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Q. 8.2

A helical gear has a transverse diametral pitch of 12, a transverse pressure angle of 14(1/2)°, 28 teeth, a face width of 1.25 in, and a helix angle of 30°. Compute the transverse circular pitch, normal circular pitch, normal diametral pitch, axial pitch, pitch diameter, and the normal pressure angle. Compute the number of axial pitches in the face width.

Verified Solution

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.