1.2 b
0
1
T =
c
1
−s
1
0 0
s
1
c
1
0 0
0 0 1 17
0 0 0 1
1
2
T =
−s
2
−c
2
0 0
0 0 −1 0
c
2
−s
2
0 0
0 0 0 1
2
3
T =
−s
3
−c
3
0 17
c
3
−s
3
0 0
0 0 1 0
0 0 0 1
3
4
T =
−s
4
−c
4
0 4
0 0 −1 −11
c
4
−s
4
0 0
0 0 0 1
4
5
T =
c
5
−s
5
0 0
0 0 −1 0
s
5
c
5
0 0
0 0 0 1
5
6
T =
0 −1 0 0
0 0 1 9
−1 0 0 0
0 0 0 1
0
6
T =
0
1
T
1
2
T
2
3
T
3
4
T
4
5
T
5
6
T
0
P
6
is the 4
th
column of
0
6
T , can be given by,
0
P
6
=
4 c
1
s
2
s
3
− 9 c
5
(c
1
c
2
s
3
+ c
1
c
3
s
2
) − 9 s
5
(c
4
s
1
+ s
4
(c
1
c
2
c
3
− c
1
s
2
s
3
)) − 4 c
1
c
2
c
3
− 11 c
1
c
2
s
3
− 11 c
1
c
3
s
2
− 17 c
1
s
2
9 s
5
(c
1
c
4
− s
4
(c
2
c
3
s
1
− s
1
s
2
s
3
)) − 9 c
5
(c
2
s
1
s
3
+ c
3
s
1
s
2
) − 17 s
1
s
2
− 4 c
2
c
3
s
1
− 11 c
2
s
1
s
3
− 11 c
3
s
1
s
2
+ 4 s
1
s
2
s
3
17 c
2
+ 11 c
2
c
3
− 4 c
2
s
3
− 4 c
3
s
2
− 11 s
2
s
3
+ 9 c
5
(c
2
c
3
− s
2
s
3
) − 9 s
4
s
5
(c
2
s
3
+ c
3
s
2
) + 17
1
2 Inverse Kinematics
In 1.2 (b) we have calculated
0
p
6
. To work on the inverse kinematics problem, it is assumed that θ
4
≡ 0 and θ
5
≡ 0.
We have input these values in vector
0
p
6
. The reduced vector is shown in equation 1.
0
p
6
=
−9c
1
s
2+3
− c
1
(4c
2+3
+ 11s
2+3
+ 17s
2
)
−9s
1
s
2+3
− s
1
(4c
2+3
+ 11s
2+3
+ 17s
2
)
9c
2+3
+ 17c
2
+ 11c
2+3
− 4s
2+3
+ 17
Further, simplifying equation 1 we have obtained equation 2.
0
p
6
=
−c
1
(4c
2+3
+ 20s
2+3
+ 17s
2
)
−s
1
(4c
2+3
+ 20s
2+3
+ 17s
2
)
17c
2
+ 20c
2+3
− 4s
2+3
+ 17
(1)
Further, equating equation 2 to corresponding x, y, z
−c
1
(4c
2+3
+ 20s
2+3
+ 17s
2
)
−s
1
(4c
2+3
+ 20s
2+3
+ 17s
2
)
17c
2
+ 20c
2+3
− 4s
2+3
+ 17
=
x
y
z
(2)
2
