CSCI 5551 - Introduction to Intelligent Robotics
Assignment 3 Solution Key
Problem 1
Figure 1: Reachable workspace of the tip of link 2.
CS&E Department, University of Minnesota Page 1 of 6
CSCI 5551 - Introduction to Intelligent Robotics
Problem 2
Frames
Figure 2: Frame assignment. Note that we leave frame 6 oriented so that Z6 and X6 match
Z0 and X0, respectively.
CS&E Department, University of Minnesota Page 2 of 6
CSCI 5551 - Introduction to Intelligent Robotics
DH Parameter Table
i αi−1 ai−1 di θi
1 0 0 0 θ1
2 0 L 0 ω2 − 90
3 -90 0 L2 0
4 90 0 0 ω3
5 -90 0 L3 0
6 90 0 0 90
Forward Kinematics
0
1T =

cos(θ1) -sin(θ1) 0 0
sin(θ1) cos(θ1) 0 0
0 0 1 0
0 0 0 1

1
2T =

cos(ω2 − 90) -sin(ω2 − 90) 0 L
sin(ω2 − 90) cos(ω2 − 90) 0 0
0 0 1 0
0 0 0 1

2
3T =

1 0 0 0
0 0 1 L2
0 −1 0 0
0 0 0 1

3
4T =

cos(ω3) -sin(ω3) 0 0
0 0 −1 0
sin(ω3) cos(ω3) 0 0
0 0 0 1

4
5T =

1 0 0 0
0 0 1 L3
0 −1 0 0
0 0 0 1
 ; 56T =

0 −1 0 0
0 0 −1 0
1 0 0 0
0 0 0 1

0
6T =

cos(ω2 + ω3 + θ1) -sin(ω2 + ω3 + θ1) 0 L2cos(ω2 + θ1) + Lcos(θ1) + L3cos(ω2 + ω3 + θ1)
sin(ω2 + ω3 + θ1) cos(ω2 + ω3 + θ1) 0 L2sin(ω2 + θ1) + Lsin(θ1) + L3sin(ω2 + ω3 + θ1)
0 0 1 0
0 0 0 1

CS&E Department, University of Minnesota Page 3 of 6
CSCI 5551 - Introduction to Intelligent Robotics
Inverse Kinematics - Algebraic Method
For a goal x, y, and ϕ, we see our solution takes the form:
0
6T =

c123 -s123 0 x
s123 c123 0 y
0 0 1 0
0 0 0 1

So we have:
ϕ = θ1 + ω2 + ω3 (1)
x = L2cos(ω2 + θ1) + Lcos(θ1) + L3cos(ω2 + ω3 + θ1) (2)
y = L2sin(ω2 + θ1) + Lsin(θ1) + L3sin(ω2 + ω3 + θ1) (3)
Then from equation 1:
θ1 = ϕ− ω2 − ω3 (4)
Which means that in equations 2 and 3, the only unknowns are L2 and L3 (2 equations,
2 unknowns). Solving for L2 and L3 gives:
L2 =
−(yc123 − xs123 + Lc1s123 − Ls1c123)
c12s123 − c123s12 (5)
L3 =
yc12 − s12x+ Lc1s12 − Lc12s1
c12s123 − c123s12 (6)
Inverse Kinematics - Geometric Method
The same result as above can be obtained with the Geometric Method. You can use the
geometric method to find x, y and ϕ.
Jacobian
J =

∂x
∂θ1
∂x
∂L2
∂x
∂L3
∂y
∂θ1
∂y
∂L2
∂y
∂L3
∂ϕ
∂θ1
∂ϕ
∂L2
∂ϕ
∂L3

Therefore:
J =
−L2s12 − Ls1 − L3s123 c12 c123L2c12 + L3c123 + Lc1 s12 s123
1 0 0

CS&E Department, University of Minnesota Page 4 of 6
CSCI 5551 - Introduction to Intelligent Robotics
Problem 3
For any given target position in the reachable workspace at whatever elevation the d1 pa-
rameter sets the two rotary joints there are two solutions.
Figure 3: Question 3 explanation. Top view of manipulator.
CS&E Department, University of Minnesota Page 5 of 6
CSCI 5551 - Introduction to Intelligent Robotics
Problem 4
0J =
−c1d3 0 −s1−s1d3 0 c1
0 1 0

Singularities exist where det(0J) = 0, therefore:
det(0J) = det
−c1d3 0 −s1−s1d3 0 c1
0 1 0
 = −c1d3 · (0− c1)− s1 · (−s1d3 − 0) = 0
∴ d3 · (c21 + s21) = d3 ∗ 1 = 0
Therefore singularities exist only where d3 = 0.
CS&E Department, University of Minnesota Page 6 of 6