Introduction to Robotics
Kinematics
Link
Description
Kinematics Function of a link
Link length
Link twist
What are the kinematics functions of this link?
• a=7
•  = 450
• Link offset d
• Joint angle 
•
• Describe the connection between two links
Summary of the link parameters in
terms of link frames.
•
•
•
•
•
•
ai = the distance from Zi to Zi+1 measured along Xi
i = the angle between Zi and Zi+1 measured about Xi
di = the distance from Xi-1 to Xi measured along Zi
i = the angle between Xi-1 and Xi measured about Zi
We usually choose ai > 0 since it corresponds to a
distance;
• However, i , di , i are signed quantities.
There is no unique attachment of frames to links:
• 1.
When we align Zi axis with joint axis i,
two choices of the Zi direction.
• 2.
When we have intersecting joint axes
(ai=0), two choices of the Xi direction,
corresponding to choice of signs for the normal
to the plane containing Zi and Zi+1.
• 3.
When axes i and i+1 are parallel, the
choice of origin location for {i} is arbitrary
(generally chosen in order to cause di to be zero).
Three link Arm (RRR)
• Schematic
• Parallel axes
•
• Find coordinate
systems and a, , d, 
of all the three
accesses
• z is overlapping the joint’s
axis
• x is perpendicular to the two
joint’s axis
• y is …?
•
•
•
•
0 = 1 = 2 = 0
a0 = 0; a1 = L1; a2 = L2
d1 = d 2 = d3 = 0
i = i
Three link Arm : RPR mechanism
• “Cylindrical” robot – 2 joints analogous to polar
coordinates when viewed from above.
• Schematic: point – axes intersection; prismatic joint
at minimal extension
• Find coordinate systems and a, , d,  (i=3)
•
•
•
•
0 = 0; 1 = 90; 2 = 0
a0 = 0; a1 = 0; a2 = 0
d1 = 0; d2 = d2; d3 = L2;
1 = 1; 2 = 0; 3 = 3;
Schematic RRR
• Parallel / Intersect (orthogonal) axes
• Find coordinate systems and a, , d,  of all joints
• Two possible frame assignments and corresponding
parameters for the two possible choices of Z and X
directions.
Option 1
1 = -90; 2 = 0
1 = 90; 2 = 0
a1 = 0;
a2 = L2
a1 = 0;
a2 = L2
d1 = 0;
d2 = L1
d1 = 0;
d2 = -L1
1 = 1;
2 = -90+2
1 = 1; 2 = 90+2
Option 2
1 = 90; 2 = 0
1 = -90; 2 = 0
a1 = 0;
a2 = L2
a1 = 0;
a2 = L2
d1 = 0;
d2 = L1;
d1 = 0;
d2 = -L1
1 = 1;
2 = 90+2
1 = 1;
2 = -90+2
i 1
i 1
i
i 1
R
P T P T T T T P
i
R Q P i
Q P i
i 1
i
i 1
R
T T T T T
R
Q
P R
R i
T  RX  i 1 DX ai 1 RZ i DZ di 
i 1
i
T  Screw X  i 1 , ai 1 Screw Z di ,i 
i 1
i
cos( i )

sin(  ) cos( )
i
i 1
i 1

T

i
 sin(  i ) sin(  i 1 )

0

 sin(  i )
0
cos( i ) cos( i 1 )
 sin(  i 1 )
cos( i ) sin(  i 1 )
0
cos( i 1 )
0

 sin(  i 1 )di 

cos( i 1 )di 

1

ai 1
•
•
•
•
0 = 0; 1 = 90; 2 = 0
a0 = 0; a1 = 0; a2 = 0
d1 = 0; d2 = d2; d3 = L2;
1 = 1; 2 = 0; 3 = 3;
cos(1 )  sin( 1 )
 sin(  ) cos( )
1
1
0

1T 
 0
0

0
 0
0 0
0 0

1 0

0 1
1
0
1

T

2
0

0
0 

0  1  d2

1 0
0 

0 0
1 
0
0
cos( 3 )  sin(  3 )
 sin(  ) cos( )
3
3
2

T

3
 0
0

0
 0
0
0 0

1 L2 

0 1
0
T  T T T ...
0
N
0
1
1 2
2 3
N 1
N
T