Chapter 34 : Transmission of Power by Gear Trains „„„„„ 705

34.12.DESIGN OF SPUR WHEELS

Sometimes, the spur wheels (i.e. driver and follower) are to be designed for the given velocity
ratio and distance between centres of their shafts.

Let l= Distance between the centres of the two shafts,

T 1 = No. of teeth on the driver,

d 1 = Diameter of the pitch circle of the driver,

T 2 , d 2 = Corresponding values for the follower, and

p= Pitch of the teeth.

A little consideration will show, that the distance between the centres of two shafts,

l= 12

1

()

2

dd+ ...(i)

and V. R. =

11

22

dT

dT

= ...(ii)

From the above equations, we can conveniently find out the values of d 1 and d 2 [or T 1 and T 2
and pitch (p)]. The values of T 1 and T 2 as obtained above, may or may not be whole numbers. But in
a wheel, since the no. of its teeth is always a whole number, therefore a slight alteration must be
made in the values of l, d 1 and d 2 , so that the number of teeth in the two wheels may be a complete
number.

Example 34.4. Two parallel shafts, about 600 mm apart are to be connected by spur
wheels. One shaft is to run at 360 r.p.m. and the other at 120 r.p.m. Design the wheels, if the pitch
of the teeth is to be 25 mm.

Solution. Given: Distance between the centres of the two shafts (l) = 600 mm; Speed of the
first shaft (N 1 ) = 120 and pitch of the wheel (p) = 25 mm.

Let d 1 = Diameter of the first wheel,

T 1 = No. of teeth on the first wheel, and

d 2 , T 2 = Corresponding values for the follower,

We know that the distance between the shafts (l),

600 = 12

1

()

2

dd+

∴ d 1 + d 2 = 600 × 2 = 12 0 0 ...(i)

and^12
21

360

3

120

dN

dN

== =

or d 1 =3d 2 ...(ii)

From equations (i) and (ii), we find that

d 1 = 900 mm and d 2 = 300 mm

∴ No. of teeth on the first wheel,

T 1 =

(^1900) 113·1
25
d
p
π π×

Similarly T 2 =
(^2300) 37·7
25
d
p
π π×

Since the no. of teeth on both the wheels are to be in complete numbers, therefore let us
make the no. of teeth on the second wheel as 38 instead of 37.7. Therefore for a velocity ratio of 3
[as obtained in equation (ii)] the no. of teeth on the first wheel should be 38 × 3 = 114.