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