamics
485
yul
yur
z
zi
zu
{> |> }> x
z=rs
speed of the inner rear wheel
speed of the outer rear wheel
track
front track
rear track
displacement
position vector of a trailer relative to the car
1 = i o
2 = i u
Df
i o
i u
l
uo
uu
r
V
= 2 Df
$
$ l = $ ul
$ r = $ ur
arm angle in trapezoidal steering mechanism
cot-average of the inner and outer steer angles
front left wheel steer angle
front right wheel steer angle
steer angle based on Ackerman condition
front left wheel steer angle
front right wheel steer angle
inner wheel
rear left wheel steer angle
rear right wheel steer angle
outer wheel
steer command
steer angle dierence
angle between trailer and vehicle longitudinal axes
curvature of a road
eigenvalue
angular velocity
angular velocity of the rear inner wheel
angular velocity of the rear outer wheel
486
7. Steering Dynamics
Exercises
1. Bicycle model and radius of rotation.
Mercedes-Benz JO450W P has the following dimensions.
o = 121=1 in
zi = 65=0 in
zu = 65=1 in
U = 39=7 ft
Assume d1 = d2 and use an average track to determine the maximum
steer angle for a bicycle model of the car.
2. Radius of rotation.
Consider a two-axle truck that is oered in dierent wheelbases.
o = 109 in
o = 132=5 in
o = 150=0 in
o = 176=0 in
If the front track of the vehicles is
z = 70 in
and d1 = d2 , calculate the radius of rotations for = 30 deg.
3. Required space.
Consider a two-axle vehicle with the following dimensions.
o = 4m
z = 1=3 m
j = 1=2 m
Determine Uplq , UPd{ , and U for = 30 deg.
4. Rear wheel steering lift truck.
A battery powered lift truck has the following dimensions.
o = 55 in
z = 30 in
Calculate the radius of rotations if = 55 deg and d1 = d2 .
5. Wheel angular velocity.
Consider a two-axle vehicle with
o = 2=7 m
z = 1=36 m
What is the angular velocity ratio of $ r @$ l as a function of the steer
angle ?
6. A three-axle vehicle.
A three-axle vehicle is shown in Figure 7.21. Find the relationship
between 2 and 3 , and also between 1 and 6 .
7. Steering Dynamics
487
7. A three-axle truck.
Consider a three-axle truck that has only one steerable axle in front.
The dimensions of the truck are
d1
z
d2 = 300 mm
= 5300 mm
= 1800 mm
d3 = 1500 mm
Determine maximum steer angles of the front wheels if the truck is
supposed to be able to turn with U = 11 m.
8. F Three-wheel vehicle steering condition.
Figure 7.79 illustrates a three-wheel vehicle with all wheels steerable.
Determine the kinematic steering condition.
Gf
A
a1
Gor
Gir
l
C
R
c1
a2
D
c2
Gf
O
C
Gor
Gir
R1
wr
G
E
F
H
FIGURE 7.79. A three-wheel vehicle with all wheels steerable.
(a) Determine the kinematic steering condition between the three
wheels.
(b) Determine U, f1 , f2 , and U1 .
(c) Determine the required condition on the kinematic steering of
I Z V vehicles to be able to simplify them to the three-wheel
steering condition.
488
7. Steering Dynamics
9. Optimization of trapezoidal steering mechanism.
Use the ideal kinematic steering condition
cot Dr cot l =
z
o
and the equation of the trapezoidal steering mechanism
2
(z 2g sin )
2
= (z g sin ( Gr ) g sin ( + l ))
+ (g cos ( Gr ) g cos ( + l ))2
to dene the function h () = Gr Dr . For a set of given z> o> g> l
determine the minimum of h using derivative method. Is there any
solution? Is the answer (if any) unique?
10. A vehicle with a one-axle trailer.
Determine the steady state angle between the trailer and vehicle with
the following dimensions.
d1
e1
zw
j
=
=
=
=
1000 mm
1200 mm
1100 mm
800 mm
d2 = 1300 mm
e2 = 1800 mm
zy = 1500 mm
l = 12 deg
What is the rotation radius of the trailer Uw , and the vehicle U?
Determine minimum radius Umin , maximum UPd{ , and dierence
radius U?
11. Best g for trapezoidal steering mechanism.
A trapezoidal steering mechanism is shown in Figure 7.80.Assume
R
d
E
w
E
FIGURE 7.80. A trapezoidal steering mechanism.
7. Steering Dynamics
z = 1=66 m
o = 2=93 m
489
19=5 deg
and determine the best g by minimization of the UP V error function
h
v
u q
u1 X
( Gr Dr )2
h=t
q l=1
where q is the number of subdivision of l in the working range
30 l 30.
12. F Coordinates of the turning center.
Determine the coordinates of the turning center for the vehicle in
Exercise 16 if i o = 5 deg and f1 = 1300 mm.
13. F Steer angle for given curvature center.
Consider a car with
o = 2=8 m
zi = 1=35 m
zu = 1=4 m
d1 = d2
and the coordrinates of the curvature center in body coordinate frame
{R = 60=336
|R = 118=3
Determine the required steer angles of a 4Z V vehicle.
14. F Motion of the turning center.
Consider a stationary 4Z V vehicle as is shown in Figure 7.81.
Gi
wf
Go
Center of
rotation
a1
R
O
T
l
C
a2
l
Gi
wr
Go
FIGURE 7.81. A stationary 4Z V vehicle with a moving turning center.
490
7. Steering Dynamics
Assume the inner front and rear steer angles are equal at the moment
and
U = 3o
(a) Determine the turning center of rotation in the vehicle body
coordinate frame.
(b) Determine the steer angles of the wheels at the moment.
(c) The kinematic turning center of the vehicle is moving on a circle
with radius equal to the wheelbase of the car, o. Determine the
steer angles as functions of .
15. Local curvature center.
Calculate the coordinates of the curvature center in the body coordinate frame of a car that is moving tangent to the road in Example
304.
16. F Turning radius of a 4Z V vehicle.
Consider a I Z V vehicle with the following dimensions.
o
= 2300 mm
zi
= 1457 mm
d1
38
=
d2
62
zu = 1607 mm
Determine the turning radius of the vehicle for i o = 5 deg.
What should be the steer angles of the rear wheels to decrease 10%
of the turning radius, if we make the vehicle 4Z V?
17. F Dierent front and rear tracks.
Lotus 2-ElevenW P is a UZ G sportscar with the following specications.
o = 2300 mm
Front tire = 195@50U16
I}1
38
=
I}2
62
zu = 1607 mm
zi = 1457 mm
Rear tire = 225@45U17
Determine the angular velocity ratio of $ r @$ l , U, l , and r for =
5 deg.
18. F Coordinates of turning center.
Determine the coordinates of turning center of a 4Z V vehicle in terms
of outer steer angles ri and ru .
7. Steering Dynamics
491
19. F Turning radius.
Determine the turning radius of a 4Z V vehicle in terms of u .
1
(cot lu + cot ru )
2
cot u =
20. Curvature radius.
Consider a road of motion as
\ = sin
2
[
100
(a) Determine curvature of the road as a function of [.
(b) Determine radius curvature of the road as a function of [.
(c) Determine the path of the curvature center.
21. F A three-axle car.
Consider a three-axle o-road pick-up car. Assume
d1
z
= 1100 mm
= 1457 mm
d2 = 1240 mm
d3 = 1500 mm
and determine r , U1 , Ui , and U if l = 10 deg.
22. F A 3G road.
Consider a spatial road with the parametric equations
{ = (d + e sin ) cos | = (d + e sin ) sin } = e + e cos (a) Show that the curvature of the road as a function of the parameter is
2
= s
4d2 + 6e2 2e2 cos 2 + 8de sin (b) Determine the path of the road curvature center for d = 200,
e = 150.
23. F Steering mechanism optimization.
Find the optimum length { for the multi-link steering mechanism
shown in Figure 7.82 to operate as close as possible to the kinematic
steering condition. The vehicle has a track z = 2=64 m and a wheelbase o = 3=84 m, and must be able to turn the front wheels within a
working range equal to 22 deg 22 deg. Use d1 = 380 mm.
492
7. Steering Dynamics
54.6q
b
R
c
0.24 m
A
M
P
c
B
x
D
C
b
a
N
d=1.32 m
FIGURE 7.82. A multi-link steering mechanism tha must be optimized by varying
{.
24. A car on an ellipse.
Consider a 4Z V vehicle that is moving on an elliptic path.
[ = d cos \ = e sin (a) Determine the global coordinates of the curvature center.
(b) Determine the coordinates of the curvature center in the vehicle
body coordinate.
(c) Determine the steer angle of the wheels if the car is always moving tangent to the road.
25. F Road connection.
Design a clothoid connection road to connect the [-axis to the shown
line in Figure 7.83. How many design are possible?
26. F Curvature of a road.
The curvature of a planar road can be calculated by
¯
¯
¯
¯
¯
¯
{0 | 00 | 0 {00
g2 |@g{2
¯
¯
= ¯³
¯=
´
3@2
¯ ({02 + | 02 )3@2
¯
¯
¯ 1 + (g|@g{)2
(a) Consider a road with equation
[ (w) =
Z w
0
cos
μ 2¶
x
gx
2
\ (w) =
Z w
and verify that the curvature of the road is
=w
0
sin
μ 2¶
x
gx
2
7. Steering Dynamics
493
Y
100
80
60
40
20
0
X
40
80
120
160
200
FIGURE 7.83. A road connecting (0=100) to (200> 0).
(b) Consider a road with equation
μ 3¶
Z w
x
[ (w) =
gx
cos
3
0
μ 3¶
x
\ (w) =
gx
sin
3
0
Z w
and verify that the curvature of the road is
= w2
27. F A loop road.
Figure 7.84 illustrates two straight roads tangentially connected to
two circular paths.
(a) Design proper clothoid transitions to connect the straight roads
to the circular paths.
(b) Determine and plot the loci of the curvature center of the designed loop.
28. Steering mechanism design.
Figure 7.85 illustrates a steering mechanism. Suppose the middle box
can control the length of { and |.
(a) Determine { and | as functions of o> z> g> such that the mechanism follows the Ackerman kinematic condition.
(b) Plot {, |, { |, and {@| for a symmetric range of inner steering
angle l = ±35 deg if
z = 1=66 m
o = 2=93 m
g = 0=4
= 19=5 deg
494
7. Steering Dynamics
Y
100
80
60
40
20
X
0
40
80
120
160
200
FIGURE 7.84. A loop road by two straight paths tangentially connected to two
circular paths.
R
y
x
d
E
w
E
FIGURE 7.85. A steering mechanism with controlled length of { and |.
29. F Road design.
Assume we are asked to nd a clothoid transition road to connect a
straight road to a circle of radius U = 58=824 m. Assume the clothoid
begins from (0> 0) and determine the center of the circular path and
the steer angles of a 4Z V car as a function of v for
(a) d = 250 m
(b) d = 210 m
(c) d = 180 m
(d) d = 150 m
(e) d = 120 m
7. Steering Dynamics
495
30. Clothoid.
Show that
g[
g[
=d
gw
gv
g\
g\
=d
gw
gv
31. Figure 8 road.
The problem in example 317 is that the curvature of the road at the
ends of clothoids are not equal to the curvature of the connecting
circular paths. Find the curvatures at the connection points.
8
Suspension Mechanisms
The suspension is what links the wheels to the vehicle chassis and allows
relative motion. This chapter covers the suspension mechanisms, and discusses the possible relative motions between the wheel and the vehicle
chassis. The wheels, through the suspension linkage, must propel, steer,
and stop the vehicle, and support the associated forces.
8.1 Solid Axle Suspension
The simplest way to attach a pair of wheels to a chassis is to mount them
at two ends of a solid axle, such as the one that is shown in Figure 8.1.
FIGURE 8.1. A solid axle with leaf spring suspension.
The solid axle must be attached to the body such that an up and down
motion in the }-direction, as well as a roll rotation about the {-axis, are
possible. Therefore„ no forward and lateral translation, and also no rotation
about the axle and the }-axis, are allowed. There are many combinations of
links and springs that can provide the kinematic and dynamic requirements.
The simplest design is to clamp the axle to the middle of two leaf springs
with their ends tied and shackled to the chassis as shown schematically
in Figure 8.1. A side view of a multi-leaf spring and solid axle is shown in
Figure 8.2. A suspension with a solid axle between the left and right wheels
is called dependent suspension.
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_8, © Springer Science+Business Media New York 2014
497
498
8. Suspension Mechanisms
FIGURE 8.2. A side view of a multi-leaf spring and solid axle suspension.
The performance of a solid axle with leaf springs suspension can be
improved by adding a linkage to guide the axle kinematically and provide
dynamic support to carry the non }-direction forces.
The solid axle with leaf spring combination came to vehicle industry from
horse-drawn vehicles.
Example 328 Hotchkiss drive.
When a live solid axle is connected to the chassis with nothing but two
leaf springs, it is called the Hotchkiss drive, which is the name of the
car that used it rst. Figure 8.2 illustrates a Hotchkiss suspension. The
main problems of Hotchkiss drive are locating the axle under lateral and
longitudinal forces, and having a low mass ratio % = pv @px , where pv is
the sprung mass and px is the unsprung mass.
Sprung mass refers to all masses that are supported by the spring, such
as vehicle body. Unsprung mass refers to all masses that are attached to
and not supported by the spring, such as wheel, axle, or brakes.
Example 329 Leaf spring suspension and exibility problem.
The solid axle suspension systems with longitudinal leaf springs have
many drawbacks. The main problem lies in the fact that springs themselves
should act as locating members. Springs are supposed to ex under load, but
their exibility is needed in only one direction. However, it is the nature
of leaf springs to twist and bend laterally and hence, ex also and twist in
planes other than the tireplane. Leaf springs are not suited for taking up the
driving and braking traction forces. These forces tend to push the springs
into an V-shaped prole, as shown in Figure 8.3. The driving and braking exibility of leaf springs may generate a negative caster and increase
instability.
Long springs provide better ride. However, long sprigs exaggerate their
bending and twisting under dierent load conditions.
Example 330 Leaf spring suspension and exibility solution.
To reduce the eect of a horizontal force and V-shaped prole appearance
in a solid axle with leaf springs, the axle may be attached to the chassis by a
8. Suspension Mechanisms
499
(a) Acceleration
(b) Braking
FIGURE 8.3. A driving and braking trust, force leaf springs into an V shaped
prole.
longitudinal bar as Figure 8.4(d) shows. Such a bar is called an anti-tramp
bar, and the suspension is the simplest cure for longitudinal problems of a
Hotchkiss drive.
A solid axle with an anti-tramp bar may kinematically be approximated
by a four-bar linkage, as shown in Figure 8.4(e). Although an anti-tramp
bar may control the shape of the leaf spring, it introduces a twisting angle
problem when the axle is moving up and down, as shown in Figure 8.5.
Twisting the axle and the wheel about the axle is called caster.
The solid axle is frequently used to help keeping the wheels perpendicular
to the road.
Example 331 Leaf spring location problem.
The front wheels need room to steer left and right. Therefore, leaf springs
cannot be attached close to the wheel hubs, and must be placed closer to
the middle of the axle. That gives a narrow spring-base, which means that
a small side force can sway or tilt the body relative to the axle through a
considerable roll angle. This is uncomfortable for the vehicle passengers,
and may also produce unwanted steering.
The solid axle positively prevents the camber change by body roll. The
wheels remain almost upright and hence, do not roll on a side. However, a
solid axle shifts laterally from its static plane and its center does not remain
on the vehicle’s longitudinal axis under a lateral force.
A solid axle produces bump-camber when single-wheel bump occurs. If
500
8. Suspension Mechanisms
(a)
M
A
N
B
(b)
FIGURE 8.4. (d) Adding an anti-tramp bar to guide a solid axle. (e) Equivalent
kinematic model.
A
M
B
N
M
N
A
B
(a)
(b)
FIGURE 8.5. An anti-tramp bar introduces a twistng angle problem. (d) The
wheel moves up and (e) The wheel moves down.
8. Suspension Mechanisms
501
FIGURE 8.6. A solid axle suspension with a triangulated linkage.
the right wheel goes over a bump, the axle is raised at its right end, and
that tilts the left wheel hub, putting the left wheel at a camber angle for the
duration of de ection.
Example 332 Triangular linkage.
A triangulated linkage, as shown in Figure 8.6, may be attached to a solid
axle to provide lateral resistance in turning maneuvers, and twist resistance
during acceleration and braking.
Example 333 Panhard arm and lateral displacement.
High spring rate is a problem of leaf springs. Reducing their stiness by
narrowing them and using fewer leaves, reduces the lateral stiness and increases the directional stability of the suspension signicantly. A Panhard
arm is a bar that attaches a solid axle suspension to the chassis laterally.
Figure 8.7 illustrates a solid axle and a Panhard arm to limit the lateral displacement of the axle. Figure 8.8 shows a triangular linkage and a Panhard
arm combination for guiding a solid axle.
A double triangle mechanism, as shown in Figure 8.9, is an alternative
design to guide the axle and support it laterally.
Example 334 Straight line linkages.
There are many mechanisms that provide a straight line motion to be
used in suspension design. The simplest mechanisms are four-bar linkages
with a coupler point moving straight. Some of the most applied and famous
linkages are shown in Figure 8.10. By having proper lengths, the Watt,
Robert, Chebyshev, and Evance linkages can make the coupler point F move
on a straight line vertically. Such a mechanism and straight motion can be
used to guide a solid axle.
Two Watt suspension mechanisms with a Panhard arm are shown in
Figures 8.11 and 8.12.
502
8. Suspension Mechanisms
FIGURE 8.7. A solid axle and a Panhard arm to guide the axle.
FIGURE 8.8. A triangle mechanism and a Panhard arm to guide a solid axle.
8. Suspension Mechanisms
FIGURE 8.9. Double triangle suspension mechanism.
C
C
(a) – Watt linkage
(b) – Robert linkage
C
C
(b) – Chebyshev linkage
(d) – Evans linkage
FIGURE 8.10. Some linkages with straight line motion.
503
504
8. Suspension Mechanisms
FIGURE 8.11. A Watt suspension mechanisms with a Panhard arm.
FIGURE 8.12. A Watt suspension mechanisms with a Panhard arm.
8. Suspension Mechanisms
505
FIGURE 8.13. A Robert suspension mechanism with a Panhard arm.
FIGURE 8.14. A Robert suspension mechanism with a Panhard arm.
Figures 8.13, 8.14, and 8.15 illustrate three combinations of Robert suspension linkages equipped with a Panhard arm.
Example 335 Solid axle suspension and unsprung mass problem.
Solid axles are usually heavier than the same size independent suspensions. A solid axle is a part of unsprung members, and hence, the unsprung
mass is increased where using solid axle suspension. A heavy unsprung mass
ruins both, the ride and handling quality of a vehicle. Lightening the solid
axle makes it weaker and increases the most dangerous problem in vehicles:
axle breakage. The solid axle must be strong enough to make sure it will
not break under any loading conditions at any age. As a rough estimate,
90% of the leaf spring mass may also be counted as unsprung mass, which
makes the problem worse.
506
8. Suspension Mechanisms
FIGURE 8.15. A Robert suspension mechanism with a Panhard arm.
The unsprung mass problem is worse in front, and it is the main reason
that they are no longer being used in street cars. However, front solid axles
are still common on trucks and buses. These are heavy vehicles and solid
axle suspension does not reduce the mass ratio % = pv @px very much.
When a vehicle is rear-wheel-drive and a solid axle suspension is used
in the back, the suspension is called live axle. A live axle is a casing that
contains a dierential, and two drive shafts. The drive shafts are connected
to the wheel hubs. A live axle can be three to four times heavier than a dead
I-beam axle. It is called live axle because of rotating gears and shafts inside
the axle.
Example 336 Solid axle and coil spring.
To decrease the unsprung mass and increase vertical exibility of solid
axle suspensions, it is possible to equip them with coil springs instead of
leaf springs. A sample of a solid axle suspension with coil spring is shown
in Figure 8.16. The suspension mechanism is made of four longitudinal
bars between the axle and chassis. The springs may have some lateral or
longitudinal angle to introduce some lateral or longitudinal compliance as
well.
Example 337 De Dion axle.
When a solid axle is a dead axle with no driving wheels, the connecting
beam between the left and right wheels may have dierent shapes to do
dierent jobs, usually to give the wheels independent exibility. We may
also modify the shape of a live axle to attach the dierential to the chassis
and reduce the unsprung mass.
De Dion design is a modication of a beam axle that may be used as
a dead axle or to attach the dierential to the chassis and transfer the
driving power to the drive wheels by employing universal joints and split
shafts. Figure 8.17 illustrates a De Dion suspension.
8. Suspension Mechanisms
FIGURE 8.16. A solid axle suspension with coil springs.
FIGURE 8.17. A De Dion suspension.
507
508
8. Suspension Mechanisms
Upper A-arm
Steer arm
Kingpin
Lower A-arm
FIGURE 8.18. A double D-arm suspension.
8.2 Independent Suspension
Independent suspensions are introduced to allow a wheel to move up and
down without aecting the opposite wheel. There are many forms and designs of independent suspensions. However, double A-arm and McPherson
strut suspensions are the simplest and the most common designs. Figure 8.18 illustrates a sample of a double D-arm and Figure 8.19 shows a
McPherson suspension.
Kinematically, a double D-arm suspension mechanism is a four-bar linkage with the chassis as the ground link, and coupler as the wheel carrying
link. A McPherson suspension is an inverted slider mechanism that has the
chassis as the ground link and the coupler as the wheel carrying link. A
double D-arm and a McPherson suspension mechanism are schematically
shown in Figures 8.20 and 8.21 respectively.
Double D-arm, is also called double wishbone, or short/long arm suspension. McPherson also may be written as MacPherson.
Example 338 Double D-arm suspension and spring position.
Consider a double D-arm suspension mechanism. The coil spring may
be between the lower arm and the chassis, as shown in Figure 8.18. It is
also possible to install the spring between the upper arm and the chassis, or
between the upper and lower arms. In either case, the lower or the upper
arm, which supports the spring, is made stronger and the other arm acts
as a connecting arm.
Example 339 Multi-link suspension mechanism.
When the two side bars of an D-arm are attached to each other with a
joint, as shown in Figure 8.22, then the double D-arm is called a multi-link
mechanism. Such a multi-link suspension is a six-bar mechanism that may
have a better coupler motion than a double D-arm mechanism. However,
multi-link suspensions are more expensive, less reliable, and more compli-
8. Suspension Mechanisms
509
Coil spring
Shock absorber
Strut
Steer arm
Kingpin
Lower A-arm
FIGURE 8.19. A McPherson suspension.
FIGURE 8.20. A double D-arm suspension mechanism on the left and right
wheels.
510
8. Suspension Mechanisms
FIGURE 8.21. A McPherson suspension mechanism on the left and right wheels.
cated compare to a double D-arm four-bar linkage. There are vehicles with
more than six-link suspension with possibly better kinematic performance.
Example 340 Swing arm suspension.
An independent suspension may be as simple as a triangle shown in Figure 8.23. The base of the triangle is jointed to the chassis and the wheel
to the tip point. The base of the triangle is aligned with the longitudinal
axis of the vehicle. Such a suspension mechanism is called a swing axle
or swing arm.
Compared to the other suspension mechanisms, a swing arm suspension
has the maximum camber angle variation.
Example 341 Trailing arm suspension.
Figure 8.24 illustrates a trailing arm suspension that is a longitudinal
arm with a lateral axis of rotation. The camber angle of the wheel, supported
by a trailing arm, will not change during the up and down motion.
Trailing arm suspension has been successfully using in a variety of frontwheel-drive vehicles, to suspend their rear wheels.
Example 342 Semi-trailing arm
Semi-trailing arm suspension, as shown in Figure 8.25, is a compromise between the swing arm and trailing arm suspensions. The joint axis
may have any angle, however an angle not too far from 45 deg is more
applied. Such suspensions have acceptable camber angle change, while they
can handle both, the lateral and longitudinal forces. Semi-trailing design has
successfully applied to a series of rear-wheel-drive cars for several decades.
8. Suspension Mechanisms
FIGURE 8.22. A multi-link suspension mechanism.
FIGURE 8.23. A swing arm suspension.
511
512
8. Suspension Mechanisms
FIGURE 8.24. A trailing arm suspension.
FIGURE 8.25. A semi-trailing arm suspension.
8. Suspension Mechanisms
513
Antiroll bar
FIGURE 8.26. An antiroll bar attached to a solid axle with coil springs.
Example 343 Antiroll bar and roll stiness.
Coil springs are used in vehicles because they are less sti with better
ride comfort compared to leaf springs. The roll stiness of the vehicle with
coil springs is usually less than vehicles with leaf springs. To increase the
roll stiness of such suspensions, an antiroll bar should be used. Leaf
springs with reduced layers, uni-leaf, trapezoidal, or nonuniform thickness
may also need an antiroll bar to compensate their reduced roll stiness. The
antiroll bar is also called a stabilizer. Figure 8.26 illustrates an antiroll
bar attached to a solid axle with coil springs.
Example 344 Need for longitudinal compliance.
When a vehicle goes over a bump, the rst action is a displacement that
tends to push the wheel backward relative to the rest of the vehicle. So,
the lifting force has a longitudinal component, which will be felt inside the
vehicle even if the suspension system has some horizontal compliance.
There are situations in which the horizontal component of the force is
even higher than the vertical component. Leaf springs can somewhat absorb
this horizontal force by attening out and stretching the distance from the
forward spring anchor and the axle. Such a stretch is usually less than
1@2 in 1 cm.
8.3 Roll Center and Roll Axis
The roll axis is the instantaneous line about which the body of a vehicle
rolls. Roll axis is found by connecting the roll center of the front and rear
suspensions of the vehicle. Assume we cut a vehicle laterally to disconnect
the front and rear halves of the vehicle. Then, the roll center of the front
or rear suspension is the instantaneous center of rotation of the body with
respect to the ground.
514
8. Suspension Mechanisms
(a)
Wheel
5
4
Wheel
Body
2
8
6
3
7
1
(b)
FIGURE 8.27. (d) An example of a double A-arm front suspensions. (e) Kinematically equivalent mechanism for the front half of the double A-arm suspension.
Figure 8.27(d) illustrates a sample of suspension of a car with a double
D-arm mechanism. To nd the roll center of the body with respect to the
ground, we analyze the two-dimensional kinematically equivalent mechanism shown in Figure 8.27(e). The center of tireprint is the instant center of
rotation of the wheel with respect to the ground, so the wheels are jointed
links to the ground at their center of tireprints. In this gure, the relatively
moving bodies are numbered staring from 1 for the ground and ending with
number 8 for the chassis.
The instant center L18 is the roll center of the body with respect to the
ground. To nd L18 , we apply the Kennedy theorem and nd the intersection of the line L12 L28 and L13 L38 as shown in Figure 8.28.
The point L28 and L38 are the instant center of rotation for the wheels with
respect to the body. The instant center of rotation of a wheel with respect
to the body is called suspension roll center or simply suspension center. So,
to nd the roll center of the front or rear half of a car, we should determine
the suspension roll centers, and nd the intersection of the lines connecting
the suspension roll centers to the center of their associated tireprints.
The Kennedy theorem states that the instant center of every three relatively moving objects are colinear.
8. Suspension Mechanisms
Wheel
515
Wheel
5
4
I28
I38
Body
2
8
3
6
I12
7
I13
1
I18
FIGURE 8.28. The roll center L18 is at the intersection of lines L12 L28 and L13 L38 .
I13
I12
3
1
I14
I23
I34
4
2
I24
FIGURE 8.29. Instant center or rotation for an example of inverted slider crank
mechanism.
Example 345 McPherson suspension roll center.
A McPherson suspension is an inverted slider crank mechanism. The
instant centers of an example of inverted slider crank mechanism are shown
in Figure 8.29. The point L12 is the suspension roll center, which is the
instant center of rotation for the wheel link number 2 with respect to the
chassis link number 1.
A car with a McPherson suspension is shown in Figure 8.30(d). The
kinematic equivalent mechanism is depicted in Figure 8.30(e). Suspension
roll centers along with the body roll center are shown in Figure 8.31. To
nd the roll center of the front or rear half of a car, we determine each
suspension roll center and then nd the intersection of the lines connecting
the suspension roll centers to the center of the associated tireprint.
516
8. Suspension Mechanisms
(a)
Body
Wheel
4
5
2
Wheel
3
8
6
7
1
(b)
FIGURE 8.30. (d) McPherson suspension system. (e) Kinematics model of the
McPherson suspension.
Example 346 Roll center of double D-arm suspension.
The roll center of an independent suspension such as a double D-arm
can be internal or external. The kinematic model of a double D-arm suspension for the front left wheel of a car is illustrated in Figure 8.32. The
suspension roll center in Figure 8.32(d) is internal, and in Figure 8.32(e)
is external. An internal suspension roll center is toward the vehicle body,
while an external suspension roll center goes away from the vehicle body.
A suspension roll center may be on, above, or below the road surface, as
shown in Figure 8.33(d)-(f) for an external suspension roll center. When
the suspension roll center is on the ground, above the ground, or below the
ground, the vehicle roll center would be on the ground, below the ground,
and above the ground, respectively.
Example 347 Roll axis, roll height, roll torque.
Both the front and rear suspensions of a car have roll centers. They are
8. Suspension Mechanisms
Wheel
I38
517
Wheel
Body
4
I28
5
2
3
8
I38
I38
6
7
I18
I28
I28
1
FIGURE 8.31. Roll center of a car with a McPherson suspension system.
(a)
(b)
FIGURE 8.32. The kinematic model of the suspension of a front left wheel: (d)
an internal roll center, and (e) an external roll center.
518
8. Suspension Mechanisms
(a)
Suspension roll
center
(b)
Suspension roll
center
(c)
Suspension roll
center
FIGURE 8.33. A suspension roll center at (d) on, (e) above, and (f) below the
road surface.
in general at dierent heights. The line connecting the front and rear roll
centers is the roll axis of the car about which the car will roll. The vertical
distance of the mass center and the roll axis is called the roll height, ku .
Figure 8.34 illustrates the roll axis and roll height of a car. When the car is
turning in a curved path, the centripetal force i| = py 2 @U is the eective
lateral force at the mass center that generates a roll torque P{ about the
roll center.
py 2
ku
(8.1)
P{ =
U
It is common that the mass center locate above the roll axis. As a result
the car will roll outward of the turning path. However, we can technically
design the suspensions of a car such that the roll axis pass above the mass
center. Such a car will roll inward in a turning path and acts similar to a
boat. If the roll axis passes through the mass center, the car will not roll on
a curved path.
The roll torque, and hence, the roll angle of a car increases by the roll
height ku , and square of the speed y2 . Therefore, doubling the speed needs
to have four times shorter roll height for the same roll angle.
8. Suspension Mechanisms
519
z
a2
a1
g
C
x
Roll height hr
Roll axis
Rear roll center
Front roll center
FIGURE 8.34. The roll axis is the line connecting the front and rear suspension
roll centers.
z
b2
b1
g
C
M
h
y
hr
Roll center
2Fz2
2Fy2
2Fz1
2Fy1
mg
FIGURE 8.35. A car in a turning path about the }-axis.
Example 348 Vertical tireprint forces.
The roll torque makes the car body roll until it is balanced by the moment of the forces of the de ected springs. The resultant moment of the
spring forces for one radian or one degree roll of the body is called the roll
stiness n* .
(8.2)
P{ = n* *
Any roll angle causes a proportional de ection in the main springs of the
car and as a result decreases the vertical force of the tireprint of the inner
tires and increases the vertical force of the outer tires. Consider a car in a
turning path about the }-axis as is shown in Figure 8.35.
Assuming similar springs with stiness n and a horizontal roll axis, the
roll angle * provides us with
Pu = 2n (e1 + e2 ) *
(8.3)
520
8. Suspension Mechanisms
indicating that the roll stiness of the car is:
n* = 2n (e1 + e2 )
(8.4)
Combining with Equation (8.1), yields:
*=
py2
ku
py2 ku
=
U n*
U 2n (e1 + e2 )
(8.5)
To decrease the roll angle, we need to increase the roll stiness n* and
decrease the roll height ku . As the spring rates are generally designed based
on vertical ride comfort, adding anti-roll bar is a common approach to
increase n* if needed. To design the roll stiness of a car, the roll natural
frequency of the car must also be considered.
If the spring rate of the front and rear suspensions are dierent then, the
roll stiness of the car would be
n* = ni (e1 + e2 ) + nu (e1 + e2 )
(8.6)
where, nu is the stiness of the front suspension and nu is the stiness of
the rear suspension.
Assume that the wheel load on the left and right sides of the car are equal
at the rest position,
I}1
z
1
e2
pj
2
z
= e1 + e2
=
I}2 =
1
e1
pj
2
z
(8.7)
(8.8)
we may calculate the wheel loads for a rolled car as
I}1 =
1
e2
pj ne1 *
2
z
I}2 =
1
e1
pj + ne2 *
2
z
(8.9)
The lateral force of a tire is proportional to its vertical load. Therefore,
there is a critical roll angle *f at which the vertical force and hence, the
lateral force of the tire disappear. This is the rst step of instability and
roll over of a car in a turn. As a result, the roll angle must be limited to
have all tires on the road.
1 pj e2
1 pj e1
*
2 nz e2
2 nz e1
(8.10)
As a reasonable approximation, we may assume e1 e2 and calculate the
critical roll angle.
1 pj
*f ±
(8.11)
2 nz
8. Suspension Mechanisms
z
521
b1
y
b2
hr
C
h
I
Roll
2F y 2
2F z2
c
enter
2F y1
2Fz1
g
mg
FIGURE 8.36. A car with a roll angle * on a road with the bank angle !.
Consider a car is turning on a banked road such as shown in Figure 8.36.
If the bank angle is ! while the roll angle is *, the vertical force under the
each left and right tires are:
1 pj
(e2 cos ! k sin !) ne1 *
2 z
1 pj
(e1 cos ! + k sin !) + ne1 *
I}2 =
2 z
I}1 =
(8.12)
(8.13)
Example 349 F Camber variation of double D-arm suspension.
When a wheel moves up and down with respect to the vehicle body, depending on the suspension mechanism, the wheel may camber. Figure 8.37
illustrates the kinematic models a double D-arm suspension mechanism.
The mechanism is equivalent to a four-bar linkage with the ground link
as the vehicle chassis. The wheel is attached to a coupler point F of the
mechanism. We set a local suspension coordinate frame ({> |) with the {axis indicating the ground link P Q . The {-axis makes a constant angle 0
with the vertical direction. The suspension mechanism has a length d for
the upper D-arm, e for the coupler link, f for the lower D-arm, and g for
the ground link. The conguration of the suspension is determined by the
angles 2 , 3 , and 4 , all measured from the positive direction of the {-axis.
When the suspension is at its equilibrium position, the links of the double
D-arm suspension make initial angles 20 30 , and 40 with the {-axis. The
equilibrium position of a suspension is called the rest position.
To determine the camber angle during the uctuation of the wheel, we
should determine the variation of the coupler angle 3 , as a function of
vertical motion } of the coupler point F.
Using 2 as a parameter, we can nd the coordinates ({F > |F ) of the
522
8. Suspension Mechanisms
e
A
a
M
T2
d
z
C
b
z0
c
N
B
T4
T0
D
T3
x
y
FIGURE 8.37. The kinematic model for a double A-arm suspension mechanism.
coupler point F in the suspension coordinate frame ({> |) as
{F
|F
= d cos 2 + h cos (s t + )
= d sin 2 + h sin (s t + )
(8.14)
(8.15)
where,
d sin 2
g d cos 2
q
2
4e2 i 2 (e2 + i 2 f2 )
s = tan1
e2 + i 2 f2
p
i =
d2 + g2 2dg cos 2
t
= tan1
(8.16)
(8.17)
(8.18)
The position vector of the coupler point is uF
uF = {F ˆ~ + |F ˆ
(8.19)
and the unit vector in the tire displacement }-direction is
x̂} = cos 0ˆ~ sin 0 ˆ
(8.20)
Therefore, the displacement } in terms of {F and |F is:
} = uF · x̂} = {F cos 0 |F sin 0
(8.21)
8. Suspension Mechanisms
523
The initial coordinates of the coupler point F and the initial value of }
are:
= d cos 20 + h cos (s0 t0 + )
= d sin 20 + h sin (s0 t0 + )
= {F0 cos 0 |F0 sin 0
{F0
|F0
}0
(8.22)
(8.23)
(8.24)
and hence, the vertical displacement of the wheel center can be calculated
by
(8.25)
k = } }0
The initial angle of the coupler link with the vertical direction is 0 30 .
Therefore, the camber angle of the wheel would be
= (0 3 ) (0 30 ) = 30 3
(8.26)
The angle of the coupler link with respect to the {-direction is equal to
Ã
!
s
H ± H 2 4GI
1
3 = 2 tan
(8.27)
2G
where,
G
H
I
= M5 M1 + (1 + M4 ) cos 2
= 2 sin 2
= M5 + M1 (1 M4 ) cos 2
(8.28)
(8.29)
(8.30)
and
M1
=
M4
=
g
d2 e2 + f2 + g2
M3 =
f
2df
f2 g2 d2 e2
M5 =
2de
g
d
g
e
M2 =
(8.31)
Substituting (8.27) and (8.26), and then, eliminating 2 between (8.26) and
(8.21) provides us with the relationship between the vertical motion of the
wheel, }, and the camber angle .
Example 350 F Camber angle and wheel uctuations.
Consider the double D-arm suspension that is shown in Figure 8.37. The
dimensions of the equivalent kinematic model are:
d = 22=4 cm
g = 17=4 cm
e = 22=1 cm
0 = 24=3 deg
f = 27=3 cm
(8.32)
The coupler point F is at:
h = 14=8 cm
= 54=8 deg
(8.33)
524
8. Suspension Mechanisms
a=22.4 cm
b=22.1 cm
c=27.3 cm
d=17.4 cm
e=14.8 cm
h [ cm]
24.3q
T0
T20
D
121.5q
58.8q
J >deg @
FIGURE 8.38. The vertical displacement of the wheel center k as a function of
the camber angle .
If the angle 2 at the rest position is at
20 = 121=5 deg
(8.34)
then the initial angle of the other links are:
30 = 18=36 deg
40 = 107=32 deg
(8.35)
At the rest position, the coupler point is at:
{F0 = 7=417 cm
|F0 = 33=26 cm
}0 = 6=929 cm
(8.36)
We may calculate k and by varying the parameter 2 . Figure 8.38 illustrates k as a function of the camber angle . The vertical tire displacement
k versus 2 , 3 , 3 are respectively shown in Figures 8.39, 8.40, 8.41. Figure
8.42 depicts the angle change of the upper arm 2 20 versus the camber
angle .
For this suspension mechanism, the wheel gains a positive camber when
the wheel moves up, and gains a negative camber when the tire moves down.
The mechanism is shown in Figure 8.43, when the wheel is at the rest
position and has a positive or a negative displacement.
8.4 F Car Tire Relative Angles
There are four major wheel alignment parameters that aect the dynamics
of a vehicle: toe, camber, caster, and thrust angle.
8. Suspension Mechanisms
h [ cm]
a=22.4 cm
b=22.1 cm
c=27.3 cm
d=17.4 cm
e=14.8 cm
T0
24.3q
T20
121.5q
D
58.8q
T2 > deg @
FIGURE 8.39. The vertical displacement of the wheel center k versus 2 .
h [ cm]
a=22.4 cm
b=22.1 cm
c=27.3 cm
d=17.4 cm
e=14.8 cm
T0
24.3q
T20
121.5q
D
58.8q
T3 > deg @
FIGURE 8.40. The vertical displacement of the wheel center k versus 3 .
525
526
8. Suspension Mechanisms
a=22.4 cm
b=22.1 cm
c=27.3 cm
d=17.4 cm
e=14.8 cm
h [ cm]
T0
24.3q
T20
121.5q
D
58.8q
T4 > deg @
FIGURE 8.41. The vertical displacement of the wheel center k versus 4 .
T2 T20 > deg @
a=22.4 cm
b=22.1 cm
c=27.3 cm
d=17.4 cm
e=14.8 cm
T0
24.3q
T20
121.5q
D
58.8q
J > deg @
FIGURE 8.42. The angle change of the upper arm 2 3 20 versus the camber
angle .
8. Suspension Mechanisms
527
J
J
h
h
(a)
(b)
A
a
M
d
C
b
c
N
B
(c)
FIGURE 8.43. A double D-arm suspension mechanism when the wheel is at: (d)
a positive displacement, (e) a negative displacement, and (f) rest position.
8.4.1 F Toe
When a pair of wheels is set such that their leading edges are pointed toward
each other, the wheel pair is said to have toe-in. If the leading edges point
away from each other, the pair is said to have toe-out. Toe-in and toe-out
of front wheel congurations of a car are illustrated in Figure 8.44.
The amount of toe can be expressed in degrees of the angle to which the
wheels are not parallel. However, it is more common to express the toein and toe-out as the dierence between the track widths as measured at
the leading and trailing edges of the tires. Toe settings aect three major
performances: tire wear, straight-line stability, and corner entry handling.
For minimum tire wear and power loss, the wheels on a given axle of
a car should be neutral and point directly ahead when the car is running
in a straight line. Excessive toe-in causes accelerated wear at the outboard
528
8. Suspension Mechanisms
Toe-in
Nutral
Toe-out
FIGURE 8.44. Toe-in, nutral, and toe-out conguration on the front wheels of a
car.
edges of the tires, while too much toe-out causes wear at the inboard edges.
Toe-in increases the directional stability of the vehicle, and toe-out increases the steering response. Hence, a toe-in setting makes the steering
function lazy, while a toe-out makes the vehicle unstable. As a result, most
street cars are set to have a few angles toe-in and race cars are set to have
a few angles toe-out in their front wheels.
With four wheel independent suspensions, the toe may also be set at
the rear of the car. Toe settings at the rear have the same eect on wear,
directional stability, and turn-in response as they do on the front. However,
we usually do not set a rear-drive race car toed out in the rear, because of
excessive instability.
When driving torque is applied to the wheels, they pull themselves forward and try to create toe-in. On the other hand, a non-driven wheel or a
braking wheel will tend to toe-out.
Example 351 Toe-in and directional stability.
Toe settings have an impact on directional stability. When the steering
wheel is centered, toe-in causes the wheels to tend to move along paths
that intersect each other in front of the vehicle. However, if the wheels’
angle are equal, the wheels’ lateral forces are in balance and no turn results.
Toe-in setup can increase the directional stability caused by little steering
uctuations and keep the car moving straight. Steering uctuations may be
a result of road disturbances, wind, bank angle, bump steering and so on.
Little steering disturbance, to the left of example, increases the directional
angle and the lateral force of the left wheel and decreases the directional
angle and the lateral force of the right wheel. Therefore a rightward resultant
8. Suspension Mechanisms
529
lateral force will be generated that opposes the disturbance.
If a car is set up with toe-out, the front wheels are aligned so that slight
disturbances cause the wheel pair to assume rolling directions that approach
a turn. Therefore, toe-out encourages the initiation of a turn, while toe-in
discourages it. Toe-out makes the steering quicker. So, it may be used in
vehicles for a faster response. The toe setting on a particular car becomes a
trade-o between the straight-line stability aorded by toe-in and the quick
steering response by toe-out. Toe-out is not desirable for street cars, however, race car drivers are willing to drive a car with a little directional
instability, for sharper turn-in to the corners. So street cars are generally
set up with toe-in, while race cars are often set up with toe-out.
Example 352 Toe-in and toe-out in the front and rear axles.
Front toe-in: slower steering response, more straight-line stability, greater
wear at the outboard edges of the tires.
Front toe-zero: medium steering response, minimum power loss, minimum tire wear.
Front toe-out: quicker steering response, less straight-line stability, greater
wear at the inboard edges of the tires.
Rear toe-in: straight-line stability, traction out of the corner, more steerability, higher top speed.
8.4.2 F Caster Angle
Caster is the angle to which the steering axis is tilted forward or rearward
from vertical as viewed from the side in the direction of the |-axes. The
steering axis is the line about which the wheel will turn when steered.
Assume the wheel is straight to have the body frame and the wheel frame
coincident. If the steering axis is turned about the wheel |z -axis then the
wheel has positive caster. If the steering axis is turned about the wheel
|z -axis, then the wheel has negative caster. Positive and negative caster
congurations on the front wheel of a car are shown in Figure 8.45.
Negative caster aids in centering the steering wheel after a turn and
makes the front tires straighten quicker. Most street cars are made with
4 deg to 6 deg negative caster. Negative caster tends to straighten the wheel
when the vehicle is traveling forward, and thus is used to enhance straightline stability. It is caused by the position of the backward friction force in
the tireprint and the steering axis point on the ground, about which the
wheel steers. If the tireprint is behind the point, then we have a stable
system similar to a hanging pendulum under the gravitational force being
lower than the hanging fulcrum.
Example 353 Negative caster of shopping carts.
The steering axis of a shopping cart wheel is set to be in front of where
the wheel contacts the ground. As the cart is pushed forward, the steering
530
8. Suspension Mechanisms
Steering axis
zw
zw
Steering axis
xw
Negative caster
xw
Positive caster
FIGURE 8.45. A positive and negative caster conguration on front wheel of a
car.
axis pulls the wheel along. Because the wheel drags along the ground, it falls
directly in line behind the steering axis. The force that causes the wheel to
follow the steering axis is proportional to the distance between the steering
axis and the wheel-to-ground contact point, if the caster is small. This distance is referred to as trail. The cars’ steering axis intersects the ground
at a point in front of the tireprint, and thus the same eect as seen in the
shopping cart casters is achieved.
While greater caster angles improves straight-line stability, they cause an
increase in steering eort.
Example 354 Characteristics of caster in front axle.
Zero castor provides: easy steering into the corner, low steering out of
the corner, low straight-line stability.
Negative caster provides: lazy steering into the corner, easy steering out
of the corner, more straight-line directional stability, high tireprint area
during turn, good steering feel.
When a negative castered wheel rotates about the steering axis, the wheel
gains camber. This camber is generally favorable for cornering.
8.4.3 F Camber
Camber is the lateral angle of a wheel about the {-axis relative to the
vertical line to the road. Figure 8.46 illustrates the wheel number 1 of a
vehicle. If the wheel plane is turned about the {-axis and leans toward the
chassis, it has a positive camber. The wheel has a negative camber if its
plane is turned about {-axis.
The cornering force that a tire can develop is a function of its angles
relative to the road surface. The wheel camber has a signicant eect on
the lateral force development and road holding of a car. A tire develops
its maximum lateral force at a small camber angle. This fact is due to the
contribution of camber thrust, which is an additional lateral force generated by elastic deformation as the tread rubber pulls through the tire/road
8. Suspension Mechanisms
z
z
J
J
y
Positive camber
531
y
Negative camber
FIGURE 8.46. A positive and negative camber on the front wheel of a car.
interface.
To optimize a tire’s performance in a turn, the suspension should provide
a slight camber angle in the direction of rotation. As the body rolls in
a turn, the suspension de ects vertically. The wheel is connected to the
chassis by suspension mechanism, which must rotate to allow for the wheel
de ection. Therefore, the wheel can be subject to some camber changes as
the suspension moves up and down. So, the more the wheel de ects from
its static position, the more di!cult it is to maintain an ideal camber angle.
To provide a good ride comfort in passenger cars, a relatively large wheel
travel and soft roll stiness are needed. As a result, suspension design to
provide a good camber presents a design challenge, while the small wheel
travel and high roll stiness inherent in racing cars reduces the problem.
Example 355 Castor versus camber.
Camber doesn’t improve turn-in as the positive caster does. Camber is
not generally good for tire wear. Camber in one wheel does not improve
directional stability. Camber adversely aects braking and acceleration efforts.
Example 356 Camber versus sideslip.
Cars usually use sideslip angle to generate the required lateral force for
turning. Bikes usually use camber angle to generate the required lateral force
to turn. Because a tire needs camber angle ten times of the sideslip angle
to produce the same amount of lateral force, roll of bikes in a turn is an
observable motion, while sideslip angle is hard to see.
532
8. Suspension Mechanisms
Thrust angle
X
A
B
C
lr
ll
D
C
FIGURE 8.47. Thrust angle.
8.4.4 F Thrust Angle
The thrust angle is the angle between vehicle’s centerline and perpendicular to the rear axle. It compares the direction that the rear axle is aimed
with the centerline of the vehicle. A nonzero angle conguration is shown
in Figure 8.47.
Zero angle conrms that the rear axle is parallel to the front axle, and the
wheelbase on both sides of the vehicle are the same. A reason for nonzero
thrust angle would have unequal toe-in or toe-out on both sides of the axle.
Example 357 Torque reaction.
There are two kinds of torque reactions in rear-wheel-drive vehicles: 1=
the reaction of the axle housing to rotate in the opposite direction of the
crown wheel rotation, and 2= the reaction of axle housing to spin about its
own center, opposite to the direction of pinion’s rotation.
The rst reaction leads to a lifting force in the dierential causing a
wind-up in springs. The second reaction leads to a lifting force on the right
wheels.
8. Suspension Mechanisms
533
zw
G Steer angle
a
Tire
xis yw
M Spin angle
J Camber angle
xw
Grou
nd s
urfac
e
FIGURE 8.48. Six degrees of freedom of a wheel with respect to a vehicle body.
8.5 F Suspension Requirements and Coordinate
Frames
The suspension mechanism should allow some relative motions between the
wheel and the vehicle body. The relative motions are needed to pass the
road irregularities and steering. Therefore, to function properly, a suspension mechanism should have some kinematic and dynamics requirements.
8.5.1 Kinematic Requirements
To express the motions of a wheel, we attach a wheel coordinate system
Z (r{z |z }z ) to the center of the wheel. A wheel, as a rigid body, has vl{
degrees-of-freedom (DOF) with respect to the vehicle body: three translations and three rotations, as shown in Figure 8.48.
The axes {z , |z , and }z indicate the direction of forward, lateral, and
vertical translations and rotations. In the position shown in the gure, the
rotation about the {z -axis is the camber angle, about the |z -axis is the
spin, and about the }z -axis is the steer angle.
Consider a non-steerable wheel. Translation in }z -direction and spin
about the |z -axis are the only wzr DOF allowed. So, we need to take
four DOF. If the wheel is steerable, then translation in the }z -direction,
spin about the |z -axis, and steer rotation about the }z -axis are the three
DOF allowed and we must take three DOF of a steerable wheel.
Kinematically, non-steerable and steerable wheels should be supported
534
8. Suspension Mechanisms
z
y
M Spin angle
FIGURE 8.49. A non-steerable wheel must have two DOF.
as shown in Figures 8.49 and 8.50 respectively. Providing the required freedom, as well as taking the non-allowed DOF, are the kinematic requirements
of a suspension mechanism.
8.5.2 Dynamic Requirements
Wheels should be able to propel, steer, and stop the vehicle. So, the suspension system must transmit the driving traction and deceleration braking
forces between the vehicle body and the ground. The suspension members
must also resist lateral forces acting on the vehicle. Hence, the wheel suspension system must make the wheel rigid for the taken DOF. However,
there must also be some compliance members to limit the untaken DOF.
The most important compliant members are spring and dampers to provide
returning and resistance forces in the }-direction.
8.5.3 Wheel, wheel-body, and tire Coordinate Frames
Three coordinate frames are employed to express the orientation of a tire
and wheel with respect to the vehicle: the wheel frame Z , wheel-body
frame F, and tire frame W . The wheel coordinate frame Z ({z > |z > }z ) is
attached to the center of the wheel. It follows every translation and rotation
of the wheel except the spin. Hence, the {z and }z axes are always in the
8. Suspension Mechanisms
z
535
G Steer angle
y
M Spin angle
FIGURE 8.50. A steerable wheel must have three DOF.
tire-plane, while the |z -axis is always parallel to the spin axis. A wheel
coordinate frame is shown in Figure 8.48.
When the wheel is straight and upright and the Z frame is parallel
to the vehicle coordinate frame, we attach a wheel-body coordinate frame
F ({f > |f > }f ) at the center of the wheel parallel to the vehicle coordinate
axes. The wheel-body frame F is motionless with respect to the vehicle
coordinate and does not follow any motion of the wheel.
The tire coordinate frame W ({w > |w > }w ) is set at the center of the tireprint.
The }w -axis is always perpendicular to the ground. The {w -axis is along the
intersection line of the tire-plane and the ground. The tire frame does not
follows the spin and camber rotations of the tire however, it follows the
steer angle rotation about the }f -axis.
Figure 8.51 illustrates a tire and a wheel coordinate frames.
Example 358 Visualization of the wheel, tire, and wheel-body frames.
Figure 8.52 illustrates the relative conguration of a wheel-body frame
F, a tire frame W , and a wheel frame Z . If the steering axis is along the
}f -axis then, the rotation of the wheel about the }f -axis is the steer angle .
Rotation about the {w -axis is the camber angle .
Generally speaking, the steering axis may have any angle and may go
through any point of the ground plane.
Example 359 Wheel to tire coordinate frame transformation.
If W dZ indicates the W -expression of the position vector of the wheel
536
8. Suspension Mechanisms
zw
W
zc
zt
J
Vertical plane
yw
Tire plane
yc
C
Ground plane
yt
T
xc
xw
xt
FIGURE 8.51. Illustration of tire and wheel coordinate frames.
frame origin relative to the tire frame origin, then having the coordinates
of a point S in the wheel frame, we can nd its coordinates in the tire frame
by:
W
rS = W UZ Z rS + W dZ
(8.37)
If Z rS indicates the position vector of a point S in the wheel frame,
5
6
{S
Z
rS = 7 |S 8
}S
(8.38)
then the coordinates of the point S in the tire frame W rS are
W
rS
UZ Z rS + W d = W UZ Z rS + W UZ Z
W dZ
6
{S
= 7 |S cos Uz sin }S sin 8
Uz cos + }S cos + |S sin =
W
5
(8.39)
where, Z
W dZ is the Z -expression of the position vector of the wheel frame
in the tire frame, Uz is the radius of the tire, and W UZ is the rotation
8. Suspension Mechanisms
zw
W
zc
537
zt
J
Vertical plane
yw
Tire plane
C
yc
Ground plane
yt
T
xw
xc
xt
G
FIGURE 8.52. Illustration of tire, wheel, and body coordinate frames.
matrix to transfer from the wheel frame Z to the tire frame W .
5
6
1
0
0
W
UZ = 7 0 cos sin 8
0 sin cos 6
5
0
Z
7
0 8
d
=
Z
W
Uz
(8.40)
(8.41)
As an example, the center of the wheel Z rS = Z rr = 0 is the origin of
the wheel frame Z , that is at W rr in the tire coordinate frame W .
5
6
0
W
7 Uz sin 8
rr = W dZ = W UZ Z
(8.42)
W dZ =
Uz cos Example 360 F Tire to wheel coordinate frame transformation.
If rS indicates the position vector of a point S in the tire coordinate
frame,
5
6
{S
W
rS = 7 |S 8
(8.43)
}S
538
8. Suspension Mechanisms
then the position vector Z rS of the point S in the wheel coordinate frame
is
Z
rS
UW W rS Z
W dZ
6
{S
8
|S cos + }S sin = 7
}S cos Uz |S sin =
Z
5
(8.44)
because
5
1
0
Z
UW = 7 0 cos 0 sin 6
0
sin 8
cos 5
6
0
Z
dW = 7 0 8
Uz
(8.45)
W
and we may multiply both sides of Equation (8.37) by W UZ
to get
W
W W
UZ
rS
Z
rS
=
=
Z
W W
rS + W UZ
dZ = Z rS + Z
W dZ
Z
Z
W
UW rS W dZ
(8.46)
(8.47)
As an example, the center of tireprint in the wheel frame is at
5
1
0
Z
rS = 7 0 cos 0 sin 6W 5 6 5
6 5
6
0
0
0
0
sin 8 7 0 8 7 0 8 = 7 0 8
cos 0
Uz
Uz
(8.48)
Example 361 F Wheel to tire homogeneous transformation matrices.
The transformation from the wheel to tire coordinate frame may also be
expressed by a 4 × 4 homogeneous transformation matrix W WZ ,
W
¸
UZ W dZ
W
Z
rS = W WZ Z rS =
rS
(8.49)
0
1
where
5
1
0
9
0
cos
W
WZ = 9
7 0 sin 0
0
0
sin cos 0
6
0
Uz sin :
:
Uz cos 8
1
(8.50)
The corresponding homogeneous transformation matrix Z WW from the tire
to wheel frame would be
6
5
1
0
0
0
Z
¸
9 0 cos sin UW Z dW
0 :
Z
:
WW =
(8.51)
=9
7 0 sin 0
1
cos Uz 8
0
0
0
1
8. Suspension Mechanisms
539
1
It can be checked that Z WW = W WZ
, using the inverse of a homogeneous
transformation matrix rule.
¸1 W W
¸
W
W W
UZ W dZ
UZ W UZ
dZ
W 1
WZ =
=
0
1
0
1
¸
Z
Z
W
UW UW dZ
(8.52)
=
0
1
Example 362 F Tire to wheel-body frame transformation.
The origin of the tire frame is at F dW in the wheel-body frame.
6
5
0
F
dW = 7 0 8
Uz
(8.53)
The tire frame can steer about the }f -axis with respect to the wheel-body
frame. The associated transformation matrix is
5
6
cos sin 0
F
cos 0 8
UW = 7 sin (8.54)
0
0
1
Therefore, the transformation between the tire and wheel-body frames can
be expressed by
F
r = F UW W r + F dW
(8.55)
or equivalently, by a homogeneous transformation matrix F WW .
5
6
cos sin 0
0
¸
F
9 sin cos 0
0 :
UW F dW
F
:
=9
WW =
7 0
0
1
0
1 Uz 8
0
0
0
1
(8.56)
As an example, the wheel-body coordinates of the point S on the tread of a
negatively steered tire at the position shown in Figure 8.53, are:
F
r = F WW W rS
5
cos sin 9 sin cos = 9
7
0
0
0
0
65
6 5
6
0
0
Uz
Uz cos 9
: 9
:
0
0 :
: 9 0 : = 9 Uz sin : (8.57)
8
1 Uz 8 7 Uz 8 7
0
0
1
1
1
The homogeneous transformation matrix for tire to wheel-body frame W WF
is:
¸1 F W
¸
F
UW F dW
UW F UWW F dW
W
WF = F WW1 =
=
0
1
0
1
5
6
cos sin 0 0
F W
¸
9 sin cos 0 0 :
UW WF dW
:
=
=9
(8.58)
7
0
1
0
0
1 Uz 8
0
0
0 1
540
8. Suspension Mechanisms
zt
zc
zw
C
xc
P
G
xw
W
Rw
xt
T
FIGURE 8.53. The tire, wheel, and wheel body frames of a steered wheel.
Example 363 F Cycloid.
Assume that the wheel in Figure 8.53 is turning with angular velocity $
and has no slip on the ground. If the point S is at the center of the tireprint
when w = 0,
6
5
0
P
(8.59)
rS = 7 0 8
Uz
then we can nd its position in the wheel frame at a time w by employing
another coordinate frame P . The frame P is called the rim frame and is
stuck to the wheel at its center. Because of spin, the P frame turns about
the |z -axis, and therefore, the rotation matrix to go from the rim frame to
the wheel frame is:
5
6
cos $w 0 sin $w
Z
0
1
0 8
UP = 7
(8.60)
sin $w 0 cos $w
So the coordinates of S in the wheel frame are:
6
5
Uz sin w$
Z
8
0
rS = Z UP P rS = 7
Uz cos w$
(8.61)
J
The
center of the
£
¤ wheel is moving with speed y{ = Uz $ and it is at r =
y{ w 0 Uz in the global coordinate frame J on the ground. Hence,
the coordinates of point S in the global frame J, would be
6
5
6 5
y{ w
Uz ($w sin w$)
J
8
0
(8.62)
rS = Z rS + 7 0 8 = 7
Uz
Uz (1 cos w$)
8. Suspension Mechanisms
P
Z
zw
zt
xw
xm
xt
zm
Z
541
(a)
X
d
(b)
X
Z
d
(c)
X
FIGURE 8.54. A cycloid (d), curtate cycloid (e), and prolate cycloid (f).
The path of motion of point S in the ([> ])-plane can be found by eliminating w between [ and ] coordinates. However, it is easier to expressed
the path by using $w as a parameter. Such a path is called cycloid.
In general case, the interested point S can be at any distance from the
center of the rim frame. If the point is at a distance g 6= Uz , then its path
of motion is called the trochoid. A trochoid is called a curtate cycloid
when g ? Uz and a prolate cycloid when g A Uz . Figure 8.54(d)-(f)
illustrate a cycloid, curtate cycloid, and prolate cycloid respectively.
Example 364 F Wheel to wheel-body frame transformation.
The homogeneous transformation matrix F WZ to go from the wheel frame
to the wheel-body frame will be found by combined transformation.
F
WZ
F
WW W WZ
(8.63)
65
6
f v 0
0
1 0
0
0
9 v f 0
9
:
0 :
: 9 0 f v Uz sin :
= 9
7 0
8
7
0
1 Uz
0 v f
Uz cos 8
0
0
0
1
0 0
0
1
5
6
cos cos sin sin sin Uz sin sin 9 sin cos cos cos sin Uz cos sin :
:
= 9
7 0
sin cos Uz cos Uz 8
0
0
0
1
=
5
If rS indicates the position vector of a point S in the wheel coordinate
542
8. Suspension Mechanisms
frame,
6
{S
Z
rS = 7 |S 8
}S
5
(8.64)
then the homogeneous position vector F rS of the point S in the wheel-body
coordinate frame would be:
F
rS
F
WZ Z rS
6
{S cos |S cos sin + (Uz + }S ) sin sin 9 {S sin + |S cos cos (Uz + }S ) cos sin :
: (8.65)
= 9
7
8
Uz + (Uz + }S ) cos + |S sin 1
=
5
The position of the wheel center Z r = 0, for a cambered and steered wheel
is at
5
6
Uz sin sin 9 Uz cos sin :
F
:
r = F WZ Z r = 9
(8.66)
7 Uz (1 cos ) 8
1
The }f = Uz (cos 1) indicates how much the center of the wheel comes
down when the wheel cambers.
If the wheel is not steerable, then = 0 and the transformation matrix
F
WZ reduces to
5
6
1
0
0
0
9 0 cos sin Uz sin :
F
:
WZ = 9
(8.67)
7 0 sin cos Uz (cos 1) 8
0
0
0
1
that shows
F
rS
=
F
5
WZ Z rS
6
{S
9
:
|S cos Uz sin }S sin :
= 9
7 }S cos + |S sin + Uz (cos 1) 8
1
(8.68)
Example 365 F Tire to vehicle coordinate frame transformation.
Figure 8.55 illustrates the rst and fourth tires of a 4-wheel vehicle. There
is a body coordinate frame E ({> |> }) attached to the mass center F of
the vehicle. There are also two tire coordinate frames W1 ({w1 > |w1 > }w1 ) and
W4 ({w4 > |w4 > }w4 ) attached to the tires 1 and 4 at the center of their tireprints.
The origin of the tire coordinate frame W1 is at E d1
6
5
d1
E
dW1 = 7 e1 8
k
(8.69)
8. Suspension Mechanisms
543
z
x
T1
b1
B
zt1
xt1
T4
C
zt4
a1
l
yt1
xt4
h
y
yt4
FIGURE 8.55. The coordinate frames of the rst and fourth tires of a 4-wheel
vehicle with respect to the body frame.
where, d1 is the longitudinal distance between F and the front axle, e1 is
the lateral distance between F and the tireprint of the tire 1, and k is the
height of F from the ground level. If S is a point in the tire frame at W1 rS
5
6
{S
W1
rS = 7 |S 8
(8.70)
}S
then its coordinates in the body frame are
E
rS
UW1 W1 rS + E dW1
6
{S cos 1 |S sin 1 + d1
= 7 |S cos 1 + {S sin 1 + e1 8
}S k
=
E
5
(8.71)
The transformation matrix E UW1 is a result of steering about the }1 -axis.
5
6
cos 1 sin 1 0
E
cos 1 0 8
UW1 = 7 sin 1
(8.72)
0
0
1
Employing Equation (8.37), we may examine a wheel point S at Z rS
5
6
{S
Z
rS = 7 |S 8
(8.73)
}S
544
8. Suspension Mechanisms
and nd the body coordinates of the point
E
rS
=
=
E
=
=
E
E
E
UW1 W1 rS + E dW1
¡
¢
UW1 W1 UZ Z rS + W1 dZ + E dW1
UW1 W1 UZ Z rS + E UW1 W1 dZ + E dW1
UZ Z rS + E UW1 W1 dZ + E dW1
5
(8.74)
6
{S cos 1 |S cos sin 1 + (Uz + }S ) sin sin 1 + d1
rS = 7 {S sin 1 + |S cos cos 1 (Uz + }S ) cos 1 sin + e1 8
(Uz + }S ) cos + |S sin k
(8.75)
where,
E
E
UZ
UW1 W1 UZ
cos 1 cos sin 1
cos cos 1
= 7 sin 1
0
sin =
E
5
6
sin sin 1
cos 1 sin 8
cos 6
0
W1
dZ = 7 Uz sin 8
Uz cos (8.76)
5
(8.77)
Example 366 F Wheel-body to vehicle transformation.
The wheel-body coordinate frames is always parallel to the vehicle frame.
The origin of the wheel-body coordinate frame of the wheel number 1 is at
5
6
d1
E
8
e1
dZ1 = 7
(8.78)
k + Uz
Hence the transformation between the two frames is only a displacement.
E
r = E IZ1 Z1 r + E dZ1
(8.79)
8.6 F Caster Theory
The steer axis of a wheel may have any angle and be at any location
with respect to the wheel-body coordinate frame. The wheel-body frame
F ({f > |f > }f ) is a frame at the center of the wheel at its rest position, and is
parallel to the vehicle coordinate frame. The frame F does not follow any
motion of the wheel. The steer axis is the kingpin axis of rotation.
Figure 8.56 illustrates the front and side views of a wheel and its steering
axis. The steering axis has angle * with (|f > }f ) plane, and angle with
({f > }f ) plane. The angles * and are measured about the |f and {f axes,
8. Suspension Mechanisms
zc
545
zc
M
T
Steering axis
yc
xc
Steering axis
Rw
sb
sa
Front view
Side view
FIGURE 8.56. The front and side views of a wheel and its steering axis.
respectively. The angle * is the caster angle of the wheel, while the angle
is the lean angle. The steering axis of the wheel, as shown in Figure
8.56, is at positive caster and lean angles. The steering axis intersects the
ground plane at a point that has coordinates (vd > ve > Uz ) in the wheelbody coordinate frame.
If we indicate the steering axis by the unit vector x̂, then the components
of x̂ are functions of the caster and lean angles.
6
5
5
6
x1
cos sin *
1
F
7 cos * sin 8
x̂ = 7 x2 8 = p
(8.80)
cos2 * + cos2 sin2 *
x3
cos cos *
The position vector of the point that x̂ intersects the ground plane, is
called the location vector s that in the wheel-body frame has the following
coordinates:
6
5
vd
F
(8.81)
s = 7 ve 8
Uz
We express the rotation of the wheel about the steering axis x̂ by a zero
pitch screw motion v.
F
WZ
F
vZ (0> > x̂> s)
F
¸ F
UZ F s F UZ F s
UZ
=
=
0
0
1
=
F
dZ
1
¸
(8.82)
Proof. The steering axis is at the intersection of the caster plane F and
the lean plane O , both expressed in the wheel-body coordinate frame. The
546
8. Suspension Mechanisms
two planes can be indicated by their normal unit vectors q̂1 and q̂2 .
5
6
0
F
q̂1 = 7 cos 8
(8.83)
sin 5
6
cos *
F
8
0
q̂2 = 7
(8.84)
sin *
The unit vector x̂ on the intersection of the caster and lean planes can be
found by
q̂1 × q̂2
x̂ =
(8.85)
|q̂1 × q̂2 |
where,
5
and therefore,
6
cos sin *
q̂1 × q̂2 = 7 cos * sin 8
cos cos *
q
|q̂1 × q̂2 | =
cos2 * + cos2 sin2 *
6
cos sin *
p
6 9
5
cos2 * + cos2 sin2 * :
:
9
x1
:
9
cos * sin F
:
9
8
7
p
x̂ = x2 = 9
:
2
2
2
cos
*
+
cos
sin
*
:
9
x3
8
7
cos cos *
p
2
2
2
cos * + cos sin *
(8.86)
(8.87)
5
(8.88)
Steering axis does not follow any motion of the wheel except the wheel
hop in the }-direction. Therefore, it is assumed that the steering axis is a
xed line with respect to the vehicle, and the steer angle is the rotation
angle about x̂.
The intersection point of the steering axis and the ground plane denes
the location vector s.
5
6
vd
F
s = 7 ve 8
(8.89)
Uz
The components vd and ve are called the forward and lateral locations
respectively.
Using the axis-angle rotation (x̂> ), and the location vector s, we can
dene the steering of the wheel by a screw motion v with zero pitch. Employing Equations (5.460)-(5.464), we nd the transformation screw for
8. Suspension Mechanisms
547
wheel frame Z to wheel-body frame F.
F
WZ
F
vZ (0> > x̂> s)
F
¸ F
UZ F s F UZ F s
UZ
=
=
0
1
0
=
F
UZ
F
dZ
= I cos + x̂x̂W vers + x̃ sin ¡¡
¢
¢
=
I x̂x̂W vers x̃ sin F s
5
0
x̃ = 7 x3
x2
vers = 1 cos x3
0
x1
6
x2
x1 8
0
F
d
1
¸
(8.90)
(8.91)
(8.92)
(8.93)
(8.94)
Direct substitution shows that F UZ and F dZ are:
5
6
x21 vers + f
x1 x2 vers x3 v x1 x3 vers + x2 v
F
x22 vers + f
x2 x3 vers x1 v 8
UZ = 7 x1 x2 vers + x3 v
x1 x3 vers x2 v x2 x3 vers + x1 v
x23 vers + f
(8.95)
5
6
(v1 x1 (v3 x3 + v2 x2 + v1 x1 )) vers + (v2 x3 v3 x2 ) sin F
dZ = 7 (v2 x2 (v3 x3 + v2 x2 + v1 x1 )) vers + (v3 x1 v1 x3 ) sin 8
(v3 x3 (v3 x3 + v2 x2 + v1 x1 )) vers + (v1 x2 v2 x1 ) sin (8.96)
The vector F dZ indicates the position of the wheel center with respect to
the wheel-body frame. The matrix F WZ is the homogeneous transformation
from wheel frame Z to wheel-body frame F, when the wheel is steered by
the angle about the steering axis x̂.
Example 367 F Zero steer angle.
To examine the screw transformation, we may check the zero steering.
Substituting = 0 simplies the rotation matrix F UZ and the position
vector F dZ to I and 0 respectively
5
6
5 6
1 0 0
0
F
F
UZ = 7 0 1 0 8
dZ = 7 0 8
(8.97)
0 0 1
0
indicating that at zero steering, the wheel frameZ and wheel-body frame F
are coincident.
Example 368 F Steer angle transformation for zero lean and caster.
Consider a wheel with a steer axis coincident with }z . Such a wheel has
no lean or caster angle. When the wheel is steered by the angle , we can
nd the coordinates of a wheel point S in the wheel-body coordinate frame
548
8. Suspension Mechanisms
zt
zw
xw xt
G
G
x
W
W
B
T
yw
yt yw
y
y
yt
xw
x
G
(a)
B
T
xt
Steer angle
(b)
FIGURE 8.57. A steered wheel by the angle and with a steer axis coincident
with }z . (d) A 3G illustration of the wheel. (e) Top view of the wheel.
using transformation method. Figure 8.57(d) illustrates a 3G view, and
Figure 8.57(e) a top view of such a wheel.
W
Assume Z rS = [{z > |z > }z ] is the position vector of a wheel point, then
its position vector in the wheel-body coordinate frame F is
F
rS
F
UZ Z rS = U}> Z rS
(8.98)
6 5
6
65
{z cos |z sin cos sin 0
{z
= 7 sin cos 0 8 7 |z 8 = 7 |z cos + {z sin 8
}z
}z
0
0
1
=
5
We assumed that the wheel-body coordinate is installed at the center of
the wheel and is parallel to the vehicle coordinate frame. Therefore, the
transformation from the frame Z to the frame F is a rotation about the
wheel-body }-axis. There would be no camber angle when the lean and caster
angles are zero and steer axis is on the }z -axis.
Example 369 F Zero lean, zero lateral location.
The case of zero lean, = 0, and zero lateral location, ve = 0, is important in caster dynamics of cars bicycle model. The screw transformation
for this case will be simplied to
5
5
6 5
6
6
x1
vd
sin *
F
F
x̂ = 7 x2 8 = 7 0 8
s=7 0 8
(8.99)
x3
cos *
Uz
8. Suspension Mechanisms
5
549
6
sin * cos * vers 8
sin * sin 2
cos * vers + cos (8.100)
6
5
cos * (vd cos * + Uz sin *) vers :
9
(vd cos * + Uz sin *) sin F
(8.101)
d =7
8
1
(Uz Uz cos 2* + vd sin 2*) vers 2
Example 370 F Position of the tireprint.
The center of tireprint in the wheel coordinate frame is at rW
6
5
0
Z
(8.102)
rW = 7 0 8
Uz
sin2 * vers + cos F
7
UZ =
cos * sin sin * cos * vers cos * sin cos sin * sin If we assume the width of the tire is zero and the wheel is steered, the center
of tireprint would be at
5
6
{W
F
rW = F WZ Z rW = 7 |W 8
(8.103)
}W
where
{W
|W
}W
or
{W
|W
¡
¢
1 x21 (1 cos ) vd + (x3 sin x1 x2 (1 cos )) ve (8.104)
¡
¢
= (x3 sin + x1 x2 (1 cos )) vd + 1 x22 (1 cos ) ve(8.105)
= (x2 sin x1 x3 (1 cos )) vd
(x1 sin + x2 x3 (1 cos )) ve Uz
(8.106)
=
!
1 sin 2 sin 2* (1 cos )
+
= ve p
2
cos2 sin2 * + cos2 * 4 cos2 sin * + cos2 *
¶
μ
cos2 sin2 *
+vd 1 (1 cos )
(8.107)
cos2 sin2 * + cos2 *
Ã
cos cos * sin !
1 sin 2 sin 2* (1 cos )
= vd p
2
cos2 sin2 * + cos2 * 4 cos2 sin * + cos2 *
¶
μ
cos2 * sin2 +ve 1 (1 cos )
(8.108)
cos2 sin2 * + cos2 *
Ã
}W
cos cos * sin ve cos sin * + vd cos * sin p
sin cos2 sin2 * + cos2 *
1 ve cos2 * sin 2 vd cos2 sin 2*
(1 cos )
+
2
cos2 sin2 * + cos2 *
= Uz (8.109)
550
8. Suspension Mechanisms
Example 371 F Wheel center drop.
The }W coordinate in (8.106) or (8.109) indicates the height that the
center of the tireprint will move vertically with respect to the wheel-body
frame when the wheel is steering. If the steer angle is zero, = 0, then }W
is at
(8.110)
}W = Uz
Because the center of tireprint must be on the ground, K = Uz }W
indicates the height that the center of the wheel will drop during steering.
K
= Uz }W
ve cos sin * + vd cos * sin p
=
sin cos2 sin2 * + cos2 *
1 ve cos2 * sin 2 vd cos2 sin 2*
(1 cos )
2
cos2 sin2 * + cos2 *
(8.111)
The }W coordinate of the tireprint may be simplied for dierent designs:
1= If the lean angle is zero, = 0, then }W is at
1
}W = Uz vd sin 2* (1 cos ) ve sin * sin 2
(8.112)
2= If the lean angle and lateral location are zero, = 0, ve = 0, then }W
is at
1
}W = Uz vd sin 2* (1 cos )
(8.113)
2
In this case, the wheel center drop may be expressed by a dimensionless
equation.
K
1
(8.114)
= sin 2* (1 cos )
vd
2
Figure 8.58 illustrates K@vd for the caster angle * = 5 deg, 0 deg, 5 deg,
10 deg, 15 deg, 20 deg, and the steer angle in the range 10 deg ?
? 10 deg. For street cars, we set the steering axis with a positive longitudinal location vd A 0, and a few degrees negative caster angle * ? 0. In
this case the wheel center drops as is shown in the gure.
3= If the caster angle is zero, * = 0, then }W is at
1
}W = Uz + ve sin 2 (1 cos ) vd sin sin 2
(8.115)
4= If the caster angle and lateral location are zero, * = 0, ve = 0, then
}W is at
(8.116)
}W = Uz vd sin sin In this case, the wheel center drop may be expressed by a dimensionless
equation.
K
= sin sin (8.117)
vd
8. Suspension Mechanisms
551
M 5q
H
sa
M 0
T 0
M
M
M
M
5q
10q
15q
20q
G >deg @
FIGURE 8.58. K@vd for the caster angle * = 5 deg 0, 35 deg, 310 deg, 315 deg,
320 deg and the steer angle in the range 310 deg ? ? 10 deg.
Figure 8.59 illustrates K@vd for the lean angle = 5 deg, 0, 5 deg, 10 deg,
15 deg, 20 deg and the steer angle in the range 10 deg ? ? 10 deg.
The steering axis of street cars is usually set with a positive longitudinal
location vd A 0, and a few degrees positive lean angle A 0. In this case
the wheel center lowers when the wheel number 1 turns to the right, and
elevates when the wheel turns to the left.
Comparison of Figures 8.58 and 8.59 shows that the lean angle has much
more aect on the wheel center drop than the caster angle.
5= If the lateral location is zero, ve = 0, then }W is at
}W
cos * sin = Uz vd p
sin 2
cos sin2 * + cos2 *
1
cos2 sin 2*
vd 2
(1 cos )
2 cos sin2 * + cos2 *
(8.118)
and the wheel center drop, K, may be expressed by a dimensionless equation.
1 cos2 sin2 * (1 cos )
cos * sin sin K
=
p
vd
2 cos2 sin2 * + cos2 *
cos2 sin2 * + cos2 *
(8.119)
Example 372 F Position of the wheel center.
As given by Equation (8.96), the wheel center is at F dZ with respect to
the wheel-body frame.
5
6
{Z
F
dZ = 7 |Z 8
(8.120)
}Z
Substituting for x̂ and s from (8.80) and (8.81) in (8.96) provides us with
552
8. Suspension Mechanisms
T 20q
T 15q
T 10q
T 5q
T 0
T 5q
H
sa
M 0
G >deg @
FIGURE 8.59. K@vd for the lean angle = 5 deg 0, 35 deg, 310 deg, 315 deg,
320 deg and the steer angle in the range 310 deg ? ? 10 deg.
the coordinates of the wheel center in the wheel-body frame as
{Z
|Z
}Z
= (vd x1 (Uz x3 + ve x2 + vd x1 )) (1 cos )
+ (ve x3 + Uz x2 ) sin (8.121)
= (ve x2 (Uz x3 + ve x2 + vd x1 )) (1 cos )
(Uz x1 + vd x3 ) sin (8.122)
= (Uz x3 (Uz x3 + ve x2 + vd x1 )) (1 cos )
+ (vd x2 ve x1 ) sin (8.123)
or
{Z
|Z
= vd (1 cos )
¶
μ
1
1
Uz sin 2* vd sin2 * cos2 + ve sin 2 sin 2*
2
4
(1 cos )
+
cos2 * + cos2 sin2 *
(ve cos Uz sin )
cos * sin (8.124)
+p
cos2 * + cos2 sin2 *
= ve (1 cos )
¢
1
1¡
Uz sin 2 + ve sin2 cos2 * vd sin 2 sin 2*
2
4
(1 cos )
cos2 * + cos2 sin2 *
Uz sin * + vd cos *
cos sin (8.125)
p
cos2 * + cos2 sin2 *
8. Suspension Mechanisms
}Z
553
= Uz (1 cos )
¶
μ
1
1
Uz cos2 + ve sin 2 cos2 * vd cos2 sin 2*
2
2
(1 cos )
+
cos2 * + cos2 sin2 *
vd cos * sin + ve cos sin *
sin (8.126)
p
cos2 * + cos2 sin2 *
The }Z coordinate indicates how the center of the wheel moves vertically
with respect to the wheel-body frame, when the wheel is steering. It shows
that }Z = 0, as long as = 0.
The }Z coordinate of the wheel center may be simplied for dierent
designs:
1= If the lean angle is zero, = 0, then }Z is at
¢
¡
}Z = Uz 1 cos2 * (1 cos ) ve sin * sin 1
(8.127)
vd sin 2* (1 cos )
2
2= If the lean angle and lateral location are zero, = 0, ve = 0, then }Z is
at
¡
¢
1
}Z = Uz 1 cos2 * (1 cos ) vd sin 2* (1 cos )
2
3= If the caster angle is zero, * = 0, then }Z is at
¢
¡
}Z = Uz 1 cos2 (1 cos ) vd sin sin 1
+ ve sin 2 (1 cos )
2
(8.128)
(8.129)
4= If the caster angle and lateral location are zero, * = 0, ve = 0, then
}Z is at
¢
¡
}Z = Uz 1 cos2 (1 cos ) vd sin sin (8.130)
5= If the lateral location is zero, ve = 0, then }W is at
}Z
vd cos * sin sin = Uz (1 cos ) p
cos2 * + cos2 sin2 *
1
Uz cos2 cos2 * vd cos2 sin 2*
2
+
(1 cos )
cos2 * + cos2 sin2 *
(8.131)
In each case of the above designs, the height of the wheel center with
respect to the ground level can be found by adding K to }Z . The equations
for calculating K are given in Example 372.
554
8. Suspension Mechanisms
Example 373 F Camber theory.
Having a non-zero lean and caster angles causes a camber angle for
a steered wheel. To nd the camber angle of an steered wheel, we may determine the angle between the camber line and the vertical direction }f .
The camber line is the line connecting the wheel center and the center of
tireprint.
The coordinates of the center of tireprint ({W > |W > }W ) are given in Equations (8.107)-(8.109), and the coordinates of the wheel center ({Z > |Z > }Z )
are given in Equations (8.124)-(8.126). The line connecting ({W > |W > }W ) to
({Z > |Z > }Z ) may be indicated by the unit vector ˆof
{W ) Lˆ + (|Z |W ) Mˆ + (}Z }W ) N̂
ˆof = ({
qZ
2
2
2
({Z {W ) + (|Z |W ) + (}Z }W )
(8.132)
ˆ M>
ˆ N̂, are the unit vectors of the wheel-body coordinate frame F.
in which L>
The camber angle is the angle between ôf and N̂, which can be found by
the inner vector product.
³
´
= arccos ˆof · N̂
(}Z }W )
(8.133)
= arccos q
2
2
2
({Z {W ) + (|Z |W ) + (}Z }W )
As an special case, let us determine the camber angle when the lean angle
and lateral location are zero, = 0, ve = 0. In this case, we have
¢
¡
(8.134)
{W = vd 1 sin2 * (cos 1)
|W = vd cos * sin (8.135)
1
(8.136)
}W = }W = Uz vd sin 2* (1 cos )
2
{Z
|Z
}Z
μ
¶
1
vd + Uz sin 2* vd sin2 * (1 cos )
2
= ve (1 cos ) Uz sin * + vd cos * sin μ
¶
¢ 1
¡ 2
=
Uz cos * 1 vd sin 2* (1 cos ) =
2
=
(8.137)
(8.138)
(8.139)
8.7 Summary
There are two general types of suspensions: dependent, in which the left and
right wheels on an axle are rigidly connected, and independent, in which
the left and right wheels are disconnected. Solid axle is the most common
8. Suspension Mechanisms
555
dependent suspension, while McPherson and double D-arm are the most
common independent suspensions.
The roll axis is the instantaneous line about which the body of a vehicle
rolls. Roll axis is found by connecting the roll center of the front and rear
suspensions of the vehicle. The instant center of rotation of a wheel with
respect to the body is called the suspension roll center. To nd the roll
center of the front or rear half of a car, we should determine the suspension
roll centers, and nd the intersection of the lines connecting the suspension
roll centers to the center of their associated tireprints.
Three coordinate frames are employed to express the orientation of a
tire and wheel with respect to the vehicle body: the wheel frame Z , wheelbody frame F, and tire frame W . A wheel coordinate frame Z ({z > |z > }z )
is attached to the center of a wheel. It follows every translation and rotation
of the wheel except the spin. Hence, the {z and }z axes are always in the
tire-plane, while the |z -axis is always along the spin axis. When the wheel
is straight and the Z frame is parallel to the vehicle coordinate frame,
we attach a wheel-body coordinate frame F ({f > |f > }f ) at the center of the
wheel parallel to the vehicle coordinate axes. The wheel-body frame F is
motionless with respect to the vehicle coordinate and does not follow any
motion of the wheel. The tire coordinate frame W ({w > |w > }w ) is set at the
center of the tireprint. The }w -axis is always perpendicular to the ground.
The {w -axis is along the intersection line of the tire-plane and the ground.
The tire frame does not follow the spin and camber rotations of the tire;
however, it follows the steer angle rotation about the }f -axis.
We dene the orientation and position of a steering axis by the caster
angle *, lean angle , and the intersection point of the axis with the ground
surface at (vd > ve ) with respect to the center of tireprint. Because of these
parameters, a steered wheel will camber and generates a lateral force. This
is called the caster theory. The camber angle of a steered wheel for = 0,
and ve = 0 is:
´
³
= cos1 ˆof · N̂
(}Z }W )
(8.140)
= cos1 q
({Z {W )2 + (|Z |W )2 + (}Z }W )2
where
{W
|W
}W
¡
¢
= vd 1 sin2 * (cos 1)
= vd cos * sin 1
= }W = Uz vd sin 2* (1 cos )
2
(8.141)
(8.142)
(8.143)
556
8. Suspension Mechanisms
{Z
|Z
}Z
μ
¶
1
vd + Uz sin 2* vd sin2 * (1 cos )
2
= ve (1 cos ) Uz sin * + vd cos * sin μ
¶
¡ 2
¢ 1
=
Uz cos * 1 vd sin 2* (1 cos )
2
=
(8.144)
(8.145)
(8.146)
8. Suspension Mechanisms
557
8.8 Key Symbols
d> e> f> g
dl
D> E> · · ·
e1 > e2
E ({> |> })
F
F ({f > |f > }f )
F
W dZ
g
h> j
g
k
k = } }0
ku
K
Llm
Llm Lpq
ˆ M>
ˆ N̂
L>
I
M1 > M2 > · · ·
n
ni
nl
nu
n*
o
ôf
p
pv
px
P
q̂1
q̂2
S
t> s> i
r
U
Uz
W
UZ
s
vd
lengths of the links of a four-bar linkage
distance of the axle number l from the mass center
coe!cients in equation for calculating 3
distance of left and right wheels from mass center
vehicle coordinate frame
mass center, coupler point
wheel-body coordinate frame
F expression of the position of Z with respect to W
radial distance from center of a wheel
polar coordinates of a coupler point
overhang, gravitational acceleration
gravitational acceleration vector
height of mass center
vertical displacement of the wheel center
roll height
wheel center drop
instant center of rotation between link l and link m
a line connecting Llm and Lpq
unit vectors of the wheel-body frame F
identity matrix
length function for calculating 3
stiness, speing stiness
spring stiness of the front wheels
spring stiness of the wheel number l
spring stiness of the rear wheel number l
roll stiness
wheelbase
unit vector on the line ({W > |W > }W ) to ({Z > |Z > }Z )
mass
sprung mass
unsprung mass
moment
normal unit vectors to O
normal unit vectors to F
point
parameters for calculating couple point coordinate
position vector
curvature radius of the road, radius of rotation
tire radius
rotation matrix to go from Z frame to W frame
position vector of the steer axis
forward location of the steer axis
558
8. Suspension Mechanisms
ve
vZ (0> > x̂> s)
W ({w > |w > }w )
W
WZ
x1 > x2 > x3
x̂
x̃
uF
x̂}
y> y{
vers z
Z ({z |z }z )
{> |
{F > |F
{W > |W > }W
{Z > |Z > }Z
Z ({z |z }z )
}
}0
lateral location of the steer axis
zero pitch screw about the steer axis
tire coordinate system
homogeneous transformation to go from Z to W
components of x̂
steer axis unit vector
skew symmetric matrix associated to x̂
position vector of the coupler point
unit vector in the }-direction
forward speed
1 cos track
wheel coordinate system
suspension coordinate frame
coordinate of a couple point
wheel-body coordinates of the origin of W frame
wheel-body coordinates of the origin of Z frame
wheel coordinate system
vertical position of the wheel center
initial vertical position of the wheel center
% = pv @px
0
l
2
3
4
l0
F
O
angle of a coupler point with upper D-arm
camber angle
steer angle
sprung to unsprung mass ratio
lean angle
angle between the ground link and the }-direction
angular position of link number l
angular position of the upper D-arm
angular position of the coupler link
angular position of link lower D-arm
initial angular position of l
caster plane
lean plane
thrust angle
roll angle, caster angle, spin angle
bank angle
angular velocity
critical roll angle
*
!
$
*f
8. Suspension Mechanisms
559
Exercises
1. Roll center.
Determine the roll center of the kinematic models of vehicles shown
in Figures 8.60 to 8.63.
5
4
Body
8
2
3
6
7
1
FIGURE 8.60.
4
Body
5
8
2
3
6
7
1
FIGURE 8.61.
2. Upper D-arm and roll center.
Design the upper D-arm for the suspensions that are shown in Figures
8.64 to 8.66, such that the roll center of the vehicle is at point S .
3. Lower arm and roll center.
Design the lower arm for the McPherson suspensions that are shown
in Figures 8.67 to 8.69, such that the roll center of the vehicle is at
point S .
560
8. Suspension Mechanisms
4
5
Body
8
2
3
6
7
1
FIGURE 8.62.
4
Body
5
8
2
3
6
7
1
FIGURE 8.63.
P
FIGURE 8.64.
P
FIGURE 8.65.
8. Suspension Mechanisms
P
FIGURE 8.66.
P
FIGURE 8.67.
P
FIGURE 8.68.
P
FIGURE 8.69.
561
562
8. Suspension Mechanisms
4. F Position of the roll center and mass center.
Figure 8.70 illustrates the wheels and mass center F of a vehicle.
Design a double D-arm suspension such that the roll center of the
C
FIGURE 8.70.
vehicle is
(a) above F.
(b) on F.
(c) below F.
(d) Is it possible to make street cars with a roll center on or above
F?
(e) What would be the advantages or disadvantages of a roll center
on or above F.
5. Asymmetric position of the roll center.
Design double D-arm suspensions for the vehicle shown in 8.71, such
that the roll center of the vehicle is at point S . What would be the
advantages or disadvantages of an asymmetric roll center?
P
FIGURE 8.71.
6. F Camber angle variation.
Consider a double D-arm suspension such that is shown in Figure
8.72. Assume that the dimensions of the equivalent kinematic model
8. Suspension Mechanisms
563
e
D
a
M
z
T2
d
C
b
c
N
x
A
y
B
T4
T0
T3
FIGURE 8.72. A double D-arm suspension kinematics.
are:
d = 22=57 cm
0 = 23=5 deg
e = 18=88 cm
f = 29=8 cm
g = 24=8 cm
and the coupler point F is at:
h = 14=8 cm
= 56=2 deg
Draw a graph to show the variation of the camber angle, when the
wheel is moving up and down.
7. F Steer axis unit vector.
Determine the F expression of the unit vector x̂ on the steer axis, for
a caster angle * = 15 deg, and a lean angle = 8 deg.
8. F Location vector and steer axis.
Determine the location vector s, if the steer axis is going through
the wheel center. The caster and lean angles are * = 10 deg and
= 0 deg.
9. F Homogeneous transformation matrix F WZ .
Determine F WZ for * = 8 deg, = 12 deg, and the location vector
F
s
£
¤W
F
s = 3=8 cm 1=8 cm Uz
(a) The vehicle uses a tire 235@35]U19.
(b) The vehicle uses a tire S 215@65U15 96K.
564
8. Suspension Mechanisms
10. F Wheel drop.
Find the coordinates of the tireprint for
£
¤
F
s = 3=8 cm 1=8 cm 38 cm
* = 10 deg
= 10 deg
if = 18 deg. How much is the wheel drop K.
11. F Wheel drop and steer angle.
Draw a plot to show the wheel drop K at dierent steer angle for
the given data in Exercise 10.
12. F Camber and steering.
Draw a plot to show the camber angle at dierent steer angle for
the following characteristics:
£
¤
F
s = 3=8 cm 0 cm 38 cm
* = 10 deg
= 0 deg
13. F Parallelogram suspension.
If the double D-arm suspension is geometrically made by a parallelogram then, the wheel will always remain perpendicular to the road.
This is the best design to have uniform tire wear. A parallelogram
suspension however, has a variable track when the load of the car
changes or the wheel moves up and down. The maximum track of
a vehicle with such a suspension occurs when the upper and lower
arms of the suspension are parallel to the ground such as the one
illustrated in Figure 8.73.
Calculate and show the track changes of the suspension for } =
±25 cm.
14. F Track change suspension.
If the rest position a parallelogram suspension is as shown in Figure
8.74 then, the track of the car will increase by increasing the load of
the car. Having a wider track, increases the stability of the car, so
such a suspension works well. Calculate and show the track changes of
the suspension for } = ±25 cm if the initial value of 2 = 4 = 75 deg.
8. Suspension Mechanisms
e
D
A
a
M
z
T2
d
b
y
C
c
N
B
T4
T3
x
FIGURE 8.73. A parallelogram double D-arm suspension.
e
M
D
a
T2
d
y
A
z
N
b
c
C
T4
B
x
T3
FIGURE 8.74. A tilted parallelogram double D-arm suspension.
565
Part III
Vehicle Dynamics
9
F Applied Dynamics
Dynamics of a rigid vehicle may be considered as the motion of a rigid
body with respect to a xed global coordinate frame. The principles of
Dynamics as well as Newton and Euler equations of motion that describe
the translational and rotational motion of the rigid body are reviewed in
this chapter.
z
x
1
y
B
2
4
C
My2
Mz2
Fx2 Mx2
3
Fx4
Mx4
Fx3
Mx3
Mz3
Fy2
Fz2
My3
Fy3
Fz3
FIGURE 9.1. The force system of a vehicle is the applied forces and moments at
the tireprints.
9.1 Elements of Dynamics
In this section, we will review the denition of the elements that are being
used in Dynamics.
9.1.1 Force and Moment
In Newtonian dynamics, the forces acting on a system of connected rigid
bodied can be divided into internal and external forces. Internal forces are
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_9, © Springer Science+Business Media New York 2014
569
570
9. F Applied Dynamics
acting between connected bodies, and external forces are acting from outside of the system. An external force can be a contact force, such as traction
force at tireprint of a driving wheel, or a body force, such as gravitational
force on vehicle’s body.
The external forces and moments are called load, and a set of forces and
moments acting on a rigid body, such as forces and moments on the vehicle
shown in Figure 9.1, is called a force system. The resultant or total force
F is the vectorial sum of all the external forces acting on a body, and the
resultant or total moment M is the vectorial sum of all the moments of the
external forces.
X
Fl
(9.1)
F =
l
M =
X
Ml
(9.2)
l
Consider a force F acting on a point S at rS . The moment of the force
about a directional line o with unit vector x̂ passing through the origin is
Mo = x̂ · (rS × F)
(9.3)
The moment of the force F, about a point T at rT is
MT = (rS rT ) × F
(9.4)
so, the moment of F about the origin is
M = rS × F
(9.5)
The moment of a force may also be called torque or moment.
The eect of a force system is equivalent to the eect of the resultant
force and resultant moment of the force system. Any two force systems
are equivalent if their resultant forces and resultant moments are equal. If
the resultant force of a force system is zero, the resultant moment of the
force system is independent of the origin of the coordinate frame. Such a
resultant moment is called couple.
When a force system is reduced to a resultant FS and MS with respect
to a reference point S , we may change the reference point to another point
T and nd the new resultants as
FT
MT
= FS
= MS + (rS rT ) × FS = MS + T rS × FS
(9.6)
(9.7)
9.1.2 Momentum
The momentum of a moving rigid body is a vector quantity equal to the
total mass of the body times the translational velocity of the mass center
of the body.
p = pv
(9.8)
9. F Applied Dynamics
571
The momentum p is also called translational momentum or linear momentum.
Consider a rigid body with momentum p. The moment of momentum,
L, about a directional line o with directional unit vector x̂ passing through
the origin is
Lo = x̂ · (rF × p)
(9.9)
where rF is the position vector of the mass center F. The moment of
momentum about the origin is
L = rF × p
(9.10)
The moment of momentum L is also called angular momentum.
9.1.3 Vectors
A vector expresses any physical quantity that can be represented by a
directed section of a line with a start point, such as R, and an end point,
such as S . A vector can be shown by an ordered pair of points with an
$
$
arrow, such as RS . Therefore, a sign S S indicates a zero vector at point
S . A vector may have up to ve characteristics: length, axis, end point,
direction, physical quantity. However, the length and direction are necessary
for every vector.
1. Length. The length of a vector corresponds to the magnitude of the
physical quantity that the vector is representing.
2. Axis. The straight line on which the vector sits, is the axis of the
vector. The vector axis is also called the line of action.
3. End point. A start or an end point, called the aecting point, indicates
the point at which the vector is applied.
4. Direction. The direction of a vector indicates at what direction on
the axis the vector is pointing.
5. Physical quantity. Any vector represents a physical quantity. If a physical quantity can be represented by a vector, it is called a vectorial
physical quantity. The magnitude of the quantity is proportional to
the length of the vector. Although a vector may be dimensionless, a
vector that represents no physical quantity is meaningless, .
Depending on the physical quantity and application, there are seven
types of vectors: vecpoint, vecline, vecface, vecfree, vecpoline, vecpoface, pecporee.
572
9. F Applied Dynamics
1. Vecpoint. When all of the vector characteristics; length, axis, end
point, direction, and physical quantity; are specied, the vector is
called a bounded vector, point vector, or vecpoint. Such vecpoint is
xed at a point with no movability.
2. Vecline. If the aecting point of a vector are not xed on the axis,
the vector is called a sliding vector, line vector, or vecline. A vecline
is free to slide on its axis.
3. Vecface. If the aecting point of a vector can move on a surface while
the vector displaces parallel to itself, the vector is called a surface
vector or vecface. If the surface is a plane, then the vector is a plane
vector or veclane.
4. Vecfree. If the axis of a vector is not xed, the vector is called a free
vector, direction vector, or vecfree. A vecfree can move parallel to
itself to any point of a specied space while keeping its direction.
5. Vecpoline. If the start point of a vector is xed while the end point
can slide on a line, the vector is a point-line vector or vecpoline. A
vecpoline has a constraint variable length and orientation. However,
if the start and end points of a vecpoline are on the sliding line, its
orientation is constant.
6. Vecpoface. If the start point of a vector is xed, while the end point
can slide on a surface, the vector is a point-surface vector or vecpoface. A vecpoface has a constraint variable length and orientation.
If the surface is a plane, the vector is called a point-plane vector or
vecpolane. The start point of a vecpoface may also be on the sliding
surface.
7. Vecporee. When the start point of a vector is xed and the end point
can move anywhere in a specied space, the vector is called a pointfree vector or vecporee. A vecporee has a variable length and orientation.
Two vectors are comparable only if they represent the same physical
quantity and are expressed in the same coordinate frame. Two vectors are
equal if they are comparable and are the same type and have the same
characteristics. Two vectors are equivalent if they are comparable and the
same type and can be substituted with each other.
Vectors can only be added if they are coaxial. In case the vectors are
not coaxial, the decomposed expression of vectors must be used to add the
vectors.
Force is a sliding vector and couple and moment are free vectors.
9. F Applied Dynamics
573
9.1.4 Equation of Motion
The application of a force system is emphasized by Newton’s second and
third laws of motion. The second law of motion, also called the Newton’s
equation of motion, states that the global rate of change of linear momentum is proportional to the global applied force.
J
J ¡
¢
gJ
g
p=
p Jv
gw
gw
J
F=
(9.11)
The third Newton’s law of motion states that the action and reaction forces
acting between two bodies are equal and opposite.
The second law of motion can be expanded to include rotational motions.
Hence, the second law of motion also states that the global rate of change
of angular momentum is proportional to the global applied moment.
J
J
M=
gJ
L
gw
(9.12)
Proof. Dierentiating from moment of momentum (9.10) shows that
μJ
¶
J
J
J
gJ
g
grF
gp
L =
(rF × p) =
× p + rF ×
gw
gw
gw
gw
J
gp
(9.13)
= J rF × J F = J M
= J rF ×
gw
9.1.5 Work and Energy
Kinetic energy N of a moving body point S with mass p at a position
J
rS , and having a velocity J vS , is
´2
1
1 ³
E
N = p J vS2 = p J db E + E vS + E
rS
(9.14)
J $E ×
2
2
where, J indicates the global coordinate frame in which the velocity vector
vS is expressed. The work done by the applied force J F on p in moving
from point 1 to point 2 on a path, indicated by a vector J r, is
Z 2
J
F · g Jr
(9.15)
1 Z2 =
1
However,
Z 2
1
J
Z 2 J
1
gJ
F·g r = p
v · J vgw = p
gw
2
1
¢
1 ¡ 2
=
p y2 y12 = N2 N1
2
J
Z 2
1
g 2
y gw
gw
(9.16)
574
9. F Applied Dynamics
that shows 1 Z2 is equal to the dierence of the kinetic energy between
terminal and initial points.
1 Z2 = N2 N1
(9.17)
Equation (9.17) is called principle of work and energy. If there is a scalar
potential eld function Y = Y ({> |> }) such that
μ
¶
CY
CY
gY
CY
=
ˆ~ +
ˆ +
F = uY = n̂
(9.18)
gr
C{
C|
C}
then the principle of work and energy simplies to the principle of conservation of energy,
(9.19)
N1 + Y1 = N2 + Y2
The value of the potential eld function Y = Y ({> |> }) is the potential
energy of the system.
Example 374 Position of mass center.
The position of the mass center of a rigid body in a coordinate frame is
indicated by E rF and is usually measured in a body coordinate frame E.
Z
1
E
E
rF =
r gp
(9.20)
p E
5 1 R
6
5
6
p E { gp
{F
9
:
R
:
7 |F 8 = 9 1
(9.21)
7 p E | gp 8
R
}F
1
p E } gp
Applying the mass center integral on the symmetric and uniform O-section
rigid body with = 1 shown in Figure 9.2 provides us with the position of
mass center F of the section. The { position of F is
Z
Z
1
1
e2 + de d2
{ gp =
{ gD = (9.22)
{F =
p E
D E
4de + 2d2
and because of symmetry, we have
|F = {F =
e2 + de d2
4de + 2d2
(9.23)
When d = e, the position of F reduces to
|F = {F =
1
e
2
(9.24)
Example 375 F Every force system is equivalent to a wrench.
The Poinsot theorem states: Every force system is equivalent to a single
force, plus a moment parallel to the force. Let F and M be the resultant force
9. F Applied Dynamics
y1
575
x
Im
I ma
a
in
y
B1
45 deg
x1
C
b
x
B
FIGURE 9.2. Principal coordinate frame for a symmetric L-section.
and moment of a force system. We decompose the moment into parallel and
perpendicular components, Mk and MB , to the force axis. The force F and
the perpendicular moment MB can be replaced by a single force F0 parallel
to F. Therefore, the force system is reduced to a force F0 and a moment
Mk parallel to each other. A force and a moment about the force axis is
called a wrench.
The Poinsot theorem is similar to the Chasles theorem that states: Every
rigid body motion is equivalent to a screw, which is a translation plus a
rotation about the axis of translation.
There is no simple relationship between screw and wrench similar to Newton equation of motion. If there was such relation as the second time derivative of screw is proportional to the applied wrench then we would have an
equation of motion as a combination of Newton and Euler equation.
Example 376 F Motion of a moving point in a moving body frame.
The velocity and acceleration of a moving point S as shown in Figure
5.9 are found in Example 216.
¡
¢
J
E
(9.25)
vS = J db E + J UE E vS + E
rS
J $E ×
¡E
¢
J
J
J
E
E
E
E
d̈E + UE
aS =
aS + 2 J $ E × vS + J $
b E × rS
¡E
¢¢
¡E
J
E
(9.26)
+ UE J $ E × J $ E × rS
Therefore, the equation of motion for the point mass S is
J
F = p J aS
³
¡
¢´
E
E
E
= p J d̈E + J UE E aS + 2 E
$
×
v
+
$
b
×
r
E
S
E
S
J
J
¡E
¢¢
¡E
J
E
(9.27)
+p UE J $ E × J $ E × rS
576
9. F Applied Dynamics
Example 377 Newton’s equation in a rotating frame.
Consider a spherical rigid body, such as Earth, with a xed point that is
rotating with a constant angular velocity $. The equation of motion for a
moving point vehicle S on the rigid body is found by setting J d̈E = E
bE =
J$
0 in the equation of motion of a moving point in a moving body frame (9.27)
¡E
¢
E
E
E
F = p E aS + p E
rS + 2p E
rb S (9.28)
J $E × J $E ×
J $E ×
6= p E aS
which shows that the Newton’s equation of motion F = p a must be modied
for rotating frames.
Example 378 Coriolis force.
The equation of motion of a moving vehicle point on the surface of the
Earth is
¡E
¢
E
E
E
F = p E aS + p E
rS + 2p E
vS
(9.29)
J $E × J $E ×
J $E ×
which can be rearranged to
¡E
¢
E
E
E
F pE
rS 2p E
vS = p E aS
J $E × J $E ×
J $E ×
(9.30)
Equation (9.30) is the equation of motion for an observer in the rotating
frame, which in this case is an observer on the Earth. The left-hand side
of this equation is called the eective force Fhi i ,
¡E
¢
E
E
rS 2p E
vS
(9.31)
Fhi i = E F p E
J $E × J $E ×
J $E ×
because it seems that the particle is moving in the body coordinate E under
the in uence of this force.
The second term is negative of the centrifugal force and pointing outward.
The maximum value of this force on the Earth is on the equator which is
about 0=3% of the acceleration of gravity.
u$
2
3
= 6378=388 × 10 ×
μ
366=25
2
24 × 3600 365=25
= 3=3917 × 102 m@ s2
¶2
(9.32)
If we add the variation of the gravitational acceleration because of a change
of radius from U = 6356912 m at the pole to U = 6378388 m on the equator,
then the variation of the acceleration of gravity becomes 0=53%. So, generally speaking, a sportsman such as a pole-vaulter who has practiced in the
north pole can show a better record in a competition held on the equator.
The third term of the eective force is called the Coriolis force or
Coriolis eect, IF , which is perpendicular to both $ and E vS . For a mass
p moving on the north hemisphere at a latitude towards the equator, we
9. F Applied Dynamics
577
should provide a lateral eastward force equal to the Coriolis eect to force
the mass, to keep its direction relative to the ground.
E
vp = 1=4584 × 104 E pp cos kg m@ s2
IF = 2p E
J $E ×
(9.33)
The Coriolis eect is the reason why the west side of railways, roads, and
rivers wear faster than east side. The lack of providing the Coriolis force is
the reason for turning the direction of winds, projectiles, ood, and falling
objects.
Example 379 Work, force, and kinetic energy in a unidirectional motion.
A vehicle with mass p = 1200 kg has an initial kinetic energy N =
6000 J. The mass is under a constant force F = I Lˆ = 4000Lˆ and moves
from [(0) = 0 to [(wi ) = 1000 m at a terminal time wi . The work done
by the force during this motion is
Z r(wi )
Z 1000
Z =
F · gr =
4000 g[ = 4 × 106 N m = 4 MJ
(9.34)
r(0)
0
The kinetic energy at the terminal time is
N(wi ) = Z + N(0) = 4006000 J
which shows that the terminal speed of the mass is
r
2N(wi )
y2 =
81=7 m@ s 22=694 km@ h
p
(9.35)
(9.36)
Example 380 Direct and inverse dynamics dynamics.
When the applied force is time varying and is a known function, then,
F(w) = p r̈
(9.37)
The general solution for the equation of motion can be found by integration.
Z
1 w
F(w)gw
(9.38)
rb (w) = rb (w0 ) +
p w0
Z Z
1 w w
r(w) = r(w0 ) + rb (w0 )(w w0 ) +
F(w)gw gw
(9.39)
p w0 w0
This kind of problem is called direct or forward dynamics.
If the path of motion r(w) is a known function of time, then the required
force to move the system on the path would be found by dierentiation.
F(w) =
g2
(p r̈)
gw2
This kind of problem is called indirect or inverse dynamics.
(9.40)
578
9. F Applied Dynamics
Example 381 F Force function in equation of motion.
By denition, force is qualitatively whatever that changes the motion,
and quantitatively, is whatever that is equal to mass times acceleration.
Mathematically, the equation of motion provides us with a vectorial secondorder dierential equation
pr̈ = F (br> r> w)
(9.41)
It is assumed that the force function is generally only a function of time w,
position r, and velocity rb . In other words, the Newton equation of motion
is correct as long as we can show that the force is only a function of rb > r> w.
If there is an applied force that depends on the acceleration, jerk, or other
variables that cannot be reduced to rb > r> w, the system is not Newtonian.
There is no known equation of motion for non-Newtonian dynamic systems,
because
...
F (r> rb > r̈> r > · · · > w) 6= pr̈
(9.42)
In Newtonian mechanics, we assume that force can only be a function of
rb > r> w. In real world, force may be a function of everything: however, we
always ignore any other variables than rb > r> w.
Because Equation (9.41) is a linear equation for force F, it accepts the
force superposition principle. When a mass p is aected by several forces
F1 , F2 , F3 , · · ·, we may calculate their summation vectorially
F = F1 + F2 + F3 + · · ·
(9.43)
and apply the resultant force on p. Therefore, if a force F1 provides us
with acceleration r̈1 , and F2 provides us with r̈2 ,
pr̈1 = F1
pr̈2 = F2
(9.44)
then the resultant force F3 = F1 + F2 provides us with the acceleration r̈3
such that
r̈3 = r̈1 + r̈2
(9.45)
To see that the Newton equation of motion is not correct when the force
is not only a function of rb > r> w, let us assume that a particle with mass p
is under two acceleration dependent forces I1 (¨
{) and I2 (¨
{) on {-axis.
{1 )
p¨
{1 = I1 (¨
p¨
{2 = I2 (¨
{2 )
(9.46)
The acceleration of p under the action of both forces would be {
¨3
p¨
{3 = I1 (¨
{3 ) + I2 (¨
{3 )
(9.47)
{
¨3 = {
¨1 + {
¨2
(9.48)
¨2 ) = I1 (¨
{1 + {
¨2 ) + I2 (¨
{1 + {
¨2 )
p (¨
{1 + {
6= I1 (¨
{1 ) + I2 (¨
{2 )
(9.49)
however, we must have
while we have:
9. F Applied Dynamics
579
z
Z
x
B
o
B
rP
dm
df
P
G
dB
G
rP
y
O
G
X
Y
FIGURE 9.3. A body point mass moving with velocity J vS and acted on by
force gf .
9.2 Rigid Body Translational Dynamics
Figure 9.3 depicts a moving body E in a global coordinate frame J. Assume
that the body frame is attached at the mass center of the body. Point S
indicates an innitesimal sphere of the body, which has a very small mass
gp. The point mass gp is acted on by an innitesimal force gf and has a
global velocity J vS .
According to Newton’s law of motion we have
gf = J aS gp
(9.50)
However, the equation of motion for the whole body in the global coordinate
frame is
J
F = p J aE
(9.51)
which can be expressed in the body coordinate frame as
E
E
F = pE
aE + p E
vE
J $E ×
6
6
5J
pd{ + p ($ | y} $ } y| )
I{
7 I| 8 = 7 pd| p ($ { y} $ } y{ ) 8
I}
pd} + p ($ { y| $ | y{ )
5
(9.52)
(9.53)
In these equations, aE is the acceleration vector of the body mass center F
in the global frame, p is the total mass of the body, and F is the resultant
of the external forces acted on the body at F.
Proof. A body coordinate frame at the mass center is called a central
frame. If frame E is central, then the center of mass, F, is dened such
580
9. F Applied Dynamics
that
Z
E
rgp gp = 0
(9.54)
E
The global position vector of gp is related to its local position vector by
J
rgp = J dE + J UE E rgp
(9.55)
J
where dE is the global position vector of the central body frame, and
therefore,
Z
Z
Z
J
J
J
E
rgp gp =
dE gp + UE
rgp gp
E
E
p
Z
Z
J
dE gp = J dE
gp = p J dE
(9.56)
=
E
E
A time derivative of both sides shows that
Z
Z
J
rb gp gp =
p J db E = p J vE =
E
and another derivative is
p J vb E = p J aE =
Z
J
J
vgp gp
(9.57)
E
vb gp gp
(9.58)
E
However, we have gf = J vb S gp and therefore,
Z
gf
p J aE =
(9.59)
E
The integral on the right-hand side is the resultant of all the forces acting
on the body. The internal forces cancel one another out, so the net result
is the vector sum of all the externally applied forces, F, and therefore,
J
F = p J aE = p J vb E
(9.60)
In the body coordinate frame we have
E
F = E UJ J F = p E UJ J aE = p E
J aE
E
E
E
= p aE + p J $ E × vE
(9.61)
The expanded form of the Newton’s equation in the body coordinate
frame is then equal to
E
F = p E aE + p E
$ E × E vE
6
5
6 J 5
6 5
6
I{
d{
${
y{
7 I| 8 = p 7 d| 8 + p 7 $ | 8 × 7 y| 8
I}
d}
$}
y}
5
6
pd{ + p ($ | y} $ } y| )
= 7 pd| p ($ { y} $ } y{ ) 8
pd} + p ($ { y| $ | y{ )
5
(9.62)
(9.63)
9. F Applied Dynamics
Y
T
x
O
G
581
X
y
B
T
R
mg
FIGURE 9.4. A ribbon of negligible weight and thickness that is wound tightly
around a uniform massive disc.
Example 382 A wound ribbon.
Figure 9.4 illustrates a ribbon of negligible weight and thickness that is
wound tightly around a uniform massive disc of radius U and mass p. The
ribbon is fastened to a rigid support, and the disc is released to roll down
vertically. There are two forces acting on the disc during the motion, its
weight pj and the tension of the ribbon W . The translational equation of
motion of the disc is expressed easier in the global coordinate frame:
X
(9.64)
I\ = pj + W = p\¨
The rotational equation of motion is simpler if expressed in the body coordinate frame:
X
E E
P} = W U = E L E
bE + E
L J $ E = L ¨
(9.65)
J$
J $E ×
There is a constraint between the coordinates \ and :
\ = \0 U
(9.66)
To solve the motion, let us eliminate W between (9.64) and (9.65) to obtain
L¨
U
(9.67)
and use the constraint to eliminate \¨ :
¨ = μ pj ¶
L
+ pU
U
(9.68)
p\¨ = pj +
9. F Applied Dynamics
582
Now, we can nd W , \¨ and \ :
W
\¨
L¨
L
=
pj
2
U
pU + L
pU2
pU2
= j
\
=
jw2 + \b (0) w + \ (0)
pU2 + L
pU2 + L
=
(9.69)
(9.70)
For a point mass with L = 0, the falling acceleration is the same as the free
fall of a particle. However, being a rigid body and having L 6= 0, the falling
acceleration of the disc will be less. This is because the kinetic energy of the
disc splits between rotation and translation.
9.3 Rigid Body Rotational Dynamics
The rigid body rotational equation of motion is the Euler equation
E
J
M =
=
g E
E
b E
L = E L+
L
J $E ×
gw
¡
¢
E E
E E
L J$
bE + E
L J $E
J $E ×
(9.71)
where L is the angular momentum
E
L = EL E
J $E
and L is the mass moment of the rigid body.
5
6
L{{ L{| L{}
L = 7 L|{ L|| L|} 8
L}{ L}| L}}
(9.72)
(9.73)
The elements of L are functions of the mass distribution of the rigid body
and can be dened by
Z
¡ 2
¢
Llm =
ul pq {lp {mq gp
l> m = 1> 2> 3
(9.74)
E
where lm is Kronecker’s delta.
pq =
½
1 li
0 li
p=q
p 6= q
(9.75)
The expanded form of the Euler equation (9.71) is
P{
= L{{ $b { + L{| $b | + L{} $b } (L|| L}} ) $ | $ }
¡
¢
L|} $ 2} $ 2| $ { ($ } L{| $ | L{} )
(9.76)
9. F Applied Dynamics
P|
P}
= L|{ $b { + L|| $b | + L|} $b } (L}} L{{ ) $ } $ {
¡
¢
L{} $ 2{ $ 2} $ | ($ { L|} $ } L{| )
= L}{ $b { + L}| $b | + L}} $b } (L{{ L|| ) $ { $ |
¡
¢
L{| $ 2| $ 2{ $ } ($ | L{} $ { L|} )
583
(9.77)
(9.78)
which can be reduced to
P1
P2
P3
= L1 $b 1 (L2 L2 ) $ 2 $ 3
= L2 $b 2 (L3 L1 ) $ 3 $ 1
= L3 $b 3 (L1 L2 ) $ 1 $ 2
(9.79)
in a special Cartesian coordinate frame called the principal coordinate
frame. The principal coordinate frame is denoted by numbers 123 to indicate the rst, second, and third principal axes. The parameters Llm > l 6= m
are zero in the principal frame. The body and principal coordinate frame
sit at the mass center F.
Kinetic energy of a rotating rigid body is
N
=
=
¢
1¡
L{{ $ 2{ + L|| $ 2| + L}} $ 2} L{| $ { $ | L|} $ | $ } L}{ $ } $ {
2
1
1
(9.80)
$ · L = $W L $
2
2
that in the principal coordinate frame reduces to
N=
¢
1¡
L1 $ 21 + L2 $ 22 + L3 $ 23
2
(9.81)
Proof. Let pl be the mass of the lth particle of a rigid body E, which is
made of q particles and
£
¤W
(9.82)
rl = E rl = {l |l }l
is the Cartesian position vector of pl in a central body xed coordinate
frame R{|}. Assume that
£
¤W
${ $| $}
(9.83)
$=E
J $E =
is the angular velocity of the rigid body with respect to the ground, expressed in the body coordinate frame.
The angular momentum of pl is
Ll
= rl × pl rb l = pl [rl × ($ × rl )]
= pl [(rl · rl ) $ (rl · $) rl ]
= pl ul2 $ pl (rl · $) rl
(9.84)
584
9. F Applied Dynamics
Hence, the angular momentum of the rigid body would be
L=$
q
X
pl ul2 l=1
q
X
l=1
pl (rl · $) rl
(9.85)
Substitution for rl and $ provides us with
q
³
´X
¡
¢
pl {2l + |l2 + }l2
$ {ˆ~ + $ | ˆ + $ } n̂
L =
l=1
q
X
l=1
³
´
pl ({l $ { + |l $ | + }l $ } ) · {lˆ~ + |l ˆ + }l n̂
(9.86)
and therefore,
L =
q
X
l=1
+
q
X
¡
¢
¡
¢
pl {2l + |l2 + }l2 $ {ˆ~ +
pl {2l + |l2 + }l2 $ | ˆ
l=1
q
X
l=1
q
X
l=1
q
X
¡
¢
pl {2l + |l2 + }l2 $ } n̂
pl ({l $ { + |l $ | + }l $ } ) {lˆ~ q
X
pl ({l $ { + |l $ | + }l $ } ) |l ˆ
l=1
pl ({l $ { + |l $ | + }l $ } ) }l n̂
(9.87)
l=1
or
L =
q
X
pl
l=1
+
q
X
£¡ 2
¢
¤
{l + |l2 + }l2 $ { ({l $ { + |l $ | + }l $ } ) {l ˆ~
pl
l=1
+
q
X
l=1
pl
£¡ 2
¢
¤
{l + |l2 + }l2 $ | ({l $ { + |l $ | + }l $ } ) |l ˆ
£¡ 2
¢
¤
{l + |l2 + }l2 $ } ({l $ { + |l $ | + }l $ } ) }l n̂
(9.88)
9. F Applied Dynamics
585
which can be rearranged as
q
q
X
X
£ ¡ 2
¢¤
£ ¡ 2
¢¤
L =
pl |l + }l2 $ {ˆ~ +
pl }l + {2l $ | ˆ
l=1
l=1
q
X
£ ¡ 2
¢¤
pl {l + |l2 $ } n̂
+
l=1
à q
X
l=1
à q
X
l=1
à q
X
(pl {l |l ) $ | +
(pl |l }l ) $ } +
l=1
q
X
(pl }l {l ) $ { +
l=1
!
q
X
(pl {l }l ) $ } ˆ~
!
(pl |l {l ) $ { ˆ
l=1
q
X
(pl }l |l ) $ |
l=1
!
n̂
(9.89)
By introducing the mass moment matrix L with the following elements,
q
X
£ ¡ 2
¢¤
L{{ =
pl |l + }l2
(9.90)
l=1
L||
L}}
q
X
£ ¡ 2
¢¤
=
pl }l + {2l
=
l=1
q
X
l=1
L{|
L|}
L}{
£ ¡ 2
¢¤
pl {l + |l2
= L|{ = = L}| = = L{} = q
X
l=1
q
X
l=1
q
X
(9.91)
(9.92)
(pl {l |l )
(9.93)
(pl |l }l )
(9.94)
(pl }l {l )
(9.95)
l=1
we can write the angular momentum L in a concise form
O{
O|
O}
= L{{ $ { + L{| $ | + L{} $ }
= L|{ $ { + L|| $ | + L|} $ }
= L}{ $ { + L}| $ | + L}} $ }
or in a matrix form
5
6
5
O{
L{{
7 O| 8 = 7 L|{
O}
L}{
L = L ·$
L{|
L||
L}|
65
6
${
L{}
L|} 8 7 $ | 8
L}}
$}
(9.96)
(9.97)
(9.98)
(9.99)
(9.100)
586
9. F Applied Dynamics
For a rigid body that is a continuous solid, the summations must be replaced by integrations over the volume of the body as in Equation (9.74).
The Euler equation of motion for a rigid body is
E
J
M=
g E
L
gw
(9.101)
where E M is the resultant of the external moments applied on the rigid
body. The angular momentum vector E L is dened in the body coordinate
frame E. Hence, its time derivative in the global coordinate frame is
J
g EL
E
b E
L
= E L+
J $E ×
gw
(9.102)
Therefore,
E
M=
gL
b + $ × L = L$
=L
b + $× (L$)
gw
(9.103)
or in expanded form
E
M = (L{{ $b { + L{| $b | + L{} $b } ) ˆ~ + $ | (L{} $ { + L|} $ | + L}} $ } ) ˆ~
$ } (L{| $ { + L|| $ | + L|} $ } ) ˆ~
+ (L|{ $b { + L|| $b | + L|} $b } ) ˆ + $ } (L{{ $ { + L{| $ | + L{} $ } ) ˆ
$ { (L{} $ { + L|} $ | + L}} $ } ) ˆ
+ (L}{ $b { + L}| $b | + L}} $b } ) n̂ + $ { (L{| $ { + L|| $ | + L|} $ } ) n̂
$ | (L{{ $ { + L{| $ | + L{} $ } ) n̂
(9.104)
and therefore, the most general form of the Euler equations of motion for
a rigid body in a body frame attached to F are
P{
P|
P}
= L{{ $b { + L{| $b | + L{} $b } (L|| L}} ) $ | $ }
¡
¢
L|} $ 2} $ 2| $ { ($ } L{| $ | L{} )
= L|{ $b { + L|| $b | + L|} $b } (L}} L{{ ) $ } $ {
¡
¢
L{} $ 2{ $ 2} $ | ($ { L|} $ } L{| )
= L}{ $b { + L}| $b | + L}} $b } (L{{ L|| ) $ { $ |
¡
¢
L{| $ 2| $ 2{ $ } ($ | L{} $ { L|} )
(9.105)
(9.106)
(9.107)
Assume that we are able to rotate the body frame about its origin to nd
an orientation that makes Llm = 0, for l 6= m. In such a coordinate frame,
which is the principal frame, the Euler equations reduce to
P1
P2
P3
= L1 $b 1 (L2 L2 ) $ 2 $ 3
= L2 $b 2 (L3 L1 ) $ 3 $ 1
= L3 $b 3 (L1 L2 ) $ 1 $ 2
(9.108)
(9.109)
(9.110)
9. F Applied Dynamics
Y
y
587
B
T
G
A
B
x
FA
Z
X
FB
l
FIGURE 9.5. A disc with mass p and radius u, mounted on a massless turning
shaft.
The kinetic energy of a rigid body may be found by the integral of the
kinetic energy of a mass element gp, over the whole body.
Z
Z
1
1
vb 2 gp =
($ × r) · ($ × r) gp
N =
2 E
2 E
Z
Z
Z
¡ 2
¢
¡ 2
¢
¡ 2
¢
$ 2|
$2
$ 2{
| + } 2 gp +
} + {2 gp + }
{ + | 2 gp
=
2 E
2 E
2 E
Z
Z
Z
$ { $ |
{| gp $ | $ }
|} gp $ } $ {
}{ gp
E
E
¢
1¡
L{{ $ 2{ + L|| $ 2| + L}} $ 2}
=
2
L{| $ { $ | L|} $ | $ } L}{ $ } $ {
E
(9.111)
The kinetic energy can be rearranged to a matrix multiplication form
N=
1
1 W
$ L$ = $·L
2
2
(9.112)
When the body frame is principal, the kinetic energy will simplify to
N=
¢
1¡
L1 $ 21 + L2 $ 22 + L3 $ 23
2
(9.113)
Example 383 A tilted disc on a massless shaft.
Figure 9.5 illustrates a disc with mass p and radius u, mounted on a
massless shaft. The shaft is turning with a constant angular speed $. The
disc is attached to the shaft at an angle . Because of , the bearings at D
and E must support a rotating force.
588
9. F Applied Dynamics
To analyze the system, we attach a principal body coordinate frame at
the disc center as shown in the gure. The angular velocity vector in the
body frame is
E
~ + $ sin ˆ
(9.114)
J $ E = $ cos ˆ
and the mass moment matrix is
5
2pu2
1
E
7
0
L=
4
0
0
pu2
0
6
0
0 8
pu2
(9.115)
Substituting (9.114) and (9.115) in (9.108)-(9.110), with 1 {, 2 |,
3 }, yields
P| = 0
P{ = 0
P} =
pu2
$ cos sin 4
(9.116)
Therefore, the bearing reaction forces ID and IE are
ID = IE = P}
pu2
=
$ cos sin o
4o
(9.117)
Example 384 Steady rotation of a freely rotating rigid body.
The Newton-Euler equations of motion for a rigid body are
J
E
F = p J vb
E
M = L E
bE + E
L
J$
J $E ×
(9.118)
(9.119)
Consider a situation in which the resultant applied force and moment on
the body are zero.
J
F = EF = 0
J
M = EM = 0
(9.120)
Based on the Newton’s equation, the velocity of the mass center will be
constant in the global coordinate frame. However, the Euler equation reduces
to
$b 1
=
$b 2
=
$b 3
=
L2 L3
$2 $3
L1
L3 L1
$3 $1
L22
L1 L2
$1 $2
L3
(9.121)
(9.122)
(9.123)
that show the angular velocity can be constant if
L1 = L2 = L3
(9.124)
or if two principal moments of inertia, say L1 and L2 , are zero and the third
angular velocity, in this case $ 3 , is initially zero, or if the angular velocity
vector is initially parallel to a principal axis.
9. F Applied Dynamics
589
Z
zA
.
M
A
X
G
.
\
yA
xA z
r
B
Y
xB
B
FIGURE 9.6. A two-link manipulator.
Example 385 Angular momentum of a two-link manipulator.
A two-link manipulator is shown in Figure 9.6. Link D rotates with angular velocity *b about the }-axis of its local coordinate frame. Link E is
attached to link D and has angular velocity #b with respect to D about the
{D -axis. We assume that D and J were coincident at * = 0, therefore, the
rotation matrix between D and J is
5
6
cos *(w) sin *(w) 0
J
UD = 7 sin *(w) cos *(w) 0 8
0
0
1
(9.125)
Frame E is related to frame D by Euler angles * = 90 deg, = 90 deg, and
#, hence,
5
D
UE
ff# fvv# fv# ff#v
= 7 f#v + ffv# vv# + fff#
vv#
vf#
6
5
cos # sin # 0
7 sin #
cos # 0 8
0
0
1
6
vv
fv 8
f
(9.126)
590
9. F Applied Dynamics
and therefore,
J
UE
J
UD D UE
cos * cos # sin * sin #
= 7 cos * sin # cos # sin *
0
=
5
(9.127)
6
cos * sin # cos # sin * 0
cos * cos # + sin * sin # 0 8
0
1
The angular velocity of D in J, and E in D are
b
b N̂
J $D = *
~D
D $ E = #ˆ
(9.128)
Moment of inertia matrices for the arms D and E can be dened as
5
5
6
6
LD1
LE1
0
0
0
0
D
E
0 8
0 8
LD = 7 0 LD2
LE = 7 0 LE2
(9.129)
0
0 LD3
0
0
LE3
These moments of inertia must be transformed to the global frame
J
W
LD = J UE D LD J UD
J
W
LE = J UE E LE J UE
(9.130)
The total angular momentum of the manipulator is
J
L = J LD + J LE
(9.131)
LD J $ D
¡
¢
LE J $ E = J LE J
D $E + J $D
(9.132)
(9.133)
where
J
LD
J
LE
=
=
J
J
Example 386 F Poinsot’s construction.
Consider a freely rotating rigid body with an attached principal coordinate
frame. Having M = 0 provides a motion under constant angular momentum
and constant kinetic energy
L = L $ = fwh
1 W
N =
$ L $ = fwh
2
(9.134)
(9.135)
Because the length of the angular momentum L is constant, the equation
O2 = L · L = O2{ + O2| + O2} = L12 $ 21 + L22 $ 22 + L32 $ 23
(9.136)
introduces an ellipsoid in the ($ 1 > $ 2 > $ 3 ) coordinate frame, called the momentum ellipsoid. The tip of all possible angular velocity vectors must lie
on the surface of the momentum ellipsoid. The kinetic energy also denes
an energy ellipsoid in the same coordinate frame so that the tip of the
angular velocity vectors must also lie on its surface.
N=
¢
1¡
L1 $ 21 + L2 $ 22 + L3 $ 23
2
(9.137)
9. F Applied Dynamics
L3
591
Energy ellipsoid
Momentum ellipsoid
L1
L2
FIGURE 9.7. Intersection of the momentum and energy ellipsoids.
In other words, the dynamics of moment-free motion of a rigid body requires
that the corresponding angular velocity $(w) satisfy both Equations (9.136)
and (9.137) and therefore lie on the intersection of the momentum and
energy ellipsoids.
For clarity, we may dene the ellipsoids in the (O{ > O| > O} ) coordinate
system as
O2{ + O2| + O2}
O2{
2L1 N
+
O2|
2L2 N
+
O2}
2L3 N
= O2
(9.138)
= 1
(9.139)
Equation
(9.138) is a sphere and Equation (9.139) is an ellipsoid with
s
2Ll N as semi-axes. To have a meaningful motion, these two shapes must
intersect. The intersection forms a trajectory for the tip point of L, as
shown in Figure 9.7.
It can be deduced that for a certain value of angular momentum there are
maximum and minimum limit values for acceptable kinetic energy. Assuming
(9.140)
L1 A L3 A L3
the limits of possible kinetic energy are
Nmin =
O2
2L1
Nmax =
O2
2L3
(9.141)
and the corresponding motions are turning about the axes L1 and L3 respectively.
Example 387 F Alternative derivation of Euler equations of motion.
Assume that the moment of the small force gf is shown by gm and a
mass element is shown by gp, then,
gm = J rgp × gf = J rgp × J vb gp gp
(9.142)
9. F Applied Dynamics
592
The global angular momentum gl of gp is equal to
gl = J rgp × J vgp gp
(9.143)
and according to (9.12) we have
J
gm =
J
rgp × gf
=
g
gl
gw
J ¡
¢
g J
rgp × J vgp gp
gw
(9.144)
(9.145)
Integrating over the body results in
Z
J
E
rgp × gf
=
=
Z
¢
g ¡J
rgp × J vgp gp
gw
E
J Z ¡
¢
g
J
rgp × J vgp gp
gw E
J
(9.146)
However, we have
J
rgp = J dE + J UE E rgp
(9.147)
where J dE is the global position vector of the central body frame, can simplify the left-hand side of the integral to
Z
Z
¡J
¢
J
rgp × gf =
dE + J UE E rgp × gf
E
Z
ZE
J
J
dE × gf +
=
E rgp × gf
=
E
J
E
dE × J F + J MF
(9.148)
where MF is the resultant external moment about the body mass center F.
The right-hand side of Equation (9.146) is
J
g
gw
Z
¡J
¢
rgp × J vgp gp
E
J Z ¡¡
¢
¢
g
J
dE + J UE E rgp × J vgp gp
=
gw E
J Z ¡
J Z ¡
¢
¢
g
g
J
J
J
J
dE × vgp gp +
vgp gp
=
E rgp ×
gw E
gw E
¶ J
μ
Z
J
g J
g
J
dE ×
vgp gp +
=
LF
gw
gw
E
Z
Z
g
J
J
= J db E ×
vgp gp + J dE ×
vb gp gp + LF (9.149)
gw
E
E
9. F Applied Dynamics
593
We use LF for moment of momentum about the body mass center. Because
the body frame is at the mass center, we have
Z
J
rgp gp = p J dE = p J rF
(9.150)
E
Z
J
vgp gp = p J db E = p J vF
(9.151)
E
Z
J
vb gp gp = p J d̈E = p J aF
(9.152)
E
and therefore,
J Z ¡
J
¢
g
gJ
J
rgp × J vgp gp = J dE × J F +
LF
gw E
gw
(9.153)
Substituting (9.148) and (9.153) in (9.146) provides us with the Euler equation of motion in the global frame, indicating that the resultant of externally
applied moments about F is equal to the global derivative of angular momentum about F.
J
gJ
J
MF =
LF
(9.154)
gw
The Euler equation in the body coordinate can be found by transforming
(9.154)
E
MF
=
=
J
J
g
gJ W
gE
UE LF =
LF
LF =
gw
gw
gw
Eb
E
LF + E
LF
(9.155)
J $E ×
J
W J
W
UE
MF = J UE
J
9.4 Mass Moment Matrix
In analyzing the motion of rigid bodies, two types of integrals arise that
belong to the geometry of the body. The rst type denes the center of mass
and is important when the translation motion of the body is considered.
The second is the mass moment that appears when the rotational motion of
the body is considered. The mass moment is also called moment of inertia,
centrifugal moments, or deviation moments. Every rigid body has a 3 × 3
moment of inertia matrix L, which is denoted by
5
6
L{{ L{| L{}
L = 7 L|{ L|| L|} 8
(9.156)
L}{ L}| L}}
The diagonal elements Llm > l = m are called polar moments of inertia
Z
¡ 2
¢
L{{ = L{ =
| + } 2 gp
(9.157)
E
594
9. F Applied Dynamics
L|| = L| =
Z
E
L}} = L} =
Z
E
¡ 2
¢
} + {2 gp
¡ 2
¢
{ + | 2 gp
(9.158)
(9.159)
and the o-diagonal elements Llm > l 6= m are called products of inertia
Z
L{| = L|{ = {| gp
(9.160)
E
L|} = L}| = Z
|} gp
(9.161)
}{ gp
(9.162)
E
L}{ = L{} = Z
E
The elements of L for a rigid body, made of discrete point masses, are
dened in Equation (9.74).
The elements of L are calculated about a body coordinate frame attached
to the mass center F of the body. Therefore, L is a frame-dependent quantity and must be written like E L to show the frame it is computed in.
E
L
5
Z
|2 + }2
{|
7
{|
} 2 + {2
=
E
}{
|}
Z
¡ 2
¢
=
u I r rW gp
E
Z
ũ ũ gp
=
6
}{
|} 8 gp
2
{ + |2
(9.163)
(9.164)
(9.165)
E
Mass moments can be transformed from a coordinate frame E1 to another
coordinate frame E2 , both installed at the mass center of the body, according to the rule of the rotated-axes theorem
E2
W
L = E2 UE1 E1 L E2 UE
1
(9.166)
Transformation of the mass moment from a central frame E1 located at
rF to another frame E2 , which is parallel to E1 , is, according to the rule
of parallel-axes theorem,
E2
E2
W
L = E1 L + p ũF ũF
(9.167)
If the local coordinate frame R{|} is located such that the products of
inertia vanish, the local coordinate frame is called the principal coordinate
frame and the associated moments of inertia are called principal moments
9. F Applied Dynamics
595
of inertia. Principal axes and principal moments of inertia can be found by
solving the following equation for L:
¯
¯
¯ L{{ L
L{|
L{} ¯¯
¯
¯ L|{
L|| L
L|} ¯¯ = 0
(9.168)
¯
¯ L}{
L}|
L}} L ¯
det ([Llm ] L [ lm ]) = 0
(9.169)
Since Equation (9.169) is a cubic equation in L, we obtain three eigenvalues
L1 = L{
L2 = L|
L3 = L}
(9.170)
that are the principal moments of inertia.
Proof. Consider two coordinate frames with a common origin at the mass
center of a rigid body as shown in Figure (d). The angular velocity and
angular momentum of a rigid body transform from the frame E1 to the
frame E2 by vector transformation rule
E2
$ =
E2
L =
E2
UE1 E1 $
E2
UE1 E1 L
(9.171)
(9.172)
However, L and $ are related according to Equation (9.72)
E1
L = E1 L E1 $
(9.173)
W E2
L = E2 UE1 E1 L E2 UE
$ = E2 L E2 $
1
(9.174)
and therefore,
E2
which shows how to transfer the moment of inertia from the coordinate
frame E1 to a rotated frame E2
E2
W
L = E2 UE1 E1 L E2 UE
1
(9.175)
Now consider a central frame E1 , shown in Figure 9.8(e), at E2 rF , which
rotates about the origin of a xed frame E2 such that their axes remain
parallel. The angular velocity and angular momentum of the rigid body
transform from frame E1 to frame E2 by
E2
E2
$ =
L =
E1
E1
$
L + (rF × pvF )
(9.176)
(9.177)
Therefore,
E2
¢
¡
L + p E2 rF × E2 $× E2 rF
¢ E2
¡
W
= E1 L + p E2 ũF E2 ũF
$
¢
¡E1
E2 W E2
E2
L + p ũF ũF
$
=
L =
E1
(9.178)
9. F Applied Dynamics
596
z1
z1
z2
B2
z2
B1
B2
o2
x2
x1
B1
y2
y1
x2
rC
C
x1
( a)
y2
( b)
FIGURE 9.8. (d) Two coordinate frames with a common origin at the mass center
of a rigid body. (e) A central coordinate frame E1 and a translated frame E2 .
which shows how to transfer the mass moment from frame E1 to a parallel
frame E2
W
E2
L = E1 L + p ũF ũF
(9.179)
The parallel-axes theorem is also called the Huygens-Steiner theorem.
Referring to Equation (9.175) for transformation of the mass moment to
a rotated frame, we can always nd a frame in which E2 L is diagonal. In
such a frame, we have
E2
UE1 E1 L = E2 L E2 UE1
(9.180)
or
5
u11
7 u21
u31
u12
u22
u32
65
L{{
u13
u23 8 7 L|{
u33
L}{
5
L1 0
= 7 0 L2
0 0
6
L{}
L|} 8
L}}
65
u11 u12
0
0 8 7 u21 u22
L3
u31 u32
L{|
L||
L}|
6
u13
u23 8
u33
(9.181)
which shows that L1 , L2 , and L3 are eigenvalues of E1 L. These eigenvalues
can be found by solving the following equation for :
¯
¯
¯ L{{ L{|
L{} ¯¯
¯
¯ L|{
L|| L|} ¯¯ = 0
(9.182)
¯
¯ L}{
L}|
L}} ¯
The eigenvalues L1 , L2 , and L3 are principal mass moments, and their associated eigenvectors are called principal directions. The coordinate frame
y1
9. F Applied Dynamics
597
made by the eigenvectors is the principal body coordinate frame. In the
principal coordinate frame, the rigid body angular momentum is
5
6 5
65
6
O1
L1 0 0
$1
7 O2 8 = 7 0 L2 0 8 7 $ 2 8
(9.183)
0 0 L3
O3
$3
Example 388 Principal moments of inertia.
Consider the inertia matrix L
5
6
20 2 0
L = 7 2 30 0 8
0
0 40
(9.184)
We set up the determinant (9.169)
¯
¯
¯
¯ 20 2
0
¯
¯
¯=0
¯ 2
30
0
¯
¯
¯
0
0
40 ¯
(9.185)
which leads to the following characteristic equation.
(20 ) (30 ) (40 ) 4 (40 ) = 0
(9.186)
Three roots of Equation (9.186) are
L1 = 30=385
L2 = 19=615
L3 = 40
and therefore, the principal moment of inertia matrix is
5
6
30=385
0
0
0
19=615 0 8
L=7
0
0
40
Example 389 Principal coordinate frame.
Consider the inertia matrix L
5
6
20 2 0
L = 7 2 30 0 8
0
0 40
the direction of a principal axis {l is established by solving
5
65
6 5 6
L{{ Ll
cos l
0
L{|
L{}
7 L|{
8 7 cos l 8 = 7 0 8
L|| Ll
L|}
0
L}{
L}|
L}} Ll
cos l
(9.187)
(9.188)
(9.189)
(9.190)
598
9. F Applied Dynamics
for direction cosines, which must also satisfy
cos2 l + cos2 l + cos2 l = 1
(9.191)
For the rst principal moment of inertia L1 = 30=385 we have
65
6 5 6
5
cos 1
0
20 30=385
2
0
8 7 cos 1 8 = 7 0 8
7
2
30 30=385
0
0
0
40 30=385
cos 1
0
(9.192)
or
10=385 cos 1 2 cos 1 + 0 = 0
2 cos 1 0=385 cos 1 + 0 = 0
0 + 0 + 9=615 cos 1 = 0
(9.193)
(9.194)
(9.195)
and we obtain
1 = 79=1 deg
1 = 169=1 deg
1 = 90=0 deg
Using L2 = 19=615 for the second principal axis
65
6 5 6
5
cos 2
0
20 19=62
2
0
8 7 cos 2 8 = 7 0 8
7
2
30 19=62
0
0
0
40 19=62
cos 2
0
(9.196)
(9.197)
we obtain
2 = 10=9 deg
2 = 79=1 deg
2 = 90=0 deg
The third principal axis is for L3 = 40
65
6 5 6
5
cos 3
0
20 40
2
0
8 7 cos 3 8 = 7 0 8
7 2
30 40
0
0
0
40 40
cos 3
0
(9.198)
(9.199)
which leads to
3 = 90=0 deg
3 = 90=0 deg
3 = 0=0 deg
(9.200)
Example 390 Mass moment of a rigid rectangular bar.
Consider a homogeneous rectangular link with mass p, length o, width
z, and height k, as shown in Figure 9.9(d).
The local central coordinate frame is attached to the link at its mass
center. The moments of inertia matrix of the link can be found by the
integral method. We begin with calculating L{{
Z
Z
Z
¡ 2
¢
¡ 2
¢
¡ 2
¢
p
| + } 2 gp =
| + } 2 gy =
| + } 2 gy
L{{ =
ozk y
E
y
Z k@2 Z z@2 Z o@2
¡
¢
¢
p
p¡ 2
=
| 2 + } 2 g{ g| g} =
z + k2 (9.201)
ozk k@2 z@2 o@2
12
9. F Applied Dynamics
599
z'
z
z
h
h
y'
y
y
l
l
x
w
w
(a)
x
x'
(b)
FIGURE 9.9. (d) A homogeneous rigid rectangular link. (e) A homogeneous rigid
rectangular link in the principal and non principal frames.
The L|| and L}} can be calculated similarly
¢
¢
p¡ 2
p¡2
L}} =
(9.202)
k + o2
o + z2
L|| =
12
12
The coordinate frame is central and therefore, the products of inertia
must be zero. To show this, we examine L{| .
Z
Z
L{| = L|{ = {| gp = {|gy
E
=
p
ozk
Z k@2 Z z@2 Z o@2
k@2
z@2
y
{| g{ g| g} = 0
(9.203)
o@2
Therefore, the mass moment for the rigid rectangular bar in its central
frame is
¢
5 p¡ 2
6
2
0
0
12 z + k
¡
¢
p
2
2
8
0
L =7
(9.204)
12 k + o
¡ 20 2 ¢
p
o
+
z
0
0
12
Example 391 Translation of the inertia matrix.
The moment of inertia matrix of the rigid body shown in Figure 9.9(e),
in the principal frame E(r{|}) is given in Equation (9.204). The mass
moment matrix in the non-principal frame E 0 (r{0 | 0 } 0 ) can be found by
applying the parallel-axes transformation formula (9.179).
E0
The mass center is at
0
0
W
L = E L + p E ũF E ũF
5
6
o
1
E
rF = 7 z 8
2
k
0
(9.205)
(9.206)
600
9. F Applied Dynamics
and therefore,
5
0
1
E
ũF = 7 k
2
z
0
that provides
5 1 2
1
2
3 k p + 3 pz
0
E
1
7
4 opz
L =
14 kop
6
k z
0 o 8
o
0
6
14 kop
8
14 kpz
1 2
1
2
3 o p + 3 pz
14 opz
1 2
1 2
3k p + 3o p
1
4 kpz
Example 392 Principal rotation matrix.
Consider a body mass moment matrix as
6
5
2@3 1@2 1@2
L = 7 1@2 5@3 1@4 8
1@2 1@4 5@3
The eigenvalues and eigenvectors of L are
£
¤W
L1 = 0=2413
w1 = 2=351 1 1
£
¤W
w2 = 0=851 1 1
L2 = 1=8421
£
¤W
w3 = 0 1 1
L3 = 1=9167
(9.207)
(9.208)
(9.209)
(9.210)
(9.211)
(9.212)
The normalized eigenvector matrix Z is equal to the transpose of the required transformation matrix to make the inertia matrix diagonal
¤
£
w1 w2 w3 = 2 U1W
Z =
5
6
0=856 9 0=515 6
0=0
= 7 0=364 48 0=605 88 0=707 11 8
(9.213)
0=364 48 0=605 88
0=707 11
We may verify that
2
L
2
U1 1 L 2 U1W = Z W 1 L Z
5
0=2413
1 × 104
4
7
1 × 10
1=842 1
=
0=0
0=0
r
6
0=0
1 × 1019 8
1=916 7
(9.214)
Example 393 F Relative diagonal mass moments.
Using the denitions for mass moments (9.157), (9.158), and (9.159) it
is seen that the inertia matrix is symmetric, and
Z
¡ 2
¢
1
(9.215)
{ + | 2 + } 2 gp = (L{{ + L|| + L}} )
2
E
and also
L{{ + L|| L}}
L|| + L}} L{{
L}} + L{{ L||
(9.216)
9. F Applied Dynamics
601
Noting that
(| })2 0
it is evident that
and therefore
¢
¡ 2
| + } 2 2|}
L{{ 2L|}
(9.217)
(9.218)
(9.219)
and similarly
L|| 2L}{
L}} 2L{|
Example 394 F Coe!cients of the characteristic equation.
The determinant (9.182)
¯
¯
¯ L{{ L{|
L{} ¯¯
¯
¯ L|{
L|| L|} ¯¯ = 0
¯
¯ L}{
L}|
L}} ¯
(9.220)
(9.221)
for calculating the principal moments of inertia, leads to a third-degree
equation for , called the characteristic equation.
3 d1 2 + d2 d3 = 0
(9.222)
The coe!cients of the characteristic equation are called the principal invariants of [L]. The coe!cients of the characteristic equation can directly
be found from the following equations:
d1 = L{{ + L|| + L}} = tr [L]
d2
d3
2
2
2
= L{{ L|| + L|| L}} + L}} L{{ L{|
L|}
L}{
¯ ¯
¯ ¯
¯
¯
¯ L{{ L{| ¯ ¯ L|| L|} ¯ ¯ L{{ L{} ¯
¯
¯
¯
¯
¯
¯
+
+
= ¯
L|{ L|| ¯ ¯ L}| L}} ¯ ¯ L}{ L}} ¯
£ ¤¢
1¡ 2
=
d1 tr L 2
2
(9.223)
(9.224)
= L{{ L|| L}} + L{| L|} L}{ + L}| L|{ L{}
(L{{ L|} L}| + L|| L}{ L{} + L}} L{| L|{ )
¡
¢
2
2
2
= L{{ L|| L}} + 2L{| L|} L}{ L{{ L|}
+ L|| L}{
+ L}} L{|
= det [L]
(9.225)
Example 395 F The principal mass moments are coordinate invariants.
The roots of the inertia characteristic equation are the principal mass moments. They are all real but not necessarily dierent. The principal mass
602
9. F Applied Dynamics
moments are extreme values. That is, the principal mass moments determine the smallest and the largest values of Lll . Since the smallest and largest
values of Lll do not depend on the choice of the body coordinate frame, the
solution of the characteristic equation is not dependent of the coordinate
frame.
In other words, if L1 , L2 , and L3 are the principal moments of inertia for
E1
L, the principal moments of inertia for E2 L are also L1 , L2 , and L3 , where
E2
W
L = E2 UE1 E1 L E2 UE
1
(9.226)
We conclude that L1 , L2 , and L3 are coordinate invariants of the matrix [L],
and therefore any quantity that depends on L1 , L2 , and L3 is also coordinate
invariant. The matrix [L] has only three independent invariants and every
other invariant can be expressed in terms of L1 , L2 , and L3 .
Since L1 , L2 , and L3 are the solutions of the characteristic equation of [L]
given in (9.222), we may write the determinant (9.182) in the form
( L1 ) ( L2 ) ( L3 ) = 0
(9.227)
The expanded form of this equation is
3 (L1 + L2 + L3 ) 2 + (L1 L2 + L2 L3 + L3 L1 ) d2 L1 L2 L3 = 0
(9.228)
By comparing (9.228) and (9.222) we conclude that
d1
d2
d3
=
=
=
=
=
L{{ + L|| + L}} = L1 + L2 + L3
(9.229)
2
2
2
L{{ L|| + L|| L}} + L}} L{{ L{| L|} L}{
L1 L2 + L2 L3 + L3 L1
(9.230)
¡
¢
2
2
2
L{{ L|| L}} + 2L{| L|} L}{ L{{ L|}
+ L|| L}{
+ L}} L{|
L1 L2 L3
(9.231)
Being able to express the coe!cients d1 , d2 , and d3 as functions of L1 , L2 ,
and L3 determines that the coe!cients of the characteristic equation are
also coordinate-invariant.
Example 396 F Short notation for the elements of inertia matrix.
Taking advantage of the Kronecker’s delta (5.133) we may write the elements of the moment of inertia matrix Llm in short notation forms
Z
¡¡ 2
¢
¢
{1 + {22 + {23 lm {l {m gp
(9.232)
Llm =
E
Z
¡ 2
¢
u lm {l {m gp
(9.233)
Llm =
E
Ã
!
Z
3
X
(9.234)
Llm =
{n {n lm {l {m gp
E
n=1
9. F Applied Dynamics
603
where we adopted the following notations:
{1 = {
{2 = |
{3 = }
(9.235)
Example 397 F Mass moment with respect to plane, line, and point.
The mass moments of a system of particles may be dened with respect
to a plane, a line, or a point as the sum of the products of the mass of the
particles into the square of the perpendicular distance from the particle to
the plane, line, or point. For a continuous body, the sum would be denite
integral over the volume of the body.
The mass moments with respect to the {|, |}, and }{-plane are
Z
Z
Z
L}2 =
} 2 gp
L|2 =
| 2 gp
L{2 =
{2 gp
(9.236)
E
E
E
The mass moments with respect to the {, |, and } axes are
Z
¡ 2
¢
| + } 2 gp
L{ =
ZE
¡ 2
¢
} + {2 gp
L| =
ZE
¡ 2
¢
{ + | 2 gp
L} =
(9.237)
(9.238)
(9.239)
E
and therefore,
L{ = L|2 + L}2
L| = L}2 + L{2
L} = L{2 + L|2
(9.240)
The mass moment with respect to the origin is
Z
¡ 2
¢
1
Lr =
{ + | 2 + } 2 gp = L{2 + L|2 + L}2 = (L{ + L| + L} ) (9.241)
2
E
Because the choice of the coordinate frame is arbitrary, we can say that the
mass moments with respect to a line is the sum of the mass moments with
respect to any two mutually orthogonal planes that pass through the line.
The mass moments with respect to a point have similar meaning for three
mutually orthogonal planes intersecting at the point.
9.5 Lagrange’s Form of Newton’s Equations of
Motion
Newton’s equation of motion can be transformed to
¶
μ
g CN
CN
= Iu
u = 1> 2> · · · q
gw C tbu
Ctu
(9.242)
9. F Applied Dynamics
604
where
¶
q μ
X
Cil
Cjl
Ckl
Il{
+ Il|
+ Il}
Iu =
Ct1
Ct2
Ctq
l=1
(9.243)
Equation (9.242) is called the Lagrange equation of motion, where N is the
kinetic energy of the q degree-of-freedom (GRI ) system, tu , u = 1> 2> · · · > q
¤W
£
is
are the generalized coordinates of the system, F = Il{ Il| Il}
the external force acting on the lth particle of the system, and Iu is the
generalized force associated to tu .
Proof. Let pl be the mass of one of the particles of a system and let
({l > |l > }l ) be its Cartesian coordinates in a globally xed coordinate frame.
Assume that the coordinates of every individual particle are functions of
another set of coordinates t1 > t2 > t3 > · · · > tq , and possibly time w.
{l
|l
}l
= il (t1 > t2 > t3 > · · · > tq > w)
= jl (t1 > t2 > t3 > · · · > tq > w)
= kl (t1 > t2 > t3 > · · · > tq > w)
(9.244)
(9.245)
(9.246)
If I{l > I|l > I}l are components of the total force acting on the particle
pl , then the Newton equations of motion for the particle would be
I{l
I|l
I}l
= pl {
¨l
= pl |¨l
= pl }¨l
(9.247)
(9.248)
(9.249)
We respectively multiply both sides of these equations by
Cil
Ctu
Cjl
Ctu
Ckl
Ctu
(9.250)
respectively, and add them up for all the particles to have
μ
¶ X
¶
q μ
Cil
Cjl
Ckl
Cil
Cjl
Ckl
pl {
+ |¨l
+ }¨l
+ I|l
+ I}l
¨l
=
I{l
Ctu
Ctu
Ctu
Ctu
Ctu
Ctu
l=1
l=1
(9.251)
where q is the total number of particles.
Taking a time derivative of Equation (9.244),
q
X
{b l =
Cil
Cil
Cil
Cil
Cil
tb1 +
tb2 +
tb3 + · · · +
tbq +
Ct1
Ct2
Ct3
Ctq
Cw
(9.252)
we nd
C
C {b l
=
C tbu
C tbu
μ
Cil
Cil
Cil
Cil
tb1 +
tb2 + · · · +
tbq +
Ct1
Ct2
Ctq
Cw
¶
=
Cil
Ctu
(9.253)
9. F Applied Dynamics
605
and therefore,
{
¨l
g
Cil
C {b l
={
¨l
=
Ctu
C tbu
gw
¶
¶
μ
μ
C {b l
g C {b l
{b l
{b l
C tbu
gw C tbu
(9.254)
However,
{b l
g
gw
μ
C {b l
C tbu
¶
¶
Cil
Ctu
¶
C 2 il
C 2 il
C 2 il
= {b l
tb1 + · · · +
tbq +
Ct1 Ctu
Ctq Ctu
CwCtu
μ
¶
C
Cil
Cil
Cil
Cil
= {b l
tb1 +
tb2 + · · · +
tbq +
Ctu Ct1
Ct2
Ctq
Cw
C {b l
(9.255)
= {b l
Ctu
= {b l
g
gw
μ
μ
and we have
{
¨l
C {b l
g
=
C tbu
gw
¶
μ
C {b l
C {b l
{b l
{b l
C tbu
Ctu
(9.256)
which is equal to
μ
¶¸
μ
¶
1 2
1 2
C
{b l
g
C
{
¨l =
{b
{b
tbu
gw C tbu 2 l
Ctu 2 l
(9.257)
Now substituting (9.254) and (9.257) in the left-hand side of (9.251) leads
to
μ
¶
q
X
Cil
Cjl
Ckl
pl {
+ |¨l
+ }¨l
¨l
Ctu
Ctu
Ctu
l=1
μ
¶¸
q
X
1 2 1 2 1 2
g
C
pl
=
{b l + |b l + }bl
gw C tbu 2
2
2
l=1
μ
¶
q
X
1 2 1 2 1 2
C
pl
{b + |b + }bl
Ctu 2 l 2 l
2
l=1
¸
q
¢
g
C ¡ 2
1X
pl
{b l + |bl2 + }bl2
=
2 l=1
gw C tbu
q
where
¢
C ¡ 2
1X
pl
{b l + |b l2 + }bl2
2 l=1
Ctu
q
¡
¢
1X
pl {b 2l + |bl2 + }bl2 = N
2 l=1
(9.258)
(9.259)
606
9. F Applied Dynamics
is the kinetic energy of the system. Therefore, the Newton equations of
motion (9.247), (9.248), and (9.249) are converted to
g
gw
μ
CN
C tbu
¶
¶
q μ
X
Cil
Cjl
Ckl
CN
=
+ I|l
+ I}l
I{l
Ctu
Ctu
Ctu
Ctu
l=1
(9.260)
Because of (9.244), (9.245), and (9.246), the kinetic energy is a function
of t1 > t2 > t3 > · · · > tq and time w. The left-hand side of Equation (9.260) includes the kinetic energy of the whole system and the right-hand side is
a generalized force and shows the eect of changing coordinates from {l
to tm on the external forces. Let us assume that the coordinate tu alters
to tu + tu while the other coordinates t1 > t2 > t3 > · · · > tu1 > tu+1 > · · · > tq and
time w are unaltered. So, the coordinates of pl are changed to
{l +
Cil
tu
Ctu
|l +
Cjl
tu
Ctu
}l +
Ckl
tu
Ctu
(9.261)
Such a displacement is called virtual displacement. The work done in this
virtual displacement by all forces acting on the particles of the system is
Z =
q μ
X
l=1
Cil
Cjl
Ckl
I{l
+ I|l
+ I}l
Ctu
Ctu
Ctu
¶
tu
(9.262)
Because the work done by internal forces appears in opposite pairs, only the
work done by external forces remains in Equation (9.262). Let us denote
the virtual work by
Z = Iu (t1 > t2 > t3 > · · · > tq > w) tu
Then we have
g
gw
where
μ
CN
C tbu
¶
CN
= Iu
Ctu
¶
q μ
X
Cil
Cjl
Ckl
+ I|l
+ I}l
I{l
Iu =
Ctu
Ctu
Ctu
l=1
(9.263)
(9.264)
(9.265)
Equation (9.264) is the Lagrange form of equations of motion. This equation
is true for all values of u from 1 to q. We thus have q second-order ordinary
dierential equations in which t1 , t2 , t3 , · · · , tq are the dependent variables
and w is the independent variable. The coordinates t1 > t2 > t3 > · · · > tq are
called generalized coordinates and they can be any measurable parameters
to provide the conguration of the system. The number of equations and
the number of dependent variables are equal, therefore, the equations are
theoretically su!cient to determine the motion of all pl .
9. F Applied Dynamics
607
x
Fixed pivot
l
T
m
y
FIGURE 9.10. A simple pendulum.
Example 398 Equation of motion for a simple pendulum.
A pendulum is shown in Figure 9.10. Using { and | for the Cartesian
position of p, and using = t as the generalized coordinate, we have
{ = i () = o sin | = j() = o cos ¢ 1
2
1 ¡ 2
N =
p {b + |b 2 = po2 b
2
2
and therefore,
g
gw
μ
CN
C b
¶
CN
g
b = po2 ¨
= (po2 )
C
gw
(9.266)
(9.267)
(9.268)
(9.269)
The external force components, acting on p, are
I{ = 0
I| = pj
(9.270)
and therefore,
Ci
Cj
+ I|
= pjo sin C
C
Hence, the equation of motion for the pendulum is
I = I{
po2 ¨ = pjo sin (9.271)
(9.272)
Example 399 A pendulum attached to an oscillating mass.
Figure 9.11 illustrates a vibrating mass with a hanging pendulum. The
pendulum can act as a vibration absorber if designed properly. This system
has two degrees of freedom and therefore, needs two generalized coordinates.
Adopting { and as the generalized coordinate and starting with coordinate relationships
{P
{p
= iP = {
= ip = { + o sin |P = jP = 0
|p = jp = o cos (9.273)
(9.274)
608
9. F Applied Dynamics
M
k
x
T
l
m
y
FIGURE 9.11. A vibrating mass with a hanging pendulum.
we may nd the kinetic energy in terms of { and .
N
=
=
¢ 1 ¡ 2
¢
1 ¡ 2
2
2
+ p {b p + |bp
P {b P + |b P
2
2
´
2
1
1 ³ 2
2
P {b + p {b + o2 b + 2o{b b cos 2
2
The left-hand side of the Lagrange equations are
μ
¶
2
g CN
CN
= (P + p)¨
{ + po¨ cos pob sin gw C {b
C{
¶
μ
g CN
CN
{ cos = po2 ¨ + po¨
gw C b
C
(9.275)
(9.276)
(9.277)
The external forces acting on P and p are
I{P = n{
I|P = 0
I{p = 0
I|p = pj
(9.278)
Therefore, the generalized forces are
I{
I
= I{P
CiP
CjP
Cip
Cjp
+ I|P
+ I{p
+ I|p
C{
C{
C{
C{
= n{
(9.279)
CiP
CjP
Cip
Cjp
+ I|P
+ I{p
+ I|p
C
C
C
C
= pjo sin (9.280)
= I{P
and nally the Lagrange equations of motion are
2
(P + p)¨
{ + po¨ cos pob sin po2 ¨ + po¨
{ cos = n{
= pjo sin (9.281)
(9.282)
9. F Applied Dynamics
609
Example 400 Kinetic energy of the Earth.
Earth is approximately a rotating rigid body about a xed axis. The two
motions of the Earth are called revolution about the sun, and rotation
about an axis approximately xed in the Earth. The kinetic energy of the
Earth due to its rotation is
N1
=
=
1 2
L$
2 1
μ
¶ μ
¶2
¢ 6356912 + 6378388 2
12¡
2
366=25
24
5=9742 × 10
25
2
24 × 3600 365=25
= 2=5762 × 1029 J
(9.283)
where L = 2P U2 @5 is the mass moment of the Earth, and $ 1 is the angular
speed about the Earth’s axis. The kinetic energy of the Earth due to its
revolution is
N2
=
=
1
P u2 $ 22
2
μ
¶2
¢¡
¢
1¡
2
1
24
11 2
1=49475 × 10
5=9742 × 10
2
24 × 3600 365=25
= 2=6457 × 1033 J
(9.284)
where u is the distance from the sun, and $ 2 is the angular speed about the
sun. The total kinetic energy of the Earth is N = N1 + N2 . However, the
ratio of the revolutionary to rotational kinetic energies is
N2
2=6457 × 1033
=
r 10000
N1
2=5762 × 1029
(9.285)
Therefore, rotation of the Earth contribute only 0=1% of the total kinetic
energy of the Earth.
Example 401 F Explicit form of Lagrange equations.
Assume the coordinates of every particle are functions of the coordinates
t1 > t2 > t3 > · · · > tq but not the time w. The kinetic energy of the system made
of q massive particles can be written as
q
q
q X
¡ 2
¢ 1X
1X
2
2
N=
pl {b l + |bl + }bl =
dmn tbm tbn
2 l=1
2 m=1
(9.286)
n=1
where the coe!cients dmn are functions of t1 , t2 , t3 , · · · , tq and
dmn = dnm
The Lagrange equations of motion
¶
μ
g CN
CN
= Iu
gw C tbu
Ctu
(9.287)
u = 1> 2> · · · q
(9.288)
610
9. F Applied Dynamics
are then equal to
q
q
q
1 X X dmn
g X
dpu tbp tbm tbn = Iu
gw p=1
2 m=1
Ctu
(9.289)
n=1
or
q
X
dpu ẗp +
p=1
q X
q
X
un>q tbn tbq = Iu
(9.290)
n=1 q=1
where lm>n is called the Christoel operator
1
lm>n =
2
μ
Cdlm
Cdln
Cdnm
+
Ctn
Ctm
Ctl
¶
(9.291)
9.6 Lagrangian Mechanics
£
¤W
Assume for some forces F = Il{ Il| Il}
there is a function Y ,
called potential energy, such that the force is derivable from Y
F = uY
(9.292)
Such a force is called potential or conservative force. Then, the Lagrange
equation of motion can be written as
¶
μ
g CL
CL
= Tu
u = 1> 2> · · · q
(9.293)
gw C tbu
Ctu
where
L=N Y
(9.294)
is the Lagrangean of the system and Tu is the nonpotential generalized
force.
£
¤W
Proof. Assume the external forces F = I{l I|l I}l
acting on the
system are conservative.
F = uY
(9.295)
The work done by these forces in an arbitrary virtual displacement t1 ,
t2 , t3 , · · · , tq is
CZ = CY
CY
CY
t1 t2 · · ·
tq
Ct1
Ct2
Ctq
then the Lagrange equation becomes
¶
μ
g CN
CY
CN
=
gw C tbu
Ctu
Ct1
u = 1> 2> · · · q=
(9.296)
(9.297)
9. F Applied Dynamics
611
Z
C
X
M
l
T
m
Y
FIGURE 9.12. A spherical pendulum.
Introducing the Lagrangean function L = N Y converts the Lagrange
equation to
μ
¶
CL
g CL
=0
u = 1> 2> · · · q
(9.298)
gw C tbu
Ctu
for a conservative system. The Lagrangean L is also called kinetic potential.
If a force is not conservative, then the virtual work done by the force is
¶
q μ
X
Cil
Cjl
Ckl
Z =
I{l
tu = Tu tu
+ I|l
+ I}l
Ctu
Ctu
Ctu
l=1
and the equation of motion would be
μ
¶
CL
g CL
= Tu
gw C tbu
Ctu
u = 1> 2> · · · q
(9.299)
(9.300)
where Tu is the nonpotential generalized force doing work in a virtual
displacement of the uth generalized coordinate tu .
Example 402 Spherical pendulum.
A pendulum analogy is utilized in modeling of many dynamical problems.
Figure 9.12 illustrates a spherical pendulum with mass p and length o. The
angles * and may be used as describing coordinates of the system.
The Cartesian coordinates of the mass p as a function of the generalized
coordinates are
5
6 5
6
[
u cos * sin 7 \ 8 = 7 u sin sin * 8
(9.301)
]
u cos 612
9. F Applied Dynamics
Y
O
Q
l
X
T
m, I
C
FIGURE 9.13. A controlled compound pendulum.
Having the constraint
[b 2 + \b 2 + ]b 2 u2 = 0
(9.302)
makes the three coordinates u, , * dependent with only two coordinates
independent. Taking , * as the two generalized coordinates, the kinetic
and potential energies of the pendulum would be
´
1 ³ 2 b2
p o + o2 *b 2 sin2 (9.303)
N =
2
Y = pjo cos (9.304)
The kinetic potential function of this system is then equal to
´
1 ³ 2
L = p o2 b + o2 *b 2 sin2 + pjo cos 2
(9.305)
which leads to the following equations of motion:
¨ *b 2 sin cos + j sin o
*
¨ sin2 + 2*b b sin cos = 0
(9.306)
= 0
(9.307)
Example 403 Controlled compound pendulum.
A massive arm is attached to a ceiling at a pin joint R as illustrated
in Figure 9.13. Assume that there is viscous friction in the joint where an
ideal motor can apply a torque T to move the arm. The rotor of an ideal
motor has no mass moment by assumption.
The kinetic and potential energies of the manipulator are
N
Y
¢ 2
1 b2 1 ¡
LF + po2 b
L =
2
2
= pjo cos =
(9.308)
(9.309)
9. F Applied Dynamics
x1
y1
613
y2
x2
T
l1
m2
Y
Q2
m1
l2
T
Q1
X
FIGURE 9.14. A model for a 2U planar manipulator.
where p is the mass and L is the mass moment of the pendulum about R.
The Lagrangean of the manipulator is
L=N Y =
1 b2
L + pjo cos 2
and therefore, the equation of motion of the pendulum is
¶
μ
CL
g CL
P=
= L ¨ + pjo sin gw C b
C
(9.310)
(9.311)
The generalized force P is the contribution of the motor torque T and
b Hence, the equation of motion of the mathe viscous friction torque f.
nipulator is
T = L ¨ + fb + pjo sin (9.312)
Example 404 An ideal 2U planar manipulator dynamics.
An ideal model of a 2U planar manipulator is illustrated in Figure 9.14.
It is called ideal because we assume the links are massless and there is no
friction in the system. There is a motor on the ground that applies a torque
T1 on the link one. The second link has the mass p1 and runs the second
link and the load p2 at the endpoint. We take the absolute angle 1 and the
relative angle 2 as the generalized coordinates to express the conguration
of the manipulator.
The global positions of p1 and p2 are
¸
¸
o1 cos 1
[1
=
(9.313)
\2
o1 sin 1
¸
¸
[2
o1 cos 1 + o2 cos (1 + 2 )
=
(9.314)
\2
o1 sin 1 + o2 sin (1 + 2 )
614
9. F Applied Dynamics
and therefore, the global velocity of the masses are
¸
¸
[b 1
o1 b 1 sin 1
=
\b 1
o1 b 1 cos 1
³
´
5
6
¸
b
b
b
sin
o
+
+
)
sin
(
o
b
1 1
1
2
1
2
1
2
[2
8
³
´
= 7
b
b
b
\b 2
o1 1 cos 1 + o2 1 + 2 cos (1 + 2 )
(9.315)
(9.316)
The kinetic energy of this manipulator is made of kinetic energy of the
masses and is equal to
³
´ 1
³
´
1
N = N1 + N2 = p1 [b 12 + \b 12 + p2 [b 22 + \b 22
2
2
1
2 b2
=
p1 o1 1
2
μ
¶
³
´2
³
´
1
2 b2
2 b
b
b
b
b
+ p2 o1 1 + o2 1 + 2 + 2o1 o2 1 1 + 2 cos 2 (9.317)
2
The potential energy of the manipulator is
Y
= Y1 + Y2 = p1 j\1 + p2 j\2
= p1 jo1 sin 1 + p2 j (o1 sin 1 + o2 sin (1 + 2 ))
(9.318)
The Lagrangean is then obtained from Equations (9.317) and (9.318)
L = N Y
(9.319)
2
1
=
p1 o12 b 1
2
μ
¶
³
´2
³
´
1
2 b2
2 b
b
b
b
b
+ p2 o1 1 + o2 1 + 2 + 2o1 o2 1 1 + 2 cos 2
2
(p1 jo1 sin 1 + p2 j (o1 sin 1 + o2 sin (1 + 2 )))
which provides the required partial derivatives as follows:
CL
C1
CL
C b 1
g
gw
= (p1 + p2 ) jo1 cos 1 p2 jo2 cos (1 + 2 )
³
´
= (p1 + p2 ) o12 b 1 + p2 o22 b 1 + b 2
³
´
+p2 o1 o2 2b 1 + b 2 cos 2
μ
CL
C b 1
¶
³
´
= (p1 + p2 ) o12 ¨1 + p2 o22 ¨1 + ¨2
³
´
+p2 o1 o2 2 ¨1 + ¨2 cos 2
³
´
p2 o1 o2 b 2 2b 1 + b 2 sin 2
(9.320)
(9.321)
(9.322)
9. F Applied Dynamics
³
´
= p2 o1 o2 b 1 b 1 + b 2 sin 2 p2 jo2 cos (1 + 2 ) (9.323)
³
´
= p2 o22 b 1 + b 2 + p2 o1 o2 b 1 cos 2
(9.324)
CL
C2
CL
C b 2
g
gw
μ
CL
C b 2
¶
615
³
´
= p2 o22 ¨1 + ¨2 + p2 o1 o2 ¨1 cos 2 p2 o1 o2 b 1 b 2 sin 2 (9.325)
Therefore, the equations of motion for the 2U manipulator are
¶
μ
g CL
CL
T1 =
gw C b 1
C1
³
´
= (p1 + p2 ) o12 ¨1 + p2 o22 ¨1 + ¨2
³
´
³
´
+p2 o1 o2 2 ¨1 + ¨2 cos 2 p2 o1 o2 b 2 2b 1 + b 2 sin 2
+ (p1 + p2 ) jo1 cos 1 + p2 jo2 cos (1 + 2 )
T2
(9.326)
¶
CL
CL
C2
C b 2
³
´
= p2 o22 ¨1 + ¨2 + p2 o1 o2 ¨1 cos 2 p2 o1 o2 b 1 b 2 sin 2
³
´
+p2 o1 o2 b 1 b 1 + b 2 sin 2 + p2 jo2 cos (1 + 2 )
(9.327)
=
g
gw
μ
The generalized forces T1 and T2 are the required forces to drive the
generalized coordinates. In this case, T1 is the torque at the base motor
and T2 is the torque of the motor at p1 .
The equations of motion can be rearranged to have a more systematic
form
¡
¢
T1 = (p1 + p2 ) o12 + p2 o2 (o2 + 2o1 cos 2 ) ¨1
+p2 o2 (o2 + o1 cos 2 ) ¨2
2
2p2 o1 o2 sin 2 b 1 b 2 p2 o1 o2 sin 2 b 2
+ (p1 + p2 ) jo1 cos 1 + p2 jo2 cos (1 + 2 )
T2
(9.328)
2
= p2 o2 (o2 + o1 cos 2 ) ¨1 + p2 o22 ¨2 + p2 o1 o2 sin 2 b 1
+p2 jo2 cos (1 + 2 )
(9.329)
Example 405 Mechanical energy.
If a system of masses pl are moving in a potential force eld
Fpl = ul Y
(9.330)
616
9. F Applied Dynamics
r
A
R r 1 cos T
B
Y
R+r
R
T
X
FIGURE 9.15. A wheel turning, without slip, over a cylindrical hill.
their Newton equations of motion will be
pl r̈l = ul Y
l = 1> 2> · · · q
(9.331)
The inner product of equations of motion with rb l and adding the equations
q
X
l=1
pl rb l · r̈l = q
X
l=1
rb l · ul Y
(9.332)
and then, integrating over time
q
1X
pl rb l · rb l = 2 l=1
Z X
q
l=1
rl · ul Y
(9.333)
¶
(9.334)
shows that
N=
Z X
q μ
l=1
CY
CY
CY
{l +
|l +
}l
C{l
C|l
C}l
= Y + H
where H is the constant of integration. H is called the mechanical energy
of the system and is equal to kinetic plus potential energies.
H =N +Y
(9.335)
Example 406 Falling wheel.
Figure 9.15 illustrates a wheel rolling over a cylindrical hill. We may use
the conservation of mechanical energy to nd the angle at which the wheel
leaves the hill.
At the initial instant of time, the wheel is at point D. We assume the
initial kinetic and potential, and hence, the mechanical energies are zero.
When the wheel is turning over the hill, its angular velocity, $, is
$=
y
u
(9.336)
9. F Applied Dynamics
617
where y is the speed at the center of the wheel. At any other point E, the
wheel achieves some kinetic energy and loses some potential energy. At a
certain angle, where the normal component of the weight cannot provide
more centripetal force,
py 2
(9.337)
pj cos =
U+u
the wheel separates from the surface. Employing the conservation of energy,
we have
HD
ND + YD
= HE
= NE + YE
(9.338)
(9.339)
The kinetic and potential energy at the separation point E are
NE
YE
1
1
py 2 + LF $ 2
2
2
= pj (U + u) (1 cos )
=
(9.340)
(9.341)
where LF is the mass moment of the wheel about its center. Therefore,
1
1
py2 + LF $ 2 = pj (U + u) (1 cos )
2
2
and substituting (9.336) and (9.337) provides
μ
¶
LF
1+
(U + u) j cos = 2j (U + u) (1 cos )
pu2
(9.342)
(9.343)
and therefore, the separation angle is
= cos1
2pu2
LF + 3pu2
(9.344)
Let us examine the equation for a disc wheel with
LF =
1 2
pu
2
(9.345)
and nd the separation angle.
= cos1
4
0=96 rad 55=15 deg
7
(9.346)
Example 407 Turning wheel over a step.
Figure 9.16 illustrates a wheel of radius U rolling with speed y to go over
a step with height K ? U.
We may use the principle of energy conservation and nd the speed of the
wheel after getting across the step. Employing the conservation of energy,
618
9. F Applied Dynamics
B
R
A
v2
v1
H
FIGURE 9.16. A rolling wheel moving up a step.
we have
= HE
= NE + YE
1
1
1
1
py12 + LF $ 21 + 0 =
py22 + LF $ 22 + pjK
2
2
2¶
¶
μ
μ 2
LF
LF
2
p + 2 y22 + 2pjK
p + 2 y1 =
U
U
HD
ND + YD
and therefore,
v
u
2
y2 = u
ty1 2jK
LF
1+
pU2
The condition for having a real y2 is
v
u 2jK
y1 A u
t
LF
1+
pU2
(9.347)
(9.348)
(9.349)
(9.350)
(9.351)
(9.352)
The second speed (9.351) and the condition (9.352) for a solid disc are
r
4
y12 Kj
y2 =
(9.353)
3
r
4
Kj
(9.354)
y1 A
3
because we assumed that
LF =
1
pU2
2
(9.355)
Example 408 Trebuchet.
A trebuchet, shown schematically in Figure 9.17, is a shooting weapon of
war powered by a falling massive counterweight p1 . A beam DE is pivoted
to the chassis with two unequal sections d and e.
9. F Applied Dynamics
619
y
b
B
c
J
M
x
a
T
m1
d
m2
D
A
l
FIGURE 9.17. A trebuchet at starting position.
The gure shows a trebuchet at its initial conguration. The origin of
a global coordinate frame is set at the pivot point. The counterweight p1
is at ({1 > |1 ) and is hinged at the shorter arm of the beam at a distance f
from the end E. The mass of the projectile is p2 and it is at the end of a
massless sling with a length o attached to the end of the longer arm of the
beam. The three independent variable angles , , describe the motion of
the device. We consider the parameters d, e, f, g, o, p1 , p2 constant, and
determine the equations of motion by the Lagrange method.
Figure 9.18 illustrates the trebuchet when it is in motion. The position
coordinates of masses p1 and p2 are
{1
|1
= e sin f sin ( + )
= e cos + f cos ( + )
(9.356)
(9.357)
{2
|2
= d sin o sin ( + )
= d cos o cos ( + )
(9.358)
(9.359)
and
Taking a time derivative provides the velocity components
³
´
{b 1 = eb cos f b + b cos ( + )
³
´
|b 1 = eb sin f b + b sin ( + )
{b 2
|b 2
= o (f )
b cos ( ) db cos ()
³
´
= db sin o b b sin ( )
(9.360)
(9.361)
(9.362)
(9.363)
9. F Applied Dynamics
620
which shows that the kinetic energy of the system is
N
¡
¢ 1
¡
¢
1
1
1
p1 y12 + p2 y22 = p1 {b 21 + |b 12 + p2 {b 22 + |b 22
2
2
2
2
³¡
´
¢
1
2
2 b2
2 2
2b
=
p1 e + f + f b + 2f b
2
³
´
³¡
´
¢ 2
1
p1 efb b + b cos + p2 d2 + o2 b + o2 b 2 2o2 b b
2
³
´
b
b
p2 do b cos (2 )
(9.364)
=
The potential energy of the system can be calculated by |-position of the
masses.
Y
= p1 j|1 + p2 j|2
= p1 j (e cos + f cos ( + ))
+p2 j (d cos o cos ( + ))
(9.365)
Having the energies N and Y , we can set up the Lagrangean L.
L=N Y
(9.366)
Using the Lagrangean, we are able to nd the three equations of motion.
¶
μ
CL
g CL
= 0
(9.367)
b
gw C C
μ
¶
CL
g CL
= 0
(9.368)
gw C b
C
μ
¶
CL
g CL
= 0
(9.369)
gw C b
C
The trebuchet appeared in 500 to 400 B.C. in China and was developed
by Persian armies around 300 B.C. It was used by the Arabs against the
Romans during 600 to 1200 A.D. The trebuchet may also be called the
manjaniq, catapults, or onager. The Persian word "Manganic" is the
root manjaniq as well as the words "mechanic" and "machine".
9.7 Summary
The translational and rotational equations of motion for a rigid body, expressed in the global coordinate frame J, are
J
J
J
F =
M =
gJ
p
gw
J
gJ
L
gw
(9.370)
(9.371)
9. F Applied Dynamics
621
y
a
A
b
D
x
M
T J
l
B
c
m2
d
m1
FIGURE 9.18. A trebuchet in motion.
where J F and J M indicate the resultant of the external forces and moments applied on the rigid body, measured at the mass center F.
The vector J p is the momentum and J L is the moment of momentum
for the rigid body at F
p = pv
L = rF × p
(9.372)
(9.373)
The expression of the equations of motion in the body coordinate frame
are
E
F =
E
M =
J
E
E
pb + E
p = p E aE + p E
vE
(9.374)
J $E ×
J $E ×
¡
¢
Eb
E
E
E E
E
E E
b E + J $E ×
L J $ E (9.375)
L+ J $ E × L = L J $
where L is the mass moment of the rigid body.
6
5
L{{ L{| L{}
L = 7 L|{ L|| L|} 8
L}{ L}| L}}
(9.376)
The elements of L are functions of the mass distribution of the rigid body
and are dened by
Z
¡ 2
¢
Llm =
ul pq {lp {mq gp
l> m = 1> 2> 3
(9.377)
E
where lm is Kronecker’s delta.
Every rigid body has a principal body coordinate frame in which the
mass moment matrix is diagonal
5
6
L1 0 0
E
L = 7 0 L2 0 8
(9.378)
0 0 L3
622
9. F Applied Dynamics
The rotational equation of motion in the principal coordinate frame simplies to
P1
P2
P3
= L1 $b 1 (L2 L2 ) $ 2 $ 3
= L2 $b 2 (L3 L1 ) $ 3 $ 1
= L3 $b 3 (L1 L2 ) $ 1 $ 2
(9.379)
The equations of motion for a mechanical system having q DOF can also
be found by the Lagrange equation
¶
μ
CL
g CL
= Tu
u = 1> 2> · · · q
(9.380)
gw C tbu
Ctu
L = N Y
(9.381)
where L is the Lagrangean of the system, N is the kinetic energy, Y is the
potential energy, and Tu is the nonpotential generalized force.
Tu =
¶
q μ
X
Cil
Cjl
Ckl
+ Tl|
+ Tl}
Tl{
Ct1
Ct2
Ctq
l=1
(9.382)
The parameters tu , u = 1> 2> · · · > q are the generalized coordinates of the
¤W
£
is the external force acting on the lth
system, Q = Tl{ Tl| Tl}
particle of the system, and Tu is the generalized force associated to tu .
When ({l > |l > }l ) are Cartesian coordinates in a globally xed coordinate
frame for the particle pl , then its coordinates may be functions of another
set of generalized coordinates t1 > t2 > t3 > · · · > tq and possibly time w.
{l
|l
}l
= il (t1 > t2 > t3 > · · · > tq > w)
= jl (t1 > t2 > t3 > · · · > tq > w)
= kl (t1 > t2 > t3 > · · · > tq > w)
(9.383)
(9.384)
(9.385)
9. F Applied Dynamics
9.8 Key Symbols
d> e> z> k
a
F
d
gf
gp
gm
H
F
IF
j
K
L
L1 > L2 > L3
N
o
L
L=NY
p
M
p
S> T
T
u
r
U
U
w
W
x̂
y {>
b v
Y
w
Z
{> |> }> x
length
acceleration
mass center
position vector of the body coordinate frame
innitesimal force
innitesimal mass
innitesimal moment
mechanical energy
force
Coriolis force
gravitational acceleration
height
mass moment matrix
principal moment of inertia
kinetic energy
directional line
moment of momentum
Lagrangean
mass
moment
momentum
points in rigid body
torque
radius of disc
position vector
radius
rotation matrix
time
tension force
unit vector to show the directional line
velocity
potential energy
eigenvector
work, eigenvector matrix
displacement
lm
lm>n
*> > #
$> $
k
B
u
Kronecker’s delta
Christoel operator
eigenvalue
Euler angles
angular velocity
parallel
orthogonal
gradient
623
624
9. F Applied Dynamics
Exercises
1. Kinetic energy of a rigid link.
Consider a straight and uniform bar as a rigid bar. The bar has a
mass p. Show that the kinetic energy of the bar can be expressed as
N=
1
p (v1 · v1 + v1 · v2 + v2 · v2 )
6
where v1 and v2 are the velocity vectors of the endpoints of the bar.
2. Discrete particles.
There are three particles p1 = 10 kg, p2 = 20 kg, p3 = 30 kg, at
5
6
5
6
5
6
1
1
2
r1 = 7 1 8 r1 = 7 3 8 r1 = 7 1 8
1
2
3
Their velocities are
5
6
2
v1 = 7 1 8
1
5
6
1
v1 = 7 0 8
2
5
6
3
v1 = 7 2 8
1
Find the position and velocity of the system at F. Calculate the system’s momentum and moment of momentum. Calculate the system’s
kinetic energy and determine the rotational and translational parts
of the kinetic energy.
3. Newton’s equation of motion in the body frame.
Show that Newton’s equation of motion in the body frame is
5
6
5
6 5
6
65
I{
d{
0
$ } $ |
y{
7 I| 8 = p 7 d| 8 + 7 $ }
0
$ { 8 7 y| 8
I}
d}
$ | $ {
0
y}
4. Work on a curved path.
A particle of mass p is moving on a circular path given by
J
rS = cos Lˆ + sin Mˆ + 4 N̂
Calculate the work done by a force J F when the particle moves from
= 0 to = 2 .
(a)
J
F=
} 2 | 2 ˆ | 2 {2 ˆ {2 | 2
2 L +
2 M +
2 N̂
({ + |)
({ + |)
({ + })
9. F Applied Dynamics
625
(b)
J
F=
}2 |2 ˆ
2| ˆ {2 | 2
L
+
M+
2
2 N̂
{+|
({ + |)
({ + })
5. Principal moments of inertia.
Find the principal moments of inertia and directions for the following
inertia matrices:
(a)
5
6
3 2 2
[L] = 7 2 2 0 8
2 0 4
(b)
6
3 2 4
[L] = 7 2 0 2 8
4 2 3
(c)
s
6
100
s 20 3 0
[L] = 7 20 3
60
0 8
0
0
10
5
5
6. Rotated moment of inertia matrix.
A principal moment of inertia matrix E2 L is given as
6
5
3 0 0
[L] = 7 0 5 0 8
0 0 4
The principal frame was achieved by rotating the initial body coordinate frame 30 deg about the {-axis, followed by 45 deg about the
}-axis. Find the initial moment of inertia matrix E1 L.
7. Rotation of moment of inertia matrix.
Find the required rotation matrix that transforms the moment of
inertia matrix [L] to an diagonal matrix.
5
6
3 2
2
[L] = 7 2 2 0=1 8
2 0=1 4
8. F Cubic equations.
The solution of a cubic equation
d{3 + e{2 + f{ + g = 0
626
9. F Applied Dynamics
where d 6= 0, can be found in a systematic way.
Transform the equation to a new form with discriminant 4s3 + t 2 ,
| 3 + 3s| + t = 0
e
using the transformation { = | 3d
, where,
2e3 9def + 27d2 g
3df e2
t
=
9d2
27d3
The solutions are then
p
s
|1 = 3 3 2l s
4l p
4l s
2l p
|3 = h 3 3 h 3 3 |2 = h 3 3 h 3 3 s=
where,
p
t + t 2 + 4s3
t 2 + 4s3
=
=
2
2
For real values of s and t, if the discriminant is positive, then one root
is real, and two roots are complex conjugates. If the discriminant is
zero, then there are three real roots, of which at least two are equal. If
the discriminant is negative, then there are three unequal real roots.
t +
p
Apply this theory for the characteristic equation of the matrix [L] and
show that the principal moments of inertia are real.
9. Kinematics of a moving car on the Earth.
The location of a vehicle on the Earth is described by its longitude *
from a xed meridian, say, the Greenwich meridian, and its latitude
from the equator, as shown in Figure 9.19. We attach a coordinate
frame E at the center of the Earth with the {-axis on the equator’s
plane and the |-axis pointing to the vehicle. There are also two coordinate frames H and J where H is attached to the Earth and J is
the global coordinate frame. Show that the angular velocity of E and
the velocity of the vehicle are
E
J $E
E
J vS
= b ˆ~E + ($ H + *)
b sin ˆE + ($ H + *)
b cos n̂
b
= u ($ H + *)
b cos ˆ~E + u n̂
Calculate the acceleration of the vehicle.
10. Global dierential of angular momentum.
Convert the moment of inertia E L and the angular velocity E
J $ E to
the global coordinate frame and then nd the dierential of angular
momentum. It is an alternative method to show that
J
J
¢
g E
g ¡E E
E
b E
L=
L J $ E = E L+
L = L$
b + $× (L$)
J $E ×
gw
gw
9. F Applied Dynamics
Z
Z
zE
ȦE
ȦE
E
G
P
z
X
627
P
y
y
r
z
T
T
yE
x
M
xE
B
yE
Y
(a)
(b)
FIGURE 9.19. The location on the Earth is dened by longitude * and latitude
.
11. Lagrange method and nonlinear vibrating system.
Use the Lagrange method and nd the equation of motion for the
pendulum shown in Figure 9.20. The stiness of the linear spring is
n.
12. Forced vibration of a pendulum.
Figure 9.21 illustrates a simple pendulum having a length o and a bob
with mass p. Find the equation of motion if
(a) the pivot R has a dictated motion in [ direction
[R = d sin $w
(b) the pivot R has a dictated motion in \ direction
\R = e sin $w
(c) the pivot R has a uniform motion on a circle
rR = U cos $w Lˆ + U sin $w Mˆ
13. Equations of motion from Lagrangean.
Consider a physical system with a Lagrangean as
L=
1
1
p (d{b + e|)
b 2 n (d{ + e|)2
2
2
628
9. F Applied Dynamics
Y
l
O
m
a
X
C
T
k
FIGURE 9.20. A compound pendulum attached with a linear spring at the tip
point.
Y
l
O
X
T
m
FIGURE 9.21. A pendulum with a vibrating pivot.
9. F Applied Dynamics
629
and nd the equations of motion. The coe!cients p, n, d, and e are
constant.
14. Lagrangean from equation of motion.
Find the Lagrangean associated to the following equations of motions:
(a)
pu2 ¨ + n1 o1 + n2 o2 + pjo = 0
(b)
2
ü u b = 0
u2 ¨
+ 2u ub b = 0
15. Trebuchet.
Derive the equations of motion for the trebuchet shown in Figure
9.17.
16. Simplied trebuchet.
Three simplied models of a trebuchet are shown in Figures 9.22 to
9.24. Derive and compare their equations of motion.
y
b
B
J
M
m1
x
a
T
d
m2
D
A
l
FIGURE 9.22. A simplied models of a trebuchet.
9. F Applied Dynamics
y
b
B
c
J
M
x
a
T
m1
m2
d
D
A
FIGURE 9.23. A simplied models of a trebuchet.
y
b
B
J
M
a
630
m1
x
T
d
m2
D
A
FIGURE 9.24. A simplied models of a trebuchet.
10
Vehicle Planar Dynamics
In this chapter we develop a dynamic model for a rigid vehicle in a planar
motion. The planar model is applicable whenever the forward, lateral and
yaw velocities are important and are enough to examine the behavior of a
vehicle.
z
y q
Fy
My
Fz
Mz
r
\
Mx
Fx x
p
M
T
C
B
FIGURE 10.1. Vehicle body coordinate frame E(F{|}).
10.1 Vehicle Coordinate Frame
The equations of motion in vehicle dynamics are usually expressed in a set
of vehicle coordinate frame E(F{|}), attached to the vehicle at the mass
center F, as shown in Figure 10.1. The {-axis is a longitudinal axis passing
through F and directed forward. The |-axis goes laterally to the left from
the driver’s viewpoint. The }-axis makes the coordinate system a righthand triad. When the car is parked on a at horizontal road, the }-axis is
perpendicular to the ground, opposite to the gravitational acceleration g.
To show the vehicle orientation, we use three angles: roll angle * about
the {-axis, pitch angle about the |-axis, and yaw angle # about the }axis. Because the rate of the orientation angles are important in vehicle
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_10, © Springer Science+Business Media New York 2014
631
632
10. Vehicle Planar Dynamics
dynamics, we usually show them by a special character and call them roll
rate, pitch rate, and yaw rate respectively.
*b = s
b = t
#b = u
(10.1)
(10.2)
(10.3)
The resultant of external forces and moments, that the vehicle receives
from the ground and environment, makes the vehicle force system (F> M).
This force system will be expressed in the body coordinate frame.
E
E
F = I{ˆ~ + I| ˆ + I} n̂
(10.4)
M = P{ˆ~ + P| ˆ + P} n̂
(10.5)
The individual components of the 3D vehicle force system are shown in
Figure 10.1. These components have special names and importance.
1. Longitudinal force I{ . It is a force acting along the {-axis. The resultant I{ A 0 if the vehicle is accelerating, and I{ ? 0 if the vehicle
is braking. Longitudinal force is also called forward force, or traction
force.
2. Lateral force I| . It is an orthogonal force to both I{ and I} . The
resultant I| A 0 if it is leftward from the driver’s viewpoint. Lateral
force is usually a result of steering and is the main reason to generate
a yaw moment and turn a vehicle.
3. Normal force I} . It is a vertical force, normal to the ground plane.
The resultant I} A 0 if it is upward. Normal force is also called
vertical force or vehicle load.
4. Roll moment P{ . It is a longitudinal moment about the {-axis. The
resultant P{ A 0 if the vehicle tends to turn about the {-axis. The roll
moment is also called the bank moment, tilting torque, or overturning
moment.
5. Pitch moment P| . It is a lateral moment about the |-axis. The resultant P| A 0 if the vehicle tends to turn about the |-axis and move
the head down.
6. Yaw moment P} . It is an upward moment about the }-axis. The
resultant P} A 0 if the tire tends to turn about the }-axis. The yaw
moment is also called the aligning moment.
The position and orientation of the vehicle coordinate frame E(F{|}) is
measured with respect to a grounded xed coordinate frame J(R[\ ]).
10. Vehicle Planar Dynamics
B
633
G
Z
z
Mz Fz
r
X
x
\
M
Fx
Mx
p
\
My
Fy
T
q
Y
y
FIGURE 10.2. Illustration of a moving vehicle, indicated by its body coordinate
frame E in a global coordinate frame J.
The vehicle coordinate frame is called the body frame or vehicle frame,
and the grounded frame is called the global coordinate frame. Analysis of
the vehicle motion is equivalent to express the position and orientation of
E(F{|}) in J(R[\ ]). Figure 10.2 shows how a moving vehicle is indicated by a body frame E in a global frame J.
The angle between the [ and { axes measured from [ to { about ]
is the yaw angle # and is called the heading angle. The velocity vector v
of the vehicle makes an angle with the body {-axis, measured from { to
v about ], which is called sideslip angle or attitude angle. The vehicle’s
velocity vector v makes an angle + # with the global [-axis, measured
from [ to v about ] that is called the cruise angle. A positive conguration
of these angles are shown in the top view of a moving vehicle in Figure 10.3.
There are many situations in which we need to number the wheels of a
vehicle. We start numbering from the front left wheel as number 1, and then
the front right wheel would be number 2. Numbering increases sequentially
on the right wheels going to the back of the vehicle up to the rear right
wheel. Then, we go to the left of the vehicle and continue numbering the
wheels from the rear left toward the front. Each wheel is indicated by a
position vector rl
E
rl = {l l + |l m + }l n
(10.6)
expressed in the body coordinate frame E. Numbering of a four wheel
vehicle is shown in Figure 10.3.
Example 409 Wheel numbers and their position vectors.
Figure 10.4 depicts a six-wheel passenger car. The wheel numbers are
indicated next to each wheel. The front left wheel is wheel number 1, and
the front right wheel is number 2. Moving to the back on the right side,
634
10. Vehicle Planar Dynamics
v
Y
G
y
1
E
r1
\ E Cruise angle
\
Yaw angle
E
x
B
2
r2
4
Sideslip
r4
C
r3
d
3
\
X
FIGURE 10.3. Top view of a moving vehicle to show the yaw angle # between
the [ and { axes, the sideslip angle between the {-axis and the velocity vector
v, and the crouse angle + # between the [-axis and the velocity vector v.
we count the wheels numbered 3 and 4. The back left wheel gets number 5,
and then moving forward on the left side, the only unnumbered wheel is the
wheel number 6.
If the global position vector of the car’s mass center is given by
¸
[F
J
d=
(10.7)
\F
and the body position vectors of the wheels are
¸
¸
d1
d1
E
E
r1 =
r2 =
z@2
z@2
¸
d2
r3 =
z@2
¸
d3
E
r5 =
z@2
E
d3
r4 =
z@2
¸
d2
E
r6 =
z@2
E
¸
(10.8)
(10.9)
(10.10)
then the global position of the wheels are
6
1
[F z sin # + d1 cos #
:
2
9
J
r1 = J d + J UE E r1 = 7
8
1
\F + z cos # + d1 sin #
2
5
(10.11)
10. Vehicle Planar Dynamics
Y
B
y
G
1
x
w
a1
a3
r1
a2
6
r5
5
635
r4
d
C
r3
l1
3
\
2
r2
r6
l2
X
4
FIGURE 10.4. A six-wheel passenger car and its wheel numbering.
6
1
[F + z sin # + d1 cos #
:
9
2
J
r2 = J d + J UE E r2 = 7
8
1
\F z cos # + d1 sin #
2
(10.12)
6
1
[F + z sin # d2 cos #
:
2
9
J
r3 = J d + J UE E r3 = 7
8
1
\F z cos # d2 sin #
2
(10.13)
6
1
[F + z sin # d3 cos #
:
9
2
J
r4 = J d + J UE E r4 = 7
8
1
\F z cos # d3 sin #
2
(10.14)
6
1
[F z sin # d3 cos #
:
2
9
J
r5 = J d + J UE E r5 = 7
8
1
\F + z cos # d3 sin #
2
(10.15)
6
1
[F z sin # d2 cos #
:
2
9
J
r6 = J d + J UE E r6 = 7
8
1
\F + z cos # d2 sin #
2
(10.16)
5
5
5
5
5
636
10. Vehicle Planar Dynamics
Y
G
y
B
v
E 16q
x
\ 15q
C
d
X
FIGURE 10.5. A car moving on a road with sideslip angle and heading angle
#.
where the rotation matrix between the global J and body coordinate E is
¸
cos # sin #
J
UE =
(10.17)
sin # cos #
Example 410 Cruise angle, attitude angle, and heading angle.
Figure 10.5 illustrates a car moving on a road with the angles
# = 15 deg
= 16 deg
(10.18)
Khdglqj dqjoh = # = 15 deg
(10.19)
The heading angle of the car is
which is the angle between a reference [-axis on the road and the car’s
longitudinal {-axis. The attitude angle of the car is
Dwwlwxgh dqjoh = = 16 deg
(10.20)
which is the angle between the body longitudinal {-axis and the direction of
the car’s motion. The cruise angle of the car is
Fuxlvh dqjoh = + # = 31 deg
(10.21)
which is the angle between the reference [-axis on the road and the car’s
direction of motion.
10. Vehicle Planar Dynamics
Y
G
B
637
v
x
y Fy
E
Fx
vx
vy
C
d
\
X
FIGURE 10.6. A rigid vehicle in a planar motion.
10.2 Rigid Vehicle Newton-Euler Dynamics
A rigid vehicle is assumed to act similar to a at box moving on a horizontal
surface. A rigid vehicle has a planar motion with three degrees of freedom
that are: translation in the { and | directions, and a rotation about the }axis. The Newton-Euler equations of motion for a rigid vehicle in the body
coordinate frame E, attached to the vehicle at its mass center F, are:
I{
I|
P}
= p yb { p$ } y|
= p yb | + p$ } y{
= $b } L}
(10.22)
(10.23)
(10.24)
Proof. Figure 10.6 illustrates a rigid vehicle in a planar motion. A global
coordinate frame J is attached to the ground and a body coordinate frame
E is attached to the vehicle at the mass center F. The ] and } axes are
parallel, and the orientation of the frame E is indicated by the heading
angle # between the [ and { axes. The global position vector of the mass
center is denoted by J d.
The velocity vector of the vehicle, expressed in the body frame, is
6
y{
E
vF = 7 y| 8
0
5
(10.25)
where y{ is the forward component and y| is the lateral component of v.
638
10. Vehicle Planar Dynamics
The rigid body equations of motion in the body coordinate frame are:
¡
¢
E
F = E UJ J F = E UJ p J aE = p E
J aE
E
= p E vb E + p E
vE
J $E ×
E
(10.26)
J
M =
=
g E
Eb
E
b
L+ E
L= E
L
J LE =
J $E ×
gw
¡
¢
E E
E E
L J$
bE + E
L J $E
J $E ×
(10.27)
The force, moment, and kinematic vectors for the rigid vehicle are:
5
5
6
6
I{
0
E
E
FF = 7 I| 8
MF = 7 0 8
(10.28)
P}
0
5
5
6
6
yb {
0
E
E
7 0 8
vb F = 7 yb | 8
(10.29)
J $E =
$}
0
5
6
0
E
bE =7 0 8
(10.30)
J$
$b }
We may assume that the body coordinate is the principal coordinate frame
of the vehicle to have a diagonal moment of inertia matrix.
5
6
L1 0 0
E
L = 7 0 L2 0 8
(10.31)
0 0 L3
Substituting the above vectors and matrices in the equations of motion
(10.26)-(10.27) provides the following equations:
E
F = p E vb E + p E
$ E × E vE
6 5
5
6 J5
6 5
6
y{
yb {
0
pyb { p$ } y|
= p 7 yb | 8 + p 7 0 8 × 7 y| 8 = 7 pyb | + p$ } y{ 8 (10.32)
0
$}
0
0
E
¡E E
¢
M = EL E
bE + E
L J $E
J$
J $E ×
5
65
6
L1 0 0
0
= 7 0 L2 0 8 7 0 8
$b }
0 0 L3
5
6 35
65
64 5
6
0
L1 0 0
0
0
+ 7 0 8 × C7 0 L2 0 8 7 0 8D = 7 0 8
$}
$}
L3 $b }
0 0 L3
(10.33)
10. Vehicle Planar Dynamics
639
The rst two Newton equations (10.32) and the third Euler equation
(10.33) are the only nonzero equations that make up the set of equations
of motion (10.22)-(10.24) for the planar rigid vehicle.
Example 411 Rigid vehicle and Lagrange method.
The kinetic energy of a rigid vehicle in a planar motion is,
N
1
1J W
vE p J vE + J $ WE J L J $ E
2
2
6W
6
5
6W 5
6
5
5
y[
0
y[
0
17
1
y\ 8 p 7 y\ 8 + 7 0 8 J L 7 0 8
2
2
$]
0
0
$]
´
³
2
1
1
1
1
1
py2 + py 2 + L3 $ 2] = p [b 2 + \b 2 + L} #b
2 [ 2 \
2
2
2
=
=
=
(10.34)
where,
J
L
=
J
5
W
UE E L J UE
65
65
6W
cos # sin # 0
L1 0 0
cos # sin # 0
= 7 sin # cos # 0 8 7 0 L2 0 8 7 sin # cos # 0 8
0
0
1
0
0
1
0 0 L3
6
5
2
2
L1 cos # + L2 sin # (L1 L2 ) sin # cos # 0
(10.35)
= 7 (L1 L2 ) sin # cos # L2 cos2 # + L1 sin2 # 0 8
0
0
L3
and
6
6 5
y[
[b
= 7 y\ 8 = 7 \b 8
0
0
5
6 5 6 5
6
0
0
0
= 7 0 8=7 0 8=7 0 8
$]
u
#b
5
J
vE
J $E
(10.36)
(10.37)
The resultant external force system are:
5
6
I[
J
FF = 7 I\ 8
0
6
0
J
MF = 7 0 8
P]
5
(10.38)
Applying the Lagrange method
g
gw
μ
CN
C tbl
¶
CN
= Il
Ctl
l = 1> 2> · · · q
(10.39)
640
10. Vehicle Planar Dynamics
and using the coordinates [, \ , and # for tl , generates the following equations of motion in the global coordinate frame:
pyb {
pyb |
L} $b }
g b
[ = I[
gw
g
= p \b = I\
gw
g
= L} #b = P]
gw
= p
(10.40)
(10.41)
(10.42)
Example 412 Transforming to the body coordinate frame.
We may nd the rigid vehicle’s equations of motion in the body coordinate
frame by expressing the global equations of motion (10.40)-(10.42) in the
vehicle’s body coordinate E, using the transformation matrix J UE .
5
6
cos # sin # 0
J
UE = 7 sin # cos # 0 8
(10.43)
0
0
1
The body expression of the velocity vector is equal to
UE E vF
65
6
y[
cos # sin # 0
y{
7 y\ 8 = 7 sin # cos # 0 8 7 y| 8
0
0
0
1
0
5
6
y{ cos # y| sin #
= 7 y| cos # + y{ sin # 8
0
5
J
E vF
6
=
J
5
and therefore, the global acceleration components are
6
´
³
´
5 ³
yb { #b y| cos # yb | + #b y{ sin #
5
6
:
9
yb [
:
9 ³
´
³
´
:
7 yb \ 8 = 9
b
b
9 yb | + # y{ cos # + yb { # y| sin # :
8
7
0
0
(10.44)
(10.45)
(10.46)
The global Newton’s equation of motion is
J
FF = p J vb F
(10.47)
and the force vector transformation is
J
FF = J UE E FF
(10.48)
therefore, the body coordinate expression for the equations of motion is
E
W J
W J
FF = J UE
FF = p J UE
vb F
(10.49)
10. Vehicle Planar Dynamics
641
Substituting the associated vectors generates the Newton equations of motion in the body coordinate frame.
5 ³
´
³
´
6
yb { #b y| cos # yb | + #b y{ sin #
6
5
9
:
I{
9 ³
:
´
³
´
W 9
:
7 I| 8 = p J UE
9 yb | + #b y{ cos # + yb { #b y| sin # :
7
8
0
0
5
6
yb { #b y|
= p 7 yb | + #b y{ 8
(10.50)
0
Applying the same procedure for moment transformation,
J
MF = J UE E MF
6
5
cos # sin #
0
7 0 8 = 7 sin # cos #
P]
0
0
5
65
6
5
6
0
0
0
0 87 0 8 = 7 0 8
1
P}
P}
(10.51)
(10.52)
we nd the Euler equation in the body coordinate frame.
P} = $b } L}
(10.53)
Example 413 Vehicle path.
When we nd the translational and rotational velocities of a rigid vehicle,
y[ , y\ , u, we may nd the path of motion for the vehicle by integration.
Z
Z
b
(10.54)
# =
# gw = # 0 + u gw
Z
Z
(10.55)
[ =
[b gw = (y{ cos # y| sin #) gw
Z
Z
(10.56)
\ =
\b gw = (y{ sin # + y| cos #) gw
Example 414 F Equations of motion using principal method.
The equations of motion for a rigid vehicle in a planar motion may also
be found by principle of dierential calculus. Consider a vehicle at time
w = 0 that has a lateral velocity y| , a yaw rate u, and a forward velocity
y{ . The longitudinal {-axis makes angle # with a xed [-axis as shown
in Figure 10.7. Point S ({> |) indicates a general point of the vehicle. The
velocity components of point S are
y S { = y{ | u
yS | = y| + { u
(10.57)
because
E
J vS
vF + E
$ E × E rS
6J 5 6 5
6
0
{
y{
= 7 y| 8 + 7 0 8 × 7 | 8
0
u
0
=
E
5
(10.58)
642
10. Vehicle Planar Dynamics
Y
G
vx Gvx
x
B
B
x vx
vy
y
P
v y Gv y
y
\ G\
P
\
r Gr
C
r
X
FIGURE 10.7. A vehicle at time w = 0 and w = gw moving with a lateral velocity
y| , a yaw rate u, and a forward velocity y{ at a heading angle #.
After an increment of time, at w = gw, the vehicle has moved to a new
position. The velocity components of point S at the second position are
yS0 {
yS0 |
= (y{ + gy{ ) | (u + gu)
= (y| + gy| ) + { (u + gu)
(10.59)
(10.60)
However,
yS { + gyS {
yS | + gyS |
= yS0 { cos g# yS0 | sin g#
= yS0 { sin g# + yS0 | cos g#
(10.61)
(10.62)
and therefore,
gyS {
gyS |
= [(y{ + gy{ ) | (u + gu)] cos g#
[(y| + gy| ) + { (u + gu)] sin g# (y{ | u) (10.63)
= [(y{ + gy{ ) | (u + gu)] sin g#
+ [(y| + gy| ) + { (u + gu)] cos g# (y| + { u) (10.64)
We simplify Equations (10.63) and (10.64) and divide them by gw.
gyS {
gw
gyS |
gw
1
([gy{ | gu] cos g#)
gw
1
([(y| + gy| ) + { (u + gu)] sin g#)
gw
1
=
([gy| + { gu] cos g#)
gw
1
+ ([(y{ + gy{ ) | (u + gu)] sin g#)
gw
=
(10.65)
(10.66)
10. Vehicle Planar Dynamics
643
When gw $ 0, then sin g# $ # and cos g# $ 1, and we may substitute
#b = u to get the acceleration components of point S .
= dS { = yb { y| u | ub + { u2
= dS | = yb | + y{ u + { ub | u2
yb S {
yb S |
(10.67)
(10.68)
Let us assume point S has a small mass gp. Multiplying gp by the
acceleration components of point S and integrating over the whole rigid
vehicle must be equal to the applied external force system.
Z
I{ =
dS { gp
(10.69)
Zp
dS | gp
(10.70)
I| =
p
Z
({ dS | | dS { ) gp
(10.71)
P} =
p
Substituting for accelerations and assuming the body coordinate frame is
the principal frame at the mass center F, we nd
Z
¡
¢
I{ =
yb { y| u | ub + { u2 gp
p
Z
Z
| gp + u2
{ gp
= p (yb { y| u) ub
p
p
= p (yb { y| u)
I|
(10.72)
Z
¡
¢
yb | + y{ u + { ub | u2 gp
p
Z
Z
{ gp u2
| gp
= p (yb | + y{ u) + ub
=
p
p
= p (yb | + y{ u)
P}
(10.73)
Z
¡ ¡
¢
¡
¢¢
{ yb | + y{ u + { ub | u2 | yb { y| u | ub + { u2 gp
p
Z
Z
¡ 2
¢
{ gp
{ + | 2 gp + (yb | + y{ u)
= ub
p
p
Z
Z
| gp 2u2
{| gp = L} ub
(10.74)
(yb { y| u)
=
p
p
because for a principal coordinate frame we have
Z
Z
Z
{ gp = 0
| gp = 0
{| gp = 0
p
p
p
(10.75)
644
10. Vehicle Planar Dynamics
x
G1
Fx1
Fxw
C
xw
Mz1
Fy1
1
yw
Fyw
B
W
C
y
FIGURE 10.8. The force system at the tireprint of tire number 1.
10.3 Force System Acting on a Rigid Vehicle
To determine the force system on a rigid vehicle, rst we dene the force
system at the tireprint of a wheel. Then, we transform and apply the tire
force system on the body of the vehicle.
10.3.1 Tire Force and Body Force Systems
Figure 10.8 depicts wheel number 1 of a vehicle. The components of the
force system in the F-frame on {|-plane, because of the forces at the
tireprint of the wheel number l, are
= I{zl cos l I|zl sin l
= I|zl cos l + I{zl sin l
= P}zl
I{l
I|l
P}l
(10.76)
(10.77)
(10.78)
Therefore, the total planar force system on the rigid vehicle in the body
coordinate frame is
X
X
X
E
I{ =
I{l =
I{z cos l I|z sin l
(10.79)
l
E
I|
=
X
l
I|l =
l
E
P}
=
X
l
X
l
I|z cos l +
l
P}l +
X
l
{l I|l X
X
I{z sin l
(10.80)
l
|l I{l
(10.81)
l
Proof. The wheel coordinate frame Z ({z > |z > }z ), shown by or Z , is a
10. Vehicle Planar Dynamics
645
local coordinate frame attached to the center of the wheel. The tire coordinate frame W ({w > |w > }w ) is set at the center of the tireprint. Let us assume
that the force system in the wheel frame is
£
¤W
Z
I{z I|z I|z
Fz =
(10.82)
£
¤W
Z
P{z P|z P|z
Mz =
(10.83)
which is because of the force system at the tireprint
£
¤W
W
I{zl I|zl I}zl
Fz =
£
¤W
W
P{zl P|zl P}zl
Mz =
(10.84)
(10.85)
The tire force in the {z -direction, I{zl , is a combination of the longitudinal force I{z1 , dened by (3.119) or (4.49), and the tire roll resistance
Iu1 dened in (3.88). The tire force in the |z -direction, I|zl , is a combination of the lateral force I|z1 dened by (3.153) and (3.177), and the tire
roll resistance Iu1 dened in (3.88). The tire moment in the }z -direction,
P}zl , is a resultant of the aligning moment P}z1 dened by (3.156) and
(3.184).
The rotation matrix between the wheel frame Z and the wheel-body
coordinate frame F, parallel to the vehicle coordinate frame E, is
¸
cos 1 sin 1
F
UZ =
(10.86)
sin 1
cos 1
and therefore, the force system at the tireprint of the wheel, parallel to the
vehicle coordinate frame, is
Fz = F UZ Z Fz
¸
¸
¸
I{1
cos 1 sin 1
I{z
=
I|1
sin 1
cos 1
I|z
¸
I{z cos 1 I|z sin 1
=
I|z cos 1 + I{z sin 1
F
F
Mz
P}1
= F UZ Z Mz
= P}z
(10.87)
(10.88)
(10.89)
(10.90)
Transforming the force system of each tire to the body coordinate frame
E, located at the body mass center F, generates the total force system
applied on the vehicle
X
X
E
F =
I{l ˆ~ +
I|l ˆ
(10.91)
l
E
M =
X
l
l
P}l n̂ +
X
l
E
rl × E Fzl
(10.92)
646
10. Vehicle Planar Dynamics
where, E rl is the position vector of the wheel number l.
E
rl = {lˆ~ + |l ˆ + }l n̂
(10.93)
Expanding Equations (10.91) and (10.92) provides us with the total planar
force system.
E
I{
X
=
I{z cos l l
E
I|
X
=
P}
X
=
I|z sin l
(10.94)
I{z sin l
(10.95)
X
(10.96)
l
I|z cos l +
l
E
X
P}l +
l
X
X
l
{l I|l l
|l I{l
l
Example 415 Dierence between tireprint and wheel frames.
To express the orientation of a wheel and the force system, three coordinate frames are needed: the wheel frame Z , wheel-body frame F, and
tire frame W . The wheel coordinate frame Z ({z > |z > }z ) is attached to the
center of the wheel and follows every motion of the wheel except the wheel
spin. Therefore, the {z and }z axes are always in the tire-plane, while the
|z -axis is always on the spin axis.
We also attach a wheel-body coordinate frame F ({f > |f > }f ) at the center
of the wheel parallel and xed with respect to the vehicle coordinate axes
E ({> |> }). The wheel-body frame F is motionless with respect to the vehicle
coordinate and does not follow any motion of the wheel. When the wheel is
upright straight, the Z -frame coincides with F-frame and becomes parallel
to the vehicle coordinate frame. The Z -frame makes the steer angle and
camber angle with respect to the F-frame.
The tire coordinate frame W ({w > |w > }w ) is set at the center of the tireprint
with the }w -axis perpendicular to the ground and parallel to the }-axis. The
{w -axis is along the intersection line of the tire-plane and the ground. The
tire frame follows the steer angle rotation about the }f -axis but it does not
follows the spin and camber rotations of the tire.
To determine the dierence between the W and Z frames, let us use W dZ
to indicates the W -expression of the position vector of the wheel frame origin
relative to the tire frame origin. Having the coordinates of a point S in the
wheel frame, we can nd its coordinates in the tire frame by:
W
rS = W UZ Z rS + W dZ
(10.97)
If Z rS indicates the position vector of a point S in the wheel frame,
Z
rS =
£
{S
|S
}S
¤W
(10.98)
10. Vehicle Planar Dynamics
647
then the coordinates of the point S in the tire frame W rS are
W
rS
UZ Z rS + W d = W UZ Z rS + W UZ Z
W dZ
6
{S
= 7 |S cos Uz sin }S sin 8
Uz cos + }S cos + |S sin =
W
5
(10.99)
because
5
1
0
W
UZ = 7 0 cos 0 sin 6
0
sin 8
cos 6
0
Z
7 0 8
W dZ =
Uz
5
(10.100)
where, Z
W dZ is the Z -expression of the position vector of the wheel frame in
the tire frame, Uz is the radius of the wheel, and W UZ is the transformation
matrix from the Z to W .
The center of the wheel Z rS = Z rr = 0 is the origin of the wheel frame
Z , that will be at W rr in the tire coordinate frame W .
5
6
0
W
7 Uz sin 8
rr = W dZ = W UZ Z
(10.101)
W dZ =
Uz cos If the camber angle is zero, = 0, then
6
5
0
W
rr = 7 0 8 = Z
W dZ
Uz
=0
(10.102)
If the force system at the tireprint is W Fz and W Mz then the force system
in the Z -frame at the center of the wheel would be
Z
W W
Fz = Z UW W Fz = W UZ
Fz
6
5
6W 5
6
I{zl
I{zl
1
0
0
7 I|z 8 = 7 0 cos sin 8 7 I|z 8
l
l
0 sin cos I}zl
I}zl
5
Z
Mz
=
W
W
UZ
¡W
Mz W rr × W Fz
¢
6
Uz I|zl cos + Uz I}zl sin 8
P}zl sin Uz I{zl
= 7
P}zl cos 5
(10.103)
(10.104)
(10.105)
where
5
6
0
W
rr = 7 Uz sin 8
Uz cos 6
0
W
Mz = 7 0 8
P}zl
5
(10.106)
648
10. Vehicle Planar Dynamics
The wheel force system at zero camber, = 0, reduces to
6
6
5
5
I{zl
Uz I|zl
Z
Fz = 7 I|zl cos + I}zl sin 8 Z Mz = 7 Uz I{zl 8 (10.107)
I}zl cos I|zl sin P}zl
10.3.2 Tire Lateral Force
Figure 10.9(d) illustrates a tire, moving along the velocity vector v at a
sideslip angle . The tire is steered by the steer angle . If the angle between
the vehicle {-axis and the velocity vector v is shown by , then
=
(10.108)
The angle is called the wheel sideslip angle while is the tire sideslip
angle. If the word "sideslip" is used individually, it always refer to the tire
sideslip angle . The lateral force, generated by a tire, is dependent on
sideslip angle that is proportional to the sideslip for small .
I| = F = F ( )
(10.109)
Proof. A wheel coordinate frame Z ({z > |z ) is attached to the wheel at
the center of wheel as shown in Figure 10.9(d). The orientation of the
wheel frame is measured with respect to the wheel-body coordinate frame
F ({f > |f ), parallel to the vehicle frame E({> |). The angle between the {
and {z axes is the wheel steer angle , measured about the }z -axis. The
wheel is moving along the tire velocity vector v. The angle between the
{z -axis and v is the tire sideslip angle , and the angle between the body
{-axis and v is called the wheel sideslip angle . The angles , , and in
Figure 10.9(d) are positive. The gure shows that
=
(10.110)
Practically, when a steered wheel is moving forward, the relationship
between the angles , , and are such that the velocity vector sits between
the { and {z axes. A practical situation is shown in Figure 10.9(e). A steer
angle will turn the heading of the wheel by a angle. However, because of
tire exibility, the velocity vector of the wheel is lazier than the heading
and turns by a angle, where ? . As a result, a positive steer angle
generates a negative sideslip angle . Analysis of Figure 10.9(e) and
using the denition for positive direction of the angles, show that under a
practical situation we have the same relation (10.108).
According to (3.154), the existence of a sideslip angle is su!cient to
generate a lateral force I| , which is proportional to when the angle is
small.
(10.111)
I| = F = F ( )
10. Vehicle Planar Dynamics
xw v
x
v
E
xw
D
649
x
D E
G
G
C
y
y
yw
yw
W
(a)
(b)
FIGURE 10.9. Angular orientation of a moving tire along the velocity vector v
at a sideslip angle and a steer angle . (d) A 0. (e) ? 0.
Example 416 Extreme velocity cases of a wheel.
Consider a wheel as is shown in Figure 10.9(e) which has a spinning
angular velocity of $ 6= 0 on a frictionless ground. Therefore, the velocity
of the wheel center would be zero, v = 0. The sideslip angle of such a wheel
would be zero, = 0.
Now consider the wheel which has a zero spinning angular velocity $ = 0
and a nonzero translational velocity v 6= 0. The sideslip angle of such a
wheel would be as is shown in Figure 10.9(e). The sideslip angle of a wheel
is not a function of the spinning angular velocity of a wheel.
10.3.3 Two-wheel Model and Body Force Components
Figure 10.10 illustrates the forces in the {|-plane acting at the wheel center
of a front-wheel-steering four-wheel vehicle. When we ignore the roll motion
of the vehicle, the body } and global ] axes are parallel and consequently
the {|-plane remains parallel to the road’s [\ -plane. Therefore, we may
use a two-wheel model for the vehicle. Figure 10.11 illustrates a two-wheel
model for a vehicle with no roll motion. The two-wheel model is also called
a bicycle model, although a two-wheel model does not act similar to a
traditional bicycle.
The force system applied on a bicycle model of vehicle, in which only the
650
10. Vehicle Planar Dynamics
x
G1
Fx1
Fx2
Mz1
Fy1
B
G2
Mz2
Fy2
1
y
C
Fx4
Mz4
Fy4
Fy3
4
2
Fx3
Mz3
3
FIGURE 10.10. A front-wheel-steering four-wheel vehicle and the forces in the
{|-plane acting at the wheel center to be considered equal to the trireprint’s
forces.
front wheel is steerable, and the steer angle is assumed to be small, are:
I{
I|
P}
I{i + I{u
I|i + I|u
d1 I|i d2 I|u
(10.112)
(10.113)
(10.114)
¢
¡
where, I{i > I|i and (I{u > I|u ) are the planar forces on the tireprint of
the front and rear wheels and we consider them at the wheel center.
The vehicle lateral force I| and moment P} depend only on the front
and rear wheels’ lateral forces I|i and I|u , which are functions of the tires
sideslip angles i and u . They can be approximated by:
μ
¶
d1
d2
I| =
Fi + Fu u (Fi + Fu ) + Fi (10.115)
y{
y{
μ 2
¶
d
d2
P} =
1 Fi 2 Fu u (d1 Fi d2 Fu ) + d1 Fi (10.116)
y{
y{
where Fi and Fu are sum of the sideslip coe!cients of the left and right
tires in front and rear, respectively.
Fi
Fu
= FiO + FiU
= FuO + FuU
(10.117)
(10.118)
10. Vehicle Planar Dynamics
651
x
Df
xw
Ef
G
vf
v
E
B
a1
C
y
R
Center of
rotation
O
G
Dr
r
vr
l
a2
R1
FIGURE 10.11. A two-wheel model for a vehicle moving with no roll.
Proof. For the two-wheel vehicle, we use the cot-average (7.4) of the outer
and inner steer angles as the only steer angle .
cot =
cot r + cot l
2
(10.119)
Furthermore, we dene a single sideslip coe!cient Fi and Fu as (10.117)
and (10.118) for the front and rear wheels.
Employing Equations (10.79)-(10.81) and ignoring the aligning moments
P}l , the applied forces on the two-wheel vehicle are:
I{
I|
P}
= I{i cos + I{u I|i sin = I|i cos + I|u + I{i sin = d1 I|i cos + d1 I{i sin d2 I|u
(10.120)
(10.121)
(10.122)
The force equations can be approximated by the following equations, if we
assume small.
I{
I|
P}
I{i + I{u
I|i + I|u
d1 I|i d2 I|u
(10.123)
(10.124)
(10.125)
652
10. Vehicle Planar Dynamics
Assume the wheel number l of a rigid vehicle is located at ({l > |l ) in the
body coordinate frame. The velocity of the wheel number l is
E
b × E rl
vl = E v + E #
(10.126)
in which E rl is the position vector of the wheel number l, E v is the velocity
b is the yaw rate of the
vector of the vehicle at its mass center F, and E #
vehicle.
E b
# = $ } n̂ = un̂
(10.127)
Expanding Equation (10.126) provides the following velocity vector for the
wheel number l expressed in the vehicle coordinate frame at F.
6
6 5
6 5 6 5
6 5
5
y{
0
{l
y{ |l u
y{l
7 y|l 8 = 7 y| 8 + 7 0 8 × 7 |l 8 = 7 y| + {l u 8
(10.128)
0
0
0
u
0
The wheel sideslip l for the wheel l, is the angle between the vehicle body
{-axis and the wheel velocity vector vl .
μ
¶
μ
¶
y|l
y| + {l u
l = arctan
= arctan
(10.129)
y{l
y{ | l u
If the wheel number l has a steer angle l then, its tire sideslip angle l ,
that generates a lateral force I|z on the tire, is
μ
¶
y| + {l u
(10.130)
l
l = l l = arctan
y{ |l u
The wheel sideslip angles l for the front and rear wheels of a two-wheel
vehicle model, i and u , are
μ
¶
μ
¶
y|i
y| + d1 u
= arctan
(10.131)
i = arctan
y{i
y{
μ
¶
μ
¶
y|u
y| d2 u
= arctan
(10.132)
u = arctan
y{u
y{
and the vehicle sideslip angle is
= arctan
μ
y|
y{
¶
(10.133)
Assuming small angles for wheel and vehicle sideslips i , u , and ,
the tire sideslip angles for the front and rear wheels, i and u , may be
approximated as
i
u
1
d1
(y| + d1 u) = + u y{
y{
1
d2
= u =
(y| d2 u) = u
y{
y{
= i =
(10.134)
(10.135)
10. Vehicle Planar Dynamics
653
For small sideslip angles, the associated lateral forces are
I|i
I|u
= Fi i
= Fu u
(10.136)
(10.137)
and therefore, the second and third equations of motion (10.113) and
(10.114) can be written as
I|
P}
= I|i + I|u = Fi i Fu u
¶
¶
μ
μ
d1
d2
= Fi + u Fu u
y{
y{
= d1 I|i d2 I|u = d1 Fi i + d2 Fu u
¶
¶
μ
μ
d1
d2
= d1 Fi + u + d2 Fu u
y{
y{
(10.138)
(10.139)
which can be rearranged to the following force system
¶
μ
d2
d1
I| =
Fu Fi u (Fi + Fu ) + Fi (10.140)
y{
y{
μ 2
¶
d
d2
P} =
1 Fi 2 Fu u + (d2 Fu d1 Fi ) + d1 Fi (10.141)
y{
y{
The parameters Fi , Fu are the sideslip coe!cient for the front and rear
tires, u is the yaw rate, is the front wheel steer angle, and is the sideslip
angle of the vehicle.
These equations are dependent on three parameters, u, , , and may be
written as
I|
P}
CI|
CI|
CI|
u+
+
Cu
C
C
= Fu u + F + F (10.142)
CP}
CP}
CP}
u+
+
Cu
C
C
= Gu u + G + G (10.143)
= I| (u> > ) =
= P} (u> > ) =
where the force system coe!cients are
Fu
=
F
=
F
=
CI|
d2
d1
= Fi + Fu
Cu
y{
y{
CI|
= Fi Fu
C
CI|
= Fi
C
(10.144)
(10.145)
(10.146)
654
10. Vehicle Planar Dynamics
Gu
=
G
=
G
=
d2
CP}
d2
= 1 Fi 2 Fu
Cu
y{
y{
CP}
= d2 Fu d1 Fi
C
CP}
= d1 Fi
C
(10.147)
(10.148)
(10.149)
The coe!cients Fu , F , F , Gu , G , G are slopes of the curves for lateral
force I| and yaw moment P} as a function of u, , and respectively.
Example 417 Physical signicance of the force system coe!cients.
Assuming a steady-state condition and constant values for u, , , Fi ,
and Fu , the lateral force I| and the yaw moment P} can be written as a
superposition of three independent forces proportional to u, , and .
I|
P}
= Fu u + F + F = Gu u + G + G (10.150)
(10.151)
Fu indicates the proportionality between the lateral force I| and the yaw
rate u. The value of Fu decreases by increasing the forward velocity of the
vehicle, y{ .
F indicates the proportionality between the lateral force I| and the vehicle sideslip angle . F is always negative and indicates the lateral stiness
for the whole vehicle. F acts similar to the sideslip coe!cient of tires F .
F indicates the proportionality between the lateral force I| and the steer
angle . F is always positive and generates greater lateral force by increasing the steer angle.
Gu indicates the proportionality between the yaw moment P} and the yaw
rate u. Gu is a negative number and is called the yaw damping coe!cient
because it is proportional to the angular velocity in the P} -direction. The
value of Gu decreases with the forward velocity of the vehicle, y{ .
G indicates the proportionality between the yaw moment P} and the vehicle sideslip angle . G indicates the under/oversteer behavior and hence,
indicates the directional stability of a vehicle. A negative G tries to align
the vehicle with the velocity vector.
G indicates the proportionality between the yaw moment P} and the
steer angle . Because is the input command to control the maneuvering
of a vehicle, G is called the control moment coe!cient. G is a positive
number and increases with d1 and Fi .
Example 418 Kinematic steering of a two-wheel vehicle.
For the two-wheel vehicle shown in Figure 10.12, we use the cot-average
(7.4) of the outer and inner steer angles as the input steer angle,
cot =
cot r + cot l
2
(10.152)
10. Vehicle Planar Dynamics
655
x
vf
xw
G
v
B
E
a1
C
y
l
R
G
Center of
rotation
r
a2
O
R1
FIGURE 10.12. A two-wheel model for a vehicle moving with no roll.
where,
tan l =
o
z
U1 2
tan r =
o
U1 +
z
2
(10.153)
The radius of rotation U for the two-wheel vehicle is given by (7.3)
U=
q
d22 + o2 cot2 (10.154)
and is measured at the mass center of the steered vehicle.
Example 419 Four-wheel rigid vehicle.
The force on each wheel of a planar model of vehicles is
5
F1
F2
6
I{1 cos 1 I|1 sin 1
= 7 I{1 sin 1 + I|1 cos 1 8
0
5
6
I{2 cos 2 I|2 sin 2
= 7 I{2 sin 2 + I|2 cos 2 8
0
5
6
I{3
F3 = 7 I|3 8
0
5
6
I{4
F4 = 7 I|4 8
0
(10.155)
(10.156)
656
10. Vehicle Planar Dynamics
The position vector of the wheels are
5
6
d1
r1 = 7 e1 8
0
5
6
d1
r2 = 7 e2 8
0
5
6
d2
r3 = 7 e1 8
0
5
6
d2
r4 = 7 e2 8
0
(10.157)
(10.158)
and therefore, the force system on a planar model of vehicles is
X
(10.159)
F =
Fl
5
6
I{1 cos 1 + I{2 cos 2 I|1 sin 1 I|2 sin 2 + I{3 + I{4
= 7 I|1 cos 1 + I|2 cos 2 + I{1 sin 1 + I{2 sin 2 + I|3 + I|4 8
0
5
6
0
X
M=
rl × Fl = 7 0 8
(10.160)
P}
P}
= d1 I|1 cos 1 + d1 I|2 cos 2 d2 I|3 d2 I|4
+e1 I|1 sin 1 e2 I|2 sin 2 + e2 I{2 cos 2 e1 I{1 cos 1
+d1 I{1 sin 1 + d1 I{2 sin 2 + e2 I{4 e1 I{3
(10.161)
Let us assume a planar dynamic model of vehicles with four wheels such
that its equation can be approximated by
I{
I|
P}
I{1 + I{2 + I{3 + I{4
I|1 + I|2 + I|3 + I|4
d1 I|1 + d1 I|2 d2 I|3 d2 I|4
(10.162)
(10.163)
(10.164)
in which, we assumed small steer angles and ignored the yew moment caused
by imbalance longitudinal forces, I{l . Knowing that
y{l = y{ |l u
and tire angles as
l
l
we nd that
1
2
y|l = y| + {l u
μ
¶
y| + {l u
= arctan
y{ | l u
μ
¶
y| + {l u
= l l = arctan
l
y{ | l u
= arctan
μ
μ
y|l
y{l
¶
¶
μ
¶
y|1
y| + d1 u
= arctan
= arctan
y{
y{ e 1 u
μ 1¶
μ
¶
y|2
y| + d1 u
= arctan
= arctan
y{2
y{ + e 2 u
(10.165)
(10.166)
(10.167)
(10.168)
(10.169)
10. Vehicle Planar Dynamics
3
4
μ
¶
μ
¶
y|3
y| d2 u
= arctan
y{
y{ + e 2 u
μ
μ 3¶
¶
y| d2 u
y|4
= arctan
= arctan
y{4
y{ e 1 u
= arctan
657
(10.170)
(10.171)
Employing the vehicle sideslip angle = arctan (y| @y{ ) and assuming small
angles for i , u , and , the tire sideslip angles would be:
d1
u
y| + d1 u
y{
1
= 1 1 =
1 =
e1
y{ e1 u
1
u
y{
d1
+
u
y| + d1 u
y{
2
= 2 2 =
2 =
e2
y{ + e2 u
1+
u
y{
+
1
2
3
4
d2
u
y{
e2
1+
u
y{
d2
u
y| d2 u
y{
= 4 =
=
e1
y{ e1 u
1
u
y{
y| d2 u
= 3 =
=
y{ + e2 u
(10.172)
(10.173)
(10.174)
(10.175)
Modeling the lateral forces as
I|l = Fl l
(10.176)
and assuming that the left and right wheels have the same sideslip coe!cients
F1 = F2
F3 = F4
(10.177)
we nd the second and third equations of motion as
I|
= I|1 + I|2 + I|3 + I|4 = F1 1 F1 2 F4 3 F4 4
3
4
3
4
d1
d1
E + y{ u
F
E + y{ u
F
F F1 E
F
= F1 E
1
2
C
D
C
D
e1
e2
1
u
1+
u
y{
y{
d2
d2
u
u
y{
y{
F4
(10.178)
F4
e2
e1
1+
u
1
u
y{
y{
658
10. Vehicle Planar Dynamics
P}
= d1 I|1 + d1 I|2 d2 I|3 d2 I|4
= d1 F1 1 d1 F1 2 + d2 F4 3 + d2 F4 4
3
4
3
4
d1
d1
+
+
u
u
E
F
E
F
y{
y{
1 F
2 F
d1 F1 E
= d1 F1 E
C
D
C
D
e1
e2
1
u
1+
u
y{
y{
d2
d2
u
u
y{
y{
+ d2 F4
(10.179)
+d2 F4
e2
e1
1+
u
1
u
y{
y{
These equations are nonlinear and therefore, they cannot be arranged at the
bicycle model.
Example 420 Linearized equations of four-wheel rigid vehicle
Using the Taylor series expansion formula of a two variable function
i (> u)
Ci
Ci
1 C2i 2 1 C2i
1 C2i
1 C2i 2
+
u + ···
+
u+
u +
u +
2
C
Cu
2 C
2 CCu
2 CuC
2 Cu2
(10.180)
we have
gi =
d1
u
y{
e1
1
u
y{
d2
u
y{
e2
1+
u
y{
+
d1
+ u
y{
d2
u
y{
d1
u
d1
y{
+ u
e2
y{
1+
u
y{
d2
u
d2
y{
u
e1
y{
1
u
y{
+
(10.181)
(10.182)
and therefore, Equations (10.178) and (10.179) become
I|
P}
μ
¶
μ
¶
d1
d1
= F1 + u 1 F1 + u 2
y{
y{
μ
μ
¶
¶
d2
d2
F4 u F4 u
(10.183)
y{
y{
μ
¶
μ
¶
d1
d1
= d1 F1 + u 1 d1 F1 + u 2
y{
y{
¶
¶
μ
μ
d2
d2
+d2 F4 u + d2 F4 u
(10.184)
y{
y{
10. Vehicle Planar Dynamics
659
Rearranged to the force system yields
μ
¶
d2
d1
I| = 2
F4 F1 u 2 (F1 + F3 ) y{
y{
+F1 ( 1 + 2 )
(10.185)
μ 2
¶
d1
d22
P} = 2
F + F u + 2 (d2 F4 d1 F1 ) y{ 1 y { 4
+d1 F1 ( 1 + 2 )
(10.186)
These equations are comparaible with (10.115) and (10.116).
This analysis indicates that adopting the bicycle model of vehicles is
equivalent to accepting a linearized equation of l as function of u and
.
10.4 Two-wheel Rigid Vehicle Dynamics
We combine the planar equations of motion (10.22)-(10.24) with the force
expressions (10.112)-(10.114) to express the motion of a two-wheel rigid
vehicle with no roll, by the following set of equations:
yb { =
¢
1
1 ¡
I{ + u y| =
I{i + I{u + u y|
p
p
(10.187)
6
6
5
Fu
F
F
¸
y{ : y|
9 py{ p
9 p :
:
:
= 9
+9
8
7 G
7 G 8 u
Gu
L}
L} y{
L}
5
6
Fi + Fu
d1 Fi + d2 Fu
y{ ¸
9
: y|
py{
py{
:
= 9
7 d F d F
8 u
d2 Fi + d22 Fu
1 i
2 u
1
L} y{
L} y{
6
5
Fi
:
9
p
:
(10.188)
+9
7 d1 Fi 8 L}
5
yb |
ub
¸
These sets of equations are linear when a vehicle is moving at a constant forward speed. Having yb { = 0, the set of Equations (10.188) become
independent of the rst Equation (10.187). Then, the lateral velocity y|
and yaw rate u of the vehicle will change according to the two coupled
Equations (10.188).
660
10. Vehicle Planar Dynamics
Assuming the steer angle is the input command, the lateral velocity y|
and the yaw rate u may be assumed as the output. Hence, we may consider
Equation (10.188) as a linear control system to write them as
qb = [D] q + u
(10.189)
in which [D] is a coe!cient matrix, q is the vector of control variables, and
u is the vector of inputs.
5
6
Fi + Fu
d1 Fi + d2 Fu
y{
9
:
py{
py{
:
[D] = 9
(10.190)
7 d F d F
8
2
2
d1 Fi + d2 Fu
1 i
2 u
L} y{
L} y{
¸
y|
u
5
Fi
9
p
u = 9
7 d1 Fi
L}
q =
(10.191)
6
:
:
8
(10.192)
Proof. The Newton-Euler equations of motion for a rigid vehicle in the
local coordinate frame E, attached to the vehicle at its mass center F, are
given in Equations (10.22)—(10.24) as
I{
I|
P}
= p yb { pu y|
= p yb | + pu y{
= ub L}
(10.193)
(10.194)
(10.195)
The approximate force system applied on a two-wheel rigid vehicle is found
in Equations (10.112)—(10.114)
I{
I|
P}
I{i + I{u
I|i + I|u
d1 I|i d2 I|u
(10.196)
(10.197)
(10.198)
and in terms of tire characteristics, in (10.115) and (10.116).
I{
=
I|
=
=
P}
=
=
Wz
(10.199)
U
μz
¶
d1
d2
Fi + Fu u (Fi + Fu ) + Fi y{
y{
Fu u + F + F (10.200)
μ 2
¶
2
d
d
1 Fi 2 Fu u (d1 Fi d2 Fu ) + d1 Fi y{
y{
Gu u + G + G (10.201)
10. Vehicle Planar Dynamics
661
Substituting (10.199)—(10.201) in (10.193)—(10.195) produces the following equations of motion:
p yb { pu y|
p yb | + pu y{
ub L}
= I{
= Fu u + F + F = Gu u + G + G (10.202)
(10.203)
(10.204)
Solving the equations for yb { , yb | , and ub yields
I{
+ u y|
p
1
=
(Fu u + F + F ) u y{
pμ
¶
1
d2
d1
=
Fi + Fu u
p
y{
y{
1
1
(Fi + Fu ) + Fi u y{
p
p
yb {
=
yb |
(10.205)
(10.206)
1
(Gu u + G + G )
L}
μ 2
¶
d2
d
1
1 Fi 2 Fu u
=
L}
y{
y{
1
1
(10.207)
(d1 Fi d2 Fu ) + d1 Fi L}
L}
The vehicle sideslip angle can be substituted by vehicle velocity components
y|
(10.208)
=
y{
So, we can nd the equations of motion in terms of the three variables, y{ ,
y| , u.
ub
=
I{
+ u y|
pμ
¶
1
F
y| + F u y{
Fu u +
yb | =
p
y{
1
=
(d1 Fi + d2 Fu ) u
py{
1
1
(Fi + Fu ) y| + Fi u y{
py{
p
¶
μ
G
1
y| + G Gu u +
ub =
L}
y{
¢
1 ¡ 2
d1 Fi d22 Fu u
=
L} y{
1
1
(d1 Fi d2 Fu ) y| + d1 Fi L} y{
L}
yb {
=
(10.209)
(10.210)
(10.211)
662
10. Vehicle Planar Dynamics
These equations are coupled. The rst equation (10.209) depends on the
yaw rate u and the lateral velocity y| , which are the output of the second
and third equations, (10.210) and (10.211). Also Equations (10.210) and
(10.211) depend on y{ which is the output of the rst equation. However,
if we assume the vehicle is moving with a constant forward speed,
y{ = frqvw
(10.212)
then Equation (10.209) becomes an algebraic equation and then Equations
(10.210) and (10.211) become independent with (10.209). So, the second
and third equations can be treated independent of the rst one.
Equations (10.210) and (10.211) can also be considered as two coupled
dierential equations describing the behavior of a dynamic system. The
dynamic system receives the steering angle as the input, and uses y{ as
a parameter to generate two outputs, y| and u.
5
yb |
ub
¸
9
= 9
7
Fi + Fu
py{
d1 Fi d2 Fu
L} y{
6
Fi
:
9
p
:
+9
7 d1 Fi 8 L}
5
6
d1 Fi + d2 Fu
y{ ¸
: y|
py{
:
8 u
d2 Fi + d22 Fu
1
L} y{
(10.213)
We can rearrange Equation (10.213) in the following form to show the
input-output relationship:
qb = [D] q + u
(10.214)
The vector q is called the control variables vector, and u is called the inputs
vector. The matrix [D] is the control variable coe!cients matrix.
Employing the force system coe!cients Fu , F , F , Gu , G , and G from
(10.144)-(10.149) for a front-wheel steering vehicle, we may write the set
of Equations (10.213) as
6
6
5
5
Fu
F
F
¸
¸
y{ : y|
9 py{ p
9 p :
yb |
:
:
(10.215)
=9
+9
8
7
7 G 8 ub
u
G
Gu
L}
L} y{
L}
Example 421 Equations of motion based on kinematic angles.
The equations of motion (10.188) can be expressed based on only the
angles , u, and , by employing (10.208).
10. Vehicle Planar Dynamics
663
Taking a time derivative from Equation (10.208) for constant y{
yb |
b =
y{
(10.216)
and substituting it in Equations (10.206) shows that we can transform the
equation for b to:
¶
μ
d2
1
d1
Fi + Fu u
y{ b =
p
y{
y{
1
1
(10.217)
(Fi + Fu ) + Fi u y{
p
p
Therefore, the set of equations of motion can be expressed in terms of the
vehicle’s angular variables , u, and .
6
5
d1 Fi + d2 Fu
Fi + Fu
1
¸
¸
: 9
py{
py{2
b
:
9
= 7
8 u
ub
d2 Fi + d22 Fu
d1 Fi d2 Fu
1
L}
L} y{
6
5
Fi
9 py{ :
:
+9
(10.218)
7 d1 Fi 8 L}
Employing the force system coe!cients Fu , F , F , Gu , G , and G for a
front-wheel steering vehicle, we can write the set of Equations (10.218) as
6
5
6
5
F
Fu
F
¸
¸
1 : 9 py{ :
9 py{ py{
b
:
9
:
+
(10.219)
=9
8 u
7 G 8 7 G
ub
Gu
L}
L}
L}
Example 422 Four-wheel-steering vehicles.
Consider a vehicle with steerable wheels in front and rear. Let us indicate
the steer angle in front and rear by i and u respectively. To nd the
equations of motion of the bicycle planar model, we begin with Equation
(10.108) for the relationship between , , and =
(10.220)
and apply the equation to the front and rear wheels.
i
u
1
d1 u
(y| + d1 u) i = +
i (10.221)
y{
y{
1
d2 u
= u u =
(y| d2 u) u = u (10.222)
y{
y{
= i i =
664
10. Vehicle Planar Dynamics
When the sideslip angles are small, the associated lateral forces are
I|i = Fi i
I|u = Fu u
(10.223)
Substituting these equations in the second and third equations of motion
(10.113) and (10.114) yields:
μ
¶
d1
d2
I| = I|i + I|u = Fi + Fu u
y{
y{
(Fi + Fu ) + Fi i + Fu u
(10.224)
μ 2
¶
2
d
d
P} = d1 I|i d2 I|u = 1 Fi 2 Fu u
y{
y{
(d1 Fi d2 Fu ) + d1 Fi i d2 Fu u
(10.225)
The Newton-Euler equations of motion for a rigid vehicle are given in
Equations (10.22)—(10.24) as
I{
I|
P}
= p yb { pu y|
= p yb | + pu y{
= ub L}
(10.226)
(10.227)
(10.228)
and therefore, the equations of motion for a four-wheel-steering vehicle are
p yb { pu y|
p yb | + pu y{
ub L}
= I{
μ
¶
d1
d2
=
Fi + Fu u
y{
y{
(Fi + Fu ) + Fi i + Fu u
μ 2
¶
d1
d22
=
Fi Fu u
y{
y{
(d1 Fi d2 Fu ) + d1 Fi i d2 Fu u
(10.229)
(10.230)
(10.231)
Solving these equations for yb { , yb | , and ub provide us with three rst order
coupled equations.
yb {
yb |
ub
I{
(10.232)
+ u y|
pμ
¶
1
d2
d1
=
Fi + Fu u
p
y{
y{
1
1
1
(10.233)
(Fi + Fu ) + Fi i + Fu u u y{
pμ
p
p
¶
d2
d2
1
1 Fi 2 Fu u
=
L}
y{
y{
1
1
1
(d1 Fi d2 Fu ) + d1 Fi i d2 Fu u (10.234)
L}
L}
L}
=
10. Vehicle Planar Dynamics
665
The three variables y{ , y| , , are related by the vehicle sideslip angle
=
y|
y{
(10.235)
Using this relation, we can replace by y| @y{ and transform the equations to the following set of three coupled rst order ordinary dierential
equations:
I{
+ u y|
p
1
=
(d1 Fi + d2 Fu ) u
py{
1
1
1
(Fi + Fu ) y| + Fi i + Fu u u y{
py{
p
p
¢
1 ¡ 2
=
d1 Fi d22 Fu u
L} y{
1
1
1
(d1 Fi d2 Fu ) y| + d1 Fi i d2 Fu u
L} y{
L}
L}
yb {
=
yb |
ub
The second and third equations may be cast in a matrix form for
yb |
ub
¸
Fi + Fu
py{
or to the following form for
b
ub
¸
(10.238)
y|
u
¤W
:
£
u
¤W
:
6
d1 Fi + d2 Fu
1
¸
: 9
py{2
:
9
= 7
8 u
d2 Fi + d22 Fu
d1 Fi d2 Fu
1
L}
L} y{
5
6
1
1
¸
F
Fu
9 py{ i
: i
py{
:
+9
(10.240)
7 1
8 u
1
d1 Fi d2 Fu
L}
L}
5
(10.237)
6
d1 Fi + d2 Fu
y{ ¸
: y|
9
py{
:
9
= 7
8 u
d1 Fi d2 Fu
d2 Fi + d22 Fu
1
L} y{
L} y{
5
6
1
1
¸
Fu : 9 p Fi
p
i
9
:
+7
(10.239)
8 u
1
1
d1 Fi d2 Fu
L}
L}
5
£
(10.236)
Fi + Fu
py{
For computerization of the equations of motion, it is better to write them
666
10. Vehicle Planar Dynamics
as
5
5
6
Fi
Fu
¸ 9 p
y{ : y|
p
9
:
+9
8
u
7 G
Gu
i
L}
L}
6
Fu
¸
p :
: i
:
Gu 8 u
L}
(10.241)
5
5 F
6
i
Fu
¸ 9 py
1 {
: py{
9
:
+9
8
u
7 G
Gu
i
L}
L
6
Fu
¸
py{ :
: i
:
u
Gu 8
L}
(10.242)
F
¸
9 py{
yb |
=9
7 G
ub
L} y{
or
b
ub
¸
F
9 py{
=9
7 G
L}
}
where
Fu
=
F
=
Fi
=
Fu
=
Gu
=
G
=
Gi
=
Gu
=
CI|
d2
d1
= Fi + Fu
Cu
y{
y{
CI|
= (Fi + Fu )
C
CI|
= Fi
C i
CI|
= Fu
C u
CP}
d2
d2
= 1 Fi 2 Fu
Cu
y{
y{
CP}
= (d1 Fi d2 Fu )
C
CP}
= d1 Fu
C i
CP}
= d2 Fu
C u
(10.243)
(10.244)
(10.245)
(10.246)
(10.247)
(10.248)
(10.249)
(10.250)
Equation (10.239) may be rearranged in the following form to show the
input-output relationship:
qb = [D] q + [E] u
The vector q is called the control variables vector,
¸
y|
q=
u
(10.251)
(10.252)
10. Vehicle Planar Dynamics
and u is called the inputs vector.
u=
i
u
¸
667
(10.253)
The matrix [D] is the control variable coe!cients matrix and the matrix
[E] is the input coe!cient matrix.
To double check, we may substitute u = 0, and i = to reduce Equations (10.239) to (10.213) for a front-wheel-steering vehicle.
Example 423 Rear-wheel-steering vehicle.
Rear wheel steering is frequently employed in lift trucks and construction
vehicles. The equations of motion for rear steering vehicles are similar to
those in front steering. To nd the equations of motion, we substitute i = 0
in Equations (10.239) to nd these equations
5
6
Fi + Fu
d1 Fi + d2 Fu
y{ ¸
¸
9
: y|
py{
py{
yb |
9
:
= 7
8 u
ub
d1 Fi d2 Fu
d2 Fi + d22 Fu
1
L} y{
L} y{
6
5
1
9 p Fu :
: u
(10.254)
+9
8
7
1
d2 Fu
L}
These equations are valid as long as the angles are very small. However,
most of the rear steering construction vehicles work at a high steer angle.
Therefore, these equations cannot predict the behavior of the construction
vehicles very well.
Example 424 F Better model for two-wheel vehicles.
Because of the steer angle, a reaction moment appears at the tireprints
of the front and rear wheels, which act as external moments Pi and Pu
on the wheels. Therefore, the total steering reaction moment on the front
and rear wheels are
Pi 2Gi P}
Pu 2Gu P}
(10.255)
where
gP}
gP}
Gu =
(10.256)
g i
g u
Figure 10.13 illustrates a two-wheel vehicle model with the reaction moments Pi and Pu . The force system on the vehicle is
Gi =
I{
I|
P}
I{i + I{u
I|i + I|u
d1 I|i d2 I|u + Pi + Pu
(10.257)
(10.258)
(10.259)
668
10. Vehicle Planar Dynamics
xw
Df
vf x
Ef
G
v
Center of
rotation
Mf
O
R
E
a1
C
y
r
Dr
B
l
vr
a2
Mr
FIGURE 10.13. A two-wheel vehicle and its force system, including the steer
moment reactions.
Example 425 F The race car 180 deg quick turn from reverse.
You have seen that stunt car drivers can turn a 180 deg quickly when
the car is moving backward. Here is how they do this. The driver moves
backward when the car is in reverse gear. To make a fast 180 deg turn
without stopping, the driver may follow these steps: 1=The driver should
push the gas pedal to gain enough speed, 2=free the gas pedal and put the
gear in neutral, 3=cut the steering wheel sharply around 90 deg, 4=change
the gear to drive, and 5=push the gas pedal and return the steering wheel to
0 deg after the car has completed the 180 deg turn.
The backward speed before step 2 may be around 20 m@ s 70 km@ h 45 mi@ h. Steps 2 to 4 should be done fast and almost simultaneously. Figure
10.14(d) illustrates the 180 deg fast turning maneuver from reverse.
This example should never be performed by the reader of this book due to
high safety risk.
Example 426 F The race car 180 deg quick turn from forward.
You have seen that stunt car drivers can turn a 180 deg quickly when the
car is moving forward. Here is how they do this. The driver moves forward
when the car is in drive or a forward gear. To make a fast 180 deg turn
without stopping, the driver may follow these steps: 1=The driver should
push the gas pedal to gain enough speed, 2=free the gas pedal and put the
gear in neutral, 3=cut the steering wheel sharply around 90 deg while pulling
10. Vehicle Planar Dynamics
669
1
1
v
v
2
2
3
3
4
4
v
5
5
v
(a)
(b)
FIGURE 10.14. (d). A 180 deg, fast turning maneuver from reverse. (e). A 180 deg,
fast turning maneuver from forward.
hard the hand brake, 4=while the rear of the car swings around, return the
steering wheel to 0 deg and put the gear into drive, and 5=push the gas pedal
after the car has completed the 180 deg turn. Figure 10.14(e) illustrates this
maneuver.
The forward speed before step 2 may be around 20 m@ s 70 km@ h 45 mi@ h. Steps 2 to 4 should be done fast and almost simultaneously. The
180 deg fast turning from forward is more di!cult than backward and can
be done because the hand brakes are connected to the rear wheels. It can be
done better when the rear of a car is lighter than the front to slide easier.
Road condition, nonuniform friction, slippery surface can cause ipping the
car and spinning out of control.
This example should never be performed by the reader of this book due to
high safety risk.
670
10. Vehicle Planar Dynamics
10.5 Steady-State Turning
The turning of a front-wheel-steering two-wheel rigid vehicle at its steadystate condition is governed by the following equations:
I{
Fu u + F + F Gu u + G + G = pu y|
= pu y{
= 0
(10.260)
(10.261)
(10.262)
or equivalently, by
I{
¡
¢ 1
F + Fu y{ p y{2
U
1
G + Gu y{
U
= p
y{ y|
U
(10.263)
= F (10.264)
= G (10.265)
The rst equation determines the required forward force to keep y{ constant
when the car is on a circular path. The second and third equations show
the steady-state values of the output variables, vehicle slip angle , and
path curvature ,
1
u
=
(10.266)
=
U
y{
for a constant steering input at a constant forward speed y{ . The steadystate output-input relationships of the vehicle are dened by the following
responses:
1= Curvature response, V
V =
1
F G F G
=
=
U
y{ (Gu F Fu G + py{ G )
(10.267)
2= Sideslip response, V
V =
G (Fu py{ ) Gu F
=
Gu F Fu G + py{ G
(10.268)
3= Yaw rate response, Vu
Vu =
u
F G F G
= y{ = V y{ =
Gu F Fu G + py{ G
(10.269)
4= Centripetal acceleration response, Vd
Vd =
y{2 @U
(F G F G ) y{
= y{2 = V y{2 =
Gu F Fu G + py{ G
(10.270)
10. Vehicle Planar Dynamics
671
5= Lateral velocity response, V| .
V| =
y|
G (Fu py{ ) Gu F
y{
= V y{ =
Gu F Fu G + py{ G
(10.271)
Proof. In steady-state conditions, all the variables are constant, and hence,
their derivatives are zero. Therefore, the equations of motion (10.202)—
(10.204) reduce to
I{
I|
P}
= pu y|
= pu y{
= 0
(10.272)
(10.273)
(10.274)
where the lateral force I| and yaw moment P} from (10.142) and (10.143)
are
I|
P}
= Fu u + F + F = Gu u + G + G (10.275)
(10.276)
Therefore, the equations describing the steady-state turning of a two-wheel
rigid vehicle are
I{
Fu u + F + F Gu u + G + G = pu y|
= pu y{
= 0
(10.277)
(10.278)
(10.279)
At steady-state turning, the mass center of the vehicle moves on a circle
with radius U at a forward speed y{ and angular velocity u, so y{ and u
are approximately related by
y{ U u
(10.280)
Substituting (10.280) in Equations (10.277)-(10.279), we may write the
equations as (10.263)-(10.265). Equation (10.263) may be used to calculate
the required traction force to keep the motion steady. However, Equations
(10.264) and (10.265) can be used to determine the steady-state responses
of the vehicle. We use the curvature denition (10.266) and write the equations in matrix form
¸
¸ ¸
F Fu y{ p y{2
F
=
(10.281)
G
Gu y{
G
Solving the equations for and shows that
¸1 ¸
¸
F
F Fu y{ p y{2
=
G
Gu y{
G
5
6
G (Fu py{ ) Gu F
9
:
Gu F Fu G + py{ G
9
:
= 9
:
7
8
F G F G
y{ (Gu F Fu G + py{ G )
(10.282)
672
10. Vehicle Planar Dynamics
Based on the solutions of (10.282) and by using Equation (10.280), we are
able to dene dierent output-input relationships as (10.267)-(10.270).
Using
y|
=
(10.283)
y{
Equations (10.278)-(10.279) will be
F
y| + F y{
G
Gu u +
y| + G y{
Fu u +
= pu y{
(10.284)
= 0
(10.285)
rearrangement in matrix form
5
6
F
Fu y{ p y{2 y ¸ F ¸
9 y{
:
|
7 G
8 = G
Gu y{
y{
(10.286)
yields:
5
y|
¸
9
9
=9
7
G (Fu py{ ) Gu F
y{
Gu F Fu G + py{ G
F G F G
y{ (Gu F Fu G + py{ G )
6
:
:
:
8
(10.287)
Then the lateral velocity response V| in (10.271) would be an outcome of
this result.
Example 427 Force system coe!cients for a car.
Consider a front-wheel-steering, four-wheel-car with the following characteristics.
FiO
FuO
pj
L}
d1
= FiU = 500 N@ deg 28648 N@ rad
112=4 lb@ deg 6440 lb@ rad
(10.288)
= FuU = 460 N@ deg 26356 N@ rad
103=4 lb@ deg 5924=4 lb@ rad
(10.289)
= 9000 N 2023 lb
p = 917 kg 62=8 slug
= 1128 kg m2 832 slug ft2
= 91 cm 2=98 ft
d2 = 164 cm 5=38 ft
(10.290)
The sideslip coe!cient of an equivalent bicycle model are
Fi
Fu
= FiO + FiU = 57296 N@ rad
= FuO + FuU = 52712 N@ rad
(10.291)
(10.292)
10. Vehicle Planar Dynamics
673
The force system coe!cients Fu , F , F , Gu , G , G are then equal to the
following in which y{ is measured in [ m@ s].
Fu
F
F
Gu
G
G
d1
d2
34308
Fi + Fu =
N s@ rad
y{
y{
y{
= (Fi + Fu ) = 110008 N@ rad
= Fi = 57296 N@ rad
= d2
d2
189221
= 1 Fi 2 Fu = N m s@ rad
y{
y{
y{
= (d1 Fi d2 Fu ) = 34308 N m@ rad
= d1 Fi = 52139 N m@ rad
(10.293)
(10.294)
(10.295)
(10.296)
(10.297)
(10.298)
The coe!cients Fu and Gu are the only force coe!cients that are functions
of the forward speed y{ . As an example Fu and Gu at
y{ = 10 m@ s = 36 km@ h 32=81 ft@ s 22=37 mi@ h
(10.299)
Fu = 3430=8 N s@ rad
Gu = 18922 N m s@ rad
(10.300)
y{ = 30 m@ s = 108 km@ h 98=43 ft@ s 67=11 mi@ h
(10.301)
Fu = 1143=6 N s@ rad
(10.302)
are
and at
are
Gu = 6307=3 N m s@ rad
Example 428 Steady state responses of an understeer vehicle.
The steady-state responses are function of the forward velocity of a vehicle. To visualize how these steady-state parameters vary when the speed
of a vehicle changes, we calculate V , V , Vu , Vd , V| for a vehicle with the
following characteristics.
FiO = FiU 3000 N@ rad
FuO = FuU 3000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=0 m
d2 = 1=5 m
μ
¶
p
d1
d2
N = 2
= 1=33 × 102
o
Fi
Fu
30
75 + y{2
V
=
Vd
= V y{2 =
V =
30y{2
75 + y{2
45 2y{2
75 + y{2
(10.303)
(10.304)
(10.305)
(10.306)
30y{
75 + y{2
45 2y{2
V| =
y{ (10.307)
75 + y{2
Vu = V y{ =
674
SN
10. Vehicle Planar Dynamics
1/ R
G
vx > m / s @
FIGURE 10.15. Curvature response, V , as a function of forward velocity y{ .
The parameter N is called the stability factor and determines if a vehicle is
understeer or oversteer. A positive N indicates a stable understeer vehicle
and considered desirable. Figures 10.15-10.19 illustrate the steady-states
variations by increasing the forward velocity.
At very slow speed, y{ 0, the steady state responses are at
V = 0=4
V = 0=6
Vu = 0
Vd = 0
(10.308)
By increasing the speed of the car, V decreases and approaches zero, which
means the radius of rotation increases for constant .
The value of V also decreases by increasing y{ and approaches a constant
negative value, V $ 2. Therefore, the center of rotation starts at a point
on the rear axis when A 0, and moves away from the car and moves
forward where ? 0.
Vu has a strange behavior which
s begins from zero and increases with y{ for
a while to a maximum at y{ = 75. Then it decreases and asymptotically
approaches zero.
Vd shows the centripetal acceleration of the car while it is turning on a
circle with a constant speed. The value of Vd starts with zero at y{ = 0 and
increases rapidly to approach Vd $ 30 asymptotically when y{ $ 4.
The lateral velocity response V| indicates the speed of the vehicle at its
mass center in the |-direction. It starts at zero and increases to a maximum
before decreasing monotonically to negative values.
Example 429 Where is V before steady state?
For a given vehicle, the geometric parameters d1 , d2 , p, Fi , Fu are set.
When the steer angle and forward velocity y{ are constant, it take some
time for the car to achieve the steady state value of V . Now let us assume remains unchanged and, we change y{ to a new constant value. The vehicle
10. Vehicle Planar Dynamics
SE
675
E
G
vx > m / s @
FIGURE 10.16. Sideslip response, V , as a function of forward velocity y{ .
Sr
r
G
vx > m / s @
FIGURE 10.17. Yaw rate response, Vu , as a function of forward velocity y{ .
676
Sa
10. Vehicle Planar Dynamics
vx2 / R
G
vx > m / s @
FIGURE 10.18. Lateral acceleration response, Vd , as a function of forward velocity y{ .
Sy
vy
G
vx > m / s @
FIGURE 10.19. Lateral velocity response, V| , as a function of forward velocity
y{ .
10. Vehicle Planar Dynamics
677
will move from the previous point on V to a new value. However, moving
from one point to the other point will not be done by following the V -curve,
as this curve is valid only at steady-state conditions, while moving between
two steady-state points is a transition.
When a set of parameters and variables is xed, approached from above
or below of the V -curve to the set point on V depends the initial conditions
of the vehicle.
Example 430 Curvature response behavior V .
Substituting the force coe!cients (10.144)-(10.149) in the curvature response, V
V =
1
F G F G
=
=
U
y{ (Gu F G (Fu py{ ))
(10.309)
oFi Fu
p (Fu d2 Fi d1 ) y{2 + o2 Fi Fu
(10.310)
yields:
V =
For a given set of parameters d1 , d2 , p, Fi , Fu , the curvature response
begins at
1
1
lim V =
=
(10.311)
y{ =0
d1 + d2
o
and ends up at
lim V = 0
(10.312)
y{ =4
If Fi = Fu = F and d1 = d2 = o@2 then V simplies to be constant
and independent of y{ .
1
V =
(10.313)
o
and therefore, the radius of rotation is also constant and velocity independent.
o
U=
(10.314)
If Fi = Fu = F and d1 6= d2 then V simplies to
V =
oF
o2 F + py{2 (d2 d1 )
(10.315)
and therefore, the radius of rotation is also velocity dependent
U=
o2 F + py{2 (d2 d1 )
oF (10.316)
At a given speed, depending on the position of the mass center, U can be
between two extreme values when d1 = 0 and d2 = 0
U
oF + py{2
oF py{2
?
?
oF
o
oF
(10.317)
678
10. Vehicle Planar Dynamics
RG
l
a1
0.4 l
a2
0.6 l
a1
0.6 l
a2
0.4 l
Critical speed
vx > m / s @
FIGURE 10.20. The limits of steady state U@o for a vehicle and the existence of
a critical speed for d1 ? d2 .
In practical conditions, we have 0=4o ? d1 ? 0=6o and 0=6o A d2 A 0=4o. Figure 10.20 illustrates the limits of U depending on the mass center position,
for the following characteristics.
Fi = 6000 N@ rad
p = 1000 kg
d1 = 1=0 m
Fu = 6000 N@ rad
L} = 1650 kg m2
d2 = 1=5 m
(10.318)
(10.319)
(10.320)
When d1 ? d2 and the front of the car is heavier than the back, the steadystate U increases monotonically by increasing y{ . However, when d1 A d2
and the back of the car is heavier than the front then, U decreases monotonically by increasing y{ starting from a positive value. Therefore, for a heavier
back and positive , a car begins to turn to left at low speed. By increasing
y{ , the steady-state value of U decreases until it becomes theoretically zero
for a critical speed yf , at which the vehicle would be unstable and uncontrollable.
s
F
o
d1 A d2
(10.321)
yf =
p (d1 d2 )
Example 431 Sideslip response behavior V .
Substituting the force coe!cients (10.144)-(10.149) in the sideslip response, V
G (Fu py{ ) Gu F
V = =
(10.322)
Gu F Fu G + py{ G
yields:
d2 oFi Fu pd1 y{2 Fi
V =
(10.323)
p (Fu d2 Fi d1 ) y{2 + o2 Fi Fu
10. Vehicle Planar Dynamics
679
For a given set of parameters d1 , d2 , p, Fi , Fu , the sideslip response
begins at
d2
d2
lim V =
=
(10.324)
y{ =0
d1 + d2
o
and ends up at
lim V =
y{ =4
d1 Fi
d1 Fi d2 Fu
(10.325)
If Fi = Fu = F and d1 = d2 = o@2 then V simplies to
V =
oF py{2
2oF
(10.326)
If Fi = Fu = F and d1 6= d2 then V simplies to
V =
d2 oF pd1 y{2
p (d2 d1 ) y{2 + o2 F
(10.327)
V is proportional to which indicates the angle of the vehicle velocity
vector measured from the body longitudinal {-axis. At very low speed, A 0
and the front wheel is turning on a bigger circle than the rear. By increasing
y{ , the vehicle sideslip angle decreases until it becomes zero at a critical
speed
r
d2 oFu
(10.328)
yf =
pd1
When y{ = yf , the vehicle is turning such that the {-axis is perpendicular
to the radius of rotation U at the mass center. When y{ A yf , then ? 0
and the rear wheel is turning on a bigger circle than the front.
Example 432 Yaw rate response behavior Vu .
Substituting the force coe!cients (10.144)-(10.149) in the yaw rate response, Vu
Vu =
u
F G F G
= y{ = V y{ =
Gu F Fu G + py{ G
(10.329)
y{ oFi Fu
p (Fu d2 Fi d1 ) y{2 + o2 Fi Fu
(10.330)
yields:
Vu =
For a given set of parameters d1 , d2 , p, Fi , Fu , the yaw rate response
begins at
g Vu
lim Vu = 0
A0
(10.331)
y{ =0
g {{
and ends up at
lim Vu = 0
y{ =4
g Vu
?0
g {{
(10.332)
680
10. Vehicle Planar Dynamics
Therefore, there must be at least a maximum for Vu at a critical speed.
Taking a derivative
o2 Fi Fu p (Fu d2 Fi d1 ) y{2
g Vu
= oFi Fu
2
g {{
(o2 Fi Fu + p (Fu d2 Fi d1 ) y{2 )
determines that Vu is at maximum when y{ is at
s
o2 Fi Fu
yf =
p (Fu d2 Fi d1 )
Therefore, the maximum yaw rate of a vehicle at y{ = yf is
s
Fi Fu
u=
2 p (Fu d2 Fi d1 )
(10.333)
(10.334)
(10.335)
If Fi = Fu = F and d1 = d2 = o@2 then Vu simplies to
Vu =
y{
o
(10.336)
If Fi = Fu = F and d1 6= d2 then Vu simplies to
Vu =
oy{ F
o2 F + p (d2 d1 ) y{2
(10.337)
Example 433 Centripetal acceleration response behavior Vd .
Examining Figure 10.21, we know that
y{
U1
= yi cos i = y cos = yu cos u
= U cos (10.338)
(10.339)
Therefore, the centripetal acceleration of the vehicle is y 2 @U or equivalently
y{2 @U1 is not proportional to the centripetal acceleration response Vd .
y{2
Vd y2
=
=
U
U cos2 cos2 (10.340)
However, considering small indicates that
Vd =
1 y 2 cos2 1 y2
1 y{2
=
U
U
U
(10.341)
Substituting the force coe!cients (10.144)-(10.149) in the centripetal acceleration response, Vd
Vd =
y{2 @U
(F G F G ) y{
= y{2 = V y{2 =
Gu F Fu G + py{ G
(10.342)
10. Vehicle Planar Dynamics
xw
Df
681
vf x
Ef
G
v
Center of
rotation
Mf
O
R
y
E
a1
E
C
R1
r
Dr
l
vr
B
a2
Mr
FIGURE 10.21. Kinematics of a moving vehicle at steady-state conditions.
yields:
Vd =
y{2 oFi Fu
p (Fu d2 Fi d1 ) y{2 + o2 Fi Fu
(10.343)
For a given set of parameters d1 , d2 , p, Fi , Fu , the centripetal acceleration response begins at
lim Vd = 0
(10.344)
y{ =0
and ends up at
lim Vd =
y{ =4
oFi Fu
p (Fu d2 Fi d1 )
(10.345)
If Fi = Fu = F and d1 = d2 = o@2 then Vd simplies to
Vd =
y{2
o
(10.346)
If Fi = Fu = F and d1 6= d2 then Vd simplies to
oy{2 F
(10.347)
o2 F + p (d2 d1 ) y{2
¡
¢
The lateral acceleration response Vd = y{2 @U @ is actually the centripetal
acceleration of the vehicle of its mass center. It is the exact lateral acceleration only when = 0 and U is perpendicular to the {-axis. It is the only
Vd =
682
10. Vehicle Planar Dynamics
time that the centripetal acceleration has no longitudinal component. However, assuming a vehicle as a massive point turning about a central point
with radius U justies to use the term of lateral acceleration.
Example 434 Lateral velocity response behavior V| .
Substituting the force coe!cients (10.144)-(10.149) in the lateral velocity
response, V|
V| =
y|
G (Fu py{ ) Gu F
y{
= V y{ =
Gu F Fu G + py{ G
(10.348)
d2 oFi Fu pd1 y{2 Fi
y{
p (Fu d2 Fi d1 ) y{2 + o2 Fi Fu
(10.349)
yields:
V| =
For a given set of parameters d1 , d2 , p, Fi , Fu , the lateral velocity
response begins at
(10.350)
lim V| = 0
y{ =0
and ends up at
lim V| =
y{ =4
½
4 d1 Fi d2 Fu ? 0
4 d1 Fi d2 Fu A 0
(10.351)
If Fi = Fu = F and d1 = d2 = o@2 then V| simplies to
V| =
oF py{2
y{
2oF
(10.352)
If Fi = Fu = F and d1 6= d2 then V| simplies to
V| =
d2 oF pd1 y{2
y{
p (d2 d1 ) y{2 + o2 F
(10.353)
V| is proportional to which indicates the angle of the vehicle velocity
vector measured from the body longitudinal {-axis. At very low speed, A 0
and the front wheel is turning on a bigger circle than the rear to make
y| positive. By increasing y{ , the vehicle sideslip angle decreases until it
becomes zero at a critical speed at which the lateral velocity also becomes
zero.
r
d2 oFu
(10.354)
yf =
pd1
When y{ = yf , the vehicle is turning such that the {-axis is perpendicular to
the radius of rotation U at the mass center. Therefore, the lateral component
of the velocity vector at the mass center becomes zero. When y{ A yf , then
? 0 and the rear wheel is turning on a bigger circle than the front. At
this moment, the velocity vector of F has a negative |-component.
10. Vehicle Planar Dynamics
683
As a result, y| is not a good indicator to show if the vehicle is turning in
|-direction or not. y| is only the |-component of the vehicle velocity vector
v at the mass center F. The velocity vector, and therefore, y| , at other
points of the longitudinal axis of the vehicle are dierent. The common
property of v at any point of the {-axis is that all of them must have the
same y{ . The best indicator of the rotation of a vehicle is the yaw rate u,
which shows how fast and in what direction the vehicle is turning.
Example 435 Steady-state center of rotation.
Having the steady-state responses Vn = 1@U@ and V = @, we are
able to determine the position of the rotation center of a vehicle in the body
coordinate frame. The coordinate of the center of rotation ({R > |R ) with
respect to the mass center are
{R
= U sin = |R
= U cos =
1
sin (V )
Vn 1
cos (V )
Vn (10.355)
(10.356)
Consider a vehicle with the following characteristics
FiO = FiU 3000 N@ rad
FuO = FuU 3000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=0 m
d2 = 1=5 m
μ
¶
d1
p
d2
= 1=33 × 102
N = 2
o
Fi
Fu
V =
30
75 + y{2
V =
45 2y{2
75 + y{2
(10.357)
(10.358)
(10.359)
(10.360)
(10.361)
The variation of {R , |R when y{ changes are shown in Figure 10.22.
μ
¶
45 2y{2
75 + y{2
sin
(10.362)
{R = U sin = 30
75 + y{2
μ
¶
75 + y{2
45 2y{2
cos
|R = U cos =
(10.363)
30
75 + y{2
{R starts from {R = d2 on the real axel at zero velocity and moves forward. |R starts from |R = o cot on the rear axel at zero velocity and moves
away from the vehicle.
Figure 10.23 illustrates the loci of the steady-state rotation center at different speeds and a constant steer angle for the given vehicle. For this understeer vehicle, by increasing the forward speed, the rotation center moves
away and forward with respect to the vehicle. This example shows that kinematic steering condition determine the rotation center only when the speed
684
10. Vehicle Planar Dynamics
[m]
yO
Coordinate
xO
vx > m / s @
yO
xO
0
FIGURE 10.22. The coordinate of the center of rotation ({R > |R ) with respect to
the mass center of a vehicle for dierent speed.
of the care is zero. By increasing the speed of, the actual steady state rotation center will be determined by dynamic parameters of the vehicle and
not the kinematic steering condition.
Figure 10.24 illustrates three steady-state conditions of the vehicle with
the circular path of motions and velocity vectors. The three conditions are
for y{ to be lower, equal, and higher than the critical speed at which = 0.
Example 436 Under steering, over steering, neutral steering.
Curvature response V indicates how the radius of turning will change
10. Vehicle Planar Dynamics
685
x
vx
15 m / s
20
14 13
12 11
10 9
8 765
42
40
20
y
Path of
rotation center
100
80
60
10
FIGURE 10.23. The loci of the steady-state rotation center at dierent speeds
and a constant steer angle for a given vehicle.
R=58.3 m
E -0.0886 rad
R=32.4 m
E 0 rad
R=25.3 m
E 0.0566 rad
vx
10 m / s
Path of
rotation center
vx
y
60
40
vx
4.74 m / s
1m/s
x
v v v
20
10
20
FIGURE 10.24. The three steady-state conditions of a vehicle for y{ to be lower,
equal, and higher than the critical speed at which = 0.
686
10. Vehicle Planar Dynamics
with a change in steer angle. V can be expressed as
V
=
N
=
1@U
1
1
=
=
o 1 + Ny{2
μ
¶
d1
d2
p
o2 Fi
Fu
(10.364)
(10.365)
where N is called the stability factor. It determines that the vehicle is
X qghuvwhhu
Q hxwudo
Ryhuvwhhu
li
li
li
NA0
N=0
N?0
(10.366)
To nd N we may rewrite V as
V
=
=
=
=
1@U
F G F G
=
=
y{ (Gu F Fu G + py{ G )
1
μ
¶
py{ G
Gu F Fu G
y{
+
F G F G
F G F G
1
1
1
=
2
G
p
o
py{ G
1+
y2
o+
o F G F G {
F G F G
1
1
(10.367)
o 1 + Ny{2
Therefore,
N=
p
G
o F G F G
(10.368)
which, after substituting the force system coe!cients (10.144)-(10.149) will
be equal to
μ
¶
d2
p
d1
N= 2
(10.369)
o
Fi
Fu
The sign of stability factor N determines if V is an increasing or decreasing
function of velocity y{ . The sign of N depends on the relative weight of
d2 @Fi and d1 @Fu , which are dependent on the position of mass center
d1 , d2 , and sideslip coe!cients of the front and rear tires Fi , Fu .
If N A 0 then
d1
d2
A
(10.370)
Fi
Fu
and V = @ and gV @gy{ ? 0. Hence, the curvature of the path = 1@U
decreases with speed for a constant . Decreasing indicates that the radius
of the steady-state circle, U, increases by increasing speed y{ . A positive
stability factor is desirable. A vehicle with N A 0 is stable and is called
10. Vehicle Planar Dynamics
687
understeer. For an understeer vehicle, we need to increase the steering
angle if we increase the speed of the vehicle to keep the same turning circle.
If N ? 0 then
d2
d1
?
(10.371)
Fi
Fu
and V = @ and gV @gy{ A 0. Hence, the curvature of the path = 1@U
increases with speed for a constant . Increasing indicates that the radius
of the steady-state circle, U, decreases by increasing speed y{ . A negative
stability factor is undesirable. A vehicle with N ? 0 is unstable and is called
oversteer. For an oversteer vehicle, we need to decrease the steering angle
when we increase the speed of the vehicle, to keep the same turning circle.
If N = 0 then
d1
d2
=
(10.372)
Fi
Fu
then V = @ is not a function of y{ because gV @gy{ = 0. Hence, the
curvature of the path, = 1@U remains constant for a constant regardless
of y{ . Having a constant indicates that the radius of the steady-state
circle, U, will not change by changing the speed y{ . A zero stability factor
is neutral and a vehicle with N = 0 is on the border of stability and is called
a neutral steer. When driving a neutral steer vehicle, we do not need to
change the steering angle if we increase or decrease the speed of the vehicle,
to keep the same turning circle.
As an example, consider a car with the following characteristics:
Fi = 57296 N@ rad
p = 917 kg 62=8 slug
d2 = 164 cm 5=38 ft
Fu = 52712 N@ rad
d1 = 91 cm 2=98 ft
(10.373)
(10.374)
(10.375)
This car has a stability factor N and a curvature response V equal to
μ
¶
p
d2
d1
= 1=602 × 103
(10.376)
N = 2
o
Fi
Fu
1
1
0=39216
V =
=
(10.377)
o 1 + Ny{2
1 + 1=602 × 103 y{2
Now assume we ll the trunk and change the car’s characteristics to a new
set.
p = 1400 kg 95=9 slug
d1 = 125 cm 4=1 ft
d2 = 130 cm 4=26 ft
(10.378)
(10.379)
The new stability factor N and curvature response V are
N
V
= 2=21 × 104
0=392 16
= 2=21 × 104 y{2 1
(10.380)
(10.381)
688
10. Vehicle Planar Dynamics
Oversteer
K 0
K
0
Neutral steer
SN
Understeer
K !0
vx > m / s @
FIGURE 10.25. Comarison of the curvature response V for a car with
N = 1=602 × 1033 , N = 32=21 × 1034 , and N = 0.
Figure 10.25 compares the curvature response V for the two situations and
a neutral steering. We assumed that increasing weight did not change the
tire characteristics, and we kept the same sideslip coe!cients.
Example 437 Critical speed yf .
If N ? 0 then V increases by increasing y{ . Therefore, the steering angle
must be decreased to maintain a constant radius path. At a critical speed yf
r
1
(10.382)
yf = N
the denominator of V becomes zero and then
V $ 4
(10.383)
At the critical speed, any decrease in steering angle cannot keep the path.
When y{ = yf , the curvature is not a function of steering angle , and any
radius of rotation is theoretically possible for a constant . The critical speed
makes the system unstable. Controlling an oversteer vehicle gets harder by
y{ $ yf and becomes uncontrollable when y{ = yf .
The critical speed of an oversteer car with characteristics
Fi = 57296 N@ rad
p = 1400 kg 95=9 slug
d1 = 125 cm 4=1 ft
is
yf =
r
Fu = 52712 N@ rad
d2 = 130 cm 4=26 ft
1
= 67=33 m@ s
N
(10.384)
(10.385)
(10.386)
(10.387)
10. Vehicle Planar Dynamics
because
p
N= 2
o
μ
d2
d1
Fi
Fu
¶
= 2=2059 × 104
689
(10.388)
Example 438 Neutral steer point.
The neutral steer point of the bicycle model of a vehicle is the point
along the longitudinal axis at which the mass center allows neutral steering.
To nd the neutral steer point SQ we dene a distance dQ from the front
axle to have N = 0
dQ
o dQ
=0
(10.389)
Fi
Fu
therefore,
dQ =
Fu
o
Fi + Fu
(10.390)
The neutral distance gQ
gQ = dQ d1
(10.391)
indicates how much the mass center can move to have neutral steering.
For example, the neutral steer point SQ for a car with characteristics
Fi
d1
= 57296 N@ rad
= 91 cm 2=98 ft
Fu = 52712 N@ rad
d2 = 164 cm 5=38 ft
(10.392)
(10.393)
is at
dQ = 1=2219 m
(10.394)
Therefore, the mass center can move a distance gQ forward and still have
an understeer car.
gQ = dQ d1 31=2 cm
(10.395)
Example 439 Steady state responses of an oversteer vehicle.
To compare the steady-state responses of under and oversteer vehicles,
we examine how the steady-state parameters V , V , Vu , Vd , V| vary when
the speed of an oversteer vehicle changes. The following characteristics are
similar to Example 428 except the position of the mass center.
FiO = FiU 3000 N@ rad
FuO = FuU 3000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=28 m
d2 = 1=22 m
μ
¶
p
d2
d1
N = 2
= 1=33 × 102
o
Fi
Fu
(10.396)
(10.397)
(10.398)
(10.399)
690
SN
10. Vehicle Planar Dynamics
1/ R
G
vx > m / s @
FIGURE 10.26. Curvature response, V , of an oversteer vehicle as a function of
forward velocity y{ .
250
625 y{2
V
=
Vu
= V y{ =
V|
=
350 21=33y{2
75 y{2
250y{2
Vd = V y{2 =
625 y{2
V =
250y{
625 y{2
350 21=33y{2
y{
75 y{2
(10.400)
Figures 10.26-10.30 illustrate the steady-states variations by increasing the
forward velocity.
By increasing the speed of the car, V increases and symptomatically
approaches innity at a critical speed, which means the radius of rotation
decreases for constant . The value of V monotonically decreases by increasing y{ .
Vu begins from zero and increases with y{ . Vd shows the centripetal acceleration of the car while it is turning on a circle with a constant speed. The
value of Vd starts from zero at y{ = 0 and increases rapidly to approach
Vd $ 4. The lateral velocity response V| indicates the speed of the vehicle
at its mass center in the |-direction. It starts at zero and increases to a
maximum before decreasing monotonically to negative values.
Example 440 F Constant lateral force and steady-state response.
Consider a situation in which there is a constant lateral force I| on
a vehicle and there is no steering angle. At steady-state conditions, the
following equations describe the motion of the vehicle.
I|
P}
= Fu u + F = pu y{
= Gu u + G = 0
(10.401)
(10.402)
10. Vehicle Planar Dynamics
SE
691
E
G
vx > m / s @
FIGURE 10.27. Sideslip response, V , of an oversteer vehicle as a function of
forward velocity y{ .
Sr
r
G
vx > m / s @
FIGURE 10.28. Yaw rate response, Vu , of an oversteer vehicle as a function of
forward velocity y{ .
692
Sa
10. Vehicle Planar Dynamics
vx2 / R
G
vx > m / s @
FIGURE 10.29. Lateral acceleration response, Vd , of an oversteer vehicle as a
function of forward velocity y{ .
Sy
vy
G
vx > m / s @
FIGURE 10.30. Lateral velocity response, V| , of an oversteer vehicle as a function
of forward velocity y{ .
10. Vehicle Planar Dynamics
693
Equation (10.401) and (10.402) may be used to dene the following steadystate responses
V|1
=
V|2
=
1@U
G
=
=
I|
I|
y{ (Fu G F Gu )
u
G
=
I|
Fu G F Gu
(10.403)
(10.404)
A constant lateral force can be a result of driving straight on a banked road,
or having a side wind. A nonzero lateral rotation response V|1 indicates that
the vehicle will turn with = 0 and I| 6= 0. The lateral rotation response
V|1 may be transformed to the following equation to be a function of the
stability factor N:
V|1
=
μ
1
¶=
d1 Fi d2 Fu
y{ Fi Fu o2
Gu
y{ Fu F
G
μ
¶
1
d1
d2
1
= 2
N
=
y{ o
Fi
Fu
py{
(10.405)
To drain the rain and water from roads, we build the straight roads with
a little bank angle from the center to the shoulder. Consider a moving car
on a straight banked road. There is a lateral gravitational force because of
the bank angle I| = pj sin pj
(10.406)
to pull the car downhill. If the car is an understeer and N A 0 then the car
will turn downhill, while an oversteer car with N ? 0 will turn uphill.
Similarly, if there is a side wind in the |-direction and the car is moving
on a at straight road then, an understeer car, N A 0, will turn about the
}-axis, while an oversteer car, N ? 0, will turn about the }-axis.
Example 441 F SAE steering denition.
SAE steer denitions for under and oversteer behaviors are as follows:
X V= A vehicle is understeer if the ratio of the steering wheel angle gradient
to the overall steering ratio is greater than the Ackerman steer gradient.
RV= A vehicle is oversteer if the ratio of the steering wheel angle gradient
to the overall steering ratio is less than the Ackerman steer gradient.
DV= Ackerman steering gradient is
VD =
o
g (o@U)
=
2
y{
g (y{2 @U)
(10.407)
Example 442 F Exact steady-state responses
The steady-state turning of a two-wheel rigid vehicle is expressed by
I{
Fu u + F + F Gu u + G + G = pu y|
= pu y{
= 0
(10.408)
(10.409)
(10.410)
694
10. Vehicle Planar Dynamics
At steady-state turning, the mass center of the vehicle moves on a circle
with radius U at a forward speed y{ and angular velocity u, so y{ and u are
related by
y{ = U u cos y = Uu
U1 = U cos (10.411)
Substituting (10.280) in Equations (10.277)-(10.279), yields
y{
+ F + F U1
y{
Gu
+ G + G U1
Fu
= p
y{2
U1
= 0
(10.412)
(10.413)
Let us dene a new curvature 1 based on U1
1 =
1
Ud
and write the equations in matrix form
¸
¸ ¸
F Fu y{ p y{2
F
=
1
G
Gu y{
G
Solving the equations for and 1 shows that
6
5
G (Fu py{ ) Gu F
¸ 9
:
Gu F Fu G + py{ G
:
9
=9
:
1
8
7
F G F G
y{ (Gu F Fu G + py{ G )
(10.414)
(10.415)
(10.416)
Now we are able to dene a more accurate steady-state responses.
1= Curvature response, V U1 = U cos V
=
=
1
1
=
=
cos U
U1 F G F G
cos y{ (Gu F Fu G + py{ G )
(10.417)
2= Sideslip response, V
V =
G (Fu py{ ) Gu F
=
Gu F Fu G + py{ G
(10.418)
3= Yaw rate response, Vu
Vu =
u
F G F G
cos = y{ = V y{ =
Gu F Fu G + py{ G
(10.419)
10. Vehicle Planar Dynamics
695
4= Centripetal acceleration response, Vd
Vd =
y{2 @U
(F G F G ) y{
cos = y{2 = V y{2 =
Gu F Fu G + py{ G
(10.420)
5= Lateral velocity response, V| .
V| =
y|
G (Fu py{ ) Gu F
y{
= V y{ =
Gu F Fu G + py{ G
(10.421)
10.6 F Linearized Model for a Two-Wheel Vehicle
When the vehicle sideslip angle is very small, the equations of motion of
bicycle model reduce to the following set of equations:
I{
= pyb puy
³
´
= py u + b + p yb
I|
P}
I|
= L} ub
= (Fi Fu ) +
(10.423)
(10.424)
1
(d2 Fu d1 Fi ) u
y
+ i Fi + u Fu
P}
(10.422)
¢
1¡ 2
= (d2 Fu d1 Fi ) d1 Fi + d22 Fu u
y
+d1 Fi i d2 Fu u
(10.425)
(10.426)
Although these equations are not linear if v is not constant, because of the
assumption ¿ 1, they are called linearized equations of motion.
When the speed of the vehicle is constant, then the equations are
I{
I|
P}
= puy
³
´
= py u + b
= L} ub
(10.427)
(10.428)
(10.429)
Proof. For a small sideslip, , we may assume that,
y{
y|
= y cos y
= y sin y
(10.430)
(10.431)
Therefore, the equations of motion (10.193)—(10.195) will be simplied to
I{
I|
P}
= pyb { puy| = pyb puy
³
´
= pyb | + puy{ = p y
b + yb + puy
= uL
b }
(10.432)
(10.433)
(10.434)
696
10. Vehicle Planar Dynamics
Substitute yb = 0 for a constant velocity, these equations will be equal to
(10.427)-(10.429).
The tire sideslip angles i and u can also be simplied to
i
u
1
d1 u
(y| + d1 u) i = +
i
y{
y
1
d2 u
= u u =
(y| d2 u) u = u
y{
y
= i i =
(10.435)
(10.436)
The front and rear tires lateral forces are
I|i
I|u
= Fi i
= Fu u
(10.437)
(10.438)
Substituting these equations in (10.432)-(10.434) and using the denitions
I{
I|
P}
I{i + I{u
I|i + I|u
d1 I|i d2 I|u
(10.439)
(10.440)
(10.441)
results in the force system
I|
= I|i + I|u
= (Fi Fu ) +
P}
1
(d2 Fu d1 Fi ) u
y
+ i Fi + u Fu
= d1 I|i d2 I|u
= (d2 Fu d1 Fi ) +d1 Fi i d2 Fu u
(10.442)
¢
1¡ 2
d1 Fi + d22 Fu u
y
(10.443)
Factorization will transform the force system to (10.425)-(10.426).
Example 443 F Front-wheel-steering and constant velocity.
In most cases, the front wheel is the only steerable wheel, hence i = and u = 0. This simplies the equations of motion for a front steering
vehicle at constant velocity to
I{
py b
L} ub
= puy
= (Fi Fu ) ¶
μ
1
+ (d1 Fi d2 Fu ) py u + Fi y
= (d2 Fu d1 Fi ) μ
¶
¢
1¡
2
2
+ Fi d1 + Fu d2 u + (d1 Fi ) y
(10.444)
(10.445)
(10.446)
10. Vehicle Planar Dynamics
697
The second and third equations may be written in a matrix form for simpler
calculation.
6
5
(Fi + Fu ) d2 Fu d1 Fi
¸
¸
1
: 9
py
py 2
b
:
9
¡
¢
= 7
ub
d2 Fu d1 Fi Fi d21 + Fu d22 8 u
L}
yL}
6
5
Fi
9 py :
:
(10.447)
+9
7 d1 Fi 8 L}
Example 444 F Steady state conditions and a linearized system.
The equations of motion for a front steering, two-wheel model of fourwheel vehicles are given in equation (10.447) for linearized angle . At a
steady-state condition we have
¸
b
=0
(10.448)
ub
and therefore,
6
5
61 5
Fi Fu
d2 Fu d1 Fi
Fi
¸
1
9
: 9 py :
py
py2
:
: 9
= 9
7 d2 Fu d1 Fi
u
Fi d21 Fu d22 8 7 d1 Fi 8
L}
L}
yL}
¡
¢
5
6
d22 Fu + d1 d2 Fu py2 d1 Fi 9 F F (d + d )2 py 2 (d F d F ) :
2
1 i
2 u
9 i u 1
:
= 9
(10.449)
:
7
8
(d1 + d2 ) yFi Fu Fi Fu (d1 + d2 )2 py 2 (d1 Fi d2 Fu )
Using Equation (10.449) we can dene the following steady-state responses:
2=Sideslip response, V
¢
¡
d22 Fu + d1 d2 Fu py2 d1 Fi
V = =
(10.450)
Fi Fu o2 py2 (d1 Fi d2 Fu )
3=Yaw rate response, Vu
Vu =
u
Fi Fu oy
=
2
Fi Fu o py 2 (d1 Fi d2 Fu )
(10.451)
At steady-state conditions we have
u=
y
U
(10.452)
698
10. Vehicle Planar Dynamics
where U is the radius of the circular path of the vehicle. Employing Equation
(10.452) we are able to dene two more steady-state responses as follows:
1=Curvature response, V
V
=
=
1
u
1
=
=
= Vu
U
y
y
oFi Fu
Fi Fu o2 py 2 (d1 Fi d2 Fu )
(10.453)
4=Lateral acceleration response, Vd
Vd
=
=
y 2 @U
= y 2 = V y 2
oFi Fu y 2
2
Fi Fu o py 2 (d1 Fi d2 Fu )
(10.454)
The above steady-state responses are comparable to the steady-states numbered 1 to 4 given in Equations (10.267)-(10.270) for a more general case.
Example 445 F Understeer and oversteer for a linearized model.
Employing the curvature response V in Equation (10.453) we may dene
V
=
=
=
where
1@U
oFi Fu
=
=
Fi Fu o2 py 2 (d1 Fi d2 Fu )
1
py 2 (d1 Fi d2 Fu )
o+
oFi Fu
1
1
1
1
=
(10.455)
o
o 1 + Ny{2
py 2 (d1 Fi d2 Fu )
1+
o2 Fi Fu
p
N= 2
o
μ
d1
d2
Fi
Fu
¶
(10.456)
This N is the same as the stability factor given in Equation (10.369).
Therefore, the stability factor remains the same for the linearized equations.
Example 446 F Source of nonlinearities.
There are three main reasons for nonlinearity in rigid vehicle equations of
motion: product of variables, trigonometric functions, and nonlinear characteristic of forces. When the steer angle and sideslip angles l and are
very small, the forces act linear, trigonometric function are approximately
linear, and product of variables are too small. As a result, it is reasonable
to ignore all kinds of nonlinearities. This is called low angles condition
driving, and is correct for low turn and normal speed driving.
10. Vehicle Planar Dynamics
699
10.7 F Transient Response
To examine the transient response of a vehicle and examine how the vehicle will respond to a steering input, the following set of coupled ordinary
dierential equations must be solved.
yb { =
yb |
=
ub
=
1
I{ + u y|
p
μ
¶
F
Fu
F
y| y{ u+
(w)
py{
p
p
Gu
G
G
y| +
u+
(w)
L} y{
L}
L}
(10.457)
(10.458)
(10.459)
The answers to this set of equations to a given time dependent steer angle
are
y{
y|
u
= y{ (w)
= y| (w)
= u (w)
(10.460)
(10.461)
(10.462)
Such a solution is called time response or transient response.
Assuming a constant forward velocity, the rst equation (10.457) simplies to an algebraic equation
I{ = pu y|
(10.463)
6
6
5
Fu
F
F
¸
¸
y{ : y|
9 py{ p
9 p :
yb |
:
:
(10.464)
=9
+9
8 u
7 G
7 G 8 (w)
ub
Gu
L}
L} y{
L}
and Equations (10.464) become independent from the rst one. The set of
Equations (10.464) can be written in the form
5
qb = [D] q + u
(10.465)
in which [D] is a constant coe!cient matrix, q is the vector of control
variables, and u is the vector of inputs.
5
6
Fi + Fu
d1 Fi + d2 Fu
y{
9
:
py{
py{
: (10.466)
[D] = 9
7 d F d F
8
2
2
d1 Fi + d2 Fu
1 i
2 u
L} y{
L} y{
5
6
F
Fu
y{
9 py{ p
:
:
= 9
(10.467)
7 G
8
Gu
L} y{
L}
700
10. Vehicle Planar Dynamics
¸
y|
u
5
Fi
9
p
u = 9
7 d1 Fi
L}
q =
(10.468)
6
5
6
F
:
9
:
: (w) = 9 p : (w)
8
7 G 8
L}
(10.469)
To solve the inverse dynamic problem and nd the vehicle response, the
steering function (w) must be given.
Example 447 F Forward and inverse dynamic problems.
Two types of dynamic problems may be dened: 1=direct or forward and,
2=indirect or inverse. In forward dynamics a set of desired functions y{ =
y{ (w), y| = y| (w), u = u (w) are given and the required (w) is asked for.
In inverse dynamics, an input function = (w) is given and the output
functions y{ = y{ (w), y| = y| (w), u = u (w) are asked for.
The forward dynamic problem needs dierentiation and the inverse dynamic problem needs integration. Generally speaking, solving an inverse
dynamic problem is more complicated than a forward dynamic problem.
Example 448 F Analytic solution to a step steer input.
Consider a vehicle with the following characteristics
Fi
p
d1
y{
=
=
=
=
60000 N@ rad
1000 kg
1=0 m
20 m@ s
Fu = 60000 N@ rad
L} = 1650 kg m2
d2 = 1=5 m
(10.470)
The force system coe!cients for the vehicle from (10.144)-(10.149) are
Fu
F
G
= 1500 N s@ rad
= 60000 N@ rad
= 30000 N m@ rad
F = 120000 N@ rad
Gu = 9750 N m s@ rad
G = 60000 N m@ rad
(10.471)
Let us assume the steering input is
(w) =
½
0=1 rad 5=73 deg
0
wA0
w0
(10.472)
The steady state responses can be used to determine the steady-state values
10. Vehicle Planar Dynamics
701
of the vehicle’s variable kinematics.
U
=
=
u
=
y{2
U
=
y|
=
Gu F Fu G + py{ G y{
1
=
= 38=33 m
(10.473)
V F G F G
G (Fu py{ ) Gu F
V =
= 0=0304 rad
(10.474)
Gu F Fu G + py{ G
F G F G
= 0=522 rad@ s
(10.475)
Vu =
Gu F Fu G + py{ G
F G F G
Vd =
y{ = 10=43 m@ s2 (10.476)
Gu F Fu G + py{ G
G (Fu py{ ) Gu F
V| =
y{ = 0=608 m@ s (10.477)
Gu F Fu G + py{ G
The equations of motion for a zero initial condition
¸ ¸
y| (0)
0
q0 =
=
u (0)
0
(10.478)
are
yb | + 6y| + 18=5u
ub 0=909y| + 5=909u
= 60 (w) = 6
= 36=363 (w) = 3=636
(10.479)
(10.480)
The solution of the equations of motion are
¸
¸ y| (w)
0=609 + h5=95w (2=347 sin 4=1w + 0=609 cos 4=1w)
=
u (w)
0=522 + h5=95w (0=129 sin 4=1w 0=522 cos 4=1w)
(10.481)
To examine the response of the vehicle to the sudden change in steer angle
while going straight at a constant speed, we plot the kinematic variables of
the vehicle. Figures 10.31 and 10.32 depict the solutions y| (w) and u (w)
respectively.
The steering input is positive and therefore, the vehicle must turn left, in
a positive direction of the |-axis. The yaw rate in Figure 10.32 is positive
and correctly shows that the vehicle is turning about the }-axis.
We can also nd the lateral velocity of the front and rear wheels, by
having y| and u.
y|i = y| + d1 u
y|u = y| d2 u
(10.482)
The lateral speed of the front and rear wheels are shown in Figures 10.33
and 10.34. The vehicle sideslip angle = y| @y{ and the radius of rotation
U = y{ @u are also shown in Figures 10.35 and 10.36.
702
10. Vehicle Planar Dynamics
1
vy >m / s@
t [s]
FIGURE 10.31. Lateral velocity response to a sudden change in steer angle.
r > rad / s @
t [s]
FIGURE 10.32. Yaw velocity response to a sudden change in steer angle.
10. Vehicle Planar Dynamics
703
vy f >m / s@
t [s]
1
FIGURE 10.33. Lateral velocity response of the front wheel to a sudden change
in steer angle.
v yr > m / s @
t [s]
FIGURE 10.34. Lateral velocity response of the rear wheel to a sudden change
in steer angle.
704
10. Vehicle Planar Dynamics
E > deg @
1
t [s]
FIGURE 10.35. Sideslip response to a sudden change in steer angle.
R > m@
t [s]
FIGURE 10.36. Radius of rotation response to a sudden change in steer angle.
10. Vehicle Planar Dynamics
705
Example 449 F Time series and free response.
The response of a vehicle to zero steer angle
(w) = 0
(10.483)
at constant speed is called the free response. The equation of motion under
free dynamics is
qb = [D] q
(10.484)
To solve the equations, let us assume
¸
d e
[D] =
f g
(10.485)
and therefore, the equations of motion are
¸ ¸
¸
d e
y|
yb |
=
ub
f g
u
(10.486)
Because the equations are linear, the solutions are exponential functions
y| = Dhw
u = Ehw
(10.487)
Substituting the solutions in equations of motion
¸ ¸
¸
d e
Dhw
Dhw
=
f g
Ehw
Ehw
shows that
d
e
f
g
¸
Dhw
Ehw
¸
=0
(10.488)
(10.489)
Therefore, the condition for functions (10.487) to be the solution of Equation (10.486) is that the exponent to be the eigenvalue of [D]. To nd ,
we may expand the determinant of the coe!cient matrix
¸
d
e
det
= 2 (d + g) + (dg ef)
(10.490)
f
g
and nd the characteristic equation
2 (d + g) + (dg ef) = 0
The solution of the characteristic equations is
q
1
1
(d g)2 + 4ef
1>2 = (d + g) ±
2
2
(10.491)
(10.492)
Having the eigenvalues 1>2 provides us with the general solution for the
free dynamics of a bicycle vehicle:
y|
u
= D1 h1 w + D2 h2 w
= E1 h1 w + E2 h2 w
(10.493)
(10.494)
706
10. Vehicle Planar Dynamics
vy >m / s@
t [s]
FIGURE 10.37. Lateral velocity response in example 449.
The coe!cients D1 , D2 , E1 , and E2 , must be found from initial conditions:
As an example, consider a vehicle with
Fi
p
d1
y{
=
=
=
=
Fu = 52712 N@ rad
L} = 1128 kg m2 832 slug ft2
d2 = 130 cm 4=26 ft
(10.495)
57296 N@ rad
1400 kg 95=9 slug
125 cm 4=1 ft
20 m@ s
which starts from
q0 =
y| (0)
u (0)
¸
=
1
0
¸
Substituting these values provide us with
¸ ¸
¸
3=929
31=051
y|
yb |
=
ub
13=716 79170=337
u
(10.496)
(10.497)
and their solutions are
y|
u
¡
¢
= 0=173 × 103 h3=92w h79170=34w
3=92w
= h
7 79170=34w
+ 0=68 × 10 × 10
h
(10.498)
(10.499)
Figures 10.37 and 10.38 illustrate the time responses. Figure 10.39 is a
magnication of Figure 10.38 to show that u does not jump to a negative
point but decreases rapidly and then approaches zero gradually.
Example 450 F Matrix exponentiation.
The exponential function h[D]w is called matrix exponentiation. This function is dened as a matrix time series.
h[D]w = L + [D] w +
[D]2 2 [D]3 3
w +
w + ···
2!
3!
(10.500)
10. Vehicle Planar Dynamics
r > rad / s@
t [s]
FIGURE 10.38. Yaw velocity response in example 449.
r > rad / s @
t [s]
FIGURE 10.39. Yaw velocity response in example 449.
707
708
10. Vehicle Planar Dynamics
This series always converges. As an example assume
¸
0=1 0=2
[D] =
0=3 0=4
(10.501)
then
h[D]w
1 0
0 1
¸
+
0=1 0=2
0=3 0=4
¸
1 + 0=1w 0=025 w2 + · · ·
0=3w 0=075 w2 + · · ·
¸2
1
0=1 0=2
w2 + · · ·
2 0=3 0=4
¸
0=2w + 0=05w2 + · · ·
(10.502)
1 + 0=4w + 0=05w2 + · · ·
w+
Example 451 F Time series and free response.
The response of a vehicle to zero steer angle
(w) = 0
(10.503)
at constant speed is the free response. The equation of motion under free
dynamics is
qb = [D] q
(10.504)
The solution of this dierential equation with the general initial conditions
q (0) = q0
(10.505)
q (w) = h[D]w q0
(10.506)
is
If the eigenvalues of [D] are negative then q (w) $ 0 for ;q0 .
The free response in a series can be expressed as
q (w) = h[D]w q0
Ã
=
[D]2 2 [D]3 3
L + [D] w +
w +
w + ···
2!
3!
!
q0
(10.507)
For example, consider a vehicle with the following characteristics:
Fi
p
d1
y{
=
=
=
=
57296 N@ rad
1400 kg 95=9 slug
125 cm 4=1 ft
20 m@ s
which start from
q0 =
Fu = 52712 N@ rad
L} = 1128 kg m2 832 slug ft2
d2 = 130 cm 4=26 ft
(10.508)
y| (0)
u (0)
¸
=
1
0
¸
Employing the vehicle’s characteristics, we have
¸
3=929
31=051
[D] =
13=716 79170=337
(10.509)
(10.510)
10. Vehicle Planar Dynamics
709
and therefore, the time response of the vehicle is
¸
¸ ¸ ¸ ¸
1 0
1
3=929
31=051
1
y| (w)
=
+
w
u (w)
0 1
0
13=716 79170=337
0
¸2 ¸
1
1
3=929
31=051
+
w2
0
2 13=716 79170=337
¸3 ¸
1
1
3=929
31=051
+
···
(10.511)
w3
0
6 13=716 79170=337
Accepting an approximate solution up to cubic degree provides the following
approximate solution:
¸ ¸
5=620 3 × 106 w3 + 220=67w2 3=929w + 1
y| (w)
u (w)
1=432 9 × 1010 w3 + 5=4298 × 105 w2 13=716w
(10.512)
Example 452 F Response of an understeer vehicle to a step input.
The response of dynamic systems to a step input is a traditional method
to examine the behavior of dynamic systems. A step input for vehicle dynamics is a sudden change in steer angle from zero to a nonzero constant
value.
Consider a vehicle with the following characteristics:
Fi
p
d1
y{
=
=
=
=
57296 N@ rad
917 kg 62=8 slug
0=91 m 2=98 ft
20 m@ s
Fu = 52712 N@ rad
L} = 1128 kg m2 832 slug ft2
d2 = 1=64 m 5=38 ft
(10.513)
and a sudden change in the steering input to a constant value
½
0=1 rad 5=7296 deg w A 0
(w) =
0
w0
The equations of motion for a zero initial condition
¸ ¸
y| (0)
0
=
q0 =
u (0)
0
(10.514)
(10.515)
are
yb | + 5=9983y| + 18=129u
ub 1=521y| + 8=387u
= 62=482 (w) = 6=2482
= 46=2228 (w) = 4=6223
(10.516)
(10.517)
The force system coe!cients for the vehicle from (10.144)-(10.149) are
d1
d2
Fu = Fi + Fu = 1715=416 N s@ rad
(10.518)
y{
y{
(10.519)
F = (Fi + Fu ) = 110008 N@ rad
F = Fi = 57296 N@ rad
(10.520)
710
10. Vehicle Planar Dynamics
Gu
G
G
d2
d2
= 1 Fi 2 Fu = 9461=05 N m s@ rad
y{
y{
= (d1 Fi d2 Fu ) = 34308=32 N m@ rad
= d1 Fi = 52139=36 N m@ rad
(10.521)
(10.522)
(10.523)
Equations (10.267)-(10.270) indicate that the steady-state response of the
vehicle, when w $ 4, are
V
=
V
=
Vu
=
Vd
=
V|
=
1
=
= 0=2390051454
U
= 0=2015419091
u
= y{ = V y{ = 4=780102908
y{2 @U
= y{2 = V y{2 = 95=60205816
y|
= V y{ = 4=030838182
(10.524)
(10.525)
(10.526)
(10.527)
(10.528)
Therefore, the steady-state characteristics of the vehicle with = 0=1 must
be
U = 41=84 m
= 0=02015 rad 1=1545 deg
u = 0=478 rad@ s
2
y{
= 9=56 m@ s2
U
(10.529)
(10.530)
(10.531)
(10.532)
Substituting the input function (10.535) and solving the equations provides the following solutions:
¸
¸ 0=4 + h7=193w (1=789 sin 5=113w + 0=403 cos 5=113w)
y| (w)
=
u (w)
0=478 + h7=193w (0=232 sin 5=113w + 0=478 cos 5=113w)
(10.533)
Figures 10.40 and 10.41 depict the solutions.
Having y| (w) and u (w) are enough to calculate the other kinematic variables as well as the required forward force I{ to maintain the constant
speed.
(10.534)
I{ = pu y|
Figures 10.42 and 10.43 show the kinematics variables of the vehicle, and
Figure 10.44 depicts how the require I{ is changing as a function of time.
Example 453 F Response of an oversteer vehicle to a step input.
Let’s assume the steering input is
½
0=1 rad 5=7296 deg w A 0
(10.535)
(w) =
0
w0
10. Vehicle Planar Dynamics
vy >m / s@
t [s]
FIGURE 10.40. Lateral velocity response in example 452.
r > rad / s @
t [s]
FIGURE 10.41. Yaw rate response in example 452.
711
712
10. Vehicle Planar Dynamics
E > deg @
1
t [s]
FIGURE 10.42. Sideslip angle response in example 452.
R > m@
t [s]
FIGURE 10.43. Radius of rotation response in example 452.
10. Vehicle Planar Dynamics
713
Fx > N @
t [s]
1
FIGURE 10.44. The required forward force I{ to keep the speed constant, in
example 452.
and the vehicle characteristics are
Fi
p
d1
y{
=
=
=
=
57296 N@ rad
1400 kg 95=9 slug
1=25 m 4=1 ft
20 m@ s
Fu = 52712 N@ rad
L} = 1128 kg m2 832 slug ft2
d2 = 1=30 m 4=26 ft
(10.536)
The equations of motion for a zero initial conditions
¸ ¸
y| (0)
0
=
q0 =
u (0)
0
(10.537)
are
yb | + 3=928857143y| + 20=11051429u
ub + 0=1371631206y| + 7=91703369u
=
=
=
=
40=92571429 (w)
4=092571429
(10.538)
63=4929078 (w)
6=34929078
(10.539)
Substituting the input function (10.535) and solving the equations provides the following solutions:
¸
¸ y| (w)
6=3h3=328w 2=943h8=518w 3=361
(10.540)
=
u (w)
0=188h3=328w 0=672h8=518w + 0=86
Figures 10.45 and 10.46 depict the solutions.
Example 454 F Standard steer inputs.
Step and sinusoidal excitation inputs are the most general input to examine the behavior of a vehicle. Furthermore, some other transient inputs
714
10. Vehicle Planar Dynamics
vy >m / s@
t [s]
FIGURE 10.45. Lateral velocity response for example 453.
r > rad / s @
t [s]
FIGURE 10.46. Yaw velocity response for example 453.
10. Vehicle Planar Dynamics
715
may also be used to analyze the dynamic behavior of a vehicle. Single sine
steering, linearly increasing steering, and half sine lane change steering are
the most common transient steering inputs.
Example 455 F Position of the rotation center.
The position of the center of rotation R in the vehicle body coordinate is
at
1
sin (V )
(10.541)
{R = U sin = Vn 1
cos (V )
(10.542)
|R = U cos =
Vn because is positive when it is about positive direction of the }-axis. Figure 10.21 illustrates a two-wheel vehicle model, the vehicle body coordinate
frame, and the center of rotation R.
At steady-state conditions the radius of rotation U can be found from the
curvature response V , and the vehicle sideslip angle can be found from
the sideslip response V .
U
y{ (Gu F Fu G + py{ G )
1
=
V
(F G F G ) G (Fu py{ ) Gu F
= V =
Gu F Fu G + py{ G
=
(10.543)
(10.544)
Therefore, the position of the actual center of rotation R is at
μ
¶
G (Fu py{ ) Gu F
y{ (Gu F Fu G + py{ G )
sin
{R = (F G F G ) Gu F Fu G + py{ G
(10.545)
μ
¶
G (Fu py{ ) Gu F
y{ (Gu F Fu G + py{ G )
cos
|R =
(F G F G ) Gu F Fu G + py{ G
(10.546)
Assuming a small , we may nd the position of R approximately.
G (Fu py{ ) Gu F
y{
F G F G
Gu F Fu G + py{ G
y{
(F G F G ) { (10.547)
|
(10.548)
Example 456 F Variable y{ .
When y{ is variable, the equations of motion are no longer linear and
therefore, a numerical integration is needed. Let us assume that a vehicle
has the characteristics of
Fi = 60000 N@ rad
p = 1000 kg
d1 = 1=0 m
Fu = 60000 N@ rad
L} = 1650 kg m2
d2 = 1=5 m
(10.549)
716
10. Vehicle Planar Dynamics
Let us assume that the vehicle changes its forward velocity from zero to a
maximum of y{ = 20 m@ s and remains constant while the steering is kept
constant at
(w) = 0=1 rad 5=73 deg
(10.550)
The force system coe!cients for the vehicle from (10.144)-(10.149) are
30000
N s@ rad
y{
Fu
=
F
= 60000 N@ rad
G
= 30000 N m@ rad
F = 120000 N@ rad
195000
N m s@ rad
y{
G = 60000 N m@ rad
(10.551)
Gu = We may assume that y{ varies linearly with time and reaches the maximum
speed y{ = 20 m@ s at w = w0 .
;
? 20
w m@ s 0 ? w ? w0
y{ =
(10.552)
w
= 200 m@ s
w ?w
0
The equations of motion for the vehicle are
¶
μ
120
30
+ 60 (w)
y| u y{ yb | = y{
y{
u
18=182
y| 118=18 + 36=364 (w)
ub =
y{
y{
(10.553)
(10.554)
The graphical illustration of the solutions are shown in Figures 10.47 and
10.48 for
(10.555)
w0 = 20 s
Example 457 F Rotation center between two steady states.
Consider a vehicle with the following characteristics
Fi = 60000 N@ rad
Fu = 60000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=0 m
d2 = 1=5 m
μ
¶
p
d2
d1
N= 2
= 1=33 × 103
o
Fi
Fu
(10.556)
The steady-state responses Vn = 1@U@ and V = @ are:
V =
300
750 + y{2
V =
450 2y{2
750 + y{2
(10.557)
10. Vehicle Planar Dynamics
vy >m / s@
717
t [s]
FIGURE 10.47. Lateral velocity of a variable forward velocity vehicle dynamics.
r > rad / s @
t [s]
FIGURE 10.48. Yaw rate of a variable forward velocity vehicle dynamics.
718
10. Vehicle Planar Dynamics
The coordinate of the center of rotation ({R > |R ) in the body coordinate
frame at the vehicle mass center are
{R
= U sin = |R
= U cos =
1
sin (V )
Vn 1
cos (V )
Vn (10.558)
(10.559)
Let us assume that the car is moving with
(w) = 0=1 rad 5=73 deg
(10.560)
at constant forward speed of
y{ = 3 m@ s
(10.561)
At the steady-state condition, the vehicle would have
V
Vu
= 0=3952569170
= 1=185770751
1
= 25=3 m
V = Vu = 0=118577 rad@ s
V = 0=5691699605
V| = 1=707509882
= V = 0=056917 rad
U =
u
(10.562)
y| = 0=170751 m@ s
(10.563)
We change y{ form y{ = 3 m@ s at w = 0 to the maximum speed y{ = 20 m@ s
at w = w0 , and keep it constant. The mathematical expression of y{ with
constant acceleration is:
y{ = 3 +
20 3
w K(w0 w) + (20 3) K(w w0 ) m@ s
w0
where K (w ) is the Heaviside function.
½
0 w?
K (w ) =
1 ?w
(10.564)
(10.565)
Assuming a very high w0 and very low acceleration, the transition from
y{ = 3 m@ s to y{ = 20 m@ s will be close to the intermediate steady-state
conditions. To have a low velocity sweep, let us change the speed in
w0 = 120 s
(10.566)
With such a long time for velocity change, the kinematics of the vehicle will
change according to their steady-state values. Figure 10.49 depicts how the
forward velocity varies, and Figure 10.50 shows the variation of the lateral
velocity y| and Figure 10.51 depicts the yaw rate u. The loci of the rotation
center ({R > |R ) is shown in Figure 10.52.
10. Vehicle Planar Dynamics
719
vx > m / s @
t [s]
FIGURE 10.49. The forward velocity y{ of a vehicle with a constant low acceleration between y{ = 3 m@ s and y{ = 20 m@ s.
vy >m / s@
t [s]
FIGURE 10.50. The lateral velocity y| of a vehicle with a constant low acceleration between y{ = 3 m@ s and y{ = 20 m@ s.
720
10. Vehicle Planar Dynamics
r > rad / s @
t [s]
FIGURE 10.51. The yaw rate u of a vehicle with a constant low acceleration
between y{ = 3 m@ s and y{ = 20 m@ s.
x
vx
20 m / s
Path of
rotation center
vx
10
3 m/s
y
40
30
20
10
FIGURE 10.52. The loci of the rotation center of a vehicle with a constant low
acceleration between y{ = 3 m@ s and y{ = 20 m@ s.
10. Vehicle Planar Dynamics
721
3
vy >m / s@
t [s]
FIGURE 10.53. The lateral velocity y| of a vehicle with a constant very high
acceleration between y{ = 3 m@ s and y{ = 20 m@ s.
If w0 is short, the acceleration is high, and the dynamics of the vehicle
will deviate from the almost steady state conditions. To have a fast velocity
sweep, let us change the speed in
w0 = 1 s
(10.567)
Figure 10.53 shows the variation of the lateral velocity y| and Figure 10.54
shows the yaw rate u. The coordinate of the rotation center ({R > |R ) are
shown in Figure 10.55. The loci of the path of rotation center ({R > |R ) is
shown in Figure 10.56 to be compared with the loci at very low acceleration
maneuver.
Example 458 F Vehicles act close to steady-state conditions.
Interestingly, vehicles act on their steady-state condition with a good approximation. In the previous example 457, a vehicle was supposed to change
its speed while it is moving at steady state conditions with y{ = 3 m@ s and
(w) = 0=1 rad to y{ = 20 m@ s.
When the acceleration of the vehicle is very low d = 0=14167 m@ s2 , it
takes 120 s for the vehicle to change its speed. The vehicle is always at
its steady state conditions while the speed changes slowly. Figure 10.5010.52 show how the dynamics of the vehicle changes. At the moment the
acceleration becomes zero, the dynamics are at their steady-state values and
they stay their since after.
When the acceleration of the vehicle is 120 times higher at a very high
value of d = 17 m@ s2 , it takes only 1 s for the vehicle to change its speed. After cutting the acceleration, it takes less than 1 s for the vehicle to achieve its
steady-state conditions, as are shown in Figures 10.53-10.55. Figure 10.57
illustrates very well how the rotation center of the vehicle deviates from the
steady-state in a high acceleration. It also shows how far its rotation center
is from its steady-state when acceleration stops.
722
10. Vehicle Planar Dynamics
r > rad / s @
t [s]
FIGURE 10.54. The yaw rate u of a vehicle with a constant very high acceleration
between y{ = 3 m@ s and y{ = 20 m@ s.
yO > m @
yO
xO
1 xO > m@
0.5
0
-0.5
-1
t > s@
FIGURE 10.55. The coordinate of the rotation center of a vehicle with a constant
very high acceleration between y{ = 3 m@ s and y{ = 20 m@ s.
10. Vehicle Planar Dynamics
Steady-state rotation
center for vx 20 m / s
Path of rotation center
for vx change in 120 sec.
723
x
Steady-state rotation
center for vx 3 m / s
10
y
30
40
20
10
Path of rotation center
for vx change in 1 sec.
FIGURE 10.56. The loci of the rotation center of a vehicle with a constant very
high acceleration between y{ = 3 m@ s and y{ = 20 m@ s.
As a result of this example, we know that vehicles act based on their
steady-state dynamics in normal operations.
Example 459 F Global position of a vehicle.
A vehicle with the following characteristics
Fi
p
d1
y{
=
=
=
=
60000 N@ rad
1000 kg
1=0 m
20 m@ s
Fu = 60000 N@ rad
L} = 1650 kg m2
d2 = 1=5 m
(10.568)
is moving straight with a constant forward velocity of
y{ = 20 m@ s
(10.569)
At w = 0, when the car is at the origin of a global coordinate frame, suddenly
we change the steer angle to
½
0=1 rad 5=73 deg w A 0
(w) =
(10.570)
0
w0
Calculating the force system coe!cients for the vehicle from (10.144)-(10.149)
Fu
F
G
= 1500 N s@ rad
= 60000 N@ rad
= 30000 N m@ rad
F = 120000 N@ rad
Gu = 9750 N m s@ rad
G = 60000 N m@ rad
(10.571)
724
10. Vehicle Planar Dynamics
Steady-state rotation
center for vx 20 m / s
Path of rotation center
for vx change in 1 sec. Steady-state path of
rotation center for
t=1.5 s
t=1.4 s
t=1.3 s
t=0.7 s
t=1.2 s
t=1.1 s
x
Steady-state rotation
center for vx 3 m / s 2
1
t=0.3 s
y
0
38
36
t=1.6 s t=0.9 s
t=0.8 s
t=1.0 s
32
t=0.6 s
30
28
t=0.4 s
t=0.1 s
t=0.2 s
t=0.5 s
26
-1
-2
t=0
FIGURE 10.57. The loci of the rotation center of a vehicle with a constant
very high acceleration between y{ = 3 m@ s and y{ = 20 m@ s, compared to the
steady-state loci.
we determine the steady-state values of the vehicle’s variable kinematics.
1
= 38=33 m
V = Vu = 0=522 rad@ s
U =
= V = 0=0304 rad
u
y| = V| = 0=608 m@ s
(10.572)
The solution of the equations of motion
yb | + 6y| + 18=5u
ub 0=909y| + 5=909u
= 60 (w) = 6
= 36=363 (w) = 3=636
(10.573)
(10.574)
for a zero initial conditions are
¸ ¸
y| (w)
0=609 + h5=95w (2=347 sin 4=1w + 0=609 cos 4=1w)
=
u (w)
0=522 + h5=95w (0=129 sin 4=1w 0=522 cos 4=1w)
(10.575)
The angular orientation of the body coordinate frame with respect to the
xed global coordinate frame is
Z w
#=
u (w) gw
(10.576)
0
and the velocity vector of the vehicle in the global frame is
¸
y{
E
v=
y| (w)
(10.577)
10. Vehicle Planar Dynamics
725
Therefore, the velocity vector of the vehicle in the global frame is
¸
¸
cos # sin #
y{
J
E
v = [U] v =
y| (w)
sin # cos #
¸
¸ y[
y{ cos # y| sin #
(10.578)
=
=
y\
y| cos # + y{ sin #
The global coordinate of the mass center of the vehicle would be
Z w
Z w
y[ gw =
(y{ cos # y| sin #) gw
[ =
0
0
Z w
Z w
\ =
y\ gw =
(y| cos # + y{ sin #) gw
0
(10.579)
(10.580)
0
When the steer angle is constant, the vehicle will eventually be turning on
a constant circular path. The position of the steady-state rotation center in
the body coordinate frame is at
{R
|R
= U sin = 1=1651 m
= U cos = 38=312 m
Therefore the global coordinate of the rotation center would be
¸
¸
{R
[R
J
= J
r
+
U
E
J E
\R
|R
(10.581)
(10.582)
(10.583)
where J
J rE is the J-expression of the position vector of the origin of the
E-frame with respect to the origin of the J-frame at any point on the steadystate circular path. Figure 10.58 illustrates the path of motion of the vehicle
and the steady-state rotation center.
Example 460 F Second-order equations.
The coupled equations of motion (10.188) may be modied to a secondorder dierential equation of only one variable. To do this, let us rewrite
the equations.
yb |
=
ub
=
¢
1 ¡
y| F + uFu y{ puy{2 + y{ F
py{
1
(y| G + uGu y{ + y{ G )
L} y{
(10.584)
(10.585)
Assuming a constant forward speed
y{ = frqvw
(10.586)
and taking a derivative from Equation (10.585) yields:
ü =
1
(yb | G + uG
b u y{ )
L} y{
(10.587)
726
10. Vehicle Planar Dynamics
x
Path of motion
40
30
Steady-state
rotation center
20
10
y
-100
-80
-60
-40
-20
-10
-20
-30
FIGURE 10.58. The path of motion of a vehicle with a step steer angle that is
moving with constant forward speed.
We substitute Equation (10.584) in (10.587)
μ
¶
¢
1 ¡
1
ü =
y| F + uFu y{ puy{2 + y{ F G
L} y{ py{
1
b { G
+
uG
b u y{ + y
(10.588)
L} y{
and then substitute for y| from (10.585) and get the following equation:
pL} y{ ü (L} F + pGu y{ ) ub + (Gu F Fu I + py{ G ) u
´
³
b { G
(10.589)
= F G F G + py
This equation is similar to the equation of motion for a force vibration
single DOF system
pht ü + fht ub + nht u = iht (w)
(10.590)
where the equivalent mass pht , damping fht , stiness nht , and force iht (w)
are:
= pL} y{
= (L} F + pGu y{ )
= Gu F Fu I + py{ G
g (w)
ih (w) = p
y{ G (F G F G ) (w)
gw
pht
fht
nht
(10.591)
(10.592)
(10.593)
(10.594)
10. Vehicle Planar Dynamics
727
We may use Equation (10.589) and determine the behavior of a vehicle
similar to analysis of a vibrating system. The response of the equation to
a step steering input may be expressed by rise time, peak time, overshoot,
and settling time.
10.8 Summary
A vehicle is eectively modeled as a rigid bicycle in a planar motion by
ignoring the roll of the vehicle. Such a vehicle has three DOF in a body
coordinate frame attached to the vehicle at F: forward motion, lateral motion, and yaw motion. If the vehicle is front wheel steering, the dynamic
equations of such a vehicle are expressed in (y{ > y| > u) variables in the following set of three coupled rst order ordinary dierential equations.
yb {
yb |
ub
I{
+ u y|
p
1
=
(d1 Fi + d2 Fu ) u
py{
1
1
(Fi + Fu ) y| + Fi u y{
py{
p
¢
1 ¡ 2
=
d1 Fi d22 Fu u
L} y{
1
1
(d1 Fi d2 Fu ) y| + d1 Fi L} y{
L}
=
(10.595)
(10.596)
(10.597)
The second and third equations may be written in a matrix form for
¤W
£
y| u
5
yb |
ub
¸
9
= 9
7
Fi + Fu
py{
d1 Fi d2 Fu
L} y{
6
Fi
:
9
p
:
+9
7 d1 Fi 8 5
L}
6
d1 Fi + d2 Fu
y{ ¸
: y|
py{
:
8 u
d2 Fi + d22 Fu
1
L} y{
(10.598)
728
10. Vehicle Planar Dynamics
or in a matrix form for
5
b
ub
¸
9
= 9
7
£
u
¤W
.
Fi + Fu
py{
d1 Fi d2 Fu
L}
6
Fi
9 py{ :
:
+9
7 d1 Fi 8 5
L}
6
d1 Fi + d2 Fu
1 ¸
2
: py{
:
8 u
d2 Fi + d22 Fu
1
L} y{
(10.599)
10. Vehicle Planar Dynamics
729
10.9 Key Symbols
d{
¨
dl
dQ
[D]
e1
e2
E(F{|})
F
F
Fi
FiO
FiU
Fu
FuO
FuU
Fu , · · · , G
Fu
F
F
Gu
G
G
d
gQ
gp
Il
I{
I|
I|i
I|u
I}
F> M
g>j
J(R[\ ])
L
N
L
L
p
P{
P|
P}
acceleration
distance of the axle number l from the mass center
force coe!cient matrix
distance of the hinge point from rear axle
distance of trailer axle from the hinge point
vehicle coordinate frame
mass center
tire sideslip coe!cient
front sideslip coe!cient
front left sideslip coe!cient
front right sideslip coe!cient
rear sideslip coe!cient
rear left sideslip coe!cient
rear right sideslip coe!cient
force system coe!cients
proportionality coe!cient between I| and u
proportionality coe!cient between I| and proportionality coe!cient between I| and proportionality coe!cient between P} and u
proportionality coe!cient between P} and proportionality coe!cient between P} and frame position vector
neutral distance
mass element
generalized force
longitudinal force, forward force, traction force
lateral force
front lateral force
rear lateral force
normal force, vertical force, vehicle load
vehicle force system
gravitational acceleration
global coordinate frame
mass moment
stability factor
moment of momentum
Lagrangean
mass
roll moment, bank moment, tilting torque
pitch moment
yaw moment, aligning moment
730
10. Vehicle Planar Dynamics
s = *b
p
SQ
t = b
q
tl
u = #b
r
U
Uz
J
UE
V = @
V = @
Vu = u@
Vd = y{2 @U@
V|
V|1 > V|2
VD = 1@y{2
w
W
Wz
u
y {>
b v
z
{> |> }> x
roll rate
momentum
neutral steer point
pitch rate
control variable vector
generalized coordinate
yaw rate
position vector
radius of rotation
tire radius
rotation matrix to go from E frame to J frame
curvature response
sideslip response
yaw rate response
centripetal acceleration response
lateral velocity responses
steady-state responses
Ackerman steering gradient
time
tire coordinate frame
wheel torque
input vector
velocity
wheelbase
displacement
+#
i
u
b = t
= 1@U
*
*b = s
#
#b = u
#
$
$
b
sideslip angle
global sideslip angle
vehicle sideslip angle, attitude angle
attitude angle
cruise angle
steer angle
front steer angle
rear steer angle
pitch angle
pitch rate
curvature
eigenvalue
roll angle
roll rate
yaw angle
yaw rate
heading angle
angular velocity
angular acceleration
10. Vehicle Planar Dynamics
731
Exercises
1. Force system coe!cients.
Consider a front-wheel-steering car with the following characteristics
FuO = FuU = 460 N@ deg
FiO = FiU = 500 N@ deg
p = 1245 kg
L} = 1328 kg m2
d1 = 110 cm
d2 = 132 cm
(a) Determine the force system coe!cients Fu , F , F , Gu , G , G .
(b) Plot Fu , Gu as functions of y{ for 0 ? y{ ? 60 m@ s.
2. Dimension of the force characteristic coe!cients.
What are the dimensions of the force system coe!cients, Fu , F , F ,
Gu , G , G ?
3. Force system and two-wheel model of a car.
Consider a front-wheel-steering car with
FuO = FuU = FiO = FiU = 500 N@ deg
d1 = 110 cm
d2 = 132 cm
p = 1205 kg
L} = 1300 kg m2
(a) determine the force system that applies on the two-wheel model
of the car.
I|
P}
= Fu u + F + F = Gu u + G + G (b) Write the equations of motion of the car as
I{
I|
P}
= p yb { pu y|
= p yb | + pu y{
= ub L}
(c) Derive the force system coe!cient that the velocity is measured
in km@ h instead of m@ s.
4. Equations of motion for a front-wheel-steering car.
Consider a front-wheel-steering car with
FuO = FuU = FiO = FiU = 500 N@ deg
732
10. Vehicle Planar Dynamics
d1 = 110 cm
p = 1245 kg
y{ = 40 m@ s
d2 = 132 cm
L} = 1328 kg m2
and develop the equations of motion
qb = [D] q + u
5. F Nonlinear tire behavior.
Let us assume that the sideslip coe!cient of the tires of a vehicle are
nonlinear such that its lateral force generation capacity drops at high
.
F = F1 F2 2
(a) Develop the force system for the vehicle.
(b) Derive the equations of motion for a front wheel steerable planar
bicycle car model.
6. No approximation for .
Do not use the approximations of
μ
¶
y|i
y|
i = arctan
i
y{i
y{i
¶
μ
y|u
y|u
u = arctan
y{u
y{u
μ ¶
y|
y|
= arctan
y{
y{
(a) Accept the linear functions of
I|i = Fi i
I|u = Fu u
and develop the force system for the planar bicycle car model.
(b) Assume that
I|i = F1i i + F2i 3i
I|u = F1u u + F2u 3u
and develop the force system for the planar bicycle car model.
7. Equations of motion in dierent variables.
Consider a car with the following characteristics
FuO
d1
p
y{
=
=
=
=
FuU = FiO = FiU = 500 N@ deg
100 cm
d2 = 120 cm
1000 kg
L} = 1008 kg m2
40 m@ s
and develop the equations of motion
10. Vehicle Planar Dynamics
733
(a) in terms of (yb { > yb | > u),
b if the car is front-wheel steering.
(b) in terms of (yb { > yb | > u),
b if the car is four-wheel steering.
´
³
b ub , if the car is front-wheel steering.
(c) in terms of yb { > >
³
´
b ub , if the car is four-wheel steering.
(d) in terms of yb { > >
8. Steady state response parameters.
Consider a car with the following characteristics
FiO = FiU = 500 N@ deg
FuO = FuU = 520 N@ deg
p = 1245 kg
L} = 1328 kg m2
d1 = 110 cm
d2 = 132 cm
(a) Determine the steady-state responses V , V , Vu , Vd , V| .
(b) Plot the steady-state responses as functions of y{ .
(c) Determine their values at y{ = 40 m@ s.
9. Steady state motion parameters.
Consider a car with the following characteristics
FiO
p
d1
y{
=
=
=
=
FiU = 600 N@ deg
FuO = FuU = 550 N@ deg
1245 kg
L} = 1128 kg m2
120 cm
d2 = 138 cm
20 m@ s
= 3 deg
(a) Determine the steady-state values of u, U, , and y{2 @U.
(b) Find the coordinates of the rotation center.
10. F Inertia and steady-state parameters.
Consider a car that is made up of a uniform solid box with dimensions
260 cm × 140 cm × 40 cm in { × } × | dimensions. If the density of the
box is = 1000 kg@ m3 , and the other characteristics are
FiO
d1
= FiU = 600 N@ deg
o
= d2 = = 1=25 m
2
FuO = FuU = 550 N@ deg
then,
(a) Determine p, L} .
(b) Determine the steady-state responses V , V , Vu , and Vd as functions of y{ .
734
10. Vehicle Planar Dynamics
(c) Determine the velocity y{ at which the car has a radius of turning equal to
U = 35 m
when
= 4 deg
(d) Determine the steady-state parameters u, U, , y{2 @U and center
of rotation at that speed.
(e) Set the speed of the car at
y{ = 20 m@ s
and plot the steady-state responses V , V , Vu , and Vd for variable .
11. Stability factor and understeer behavior.
Examine the stability factor N and
(a) determine the condition to have an understeer car, if d1 = d2 .
(b) determine the condition to have an understeer car, if Fi = Fu .
(c) show that if we use the tires in front and rear such that Fi =
Fu , then the front of the car must be heavier to have an understeer car?
(d) show that if we have a car with d1 = d2 , then we must use tires
such that Fu A Fi to have an understeer car?
12. Stability factor and mass of the car.
Find d1 and d2 in terms of I}1 , I}2 , and pj to rewrite the stability
factor N to see the eect of a car’s mass distribution.
13. Stability factor and car behavior.
Examine the stability factor of a car with the parameters
FiO
p
d1
y{
=
=
=
=
FiU = 500 N@ deg
FuO = FuU = 460 N@ deg
1245 kg
L} = 1328 kg m2
110 cm
d2 = 132 cm
30 m@ s
and
(a) determine if the car is understeer, neutral, or oversteer?
(b) determine the neutral distance gQ .
10. Vehicle Planar Dynamics
735
14. F Critical speeds of a car.
Compare the critical speeds of curvature, sideslip, and yaw rate responses and order them if possible. Determine the conditions at which
they are equal if there is any.
15. Critical speed of a car.
Consider a car with the characteristics
FiO = FiU = 700 N@ deg
FuO = FuU = 520 N@ deg
p = 1245 kg
L} = 1328 kg m2
d1 = 118 cm
d2 = 122 cm
(a) determine if the car is understeer, neutral, or oversteer?
(b) in case of an oversteer situation, determine the neutral distance
gQ and the critical speed yf of the car.
16. Steady-state U for oversteer vehicles.
Consider a car with
FiO = FiU = 500 N@ deg
FuO = FuU = 460 N@ deg
p = 1245 kg
L} = 1328 kg m2
d1 = 110 cm
d2 = 132 cm
(a) Determine and plot U versus y{ .
(b) Switch d1 and d2 such that
d2 = 110 cm
d1 = 132 cm
and determine and plot U versus y{ .
17. F Step input response at dierent speed.
Consider a car with the characteristics
FuO = FuU = 750 N@ deg
FiO = FiU = 600 N@ deg
p = 1245 kg
L} = 1328 kg m2
d1 = 110 cm
d2 = 132 cm
and a step input
(w) =
½
5 deg
0
wA0
w0
Determine the time response of the car at the following forward speeds
and plot u as a function of time.
(a) y{ = 10 m@ s
736
10. Vehicle Planar Dynamics
(b) y{ = 20 m@ s
(c) y{ = 30 m@ s
(d) y{ = 40 m@ s
(e) Determine uP d{ in each case and examine if uPd{ is linearly
proportional with y{ .
(f) Find uP d{ as a function of y{ analytically and describe the care
behavior based on the plot of uPd{ .
18. Steady-state rotation center function.
Consider a car with
FiO = FiU 30000 N@ rad
FuO = FuU 30000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=0 m
d2 = 1=5 m
μ
¶
p
d1
d2
N = 2
= 1=33 × 102 A 0
o
Fi
Fu
(a) Determine the coordinate of rotation center ({R > |R ).
(b) Check if it is possible to eliminate y{ to nd |R = i ({R ).
(c) Approximate |R = i ({R ) by a linear function.
19. Stability factor analysis.
Consider a car with
Fi = 57296 N@ rad
Fu = 52712 N@ rad
p = 917 kg
L} = 1128 kg m2
o = 2=55 m
d1 = {
d2 = o {
(a) Plot N as a function of {.
(b) F Plot a three dimensional surface of V as a function of { and
y{ .
20. F Mass center and velocity.
Let us imagine a car such that the position of its mass center shifts
based on its forward speed for 0 ? y{ ? 100 m@ s.
Fi = 57296 N@ rad
Fu = 52712 N@ rad
p = 917 kg
L} = 1128 kg m2
y{
y{
o = 2=55 m
d1 = 1 +
d2 = o 1 100
100
(a) Plot N as a function of {.
10. Vehicle Planar Dynamics
737
(b) Plot V as a function of {.
21. F Step input response for dierent steer angle.
Consider a car with the characteristics
FiO
p
d1
y{
=
=
=
=
FiU = 600 N@ deg
FuO = FuU = 750 N@ deg
1245 kg
L} = 1328 kg m2
110 cm
d2 = 132 cm
20 m@ s
Determine the time response of the car to a step input
½
wA0
(w) =
0 w0
and plot u versus y{ when
(a) = 2 deg, = 3 deg, = 4 deg, = 5 deg, = 6 deg, = 7 deg,
= 8 deg, = 9 deg, = 10 deg.
(b) Determine uP d{ in each case and examine if uPd{ is linearly
proportional with y{ .
(c) Find uP d{ as a function of y{ analytically and describe the care
behavior based on the plot of uPd{ .
22. F Eigenvalues and free response.
Consider a car with the characteristics
FiO
p
d1
y{
=
=
=
=
FiU = 600 N@ deg
FuO = FuU = 750 N@ deg
1245 kg
L} = 1328 kg m2
110 cm
d2 = 132 cm
20 m@ s
(a) Determine the eigenvalues of the coe!cient matrix [D] and nd
out if the car is stable at zero steer angle.
(b) In either case, determine the weight distribution ratio, d1 @d2 ,
such that the car is neutral stable.
(c) Recommend a condition for the weight distribution ratio, d1 @d2 ,
such that the car is stable.
23. F Time response to dierent steer functions.
Consider a car with the characteristics
FiO
p
d1
y{
=
=
=
=
FiU = 600 N@ deg
FuO = FuU = 750 N@ deg
1245 kg
L} = 1328 kg m2
110 cm
d2 = 132 cm
20 m@ s
738
10. Vehicle Planar Dynamics
Determine the time response of the car to
(a) (w) = sin 0=1w for 0 ? w ? 10 and (w) = 0 for w 0 and
w 10.
(b) (w) = sin 0=5w for 0 ? w ? 2 and (w) = 0 for w 0 and w 2.
(c) (w) = sin w for 0 ? w ? and (w) = 0 for w 0 and w .
24. F Transient between two steady-state conditions.
Consider the front-wheel-steering bicycle planar model of a vehicle
with the following parameters.
Fi = 57296 N@ rad
Fu = 52712 N@ rad
p = 917 kg
L} = 1128 kg m2
d1 = 0=91 m
d2 = 1=64 m
(a) Determine the steady state values of V for y{ = 10 m@ s and
y{ = 30 m@ s.
(b) Determine the transient behavior of the vehicle between the
steady state condition assuming the speed of the car changes
suddenly.
25. Transient path of rotation center.
Consider a vehicle with the following characteristics
FiO = FiU 3000 N@ rad
FuO = FuU 3000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=0 m
d2 = 1=5 m
(a) Determine the steady-state responses V , V , Vu , Vd for y{ =
1 m@ s and y{ = 10 m@ s.
(b) Determine the critical speed at which = 0.
(c) Calculate the coordinates of the rotation center for y{ = 1 m@ s,
the critical speed, and y{ = 10 m@ s.
(d) F Assume the vehicle is at steady state condition for y{ = 1 m@ s
and suddenly the speed changes to y{ = 10 m@ s and remains
constant. Determine the path of the rotation center between the
two steady-state conditions.
26. Linear model dynamics.
Consider a vehicle with the following characteristics
FiO = FiU 3000 N@ rad
FuO = FuU 3000 N@ rad
p = 1000 kg
L} = 1650 kg m2
d1 = 1=0 m
d2 = 1=5 m
10. Vehicle Planar Dynamics
739
and compare the transient behavior of the vehicle using the planar
model and the linearized model for a step steer change.
½
5 deg w A 0
(w) =
0
w0
11
F Vehicle Roll Dynamics
In this chapter, we develop a dynamic model for a rigid bicycle vehicle
having forward, lateral, yaw, and roll motions. The model of a rollable
rigid vehicle is more exact and more eective compared to the rigid bicycle
vehicle planar model. Using this model, we are able to analyze the roll
behavior of a vehicle as well as its maneuvering.
z
y
Fy
r
Mz
\
Fx x
Mx
p
M
C
B
l
Ro
is
l ax
FIGURE 11.1. The DOF of a roll model of rigid vehicles are: {> |> *> #.
11.1 F Vehicle Coordinate and DOF
Figure 11.1 illustrates a vehicle with a body coordinate E(F{|}) at the
mass center F. The {-axis is a longitudinal axis passing through F and directed forward. The |-axis goes laterally to the left from the driver’s viewpoint. The }-axis makes the coordinate system a right-hand triad. When
the car is parked on a at horizontal road, the }-axis is perpendicular to
the ground, opposite to the gravitational acceleration g. The equations of
motion of the vehicle should be expressed in E(F{|}).
Angular orientation and angular velocity of a vehicle are expressed by
three angles: roll *, pitch , yaw #, and their rates: roll rate s, pitch rate
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_11, © Springer Science+Business Media New York 2014
741
742
11. F Vehicle Roll Dynamics
t, yaw rate u.
s = *b
t = b
u = #b
(11.1)
(11.2)
(11.3)
The vehicle force system (F> M) is the resultant of external forces and
moments that the vehicle receives from the ground and environment. The
force system may be expressed in the body coordinate frame as:
E
E
F = I{ˆ~ + I| ˆ + I} n̂
(11.4)
M = P{ˆ~ + P| ˆ + P} n̂
(11.5)
The roll model vehicle dynamics can be expressed by four kinematic
variables: the forward motion {, the lateral motion |, the roll angle *, and
the yaw angle #. In this model, we do not consider vertical movement },
and pitch motion .
11.2 F Equations of Motion
A rigid vehicle with roll and yaw motions has a motion with four degrees
of freedom, which are translation in { and | directions, and rotation about
the { and } axes. The Newton-Euler equations of motion for such a rolling
rigid vehicle in the body coordinate frame E are:
I{
I|
P}
P{
=
=
=
=
p yb { pu y|
p yb | + pu y{
L} $b } = L} ub
L{ $b { = L{ sb
(11.6)
(11.7)
(11.8)
(11.9)
Proof. Consider the vehicle shown in Figure 11.2. A global coordinate
frame J is xed on the ground, and a local coordinate frame E is attached
to the vehicle at the mass center F. The orientation of the frame E can be
expressed by the heading angle # between the { and [ axes, and the roll
angle * between the } and ] axes. The global position vector of the mass
center is denoted by J d.
The rigid body equations of motion in the body coordinate frame are:
¡
¢
E
F = E UJ J F = E UJ p J aE = p E
J aE
E
= p E vb E + p E
vE
J $E ×
E
(11.10)
J
M =
=
g E
Eb
E
b
L+ E
L= E
L
J LE =
J $E ×
gw
¢
¡
E E
E E
L J$
bE + E
L J $E
J $E ×
(11.11)
11. F Vehicle Roll Dynamics
B
743
G
Z
z
Mz
r
X
x
\
M
Fx
Mx
p
\
My
Fy
Y
y
FIGURE 11.2. A vehicle with roll, and yaw rotations.
The velocity vector of the vehicle, expressed in the body frame, is
5
6
y{
E
vF = 7 y| 8
(11.12)
0
where y{ is the forward component and y| is the lateral component of v.
The other kinematic vectors for the rigid vehicle are:
5
6
yb {
E
vb F = 7 yb | 8
(11.13)
0
6 5 6
s
${
E
8
7
7
0 8
0
=
$
=
J E
u
$}
5
(11.14)
5
6 5 6
$b {
sb
E
7
8
7
0 8
0
$
b
=
=
J E
ub
$b }
(11.15)
We may assume that the body coordinate is the principal coordinate frame
of the vehicle to have a diagonal moment of inertia matrix.
6 5
6
5
L1 0 0
L{ 0 0
E
(11.16)
L = 7 0 L| 0 8 = 7 0 L2 0 8
0 0 L}
0 0 L3
Substituting the above vectors and matrices in the equations of motion
744
11. F Vehicle Roll Dynamics
(11.10) and (11.11) provides us with the following equations:
E
F = p E vb E + p E
$ E × E vE
6
5
6 5
6 J5
6
I{
yb {
y{
${
7 I| 8 = p 7 yb | 8 + p 7 0 8 × 7 y| 8
$}
I}
0
0
5
6
pyb { p$ } y|
= 7 pyb | + p$ } y{ 8
p$ { y|
5
(11.17)
(11.18)
¡E E
¢
M = EL E
bE + E
L J $E
(11.19)
J$
J $E ×
5
6
5
65
6
P{
L1 0 0
$b {
7 P| 8 = 7 0 L2 0 8 7 0 8
0 0 L3
$b }
P}
5
6 35
65
64
${
L1 0 0
${
+ 7 0 8 × C7 0 L2 0 8 7 0 8D
$}
0 0 L3
$}
5
6
L1 $b {
= 7 L1 $ { $ } L3 $ { $ } 8
(11.20)
L3 $b }
E
The rst two Newton equations (11.18) are the translational equations
of motion in the { and | directions.
¸ ¸
I{
pyb { p$ } y|
=
(11.21)
I|
pyb | + p$ } y{
and the rst and third Euler equations (11.20) are the rotational equations
of motion in the { and } directions.
The third Newton’s equation
p$ { y| = I}
(11.22)
provides the compatibility condition to keep the vehicle on the road. However, because of ignorance gravity and any motion in |-direction or rotation
about |-axis, the equation in }-direction is not complete.
The rst and third Euler equations (11.20) are the equations of motion
about the { and } axes.
¸ ¸
P{
L1 $b {
=
(11.23)
P}
L3 $b }
The second Euler equation
L1 $ { $ } L3 $ { $ } = P|
(11.24)
is another compatibility condition that provides the required pitch moment
condition to keep the vehicle on the road.
11. F Vehicle Roll Dynamics
745
Example 461 F Motion of a six DOF vehicle.
Consider a vehicle that moves in space. Such a vehicle has six DOF. To
develop the equations of motion of such a vehicle, we need to dene the
kinematic characteristics as follows:
6
6
5
5
y{
yb {
E
E
(11.25)
vF = 7 y| 8
vb F = 7 yb | 8
y}
yb }
5
6
5
6
${
$b {
E
E
7 $| 8
b E = 7 $b | 8
(11.26)
J $E =
J$
$}
$b }
The acceleration vector of the vehicle in the body coordinate is
5
6
yb { + $ | y} $ } y|
E
E
a = E vb E + E
vE = 7 yb | + $ } y{ $ { y} 8
J $E ×
yb } + $ { y| $ | y{
(11.27)
and therefore, the Newton equations of motion for the vehicle are
5
6
5
6
I{
yb { + $ | y} $ } y|
7 I| 8 = p 7 yb | + $ } y{ $ { y} 8
(11.28)
I}
yb } + $ { y| $ | y{
To nd the Euler equations of motion,
E
M = EL E
bE + E
J$
J $E ×
¡E
L E
J $E
¢
(11.29)
we need to dene the mass moment matrix and perform the required matrix
calculations. Assume the body coordinate system is the principal coordinate
frame. So,
¡E E
¢
E E
L J$
bE + E
L J $E
J $E ×
5
65
6 5
6 35
65
64
L1 0 0
$b {
${
L1 0 0
${
= 7 0 L2 0 8 7 $b | 8 + 7 $ | 8 × C7 0 L2 0 8 7 $ | 8D
0 0 L3
0 0 L3
$b }
$}
$}
5
6
$b { L1 $ | $ } L2 + $ | $ } L3
= 7 $b | L2 + $ { $ } L1 $ { $ } L3 8
(11.30)
$b } L3 $ { $ | L1 + $ { $ | L2
and therefore, the Euler equations of motion for the vehicle are
5
6 5
6
P{
$b { L1 $ | $ } L2 + $ | $ } L3
7 P| 8 = 7 $b | L2 + $ { $ } L1 $ { $ } L3 8
P}
$b } L3 $ { $ | L1 + $ { $ | L2
(11.31)
Equations (11.28) and (11.31) are the equations of motion of a car with
pitch degree of freedom.
746
11. F Vehicle Roll Dynamics
Example 462 F Roll rigid vehicle from general motion.
We may derive the equations of motion for a roll rigid vehicle from the
general equations of motion for a vehicle with six DOF (11.28) and (11.31).
Consider a bicycle model of a four-wheel vehicle moving on a road. Because the vehicle cannot move in }-direction and cannot turn about the
|-axis, we have
y} = 0
yb } = 0
$| = 0
$b | = 0
(11.32)
Furthermore, the resultant force in the }-direction and moment in |-direction
for such a bicycle must be zero,
I} = 0
P| = 0
(11.33)
Substitution Equations (11.32)-(11.33) in (11.28) and (11.31) results in
the force system.
6
5
6
5
yb { $ } y|
I{
7 I| 8 = p 7 yb | + $ } y{ 8
(11.34)
I}
$ { y|
5
6
5
6
P{
$b { L1
7 P| 8 = 7 $ { $ } L1 $ { $ } L3 8
(11.35)
P}
$b } L3
11.3 F Vehicle Force System
To determine the force system on a rigid vehicle, we dene the force system
at the tireprint of a wheel. The lateral force at the tireprint depends on the
sideslip angle. Then, we transform and apply the tire force system on the
rollable model of the vehicle.
11.3.1 F Tire and Body Force Systems
Figure 11.3 depicts the wheel number 1 of a vehicle. The components of the
applied force system in the F-frame, because of the forces at the tireprint
of the wheel number l are:
I{l
I|l
P{l
P}l
=
=
=
=
I{zl cos l I|zl sin l
I|zl cos l + I{zl sin l
P{zl
P}zl
(11.36)
(11.37)
(11.38)
(11.39)
where ({l > |l > }l ) are body coordinates of the wheel number l. In this analy-
11. F Vehicle Roll Dynamics
G1
Mxw
Fxw
C
747
x
Fx1
xw
Mzw
Fy1
Fyw
Myw
1
yw
Mx1
x1
B
W
Mz1
y
C
My1
y1
FIGURE 11.3. The force system at the tireprint of tire number 1, and their
resultant force system at F.
sis, we ignored the components of the tire moment at the tireprint, P|zl to
simplify the equations.
The total important force system on the rigid vehicle in the body coordinate frame to analyze the roll model of rigid vehicle is
E
I{
X
=
I{l =
l
E
I|
X
=
I|l =
P{
X
=
X
l
E
P}
X
=
I|z cos l +
X
|l I}l l
P}l +
l
I|z sin l
(11.40)
I{z sin l
(11.41)
l
l
P{l +
X
I{z cos l l
l
E
X
X
X
X
l
}l I|l
(11.42)
l
{l I|l l
X
|l I{l
(11.43)
l
Proof. To simplify the roll model of vehicle dynamics, we ignore the difference between the tire frame at the center of tireprint and wheel frame
at the wheel center for small roll angles. The tireprint, the force system
generated at the tireprint of the wheel in the wheel frame Z is
Z
Z
Fz
=
Mz
=
£
£
¤W
I{z
I|z
I}z
P{z
P|z
P}z
(11.44)
¤W
(11.45)
The rotation matrix between the wheel frame Z and the wheel-frame
748
11. F Vehicle Roll Dynamics
coordinate frame F, parallel to the vehicle body coordinate frame E, is
5
cos 1
F
UZ = 7 sin 1
0
6
0
0 8
1
sin 1
cos 1
0
(11.46)
and therefore, the force system at the tireprint of the wheel, parallel to the
vehicle coordinate frame, is
F
Fz = F UZ Z Fz
6
5
6
65
I{1
cos 1 sin 1 0
I{z
7 I|1 8 = 7 sin 1
cos 1 0 8 7 I|z 8
0
0
1
I}1
I}z
5
6
I{z cos 1 I|z sin 1
= 7 I|z cos 1 + I{z sin 1 8
I}z
5
E1
Mz = E1 UEz Ez Mz
6
5
6
65
P{1
cos 1 sin 1 0
P{z
7 P|1 8 = 7 sin 1
cos 1 0 8 7 P|z 8
0
0
1
P}1
P}z
5
6
P{z cos 1 P|z sin 1
= 7 P|z cos 1 + P{z sin 1 8
P}z
5
(11.47)
(11.48)
(11.49)
(11.50)
Ignoring P|l , we transform the force system of each tire to the body
coordinate frame E, located at the vehicle mass center F, to generate the
total force system applied on the vehicle
E
F =
X
El
Fz =
l
E
M =
X
X
I{l ˆ~ +
l
El
Mz
l
=
X
P{l ˆ~ +
l
X
l
P}l n̂ +
X
I|l ˆ
(11.51)
l
X
l
E
rl × E Fzl
(11.52)
where E rl is the position vector of the wheel number l.
E
rl = {lˆ~ + |l ˆ + }l n̂
(11.53)
P
In deriving Equation (11.51), we have used the equation l I}l pj = 0.
Expanding Equations (11.51) and (11.52) provides the total vehicle force
11. F Vehicle Roll Dynamics
749
system.
E
I{
=
X
I{z cos l l
E
I|
=
X
P{
=
X
I|z cos l +
P{l +
l
E
P}
=
X
I|z sin l
(11.54)
I{z sin l
(11.55)
X
}l I|l
(11.56)
|l I{l
(11.57)
l
l
E
X
X
X
l
|l I}l l
P}l +
l
X
l
{l I|l l
X
l
For a two-wheel vehicle model we have
{1
|1
}1
= d1
{2 = d2
= |2 = 0
= Ui
}2 = Uz
(11.58)
For such a vehicle, after ignoring the eect of tire reduces Ui and Uz , the
force system reduces to
E
I{
I|
E
P{
E
P}
E
=
=
=
=
I{1 cos 1 + I{2 cos 2 I|1 sin 1 I|2 sin 2
I|1 cos 1 + I|2 cos 2 + I{1 sin 1 + I{2 sin 2
P{1 + P{2
P}1 + P}2 + d1 I|1 d2 I|2
(11.59)
(11.60)
(11.61)
(11.62)
11.3.2 F Tire Lateral Force
If the steer angle of the steering mechanism is denoted by , then the actual
steer angle d of a rollable vehicle is
d = + *
(11.63)
where, * is the roll-steering angle.
* = F* *
(11.64)
The roll-steering angle * is proportional to the roll angle * and the coefcient F* is called the roll-steering coe!cient. The roll-steering happens
because of the suspension mechanisms that generate some steer angle when
de ected. The tire sideslip angle of each tire of a rollable vehicle is
l = l d = l l *
(11.65)
11. F Vehicle Roll Dynamics
750
where l is the angle between the velocity vector v and the vehicle body
{-axis, and is called the wheel sideslip angle.
The generated lateral force by such a wheel for a small sideslip angle, is
I|
= F l F* *l
= F ( l l F* *l ) F* *
(11.66)
and F* is the tire camber thrust coe!cient, because of the vehicle’s roll.
The wheel slip angle l of a bicycle vehicle model can be approximated by
l =
y| + {l u F l s
y{
(11.67)
to nd the tire lateral force I| in terms of the vehicle kinematic variables.
I|
= {l
=
y|
y{
F
F F
u+
s F + (F F* F* ) * + F l (11.68)
y{
y{
(11.69)
F l is wheel slip coe!cient.
Proof. When a vehicle rolls, there are some new reactions in the tires
that introduce new dynamic terms in the behavior of the tire. The most
important reactions are:
1Tire camber thrust I|* , which is a lateral force because of the vehicle
roll. Tire camber thrust is assumed to be proportional to the vehicle roll
angle *.
I|*
F*
= F* *
gI|
=
g*
(11.70)
(11.71)
2Wheel roll steering angle * , which is the wheel steer angle because
of the vehicle roll. Most suspension mechanisms provide some steer angle
when the vehicle rolls and the mechanism de ects. The wheel roll steering
is assumed to be proportional to the vehicle roll angle *.
*
F*
= F* *
g
=
g*
(11.72)
(11.73)
Therefore, the actual steer angle d of such a tire is
d = + *
Assume the wheel number l of a rigid vehicle is located at
£
¤W
E
rl = {l |l }l
(11.74)
(11.75)
11. F Vehicle Roll Dynamics
751
The velocity of the wheel number l is
E
vl = E v + E $ × E rl
(11.76)
in which E v is the velocity vector of the vehicle at its mass center F, and
$ is the angular velocity of the vehicle.
E
E
$ = *ˆ
b ~ + #b n̂ = sˆ~ + un̂
(11.77)
Expanding Equation (11.76) provides the following velocity vector for the
wheel number l expressed in the vehicle coordinate frame at E.
5
6 5
6 5
6
6 5
6 5
b l
*b
y{l
y{
y{ #|
{l
7 y|l 8 = 7 y| 8 + 7 0 8 × 7 |l 8 = 7 y| *}
b l 8 (11.78)
b l + #{
y}l
0
}l
#b
*|
b l
Consider a bicycle model for the rollable vehicle to have
|l
{1
{2
= 0
= d1
= d2
(11.79)
(11.80)
(11.81)
The wheel slip angle l for the wheel l, is dened as the angle between the
wheel velocity vector vl and the vehicle body {-axis. When the roll angle
is very small, l is
μ
¶
b l
y|l
y|
y| *}
b l + #{
l = tan1
l (11.82)
y{l
y{l
y{
If the tire number l has a steer angle l then, its sideslip angle l , that
generates a lateral force I|z on the tire, is
l = l l b l
y| *}
b l + #{
l + *l
y{
(11.83)
The wheel slip angle l for the front and rear wheels of a two-wheel
vehicle, i and u , are
μ
¶
y|i
y|
y| + d1 u }i s
1
i = tan
(11.84)
i y{i
y{i
y{
μ
¶
y| d2 u }u s
y|u
y|
(11.85)
u u = tan1
y{u
y{u
y{
and the vehicle slip angle is
= tan
1
μ
y|
y{
¶
y|
y{
(11.86)
752
11. F Vehicle Roll Dynamics
The }l coordinate of the wheels is not constant however, its variation is very
small. To show the eect of }l , we substitute it by coe!cient F l called the
tire roll rate coe!cient, and dene coe!cients F i and F u to express the
change in l because of roll rate s.
l
F l
= F l s
g l
=
gs
(11.87)
(11.88)
Therefore,
¶
y| + d1 u F i s
= tan
y{
μ
¶
y| d2 u F u s
1
= tan
y{
1
i
u
μ
(11.89)
(11.90)
Assuming small angles for slip angles i , , and u then, the tire sideslip
angles for the front and rear wheels, i and u , may be approximated as
i
u
1
(y| + d1 u }i s) *i
y{
u
s
= + d1 F i
F*i *
y{
y{
1
=
(y| d2 u }u s) *u
y{
u
s
= d2 F u
F*u *
y{
y{
=
(11.91)
(11.92)
11.3.3 F Body Force Components on a Two-wheel Model
Figure 11.4 illustrates the top view of a car and the force systems acting
at the wheel center of a front-wheel-steering four-wheel vehicle. When we
consider the roll motion of the vehicle, the {|-plane does not remain parallel
to the road’s [\ -plane, however, we may still use a two-wheel model for
the vehicle.
Figure 11.5 illustrates the force system and Figure 11.6 illustrates the
kinematics of a two-wheel model for a vehicle with roll and yaw rotations.
The rolling two-wheel model is also called the bicycle model.
The force system applied on the bicycle vehicle, with only the front wheel
steerable, is
I{
I|
=
=
2
X
l=1
2
X
l=1
(I{l cos I|l sin )
(11.93)
I|l
(11.94)
11. F Vehicle Roll Dynamics
G1
G2
x
Fx1
Mx1
1
Fx2
Mx2
2
Mz1
Mz2
Fy1
B
753
Fy2
y
C
Fx4
Fx3
Mx4
Mx3
Mz4
Fy4
Mz3
Fy3
4
3
FIGURE 11.4. Top view of a car and the forces system acting at the tireprints.
P{
P}
= P{i + P{u zfi *b zni *
(11.95)
= d1 I|i d2 I|u
(11.96)
¢
¡
¢
¡
where I{i > I{u and I|i > I|u are the planar forces on the front and rear
wheels. The force system may be approximated by the following equations,
if the steer angle is assumed small:
I{
I|
P{
P}
I{i + I{u
I|i + I|u
FWi I|i + FWu I|u n* * f* *b
d1 I|i d2 I|u
(11.97)
(11.98)
(11.99)
(11.100)
The vehicle’s lateral force I| and moment P} depend only on the front
and rear wheels’ lateral forces I|i and I|u , which are functions of the
tires’ sideslip angles i and u . They can be approximated by the following
equations:
μ
¶
μ
¶
Fi F i
Fu F u
d2
d1
I| =
Fu Fi u +
+
s
y{
y{
y{
y{
³
´
+ (Fi Fu ) + Fi F*i F*u F*i + Fu F*u *
+Fi (11.101)
11. F Vehicle Roll Dynamics
754
x
Gf
Fyf
Fxf
Mxf
Mzf
a1
C
y
Fxr
B
Mxr
Mzr
l
a2
Fyr
FIGURE 11.5. A two-wheel model for a vehicle with roll and yaw rotations.
P{
=
μ
¶
d2
d1
FWu Fu FWi Fi u
y{
y{
μ
¶
1
1
+
F i FWi Fi + F u FWu Fu f* s
y{
y{
³
³
´
´
¡
¢
+ FWi F*i Fi F*i FWu F*u Fu F*u n* *
+FWi Fi P}
=
(11.102)
μ 2
¶
μ
¶
d1
d
d2
d2
1 Fi 2 Fu u +
F i Fi F u Fu s
y{
y{
y{
y{
+ (d2 Fu d1 Fi ) ³ ¡
³
´´
¢
+ d2 F*u Fu F*u d1 F*i Fi F*i
*
+d1 Fi (11.103)
where Fi = FiO + FiU and Fu = FuO + FuU are equal to the sideslip
coe!cients of the left and right wheels in front and rear, respectively.
Fi
Fu
= FiO + FiU
= FuO + FuU
(11.104)
(11.105)
11. F Vehicle Roll Dynamics
755
xw D vf E f x
f
G
v
E
B
a1
C
y
E
O
R
R1
r
vr
Dr
l
a2
Center of
rotation
FIGURE 11.6. Kinematics of a two-wheel model for a vehicle with roll and yaw
rotations.
Proof. For a two-wheel vehicle, we use the cot-average (7.4) of the outer
and inner steer angles as the only steer angle .
cot r + cot l
(11.106)
2
Furthermore, we dene a single sideslip coe!cient Fi and Fu for the
front and rear wheels. The coe!cient Fi and Fu are equal to sum of the
left and right wheels’ sideslip coe!cients.
Employing Equations (11.40)-(11.43) the forward and lateral forces on
the rollable bicycle would be
cot =
I{
I|
= I{i cos + I{u I|i sin = I|i + I|u
(11.107)
(11.108)
The yaw moment equation does not interact with the vehicle roll. We may
also ignore the moments P}l and assume that the forward forces on the
front left and right wheels are equal, as well
P as the forward forces on the
rear left and right wheels. So, the terms l |l I{l cancel each other and
the yaw moment reduces to
P} = d1 I|i d2 I|u
(11.109)
The vehicle roll moment P{ is a summation of the slip and camber
moments on the front and rear wheels, P{i , P{u , and the moment because
756
11. F Vehicle Roll Dynamics
of change in normal loads of the left and right wheels |l I}l . Let’s assume
that the slip and camber moments are proportional to the wheels’ lateral
force and write them as
P{i
P{u
= FWi I|i
= FWu I|u
(11.110)
(11.111)
where FWi and FWu are the overall torque coe!cient of the front and rear
wheels respectively.
FWi
=
FWu
=
gP{i
gI|i
gP{u
gI|u
(11.112)
(11.113)
Roll moment because of change in normal force of the left and right
wheels is a result of force change in springs and dampers. These unbalanced
forces generate a roll stiness moment that is proportional to the vehicle’s
roll angle
P{n
P{f
= n* *
= f* *b
(11.114)
(11.115)
where n* and f* are the roll stiness and roll damping of the vehicle.
n*
f*
= zn = z (ni + nu )
= zf = z (fi + fu )
(11.116)
(11.117)
z is the track of the vehicle and n and f are sum of the front and rear
springs’ stu!ness and shock absorbers damping. The coe!cients n* and
f* are called the roll stiness and roll damping, respectively.
n = ni + nu
f = fi + fu
(11.118)
(11.119)
Therefore, the applied roll moment on the vehicle can be summarized as
P{
= P{i + P{u + P{f + P{n
= FWi I|i + FWu I|u z (fi + fu ) *b z (ni + nu ) * (11.120)
If we assume a small steer angle , the vehicle force system can be approximated by the following equations:
I{
I|
P{
P}
I{i + I{u
I|i + I|u
FWi I|i + FWu I|u n* * f* *b
d1 I|i d2 I|u
(11.121)
(11.122)
(11.123)
(11.124)
11. F Vehicle Roll Dynamics
757
Substituting for the lateral forces from (11.68), and expanding Equations
(11.121)-(11.124) provides the following force system.
I{
I|
= I{i + I{u
(11.125)
= I|i + I|u = Fi i F*i * Fu u F*u *
¶
μ
u
s
F*i * F*i *
= Fi + d1 F i
y{
y{
¶
μ
u
s
Fu d2 F u
F*u * F*u *
y{
y{
μ
¶
μ
¶
Fi F i
Fu F u
d2
d1
=
Fu Fi u +
+
s
y{
y{
y{
y{
³
´
+ (Fi Fu ) + Fi F*i F*u F*i + Fu F*u *
+Fi P{
(11.126)
= FWi I|i + FWu I|u n* * f* s
¶
¶
μ
μ
u
s
= FWi Fi + d1 F i
F*i * + F*i *
y{
y{
μ
μ
¶
¶
u
s
FWu Fu d2 F u
F*u * + F*u *
y{
y{
n* * f* s
μ
¶
d2
d1
=
FWu Fu FWi Fi u
y{
y{
μ
¶
1
1
+
F i FWi Fi + F u FWu Fu f* s
y{
y{
¡
¢
+ FWi Fi FWu Fu ³
³
´
´
¡
¢
+ FWi F*i Fi F*i FWu F*u Fu F*u n* *
+FWi Fi P}
(11.127)
= d1 I|i d2 I|u
¶
¶
μ
μ
u
s
= d1 Fi + d1 F i
F*i * + F*i *
y{
y{
μ
μ
¶
¶
u
s
+d2 Fu d2 F u
F*u * + F*u *
y{
y{
μ 2
¶
μ
¶
2
d
d
d2
d1
=
1 Fi 2 Fu u +
F i Fi F u Fu s
y{
y{
y{
y{
+ (d2 Fu d1 Fi ) ³
´´
³ ¡
¢
*
+ d2 F*u Fu F*u d1 F*i Fi F*i
+d1 Fi (11.128)
758
11. F Vehicle Roll Dynamics
The parameters Fi and Fu are the sideslip stiness for the front and
rear wheels, u is the yaw rate, s is the roll rate, * is the roll angle, is the
steer angle, and is the slip angle of the vehicle.
These equations are dependent on ve parameters: u, s, , *, , and may
be written as
I|
P{
P}
= I| (u> s> > *> )
CI|
CI|
CI|
CI|
CI|
=
u+
s+
+
*+
Cu
Cs
C
C*
C
= Fu u + Fs s + F + F* * + F (11.129)
= P{ (u> s> > *> )
CP{
CP{
CP{
CP{
CP{
=
u+
s+
+
*+
Cu
Cs
C
C*
C
= Hu u + Hs s + H + H* * + H (11.130)
= P} (u> s> > *> )
CP}
CP}
CP}
CP}
CP}
=
u+
s+
+
*+
Cu
Cs
C
C*
C
= Gu u + Gs s + G + G* * + G (11.131)
where the force system coe!cients are
Fu
=
Fs
=
F
=
F*
=
F
=
CI|
d2
d1
= Fi + Fu
Cu
y{
y{
Fi F i
Fu F u
CI|
+
=
Cs
y{
y{
CI|
= (Fi + Fu )
C
CI|
= Fu F*u + Fi F*i F*i F*u
C*
CI|
= Fi
C
(11.132)
(11.133)
(11.134)
(11.135)
(11.136)
11. F Vehicle Roll Dynamics
Hu
=
Hs
=
H
=
H*
=
H
Gu
=
Gs
=
G
=
G*
=
G
=
=
CP{
d2
d1
= FWi Fi + FWu Fu
Cu
y{
y{
CP{
1
1
= F i FWi Fi + F u FWu Fu f*
Cs
y{
y{
CP{
= FWi Fi FWu Fu
C
³
´
CP{
= FWi F*i Fi F*i n*
C*
¡
¢
FWu F*u Fu F*u
CP{
= FWi Fi
C
CP}
d2
d2
= 1 Fi 2 Fu
Cu
y{
y{
CP}
d2
d1
= F i Fi F u Fu
Cs
y{
y{
CP}
= (d1 Fi d2 Fu )
C
³
´
¡
¢
CP}
= d1 F*i Fi F*i + d2 F*u Fu F*u
C*
CP}
= d1 Fi
C
759
(11.137)
(11.138)
(11.139)
(11.140)
(11.141)
(11.142)
(11.143)
(11.144)
(11.145)
(11.146)
The force system coe!cients are slopes of the curves for lateral force I| ,
roll moment P{ , and yaw moment P} as a function of u, s, , *, and respectively.
11.4 F Two-wheel Rigid Vehicle Dynamics
We may combine the equations of motion (11.6)-(11.9) along with (11.97)(11.103) for a two-wheel rollable rigid vehicle, and express its motion by
the following set of equations:
yb { =
5
F
9 py{
9
yb |
9
9 sb : 9 H
9
:=9
7 *b 8 9 L{ y{
9 0
9
ub
7 G
L} y{
5
6
¢
1
1 ¡
(11.147)
I{ + u y| =
I{i + I{u + u y|
p
p
6
5 F 6
Fs F* Fu
y{ : 5
6 9 p :
p
p
p
: y|
:
9
:
H :
Hs H*
Hu
:9 s : 9
:
:9
:+9
: (11.148)
:7 * 8 9
L{
L{
L{
9 L{ :
:
:
9
0
1
0
0
:
u
7 G 8
8
Gs G*
Gu
L}
L}
L}
L}
760
11. F Vehicle Roll Dynamics
These sets of equations are very useful to analyze vehicle motions, especially
when they move at a constant forward speed.
Assuming yb { = 0, the rst equation (11.147) becomes an independent
algebraic equation, while the lateral velocity y| , roll rate s, roll angle *,
and yaw rate u of the vehicle will change according to the four coupled
equations (11.148). Considering the steer angle is the input command,
the other variables y| , s, *, and u may be assumed as the outputs. Hence,
we may consider Equation (11.148) as a linear control system, and write
the equations as
qb = [D] q + u
(11.149)
in which [D] is the coe!cient matrix, q is the vector of control variables,
and u is the vector of inputs.
5
F
9 py{
9
9 H
9
[D] = 9
9 L{ y{
9 0
9
7 G
L} y{
q=
u=
£
Fs
p
Hs
L{
1
Gs
L}
y|
F
p
6
Fu
y{ :
p
:
:
Hu
:
:
:
L{
:
0
:
8
Gu
L}
F*
p
H*
L{
0
G*
L}
s * u
H
L{
0
¤W
G
L}
¸W
(11.150)
(11.151)
(11.152)
Proof. The Newton-Euler equations of motion for a rigid vehicle in the
local coordinate frame E, attached to the vehicle at its mass center F, are
given in Equations (11.6)-(11.9) as
I{
I|
P}
P{
=
=
=
=
p yb { pu y|
p yb | + pu y{
L} $b } = L} ub
L{ $b { = L{ sb
(11.153)
(11.154)
(11.155)
(11.156)
The approximate force system applied on a two-wheel vehicle is found in
Equations (11.97)-(11.100)
I{
I|
P{
P}
I{i + I{u
I|i + I|u
FWi I|i + FWu I|u n* * f* *b
d1 I|i d2 I|u
(11.157)
(11.158)
(11.159)
(11.160)
11. F Vehicle Roll Dynamics
761
and in terms of tire characteristics, in (11.101)-(11.103). These equations
could be summarized in (11.129)-(11.131) as follows:
I|
P{
P}
= Fu u + Fs s + F + F* * + F = Hu u + Hs s + H + H* * + H = Gu u + Gs s + G + G* * + G (11.161)
(11.162)
(11.163)
Substituting (11.161)—(11.163) in (11.153)—(11.156) produces the following set of equations of motion:
p yb { pu y|
p yb | + pu y{
L{ sb
ub L}
=
=
=
=
I{
Fu u + Fs s + F + F* * + F Hu u + Hs s + H + H* * + H Gu u + Gs s + G + G* * + G (11.164)
(11.165)
(11.166)
(11.167)
Employing
=
y|
y{
(11.168)
we are able to transform these equations to a set of dierential equations
for y{ , y| , s, and u.
I{
+ u y|
μp
¶
Fu
F*
Fs
F y|
F
+
y{ u +
s+
*+
yb | =
p
p
p y{
p
p
¶
μ
1
y|
+ H* * + H Hu u + Hs s + H
sb =
L{
y{
¶
μ
y|
1
ub =
+ G* * + G Gu u + Gs s + G
L}
y{
yb {
=
(11.169)
(11.170)
(11.171)
(11.172)
The rst equation (11.169) depends on the yaw rate u and the lateral
velocity y| , which are the outputs of the other equations, (11.170)-(11.172).
However, if we assume the vehicle is moving with a constant forward speed,
y{ = frqvw=
(11.173)
then Equations (11.170)-(11.172) become independent with (11.169), and
may be treated independent of the rst equation.
Equations (11.170)-(11.172) may be considered as three coupled dierential equations describing the behavior of a dynamic system. The dynamic
system receives the steering angle as an input, and uses y{ as a parameter
762
11. F Vehicle Roll Dynamics
to generates four outputs: y| , s, *, and u.
5
F
6 9 py{
5
9
yb |
9
9 sb : 9 H
:=9
9
7 *b 8 9 L{ y{
9 0
9
ub
7 G
L} y{
Fs
p
Hs
L{
1
Gs
L}
F*
p
H*
L{
0
G*
L}
6
5 F 6
Fu
y{ : 5
6 9 p :
p
: y|
9
:
:
H :
Hu
:9 s : 9
9
:
:
:9
(11.174)
L{ :
: 7 * 8+9
L{
9
:
:
9
:
0
0
:
u
7 G 8
8
Gu
L}
L}
Equation (11.174) may be rearranged to show the input-output relationship.
qb = [D] q + u
(11.175)
The vector q is called the control variables vector, and u is called the inputs
vector. The matrix [D] is the control variable coe!cients matrix.
Example 463 Equations of motion based on kinematic angles.
The equations of motion (11.174) can be expressed based on the angles
, s, *, u, and , by employing (11.168).
Taking a derivative from Equation (11.168) for constant y{
yb |
b =
y{
(11.176)
and substituting in Equations (11.165) shows that we can transform the
b
equation for .
py{ b + pu y{ = Fu u + Fs s + F + F* * + F (11.177)
Therefore, the set of equations of motion can be expressed in terms of the
vehicle’s angular variables.
5
F
6 9 py{
5
9
b
9
9 sb : 9 H
:=9
9
7 *b 8 9 L{
9 0
9
ub
7 G
L}
Fs
py{
Hs
L{
1
Gs
L}
F*
py{
H*
L{
0
G*
L}
5
6
6
F
Fu
1 :5
6 9 py{ :
py{
9
: :
9
:
:
Hu
: 9 s : 9 H :
:+9
:9
:
: 7 * 8 9 L{ : L{
9
:
:
0
9 0 :
: u
7 G 8
8
Gu
L}
L}
(11.178)
11. F Vehicle Roll Dynamics
763
11.5 F Steady-State Motion
Turning the front steering of a two-wheel, rollable rigid vehicle at steadystate condition is governed by
I{
Fu u + F + F* * + F Hu u + H + H* * + H Gu u + G + G* * + G =
=
=
=
pu y|
pu y{
0
0
(11.179)
(11.180)
(11.181)
(11.182)
or equivalently, by the following equations:
I{
= p
y{ y|
U
¡
¢ 1
Fu y{ p y{2
+ F + F* * = F U
1
Hu y{ + H + H* * = H U
1
Gu y{ + G + G* * = G U
(11.183)
(11.184)
(11.185)
(11.186)
The rst equation determines the required forward force to keep y{ constant. The next three equations show the steady-state values of the output
variables, which are: path curvature ,
=
1
u
=
U
y{
(11.187)
vehicle slip angle , vehicle roll rate s, and vehicle roll angle * for a constant steering input at a constant forward speed y{ . The output-input
relationships are dened by the following responses:
1= Curvature response, V
1
]1
=
=
U
y{ ]0
(11.188)
]2
=
]0
(11.189)
u
]1
= y{ = V y{ = ]0
(11.190)
V =
2= Slip response, V
V =
3= Yaw rate response, Vu
Vu =
4= Centripetal acceleration response, Vd
Vd =
y{2 @U
y{ ]1
= y{2 = V y{2 =
]0
(11.191)
764
11. F Vehicle Roll Dynamics
5= Lateral velocity response, V|
V| =
y|
y{ ]2
= V y{ =
]0
(11.192)
*
]3
=
]0
(11.193)
6= Roll angle response, V*
V* =
]0
]1
]2
]3
= H (Gu F* Fu G* + py{ G* )
+H* (Fu G Gu F py{ G ) + Hu (F G* G F* ) (11.194)
= H (F* G y{ F G* ) H* (F G y{ F G )
+H (F G* G F* )
(11.195)
= H* (py{ G Fu G + Gu y{ F ) + Hu (F* G y{ F G* )
H (Gu F* Fu G* + py{ G* )
(11.196)
= H (py{ G Fu G + Gu y{ F ) + Hu (F G y{ F G )
H (Gu F Fu G + py{ G )
(11.197)
Proof. In steady-state conditions, all the variables are constant and hence,
their derivatives are zero. Therefore, the equations of motion (11.153)—
(11.156) reduce to
I{
I|
P{
P}
=
=
=
=
pu y|
pu y{
0
0
(11.198)
(11.199)
(11.200)
(11.201)
where the lateral force I| , roll moment P{ , and yaw moment P} from
(11.161)—(11.163) would be
I|
P{
P}
= Fu u + F + F* * + F = Hu u + H + H* * + H = Gu u + G + G* * + G (11.202)
(11.203)
(11.204)
Therefore, the equations describing the steady-state turning of a two-wheel
rigid vehicle are equal to
I{
Fu u + F + F* * + F Hu u + H + H* * + H Gu u + G + G* * + G =
=
=
=
pu y|
pu y{
0
0
(11.205)
(11.206)
(11.207)
(11.208)
11. F Vehicle Roll Dynamics
765
Equation (11.205) will be used to calculate the required traction force to
keep the motion steady. However, Equations (11.206)-(11.208) can be used
to determine the steady-state responses of the vehicle.
y{
+ F + F* * + F U
y{
+ H + H* * + H Hu
U
y{
Gu
+ G + G* * + G U
Fu
= p
y{
y{
U
(11.209)
= 0
(11.210)
= 0
(11.211)
At steady-state turning, the vehicle will move on a circle with radius U
at a speed y{ and angular velocity u, so
y{ U u
(11.212)
By substituting (11.212) in Equations (11.206)-(11.208) and employing the
curvature denition (11.187), we may write the equations in matrix form
5
65
6 5
6
F Fu y{ p y{2 F*
F
7 H
Hu y{
H* 8 7 8 = 7 H 8 (11.213)
*
G
Gu y{
G*
G
Solving the equations for , , and * and using
y|
y{
(11.214)
enables us to dene dierent output-input relationships as (11.188)-(11.193).
Example 464 Force system coe!cients for a car.
Consider a front steering, four-wheel car with the following characteristics:
FiO = FiU 28648 N@ rad
FuO = FuU 26356 N@ rad (11.215)
p = 917 kg
L{ = 300 kg m2
L} = 1128 kg m2
d1 = 0=91 m
d2 = 1=64 m
n* = 20000 N@ rad
f* = 1000 N s@ rad
(11.216)
F i
FWu
F*i
= 0=01
F u = 0=01
= 0=02
F*i = 0=01
= 2
F*u = 1
y{ = 40 m@ s
FWi = 0=02
F*u = 0
= 0=1 rad
(11.217)
(11.218)
11. F Vehicle Roll Dynamics
766
The sideslip coe!cients of an equivalent bicycle model are
= FiO + FiU = 57296 N@ rad
= FuO + FuU = 52712 N@ rad
Fi
Fu
(11.219)
(11.220)
The force system coe!cients are equal to the following if y{ is measured in
[ m@ s]:
Fu
F*
Hu
H*
Gu
G*
= 857=708
= 569=96
= 17=15416
= 19988=6008
= 4730=52532
= 521=2136
Fs = 27=502
F = 57296
F = 110008
(11.221)
Hs = 999=44996
H = 1145=92
H = 2200=16
(11.222)
Gs = 8=57708
G = 52139=36
G = 34308=32
(11.223)
The ]l parameters and the steady-state responses of the vehicle are as follows:
]0
]1
]2
]3
=
=
=
=
V
=
V
=
Vu
=
Vd
=
V*
=
0=3493155774 × 1014
0=1683862164 × 1016
0=1793365332 × 1015
0=1629203858 × 1014
1@U
=
= 1=205115283
= 5=133940334
u
= 48=20461133
y{2 @U
= 1928=184453
*
= 0=4663988563
(11.224)
(11.225)
Having the steady-state responses, we are able to calculate the steady-state
characteristic of the motion.
U = 8=298 m
= 0=5138 rad 29=415 deg
u = 4=82 rad@ s
(11.226)
(11.227)
(11.228)
11. F Vehicle Roll Dynamics
767
Example 465 F Camber thrust.
When a vehicle rolls, the wheels of almost all types of suspensions take
up a camber angle in the same sense as the roll. The wheel camber is always
less than the roll angle.
At steady-state conditions, camber at the front wheels increases the understeer characteristic of the vehicle, while camber at the rear increases the
oversteer characteristic of the vehicle. Most road vehicles are made such
that in a turn, the rear wheels remain upright and the front wheels camber. These vehicles have an increasing understeer behavior with roll and are
more stable.
Example 466 F Roll steer.
Positive roll steer means the wheel steers about the }-axis when the vehicle
rolls about the {-axis. So, when the vehicle turns to the right, a positive roll
steer wheel will steer to the left.
Positive roll steer at the front wheels increases the understeer characteristic of the vehicle, while roll steer at the rear increases the oversteer
characteristic of the vehicle. Most road vehicles’ suspension is made such
that in a turn the front wheels have positive roll steering. These vehicles
have an increasing understeer behavior with roll and are more stable.
11.6 F Time Response
The equations of motion must analytically or numerically be integrated to
analyze the time response of a vehicle and examine how the vehicle will
respond to a steering input. The equations of motion are a set of coupled
ordinary dierential equations as expressed here.
5
F
5
6 9 py{
9
yb |
9
9 sb : 9 H
9
:=9
7 *b 8 9 L{ y{
9 0
9
ub
7 G
L} y{
Fs
p
Hs
L{
1
Gs
L}
yb { =
1
I{ + u y|
p
F*
p
H*
L{
0
G*
L}
6
5 F 6
Fu
y{ : 5
6
p
:
9
p
: y|
:
9
:
9
H
Hu
:9 s : 9 :
:
:9
:
(w)
L{ :
:7 * 8 + 9
L{
:
9
:
:
9
0
0
:
u
7 G 8
8
Gu
L}
L}
(11.230)
(11.229)
768
11. F Vehicle Roll Dynamics
Their answers to a given time-dependent steer angle (w) are
y{
y|
s
*
u
=
=
=
=
=
y{ (w)
y| (w)
s (w)
* (w)
u (w)
(11.231)
(11.232)
(11.233)
(11.234)
(11.235)
Such a solution is called the time response or transient response of the
vehicle.
Assuming a constant forward velocity, the rst equation (11.229) simplies to
I{ = pu y|
(11.236)
and Equation (11.230) becomes independent from the rst one. The set of
Equations (11.230) can be written in the form
qb = [D] q + u
(11.237)
in which [D] is a constant coe!cient matrix, q is the vector of control
variables, and u is the vector of inputs.
To solve the inverse dynamics problem and nd the vehicle response, the
steering function (w) must be given.
Example 467 F Free dynamics and free response.
The response of a vehicle to zero steer angle (w) = 0 at constant speed
is called free response. The equation of motion under a free dynamics
is
qb = [D] q
(11.238)
To solve the equation, let us assume
5
d11 d12
9 d21 d22
[D] = 9
7 d31 d32
d41 d42
d13
d23
d33
d43
6
d14
d24 :
:
d34 8
d44
and therefore, the equations of motion are
6 5
6
65
5
d11 d12 d13 d14
y|
yb |
9 sb : 9 d21 d22 d23 d24 : 9 s :
: 9
:
:9
9
7 *b 8 = 7 d31 d32 d33 d34 8 7 * 8
ub
u
d41 d42 d43 d44
(11.239)
(11.240)
Because the equations are linear, the solutions are exponential functions
y|
s
*
u
=
=
=
=
D1 hw
D2 hw
D3 hw
D4 hw
(11.241)
(11.242)
(11.243)
(11.244)
11. F Vehicle Roll Dynamics
769
Substituting the solutions shows that the condition for functions (11.241)(11.244) to be the solution of the equations (11.240) is that the exponent is the eigenvalue of [D]. To nd , we may expand the determinant of the
above coe!cient matrix and nd the characteristic equation
det [D] = 0
(11.245)
Having the eigenvalues 1>2>3>4 provides the following general solution for
the free dynamics of the vehicle:
y|
s
*
u
=
=
=
=
D11 h1 w + D12 h2 w + D13 h3 w + D14 h4 w
D21 h1 w + D22 h2 w + D23 h3 w + D24 h4 w
D31 h1 w + D32 h2 w + D33 h3 w + D34 h4 w
D41 h1 w + D42 h2 w + D43 h3 w + D44 h4 w
(11.246)
(11.247)
(11.248)
(11.249)
The coe!cients Dlm must be found from initial conditions. The vehicle is
stable as long as the eigenvalues have negative real part.
As an example, consider a vehicle with the characteristics given in (11.215)(11.217), and the following steer angle and forward velocity.
y{
= 40 m@ s
= 0=1 rad
(11.250)
(11.251)
Substituting those values provide the following equations of motion for free
dynamics.
5
6 5
65
6
yb |
3
0=03
0=621 39=06
y|
9 sb : 9 0=18335 3=3315 66=63 0=05718 : 9 s :
9
: 9
:9
:
7 *b 8 = 7
87 * 8
0
1
0
0
ub
u
0=76038 0=0076 0=4621 4=194
(11.252)
The eigenvalues of the coe!cient matrix are
1
2
3
4
=
=
=
=
3=593983044 + 5=409888448l
3=593983044 5=409888448l
1=668194745 + 7=995670961l
1=668194745 7=995670961l
(11.253)
which shows the vehicle is stable because all of the eigenvalues have real
negative parts.
Substituting the eigenvalues in Equations (11.246)-(11.249) provides the
solution with unknown coe!cients. Let us examine the free dynamics behavior of the vehicle for a nonzero initial condition.
5
6 5
6
y| (0)
0
9 s (0) : 9 0=1 :
: 9
:
q0 = 9
(11.254)
7 * (0) 8 = 7 0 8
u (0)
0
770
11. F Vehicle Roll Dynamics
vy >m / s@
t [s]
FIGURE 11.7. Lateral velocity response of a vehicle free dynamics.
t [s]
p > rad / s @
2
FIGURE 11.8. Roll rate response of a vehicle free dynamics.
Figures 11.7 to 11.10 illustrate the time responses of the vehicle.
Example 468 F Response to a step input.
The response of a dynamic systems to a step input is a standard test to
examine the behavior of the systems. Step input in vehicle dynamics is a
sudden change in steer angle from zero to a nonzero constant value.
Consider a vehicle that is moving straight at a constant forward speed
y{ = 40 m@ s
(11.255)
with the characteristics given in (11.215)-(11.217) and a sudden change in
the steer angle to a constant value
½
0=2 rad 11=459 deg w A 0
(w) =
(11.256)
0
w0
11. F Vehicle Roll Dynamics
M > rad @
t [s]
FIGURE 11.9. Roll angle response of a vehicle free dynamics.
r > rad / s @
t [s]
FIGURE 11.10.
771
772
11. F Vehicle Roll Dynamics
t [s]
vy >m / s@
FIGURE 11.11. Lateral velocity response of a vehicle to a step steer angle.
The equations of motion for a non-zero initial conditions
5
6 5
6
y| (0)
0
9 s (0) : 9 0=1 :
: 9
:
q0 = 9
7 * (0) 8 = 7 0 8
u (0)
0
(11.257)
are
yb | + 6=91y| + 1=62s 10=01* + 16=99u
sb 5=6y| + 4=06s + 99=91* + =192u
*b s
ub 1=81y| + 0=3s 5=19* + 6=46u
=
=
=
=
62 (w)
69=33 (w)
0
32=11 (w)
(11.258)
(11.259)
(11.260)
(11.261)
Figures 11.11 to 11.14 depict the solutions.
Having y| (w), s(w), *(w), and u (w) are enough to calculate any other
kinematic variables as well as the required forward force I{ to maintain
the constant forward speed.
I{ = pu y|
(11.262)
Figure 11.15 illustrates the required forward force I{ (w).
Example 469 F Lane-change maneuver.
Passing and lane-change maneuvers are two other standard tests to examine a vehicle’s dynamic responses. Lane-change can be expressed by a
half-sine or a sine-squared function for steering input. Two examples of
11. F Vehicle Roll Dynamics
773
p > rad / s @
t [s]
2
FIGURE 11.12. Roll rate response of a vehicle to a step steer angle.
M > rad @
t [s]
FIGURE 11.13. Roll angle response of a vehicle to a step steer angle.
774
11. F Vehicle Roll Dynamics
r > rad / s @
t [s]
FIGURE 11.14. Yaw rate response of a vehicle to a step steer angle.
F >N @
t [s]
FIGURE 11.15. The required forward force I{ of a vehicle to a step steer angle
to maintain the speed constant.
11. F Vehicle Roll Dynamics
775
FIGURE 11.16. A lane-change maneuver.
such functions are
;
? 0 sin $w w1 ? w ? $ rad
(w) =
= 0
? w ? w1
$
;
? 0 sin2 $w w1 ? w ? $ rad
(w) =
= 0
? w ? w1
$
O
$ =
y{
(11.263)
(11.264)
(11.265)
where O is the moving length during the lane-change and y{ is the forward
speed of the vehicle. The path of a lane-change car would be similar to
Figure 11.16.
Let us examine a vehicle with the characteristics given in (11.215)-(11.217)
and a change in half-sine steering input (w).
;
? 0=2 sin O w 0 ? w ? y{
y{
O rad
(w) =
y{
= 0
?w?0
O
O = 100 m
y{ = 40 m@ s
(11.266)
(11.267)
The equations of motion for zero initial conditions are as given in (11.258)(11.261).
Figures 11.17 to 11.20 show the time responses of the vehicle for the
steering function (11.266).
776
11. F Vehicle Roll Dynamics
vy >m / s@
t [s]
FIGURE 11.17. Lateral velocity response for the steering function (11.266).
p > rad / s @
t [s]
FIGURE 11.18. Roll rate response for the steering function (11.266).
M > rad @
t [ s]
FIGURE 11.19. Roll angle response for the steering function (11.266).
11. F Vehicle Roll Dynamics
r > rad / s @
777
t [s]
FIGURE 11.20. Yaw rate response for the steering function (11.266).
vy >m / s@
t [s]
FIGURE 11.21. Lateral velocity response for the steering function (11.268).
Example 470 F Lane-change with a sine-square steer function.
A good driver should change the steer angle as smoothly as possible to
minimize undesired roll angle and roll uctuation. A sine-square steer function
;
? 0 sin2 $w w1 ? w ?
$ rad
(11.268)
(w) =
1
= 0
?w?
$
$
O
$ =
O = 100 m
y{ = 40 m@ s
(11.269)
y{
which is introduced in Equation (11.264), makes for smoother lane-change
steering. The responses of the vehicle in Example 469 to the steering (11.268)
are illustrated in Figures 11.21 to 11.24.
778
11. F Vehicle Roll Dynamics
p > rad / s @
t [s]
FIGURE 11.22. Roll rate response for the steering function (11.268).
M > rad @
t [s]
FIGURE 11.23. Roll angle response for the steering function (11.268).
r > rad / s @
t [s]
FIGURE 11.24. Yaw rate response for the steering function (11.268).
11. F Vehicle Roll Dynamics
779
Example 471 F Vehicle driving and classical feedback control.
Driving a car is similar to a problem in feedback control. The driver
compares desired direction, speed, and acceleration with actual direction,
speed, and acceleration. The driver uses the cars’s indicator and measuring
devices, as well as human sensors to sense the actual direction, speed, and
acceleration. When the actual data diers from the desired values, the driver
uses the control devices such as gas pedal, brake, steering, and gear selection
to improve the actual.
Example 472 F Hatchback, notchback, and station models of a platform.
It is common in vehicle manufacturing companies to install dierent bodies on the same chassis and platform to make dierent models easier. Consider a hatchback, notchback, and station models of a car that use the same
platform. To compare the dynamic behavior of the three models, we may
examine their response to a step steer function at a constant forward speed.
(w) = 0=1 rad
y{ = 40 m@ s
(11.270)
The common characteristics of the cars are
FiO
FuO
o
F i
FWu
F*i
n*
=
=
=
=
=
= FiU 26000 N@ rad
= FuU 32000 N@ rad
(11.271)
(11.272)
2=345 m
0=4
F u = 0=1
FWi = 0=4
0=4
F*i = 0=01
F*u = 0=01
3200
F*u = 0
26612 N m@ rad
f* = 1700 N m s@ rad
(11.273)
For the hatchback, we use
p = 838=7 kg
d1 = 0=859 m
L{ = 300 kg m2
d2 = 1=486 m
L} = 1391 kg m2
(11.274)
and for the notchback we have
p = 845=4 kg
d1 = 0=909 m
L{ = 350 kg m2
d2 = 1=436 m
L} = 1490 kg m2
(11.275)
and nally for the station model we use the following data.
p = 859 kg
d1 = 0=945 m
L{ = 400 kg m2
d2 = 1=4 m
L} = 1680 kg m2
(11.276)
780
11. F Vehicle Roll Dynamics
vy >m / s@
t [s]
Hatchback
Notchback
Station
FIGURE 11.25. Lateral velocity response for hatchback, notchback, and station
models of a car to a step steer angle.
p > rad / s @
t [s]
Station
Notchback
Hatchback
FIGURE 11.26. Roll rate response for hatchback, notchback, and station models
of a car to a step steer angle.
11. F Vehicle Roll Dynamics
M > rad @
781
t [s]
Hatchback
Notchback
Station
FIGURE 11.27. Roll angle response for hatchback, notchback, and station models
of a car to a step steer angle.
r > rad / s @
Station
Notchback
Hatchback
t [s]
FIGURE 11.28. Yaw rate response for hatchback, notchback, and station models
of a car to a step steer angle.
782
11. F Vehicle Roll Dynamics
Figure 11.25 compares the lateral velocity responses of the three models.
It shows that the steady-state value of the lateral velocity of the station
model is higher than the others.
Figure 11.26 compares the roll rate responses of the three models. The
hatchback model has the least longitudinal mass moment and hence shows
the fastest roll rate response. It will also reach zero steady-state value faster
than the other models.
Figure 11.27 compares the roll angle responses of the three models. The
station model has the largest roll angle because of the highest longitudinal
moment of inertia. It will also reach to the steady-state roll angle later than
the other models.
Figure 11.28 compares the yaw rate responses of the three models. The
station model has the highest yaw rate because of the highest vertical mass
moment. It will also reach to the steady-state yaw rate later than the other
models.
11.7 Summary
The best bicycle applied dynamic model for vehicle motion shows the yaw
and roll DOF as well as the { and | motions. Such a model is called the rigid
vehicle roll model and can be expressed by the following ve dierential
equations.
1
I{ + u y|
p
5
Fs
F
9 py{
5
6
p
9
yb |
9 H
Hs
9 sb :
9
9
:
9
7 *b 8 = 9 L{ y{ L{
9 0
1
9
ub
7 G
Gs
L} y{ L}
yb {
=
F*
p
H*
L{
0
G*
L}
6
5 F 6
Fu
y{ : 5
6 9 p :
p
: y|
:
9
:
H :
Hu
:9 s : 9
:
9
:9
:
L{ :
:7 * 8+ 9
L{
:
9
:
:
9
0
0
:
u
7 G 8
8
Gu
L}
L}
The vehicle receives the steering angle as the input to generate ve outputs y{ , y| , s, *, and u. However, keeping the forward speed constant,
y{ = fqvw=, and using it as a parameter, can uncouple the rst equation
from the others. As a result, a constant forward speed assumption is used
in many of the vehicle dynamic examinations of vehicles.
11. F Vehicle Roll Dynamics
11.8 Key Symbols
d> {
¨> a> vb
di zg
duzg
d1 = {1
d2 = {2
D> E> F
[D]
e1
e2
f
fi
fu
f* = gP{n @g*b
F
Fl = gI| @gl
F l = y{ g l @gs
F*l = g@g*l
F*l = gI| @g*l
FWl = gP{ @gI|
Fu = CI| @Cu
Fs = CI| @Cs
F = CI| @C
F* = CI| @C*
F = CI| @C
Gu = CP} @Cu
Gs = CP} @Cs
G = CP} @C
G* = CP} @C*
G = CP} @C
Hu = CP{ @Cu
Hs = CP{ @Cs
H = CP{ @C
H* = CP{ @C*
H = CP{ @C
I> F
I{
I{1
I{2
I{w
I}
I}1
I}2
I}3
acceleration
front wheel drive acceleration
rear wheel drive acceleration
distance of rst axle from mass center
distance of second axle from mass center
constant parameters
control variable coe!cient matrix
distance of left wheels from mass center
distance of right wheels from mass center
damping
damping of front suspension
damping of rear suspension
vehicle roll damping
mass center of vehicle
tire sideslip coe!cient
tire roll rate coe!cient
tire roll steering coe!cient
tire camber thrust coe!cient
tire torque coe!cient
vehicle yaw-rate lateral-force coe!cient
vehicle roll-rate lateral-force coe!cient
vehicle slip-angle lateral-force coe!cient
vehicle yaw-angle lateral-force coe!cient
vehicle steer-angle lateral-force coe!cient
vehicle yaw-rate yaw-moment coe!cient
vehicle roll-rate yaw-moment coe!cient
vehicle slip-angle yaw-moment coe!cient
vehicle yaw-angle yaw-moment coe!cient
vehicle steer-angle yaw-moment coe!cient
vehicle yaw-rate roll-moment coe!cient
vehicle roll-rate roll-moment coe!cient
vehicle slip-angle roll-moment coe!cient
vehicle yaw-angle roll-moment coe!cient
vehicle steer-angle roll-moment coe!cient
force
traction or brake force under a wheel
traction or brake force under front wheels
traction or brake force under rear wheels
horizontal force at hinge
normal force under a wheel
normal force under front wheels
normal force under rear wheels
normal force under trailer wheels
783
784
11. F Vehicle Roll Dynamics
I}w
j> g
k
K
L
L1 > L2 > L3
n
ni
nu
n* = gP{n @g*
o
O
L
p
P> M
s = *b
p
t = b
q
q0
u
u = #b
r
U
U
Ui
UK
Uu
Vd = y{2 @U@
Vu = u@
V = @
V = @
V* = *@
w
y> {>
b v
yf
y{
yf
z
}l
{> |> }
{l > |l > }l
{z > |z > }z
[> \> ]
]0 > ]1 > ]2 > ]3
normal force at hinge
gravitational acceleration
height of F
height
mass moment of inertia
principal mass moment of inertia
stiness
stiness of front suspension
stiness of rear suspension
vehicle roll stiness
wheel base
road wave length
angular momentum
car mass
moment
roll rate
translational momentum
pitch rate
control variable vector
initial condition vector
control input vector
yaw rate
position vector
tire radius
rotation matrix
front tire radius
radius of curvature
rear tire radius
lateral acceleration response
yaw rate response
sideslip response
curvature response
roll angle response
time
velocity
critical velocity
forward velocity
lateral velocity
track
de ection of axil number l
vehicle coordinate axes
coordinates of wheel number l in E
axes of a wheel coordinate frame
global coordinate axes
steady-state response parameters
11. F Vehicle Roll Dynamics
= y| @y{
i
l
u
0
z
1 > i
2 > u
d
*
= 1@U
!
!P
*
#
$
$> $
$>
b $
b
tire sideslip angle between vz and {z -axis
vehicle slip angle between v and {-axis
front tire slip angle
tire slip angle between v and {-axis
rear tire slip angle
vehicle steer angle
a constant steer angle value
tire steer angle between {z -axis and {-axis
steer angle of front wheels
steer angle of rear wheels
actual steer angle
roll-steer angle
atan2 (d> e)
pitch angle
path curvature
eigenvalue
friction coe!cient
slope angle
maximum slope angle
roll angle
yaw angle
angular frequency
angular velocity
angular acceleration
Subscriptions
g|q
i
i zg
l
O
P
u
U
uzg
vw
z
dynamic
front
front-wheel-drive
wheel number
left
maximum
rear
right
rear-wheel-drive
statics
wheel
785
786
11. F Vehicle Roll Dynamics
Exercises
1. F Force system coe!cients.
(a) Consider a front-wheel-steering car with the following characteristics and
FiO
p
d1
n*
y{
=
=
=
=
=
F i
FWu
F*i
FiU = 600 N@ deg
FuO = FuU = 560 N@ deg
1245 kg
L{ = 300 kg m2
L} = 1328 kg m2
110 cm
d2 = 132 cm
26612 N@ rad
f* = 1700 N s@ rad
30 m@ s
= 0=4
= 0=2
= 3200
F u = 0=1
F*i = 0=01
F*u = 300
FWi = 0=4
F*u = 0=01
(b) Determine the force system coe!cients Fu , Fs , F , F* , F , Hu ,
Hs , H , H* , H , Gu , Gs , G , G* , and G .
(c) Derive the equations of motion of the bicycle roll model of the
vehicle.
2. F Force system and two-wheel model of a car.
Consider a front-wheel-steering car with the following characteristics
FuU = FiO = FiU = 500 N@ deg
300 kg m2
L} = 1328 kg m2
110 cm
d2 = 132 cm
1205 kg
FuO
L{
d1
p
=
=
=
=
n*
y{
= 26612 N@ rad
= 30 m@ s
F i
FWu
= 0=4
= 0=2
F*i
F*i
f* = 1700 N s@ rad
F u = 0=1
= 0=01
= 3200
FWi = 0=4
F*u = 0=01
F*u = 300
11. F Vehicle Roll Dynamics
787
and determine the force system that applies on the two-wheel model
of the car.
I|
P{
P}
= Fu u + Fs s + F + F* * + F = Hu u + Hs s + H + H* * + H = Gu u + Gs s + G + G* * + G Then, write the equations of motion of the car as
I{
I|
P}
P{
=
=
=
=
p yb { pu y|
p yb | + pu y{
L} ub
L{ sb
3. F Equations of motion for a front-wheel-steering car.
Consider a front-wheel-steering car with the following characteristics
d1 = 110 cm
p = 1245 kg
y{ = 40 m@ s
FuO
n*
d2 = 132 cm
L} = 1328 kg m2
L{ = 300 kg m2
= FuU = FiO = FiU = 500 N@ deg
= 26612 N@ rad
f* = 1700 N s@ rad
F i
FWi
F*i
F*i
= 0=4
= 0=4
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
and develop the equations of motion
qb = [D] q + u
4. F Equations of motion in dierent variables.
Consider a car with the following characteristics
FuO
d1
p
y{
=
=
=
=
FuU = FiO = FiU = 500 N@ deg
100 cm
d2 = 120 cm
1000 kg
L} = 1008 kg m2
40 m@ s
788
11. F Vehicle Roll Dynamics
L{
n*
= 300 kg m2
= 26612 N@ rad
F i
FWi
F*i
F*i
f* = 1700 N s@ rad
= 0=4
= 0=4
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
and develop the equations of motion
b *>
b u),
b if the car is front-wheel steering.
(a) in terms of (yb { > yb | > s>
³
´
b s>
(b) in terms of yb { > >
b *>
b ub , if the car is front-wheel steering.
5. F Steady state response parameters.
Consider a car with the following characteristics
FiO = FiU = 500 N@ deg
p = 1245 kg
d1 = 110 cm
y{ = 40 m@ s
L{
n*
= 300 kg m2
= 26612 N@ rad
F i
FWi
F*i
F*i
= 0=4
= 0=4
FuO = FuU = 520 N@ deg
L} = 1328 kg m2
d2 = 132 cm
f* = 1700 N s@ rad
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
and determine the steady-state curvature response V , sideslip response V , yaw rate response, Vu , roll angle response, V* , and lateral
acceleration response Vd .
6. F Steady state motion parameters.
Consider a car with the following characteristics
FiO = FiU = 600 N@ deg
FuO = FuU = 550 N@ deg
11. F Vehicle Roll Dynamics
p = 1245 kg
d1
y{
L} = 1128 kg m2
= 120 cm
= 20 m@ s
n* = 26612 N@ rad
F i
FWi
F*i
F*i
= 0=4
= 0=4
789
L{ = 300 kg m2
d2 = 138 cm
= 3 deg
f* = 1700 N s@ rad
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
and determine the steady state values of u, U, , *, and y{2 @U.
7. F Mass moment and steady state parameters.
Consider a car that is made up of a uniform solid box with dimensions
260 cm × 140 cm × 40 cm. If the density of the box is = 1000 kg@ m3 ,
and the other characteristics are
FiO
d1
= FiU = 600 N@ deg
o
= d2 =
2
n* = 26612 N@ rad
F i
FWi
F*i
F*i
= 0=4
= 0=4
FuO = FuU = 550 N@ deg
f* = 1700 N s@ rad
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
then,
(a) determine p, L} .
(b) determine the steady-state responses V , V , Vu , and Vd as functions of y{ .
(c) determine the velocity y{ at which the car has a radius of turning
equal to
U = 35 m
when
= 4 deg =
790
11. F Vehicle Roll Dynamics
(d) determine the steady state parameters u, U, , *, and y{2 @U at
that speed.
(e) set the speed of the car at
y{ = 20 m@ s
and plot the steady-state responses V , V , Vu , and Vd for variable U.
8. F Stability factor and understeer behavior.
Dene a stability factor N for the vehicle roll model.
9. F Stability factor and mass of the car.
Find d1 and d2 in terms of I}1 , I}2 , and pj to rewrite the stability
factor N to see the eect of a car’s mass distribution.
10. F Stability factor and car behavior.
Examine the stability factor of a car with the parameters
FiO = FiU = 500 N@ deg
L} = 1328 kg m2
d2 = 132 cm
p = 1245 kg
d1 = 110 cm
y{ = 30 m@ s
n* = 26612 N@ rad
F i
FWi
F*i
F*i
FuO = FuU = 460 N@ deg
= 0=4
= 0=4
L{ = 300 kg m2
f* = 1700 N s@ rad
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
and determine if the car is understeer, neutral, or oversteer?
11. F Critical speed of a car.
Consider a car with the characteristics
FiO
FuO
p = 1245 kg
d1 = 118 cm
= FiU = 700 N@ deg
= FuU = 520 N@ deg
L} = 1328 kg m2
d2 = 122 cm
L{ = 300 kg m2
11. F Vehicle Roll Dynamics
n* = 26612 N@ rad
F i
FWi
F*i
F*i
f* = 1700 N s@ rad
= 0=4
= 0=4
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
(a) Dene a critical speed for oversteer condition.
(b) Determine if the car is understeer, neutral, or oversteer?
12. F Step input response at dierent speed.
Consider a car with the characteristics
FiO = FiU = 600 N@ deg
p = 1245 kg
d1 = 110 cm
FuO = FuU = 750 N@ deg
L} = 1328 kg m2
d2 = 132 cm
f* = 1700 N s@ rad
n* = 26612 N@ rad
F i
FWi
F*i
F*i
= 0=4
= 0=4
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
and a step input
(w) =
½
5 deg
0
wA0
w0
Determine the time response of the car at
(a) y{ = 10 m@ s.
(b) y{ = 20 m@ s.
(c) y{ = 30 m@ s.
(d) y{ = 40 m@ s.
L{ = 300 kg m2
791
792
11. F Vehicle Roll Dynamics
13. F Step input response for dierent steer angle.
Consider a car with the characteristics
FiO
p
d1
y{
n*
F i
F*i
=
=
=
=
=
=
=
FiU = 600 N@ deg
FuO = FuU = 750 N@ deg
1245 kg
L} = 1328 kg m2
L{ = 300 kg m2
110 cm
d2 = 132 cm
20 m@ s
26612 N@ rad
f* = 1700 N s@ rad
0=4
F u = 0=1
FWi = 0=4
FWu = 0=2
0=01
F*u = 0=01
F*i = 3200
F*u = 300
Determine the time response of the car to a step input
½
wA0
(w) =
0 w0
when
(a) = 3 deg.
(b) = 5 deg.
(c) = 10 deg.
14. F Eigenvalues and free response.
Consider a car with the characteristics
FiO = FiU = 600 N@ deg
p = 1245 kg
d1 = 110 cm
L} = 1328 kg m2
d2 = 132 cm
n* = 26612 N@ rad
F i
FWi
F*i
F*i
FuO = FuU = 750 N@ deg
= 0=4
= 0=4
L{ = 300 kg m2
y{ = 20 m@ s
f* = 1700 N s@ rad
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
(a) Determine the eigenvalues of the coe!cient matrix [D] and nd
out if the car is stable at zero steer angle.
(b) In either case, determine the weight distribution ratio, d1 @d2 ,
such that the car is neutral stable.
11. F Vehicle Roll Dynamics
793
(c) Recommend a condition for the weight distribution ratio, d1 @d2 ,
such that the car is stable.
15. F Time response to dierent steer functions.
Consider a car with the characteristics
FiO = FiU = 600 N@ deg
FuO = FuU = 750 N@ deg
L} = 1328 kg m2
d2 = 132 cm
p = 1245 kg
d1 = 110 cm
f* = 1700 N s@ rad
n* = 26612 N@ rad
= 0=4
= 0=4
F i
FWi
L{ = 300 kg m2
y{ = 20 m@ s
F u = 0=1
FWu = 0=2
= 0=01
F*u = 0=01
= 3200
F*u = 300
F*i
F*i
and a step input
(w) =
½
5 deg
0
wA0
w0
Determine the time response of the car to
(a) (w) = sin 0=1w for 0 ? w ? 10 and (w) = 0 for w 0 and
w 10.
(b) (w) = sin 0=5w for 0 ? w ? 2 and (w) = 0 for w 0 and w 2.
(c) (w) = sin w for 0 ? w ? and (w) = 0 for w 0 and w .
16. F Research exercise 1.
Consider a bicycle model of a car such that tires are always upright
and remain perpendicular to the road surface. Develop the equations
of motion for the roll model of the car.
17. F Research exercise 2.
Employ the tire frame W at the tireprint, wheel frame Z at the wheel
center, and wheel-body frame F that is at the point corresponding
to the wheel center at zero and zero , and remains parallel to
the vehicle frame E. Develop the Z , F, and E expressions of the
generated forces at the tireprint in W frame, and develop a better set
of equations for the roll model of a car.
18. F Research exercise 3.
Use the caster theory to nd the associated camber angle for a steer
angle , when the caster angle * and lean angle are given. Then
provide a better set of equations for the roll model of vehicles.
Part IV
Vehicle Vibration
12
Applied Vibrations
Vibration is an inevitable phenomena in vehicle dynamics. In this chapter,
we review the principles of vibrations, analysis methods, and their applications, along with the frequency and time responses of vibrating systems.
Special attention is devoted to frequency response analysis, because most
of the optimization methods for vehicle suspensions and vehicle vibrating
components are based on frequency responses.
12.1 Mechanical Vibration Elements
Mechanical vibrations is a result of continuous transformation of kinetic
energy N to potential energy Y , back and forth. When the potential energy
is at its maximum, the kinetic energy is zero and vice versa. When a periodic
uctuations of kinetic energy appears as a periodic motion of a massive
body, we call the energy transformation mechanical vibrations.
.
x
x
v
m
c
k
.
y
y
Mass
Spring
Damper
FIGURE 12.1. A mass p, spring n, and damper f.
The mechanical element that stores kinetic energy is called mass, and the
mechanical element that stores potential energy, is called spring. If the total
value of mechanical energy H = N + Y decreases during a vibration, there
is a mechanical element that dissipates energy. The dissipative element
is called damper. A mass, spring, and damper are usually illustrated by
symbols in Figure 12.1.
The amount of stored kinetic energy in a mass p is proportional to the
square of its velocity, y2 . The velocity y {b may be a function of position
R.N. Jazar, Vehicle Dynamics: Theory and Application,
DOI 10.1007/978-1-4614-8544-5_12, © Springer Science+Business Media New York 2014
797
798
12. Applied Vibrations
and time.
1
(12.1)
py 2
2
The required force ip to move a mass p is proportional to its acceleration
d{
¨.
ip = pd
(12.2)
N=
A spring is characterized by its stiness n. A force in to generate a
de ection in spring is proportional to relative displacement of its ends. The
stiness n may be a function of position and time.
in = n} = n({ |)
(12.3)
If n is time independent then, the value of stored potential energy in the
spring is equal to the work done by the spring force in during the spring
de ection.
Z
Z
Y = in g} = n} g}
(12.4)
The spring potential energy is then a function of the spring’s length change.
If the stiness of a spring, n, is not a function of displacements, it is called
linear spring. Then, its potential energy is
Y =
1 2
n}
2
(12.5)
Damping of a damper is measured by the value of mechanical energy loss
in one cycle. Equivalently, a damper may be dened by the required force
if to generate a motion in the damper. If if is proportional to the relative
velocity of its ends, it is a linear damper with a constant damping f.
if = f }b = f({b |)
b
(12.6)
Such a damping is also called viscous damping.
A vibrating motion { is characterized by period W , which is the required time for one complete cycle of vibration, starting from and ending
at ({b = 0> {
¨ ? 0). Frequency i is the number of cycles in one W .
i=
1
W
(12.7)
In theoretical vibrations, we usually work with angular frequency $ [ rad@ s],
and in applied vibrations we use cyclic frequency i [ Hz].
$ = 2i
(12.8)
When there is no applied external force or excitation on a vibrating
system, any possible motion of the system is called free vibration. A free
vibrating system will oscillate if any one of the kinematic states {, {,
b or {
¨ is
12. Applied Vibrations
x
k1
x1+x2+x3
k1
x2+x3
k2
799
x
k2
k1x1=k1x2=k3x3
x3
k3
k3
Equilibrium
In motion
Free body diagram
(a)
(b)
(c)
FIGURE 12.2. Three serial springs
not zero. If we apply any external excitation, a possible motion of the system is called forced vibration. There are four types of applied excitations:
harmonic, periodic, transient, random. The harmonic and transient excitations are more applied and more predictable than the periodic and random
types. When the excitation is a sinusoidal function of time, it is called harmonic excitation and when the excitation disappears after a while or stays
steady, it is transient excitation. A random excitation has no short term
pattern, however, we may dene some long term averages to characterize
a random excitation.
We use i to indicate a harmonically-variable force with amplitude I to
be consistent with a harmonic motion { with amplitude [. We also use i
for cyclic frequency, however, i is a force unless it is indicated that it is a
frequency.
Example 473 Serial springs and dampers.
Serial springs have the same force, and a resultant displacement equal
to the sum of individual displacements. Figure 12.2 illustrates three serial
springs attached to a massless plate and the ground.
The equilibrium position of the springs is the un-stretched conguration
in Figure 12.2(d). Applying a displacement { as shown in Figure 12.2(e)
generates the free body diagram as shown in Figure 12.2(f). Each spring
makes a force il = nl {l where {l is the length change in spring number
l. The total displacement
of the springs, {, is the sum of their individual
P
displacements, { = {l .
{ = {1 + {2 + {3
(12.9)
We may substitute a set of serial springs with only one equivalent spring,
having a stiness nht , that produces the same displacement { under the
800
12. Applied Vibrations
same force in .
in = n1 {1 = n2 {2 = n3 {3 = nht {
(12.10)
Substituting (12.10) in (12.9)
iv
iv
iv
iv
=
+
+
nht
n1 n2
n3
(12.11)
shows that the inverse of the equivalent
Pstiness of the serial springs, 1@nht ,
is the sum of their inverse stiness,
1@nl .
1
1
1
1
=
+
+
nht
n1 n2
n3
(12.12)
We assume that velocity {b has no eect on the force of a linear spring.
Serial dampers have the same force,
P if , and a resultant velocity {b equal
to the sum of individual velocities,
{b l . We may substitute a set of serial dampers with only one equivalent damping fht that produces the same
velocity {b under the same force if . For three parallel dampers, the velocity
and force balance
{b = {b 1 + {b 2 + {b 3
if = f1 {b = f2 {b = f3 {b = fht {b
(12.13)
(12.14)
shows that the equivalent damping is
1
1
1
1
=
+
+
fht
f1 f2 f3
(12.15)
We assume that displacement { has no eect on the force of a linear
damper.
Example 474 Parallel springs and dampers.
Parallel springs have the same displacement
{, with a resultant force,
P
il . Figure 12.3 illustrates three
in , equal to sum of the individual forces
parallel springs between a massless plate and the ground.
The equilibrium position of the springs is the un-stretched conguration
shown in Figure 12.3(d). Applying a displacement { to all the springs in
Figure 12.3(e) generates the free body diagram shown in Figure 12.3(f).
Each spring makes a force n{ opposite to the direction of displacement.
The resultant force of the springs is
in = n1 { n2 { n3 {
(12.16)
We may substitute parallel springs with only one equivalent stiness nht
that produces the same force in under the same displacement.
in = nht {
(12.17)
12. Applied Vibrations
x
x
k1 x
k1
k2
801
k3
k1
k2
k3
k3 x
k2 x
Equilibrium
In motion
Free body diagram
(a)
(b)
(c)
FIGURE 12.3. Three parallel springs.
Therefore, the equivalent stiness of the parallel springs is sum of their
stiness.
(12.18)
nht = n1 + n2 + n3
Parallel dampers have the same speed {,
b and a resultant force if equal to
the sum of individual forces. We may substitute parallel dampers with only
one equivalent damping fht that produces the same force if under the same
velocity. Consider three parallel dampers such as is shown in Figure 12.4.
Their force balance and equivalent damping would be
if
if
fht
= f1 {b f2 {b f3 {b
= fht {b
= f1 + f2 + f3
(12.19)
(12.20)
(12.21)
Example 475 Flexible frame.
Figure 12.5 depicts a mass p hanging from a frame. The frame is exible,
so it can be modeled by some springs attached to each other, as shown in
Figure 12.6(d). If we assume that each beam is simply supported, then the
equivalent stiness for a lateral de ection of each beam at their midspan is
n5 =
48H5 L5
o53
n4 =
48H4 L4
o43
n3 =
48H3 L3
o33
(12.22)
When the mass is vibrating, the elongation of each spring would be similar
to Figure 12.6(e). Assume we separate the mass and springs, and then apply
a force i at the end of spring n1 as shown in Figure 12.6(f). Because the
802
12. Applied Vibrations
.
x
.
x x
c1
c3
c2
x
c1
c2
.
c1 x
c3
.
c2 x
.
c3 x
Equilibrium
In motion
Free body diagram
(a)
(b)
(c)
FIGURE 12.4. Three parallel dampers
k4
k3
k5
k2
k1
m
FIGURE 12.5. A mass p hanging from a exible frame.
12. Applied Vibrations
k4
k5
k4
k5
G4
G5
k3
k3
G45 G3
k2
m
f
k3
G45
k2
f
m
(b)
(a)
k2
k1
k1
x
x
k5
f
G45 G3 G2
k1
k4
803
(c)
FIGURE 12.6. Equivalent springs model for the exible frame.
springs n1 , n2 , and n3 have the same force, and their resultant displacement
is the sum of individual displacements, they are in series.
The springs n4 and n5 are neither in series nor parallel. To nd their
equivalent, let us assume that springs n4 and n5 support a force equal to
i @2. Therefore,
i
i
5 =
(12.23)
4 =
2n4
2n5
and the displacement at midspan of the lateral beam is
45 =
4 + 5
2
(12.24)
Assuming
i
n45
we can dene an equivalent stiness n45 for n4 and n5 as
¶
¶
μ
μ
1
1
1
1
1 1
1
=
+
+
=
n45
2 2n4 2n5
4 n4
n5
45 =
(12.25)
(12.26)
Now the equivalent spring n45 is in series with the series of n1 , n2 , and n3 .
Hence the overall equivalent spring nht is
1
nht
=
=
1
1
1
1
+
+
+
n1 n2 n3
n45
1
1
1
1
1
+
+
+
+
n1 n2 n3
4n4 4n5
(12.27)
804
12. Applied Vibrations
.
x
.
x
m
x
x
m
k
dz
k
.
z
z
(a)
(b)
(c)
FIGURE 12.7. A vibrating system with a massive spring.
Example 476 F Massive spring.
In modeling of vibrating systems we ignore the mass of springs and
dampers. This assumption is valid as long as the mass of springs and
dampers are much smaller than the mass of the body they support. However, when the mass of spring pv or damper pg is comparable with the
mass of body p, we may dene a new system with an equivalent mass pht
1
pht = p + pv
3
(12.28)
which is supported by massless spring and damper.
Consider a vibrating system with a massive spring as shown in Figure
12.7(d). The spring has a mass pv , and a length o, when the system is at
equilibrium. The mass of spring is uniformly distributed along its length,
so, we may dene a length density as
=
p
o
(12.29)
To show (12.28), we seek for a system with a mass pht and a massless
spring, which can keep the same amount of kinetic energy as the original
system. Figure 12.7(e) illustrates the system when the mass p is at position
{ and has a velocity {.
b The spring is between the mass and the ground. So,
the base of spring has no velocity, while the other end has the same velocity
as p. Let us dene a coordinate } that goes from the grounded base of the
spring to the end point. An element of spring at } has a length g} and a
mass gp.
gp = g}
(12.30)
12. Applied Vibrations
805
Assuming a linear velocity distribution of the elements of spring, as shown
in Figure 12.7(f), we nd the velocity }b of gp as
}b =
}
{b
o
(12.31)
The kinetic energy of the system is a summation of kinetic energy of the
mass p and kinetic energy of the spring.
Z
Z
¢ 1
1
1 o¡
1 o ³ } ´2
N =
gp }b 2 = p{b 2 +
p{b 2 +
{b g}
2
2 0
2
2 0
o
μ ¶
Z
1
1 2 o 2
1
1 2 1 3
2
2
=
} g} = p{b + 2 {b
p{b + 2 {b
o
2
2o
2
2o
3
μ
¶0
μ
¶
1
1 1
1
1
p{b 2 +
o {b 2 =
p + pv {b 2
=
2
2 3
2
3
1
pht {b 2
=
(12.32)
2
Therefore, an equivalent system should have a massless spring and a mass
pht = p + 13 pv to keep the same amount of kinetic energy.
12.2 Newton’s Method and Vibrations
Every vibrating system can be modeled as a combination of masses pl ,
dampers fl , and springs nl . Such a model is called a discrete or lumped
model of the system. A one degree of freedom (DOF) vibrating system,
with the following equation of motion, is shown in Figure 12.8.
pd = fy n{ + i ({> y> w)
(12.33)
To apply Newton’s method and nd the equations of motion, we assume
all the masses pl are out of the equilibrium at positions {l with velocities
{b l . Such a situation is shown in Figure 12.8(e) for a one DOF system. The
free body diagram that is shown in Figure 12.8(f) illustrates the applied
forces and then, Newton’s equation (9.11)
J
J ¡
¢
gJ
g
p Jv
p=
gw
gw
J
F=
(12.34)
generates the equations of motion.
The equilibrium position of a vibrating system is where the potential
energy of the system, Y , is extremum.
CY
=0
C{
(12.35)
806
12. Applied Vibrations
f
f
f
m
m
x
x
m
kx
k
c
k
.
cx
c
Equilibrium
In motion
Free body diagram
(a)
(b)
(c)
FIGURE 12.8. A one DOF vibrating systems.
We usually set Y = 0 at the equilibrium position. Linear systems with
constant stiness have only one equilibrium or innity equilibria, while
nonlinear systems may have multiple equilibria. An equilibrium is stable if
C2Y
A0
C{2
and is unstable if
(12.36)
C2Y
?0
(12.37)
C{2
The arrangement and the number of employed elements can be used to
classify discrete vibrating systems. The number of masses, times the DOF
of each mass, makes the total DOF of the vibrating system q. The nal
set of equations would be q second-order dierential equations to be solved
for q generalized coordinates. When each mass has one DOF, then the
system’s DOF is equal to the number of masses. The DOF may also be
dened as the minimum number of independent coordinates that denes
the conguration of a system.
A one, two, and three DOF model for analysis of vertical vibrations of
a vehicle are shown in Figure 12.9(d)-(f). The system in Figure 12.9(d) is
called the quarter car model, which pv represents a quarter mass of the
body, and px represents a wheel. The parameters nx and fx are models
for tire stiness and damping. Similarly, nv and fx are models for the main
suspension of the vehicle. Figure 12.9(f) is called the 1@8 car model which
does not show the wheel of the car, and Figure 12.9(e) is a quarter car with
a driver pg and the driver’s seat modeled as ng and fg .
12. Applied Vibrations
xd
md
cd
kd
xs
xs
ms
ks
ms
cs
ks
cs
xs
ms
xu
xu
mu
mu
ks
cu y
ku
(a)
807
cs
cu y
ku
(b)
y
(c)
FIGURE 12.9. Two, three, and one DOF models for vertical vibrations of vehicles.
Example 477 1@8 car model.
Figure 12.9(f) and 12.10(d) show the simplest model for vertical vibrations of a vehicle. This model is sometimes called 1@8 car model. The mass
pv represents one quarter of the car’s body, which is mounted on a suspension made of a spring nv and a damper fv . When pv is vibrating at
a position such as in Figure 12.10(e), its free body diagram is as Figure
12.10(f) shows.
Applying Newton’s method, the equation of motion would be
¨ = nv ({v |) fv ({b v |)
b
pv {
(12.38)
which can be simplied to the following equation, when we separate the
input | and output { variables.
¨ + fv {b v + nv {v = nv | + fv |b
pv {
(12.39)
Example 478 Equivalent mass and spring.
Figure 12.11(d) illustrates a pendulum made by a point mass p attached
to a massless bar with length o. The coordinate shows the angular position
of the bar. The equation of motion for the pendulum can be found by using
the Euler equation and employing the free-body-diagram shown in Figure
12.11(e).
(12.40)
po2 ¨ = pjo sin 808
12. Applied Vibrations
ms
xs
ms
xs
ms
ks
ks
. .
ks(x-y) cs(x-y)
cs
cs
y
y
(b)
(a)
(c)
FIGURE 12.10. A 1@8 car model and its free body diagram.
A
T
y
T
l
l
g
Ft
m
mg
(a)
(b)
(c)
FIGURE 12.11. Equivalent mass-spring vibrator for a pendulum.
12. Applied Vibrations
ms
xs
ms
xs
ms
ks
ks
cs
s
mu
s
u
mu
xu
mu
cu
ku
c
cu y
ku
. .
ks(xs-xu) cs(xs-xu)
ks(xs-xu) c (x. -x. )
cs
xu
809
. .
cu(xu-y)
y
(a)
(b)
(c)
FIGURE 12.12. A 1@4 car model and its free body diagram.
Simplifying the equation of motion and assuming a very small swing angle
yields:
o¨ + j = 0
(12.41)
This equation is equivalent to an equation of motion for a mass-spring
system made by a mass p o, and a spring with stiness n j. The
displacement of the mass would be { . Figure 12.11(f) depicts such an
equivalent mass-spring system.
Example 479 Force proportionality.
The equation of motion for a vibrating system is a balance between four
dierent forces. A force proportional to displacement, n{, a force proportional to velocity, fy, a force proportional to acceleration, pd, and an
applied external force i ({> y> w), which can be a function of displacement,
velocity, and time. Based on Newton’s method, the force proportional to
acceleration, pd, is always equal to the sum of all the other forces.
pd = fy n{ + i ({> y> w)
(12.42)
Example 480 A two-DOF base excited system.
Figure 12.12(d)-(f) illustrate the equilibrium, motion, and free body diagram of the two-DOF system shown in 12.9(d). The free body diagram is
plotted based on the assumption
{v A {x A |
(12.43)
810
12. Applied Vibrations
y
m
k
B
k
mg
kaT
l
B
kaT
T
T
a
Fx
A
A
Fy
(a)
(b)
FIGURE 12.13. An inverted pendulum with a tip mass p and two supportive
springs.
Applying Newton’s method provides two equations of motion as follows
pv {
¨v
px {
¨x
= nv ({v {x ) fv ({b v {b x )
= nv ({v {x ) + fv ({b v {b x )
nx ({x |) fx ({b x |)
b
(12.44)
(12.45)
The assumption (12.43) is not necessarily fullled. We can nd the same
Equations (12.44) and (12.45) using any other assumption, such as {v ?
{x A |, {v A {x ? |, or {v ? {x ? |. However, having an assumption
helps to make a consistent free body diagram.
We usually rearrange the equations of motion for a linear system in a
matrix form to take advantage of matrix calculus.
[P ] xb + [f] xb + [n] x = F
(12.46)
Rearrangement of Equations (12.44) and (12.45) results in the following
set of equations:
¸
¸ ¸
¸
{
¨v
fv
{b v
fv
pv 0
+
+
{
¨x
fv fv + fx
{b x
0 px
¸
¸ ¸
nv
{v
0
nv
=
(12.47)
nx | + fx |b
nv nv + nx
{x
Example 481 F Inverted pendulum.
Figure 12.13(d) illustrates an inverted pendulum with a tip mass p and
a length o. The pendulum is supported by two identical springs attached
12. Applied Vibrations
811
to point E at a distance d ? o from the pivot D. A free body diagram of
the pendulum is shown in Figure 12.13(e). The equation of motion may be
found by taking a moment about D.
X
PD = LD ¨
(12.48)
pj (o sin ) 2nd (d cos ) = po2 ¨
(12.49)
To derive Equation (12.49) we assumed that the springs are long enough
to remain almost straight when the pendulum oscillates. Rearrangement
and assuming a very small shows that the nonlinear equation of motion
(12.49) can be approximated by
¢
¡
(12.50)
po2 ¨ + pjo 2nd2 = 0
which is equivalent to a linear oscillator
pht ¨ + nht = 0
(12.51)
with an equivalent mass pht and and equivalent nht .
pht = po2
nht = pjo 2nd2
(12.52)
The potential energy of the inverted pendulum can be expressed as
Y = pjo (1 cos ) + nd2 2
(12.53)
which has a zero value at = 0. The potential energy Y is approximately
equal to the following equation if is very small
1
Y pjo2 + nd2 2
2
(12.54)
because
¡ ¢
1
cos 1 2 + R 4
(12.55)
2
To nd the equilibrium positions of the system, we should solve the following
equation for any possible .
CY
= 2pjo + 2nd2 = 0
C{
(12.56)
The solution of the equation is
=0
(12.57)
that shows the upright vertical position is the only equilibrium of the inverted pendulum as long as is very small. However, if
pjo = nd2
(12.58)
812
12. Applied Vibrations
then any around = 0 can be an equilibrium position and hence, the
inverted pendulum would have innity equilibria.
A second derivative of the potential energy
C2Y
= 2pjo + 2nd2
C{2
(12.59)
indicates that the equilibrium position = 0 is stable if
nd2 A pjo
(12.60)
A stable equilibrium pulls the system back, if it deviates from the equilibrium, while an unstable equilibrium repels the system. Vibration happens
when the equilibrium is stable.
12.3 Frequency Response of Vibrating Systems
Frequency response is the steady-state solution of equations of motion,
when the system is harmonically excited. Steady-state response refers to a
constant amplitude oscillation at a given frequency, after the eect of initial
conditions dies out. A harmonic excitation is any combination of sinusoidal
functions that applies on a vibrating system. If the system is linear, then
a harmonic excitation generates a harmonic response with a frequencydependent amplitude. In frequency response analysis, we are looking for
the steady-state amplitude of oscillation as a function of the excitation
frequency.
A vast amount of vibrating systems in vehicle dynamics can be modeled
by a one DOF system. Consider a one DOF mass-spring-damper system.
There are four types of one DOF harmonically excited systems:
1= base excitation,
2= eccentric excitation,
3= eccentric base excitation,
4= forced excitation.
These four systems are shown in Figure 12.14 symbolically.
Base excitation is the most practical model for vertical vibration of vehicles. Eccentric excitation is a model for every type of rotary motor on
a suspension, such as engine on engine mounts. Eccentric base excitation
is a model for vibration of any equipment mounted on an engine. Forced
excitation has almost no practical application, however, it is the simplest
model for forced vibrations, with good pedagogical use.
For simplicity, we rst examine the frequency response of a harmonically
forced vibrating system.
12. Applied Vibrations
x
m
k
c
Z
f
me
e
x
c
x
m
m-me
k
k
c
k
me
(d)
(b)
x
m
y
(a)
813
c
mb-me
Z
e
(c)
y
FIGURE 12.14. The four practical types of one DOF harmonically excited systems: d=base excitation, e=eccentric excitation, f=eccentric base excitation, g=forced
excitation.
f
F sin Zt
x
m
k
c
FIGURE 12.15. A harmonically forced excitated, single-DOF system.
12.3.1 Forced Excitation
Figure 12.15 illustrates a one DOF vibrating mass p supported by spring n
and a damper f. The absolute motion of p with respect to its equilibrium
position is measured by the coordinate {. A sinusoidal excitation force
i = I sin $w
(12.61)
is applied on p and makes the system vibrate.
The equation of motion for the sys