INTRODUCTION
TO
DYNAMICS
ANALYSIS
OF
ROBOTS
(Part 3)
Introduction to Dynamics Analysis of Robots (3)
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 acceleration tensor and angular acceleration tensor
•Derive the principles of relative motion between bodies in
terms of acceleration analysis
Summary of previous lectures
Velocity tensor and angular velocity vector
 11  12  13 
 (t )  R RT (t )   21  22  23 


 31  32  33 




vQ (t )  vP (t )   (t )Q(t )  P(t )
 1   32   23 

(t )   2    13     31 
    

3   21    12 





vQ (t )  vP (t )  (t )  Q(t )  P(t )



a
PQ / a  Pob / a b RPQ / b 
Velocity and moving FORs




a
a 
vQ / a  vob / a  b / a  b RPQ / b b RvQ / b
Relative Angular Velocity
Consider 3 FORs {a}, {b} and {c}.
frame {b} w.r.t. frame {a}. Let
a
b
R
is the rotation of

b / a = relative angular velocity of frame {b} w.r.t. frame {a}

a
 c / b = relative angular velocity of frame {c} relative to frame {b}
w.r.t. frame {a}


a
a
 c / b  b R ( c / b )

a
 c / a = relative angular velocity of frame {c} w.r.t. frame {a}



a
a
a
c / a  b / a   c / b



a
 c / a   b / a  b R ( c / b )
a
Example: Relative Angular Velocity
A=3
B=2
Z0, Z1
Y2
Example:
The 3 DOF
RRR Robot:
Y3
Y0, Y 1
X0, X1
X2
Z2
What is
C=1
X3
P
Z3

 3 / 0 after 1 second if all the joints are rotating at
i 
t
6
,
i  1,2,3
Example: Relative Angular Velocity
Solution: We re-used the following data obtained from the
previous lecture
0.866  0.5 0
 0 

0
 0.5 0.866 0


R

1
1/ 0 
0




0
1
 0
0.5236
 0 

 2 /1   0.5236


 0 
3 / 2
 0 
 0 


0.5236
0.866  0.5 0 
1
 0

R

0

1
2


 0.5 0.866 0 
0.866  0.5 0
2
 0.5 0.866 0
R

3


0
1
 0
Example: Relative Angular Velocity



1
 3 / 1   2 / 1  2 R ( 3 / 2 )



0
3 / 0  1/ 0 1 R(3 /1 )



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 



0
3 / 0  1/ 0 1 R(3 /1 )

3 / 0
 0  0.866  0.5 0  0   0.5236 
  0    0.5 0.866 0  1.0472   0.9068

 

 

0
1  0   0.5236 
0.5236  0
Example: Relative Angular Velocity

3 / 0
 0.5236 
  0.9068


 0.5236 
You should get the same answer from the overall rotational
matrix and its derivative, i.e.
R 10R 21R 32R
0  0 1 2
0 12
0 1 2
R

R
R
R

R
R
R

3
1 2 3
1 2 3
1R 2 R 3 R
0
3

 3 / 0  R R   3 / 0
0
3
0
3
T
 3 / 0 (3,2)
  3 / 0 (1,3) 


 3 / 0 (2,1) 
Example: Relative Angular Velocity
0.433  0.75
0
0 1 2
 0.25  0.433
R

R
R
R

3
1 2 3

0.5
0.866
 0.9162
0  0 1 2
0 12
0 1 2
 0.2267
R

R
R
R

R
R
R

R
R
R

3
1 2 3
1 2 3
1 2 3

 0.5236
  0.9162  0.2267 0.4534 0.433
 3 / 0   0.2267  0.6545 0.2618  0.25


0  0.866
 0.5236  0.9069
 3/ 0
0.5 
 0.866

0 
 0.2267 0.4534
 0.6545 0.2618

 0.9069
0 
T
 0.75
0.5 
 0.433  0.866

0.5
0 
 0.5236  0.9068
 0
 0.5236 



 0.5235
0
 0.5236   3 / 0   0.9068




0
0.9068 0.5236

 0.5236 
Acceleration tensor
Consider 2 points “P” and “Q” of a rigid body:




Q(t )  P(t )   (t )Q(t )  P(t )








Q(t )  P(t )   (t )Q(t )  P(t )   (t ) Q(t )  P(t )








Q(t )  P(t )   (t )Q(t )  P(t )   (t ){ (t )Q(t )  P(t )}




2


Q(t )  P(t )   (t )   (t ) Q(t )  P(t )

Rearranging:
where




aQ (t )  aP (t )  A(t )Q(t )  P(t )
A(t )  (t )   2 (t )
A(t) is called the acceleration tensor

Example: Acceleration tensor
Given
0 0 0 
 (t )  R RT (t )  0 0   


0  0 
Find the acceleration tensor if =t2
Solution:
  t 2    2t    2
0 0 0  0 0
(t )  0 0    0 0

 
0  0  0 2
0 
 2 

0 
Example: Acceleration tensor
0
0 0 0  0 0 0  0
 2 (t )  0 0   0 0    0   2


 
0
0  0  0  0  0
0
0 
0
 2 (t )  0  4 2t 2
0 


2 2
0
 4 t 
0
0  0
0
0 0
A(t )     2  0 0  2   0  4 2t 2

 
0  0
0
0 2
0
0 
0
A(t )  0  4 2t 2
 2 


2 2
2
 4 t 
0
0 
0 

2
  



2 2
 4 t 
0
0
Angular Acceleration vector










Q(t )  P(t )   (t )Q(t )  P(t )   (t ) Q(t )  P(t )












Q(t )  P(t )   (t )  Q(t )  P(t )  (t )  v (t )  v (t )
Q
where
P
 1   32   23 

(t )   2    13     31 
    

3   21    12 
Angular
velocity vector
1  32  23 

 (t )   2   13    31 
    

 3  21   12 
Angular
acceleration
vector
Similarly:
Example: Angular Acceleration vector
Given
0 0 0 
 (t )  R RT (t )  0 0   


0  0 
Find the angular acceleration vector if =t2
Solution:
  t 2    2t    2
0 0 0  0 0
(t )  0 0    0 0

 
0  0  0 2
0 
 2 

0 
32  2 

 (t )  13    0 
   
21   0 
Acceleration and moving FORs




a
a
vQ / a  vob / a   (b RPQ / b ) b R PQ / b






a
a 
a 
a 
a

 vob / a   (b RPQ / b )   (b RPQ / b )   (b RPQ / b ) b R PQ / b  b R PQ / b
 

vQ / a
   
T
T 1


RR    R   R 
R  R  R
T

R   R 
1
T 1
T 1
 R

1 1
R
 R
   
   
 







a
a 
a 
a 
a

vQ / a  vob / a   ( b RPQ / b )   (b RPQ / b )   (b RPQ / b ) b R PQ / b  b R PQ / b







a
a
a 
a
a

vQ / a  vob / a   (b RPQ / b )   ( bRPQ / b )   (b RPQ / b )  bR PQ / b  b R PQ / b






a
a
a 
a

vQ / a  vob / a   ( b RPQ / b )     b RPQ / b   2 (b RPQ / b ) b R PQ / 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
 
Acceleration and moving FORs







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
 
Let


a
Prel   b RPQ / b 
 a 
Vrel ( b RPQ / b )
 a 
arel  b R PQ / b
 










aQ / a  aob / a   b / a  ( Prel )  b / a  b / a  Prel  2b / a  (Vrel )  arel 
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 know from the previous lecture that at t=1
cos(1 )  sin(1 ) 0
 sin( ) cos( ) 0
1
1
0

T

1
0
1
 0
 0
0
0

 sin(1 )1  cos(1 )1
 cos( )
1

sin(

)

0 
1
1
1

1T 

0
0

0
0

0
0

0
1
0 0
0 0
0 0

0 0
1 

6
0.866  0.5 0
0
 0.5 0.866 0
R

1


0
1
 0
  0.51
0.866
0 
R

1
1

 0
 0.8661 0
 0.51
0
0
0
Example: Acceleration and moving FORs
 sin(1 )1
 cos( )
0 
1 1

T

1

0

0

  cos(1 )12  sin(1 )1

2

1

sin(

)


cos(

)

0 
1
1
1

1T 

0

0

 0.86612

0 
2

R


0
.
5

1
1


0
 cos(1 )1 0 0
 sin(1 )1 0 0
0
0 0

0
0 0
sin(1 )12  cos(1 )1
 cos(1 )12  sin(1 )1
0
0
0.512
 0.86612
0
0

0
0
0
0
0
0
0

0
0

0
Example: Acceleration and moving FORs
Similarly at t=1  2  
6
cos( 2 )  sin( 2 ) 0
 0
0
1
1

2T 
 sin( 2 ) cos( 2 ) 0
 0
0
0

 sin( 2 )2

0
1

2T 
 cos( 2 )2

0

 cos( 2 )2
0
 sin( 2 )2
0
A
0

0
1 
0.866  0.5 0 
1
 0

R

0

1
2


 0.5 0.866 0 
0 0
  0.52
0 0 1 R   0
2

0 0
0.8662

0 0
 0.8662
0
 0.52
0
0
0
Example: Acceleration and moving FORs
 sin( 2 )2  cos( 2 )2 0 0


0
0
0
0
1


2T 
 cos( 2 )2  sin( 2 )2 0 0


0
0
0
0


  cos( 2 )22  sin( 2 )2 sin(2 )22  cos( 2 )2

0
0
1 

2T 
 sin( 2 ) 2  cos( 2 )2  cos( 2 )22  sin( 2 )2

0
0

 0.86622

1 
R

0
2

  0.522
0.522
0
 0.86622
0

0
0
0
0
0
0
0

0
0

0
Example: Acceleration and moving FORs
At t=1,  3  
6
cos( 3 )  sin( 3 )
 sin( ) cos( )
3
3
2

T

3
0
 0
 0
0

 sin( 3 )3
 cos( )
2 
3
3

T

3

0

0

 cos( 3 )3
 sin( 3 )3
0
0
0 B
0 0

1 0
0 1 
0 0
0 0
0 0

0 0
0.866  0.5 0
2
 0.5 0.866 0
R

3


0
1
 0
  0.53
0.866
2 
R

3
3

 0
 0.8663
 0.53
0
0
0
0
Example: Acceleration and moving FORs
 sin( 3 )3  cos( 3 )3 0 0
 cos( )



sin(

)

0
0
2 
3
3
3
3


T

3

0
0
0 0


0
0
0 0

  cos( 3 )32  sin( 3 )3 sin( 3 )32  cos( 3 )3

2
2




3

sin(

)


cos(

)


cos(

)


sin(

)

2 
3
3
3
3
3
3
3

3T 

0
0

0
0

 0.86632
0.532
0


2 
2
2


R


0
.
5


0
.
866

0
3
3
3



0
0
0
0
0
0
0
0

0
0

0
Example: Acceleration and moving FORs
With 1  2  3   ;
6

T
PP / 3  1 0 0

a
We need to find P / 0
 (t )  R RT    RRT  R R T
Substitute the matrices given into the equation, we get:



0 0 T 0  0  T
1/ 0 1 R1R 1R1R  0  1/ 0  0
Similarly
2 /1  R R  R R
1
2
1
2
T
1
2
1
2
T
3 / 2 32R32RT  32R 32R T



 0   2 /1  0



 0  3/ 2  0
Example: Acceleration and moving FORs



1 / 0   2 / 1   3 / 2

0
For the data given, the following were determined in the
previous lecture:

1/ 0

vP / 2
 0 
 0 


0.5236
 0 

 2 /1   0.5236


 0 
 0.2618
 1.4304



 0.4534
vP /1   0 




 0 
 1.4304 

3 / 2

vP / 0
 0 
 0 


0.5236
 2.6084
  1.6571 


 1.4304 
Example: Acceleration and moving FORs







2
2
aP / 2  ao3 / 2   3 / 2 (3 RPP / 3 )  3 / 2  3 / 2  3 RPP / 3 


2 
2
 23 / 2 (3 RvP / 3 ) 3 RaP / 3 

ao 3 / 2  0

aP / 3  0

vP / 3  0
There is no translation acceleration between
frames {3} and {2} and no translation velocity
and acceleration of point “P” in frame {3}




2
aP / 2  3 / 2  3 / 2  3 RPP / 3 

aP / 2
 0    0  0.866  0.5 0 1   0.2374

 








0

0
 0.5 0.866 0 0    0.1371

 
 
 




0
1 0  
0
0.5236  0.5236  0

Example: Acceleration and moving FORs







1
1
aP /1  ao 2 /1   2 /1 ( 2 RPP / 2 )   2 /1   2 /1   2 RPP / 2 


1 
1
 2 2 /1 ( 2 RvP / 2 ) 2 RaP / 2 

ao 2 / 1  0
There is no translation acceleration between
frames {2} and {1}

 2 /1  0

T
aP / 2   0.2374  0.1371 0



T
2
PP / 2  Po3 / 2 3 RPP / 3   2.866 0.5 0



T
2
vP / 2  3 / 2  3 RPP / 3    0.2618 0.4534 0
Example: Acceleration and moving FORs
Substituting the values into the equation:






1
1 
1
aP /1   2 /1   2 /1   2 RPP / 2  2 2 /1 ( 2 RvP / 2 ) 2 RaP / 2 
 0
  0
 0.866  0.5 0  2.866 










a P /1   0.5236    0.5236  0
0
 1 0 .5 

 
 



 0
   0
  0.5 0.866 0   0  
 0
 0.866  0.5 0   0.2618
 2  0.5236   0
0
 1  0.4534 

 


 0
  0.5 0.866 0   0 
0.866  0.5 0   0.2374
  1.023 





 0
0
 1  0.1371  a P /1   0 





 0.5 0.866 0   0

 1.2237
Example: Acceleration and moving FORs







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 

ao1/ 0  0
There is no translation acceleration between
frames {1} and {0}

b/a  0

T
aP /1   1.023 0  1.2237



T
1
PP /1  Po 2 /1  2 RPP / 2   5.232 0 1.866



T
1
1 
vP /1  2 /1   2 RPP / 2  2 RvP / 2   1.4304 0 1.4304
Example: Acceleration and moving FORs
Substituting the values into the equation:






0
0 
0
aP / 0  1/ 0  1/ 0  1 RPP /1  21/ 0 (1 RvP /1 )1RaP /1 
 0    0  0.866  0.5 0 5.232 










aP / 0 
0

0
 0.5 0.866 0
0 

 
 



0
1 1.866 
0.5236  0.5236  0
 0   0.866  0.5 0  1.4304 







2
0
  0.5 0.866 0
0

 


0
1  1.4304  
0.5236   0
0.866  0.5 0   1.023 
  1.38

  0.5 0.866 0  0   aP / 0   2.53





0
1  1.2237
 0
  1.22 
Example: Acceleration and moving FORs
We should get the same answer if we use transformation matrix
method.
Try it at home and we’ll discuss this in the next lecture!
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:
•The acceleration tensor and angular acceleration tensor
•The principles of relative motion between bodies in terms of
acceleration analysis