Devise a gear train for a recorder for a precision measuring instrument. The input is a shaft that rotates at exactly 3600 rpm. The output speed must be exactly 11.25 rpm. Use 20°, full-depth, involute teeth; no fewer than 17 teeth; and no more than 150 teeth in any gear.

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Target [latex]TV[/latex]
[latex]TV_{nom} = 3600/11.25 = 320[/latex]
Maximum Single [latex]VR[/latex]
[latex]VR_{max} = 150/17 = 8.824[/latex]
Maximum [latex]TV[/latex] for Double Reduction
[latex]TV_{max} = (8.824)^2 = 77.8[/latex] (too low)
Maximum [latex]TV[/latex] for Triple Reduction
[latex]TV_{max} = (8.824)^3 = 687[/latex] (okay)
Design a triple-reduction gear train such as that shown in Figure 8–48. The train value is the product of the three individual velocity ratios:
[latex]TV = (VR_1)(VR_2)(VR_3)[/latex]

If we can find three factors of 320 that are within the range of the possible ratio for a single pair of gears, they can be specified for each velocity ratio.
Factors of 320
One method is to divide by the smallest prime numbers that will divide evenly into the given number, typically 2, 3, 5, or 7. For example,

320/2 = 160
160/2 = 80
80/2 = 40
40/2 = 20
20/2 = 10
10/2 = 5

Then the prime factors of 320 are 2, 2, 2, 2, 2, 2, and 5. We desire a set of three factors, which we can find by combining each set of three “2” factors into their product. That is,
(2)(2)(2) = 8
Then the three factors of 320 are
(8)(8)(5) = 320
Now let the number of teeth in the pinion of each pair be 17. The number of teeth in the gears will then be (8)(17) = 136 or (5)(17) = 85. Finally, we can specify

[latex]N_A = 17[/latex] [latex]N_C = 17[/latex] [latex]N_E = 17[/latex]
[latex]N_B = 136[/latex] [latex]N_D = 136[/latex] [latex]N_F = 85[/latex]

Question 8.12: Devise a gear train for a recorder for a precision measuring...

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