KINEMATIC
CHAINS
&
ROBOTS
(I)
Kinematic Chains and Robots (I)
This lecture continues the discussion on a body that cannot be
treated as a single particle but as a combination of a large number
of particles. Many machines can be viewed as an assemblage of
rigid bodies called kinematic chains. This lecture continues the
discussion on the analysis of kinematic chains with focus on
robots.
After this lecture, the student should be able to:
•Appreciate the concept of kinematic pairs (joints) between rigid
bodies
•Define common (lower) kinematic pairs
•Distinguish between open and closed kinematic chains
•Appreciate the concept of forward and inverse kinematics and
dynamics analysis
•Express a finite motion in terms of the transformation matrix
General Rigid-Body Motion
General Motion
=
Translation
+
Rotation
The general motion is equivalent to that of the action of a screw,
which can be described using 4 “screw” parameters:
Axis of
rotation aˆ
The displacement
component parallel
to the direction of

u
rotation s  u s aˆ
The angle of
rotation 
Axis of rotation
passes through

the point A0
General Rigid-Body Motion
Chasles Theorem:
A general finite rigid-body motion is equivalent to
•A pure translation (sliding) along, by an amount us, and
•A pure rotation about the axis of rotation, by an amount 
The general rigid-body motion can be characterized by
the screw parameters
aˆ ,
,
us ,

A0
General Rigid-Body Motion
Given R, we can solve for the eigenvector of R corresponding to =1

to get AP

i.e. solve
 R   I  AP
0
where I is the 33 identity matrix

aˆ 
AP

AP


If given AC ( t 0 ) find


AC n  AC ( t 0 )  [ AC ( t 0 )  aˆ ] aˆ

R AC
Get  from
n


( AC n )  R ( AC n )
sin(  ) 

( AC n )
2
General Rigid-Body Motion
The previous deals with rotation. In general, if we are given the


location of a point C at 2 time instance, i.e. given C ( t 2 ) and C ( t1 )
We can denote the motion of the point C due to both rotation and

translation as u c



u c  C ( t 2 )  C ( t1 )
The displacement component of the motion parallel to aˆ is

u s  aˆ  u c

u s  u s aˆ
Finally:


R C ( t1 )  C ( t 2 ) 

R  I 
A0 
2[1  cos(  )]
T
Kinematic Pair
A kinematic pair is the coupling or joint between 2 rigid bodies
that constraints their relative motion.
The kinematic pair can be classified according to the contact
between the jointed bodies:
• Lower kinematic pairs: there is surface contact between the
jointed bodies
• Higher kinematic pairs: the contact is localized to lines,
curves, or points
Lower Kinematic Pairs


s
y


Planar joint
Revolute/pin joint
s
x
Cylindrical joint


Prismatic joint
Spherical joint
Kinematic Chain
A kinematic chain is a system of rigid bodies which are joined
together by kinematic joints to permit the bodies to move relative to
one another.
Kinematic chains can be classified as:
•Open kinematic chain: There are bodies in the chain with only one
associated kinematic joint
•Closed kinematic chain: Each body in the chain has at least two
associated kinematic joints
A mechanism is a closed kinematic chain with one of the bodies
fixed (designated as the base)
In a structure, there can be no motion of the bodies relative to one
another
Open and Closed Kinematic Chains
Base (fixed)
Closed kinematic chain
Kinematic joint
Rigid bodies
Structure
Open kinematic chain
Robots
Robots for manipulation of objects are generally designed as
open kinematic chains
These robots typically contain either revolute or prismatic joints
A Simple Planar Robot
Link (2)
Link (3): Gripper
Link (1)
Link (0): Base
This simple robot will be used throughout to illustrate simple
concepts
Forward Kinematics Analysis
Consider the following motion:
Y3
?
X3
2=90°
Y3
X3
Y0
X0
Given the dimensions of the linkages and the individual relative
motion between links, how to find the position, velocity, acceleration
of the gripper? (Forward kinematics analysis problem)
Inverse Kinematics Analysis
Again consider the following motion:
Y3
X3
2=?°
Y3
X3
Y0
X0
Given the dimensions of the linkages and the desired motion of the
gripper, how to find the individual relative motion between links?
(Inverse kinematics analysis problem)
Dynamics Analysis
Again consider the following motion:
Y3
X3
2=?° and force required
Y3
X3
Y0
X0
Given the inertia and dimensions of the linkages and the desired
motion of the gripper, how to find the individual relative motion
between links and the actuator forces to achieve this motion?
(Dynamic analysis problem)
Notation
Consider the following motion. We will associate basis ( eˆ1 , eˆ 2 , eˆ3 )
vectors with frame {b} and the (X, Y, Z) axis with frame {a}
Z-axis
Z-axis
C
C
eˆ 3
“O”
A
X-axis
eˆ1
X
eˆ 2
eˆ 2
eˆ 3
B
Y-axis
“O”
A
B
eˆ1
Y-axis
X-axis
We will denote the rotation matrix “R” that brings frame {a} to
frame {b} as ba R
Notation
Example
Z-axis
C
eˆ 2
eˆ 3
“O”
A
B
Y-axis
eˆ1
X-axis
 r11
a
r
R

b
 21
 r31
r12
r22
r32
r13   0
r23    1
 
r33   0
1
0
0
0
0

1 
Notation
Z-axis
C
eˆ 2
eˆ 3
“O”
A
B
eˆ1
Y-axis
X-axis
The position
of point “C” expressed in frame {a} is denoted

by a PC

a
For example, PC / b is the position of point “C” relative to
frame {b} expressed in terms of frame {a}
Transformation matrix
Consider the 2 frames {a} and {b}. Notice that for a vector
a
Example:
 a b 
V b R V

T
PB / b  0 1 0 

T
PB / a   1 0 0 
Z-axis
C
eˆ 2
eˆ 3
“O”
A
X-axis
B

V
eˆ1
0

a
1
R
(
P
)

b
B /b

 0
Y-axis
1
0
0
0
a
1
R

b

 0
0   0    1






0 1  0  PB / a
   
1   0   0 
1
0
0
0
0

1 
Transformation matrix
So far, in the example we used the origins of the two frames are
at the same point. What if the origin of frame {b} is at a distance
T
defined by the vector P  a P   a P
a
a
P
P 
Ob / a
Ob
Ob 1
In this case, the point “B” is given by:
We can simplify the above equation to:
where
bR
a
T   
b
 0
a
a
b
 r11


a
POb / a 
r21


1   r31

0
r12
r13
r22
r23
r32
r33
0
0
POb 1 

a
POb 2

a
POb 3 

1 
Ob 2
Ob 3



a
PB / a  POb / a  b R ( PB / b )
*
*
a
PB / a  bT ( PB / b )
 PB 1 

b

 PB / b 
PB 2


 


b

1
PB 3 




 1 
b
a
*
PB / b
T is called the Transformation Matrix of frame {b} w.r.t. frame {a}

*
T
b
b
b
PB / b   PB 1
PB 2
PB 3  .
PB / b is called the augmented vector
Example: Transformation matrix
What is the transformation matrix for the case below?
Z-axis

POb / a  0
C
eˆ 2
eˆ 3
“O”
A
B
eˆ1
0
a
1
R

b

 0
Y-axis
X-axis
bR
a
T   
b
 0
a
0

1
a
POb / a 


1  0

0
1
0
0
0
0
1
0
0
0
0

0

1
0
1
0
0
0
T
0
0

1 
Example: Transformation matrix
*
T
PB / b  0 1 0 1
*
T
PB / a   1 0 0 1
Z-axis
C
eˆ 2
eˆ 3
“O”
A
B
Y-axis
eˆ1
X-axis
*
T PB / b
a
b
0
1
 
0

0
1
0
0
0
0
1
0
0
0
1
a
T  
b
0

0
1
0
0
0
0
1
0
0
0   0    1







0 1
0
 PB / a 
       PB* / a 


0  0   0 
1


   
1  1   1 
0
0

0

1
Transformation matrix and Position
*
*
a
PB / a  bT ( PB / b )
If R is the identity matrix, then there is no rotation and the
transformation matrix will represent the pure displacement of the
origins of the frames of reference:
bR
a
T   
b
 0
a
1


a
POb / a 
0


1  0

0
0
0
1
0
0
1
0
0
POb 1 

a
POb 2

a
POb 3 

1 
a
Summary
Many machines can be viewed as an assemblage of rigid bodies
called kinematic chains. This lecture continues the the discussion
on the analysis of kinematic chains with focus on robots.
The following were covered:
•The concept of kinematic pairs (joints) between rigid bodies
•Definition of common (lower) kinematic pairs
•Open and closed kinematic chains
•The concept of forward and inverse kinematics and dynamics
analysis
•Finite motion in terms of the transformation matrix