Gear trains: simple, compound and epicyclic

Why it matters

Machine-tool headstocks, feed gearboxes, hoists, automobile gearboxes and differentials, and the planetary reducers on robots and wind turbines are all gear trains. Working out the speed and direction of every member, and the torque that must be held on a fixed member, is a routine design task and a frequent numerical question.

Key ideas

Velocity ratio of a gear pair. Two meshing gears have the same pitch-line velocity, so N₁·T₁ = N₂·T₂: speed is inversely proportional to the number of teeth (or pitch diameter). External gears turn in opposite directions; an internal (ring) gear turns in the same direction as its pinion.

Train value and speed ratio. Speed ratio = N_input/N_output; train value is its reciprocal, N_output/N_input. Always say which one you mean: a "4:1 reduction" means the input turns four times for one output turn.

Simple gear train. Each shaft carries one gear. Intermediate gears are idlers: they do not change the speed ratio (N_last/N_first = T_first/T_last), only the direction and the centre distance. An odd number of idlers keeps input and output turning the same way; an even number reverses it (for external gears).

Compound gear train. At least one shaft carries two gears that turn together. The ratio is the product of the pair ratios, so a large reduction fits in a small space. A reverted compound train has input and output shafts on the same axis (lathe back gear, clock), which needs equal centre distances: r₁ + r₂ = r₃ + r₄, and with equal modules T₁ + T₂ = T₃ + T₄.

Epicyclic (planetary) train. At least one gear axis (the planet) moves, carried on an arm (carrier) that rotates about the sun axis. A common form has sun S, planets P, ring (annulus) R and arm A. Geometry with one module gives T_R = T_S + 2·T_P. Such a train has two degrees of freedom: fix one member (or drive two) to get a definite output. Fixing different members gives different ratios from the same gears, which is why automatic gearboxes and differentials use them. Planetary trains are compact and coaxial, and share the load among several planets.

Analysis methods.

• Relative velocity (Willis) formula: in any pair of gears in the train, speeds relative to the arm obey the fixed-axis ratio.
• Tabular method: (1) lock everything to the arm and give the arm +1 rev; (2) fix the arm and give one gear +x rev, filling other gears with the fixed-axis ratios; (3) add the rows and apply the known conditions.

Torques. With no losses, the power in equals the power out, and the members' torques must sum to zero: ΣT = 0 and ΣT·N = 0. The holding (fixing) torque on the stationary member is the difference that makes the sum zero.

Formulas

N₁·T₁ = N₂·T₂ — N speed (rev/min), T number of teeth.

Simple train: N_out/N_in = T_in/T_out, sign (−1)^(number of external meshes).

Compound train: N_out/N_in = (product of driver teeth)/(product of driven teeth).

Willis: (N_last − N_A)/(N_first − N_A) = ±(product of driver teeth)/(product of driven teeth), minus sign when first and last turn opposite ways with the arm fixed.

Sun–planet–ring, ring fixed, sun input, arm output: N_S/N_A = 1 + T_R/T_S.

Geometry: T_R = T_S + 2·T_P (same module).

Torque and power: P = 2π·N·T_q/60 (P in W, N in rev/min, T_q in N·m); T_q,in + T_q,out + T_q,fix = 0.

Worked examples

Example 1 (standard). A motor of 7.5 kW at 1440 rev/min drives a compound train: a 20-tooth pinion drives a 60-tooth gear; a 25-tooth pinion on the same shaft as the 60-tooth gear drives a 75-tooth output gear. Neglect losses. Find the output speed and torque.

1. N_out = N_in × (20/60) × (25/75) = 1440 × (1/3) × (1/3) = 160 rev/min.
2. Two external meshes → the output turns the same way as the motor.
3. Input torque: T_in = 60P/(2πN) = 60 × 7500/(2π × 1440) = 49.7 N·m.
4. Output torque: T_out = T_in × 9 = 447.6 N·m.
5. N_out = 160 rev/min, T_out ≈ 448 N·m

Example 2 (GATE level). A planetary train has a 24-tooth sun and a 96-tooth fixed ring. The sun is driven at 1200 rev/min clockwise with 5 kW, and the arm is the output. Find the planet teeth, the arm speed, the absolute planet speed and the holding torque on the ring.

1. Planet teeth: T_P = (T_R − T_S)/2 = (96 − 24)/2 = 36.
2. Arm speed: N_S/N_A = 1 + T_R/T_S = 1 + 96/24 = 5 → N_A = 1200/5 = 240 rev/min clockwise.
3. Planet (Willis, sun–planet external mesh): (N_P − N_A)/(N_S − N_A) = −T_S/T_P → N_P = 240 − (24/36)(1200 − 240) = 240 − 640 = −400 rev/min, i.e. 400 rev/min counter-clockwise. Check with the ring: (N_R − N_A)/(N_P − N_A) = (0 − 240)/(−640) = 0.375 = T_P/T_R = 36/96 ✓.
4. Torques: T_S = 60 × 5000/(2π × 1200) = 39.8 N·m; arm output T_A = 39.8 × 5 = 198.9 N·m (resisting).
5. ΣT = 0 → holding torque on ring = 198.9 − 39.8 = 159.2 N·m.
6. T_P = 36, N_A = 240 rev/min CW, N_P = 400 rev/min CCW, ring holding torque ≈ 159 N·m

Common mistakes

• Treating the absolute planet speed as the speed relative to the arm, or vice versa.
• Forgetting the sign change for each external mesh (internal meshes keep the sign).
• Counting an idler's teeth in the ratio.
• Assuming the holding torque equals the input or output torque; it is their difference.
• Mixing "speed ratio" and "train value" (inverse quantities).
• Using T_R = T_S + T_P instead of T_S + 2T_P.

For GATE PI

Expect epicyclic speed problems by the tabular or Willis method (often with a fixed ring or fixed sun), holding-torque questions using power balance, ratio of a compound or reverted train, and the effect of idlers on direction. Practise setting up the table quickly and checking the result with a second gear pair.

Quick check

1. A 20-tooth gear drives a 50-tooth gear through a 35-tooth idler. Input 1000 rev/min. Output speed and direction?
2. Sun 30 teeth, planet 10 teeth. How many teeth must the ring have?
3. Ring fixed, sun 30 teeth, ring 90 teeth, sun input. What is N_S/N_A?
4. Why does an epicyclic train need one member fixed?

Answers: 1. 400 rev/min, same direction as the input; 2. 50; 3. 4; 4. it has two degrees of freedom, so one more constraint gives a definite ratio.