(^672) „„„„„ A Textbook of Engineering Mechanics
Since the length of belt, that passes over the driver in one minute, is equal to the length of belt
that passes over the follower in one minute, therefore
πd 1 N 1 =πd 2 N 2
or^21
12
Nd
Nd

Note: The term N 2 /N 1 is popularly knwon as velocity ratio.
Example 33.1. It is required to drive a shaft at 620 revolutions per minute, by means of a
belt from a parallel shaft, having a pulley A 300 mm diameter on it and running at 240 revolutions
per minute. What sized pulley is required on the shaft B?
Solution. Given: Speed of the driver (N 2 ) = 620 r.p.m.; Diameter of pulley A (d 1 ) = 300
mm and speed of the pulley A (N 1 ) = 240 r.p.m.
Let d 2 = Diameter of the follower
We know that diameter of the pulley required,
1
21
2
240
300
620
N
dd
N
=× = ×
... 21
12
Nd
Nd
⎛⎞
⎜⎟=
⎜⎟
⎝⎠
Q
= 116·1 mm Ans.
33.5.VELOCITY RATIO OF A COMPOUND BELT DRIVE
Compound belt drive
Sometimes, the power is transmitted, from one shaft to another, through a number of pulleys
as shown in Fig. 33·3. Such an arrangement is known as compound belt drive.
In the figure is shown pulley 1, which drives the pulley 2. Since the pulleys 2 and 3 are keyed
to the same shaft, therefore the pulley 1 also drives the pulley 3 which, in turn, drives the pulley 4.