Lecture «Robot Dynamics»: Intro to Dynamics
151-0851-00 V
lecture:
exercise:
CAB G11
HG E1.2
Tuesday 10:15 – 12:00, every week
Wednesday 8:15 – 10:00, according to schedule (about every 2nd week)
Marco Hutter, Roland Siegwart, and Thomas Stastny
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19.09.2017
26.09.2017
Intro and Outline
Kinematics 1
Course Introduction; Recapitulation Position, Linear Velocity
Rotation and Angular Velocity; Rigid Body Formulation, Transformation
26.09.2017
Exercise 1a Kinematics Modeling the ABB
arm
03.10.2017
Kinematics 2
Kinematics of Systems of Bodies; Jacobians
03.10.2017
10.10.2017
Kinematics 3
Kinematic Control Methods: Inverse Differential Kinematics, Inverse
Kinematics; Rotation Error; Multi-task Control
10.10.2017
Exercise 1b Differential Kinematics of the
ABB arm
Exercise 1c Kinematic Control of the ABB
Arm
17.10.2017
Dynamics L1
Multi-body Dynamics
17.10.2017
24.10.2017
31.10.2017
Dynamics L2
Dynamics L3
Floating Base Dynamics
Dynamic Model Based Control Methods
24.10.2017
31.10.2017
07.11.2017
14.11.2017
21.11.2017
Legged Robot
Case Studies 1
Rotorcraft
Dynamic Modeling of Legged Robots & Control
Legged Robotics Case Study
Dynamic Modeling of Rotorcraft & Control
07.11.2017
14.11.2017
21.11.2017
28.11.2017
05.12.2017
Case Studies 2
Fixed-wing
Rotor Craft Case Study
Dynamic Modeling of Fixed-wing & Control
28.11.2017
05.12.2017
12.12.2017
Case Studies 3
Fixed-wing Case Study (Solar-powered UAVs - AtlantikSolar, Vertical
Take-off and Landing UAVs – Wingtra)
19.12.2017
Summery and Outlook Summery; Wrap-up; Exam
Exercise 2a Dynamic Modeling of the ABB
Arm
Exercise 2b Dynamic Control Methods
Applied to the ABB arm
Exercise 3
Legged robot
Exercise 4
Modeling and Control of
Multicopter
Exercise 5
Fixed-wing Control and
Simulation
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Recapitulation of Kinematics
 Kinematics = description of motions






Translations and rotations
Various representations (Euler, quaternions, etc.)
Instantaneous/Differential kinematics
Jacobians and geometric Jacobians
Inverse kinematics and control
Floating base systems (unactuated base and contacts)
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Dynamics in Robotics
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Dynamics in Robotics
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Dynamics
Outline
  b  q, q   g  q   τ  J Tc Fc
M q q
 Description of “cause of motion”
 Input τ

 Output q
Force/Torque acting on system
Motion of the system
 Principle of virtual work
 Newton’s law for particles
 Conservation of impulse and angular momentum

q
Generalized coordinates
M q
Mass matrix
b  q, q 
Centrifugal and Coriolis forces
g q
Gravity forces
τ
Generalized forces
Fc
External forces
Jc
Contact Jacobian
 3 methods to get the EoM
 Newton-Euler: Free cut and conservation of impulse & angular momentum for each body
 Projected Newton-Euler (generalized coordinates)
 Lagrange II (energy)
 Introduction to dynamics of floating base systems
 External forces
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Principle of Virtual Work
 Principle of virtual work (D’Alembert’s Principle)
 Dynamic equilibrium imposes zero virtual work
dFext external forces acting on element i

r
acceleration of element i
variational parameter
dm
mass of element i
r
virtual displacement of element i
 Newton’s law for every particle in direction it can move
r  F  r  ma
F  ma
r
d 
F  m
v   p
dt  p 
force
dm
Impulse or
linear momentum
Γ
r
dF
S
moment

d 
m
r

v

  N
dt  
N

angular momentum
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Virtual Displacements of Single Rigid Bodies
 Rigid body Kinematics
 Applied to principle of virtual work
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Impulse and angular momentum
 Use the following definitions
 Conservation of impulse and angular momentum
Newton
A free body can move
In all directions
Euler
External forces and moments
Change in impulse and angular momentum
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1st Method for EoM
Newton-Euler for single bodies
 Cut all bodies free
 Introduction of constraining force
 Apply conservation ‫ ܘ‬and ‫ ۼ‬to individual bodies
Ψ
ai
 System of equations
 6n equation
 Eliminate all constrained forces (5n)
Fi 2
Ω
Fi1
mi
vi
Fg
rOSi
 Pros and Cons
+ Intuitively clear
+ Direct access to constraining forces
− Becomes a huge combinatorial problem for large MBS
{I}
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Free Cut
Cart pendulum example
 Find the equation of motion
{I}
mc , c
g
l
m p , p
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Free Cut
Cart pendulum
Fy
x
 Impulse / angular momentum cart
mc 
xc  Fx
mc 
yc  Fy  Fl  Fr  mc g
(1)
 cc  Fr b  Fl b
(3)
m p 
x p   Fx
(4)
m p 
y p   Fy  m p g
(5)
 pp  Fx l cos  p   Fy l sin  p 
(6)
xc  x
(7)
yc  0 (constraint)
(8)
 c  0 (constraint)
(9)
x p  x  l sin  
(10a) x p  x  l cos     2l sin  
(10)
y p  l cos  
(11a ) y p  l sin     l cos  
(12a) p  
(11)
p  
2
Fx
b
(2)
 Impulse / angular momentum pendulum
 Kinematics
mc , c
Fl
mc g
Fr
Fy
Fx
{I} y
6 equations, 6 unknowns resp.
12 equations, 12 unknowns
(12)
How many dimensions does the EoM have?
g
x

l
m p , p
mp g
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Free Cut
Cart pendulum
Fy
x
 (7),(10-12) in (1) and (4-6)
mc 
x  Fx
mc , c
Fx
(13)
m p  
x  l cos     2l sin      Fx
(14)
m p l sin     2l cos      Fy  m p g
(15)
 p  Fx l cos    Fy l sin  
(16)
Fl
mc g
Fr
Fy
 From (13) and (14) remove Fx
 m p  mc  x  m pl cos     2lm p sin    0
Fx
{I}

l
 Insert (13) and (15) in (16) to remove Fx and Fy

p
 m p l 2    m p l cos   
x  glm p sin    0
g
m p , p
mp g
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Free Cut
Cart pendulum
 (7),(10-12) in (1) and (4-6)
mc 
x  Fx
m p  
x  l cos     2l sin      Fx
m p l sin     2l cos      Fy  m p g
 p  Fx l cos    Fy l sin  
(13)
(14)
mc 
xc  Fx
mc 
yc  Fy  Fl  Fr  mc g
(1)
 cc  Fr b  Fl b
(3)
m p 
x p   Fx
(4)
m p 
y p   Fy  m p g
(5)
 pp  Fx l cos  p   Fy l sin  p 
(6)
(15)
xc  x
(7)
yc  0 (constraint)
(8)
 c  0 (constraint)
(9)

x p  
x  l cos     2l sin  
(10)
(16)
 From (13) and (14) remove Fx
 m p  mc  x  m pl cos     2lm p sin    0
Fy
(2)

y p  l sin     2l cos  
(11)
p  
(12)
x
Fl
mc , c
Fx
mc g
Fr
Fy
Fx
{I}

l
 Insert (13) and (15) in (16) to remove Fx and Fy

p
 m p l 2    m p l cos   
x  glm p sin    0
g
m p , p
mp g
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Newton-Euler in Generalized Motion Directions
 For multi-body systems
 Express the impulse/angular momentum in generalized coordinates
 Virtual displacement in generalized coordinates
 With this, the principle of virtual work transforms to
0   W= qT
 q
M q
b  q, q 
g q
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Projected Newton-Euler
  b  q, q   g  q   0
M q q
 Equation of motion
 Directly get the dynamic properties of a multi-body system with n bodies
 For actuated systems, include actuation force as external force for each body
 If actuators act in the direction of generalized coordinates, ࣎ corresponds to stacked actuator
commands
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Projected Newton-Euler
Cart pendulum example
 Find the equation of motion
{I}
mc , c
g
l
m p , p
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Projected Newton-Euler
Cart pendulum example
 Kinematics cart and pendulum
rOSc
 x
q 
 
J Pc
0
0
0
0
dJ
 0 l sin   
J Pp  Pp  

dt
 0 l cos   
drOSc
1

dq
0
0
dJ
 Pc  
dt
0
J Pc 
x
 x  l sin   
rOS p  

 l cos   
drOS p 1 l cos   

J Pp 

dq
0 l sin   
 x
 
0
 Equation of motion
 p  
J Rp
 p

  0 1
q
{I}
mc , c
g
l

m p , p
 mc  m p
lm p cos   
M   J TPi mi J Pi  J TRi i J Ri  J TPc mc J Pc  J TPp m p J Pp  J TRp p J Rp  

2
lm p cos   m p l   p 


b   J mi J Pi q  J TRi Θi J Ri q  J Ri q  Θi J Ri q
T
Pi
 0 (planar system)
  2 lm p sin   

 J m p J Pp q  

0


T
Pp
0

 0  T  0  


g    J Tsi Fig   J TPc 
J
  m gl sin  
Pp 


m
g

m
g
 
p
i 1
 c 

  p
n
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3rd Method for EoM
Lagrange II
kinetic energy
 Lagrangian
potential energy
 Lagrangian equation
 Since
with
U  U q
T T
 q, q 
T
 Mq

q
gravity vector
inertial forces
U
d  T  T


τ
dt  q  q
q
 T
q

 q  1 
  M
Mq
2
 q T


1
2
T  q T Mq
M 
q
q1 
 


  b  q, q   g  q   τ

  g  M q q
M 
q
qn 
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Lagrange II
Kinetic energy
 Kinetic energy in joint space
1
2
T  q T Mq
 Kinetic energy for all bodies
 From kinematics we know that
 Hence we get
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Lagrange II
Potential energy
 Two sources for potential forces
 Gravitational forces
 Spring forces
FE  k j  r  r0  d 0 
r  r0
r  r0
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Lagrange II
Cart pendulum example
 Find the equation of motion
{I}
mc , c
g
l
m p , p
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Lagrange II
Cart pendulum example
 Kinematics cart and pendulum
 x
rOSc   
0
 x 
rSc   
0
 x  l sin   
rOS p  

 l cos   
 x  l cos   
rS p  

 l sin   
x
p  
 p  
{I}
 Kinetic and potential energy
1
1
1
1
1
T   r m r  ω  ω  m x  m x  m l   m
2
2
2
2
2
U   m p gl cos  
0-level can be chosen
T
Si
i Si
T
i
2
i
i
c
2
p
2
p
2
p
1
 cos     2
xl
2
mc , c
g
l

m p , p
 Equation of motion
d  T  T
U


0


dt  q  q
q
 mc x  m p x  m p l cos    
T


2
 cos     
q
 m p l   m p xl
x  m p 
x  m p l cos     m p l sin    2 
d  T   mc 



 sin      
xl cos    m p xl
dt  q   m p l 2  m p 
0


T





q  m p xl sin    
0

U 


m
gl

sin


q  p

 mc 
x  m p 
x  m p l cos     m p l sin    2 

0
2
xl cos      m p gl sin   
 m p l   m p 
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External Forces
 Given:
 Generalized forces are calculated as:
 Given:

Generalized forces are calculated
 For actuator torques:
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External Forces
Cart pendulum example
 Equation of motion without actuation
0
lm p cos  
 mc  m p

  2 lm p sin    

q


 lm cos 

0


2
sin
m
gl

m
l






0
p
p
p 
 
 p



M
 Add actuator for the pendulum
Tp   a
 Action on pendulum
Tc   a
 Reaction on cart
b
g
a
J Rp   0 1
J Rc   0 0
0
τ   J TR ,i Ti  J TRc Tc  J TRp Tp   
 a 
{I}
x
mc , c
g
a
l

Fx  k  x  2l sin    xs 
m p , p
l
 Add spring to the pendulum
 (world attachment point P, zero length 0, stiffness k)
Fy  k  2l cos    ys 
 Action on pendulum
P xs , y s
 x  2l sin   
r

 Fx


T
 2l cos   
  Fx 
τ

J
F


Fs  
s

 2l  Fx cos    Fy sin    

F

1
2
l
cos




r

y




Js  s  

q  0 2l sin   
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

External Forces
Cart pendulum example
 What is the external force coming from the motor
x
{I}
mc , c
M
g
l

m p , p
l
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17.10.2017
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26
External Forces
Cart pendulum example
 What is the external force coming from the motor
 Action on cart Fact
Fc  Fact
J Pc  1 0
F 
τ  J T Fc   act 
 0 
{I}
Fact
x
mc , c
g
l

m p , p
l
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