Innovative Systems Design and Engineering ISSN 2222-1727 (Paper) ISSN 2222-2871 (Online) Vol.4, No.10, 2013
www.iiste.org
b2 O
Hip
l1Sinθ1
b1 θ1
l1 θ2
l1Cosθ1
A
H Knee
l6
(l 6 + l 4 cosθ 4 − l 3 cosθ 3 − l1 cosθ1 )
l2 Ankle
(l 5 + l 4 sin θ 4 − l 3 sin θ 3 − l1 sin θ1 ) B
G
F
l3Cosθ3
l3
E
l3Cosθ3 θ3 C Crank l4 Spindle l5 D
Pedal Spindle
l4Sinθ4
θ4
l4Cosθ4
Figure 2: Space Analysis of thigh, θ1 and knee, θ2 angles
(l + l Cosθ 4 − l3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12 l 22 = 6 4 − 2[(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 )(l1Cosθ1 )] − 2[(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )(l1 Sinθ1 )]
{2[(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )(l1 Sinθ1 )] + 2[(l 6 + l 4 Cosθ 4 − l 3Cosθ 3 )(l1Cosθ1 )]}
{
= (l 6 + l 4 Cosθ 4 − l3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l3 Sinθ 3 ) 2 + l12 − l 22
}
(6)
(7)
If we let
A = (l 6 + l 4 Cosθ 4 − l 3 Cosθ 3 ) 2 + (l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 ) 2 + l12 − l 22
(8)
B = 2(l 5 + l 4 Sinθ 4 − l 3 Sinθ 3 )l1
(9)
C = 2(l 6 + l 4 Cosθ 4 − l3 Cosθ 3 )l1
(10)
Then, equation (7) becomes
CSinθ + BCosθ = A
(11)
Using trigonometry identity and let tan θ / 2 = t
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