462
K. Bhavsar et al.
2 T 3
−1 =
2 R 3
T −
2 R 3
T ∗
2 D 3
0 0 0
1
=
⎡
⎢
⎢
⎣
Cθ 3 Sθ 3 0 L 3
−Sθ 3 Cθ 3 0 0
0
0 1 0
0
0 0 1
⎤
⎥
⎥
⎦
(8)
T ∗
2 T 3
−1 =
⎡
⎢
⎢
⎣
Cθ 3 r 11 + Sθ 3 r 12 Sθ 3 r 11 + Cθ 3 r 12 r 13 −L 3 r 11 + r 14
Cθ 3 r 21 + Sθ 3 r 22 Sθ 3 r 21 − Cθ 3 r 22 r 23 −L 3 r 21 + r 24
Cθ 3 r 31 + Sθ 3 r 32 Sθ 3 r 31 − Cθ 3 r 32 r 33 −L 3 r 31 + r 34
0
0
0
1
⎤
⎥
⎥
⎦ (9)
0 T 3
2 T 3
−1 =
0 T 2 ∗
2 T 3
2 T 3
−1 = T
2 T 3
−1
0 T 2 = T ∗
2 T 3
−1
(10)
⎡
⎢
⎢
⎣
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
⎤
⎥
⎥
⎦
=
⎡
⎢
⎢
⎣
Cθ 3 r 11 + Sθ 3 r 12 Sθ 3 r 11 + Cθ 3 r 12 r 13 −L 3 r 11 + r 14
Cθ 3 r 21 + Sθ 3 r 22 Sθ 3 r 21 − Cθ 3 r 22 r 23 −L 3 r 21 + r 24
Cθ 3 r 31 + Sθ 3 r 32 Sθ 3 r 31 − Cθ 3 r 32 r 33 −L 3 r 31 + r 34
0
0
0
1
⎤
⎥
⎥
⎦
L 2 Cθ 1 Cθ 2 = −L 3 r 11 + r 14
(11)
L 2 Sθ 1 Cθ 2 = −L 3 r 21 + r 24
(12)
Squaring and adding Eqs. (11) and (12), we get
L 2 Cθ 2 = ±
(−L 3 r 11 + r 14 ) 2 + (−L 3 r 21 + r 24 ) 2
(13)
−L 2 Sθ 2 = −L 3 r 31 + r 34
(14)
θ 2 = tan
−1
−L 3 r 31 + r 34
∓
(−L 3 r 11 + r 14 ) 2 + (−L 3 r 21 + r 24 ) 2
(15)
For finding the value of θ 3 , Equating (3) and (4), we get
−S(θ 2 + θ 3 ) = r 31
(16)
−C(θ 2 + θ 3 ) = r 32
(17)