This lecture discusses the analysis of inverse kinematics for robots. It covers deriving the inverse transformation matrix between coupled links, formulating the inverse kinematics of articulated robots using transformation matrices, and solving problems of robot inverse kinematics analysis. Examples are provided to demonstrate finding the inverse of transformation matrices.

1. ENG4406 ROBOTICS AND MACHINE VISION
KINEMATICS
ANALYSIS
OF
ROBOTS
(Part 2)
PART 2 LECTURE 9

2. This lecture continues the discussion on the analysis of the
forward and inverse kinematics of robots.
After this lecture, the student should be able to:
•Put into practice the concept of inverse kinematics analysis of
robots
•Derive inverse transformation matrix between coupled links
•Formulate the inverse kinematics of articulated robots in terms of
the link transformation matrices
•Solve problems of robot inverse kinematics analysis using
transformation matrices
Kinematics Analysis of Robots II

3. Inverse Transformation


























1
0
0
0
)
cos(
)
cos(
)
cos(
)
sin(
)
sin(
)
sin(
)
sin(
)
sin(
)
cos(
)
cos(
)
sin(
)
cos(
0
)
sin(
)
cos(
1
1
1
1
1
1
1
1
1
1
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
d
d
a
T














A typical transformation matrix can be partitioned as follow:
to give








1
0
0
0
1 P
R
T
i
i
where



















)
cos(
)
cos(
)
sin(
)
sin(
)
sin(
)
sin(
)
cos(
)
cos(
)
sin(
)
cos(
0
)
sin(
)
cos(
1
1
1
1
1
1
i
i
i
i
i
i
i
i
i
i
i
i
R



























)
cos(
)
sin(
1
1
1
i
i
i
i
i
d
d
a
P


and

4. Inverse Transformation





 



1
0
0
0
1
1 P
R
R
T
T
T
i
i
The inverse transformation is given by:
1
1 

T
i
i











 

1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
0
)
sin(
)
cos(
1
1
1
1
0
1




T
For example, find the inverse of :
1
0
1

T

5. Inverse Transformation of











 

1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
0
)
sin(
)
cos(
1
1
1
1
0
1




T






















 

1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
1
1
1
1
1
1
1
1








T
R
R













































0
0
0
0
0
0
1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
0
0
0
1
1
1
1




P
R
P T
1
0
1

T

6. 










 

1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
0
)
sin(
)
cos(
1
1
1
1
0
1




T





 



1
0
0
0
1
1 P
R
R
T
T
T
i
i












1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
1
1
1
1




T
R












0
0
0
P
RT
Therefore
Inverse Transformation of
1
0
1

T















1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
0
)
sin(
)
cos(
1
1
1
1
1
0
1




T

7. 





















 

1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
2
2
2
2
2
2
2
2








T
R
R














































0
)
sin(
)
cos(
0
0
1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
0
0 2
1
2
1
1
2
2
2
2
1






A
A
A
P
R
A
P T
Find the inverse of











 

1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
2
2
1
2
2
1
2



 A
T
Inverse Transformation of
1
1
2

T

8. Therefore
















1
0
0
0
0
1
0
0
)
sin(
0
)
cos(
)
sin(
)
cos(
0
)
sin(
)
cos(
2
1
2
2
2
1
2
2
1
1
2






A
A
T












1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
2
2
2
2




T
R












0
)
sin(
)
cos(
2
1
2
1


A
A
P
RT
and
Inverse Transformation of
1
1
2

T

9. 





















 

1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
3
3
3
3
3
3
3
3








T
R
R














































0
)
sin(
)
cos(
0
0
1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
0
0 3
2
3
2
2
3
3
3
3
2






A
A
A
P
R
A
P T
Find the inverse of











 

1
0
0
0
0
1
0
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
3
3
2
3
3
2
3



 A
T
Inverse Transformation of
1
2
3

T

10. Therefore
















1
0
0
0
0
1
0
0
)
sin(
0
)
cos(
)
sin(
)
cos(
0
)
sin(
)
cos(
3
2
3
3
3
2
3
3
1
2
3






A
A
T












1
0
0
0
)
cos(
)
sin(
0
)
sin(
)
cos(
3
3
3
3




T
R












0
)
sin(
)
cos(
3
2
3
2


A
A
P
RT
and
Inverse Transformation of
1
2
3

T

11. Inverse Kinematics Problem
Robot forward kinematics involve finding the gripper position and
orientation given the angles of rotation of the linkages
Now consider the following problem for the planar robot:
The arm moves to a new location and the position along with the
orientation of the gripper w.r.t. frame {0} is known. Can you
determine the total rotations of 1, 2, and 3 required for the arm
to reach that new position?
This problem is called the robot inverse kinematics problem.

12. Inverse Kinematics of the Planar Robot
The inverse kinematics problem for the planar robot can be stated
as follow:
Given the global gripper orientation and position, i.e. given













1
0
0
0
0
3
z
z
z
z
y
y
y
y
x
x
x
x
p
a
o
n
p
a
o
n
p
a
o
n
T
Find the joint angles 1, 2, and 3 required for the arm to
reach that new position

13. Inverse Kinematics
To solve the inverse kinematics problem, we know that:
    T
T
T
T
p
a
o
n
p
a
o
n
p
a
o
n
z
z
z
z
y
y
y
y
x
x
x
x
0
3
2
3
1
2
0
1
1
0
0
0








































1
0
0
0
0
1
0
0
)
sin(
)
sin(
0
)
cos(
)
sin(
)
cos(
)
cos(
0
)
sin(
)
cos(
2
1
2
1
1
3
2
1
3
2
1
2
1
2
1
1
3
2
1
3
2
1
0
3


















A
A
A
A
T
Or
We can equate the elements in the above two matrices to try to
determine the joint angles

14. 





































1
0
0
0
0
1
0
0
)
sin(
)
sin(
0
)
cos(
)
sin(
)
cos(
)
cos(
0
)
sin(
)
cos(
1
0
0
0
2
1
2
1
1
3
2
1
3
2
1
2
1
2
1
1
3
2
1
3
2
1


















A
A
A
A
p
a
o
n
p
a
o
n
p
a
o
n
z
z
z
z
y
y
y
y
x
x
x
x
Inverse Kinematics example
Let = 1+2+3. Equate elements (1,1) and (2,1):
)
cos(
)
cos( 3
2
1 


 



x
n
)
sin(
)
sin( 3
2
1 


 



y
n
Provided that 1
)
( 2
2

 y
x n
n
We can find  using =arctan(ny, nx): 










 
x
y
n
n
1
3
2
1 tan
)
( 










)
cos(
)
sin(
)
tan(




15. Inverse Kinematics example
Next, equate elements (1,4) and (2,4):
)
cos(
)
cos( 2
1
2
1
1 

 

 A
A
px
)
sin(
)
sin( 2
1
2
1
1 

 

 A
A
py
Square both of these equations to get:
)
cos(
)
cos(
2
)
(
cos
)
(
cos 2
1
1
2
1
2
1
2
2
2
1
2
2
1
2





 



 A
A
A
A
px
)
sin(
)
sin(
2
)
(
sin
)
(
sin 2
1
1
2
1
2
1
2
2
2
1
2
2
1
2





 



 A
A
A
A
py
Adding the above two equation yields:
)
cos(
2 2
2
1
2
2
2
1
2
2

A
A
A
A
p
p y
x 



   







 



2
1
2
2
2
1
2
2
2
2
)
cos(
A
A
A
A
p
p y
x















)
cos(
)
sin(
tan
)
(
cos
1
)
sin(
)
(
cos
1
)
(
sin
2
2
1
2
2
2
2
2
2
2
2








16. Inverse Kinematics example
To solve for 1, we again reuse the following equations:
)
sin(
)
cos(
)
cos(
)
cos( 1
1
2
1
2
1
1 





 



 A
A
px
)
cos(
)
sin(
)
sin(
)
sin( 1
1
2
1
2
1
1 





 



 A
A
py
where
)
sin(
)
cos(
2
2
2
2
1




A
A
A



Let
)
sin(
)
cos(




r
r

















1
2
2
tan
r
where
 and  can be found as 2 was previously determined.
Substitute r and  into px and py to get:
)
sin(
)
sin(
)
cos(
)
cos( 1
1 


 

r
px
)
cos(
)
sin(
)
sin(
)
cos( 1
1 


 

r
py
r
p
r
p
y
x




)
sin(
)
cos(
1
1





17. Inverse Kinematics example
We can then find 1 using the known  :

























 





 1
1
1
1 tan
tan
tan
x
y
x
y
p
p
p
p




















 

x
y
p
p
1
1
1
1
1 tan
)
cos(
)
sin(
tan
)
(






Summary of inverse kinematics for the planar robot:
1) Find = 1+2+3 from











 
x
y
n
n
1
3
2
1 tan
)
( 



2) Find 2 from
    )
(
cos
1
)
sin(
,
2
)
cos( 2
2
2
2
1
2
2
2
1
2
2
2 

 








 



A
A
A
A
p
p y
x
3) Find 1 from
4) Find 3 from 3=-(1+2)

























 





 1
1
1
1 tan
tan
tan
x
y
x
y
p
p
p
p

18. Inverse Kinematics exercise 1
The gripper position and orientation for the planar robot is at:

























1
0
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
2
A
A
p
a
o
n
p
a
o
n
p
a
o
n
z
z
z
z
y
y
y
y
x
x
x
x
Where A1=3, A2=2. Find the joint angles and hence determine
the robot configuration.
  
0
0
tan
tan
)
( 1
1
3
2
1 












 

x
y
n
n




   

90
)
cos(
)
sin(
tan
1
)
(
cos
1
)
sin(
0
2
)
cos(
2
2
1
2
2
2
2
2
1
2
2
2
1
2
2
2
























 










A
A
A
A
p
p y
x

19. Inverse Kinematics exercise 1
Assuming an elbow down configuration for link 2:

90
2 


X2
Y2
2
Remember, positive 2 means clockwise rotation as follow:

20. 
0
)
( 2
1
3 


 



Results:



0
90
90
3
2
1







3
2
2
)
sin(
3
)
cos(
2
2
1
2
2
1









y
x
p
p
A
A
A
A





90
tan
tan
tan 1
1
1
1 

























 






x
y
x
y
p
p
p
p
Inverse Kinematics exercise 1
After 2 has been found:

21. Visualization of Inverse Kinematics exercise 1
To visualize the arm movement, the robot should looks like this
after rotating 1 = 90° (1 is the angle from X0 to X1 measured
along Z1).
Y1
X1
X0
Y0,
X2
Y2
X3
Y3
1=90°

22. To visualize the arm movement, the robot should looks like this
after rotating 1 = 90° (1 is the angle from X0 to X1 measured
along Z1),
and after rotating 2
= -90° (2 is the
angle from X1 to X2
measured along Z2)
Note that 3 = 0°
2=-90°
Y1
X1
X0
Y0,
X2
Y2
X3
Y3
Visualization of Inverse Kinematics exercise 1

23. The orientation of the gripper is as follow:
X3 is in the positive X0 direction
 T
n 0
0
1


Y3 is in the positive Y0 direction
 T
o 0
1
0


Z3 is in the positive Z0 direction
 T
a 1
0
0


Y1
X1
X0
Y0,
X2
Y2
X3
Y3
A1
A2
 T
A
A
p 0
1
2


Location of gripper in frame {0}

























1
0
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
2
A
A
p
a
o
n
p
a
o
n
p
a
o
n
z
z
z
z
y
y
y
y
x
x
x
x
Visualization of Inverse Kinematics exercise 1

24. Inverse Kinematics exercise 2
The gripper position and orientation for the planar robot is at:











 













1
0
0
0
0
1
0
0
5
0
0
1
0
0
1
0
1
0
0
0
z
z
z
z
y
y
y
y
x
x
x
x
p
a
o
n
p
a
o
n
p
a
o
n
Find the joint angles and hence determine the robot configuration.

90
0
1
tan
tan
)
( 1
1
3
2
1 


















 

x
y
n
n




   

0
)
cos(
)
sin(
tan
0
)
(
cos
1
)
sin(
1
2
)
cos(
2
2
1
2
2
2
2
2
1
2
2
2
1
2
2
2






















 










A
A
A
A
p
p y
x

25. 
0
)
( 2
1
3 


 



Results:



0
0
90
3
2
1






5
0
0
)
sin(
13
)
cos(
2
2
1
2
2
1








y
x
p
p
A
A
A
A





90
tan
tan
tan 1
1
1
1 

























 






x
y
x
y
p
p
p
p
Inverse Kinematics exercise 2
After 2 has been found:

26. To visualize the arm movement, the robot should finally looks
like this after rotating 1 = 90° (1 is the angle from X0 to X1
measured along Z1), 2 = 0° (2 is the angle from X1 to X2
measured along Z2) and 3 = 0°.
1=90°
Y1
X1
X0
Y0,
X2
Y2
X3
Y3
Visualization of Inverse Kinematics exercise 2

27. Inverse Kinematics – the general approach
For a more complex robot with n>3 links, it is generally not
possible to solve for all the joint variables by just equating the
elements in
To solve for the other joint angles, it may be necessary to find
T
p
a
o
n
p
a
o
n
p
a
o
n
n
z
z
z
z
y
y
y
y
x
x
x
x
0
1
0
0
0













           














 


1
0
0
0
1
0
1
2
3
1
2
2
3
1
2
0
1
1
0
1
0
3
1
0
1
z
z
z
z
y
y
y
y
x
x
x
x
p
a
o
n
p
a
o
n
p
a
o
n
T
T
T
T
T
T
T
T
T
The elements in the above equation are compared to solve for
the other joint angles. It may be necessary to repeat the process
with premultiplication of the LHS and RHS with more inverse
transformation matrices.

28. Inverse Kinematics – Summary
• Robot inverse kinematics involve finding the joint variables
given the robot arm global position and orientation
• This problem can be solved by equating the overall
transformation matrix with the matrix containing the given
information and comparing their elements
• The process may need to be repeated with premultiplication
of the above equation with inverse transformation matrices
• Although we have use a planar robot to illustrate the
concept, the approach can be applied to any robot moving
in 3-D space.

29. •The approach involves the following:
1. Solve for joint variables by comparing elements in
Inverse Kinematics – Summary
T
p
a
o
n
p
a
o
n
p
a
o
n
n
z
z
z
z
y
y
y
y
x
x
x
x
0
1
0
0
0













2. Repeat step 1 with
    ,
1
0
0
0
1
0
1
2












 
z
z
z
z
y
y
y
y
x
x
x
x
n
p
a
o
n
p
a
o
n
p
a
o
n
T
T     












 

1
0
0
0
1
0
1
1
1
2
3
z
z
z
z
y
y
y
y
x
x
x
x
n
p
a
o
n
p
a
o
n
p
a
o
n
T
T
T
etc. until all the joint variables are obtained

30. Summary
This lecture continues the discussion on the analysis of the
forward and inverse kinematics of robots.
The following were covered:
•The concept of inverse kinematics analysis of robots
•Inverse transformation matrix between coupled links
•Inverse kinematics of articulated robots in terms of the link
transformation matrices
•Robot inverse kinematics analysis using transformation matrices