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Question 15.24: A compound epicyclic gear train is shown in Fig.15.28. The g...

A compound epicyclic gear train is shown in Fig.15.28. The gears A, D and E are free to rotate on the axis P. The compound gear B and C rotate together on the axis Q at the end of arm F. All the gears have equal module. The number of teeth on the gears A, B and C are 18, 45 and 21 respectively. The gears D and E are annular gears. The gear A rotates at 120 rpm ccw and the gear D rotates at 450 rpm cw. Find the speed and direction of the arm and the gear E.

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\text { Given: } \quad z_{a}=18, z_{b}=45, z_{c}=21, n_{a}=120 rpm ccw , n_{d}=450 rpm cw .

\frac{z_{e}}{2}=\frac{z_{a}}{2}+z_{b} .

z_{e}=z_{a}+2 z_{b}=18+90=108 .

z_{e}=z_{a}+z_{b}+z_{c}=18+45+21=84 .

Table 15.22 is used to find the speed of gears.

Table 15.22
Revolutions of Operation
Gear E,108 Gear D, 84 Gear B, C, 45, 21 Gear A, 18 Arm

\frac{-z_{a}}{z_{b}} \times \frac{z_{b}}{z_{e}}

=\frac{-18}{45} \times \frac{45}{108}

=\frac{-1}{6}

 \frac{-z_{a}}{z_{b}} \times \frac{z_{c}}{z_{c}}

=\frac{-18}{45} \times \frac{21}{84}

=\frac{-1}{10}

+\frac{z_{a}}{z_{b}}=\frac{-18}{45}=\frac{-2}{5} +1 0 1. Arm
fixed, + 1
revolutions
given to gear
A, ccw
\frac{-x}{6} \frac{-x}{10} =\frac{-2 x}{5} +x 0 2. Multiply by x
y-\frac{-x}{6} y-\frac{x}{10} y-\frac{2 x}{5} y+x y 3. Add y

x+y=120 .

\frac{-x}{10}+y=-450 .

x=518.2 rpm , y=-398.2 rpm .

n_{f}=y=-398.2 rpm , \text { i.e., } cw .

n_{e}=y-\frac{x}{6}=-398.2-\frac{518.2}{6}=-484.57 rpm , \text { i.e., } cW .

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