OCTOBER 19-20, 2012 - YMCA University of Science & Technology

Proceedings of the National Conference on
Trends and Advances in Mechanical Engineering,
YMCA University of Science & Technology, Faridabad, Haryana, Oct 19-20, 2012
Appendix -A
Separating into real and imaginary parts and putting θ 1 = 0
L2 cosθ 2 + L3
cosθ3
+ L4
cosθ4
+ L1
= 0
(A.1) L
2
sin θ
2
+ L3
sin θ
3
+ L4
sin θ
4
= 0
(A.2)
Angles θ
3
and θ can be solved for input angle
4
2
Four-bar from Freudenstein’s equation:
K
1
cos θ
4
+ K
2
cos θ
2
+ K
3
= cos
Where
L1
K
1
= − , K
2
L
K
4
= −
2
L1
, K
5
L
2
= −
L1
, K
3
L
4
L1
= − , and K
6
L
3
( θ − θ )
2
2 2 2 2
L3
− L2
− L4
− L1
=
2L
L
2
4
4
2
3
(A.3)
2 2 2 2
L4
− L3
− L2
− L1
=
2L
L
θ and valid choice of vector lengths of
K
4
cos θ
3
+ K
5
cos θ
2
+ K
6
= cos
( θ − θ )
2
3
(A.4)
The solution of the equation is,
θ
θ
4 1
3 1
, 2
, 2
= 2 ta n&
= 2 tan
− 1
⎛
⎜ −
⎜
⎝
− 1
B ± B
⎛
⎜ − E ±
⎜
⎝
2
2 A
E
2 D
− 4 AC
2
− 4 DF
⎞
⎟
⎟
⎠
⎞
⎟
⎟
⎠
(A.5)
Where
A = K cosθ
− K + K + cosθ
, B= −2sinθ
, C = K + K +
2
2
2
D=
cosθ
− K
4
1
5
3
2
2
+ K cosθ
+ K , E = −2sinθ
, and F = K +
6
2
2
1
3
4
cosθ2( K2
−1)
( K5
−1) cosθ2
+ K6
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