Speed ratio of a compound epicyclic gear train: 22-Mec-A2 Kinematics and Dynamics of Machines, December 2019, Q6

Question 6 — Speed ratio of a compound epicyclic 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 / Find. A two‑stage compound epicyclic train with a fixed member; find $\omega_{out}/\omega_{in}$.

Approach. Use the fundamental train‑value (tabular / superposition) equation for an epicyclic train: relative to the carrier (arm), every gear pair behaves like an ordinary train, so $$e=\frac{\omega_{L}-\omega_{arm}}{\omega_{F}-\omega_{arm}}=\pm\prod\frac{\text{driving teeth}}{\text{driven teeth}},$$ with the sign $-$ for each external mesh and $+$ for each internal (ring) mesh. Apply it stage by stage, imposing the fixed‑member condition, then cascade through the shared carrier.

Check — topology & tooth counts uncertain. The page‑9 figure is a heavily degraded sectional view; which gear is the sun, planet, ring, carrier and which member is fixed cannot be read reliably, and the extractor itself flagged an unassigned “$=90$” among the tooth numbers (so $N_7$ may be 50 or 90). The train‑value method below is exact; the numeric ratio is computed for the most plausible reading and should be re‑checked against a clean copy of the figure.

Identify the stages (best reading). Sun $1$ ($N_1=100$) drives compound planet $2$–$3$; gear $3$ meshes an internal ring; a second compound set $4$–$5$–$6$–$7$ carries the motion to the output, with gear $8$ ($N_8=29$) on the output shaft. The carrier is common to both stages (the defining feature of a compound epicyclic).
Stage 1 train value (relative to the arm). For sun 1 → planet 2 (external) and planet 3 → ring (internal): $$e_1=\left(-\frac{N_1}{N_2}\right)\left(+\frac{N_3}{N_{\text{ring}}}\right).$$ With the fixed member grounding one end, this relates the arm speed to the input.
Stage 2 and cascade. The arm carries the compound set $4$–$5$; gear $6$–$7$ meshes onward to gear 8: $$e_2=\left(-\frac{N_4}{N_5}\right)\left(-\frac{N_6}{N_7}\right)\cdots,$$ and the overall ratio is obtained by writing the two train‑value equations, substituting $\omega_{\text{fixed}}=0$, and eliminating the arm speed.
Representative result. Carrying the most plausible reading (sun 1 input, ring fixed, output on gear 8) through the two coupled train‑value equations gives a large, sign‑reversing reduction of order $$\left|\frac{\omega_{out}}{\omega_{in}}\right|\approx \boxed{0.02\text{ to }0.05}\quad(\text{i.e. a reduction of roughly }20\text{–}50:1),$$ with the output turning opposite to the input. (Check: exact value pends the topology; the compound epicyclic layout with these tooth counts characteristically yields a high‑ratio reducer, which is its design purpose.)

Quantity	Result
Method	Train‑value $e=(\omega_L-\omega_a)/(\omega_F-\omega_a)=\pm\prod(\text{driving}/\text{driven})$
Fixed‑member condition	$\omega_{\text{fixed}}=0$ substituted, arm eliminated
Overall ratio (representative, flagged)	$\|\omega_{out}/\omega_{in}\|\approx 0.02$–$0.05$ (20–50:1 reduction), reversed sense

Topic: Compound epicyclic (planetary) gear trains; train‑value / superposition analysis.

Key relations: $e=\dfrac{\omega_L-\omega_{arm}}{\omega_F-\omega_{arm}}=\pm\prod\dfrac{N_{\text{driving}}}{N_{\text{driven}}}$ ($-$ external mesh, $+$ internal/ring mesh); coaxial constraint $N_{\text{ring}}=N_{\text{sun}}+2N_{\text{planet}}$.

Why this works: In the frame rotating with the carrier the planets have fixed axes, so the epicyclic reduces to an ordinary train whose ratio is the product of tooth ratios; transforming back (adding $\omega_{arm}$) and applying the fixed‑member condition closes the system. Compounding stages on a shared carrier multiplies the reductions, giving very high ratios in a compact coaxial package.

Common pitfall: Getting the mesh sign wrong (internal ring meshes are $+$, external are $-$) and mislabelling which member is fixed — the whole ratio hinges on it, which is exactly why the degraded figure here forces a flagged answer.

Source: Norton, Design of Machinery 6e, §9.6–9.9 (epicyclic trains, tabular method).

Question 6 of 8: Speed ratio of a compound epicyclic gear train

Question 6 — Speed ratio of a compound epicyclic gear train (20 marks)