INTRODUCTION 15
αω ωω ωω
θ
3332323 232 3 ωαω
2
2
2
(^3232)
= d ( ) = ( ) = + = + ( ) = +^2
dt
d
dt
hh
dh
dt
h
dh
d
̇ ̇hh′
αω ωω ωω
θ
4442424 242 4 ωαω
2
2
2
(^4242)
= d ( ) = ( ) = + = + ( ) = +^2
dt
d
dt
hh
dh
dt
h
dh
d
̇ ̇hh′ (1.8)
The second order kinematic coefficients hh 34 ′′, can be determined from Eq. (1.7) as follows:
dX
d
rr
d
d
2
2
2 2233
3
2
2
: – cos – cos
θ
θθ
θ
θ
⎛
⎝
⎞
⎠
- sin 33 – cos – sin = 0
(^23)
2
2 44
4
2
2
44
(^24)
2
r 2
d
d
r
d
d
r
d
d
θ
θ
θ
θ
θ
θ
θ
θ
θ
⎛
⎝⎜
⎞
⎠⎟
⎛
⎝
⎞
⎠
⎛
⎝⎜
⎞
⎠⎟
⇒– cos r 2233 θθ θ θ θ – cos r hr 32 – sin 3334 4hr′′ – cos hr 42 – sin 444 h = 0
dY
d
r r hr hr hr h
2
2
2 22333
(^233344)
4
: – sin – sin + cos – sin^2 + cos 444 = 0
θ
θθ θ θ′′θ
- sin – sin
cos cos
cos cos
sin sin
(^3)
4
334 4
3344
–1
3344
334 4
3
2
4
2
′
′
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
h
h
rr
rr
rr
rr
h
h
θθ
θθ
θθ
θθ
- sin – sin
cos cos
cos
sin
33 44
3344
–1
22
22
rr
rr
r
r
θθ
θθ
θ
θ
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥ (1.9)
Eqs. (1.7), (1.8) and (1.9) can be implemented into a computer code and the positions, velocities and
accelerations of all the linkages can be determined at each desired instant.
Example 1.2 (Slider-Crank Mechanism)
Consider the slider-crank mechanism in Figure 1.4 with the indicated vector loop. The vector loop
equation can be written as
rrrr 23
?
14
?
+ – – = 0
vvvvvI
Figure 1.4 A slider-crank mechanism
A
2
O
1
3
B
1
4
r 2
r 1
θ 2
2 π – θ 3
y
x
r 4
r 3