INTRODUCTION
TO
DYNAMICS
ANALYSIS
OF
ROBOTS
(Part 4)
Introduction to Dynamics Analysis of Robots (4)
This lecture continues the discussion on the analysis of the
instantaneous motion of a rigid body, i.e. the velocities and
accelerations associated with a rigid body as it moves from one
configuration to another.
After this lecture, the student should be able to:
•Derive the principles of relative motion between bodies in
terms of acceleration analysis
•Solve problems of robot instantaneous motion using joint
variable interpolation
•Calculate the Jacobian of a given robot
Summary of previous lecture
Acceleration tensor and angular acceleration vector
 11  12  13 
 (t )  R RT (t )   21  22  23 


 31  32  33 
 1   32   23 

(t )   2    13     31 
    

3   21    12 




vQ (t )  vP (t )   (t )Q(t )  P(t )





vQ (t )  vP (t )  (t )  Q(t )  P(t )
A(t )  (t )   2 (t )
1  32  23 

 (t )   2   13    31 
    

 3  21   12 




aQ (t )  aP (t )  A(t )Q(t )  P(t )









Q(t )  P(t )   (t )  Q(t )  P(t )  (t )  vQ (t )  vP (t )
Summary of previous lecture
Moving FORs



a
PQ / a  Pob / a b RPQ / b 





a
vQ / a  vob / a  b / a  b RPQ / b baRvQ / b







a
a
aQ / a  aob / a   b / a (b RPQ / b )   b / a   b / a   b RPQ / b 



a 
a

 2 b / a (b RPQ / b ) b R PQ / b
 
Example: Acceleration and moving FORs
A=3
B=2
Z0, Z1
Y2
Example:
The 3 DOF
RRR Robot:
C=1
Y3
Y0, Y 1
X0, X1
X2
Z2
X3
P
Z3
What is the acceleration of point “P” after 1 second if all the
joints are rotating at
t
i  ,
i  1,2,3
6
Example: Acceleration and moving FORs
We did the following:







2
2
aP / 2  ao3 / 2   3 / 2 (3 RPP / 3 )  3 / 2  3 / 2  3 RPP / 3 



 23 / 2 (32RvP / 3 ) 32RaP / 3 







1
1
aP /1  ao 2 /1   2 /1 ( 2 RPP / 2 )   2 /1   2 /1   2 RPP / 2 



 2 2 /1 ( 21RvP / 2 ) 21RaP / 2 







0
0
aP / 0  ao1/ 0  1/ 0 (1 RPP /1 )  1/ 0  1/ 0  1 RPP /1 


0 
0
 21/ 0 (1 RvP /1 )1 RaP /1 
To get

aP / 0
  1.38
  2.53


  1.22 
Example: Acceleration and moving FORs
We should get the same answer if we use transformation matrix
method. We know that
* 0 *
PP / 0  3TPP / 3
* 0   * 0 *
 vP / 0  3TPP / 3  3TvP / 3
 * 0  *
0  *
0 *
 aP / 0  3TPP / 3  2 3TvP / 3  3TaP / 3


vP* / 3  0,
aP* / 3  0
 * 0  *
aP / 0  3TPP / 3
0  0 1 2
0 1 2
0 1 2 
0 12
0 1 2 
0 12 





T

T
T
T

T
T
T

T
T
T

2
T
T
T

2
T
T
T

2
3
3 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1T 2T 3T 
For
 0.1187 1.3022  0.1371  1.2606
  1.1652 0.1187

0
.
2374

1
.
3609
0 


3T 
0
 0.2742
 0.9497  0.5483
 0

0
0
0


Example: Acceleration and moving FORs
 * 0  *
aP / 0  3TPP / 3
*
aP / 0
 0.1187 1.3022  0.1371  1.2606 1   1.38
  1.1652 0.1187
0.2374  1.3609 0  2.53
   


0
 0.2742 0   1.22 
 0.9497  0.5483
 0
 1  0 
0
0
0

  

The answer is the same as that obtained earlier:

aP / 0
  1.38
  2.53


  1.22 
Relative angular acceleration
We can differentiate the relative angular velocity to get the
relative angular acceleration:



a
 c / a   b / a  b R ( c / b )




a
a
 c / a   b / a  b R ( c / b )  b R ( c / b )




a
a
 c / a   b / a   b / a b R ( c / b )  b R ( c / b )





a
 c / a   b / a   b / a b R( c / b ) baR( c / b )

where  b / a  b / a
vb / a baR baRT ba R  vb / a (baRT ) 1  vb / a (baR)

Example: Relative Angular Acceleration
A=3
B=2
Z0, Z1
Y2
Example:
The 3 DOF
RRR Robot:
X2
Z2
What is
 3/ 0
Y3
Y0, Y 1
X0, X1

C=1
X3
P
Z3
after 1 second if all the joints are rotating at
i 
t
6
,
i  1,2,3
Example: Relative angular acceleration
Solution: We re-used the following data obtained from the
previous lecture




1 / 0   2 / 1   3 / 2  0
0.866  0.5 0
0
 0.5 0.866 0
R

1


0
1
 0

1/ 0
 0 
 0 


0.5236

0.866  0.5 0 
1
 0

R

0

1
2


 0.5 0.866 0 
 0 
 0 




 2 /1   0.5236
3 / 2   0 




 0 
0.5236




1
 3 /1   2 /1   2 /1  2 R(3 / 2 ) 21R( 3 / 2 )





0
0


 3 / 0  1/ 0  1/ 0  1 R(3 /1 ) 1 R( 3 /1 )
Example: Relative angular acceleration



1
 3 /1   2 /1   2 /1  2 R(3 / 2 ) 21R( 3 / 2 )



1
 3 / 1   2 / 1   2 R ( 3 / 2 ) 


 0   0.866  0.5 0   0   0

  






 3 /1   0.5236   0
0
1
0
 0

 

  
 0    0.5 0.866 0  0.5236  0



1
 3 / 1   2 / 1  2 R ( 3 / 2 )
 0  0.866  0.5 0   0   0 

3 /1   0.5236   0
0
 1  0    1.0472

 

 

 0   0.5 0.866 0  0.5236  0 
Example: Relative angular acceleration

3/ 0

3/ 0



0
0
 1/ 0  1/ 0  1 R(3 /1 )1 R( 3 /1 )



0
 3 / 0  1/ 0  1 R(3 /1 )

 0   0.866  0.5 0  0   0.4749

 






0
  0.5 0.866 0  1.0472   0.2742

 






0
1  0    0 
0.5236   0
You should get the same answer from the overall rotational
matrix and its derivative, i.e.

 3 / 0 30R 30RT  3 / 0 30R30RT  30R 30R T   3 / 0
3 / 0 (3,2)
 3 / 0 (1,3) 


3 / 0 (2,1) 
Example: Relative angular acceleration
Using the data from the previous example:
0
3
10R
21R 32R 10R 21R
32R 10R 21R 32R
  210 R 21R 32R   210 R 21R 32R   210 R 21R 32R 
R
 0.1187 1.3022  0.1371
0 
  1.1652 0.1187

R

0
.
2374
3


0 
 0.9497  0.5483
 0.9162  0.2267 0.4534
0  0 1 2
0 12
0 1 2
 0.2267  0.6545 0.2618
R

R
R
R

R
R
R

R
R
R

3
1 2 3
1 2 3
1 2 3


0 
 0.5236  0.9069
0.5 
0.433  0.75
0
0 1 2
 0.25  0.433  0.866
R

R
R
R

3
1 2 3


0.5
0 
0.866
Example: Relative angular acceleration
0
0.2742 
 0
3 / 0 30R30RT  30R 30R T   0
0
 0.4749


0 
 0.2742 0.4749
3 / 0 (3,2) 0.4749

  3 / 0  3 / 0 (1,3)   0.2742

 

3 / 0 (2,1)   0 
The answer is the same as that obtained earlier:

3/ 0
0.4749
 0.2742


 0 
Instantaneous motion of robots
So far, we have gone through the following exercises:
Given the robot parameters, the joint angles and their rates of
rotation, we can find the following:
1. The linear (translation) velocities w.r.t. base frame of a
point located at the end of the robot arm
2. The angular velocities w.r.t. base frame of a point located
at the end of the robot arm
3. The linear (translation) acceleration w.r.t. base frame of a
point located at the end of the robot arm
4. The angular acceleration w.r.t. base frame of a point
located at the end of the robot arm
We will now use another approach to solve the linear
velocities and linear acceleration problem.
Jacobian for Translational Velocities
In general, the position and orientation of a point at the end of the arm
can be specified using nx ox ax px 
 v11 v12 v13 v14 
n
 y
 nz

0
Note that the position
v
py  0 1
v22
n

1
n
  1T 2T  n T PT   21
pz 
v31 v32


0 0
1
0
0

P   px py
of the point w.r.t. {0} is
p x  v14  f1 (1 , 2 ,, n )
oy
oz
ay
az
v23
v33
0
pz 
v24 

v34 

1
T
p y  v24  f 2 (1 , 2 ,, n )
p z  v34  f 3 (1 , 2 ,, n )
The
  dpx dp y dpz T

velocities of the point w.r.t. frame {0} is vP / 0  P  
dt
dt 
 dt
dp v  v 
v 
v
v
v
vx  x  14 1  14 2    14 n  14 1  14 2    14 n
dt
1 t  2 t
 n t
1
 2
 n
vy 
vz 
dp y
dt

v24 1 v24  2
v 
v
v
v

   24 n  24 1  24 2    24 n
1 t  2 t
 n t
1
 2
 n
dpz v34 1 v34  2
v 
v
v
v


   34 n  34 1  34 2    34 n
dt
1 t  2 t
 n t
1
 2
 n
Jacobian for Translational Velocities
vx 
vy 
vz 

vP / 0
dpx v14  v14 
v

1 
 2    14 n
dt
1
 2
 n
dp y
dt

v24  v24 
v
1 
 2    24 n
1
 2
 n
dpz v34  v34 
v

1 
 2    34 n
dt
1
 2
 n
 v14
 
vx   1
v
 v y    24
   1
 vz   v
 34
 1
v14
v14 


1 
 2
 n  1 
 
 

v24
v24  2 

 J (T )  2 

 2
 n    
 
v34
v34   
 n 

 n 
 2
 n 
Jacobian for translational velocities
Example: Jacobian for Translational Velocities
A=3
B=2
Z0, Z1
What is the Jacobian
for translational
velocities of point
“P”?
Y2
C=1
Y3
Y0, Y 1
X0, X1
X2
Z2
X3
P
Z3
Given:
cos(1 ) cos( 2   3 )  cos(1 ) sin(  2   3 ) sin( 1 ) ( A  B cos( 2 )) cos(1 )
 sin(  ) cos(   )  sin(  ) sin(    )  cos( ) ( A  B cos( )) sin(  ) 
1
2
3
1
2
3
1
2
1
0


3T 
sin(  2   3 )
cos( 2   3 )
0
B sin(  2 )




0
0
0
1


Example: Jacobian for Translational Velocities
The transformation matrix of point “P” w.r.t. frame {3} is
1
0
3

PT 
0

0
 nx
n
 y
 nz

0
ox
ax
oy
oz
ay
az
0
0
0 0 C
1 0 0

0 1 0

0 0 1
px 
py  0 1 2 3
   1T 2T  3T  pT  P0T 
pz 

1
cos(1 ) cos( 2   3 )  cos(1 ) sin(  2   3 ) sin( 1 ) ( A  B cos( 2 )  C cos( 2   3 )) cos(1 )
 sin(  ) cos(   )  sin(  ) sin(    )  cos( ) ( A  B cos( )  C cos(   )) sin(  ) 
1
2
3
1
2
3
1
2
2
3
1


sin(  2   3 )
cos( 2   3 )
0
B sin(  2 )  C sin(  2   3 )




0
0
0
1


Example: Jacobian for Translational Velocities
v14  ( A  B cos( 2 )  C cos( 2   3 )) cos(1 )
v24  ( A  B cos( 2 )  C cos( 2   3 )) sin( 1 )
v34  B sin(  2 )  C sin(  2   3 )
v14
 ( A  B cos( 2 )  C cos( 2   3 ))(  sin( 1 )
1
v14
 cos(1 )(  B sin(  2 )  C sin(  2   3 ))
 2
v14
 C cos(1 ) sin(  2   3 )
 3
v24
 ( A  B cos( 2 )  C cos( 2   3 ))(cos(1 )
1
v24
 sin( 1 )(  B sin(  2 )  C sin(  2   3 ))
 2
v24
 C sin( 1 ) sin(  2   3 )
 3
v34
0
1
v34
 B cos( 2 )  C cos( 2   3 )
 2
v34
 C cos( 2   3 )
 3
Example: Jacobian for Translational Velocities
J (T )
J (T )
 v14
 
 1
v
  24
 1
 v
 34
 1
v14
v14 

 2
 n 

v24
v24 

 2
 n 
v34
v34 


 2
 n 
 ( A  B cos( 2 )  C cos( 2   3 )) sin(1 )  ( B sin( 2 )  C sin( 2   3 )) cos(1 )  C cos(1 ) sin( 2   3 )
  ( A  B cos( 2 )  C cos( 2   3 )) cos(1 )  ( B sin( 2 )  C sin( 2   3 )) sin(1 )  C sin(1 ) sin( 2   3 ) 




0
B cos( 2 )  C cos( 2   3 )
C cos( 2   3 )
What is the velocity of point “P” after 1 second if all the joints
are rotating at
t
i  ,
i  1,2,3
6
Example: Jacobian for Translational Velocities
Given a=3, B=2, C=1.
At t=1, 1  
6
J (T )
2 

6
3 

6




1   2   3 
6
 ( A  B cos( 2 )  C cos( 2   3 )) sin(1 )  ( B sin( 2 )  C sin( 2   3 )) cos(1 )  C cos(1 ) sin( 2   3 )
  ( A  B cos( 2 )  C cos( 2   3 )) cos(1 )  ( B sin( 2 )  C sin( 2   3 )) sin(1 )  C sin(1 ) sin( 2   3 ) 




0
B cos( 2 )  C cos( 2   3 )
C cos( 2   3 )
1   2.616  1.616  0.75  0.5236  2.61
vx 

v  v y   J (T ) 2    4.5311  0.9330  0.433 0.5236   1.66 
 


 

3   0
2.231
0.5  0.5236  1.43 
 v z 
The answer is similar to that obtained previously using another
approach! (refer to velocity and moving FORs)
Getting the Translational Acceleration

vP / 0
1 
1 
1 
vx 
ax 
 
 
 

 v y   J (T )  2   aP / 0  a y   J (T )  2   J (T )  2 
 
 



 v z 
 a z 
 
 
  


 n
 n
 n 
If the angular acceleration for 1, 2, …, n are 0s then

aP / 0
1 
ax 
 
 a y   J (T )  2 
 

 a z 
 
 n 
Example: Getting the Translational Acceleration
A=3
B=2
Z0, Z1
Y2
Example:
The 3 DOF
RRR Robot:
C=1
Y3
Y0, Y 1
X0, X1
X2
Z2
X3
P
Z3
What is the acceleration of point “P” after 1 second if all the
joints are rotating at
t
i  ,
i  1,2,3
6
Example: Getting the Translational Acceleration
J (T )
 ( A  B cos( 2 )  C cos( 2   3 )) sin(1 )  ( B sin( 2 )  C sin( 2   3 )) cos(1 )  C cos(1 ) sin( 2   3 )
  ( A  B cos( 2 )  C cos( 2   3 )) cos(1 )  ( B sin( 2 )  C sin( 2   3 )) sin(1 )  C sin(1 ) sin( 2   3 ) 




0
B cos( 2 )  C cos( 2   3 )
C cos( 2   3 )
d (T )
J (1,1)  c11 ( A  Bc 2  Cc23 )  s1 ( Bs 22  Cs23 (2  3 ))
dt
d (T )
J (1,2)  s11 ( Bs 2  Cs23 )  c1 ( Bc 22  Cc23 (2  3 ))
dt
d (T )
J (1,3)  C[ s11 ( s23 )  c1c23 (2  3 )]
dt
d (T )
J (2,1)   s11 ( A  Bc 2  Cc23 )  c1 ( Bs 22  Cs23 (2  3 ))
dt
d (T )
J (2,2)  c11 ( Bs 2  Cs23 )  s1 ( Bc 22  Cc23 (2  3 ))
dt
d (T )
J (2,3)  C[c11 ( s23 )  s1c23 (2  3 )]
dt
Example: Getting the Translational Acceleration
J (T )
 ( A  B cos( 2 )  C cos( 2   3 )) sin(1 )  ( B sin( 2 )  C sin( 2   3 )) cos(1 )  C cos(1 ) sin( 2   3 )
  ( A  B cos( 2 )  C cos( 2   3 )) cos(1 )  ( B sin( 2 )  C sin( 2   3 )) sin(1 )  C sin(1 ) sin( 2   3 ) 




0
B cos( 2 )  C cos( 2   3 )
C cos( 2   3 )
d (T )
J (3,1)  0
dt
d (T )
J (3,2)   Bs22  Cs23 (2  3 )
dt
d (T )
J (3,3)  Cs23 (2  3 )
dt
Given a=3, B=2, C=1.






At t=1, 1  
3 
1   2   3 
2 
6
6
6
6
  1.6572  0.7503  0.2267
J (T )   2.6086  1.5614  0.6545


 1.4305  0.9069
 0
Example: Getting the Translational Acceleration
All the angular acceleration for 1, 2, …, n are 0s:

aP / 0
1 
ax 
 

a P / 0  a y   J (T )  2 
 

 a z 
 
 n 
1    1.6572  0.7503  0.2267   1.38
 J (T ) 2    2.6086  1.5614  0.6545   2.53

 

3  
0
 1.4305  0.9069   1.22 
The answer is the same as that obtained earlier:
  1.38

aP / 0   2.53


  1.22 
Summary
This lecture continues the discussion on the analysis of the
instantaneous motion of a rigid body, i.e. the velocities and
accelerations associated with a rigid body as it moves from one
configuration to another.
The following were covered:
•Principles of relative motion between bodies in terms of
acceleration analysis
•Robot instantaneous motion using joint variable interpolation
•Jacobian of a robot