Determine the train value for the train shown in Figure 8–44. If the shaft carrying gear A rotates at 1750 rpm clockwise, compute the speed and the direction of the shaft carrying gear E

Question

Determine the train value for the train shown in Figure 8–44. If the shaft carrying gear [latex]A[/latex] rotates at 1750 rpm clockwise, compute the speed and the direction of the shaft carrying gear [latex]E[/latex]

Accepted Answer

Look first at the direction of rotation. Remember that a gear pair is defined as any two gears in mesh (a driver and a follower). There are actually three gear pairs:
Gear [latex]A[/latex] drives gear [latex]B: A[/latex] rotates clockwise; [latex]B[/latex], counterclockwise.
Gear [latex]C[/latex] drives gear [latex]D: C[/latex] rotates counterclockwise; [latex]D[/latex], clockwise.
Gear [latex]D[/latex] drives gear [latex]E: D[/latex] rotates clockwise; [latex]E[/latex], counterclockwise.
Because gears [latex]A[/latex] and [latex]E[/latex] rotate in opposite directions, the train value is negative. Now

[latex]TV = -(VR_1)(VR_2)(VR_3)[/latex]

In terms of the number of teeth,

[latex]TV= -\frac{N_B}{N_A} \frac{N_D}{N_C}\frac{N_E}{N_D}[/latex]

Note that the number of teeth in gear [latex]D[/latex] appears in both the numerator and the denominator and thus can be canceled. The train value then becomes

[latex]TV= -\frac{N_B}{N_A}.\frac{N_E}{N_C}= -\frac{70}{20}.\frac{54}{18}= -\frac{3.5}{1}.\frac{3}{1}= -10.5[/latex]

Gear [latex]D[/latex] is called an idler. As demonstrated here, it has no effect on the magnitude of the train value, but it does cause a direction reversal. The output speed is then found from

[latex]TV = n_A/n_E[/latex]
[latex]n_E = n_A/TV = (1750[/latex] rpm[latex])/(-10.5) = -166.7[/latex] rpm (counterclockwise)

Question 8.8: Determine the train value for the train shown in Figure 8–44...

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