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

A gearbox is needed to provide a 30:1 ( ±1 percent) increase in speed, while minimizing the overall gearbox size. Specify appropriate tooth numbers.

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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).

N_P=\frac{2k}{(1+2m)\sin ^2\phi }(m+\sqrt{m^2+(1+2m)\sin ^2\phi } )                      (13.11)

The number of teeth necessary for the mating gears is

16\sqrt{30}=87.64 \approx 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.

13.28

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