Transcript Robot Kinetics * Slide Set 10

Robot Dynamics – The Action of
a Manipulator When Forced
ME 4135
Fall 2012
R. R. Lindeke, Ph. D.
We will examine two approaches
to this problem

Euler – Lagrange Approach:
–

Develops a “Lagrangian Function” which relates Kinetic
and Potential Energy of the manipulator thus dealing
with the manipulator “As a Whole” in building
force/torque equations
Newton – Euler Approach:
–
This approach tries to separate the effects of each link
by writing down its motion as a linear and angular
motion. But due to the highly coupled motions it
requires a forward recursion through the manipulator for
building velocity and acceleration models followed by a
backward recursion for force and torque
Euler – Lagrange approach

Employs a Denavit-Hartenberg structural analysis to
define “Generalized Coordinates” as common
structural models.


It provides good insight into controller design related to STATE
SPACE
It provides a closed form interpretation of the various
components in the dynamic model:
–
–
–
–
Inertia
Gravitational Effects
Friction (joint/link/driver)
Coriolis Forces relating motion of one link to coupling effects of
other link motion
– Centrifugal Forces that cause the link to ‘fly away’ due to coupling
to neighboring links
Newton-Euler Approach

A computationally ‘more efficient’
approach to force/torque determination


It starts at the “Base Space” and moves forward
toward the “End Space” computing trajectory,
velocity and acceleration
Using this forward velocity type information it
computes forces and moments starting at the “End
Space” and moving back to the “Base Space”
We’ll start with the E-L method by
Defining the Manipulator
Lagrangian:
L ( q, q )  T ( q, q )  U ( q )
here
T (q, q)  Kinetic energy of the
manipulator
U (q)  Potential energy of the
manipulator
Generalized Equation of Motion/
Force of the Manipulator:
 
d 
Fi  
L  q, q    
L  q, q  1in


dt  qi
 qi
i is a link of the manipulator
Fi is the Generalized Force acting on Link i
Starting Generalized
Equation Solution


We begin with focus on the Kinetic energy term
(the hard one!)
Remembering from physics:


K. Energy = ½ mV2
Lets define, for the Center of Mass of a Link ‘K’:
 k as Linear Velocity
k as Angular Velocity
Rewriting the Kinetic Energy
Term:

 
T
T

n
 K  mK K  K  DK K

T  q, q    
2
K 1 



 


mK is Link Mass
DK is a 3x3 Inertial Tensor of Link K about its center of mass
expressed WRT the base frame – this tensor characterizes
mass distribution of a rigid object
Focusing on DK:
Looking at a(ny) link
For this (any) Link: DC is its Inertial
Tensor About it Center of Mass

In General:

2
2
y

z
  dV
 
V
DC  mK    xy  dV

V

   xz  dV

V
  xy  dV
V
2
2
x

z
  dV

V
  yz  dV
V

  xz  dV 
V

  yz  dV 

V

2
2
x

y

  dV 
V

Defining the terms:



The Diagonal terms are the “Moments of Inertia” of
the link
The three distinct off diagonal terms are the
Products of Inertia
If the axes used to define the pose of the center of
mass are aligned with the x and z axes of the link
defining frames (i-1 & i) then the products of inertia
are zero and the diagonal terms form the
“Principal Moments of Inertia”
Continuing after this
simplification:

2
2
y

z
  dV
 
V
DC  mK 
0



0

0
x
2
 z 2   dV
V
0

0



0


2
2
x

y

  dV 
V

If the Link is a Rectangular
Rod (of uniform mass):
 b2  c2
 12


DC  mK  0

 0


0
a2  c2
12
0

0 


0 

2
2
a b 
12 
This is a reasonable approximation for many arms!
If the Link is a Thin Cylindrical
Shell of Radius r and length L:
r 2


DC  mK 0


 0

0
0

2
2

r L
0
2
12

2
2

r
L
0

2
12
Some
General
Link
Shape
Moments
of Inertia:
From: P.J. McKerrow, Introduction
to Robotics, Addison-Wesley, 1991.
We must now Transform each
link’s Dc

Dc must be defined in the Base Space To add to the
Lagrangian Solution for kinetic energy (we will call it
DK):
 



T
K
DK K

Where: DK = [R0K*Dc *(R0K)T]
Here R0K is the rotational sub-matrix defining the Link
frame K (at the end of the link!) to the base space –
thinking back to the DH ideas
Defining the Kinetic Energy
due to Rotation (contains DK)


K .E . 

T
K

2
 K 
T
K .E . 
DK K
  R DC  R

2
K
0

K T
0

 K
Completing our models of
Kinetic Energy:

Remembering:

 
T
T

n
 K  mK K  K  DK K

T  q, q    
2
K 1 






Velocity terms are from
Jacobians:

We will define the velocity terms as parts of a “slightly” –
(really mightily) – modified Jacobian Matrix:
c1
J K (q)   q1

 1Z 0



cK
qK
 K Z K 1
AK is linear velocity effect
BK is angular velocity effect
I is 1 for revolute, 0 for prismatic
joint types
0


0
 AK ( q ) 
 K 
 B (q) 
Velocity Contributions of
all links beyond K are
ignored (this could be
up to 5 columns!)
Focusing on cK in the
modified jacobian


This is a generalized coordinate of the center of
mass of a link
K
K
c

H

T
(
q
)


c
It is given by:
K
1
0
here :
A Matrix that
essentially
strips off the
bottom row
of the
solution
c K is a vector from frame k
(at the end of link K) to the
Center of Mass of Link K
 lK

and is:  ,0,0,1
 2

Re-Writing K. Energy for the ARM:



T
K
K
K
K

A q  mK A q   B q  DK B q
n


T  q, q    
2
K 1






Factoring out the Joint Velocity
Terms
 qT
n

T  q, q    
K 1

 A 
K T
mK A
K

qq
T
 B 
K T
2

DK B K q 



Simplifies to:

n
T  q, q   q  K 1
T
A

K T
mK A
K
 
 B
2

K T
DK B K
  q
Building an Equation for Potential
Energy:
n
U (q )   mK g cK (q )
T
K 1
Generalized coordinate
of centers of mass
(from earlier)
g is acceleration due to gravity and
Introducing a new term:
n
c (q)   mK cK (q )
K 1
leads to:
U (q)   g T c (q)
This is a weighted sum of
the centers of mass of the
links of the manipulator
Finally: The Manipulator
Lagrangian:
L(q, q)  T (q, q)  U (q)
Which means:

n
L  q, q    qT   K 1
A

K T
mK A
K
 
 B
2

K T
DK B K
   q  g c (q )
 
T
Introducing a
‘Simplifying’ Term D(q):
Dq
n
{ A
K 1
Then:
K
 q  mK A
T
K
 q    B
K
K

q
D
q
B
  K    q }
T
T
T
1
L  q, q   [ q   D  q    q ]  g c (q)
2
Considering “Generalized
Forces” in robotics:



We say that a generalized force is an residual force
acting on a arm after kinetic and potential energy are
removed!?!*!
The generalized forces are connected to “Virtual
Work” through “Virtual Displacements” (instantaneous
infinitesimal displacements of the joints q),
Thus we say that the Virtual Displacement is a
Displacement that is done without the physical
constraints of time
Generalized Forces on a
Manipulator

We will consider in detail two (of
the readily identified three):

Actuator Force (torque) →
W1   T  q

Frictional Effects →
 W2  b  q   q

Tool Forces →
T
F Tool will be taken  0  in general 
Examining Friction – in detail

Defining a Generalized Coefficient of Friction for a
link:
 qK
 d
v
s
d
 
bk (q)  bK  qK   SGN bK  bK  bK e




C. Viscous
Friction
C. Dynamic
Friction

C. Static
Friction
Combining these components of
Virtual Work:
 W   W1   W2    b  q    q
T
leads to the manipulator Generalized Force:
F   b  q 
Building a General L-E
Dynamic Model

Remembering:
 
d 
Fi  
L  q, q    
L  q, q  1in


dt  qi
 qi
i is a link of manipulator
Starting with
this term
Partial of Lagrangian w.r.t. joint
velocity
  L  q, q  
qi

 T  q, q  
q
It can be ‘shown’ that
this term equals:
   Dij  q   q j 
n
j 1
Completing the 1st Term:
n



L
q
,
q






d
d

    Dij  q  q j 
dt 
qi
dx
 j 1


This is found to equal:
Completing this 1st term of
the L-E Dynamic Model:
 Dij  q  
  Dij  q q j   
qk q j
qk 
j 1
k 1 j 1 
n
n
n
Looking at the 2nd Term:



L  q, q  
T  q, q  
U q
qi
qi
qi
This term can be shown to be:
 Dkj (q ) 

 q  qk q j
3
n
k 1 j 1 
j
i


  g k m j Aki (q )
2
k 1 j i
n
n
Before Summarizing the L-E
Dynamical Model we introduce:

A Velocity Coupling Matrix (4x4)
i
kj
C

q


1
Dij  q  
Dkj  q  for 1  i, j , k  n
2 q
qk
i
A ‘Gravity’ Loading Vector (nx1)
3
n
hi  q   g k m j A  q 
k 1 j i
j
ki
The L-E (Torque)
Dynamical Model:
n
n
n
i   Dij  q q j   C  q qk q j  hi  q   bi  q 
j 1
Inertial
Forces
k 1 j 1
i
kj
Coriolis &
Centrifugal
Forces
Gravitational
Forces
Frictional
Forces