A gearset consists of a 16-tooth pinion driving a 40-tooth gear. The diametral pitch is 2, and the addendum and dedendum are 1/P and 1.25/P, respectively. The gears are cut using a pressure angle of 20◦.
(a) Compute the circular pitch, the center distance, and the radii of the base circles.
(b) In mounting these gears, the center distance was incorrectly made \frac{1}{4} in larger. Compute the new values of the pressure angle and the pitch-circle diameters.
(a)
p=\frac{\pi}{P}=\frac{\pi}{2}=1.57 \text { in }
The pitch diameters of the pinion and gear are, respectively,
d_{P}=\frac{16}{2}=8 \text { in } \quad d_{G}=\frac{40}{2}=20 \text { in }
Therefore the center distance is
\frac{d_{P}+d_{G}}{2}=\frac{8+20}{2}=14 in
Since the teeth were cut on the 20◦ pressure angle, the base-circle radii are found to be, using r_{b}=r \cos \phi,
r_{b} \text { (pinion) }=\frac{8}{2} \cos 20^{\circ}=3.76 \text { in }
r_{b}(\text { gear })=\frac{20}{2} \cos 20^{\circ}=9.40 \text { in }
(b) Designating d_{P}^{\prime} \text { and } d_{G}^{\prime} as the new pitch-circle diameters, the \frac{1}{4}-in increase in the center distance requires that
\frac{d_{P}^{\prime}+d_{G}^{\prime}}{2}=14.250 (1)
Also, the velocity ratio does not change, and hence
\frac{d_{P}^{\prime}}{d_{G}^{\prime}}=\frac{16}{40} (2)
Solving Eqs. (1) and (2) simultaneously yields
d_{P}^{\prime}=8.143 \text { in } \quad d_{G}^{\prime}=20.357 \text { in }
Since r_{b}=r \cos \phi, the new pressure angle is
\phi^{\prime}=\cos ^{-1} \frac{r_{b} \text { (pinion) }}{d_{P}^{\prime} / 2}=\cos ^{-1} \frac{3.76}{8.143 / 2}=22.56^{\circ}