Question 3 of 6: Two-stage planetary gear train — planet count and speed ratios

Question 3: Two-stage planetary gear train — planet count and speed ratios (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. A two-stage epicyclic train, read from the printed half-section. The input shaft carries both carriers — carrier 1 (arm to planet 3) and carrier 2 (arm to planet 6). Stage 1: sun 2 ($N_2=16$), planet 3, ring 4 ($N_4=50$), where ring 4 hangs on its own free drum and brake C stops it. Stage 2: sun 5 ($N_5=30$), planet 6, ring 7 ($N_7=80$). Sun 2 and ring 7 are rigidly joined on a second free sleeve, so $\omega_2=\omega_7$, and brake D stops that sleeve. The output shaft carries sun 5. $n_{in}=800\ \text{rpm}$ CCW (viewed from the output end).

Find. (i) max equally-spaced planets per stage; (ii)–(iii) output speed for each brake.

[Figure not reproduced: Figure 3.1 — Skeleton of the two-stage train as printed in the source half-section. The input shaft rises at two axial stations into carrier 1 (arm to planet 3) and carrier 2 (arm to planet 6). Ring 4 hangs on a free drum stopped by brake C; sun 2 and ring&nb. See the official exam paper.]

Approach. Planet tooth counts come from the coaxial (concentric) condition $N_{ring}=N_{sun}+2N_{planet}$. The maximum planet count is the largest integer satisfying both the equal-spacing/assembly condition $(N_{sun}+N_{ring})/n=$ integer and the adjacent-planet clearance $\sin(\pi/n)>(N_{planet}+2)/(N_{sun}+N_{planet})$; both depend on tooth numbers only, not on how the stages are coupled. For the speeds, write the fundamental epicyclic relation for each stage, $N_s\omega_s+N_r\omega_r=(N_s+N_r)\omega_c$, with $\omega_{c1}=\omega_{c2}=n_{in}$ (both carriers are on the input shaft) and the sleeve coupling $\omega_2=\omega_7$. Each brake supplies the one remaining constraint, so the output sun 5 is determinate with either brake acting alone.

1. Planet tooth counts (coaxial condition). $$N_3=\tfrac{N_4-N_2}{2}=\tfrac{50-16}{2}=17,\qquad N_6=\tfrac{N_7-N_5}{2}=\tfrac{80-30}{2}=25.$$
2. Stage 1 — maximum planets. Clearance: $\sin(\pi/n)>\dfrac{N_3+2}{N_2+N_3}=\dfrac{19}{33}=0.576\Rightarrow n<5.1$. Assembly: $(N_2+N_4)/n=66/n$ integer $\Rightarrow n\in\{1,2,3,6,\dots\}$. The largest value satisfying both is $$\boxed{n_{1}=3\ \text{planets}}\quad(66/3=22).$$
3. Stage 2 — maximum planets. Clearance: $\sin(\pi/n)>\dfrac{N_6+2}{N_5+N_6}=\dfrac{27}{55}=0.491\Rightarrow n<6.1$. Assembly: $(N_5+N_7)/n=110/n$ integer $\Rightarrow n\in\{1,2,5,10,11,\dots\}$. Largest satisfying both: $$\boxed{n_{2}=5\ \text{planets}}\quad(110/5=22).$$
4. Brake C activated ($\omega_4=0$). Stage 1 is then carrier-in / ring-fixed / sun-out, and its sun drives the sleeve: $$N_2\omega_2+N_4(0)=(N_2+N_4)\,n_{in}\;\Rightarrow\;\omega_2=\omega_7=800\cdot\frac{66}{16}=3300\ \text{rpm (CCW)}.$$ Stage 2 now has ring 7 turning at 3300 rpm while its carrier turns at 800 rpm, so the output sun runs backwards: $$N_5\omega_5+N_7\omega_7=(N_5+N_7)\,n_{in}\;\Rightarrow\;\omega_5=\frac{110(800)-80(3300)}{30}=\boxed{-5867\ \text{rpm}\ \ (5867\ \text{rpm CW})}.$$
5. Brake D activated ($\omega_7=0$). The sleeve also carries sun 2, so holding gear 7 holds sun 2 as well ($\omega_2=0$). Stage 2 is carrier-in / ring-fixed / sun-out: $$\omega_5=\frac{N_5+N_7}{N_5}\,n_{in}=800\cdot\frac{110}{30}=\boxed{2933\ \text{rpm (CCW)}}.$$ Stage 1 is then unloaded: with its sun held and its carrier driven, the free ring 4 simply idles at $\omega_4=800(66)/50=1056$ rpm CCW.