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 {2k}{(1 + 2m) sin^{2}  \phi}\left( m +\sqrt {m^{2} + (1 + 2m) sin^{2} \phi}   \right)                   (13.11)

16\sqrt{30} = 87.64 \dot{=} 88

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

e =\frac {product  of  driving  tooth  numbers}{product  of  driven  tooth  numbers}                         (13.30)

e = (88/16)(88/16) = 30.25

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.