A stock helical gear has a normal pressure angle of 20◦, a helix angle of 25◦, and a transverse diametral pitch of 6 teeth/in, and has 18 teeth. Find:
(a) The pitch diameter
(b) The transverse, the normal, and the axial pitches
(c) The normal diametral pitch
(d) The transverse pressure angle
(a)
d=\frac{N}{P_{t}}=\frac{18}{6}=3 in
(b)
p_{t}=\frac{\pi}{P_{t}}=\frac{\pi}{6}=0.5236 \text { in }
p_{n}=p_{t} \cos \psi=0.5236 \cos 25^{\circ}=0.4745 \text { in }
p_{x}=\frac{p_{t}}{\tan \psi}=\frac{0.5236}{\tan 45^{\circ}}=1.123 \text { in }
(c)
P_{n}=\frac{P_{t}}{\cos \psi}=\frac{6}{\cos 25^{\circ}}=6.620 \text { teeth } / in
(d)
\phi_{t}=\tan ^{-1}\left(\frac{\tan \phi_{n}}{\cos \psi}\right)=\tan ^{-1}\left(\frac{\tan 20^{\circ}}{\cos 25^{\circ}}\right)=21.88^{\circ}