Question 8.2: A helical gear has a transverse diametral pitch of 12, a tra...

Transverse Circular Pitch
Use Equation (8-12):
p_{t}=\pi / P_{d}=\pi / 12=0.262 \mathrm{in}
Normal Circular Pitch
Use Equation (8-13):
p_{n}=p_{t} \cos \psi=(0.262) \cos (30)=0.227 \text { in }
Normal Diametral Pitch
Use Equation (8-16)
P_{\text {nd }}=P_{d}/ \cos \psi=12 / \cos (30)=13.856
Axial Pitch
Use Equation (8-14):
p_{x}=p_{t} / \tan \psi=0.262 / \tan (30)=0.453 \text { in }
Pitch Diameter
Use Equation (8-15):
D=N / P_{d}=28 / 12=2.333 \text { in }
Normal Pressure Angle
Use Equation (8-11):
\begin{gathered}\phi_{n}=\tan ^{-1}\left(\tan \phi_{t} \cos \psi\right) \\\phi_{n}=\tan ^{-1}\left[\tan \left(14 \frac{1}{2}\right) \cos (30)\right]=12.62^{\circ}\end{gathered}
Number of Axial Pitches in the Face Width
F / p_{x}=1.25 / 0.453=2.76 \text { pitches }
Since this is greater than 2.0, there will be full helical action.