Introduction to Robotics
Kinematics
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Link Description
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Kinematics Function of a link :
Link length
Link twist
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What are the kinematics functions
of this link?
a=7
 = 450
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Link offset d
Joint angle 
Describe the
connection
between two
links
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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
ai > 0 since it corresponds to a distance;
However, i , di , i are signed quantities.
We usually choose
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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).
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‫קובץ שקפים ‪2‬‬
‫‪8‬‬
Three link Arm
(RRR)
Schematic
Parallel axes
Find coordinate
systems and
a, , d,  of all
the three accesses
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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 = d2 = d3 = 0
i = i
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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)
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0 = 0; 1 = 90; 2 = 0
a0 = 0; a1 = 0; a2 = 0
d1 = 0; d2 = d2; d3 = L2;
1 = 1; 2 = 0; 3 = 3;
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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.
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1 = -90;
a1 = 0;
d1 = 0;
1 = 1;
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2 = 0
a2 = L2
d2 = L1;
2 = -90+2
1 = 90;
a1 = 0;
d1 = 0;
1 = 1;
2 = 0
a2 = L2
d2 = -L1
2 = 90+2
2 ‫קובץ שקפים‬
1 = 90;
a1 = 0;
d1 = 0;
1 = 1;
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2 = 0
a2 = L2
d2 = L1;
2 = 90+2
1 = -90;
a1 = 0;
d1 = 0;
1 = 1;
2 = 0
a2 = L2
d2 = -L1
2 = -90+2
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i 1
16
i 1
i
P
i 1
R
T P
i
R
Q
Q
P
P
i
i
T T T TP
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i 1
i
i 1
R
T T T T T
R
Q
P
R
R
i
T  R X  i 1 D X ai 1 RZ  i DZ d i 
i 1
i
T  Screw X i 1 , ai 1 ScrewZ d i , i 
i 1
i
cos( i )
 sin(  i )

sin(  ) cos( ) cos( ) cos( )
i
i 1
i
i 1
i 1

iT 
 sin(  i ) sin(  i 1 ) cos( i ) sin(  i 1 )

0
0

17
0
 sin(  i 1 )
cos( i 1 )
0

 sin(  i 1 )d i 

cos( i 1 )d i 

1

ai 1
2 ‫קובץ שקפים‬
0 = 0; 1 = 90; 2 = 0
a0 = 0; a1 = 0; a2 = 0
d1 = 0; d2 = d2; d3 = L2;
1 = 1; 2 = 0; 3 = 3;
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cos(1 )  sin( 1 )
 sin(  ) cos( )
1
1
0

1T 
0
 0

0
 0
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0 0
0 0

1 0

0 1
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1
0
1

2T 
0

0
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0 

0  1  d2

1 0
0 

0 0
1 
0
0
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cos( 3 )  sin(  3 )
 sin(  ) cos( )
3
3
2

3T 
0
 0

0
 0
21
0
N
0
0 0

1 L2 

0 1
0
T  T T T ...
0
1
1
2
2
3
N 1
N
T
2 ‫קובץ שקפים‬