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

The governing equations are

N_2/N_3 = 6
N_4/N_5 = 5
N_2 / N_3 = N_4 + N_5

With three equations and four unknown numbers of teeth, only one free choice is available. Of the two smaller gears, N_3 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

N_2 = 6N_3 = 6(16) = 96
N_2 + N_3 = 96 + 16 = 112 = N_4 + N_5

Substituting N_4 = 5N_5 gives

112 = 5N_5 + N_5 = 6N_5
N_5 = 112/6 = 18.67

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

N_2 = 6N_3 = 6(1) = 6
N_2 + N_3 = 6 + 1 = 7 = N_4 + N_5

Substituting N_4 = 5N_5, we find

7 = 5N_5 + N_5 = 6N_5
N_5 = 7/6

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

N_2 = 6N_3 = 6(18) = 108
N_2 + N_3 = 108 + 18 = 126 = N_4 + N_5
126 = 5N_5 + N_5 = 6N_5
N_5 = 126/6 = 21
N_4 = 5N_5 = 5(21) = 105

Thus,

N_2 = 108
N_3 = 18
N_4 = 105
N_5 = 21

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

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

geometry constraint for the in-line requirement, we calculate
N_2 + N_3 = N_4 + N_5
108 + 18 = 105 + 21
126 = 126