An epicyclic gear train is shown in Fig.15.30. The main driving shaft G has a gear S1 integrally mounted and driving the internal gear A1 on the casing through two intermediate gears P1 mounted on either side. The gears P, are free to revolve on arms R, which are integral with gear S2 which in turn

Question

An epicyclic gear train is shown in Fig.15.30. The main driving shaft G has a gear [latex] S_{1} [/latex] integrally mounted and driving the internal gear [latex] A_{1} [/latex] on the casing through two intermediate gears [latex] P_{1} [/latex] mounted on either side. The gears P, are free to revolve on arms R, which are integral with gear [latex] S_{2} [/latex] which in turn drives the internal gear [latex] A_{2} [/latex] on another casing through two gears [latex] P_{2} [/latex] . The driven shaft H is integral with the casing carrying the internal gear A1 and arms [latex] R_{2} [/latex] on which the gears [latex] P_{2} [/latex] are free to rotate. The casing [latex] A_{2} [/latex] and gear [latex] S_{2} [/latex] are free to rotate on shaft G.
Calculate the speed of shaft H when G rotates at 1000 rpm anticlockwise when (a) [latex]A_{2} [/latex] is stationary;
(b) [latex] A_{2} [/latex] rotates at 500 rpm clockwise.
The number of teeth on gears are: [latex] : S_{1}=S_{2}=30, A_{1}=A_{2}=90 [/latex].

Accepted Answer

Table 15.24 is used to find the speed of gears.

Table 15.24
Revolutions of							Operation
[latex]A_{2}, 90[/latex]	[latex]P_2[/latex]	[latex]S_{2}, 30[/latex]	[latex]A_{1}, 90[/latex]	[latex]P_1[/latex]	[latex]S_{1}, 30[/latex]	Arm, R	Operation
[latex]\frac{1}{3} imes \frac{1}{3}=\frac{1}{9}[/latex]	[latex]\frac{-1}{3} imes \frac{30}{z_{p 2}}[/latex] [latex]=\frac{-10}{z_{p 2}}[/latex]	[latex]\frac{-1}{3}[/latex]	[latex]\frac{-30}{90}=\frac{-1}{3}[/latex]	[latex]\frac{-30}{z_{p 1}}[/latex]	+1	0	1. Arm [latex]R_1[/latex] fixed, + 1 revolutions given to [latex]S_1[/latex], ccw
[latex]\frac{x}{9}[/latex]	[latex]\frac{-10 x}{z_{p 2}}[/latex]	[latex]\frac{-x}{3}[/latex]	[latex]\frac{-x}{3}[/latex]	[latex]\frac{-30 x}{z_{p 1}}[/latex]	+x	0	2. Multiply by x
[latex]y+\frac{x}{9}[/latex]	[latex]y-\frac{10 x}{z_{p 2}}[/latex]	[latex]y-\frac{-x}{3}[/latex]	[latex]y-\frac{-x}{3}[/latex]	[latex]y-\frac{30 x}{z_{p 1}}[/latex]	y+x	y	3. Add y

(a) [latex] A_{2} ext { fixed: } \frac{x}{9}+y=0 [/latex].

[latex] x+y=1000 [/latex].

y = -125.

Speed of shaft H = y - 125 rpm cw.

(b) [latex] \frac{x}{9}+y=-500 [/latex]

[latex] \frac{x}{9}+y=-500 [/latex].

x + y = 1000.

x = 1687.5 rpm, y = -687.5 rpm
Speed of shaft H = y = 687.5 rpm cw.

Table 15.24
Revolutions of							Operation
$A_{2}, 90$	$P_2$	$S_{2}, 30$	$A_{1}, 90$	$P_1$	$S_{1}, 30$	Arm, R	Operation
$\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$	$\frac{-1}{3} \times \frac{30}{z_{p 2}}$ $=\frac{-10}{z_{p 2}}$	$\frac{-1}{3}$	$\frac{-30}{90}=\frac{-1}{3}$	$\frac{-30}{z_{p 1}}$	+1	0	1. Arm $R_1$ fixed, + 1 revolutions given to $S_1$ , ccw
$\frac{x}{9}$	$\frac{-10 x}{z_{p 2}}$	$\frac{-x}{3}$	$\frac{-x}{3}$	$\frac{-30 x}{z_{p 1}}$	+x	0	2. Multiply by x
$y+\frac{x}{9}$	$y-\frac{10 x}{z_{p 2}}$	$y-\frac{-x}{3}$	$y-\frac{-x}{3}$	$y-\frac{30 x}{z_{p 1}}$	y+x	y	3. Add y

Question 15.26: An epicyclic gear train is shown in Fig.15.30. The main driv...

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