6CCE3ROS/7CCSMROB ROBOTIC SYSTEM
King’s College London
Dr Shan Luo
TUTORIAL 6

1. Get the Jacobian of the three-link manipulator below. Write in terms of a frame {4} located at
the tip of the hand and having the same orientation as frame {3}.

Fig. 1. A 3R nonplanar arm
Ans:
Solution 1:
First, we calculate 4
0 , and then differentiating
0 . Finally, (4 Θ) can be calculated as
4
0 (0 Θ).
−1 −1
0 0 0
90° 1 0
0 2 0
0 3 0
1
0 = [
1 −1 0 0
1 1 0 0
0 0 1 0
0 0 0 1
]
2
1 = [
2 −2 0 1
0 0 −1 0
2 2 0 0
0 0 0 1
]
3
2 = [
3 −3 0 2
3 3 0 0
0 0 1 0
0 0 0 1
]
4
3 = [
1 0 0 3
0 1 0 0
0 0 1 0
0 0 0 1
]

4
0 = 1
0 2
1 4
3
3
2 = [
123 −123 1 11 + 212 + 3
123 −123 −1 11 + 212
23 23 0 22
0 0 0 1
]


Solution 2:
Velocity propagation.
1 1 = [
0
0
̇1
] 1 1 = [
0
0
0
]

2 2 = 1
2 1 1 + [
0
0
̇2
] = [
2 0 2
−2 0 2
0 −1 0
] [
0
0
̇1
] + [
0
0
̇2
] = [
2̇1
2̇1
̇2
]

2 2 = 1
2 ( 1 1 +
1
1 ×
1
2) = [
2 0 2
−2 0 2
0 −1 0
] ([
0
0
0
] + [
0
1̇1
0
]) = [
0
0
−1̇1
]

3 3 = 2
3 2 2 + [
0
0
̇3
] = [
3 3 0
−3 3 0
0 0 1
] [
2̇1
2̇1
̇2
] + [
0
0
̇3
] = [
23̇1
23̇1
̇2 + ̇3
]

3 3 = 2
3 ( 2 2 +
2
2 ×
2
3) = [
3 3 0
−3 3 0
0 0 1
] ([
0
0
−1̇1
] + [
0
2̇2
−22̇1
]) = [
32̇2
32̇2
−1̇1 − 22̇1
]
4 4 =
3
3
4 4 = 3
4 ( 3 3 +
3
3 ×
3
) =
3
3 +
3
3 ×
3
4 = [
32̇2
32̇2
−1̇1 − 22̇1
] + [
0
−3(̇2 + ̇3)
−323̇1
]
= [
32̇2
32̇2+3(̇2 + ̇3)
−1̇1 − 22̇1 − 323̇1
]


(4 Θ) = [
0 32 0
0 32 3
−1 − 22 − 323 0 0
]






2. A certain two-link manipulator has the following Jacobian:
(Θ) = [
−11 − 212 −212
11 + 212 212
] .0
Ignoring gravity, what are the joint torques required in order that the manipulator will apply a static
force vector = 10̂0
0 ?
Ans:
= (Θ) 00
= [
−11 − 212 11 + 212
−212 212
] [
10
0
]
1 = −1011 − 10212
2 = −10212