Question 13.5: A gearbox is needed to provide an exact 30:1 increase in spe...

The governing equations are

With three equations and four unknown numbers of teeth, only one free choice is available. Of the two smaller gears, N_{3} \text { and } N_{5}, the free choice should be used to minimize N_{3} since a greater gear ratio is to be achieved in this stage. To avoid interference, the minimum for N_{3} is 16.

Applying the governing equations yields

Substituting N_{4}=5 N_{5} gives

If the train value need only be approximated, then this can be rounded to the nearest integer. But for an exact solution, it is necessary to choose the initial free choice for N_{3} such that solution of the rest of the teeth numbers results exactly in integers. This can be done by trial and error, letting N_{3}=17, then 18, etc., until it works. Or, the problem can be normalized to quickly determine the minimum free choice. Beginning again, let the free choice be N_{3}=1. Applying the governing equations gives

Substituting N_{4}=5 N_{5}, we find

This fraction could be eliminated if it were multiplied by a multiple of 6. The free choice for the smallest gear N_{3} should be selected as a multiple of 6 that is greater than the minimum allowed to avoid interference. This would indicate that N_{3}=18. Repeating the application of the governing equations for the final time yields

Thus,

Checking, we calculate e = (108/18)(105/21) = (6)(5) = 30.

And checking the geometry constraint for the in-line requirement, we calculate