2
Solution to Problem 1. (30 points).
(i) 14 Points
. The diametral pitch is specified as
P
= 2.5 teeth/inch
(1)
From Table 6.2, see page 258, the addendum for the gear and the pinion are
11
0.40 in
2.5
==
=
a
P
(2)
Also, from Table 6.2, see page 258, the dedendum for the gear and the pinion are
1.25
1.25
0.50 in
2.5
===
d
P
(3)
From Eq. (6.4), see page 254, the circular pitch is
1.2566 in/tooth
2.5
=
p
P
π
(4)
The base pitch, see Eq. (6.6), page 261, is
b
=c
o
s
φ
pp
(5a)
which can be written as
b
1.2566 cos 23
1.1567 in/tooth
o
p
(5b)
(ii) Note that the gear is the driver and will be denoted as gear 2. The pinion is the driven gear and will
be denoted as gear 3. This choice is in agreement with Figure 6.20, page 269.
Therefore, the radius of the pitch circle of gear 2, see Eq. (6.1), page 252, is
2
2
48
9.6 in
22
x
2
.
5
=
N
R
P
(6a)
The radius of the pitch circle of pinion 3, see Eq. (6.1), page 252, is
3
3
22
4.4 in
x
2
.
5
=
N
R
P
(6b)
The radius of the base circle of gear 2, see Eq. (6.15), page 272, is
cos
9.6 cos23
8.8368 in
=φ
=
=
o
rR
(7a)
The radius of the base circle of pinion 3, see Eq. (6.15), page 272, is
33
cos
4.4 cos23
4.0502 in
=
=
o
(7b)