Compound planetary gear train with two ring brakes: 22-Mec-A2 Kinematics and Dynamics of Machines, May 2018, Q4

Question 4: Compound planetary gear train with two ring brakes (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two coaxial simple-planetary stages on a common carrier. Stage 1: sun $B$ (input, $N_B=60$), planet $C$ ($N_C=31$), ring $D$ (held by its brake). Stage 2: sun $F$ (output, $N_F=25$), planet $G$ ($N_G=48$), ring $H$ (held). Both rings grounded; carrier free and common. $\omega_{in}=1450$ rpm CCW.

Find. $\omega_{out}$ (sense and magnitude).

Compound planetary: input sun $B$ and output sun $F$ on the common axis, rings $D$ and $H$ grounded by friction pads, planets $C$ and $G$ on one free common carrier. With equal module, $N_D=N_B+2N_C=122$ and $N_H=N_F+2N_G=121$.

Approach. With equal module the ring teeth follow from the coaxial (radii-add) condition, $N_{\text{ring}}=N_{\text{sun}}+2N_{\text{planet}}$. For each stage use the fixed-carrier train value $e=-N_{\text{sun}}/N_{\text{ring}}$ in the epicyclic equation $(\omega_{\text{ring}}-\omega_c)/(\omega_{\text{sun}}-\omega_c)=e$. The common carrier speed links the two stages.

Ring tooth counts. Equal module ⇒ pitch radius $\propto$ teeth, and ring $=$ sun $+$ 2 planet: $$N_D=N_B+2N_C=60+2(31)=122,\qquad N_H=N_F+2N_G=25+2(48)=121.$$
Stage 1: carrier speed from the fixed ring $D$. Train value sun→planet→ring $e_1=-N_B/N_D$. With $\omega_D=0$: $$\frac{0-\omega_c}{\omega_B-\omega_c}=-\frac{N_B}{N_D}\;\Rightarrow\;\omega_c=\omega_B\,\frac{N_B}{N_B+N_D}=1450\cdot\frac{60}{182}=\boxed{478.0\ \text{rpm (CCW)}}.$$
Stage 2: output sun $F$ from the same carrier and fixed ring $H$. $$\frac{0-\omega_c}{\omega_F-\omega_c}=-\frac{N_F}{N_H}\;\Rightarrow\;\omega_F=\omega_c\,\frac{N_F+N_H}{N_F}=478.0\cdot\frac{146}{25}.$$
Combine. $$\omega_{out}=\omega_F=1450\cdot\frac{N_B}{N_B+N_D}\cdot\frac{N_F+N_H}{N_F}=1450\cdot\frac{60}{182}\cdot\frac{146}{25}=\boxed{2.79\times10^{3}\ \text{rpm}}.$$

Both stages preserve sense (a fixed-ring planetary turns its carrier and its sun the same way), so the output is CCW, the same sense as the input. The device is a step-up: the reduction sun $B\!\to$ carrier ($\times0.330$) is over-run by the amplification carrier $\to$ sun $F$ ($\times5.84$), netting $\omega_{out}/\omega_{in}=1.93$.

Check: the exam schematic is a bare section skeleton and does not print $N_D,N_H$; they are inferred from the equal-module coaxial condition ($N_{\text{ring}}=N_{\text{sun}}+2N_{\text{planet}}$), and $B,F$ are read as the on-axis suns (input/output shafts) with $D,H$ the braked rings. If instead each stage’s output were taken at its carrier, the train would be a step-down ($\approx82$ rpm); the sun-to-sun reading used here matches the input/output shafts drawn on the common axis.

Quantity	Value
Ring teeth $N_D,\ N_H$	$122,\ 121$
Common carrier speed $\omega_c$	$478\ \text{rpm}$ (CCW)
Output speed $\omega_{out}$	$\approx 2792\ \text{rpm}$, CCW (same sense as input)

Topic: Epicyclic (planetary) gear analysis by the train-value / tabular method with two grounded rings on a shared carrier.

Key relations: coaxial condition $N_{\text{ring}}=N_{\text{sun}}+2N_{\text{planet}}$; fixed-carrier train value $e=\pm\prod(\text{driving}/\text{driven})$ (sun→planet→ring gives $-N_s/N_r$); epicyclic equation $(\omega_L-\omega_c)/(\omega_F-\omega_c)=e$.

Why this works: the epicyclic equation is just the fixed-carrier gear ratio written in the carrier’s rotating frame; grounding a ring supplies the extra equation that makes each stage’s two remaining speeds proportional. The common carrier is the single variable shared between stages, so eliminating it chains the input sun to the output sun.

Common pitfall: using $N_{\text{ring}}=N_{\text{sun}}+N_{\text{planet}}$ (correct is $+2N_{\text{planet}}$), and dropping the internal-gear sign so the sense comes out wrong. Always confirm the coaxial tooth condition before trusting an inferred ring count.

Question 4 of 7: Compound planetary gear train with two ring brakes

Question 4: Compound planetary gear train with two ring brakes (20 marks)