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.

Question

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 [latex]\frac{1}{4}[/latex] in larger. Compute the new values of the pressure angle and the pitch-circle diameters.

Accepted Answer

(a)

[latex]p=\frac{\pi}{P}=\frac{\pi}{2}=1.57 ext { in }[/latex]

The pitch diameters of the pinion and gear are, respectively,

[latex]d_{P}=\frac{16}{2}=8 ext { in } \quad d_{G}=\frac{40}{2}=20 ext { in }[/latex]

Therefore the center distance is

[latex]\frac{d_{P}+d_{G}}{2}=\frac{8+20}{2}=14 in[/latex]

Since the teeth were cut on the 20◦ pressure angle, the base-circle radii are found to be, using [latex]r_{b}=r \cos \phi[/latex],

[latex]r_{b} ext { (pinion) }=\frac{8}{2} \cos 20^{\circ}=3.76 ext { in }[/latex]

[latex]r_{b}( ext { gear })=\frac{20}{2} \cos 20^{\circ}=9.40 ext { in }[/latex]

(b) Designating [latex]d_{P}^{\prime} ext { and } d_{G}^{\prime}[/latex] as the new pitch-circle diameters, the [latex]\frac{1}{4}[/latex]-in increase in the center distance requires that

[latex]\frac{d_{P}^{\prime}+d_{G}^{\prime}}{2}=14.250[/latex] (1)

Also, the velocity ratio does not change, and hence

[latex]\frac{d_{P}^{\prime}}{d_{G}^{\prime}}=\frac{16}{40}[/latex] (2)

Solving Eqs. (1) and (2) simultaneously yields

[latex]d_{P}^{\prime}=8.143 ext { in } \quad d_{G}^{\prime}=20.357 ext { in }[/latex]

Since [latex]r_{b}=r \cos \phi[/latex], the new pressure angle is

[latex]\phi^{\prime}=\cos ^{-1} \frac{r_{b} ext { (pinion) }}{d_{P}^{\prime} / 2}=\cos ^{-1} \frac{3.76}{8.143 / 2}=22.56^{\circ}[/latex]

Question 13.1: A gearset consists of a 16-tooth pinion driving a 40-tooth g...

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