Question 13.3: A gearbox is needed to provide a 30:1 (± 1 percent) increase...

Since the ratio is greater than 10:1, but less than 100:1, a two-stage compound gear train, such as in Figure 13–28, is needed. The portion to be accomplished in each stage is \sqrt{30}=5.4772. For this ratio, assuming a typical 20° pressure angle, the minimum number of teeth to avoid interference is 16, according to Eq. (13–11). The number of teeth necessary for the mating gears is

N_{P}=\frac{2 k}{(1+2 m) \sin ^{2} \phi}\left(m+\sqrt{m^{2}+(1+2 m) \sin ^{2} \phi}\right) (13–11)

From Eq. (13–30), the overall train value is

e=\frac{\text { product of driving tooth numbers }}{\text { product of driven tooth numbers }} (13–30)

This is within the 1 percent tolerance. If a closer tolerance is desired, then increase the pinion size to the next integer and try again.