Simple, compound and epicyclic gear trains

Why it matters

A manual gearbox is a set of compound trains, an automatic transmission is built from epicyclic (planetary) sets, and the differential that lets the outer wheel turn faster on a bend is an epicyclic train of bevel gears. Being able to find the speed, direction and torque of every member, and which member must be held, is the core skill for transmission design and a regular exam item.

Key ideas

Train value and speed ratio. For any gear train, the speed ratio is N_input / N_output and the train value is its inverse. For each external mesh the direction reverses; an internal (ring) gear mesh keeps the direction.

Simple gear train. Each shaft carries one gear. Intermediate gears (idlers) do not change the speed ratio, which depends only on the first and last gears: N₁ / N_last = T_last / T₁. Each idler reverses the direction, so an odd number of idlers keeps input and output in the same sense. Idlers are used to bridge a centre distance or to reverse direction (the reverse gear of a manual gearbox).

Compound gear train. At least one shaft carries two gears that rotate together. The speed ratio is the product of the individual mesh ratios, so large reductions fit in a small space: N_first / N_last = (product of driven teeth) / (product of driver teeth).

Reverted gear train. A compound train whose input and output shafts are coaxial (as in the countershaft gearbox of a car and a clock's hour and minute hands). The two centre distances must be equal: m₁(T₁ + T₂) = m₂(T₃ + T₄).

Epicyclic (planetary) gear train. At least one gear axis (the planet) moves around another (the sun) on an arm or carrier. A simple planetary set has a sun S, planets P, an internal ring (annulus) R and a carrier (arm) A. It has two degrees of freedom: fix one member (or drive two) and the others are determined. For coaxial sun, ring and planets of the same module: T_R = T_S + 2T_P.

How to solve epicyclic trains.

• Tabular method: (1) fix the arm and give one gear +1 revolution; write the resulting revolutions of every gear; (2) add y revolutions to everything (lock the train and turn it as a block); (3) apply the given conditions (a member fixed, a known speed) to find y and the unknown.
• Relative-velocity (Willis) method: relative to the arm, the train is an ordinary train, so (N_last − N_A) / (N_first − N_A) = e, where e is the train value with the arm fixed (negative for an external mesh, positive for an internal one).

Torques. If friction is neglected and the speeds are steady, the input power equals the output power and the sum of the external torques on the train is zero: T_in + T_out + T_fixed = 0. The torque needed to hold the fixed member (the brake band reaction in an automatic gearbox) follows directly.

Common planetary ratios (ring fixed, sun input, carrier output): N_S / N_A = 1 + T_R / T_S, the largest single-set reduction. Sun fixed, ring input, carrier output: N_R / N_A = 1 + T_S / T_R, a small reduction. Carrier fixed: the set is an ordinary train, N_S / N_R = −T_R / T_S (reverse). Two members locked together: direct drive, 1 : 1.

Differential. A bevel-gear epicyclic train in which the crown wheel carries the planet pinions (the arm). The two side (sun) gears drive the wheels, so N_left + N_right = 2·N_crown. When one wheel is held, the other turns at twice the crown speed. An open differential sends equal torque to both wheels, which is why one wheel on ice can stop the car.

Formulas

N₁ / N_n = T_n / T₁ (simple train)

• N: speed (rpm); T: number of teeth. Idlers cancel.

N_first / N_last = (product of driven teeth) / (product of driving teeth) (compound train)

m₁(T₁ + T₂) = m₂(T₃ + T₄) (reverted train)

• m: module of each stage (mm).

(N_last − N_A) / (N_first − N_A) = e (epicyclic, Willis)

• N_A: arm speed (rpm); e: train value with the arm held, with sign (− for each external mesh, + for internal).

T_R = T_S + 2·T_P

• Teeth on ring, sun and planet of a coaxial simple planetary set.

N_S / N_A = 1 + T_R / T_S (ring fixed)

T_in + T_out + T_fixed = 0 and T_in·N_in + T_out·N_out = 0

• External torques on the train (N·m), with sign; no friction, steady speed.

N_L + N_R = 2·N_C (differential)

• N_L, N_R: speeds of the two side gears (wheels); N_C: crown wheel (cage) speed (rpm).

Worked examples

Example 1 (standard: compound and reverted train). A motor at 1440 rpm drives gear A (20 teeth), which meshes with B (60 teeth). Gear C (18 teeth) is keyed to B's shaft and meshes with D (72 teeth). Find the speed and direction of D, and the module of the second stage if the train is reverted and the first stage has m = 4.5 mm.

1. N_D = 1440 × (20 / 60) × (18 / 72) = 1440 × 0.3333 × 0.25 = 120 rpm.
2. Two external meshes, so D turns in the same direction as A.
3. Reverted: m₁(T_A + T_B) = m₂(T_C + T_D) gives 4.5 × 80 = m₂ × 90, so m₂ = 360 / 90 = 4.0 mm.

Example 2 (GATE level: planetary set and holding torque). A planetary set has a 30-tooth sun and 30-tooth planets; the ring is held fixed. The sun is driven at 1200 rpm clockwise with an input torque of 100 N·m. Find the number of ring teeth, the carrier speed, the planet speed and the torque on the ring.

1. T_R = T_S + 2T_P = 30 + 60 = 90 teeth.
2. With the arm fixed, sun to ring: e = (−T_S / T_P) × (+T_P / T_R) = −T_S / T_R = −30 / 90 = −1/3.
3. Willis: (N_R − N_A) / (N_S − N_A) = −1/3 with N_R = 0 and N_S = 1200: −N_A = −(1200 − N_A) / 3, so 3N_A = 1200 − N_A and N_A = 300 rpm clockwise. Check: 1 + T_R / T_S = 4 = 1200 / 300.
4. Planet: (N_P − N_A) / (N_S − N_A) = −T_S / T_P = −1, so N_P − 300 = −900 and N_P = −600 rpm (600 rpm anticlockwise).
5. Output torque (no losses): T_out = −T_in·N_in / N_out = −100 × 1200 / 300 = −400 N·m, i.e. 400 N·m resisting at the carrier.
6. Holding torque: T_fixed = −(T_in + T_out) = −(100 − 400) = 300 N·m on the ring, which the brake band must supply.

Example 3 (differential). A car turns with the outer wheel at 400 rpm and the inner wheel at 360 rpm. Crown wheel speed = (400 + 360) / 2 = 380 rpm.

Common mistakes

• Counting idler teeth in the ratio of a simple train. Only the first and last gears matter; idlers only change direction.
• Forgetting the sign of e in the Willis equation: each external mesh is negative, each internal mesh positive.
• Writing the ring-fixed ratio as T_R / T_S instead of 1 + T_R / T_S.
• Assuming the fixed member carries no torque. It carries the difference between output and input torque.
• Using T_R = T_S + T_P instead of T_S + 2T_P for a coaxial set.
• For a reverted train, equating tooth sums when the modules differ. Equate centre distances.

For GATE ME

Expect speed and direction of a member in a compound or epicyclic train (often with the arm or the ring fixed), holding torque on a fixed member, the number of teeth on a ring from coaxiality, and differential speed relations. Practise the tabular method on two or three sun-planet-ring layouts, then switch to the Willis equation for speed; check every answer with T_in + T_out + T_fixed = 0.

Quick check

1. A 20-tooth gear drives a 50-tooth gear through a 35-tooth idler. Speed ratio?
2. Ring 80 teeth, sun 20 teeth, ring fixed. N_sun / N_carrier?
3. Sun 24 teeth and planet 18 teeth on a coaxial set: ring teeth?
4. One driving wheel of a car is jacked up and spun while the crown wheel is held. How do the wheels turn?
5. Input 50 N·m at 2000 rpm; output at 500 rpm, no losses. Holding torque on the fixed member?

Answers: 1. 50 / 20 = 2.5 (the idler does not count). 2. 1 + 80 / 20 = 5. 3. 24 + 36 = 60. 4. The other wheel turns at the same speed in the opposite direction (N_L + N_R = 0). 5. Output 200 N·m, so the fixed member carries 200 − 50 = 150 N·m.