Position Analysis: Review (Chapter 2)
Objective: Given the geometry of a
mechanism and the input motion, find the
output motion
 Graphical approach
 Algebraic position analysis
Example of graphical analysis of linkages,
four bar linkage. Given a-d and 2 find 3 and
4
Open
b
3
c
a
2
4
d
Crossed
Modified Problem:
Given: a-d and θ3
Find: position of 4-bar linkage
B
A
b
θ3
c
a
O2
d
2
Observation: Joint B is on a circle obtained by shifting the
path of the tip of the crank by vector O2O2’
B
b
A
b
a
O2’
θ3
d
O2
b
3
Solution:
 Draw path of tip of crank, A. This is a circle with radius a and
center the pivot of the crank.
 Swift the circle describing the path of the crank by O2O2’.
 Draw circle with center O4 and radius the length of the rocker C.
 Find the location of joint B by finding the intersection of the two
circles on which B is located.
B
b
A
c
b
a
O2’
θ3
d
O2
O4
b
4
Algebraic position analysis
Use trigonometry to find positions of links
and joints.
Example: four bar linkage:
A
a
O2
3
b
B
c
4
2
d
O4
Procedure:
1.Find triangle O2AO4 from a, d, 2.
2.Find triangle AO4B from b, c, AO4.
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Complex number method for position
analysis
Idea: represent links as position vectors and
represent these vectors as complex numbers.
Why complex numbers?
In many problems, it is more straightforward
to derive the equations for position analysis
using complex numbers
Complex number: Rectangular form
Im
Ry

R  Rx  jR y
Rx
Re
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Complex number: Polar form
Im

R (magnitude)
Ry
  j
R  Re
 (angle)
Rx
Re
Complex conjugate
z  x  jy
Euler’s theorem:
e j  cos  j sin
z  x  jy  z cos  j z sin rectangular form
z = z   z e j
polar form
e j  e  j
cos 
 Re( e j )
2
e j  e  j
sin 
 Im( e j )
2j
7
Complex number algebra:
z  x  jy
w  u  jv
 Equality: z = w
 x = u and y = v or z = w and angle of w = angle of v
 Addition, subtraction:
z  w  x  u  j ( y  v)
 Multiplication, Division, Powers
 Multiplication of two numbers: multiply magnitudes,
add phase angles.
 Division of two numbers: divide magnitudes, subtract
phase angles
 Raising a complex number into a power: raise
magnitude into the power, multiply phase angle by the
power
 Use rectangular form when adding or subtracting
 Use polar from when multiplying, dividing or raising into
a power
Complex equation solving:
f(z) = 0, where z and f are complex quantities. Solve for z.
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Solution
Real(f(z)) = 0 or Real[f(x+jy)] = 0
Im(f(z)) = 0 or Im[f(x+jy)] = 0
Solving the above two independent equations for x and y we
find z.
Example:
(4+j)z+5-2j = 0, where z = x + jy
Solution
Substituting z = x + jy into the equation we obtain:
(4+j)(x+jy)+5-2j = 0 or
4x+4y j +xj-y+5-2j=0 or
4x-y+5+j(4y+x-2)=0
Both real and imaginary parts should be zero:
Real part = 0 → 4x - y+5=0
Imaginary part = 0 → 4y+x-2=0
Solving the two equations for x and y we obtain the real and
imaginary parts of complex number z:
x = -1.059 and y = 0.765.
Therefore:
z = -1.059 + j0.765.
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Example: crank-rocker four bar linkage
A
R2
a
b
B
R3
2
3
R4
c
4
R1 d
O2
O4
Problem: Given a, b, c, d and 2, find 3 and 4
Solution:
Vector loop equation:




R2  R3  R4  R1  0
ae j 2  be j3  ce j 4  de j  0
Real part = 0
Imaginary part = 0
Two equations with two unknowns, 3 and 4
Important definitions:
Open Grashof mechanism: If 0   2   / 2 then the two links
adjacent to the shortest link (crank) do not cross each other.
Crossed Grashof mechanism: If 0   2   / 2 then the two
links adjacent to the shortest link (crank) cross each other.
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B 2  4 AC
)
2A
A  cos(  2 )  k1  k 2 cos(  2 )  k 3
B   2 sin  2
 4  2 tan 1 (
B
Open solution for negative
square root, crossed solution
for positive square root
C  k1  ( k 2  1) cos(  2 )  k 3
k1  d / a
k2 
d
c
a 2  c2  d 2  b2
k3 
2 ac
 E  E 2  4 DF
)
2D
D  cos( 2 )  k1  k 4 cos( 2 )  k5
 3  2 tan 1 (
E  2 sin  2
Open solution for negative
square root, crossed solution
for positive square root
F  k1  ( k 4  1) cos  2  k5
k4  d / b
k5 
c 2  d 2  a 2  b2
2ab
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Example: Slider-crank mechanism
Problem: Given a-d, 2 find b, 3, 4
Open solution first
3

R3, b
R2, a
2
R4, c
4
R1, d




Steps: R2  R3  R4  R1  0
Final result: Find angle  4 by solving numerically or
algebraically the following equation:
a cos 2 
a sin  2  c sin  4
 c cos( 4   )  d  0
tan( 4   )
Algebraic solution:
12
Open solution
 T  T 2  4 SU

1
 4  2 tan
2S
where S  R-Q, T  2 P, U  Q  R
P  a sin  2 sin   (a cos  2  d ) cos 
Q   a sin  2 cos   (a cos  2  d ) sin 
R  c sin 
Inclination angle of coupler:  3   4  
End of open solution
Crossed mechanism:  3   4  
a cos 2 
a sin  2  c sin  4
 c cos( 4   )  d  0
tan( 4   )
Algebraic solution:
2  4 SU

T

T
 4  2 tan 1
2S
Find coefficients T, S and U from the equations for the
open solution.
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Inclination angle of coupler:  3   4  
14