Question 11.4: Analysis of a Planetary Gear Train In the epicyclic gear tra......
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 N_1 teeth, the planet gear N_3 teeth, and the ring gear N_4 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: N_1 =30, N_3 =20, N_4 =80.
Assumption: The sun gear is the first gear in the train and the ring gear is the last.
Refer to Figure 11.11.
The gear value, through the use of Equation (11.17), is
e=\pm \frac{\text { Product of number of teeth on driver gears }}{\text { Product of number of teeth on driven gears }} (11.17)
e=\left(-\frac{N_1}{N_3}\right)\left(+\frac{N_3}{N_4}\right)=\left(-\frac{30}{20}\right)\left(+\frac{20}{80}\right)=-0.375
Observe the signs on the gear set ratios: one is an external set (−) and one an internal set (+). Substitution of this equation together with n_F =n_1 =60 rpm and n_L=n_4 =0 into Equation (11.18) gives
e=\frac{n_L-n_A}{n_F-n_A} (11.18)
-0.375=\frac{n_4-n_2}{n_1-n_2}=\frac{0-n_2}{60-n_2}
from which n_2 =16.4 rpm.
Comment: The sun gear rotates 3.66 times as fast and in the same direction as the arm.