Analysis of a Planetary Gear Train In the epicyclic gear train illustrated in Figure 11.11, the sun gear is driven clockwise at 60 rpm and has N1 teeth, the planet gear N3 teeth, and the ring gear N4 teeth. The sun gear is the input and the arm is the output. The ring gear is held stationary. What

Question

Analysis of a Planetary Gear Train

In the epicyclic gear train illustrated in Figure 11.11, the sun gear is driven clockwise at 60 rpm and has [latex]N_{1}[/latex] teeth, the planet gear [latex]N_{3}[/latex] teeth, and the ring gear [latex]N_{4}[/latex] teeth. The sun gear is the input and the arm is the output. The ring gear is held stationary. What is the velocity of the arm?

Given: [latex]N_{1}[/latex] = 30, [latex]N_{3}[/latex] = 20, [latex]N_{4}[/latex] = 80.

Assumption: The sun gear is the first gear in the train and the ring gear is the last.

Accepted Answer

Refer to Figure 11.11.

The gear value, through the use of Equation 11.17, is

[latex] e = \pm \frac{ ext {Product of number of teeth on driver gears}}{ ext {Product of number of teeth on driver gears}}[/latex] (11.17)

[latex] e=\left(-\frac{N_1}{N_3} ight)\left(+\frac{N_3}{N_4} ight)=\left(-\frac{30}{20} ight)\left(+\frac{20}{80} ight)=-0.375 [/latex]

Observe the signs on the gearset ratios: One is an external set (−) and one an internal set (+). Substitution of this equation together with [latex]n_{F} = n_{1}[/latex] = 60 rpm and [latex]n_{L} = n_{4}[/latex] = 0 into Equation 11.18 gives

[latex] e=\frac{n_L-n_A}{n_F-n_A} [/latex] (11.18)

[latex] -0.375=\frac{n_4-n_2}{n_1-n_2}=\frac{0-n_2}{60-n_2} [/latex]

from which [latex]n_{2}[/latex] = 16.4 rpm.

Comment: The sun gear rotates 3.66 times as fast and in the same direction as the arm.

Question 11.4: Analysis of a Planetary Gear Train In the epicyclic gear tra...

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