Question 4 of 7: Two-stage planetary gear train

Question 4: Two-stage planetary gear train (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. Input $\omega_i=1500$ rpm CCW. Stage 1: sun 1 ($N_1=21$), planet 2 ($N_2=34$), fixed ring 3 ($N_3$), carrier $H$. Stage 2: carrier $H$, planet 5 ($N_5=32$), fixed ring, output sun 4 ($N_4=81$). Equal module throughout.

Find. the general $\omega_o$ expression, its value and direction, and the maximum equally-spaced planet count per stage.

Approach. Fix the unknown ring 3 from the coaxial (equal-module) condition; apply the fixed-ring planetary reduction to stage 1 (sun in, carrier out) and the carrier-driven relation to stage 2 (carrier in, sun out); then impose the assembly (equal-spacing) and non-interference limits on the planet count.

1. Coaxial tooth counts. Equal module gives $N_3=N_1+2N_2=21+2(34)=\boxed{89}$, and the stage-2 fixed ring $N_{r2}=N_4+2N_5=81+2(32)=145$.
2. Stage 1 (fixed ring, sun in, carrier out). $$\frac{\omega_H}{\omega_i}=\frac{N_1}{N_1+N_3}=\frac{21}{110}=0.1909\quad(\text{same sense as input}).$$
3. Stage 2 (fixed ring, carrier in, sun out). From $\dfrac{\omega_4-\omega_H}{0-\omega_H}=-\dfrac{N_{r2}}{N_4}$, $$\frac{\omega_o}{\omega_H}=1+\frac{N_{r2}}{N_4}=\frac{2(N_4+N_5)}{N_4}=\frac{226}{81}=2.790\ (\text{step-up}).$$
4. (i) General expression. $$\boxed{\ \frac{\omega_o}{\omega_i}=\frac{N_1}{N_1+N_3}\cdot\frac{2(N_4+N_5)}{N_4}\ },\qquad N_3=N_1+2N_2.$$
5. (ii) Output speed. $$\omega_o=1500\left(\frac{21}{110}\right)\!\left(\frac{226}{81}\right)=\boxed{799\ \text{rpm, CCW}}$$ (same direction as the input; overall $\approx1.88{:}1$ reduction).
6. (iii) Maximum equally-spaced planets. Two limits apply. Assembly / equal spacing: $(N_{\text{sun}}+N_{\text{ring}})/n_p$ must be an integer. Non-interference: $\sin(\pi/n_p)>(N_{\text{planet}}+2)/(N_{\text{sun}}+N_{\text{planet}})$.
   Stage 1: $N_1+N_3=110$ and $(N_2+2)/(N_1+N_2)=36/55=0.655\Rightarrow n_p\le4.4$; the only divisor of $110$ that also clears interference is $n_p=\boxed{2}$.
   Stage 2: $N_4+N_{r2}=226=2\times113$ and $(N_5+2)/(N_4+N_5)=34/113=0.301\Rightarrow n_p\le10.3$; the only admissible divisor of $226$ is again $n_p=\boxed{2}$.
   So each stage takes at most two equally-spaced planets — the assembly condition, not interference, is the binding constraint for these tooth numbers.