Design and Analysis of Circular and Square Arm …
461
1 T 2 =
⎡
⎢
⎢
⎣
Cθ 2 −Sθ 2 0 L 2 Cθ 2
Sθ 2 Cθ 2 0 L 2 Sθ 2
0
0 1 0
0
0 0 1
⎤
⎥
⎥
⎦
2 T 3 =
⎡
⎢
⎢
⎣
Cθ 3 −Sθ 3 0 L 3 Cθ 3
Sθ 3 Cθ 3 0 L 3 Sθ 3
0
0 1 0
0
0 0 1
⎤
⎥
⎥
⎦
0 T 2 =
0 T 1 ∗
1 T 2 =
⎡
⎢
⎢
⎣
Cθ 1 Cθ 2 −Cθ 1 Sθ 2 −Sθ 1 L 2 Cθ 1 Cθ 2
Sθ 1 Cθ 2 −Sθ 1 Sθ 2 Cθ 1 L 2 Sθ 1 Cθ 2
−Sθ 2
−Cθ 2
0
−L 2 Sθ 2
0
0
0
1
⎤
⎥
⎥
⎦
(2)
0 T 3 =
0 T 2 ∗
2 T 3
=
⎡
⎢
⎢
⎣
Cθ 1 C(θ 2 + θ 3 ) −Cθ 1 S(θ 2 + θ 3 ) −Sθ 1 Cθ 1 (L 3 C(θ 2 + θ 3 ) + L 2 Cθ 2 )
Sθ 1 C(θ 2 + θ 3 ) −Sθ 1 S(θ 2 + θ 3 ) Cθ 1 Sθ 1 (L 3 C(θ 2 + θ 3 ) + L 2 Cθ 2 )
−S(θ 2 + θ 3 ) −C(θ 2 + θ 3 )
0
−(L 3 S(θ 2 + θ 3 ) + L 2 Sθ 2 )
0
0
0
1
⎤
⎥
⎥
⎦
(3)
6 Inverse Kinematics
T =
⎡
⎢
⎢
⎣
r 11 r 12 r 13 r 14
r 21 r 22 r 23 r 24
r 31 r 32 r 33 r 34
0 0 0 1
⎤
⎥
⎥
⎦
(4)
From Equating (3) and (4), we get
Cθ 1 (L 3 C(θ 2 + θ 3 ) + L 2 Cθ 2 ) = r 14
(5)
Sθ 1 (L 3 C(θ 2 + θ 3 ) + L 2 Cθ 2 ) = r 24
(6)
θ 1 = tan
−1 r 24
r 14
(7)
For finding the value of θ 2 , multiplying both sides by [
2 T 3 ]
−1