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Question 15.27: Fig.15.31 Shows diagrammatically a compound epicyclic gear t...

Fig.15.31 Shows diagrammatically a compound epicyclic gear train. Wheels A, D and E are free to rotate independently on spindle C, while B and C are compound and rotate together on spindle P, on the end of arm OP. All the teeth on different wheels have the same module. A has 12 teeth, B has 30 teeth and C has 14 teeth cut externally. Find the number of teeth on wheels D and E which are cut internally.
If the wheel A is driven clockwise at 1 rps while D is driven clockwise at 5 rps, determine the magnitude and direction of the angular velocities of arm OP and wheel E.

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Given:      z_{a}=12, z_{b}=30, z_{c}=14, n_{a}=1, n_{d}=5 .

z_{e}=z_{a}+2 z_{b}=12+60=72 .

z_{d}=z_{a}+z_{b}+z_{c}=12+30+14=56 .

Table 15.25 is used to find the speed of gears

x + y = -1.

\frac{-x}{10}+y=-5 .

x=\frac{40}{11}, y=-4.636 rpm .

Angular velocity of OP  =-2 \pi \times \frac{4.636}{60}=-0.486 rad / scw .

Angular velocity of wheel E:

y-\frac{x}{3}=-4.636-\frac{3.636}{3}=-5.848 rpm .

\omega_{e}=-2 \pi \times \frac{5.848}{60}=-0.612 rad / s cw .

Table 15.25
Revolutions of Operation
Gear E,
72		 Gear D,
56		 Gear B, C,
30, 14		 Gear A,
12		 Arm,
OP
\frac{-24}{30} \times \frac{30}{72}=\frac{-1}{3} \frac{-2}{30} \times \frac{14}{56}=\frac{-1}{10} -\frac{12}{30}=\frac{2}{5} +1 0 1. Arm fixed, +1
revolutions
given to gear A,
ccw
\frac{-x}{3} \frac{-x}{10} =\frac{-2 x}{5} +x 0 2. Multiply by x
y-\frac{x}{3} y-\frac{x}{10} y-\frac{2 x}{5} y+x y 3. Add y

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