Epicyclic gearboxes: Wilson and Cotal

Why it matters

Every conventional automatic transmission is built from epicyclic (planetary) gear sets, and the Wilson and Cotal gearboxes are the classic examples of how such sets give several ratios simply by holding or locking different members. Being able to find speed ratios and holding torques with the tabular method or Willis's equation is essential for automatic transmissions, overdrives, hub reductions and differentials, and is a favourite GATE gear-train problem.

Key ideas

• Simple epicyclic train. A central sun gear (S), an internally toothed ring gear or annulus (R) and several planet gears (P) that mesh with both and turn on pins in a planet carrier (C). Sun, ring and carrier are coaxial; all gears stay in constant mesh.
• Two degrees of freedom. With three coaxial members free, the train is a differential. To get a definite ratio, one member is held (by a band or multi-plate brake) and power enters at one of the others and leaves at the third. If any two members are locked together (by a clutch), the whole train turns as a solid unit: direct drive, ratio 1. If nothing is held, no torque can be transmitted: neutral.
• The six combinations (input → output, member held) give: large reduction (S → C, R held); small reduction (R → C, S held); reverse (S → R, C held); and the inverses of these, which are overdrives. Car automatics use a few of these, often in compound sets (Simpson, Ravigneaux).
• Advantages. Constant mesh (no sliding gears), so ratio changes are made by applying brakes and clutches, quickly and without interrupting drive – ideal for automatic or preselective control. The load is shared between several planets, so the unit is compact and strong for its size; input and output are coaxial.
• Tooth-number condition. For a simple train with the same module throughout: Z_R = Z_S + 2 Z_P. Equal planet spacing also requires (Z_S + Z_R)/n_p to be a whole number, where n_p is the number of planets.
• Wilson preselector gearbox. Several simple epicyclic trains (typically four) are arranged in series so that their members are interconnected. Each of the lower gears and reverse is obtained by applying a band brake to the ring (annulus) drum of one train; top gear is a direct drive obtained by locking the train with a cone clutch. The driver preselects the next gear with a small lever (usually on the steering column) and then makes the change at the chosen moment by pressing and releasing the gear-change pedal, which, through a bus bar and spring-loaded toggle linkage, applies the preselected brake and releases the previous one. It was normally used with a fluid coupling, so no separate clutch pedal was needed for take-off. Bands had automatic wear adjustment. Used in pre-war and post-war luxury cars, buses, rail cars and military vehicles.
• Cotal gearbox. A French epicyclic gearbox in which the members are held or locked by electromagnetic brakes and clutches, energised by a small selector switch on the steering column, so changes are made electrically without a gear lever. Typically two epicyclic trains in series give four forward ratios (each train either direct or in reduction), and a further reversing train lets every one of those ratios also be used in reverse – the often-quoted "four forward and four reverse speeds". A conventional clutch was retained for starting. Used in 1930s–50s French luxury cars.
• Torque balance. Because the train is in equilibrium, the input, output and holding torques sum to zero; the brake must react the difference between output and input torque.

Formulas

(N_S − N_C) / (N_R − N_C) = −Z_R / Z_S

• Willis's equation. N_S, N_R, N_C: speeds of sun, ring and carrier (rev/min, with sign for direction); Z_S, Z_R: teeth on sun and ring. The minus sign is because, with the carrier fixed, sun and ring turn in opposite directions.

Ring held, sun input, carrier output: i = 1 + Z_R / Z_S Sun held, ring input, carrier output: i = 1 + Z_S / Z_R Carrier held, sun input, ring output: i = −Z_R / Z_S (reverse) Two members locked: i = 1

• i: speed ratio, input speed ÷ output speed (–). Inverting input and output inverts i (overdrive).

Z_R = Z_S + 2 Z_P

• Z_P: teeth on each planet; same module assumed.

T_in + T_out + T_h = 0, so for ideal gearing T_out = −i · T_in and T_h = −(T_in + T_out)

• T_in, T_out, T_h: input, output and holding (brake) torques acting on the train (N·m), taken with consistent sign. Numerically the holding torque equals |T_out| − |T_in| when output turns the same way as input (and their sum when it reverses).

P_in = P_out (ideal), P = 2π · N · T / 60

Worked examples

Example 1 (standard). A simple epicyclic train has Z_S = 24 and Z_R = 72. Find the planet teeth and the ratio for (a) ring held, sun input, carrier output; (b) sun held, ring input, carrier output; (c) carrier held, sun input, ring output. For case (a) with 150 N·m input, find the output and holding torques.

1. Z_P = (Z_R − Z_S)/2 = (72 − 24)/2 = 24.
2. (a) i = 1 + 72/24 = 4.0 (reduction, same direction).
3. (b) i = 1 + 24/72 = 1.333 (small reduction).
4. (c) i = −72/24 = −3.0 (reverse).
5. Torques in (a): T_out = 4.0 × 150 = 600 N·m; holding torque on the ring = 600 − 150 = 450 N·m.

Answer: Z_P = 24; ratios 4.0, 1.333 and −3.0; output 600 N·m, ring brake reacts 450 N·m. If the ring brake drum has a radius of 0.10 m, the band must provide a tangential force of 450/0.10 = 4500 N.

Example 2 (GATE level). In the same train the sun is driven at 1200 rev/min clockwise and the ring is driven at 300 rev/min anticlockwise. Find the carrier speed and direction.

1. Take clockwise as positive: N_S = +1200, N_R = −300.
2. Willis: (N_S − N_C)/(N_R − N_C) = −72/24 = −3.
3. 1200 − N_C = −3(−300 − N_C) = 900 + 3 N_C.
4. 4 N_C = 300, so N_C = +75 rev/min.
5. Check: (1200 − 75)/(−300 − 75) = 1125/(−375) = −3. ✓

Answer: carrier turns at 75 rev/min clockwise. This two-input case is how an epicyclic set acts as a differential, and how Wilson trains interact when members are interconnected.

Common mistakes

• Using 1 + Z_R/Z_S for every case; the ratio depends on which member is held, which is input and which is output.
• Dropping the minus sign in Willis's equation, or mixing directions (take one direction as positive and keep it).
• Choosing tooth numbers that cannot mesh: check Z_R = Z_S + 2 Z_P.
• Forgetting that the brake torque is the difference (not the sum) of output and input torques when they turn the same way.
• Calling the Cotal gearbox hydraulic or the Wilson gearbox fully automatic: Wilson is mechanically preselected with band brakes and a cone clutch for top; Cotal is electromagnetically operated.
• Reading the Willis ratio as Z_S/Z_R; with the carrier fixed, sun speed ÷ ring speed is −Z_R/Z_S.

For GATE ME

The core question type is the epicyclic gear train: speed of one member given the speeds of the other two (Willis or tabular method), ratio with one member fixed, and the holding torque on a fixed member or brake. Compound trains with two sets sharing a member also appear. Descriptive questions on Wilson and Cotal gearboxes are more common in university exams; know their engaging elements and why epicyclic trains suit automatic control.

Quick check

1. Z_S = 30, Z_R = 90, ring held, sun input: what is the carrier/sun speed ratio?
2. What ratio results if any two members of a simple train are locked together?
3. Which member is held to get reverse from a simple train?
4. How many teeth must each planet have if Z_S = 20 and Z_R = 70?
5. What engages top gear in a Wilson gearbox?

Answers: 1. The sun turns 4 times per carrier turn, i = 4. 2. 1:1 (direct drive). 3. The carrier. 4. 25. 5. A cone clutch that locks the train for direct drive.