Gear trains and epicyclic gearing

In short

A gear train is two or more gears in mesh used to change speed, torque or direction between shafts. When every gear axis is fixed in the frame the train is simple, compound or reverted; when one or more gear axes themselves revolve about another axis, the train is epicyclic (planetary) and can deliver very large reductions in a compact, coaxial package. The topic is scored heavily in university Theory of Machines papers and in GATE (train value, tabular method, holding torque), and it is the working core of automotive differentials and automatic transmissions, tractor transmissions, wind-turbine gearboxes, hoists, machine-tool headstocks and robot joints. Interviews at Maruti Suzuki, Tata Motors, Mahindra, Ashok Leyland, Bharat Gears, Shanthi Gears, Elecon and Suzlon routinely ask differential and planetary-ratio questions on paper.

Explained

Classification of gear trains

• Simple train: one gear per shaft, gears in series. Each mesh reverses the direction for external gears; an internal (annular) mesh keeps the direction.
• Compound train: two gears keyed to the same intermediate shaft, so one mesh's output becomes the next mesh's input at a different tooth count. Large ratios come from a product of smaller ratios.
• Reverted train: a compound train in which the first and last shafts are coaxial - used in clocks, lathe back gears and coaxial speed reducers.
• Epicyclic / planetary train: at least one gear (the planet) turns about its own axis *and* about a second axis, carried by an arm (carrier). Standard members are the sun (central external gear), planets, the annulus/ring (internal gear) and the arm.
• Bevel epicyclic: planets are bevel gears - the automobile differential and Humpage's reduction gear are the classic cases.
• Idler gear: a gear used only to bridge a distance or to reverse direction; in a *simple* train it does not change the magnitude of the overall ratio.

Train value and velocity ratio (fixed-axis trains)

• For one mesh, pitch-line velocities are equal, so N₁T₁ = N₂T₂ and N₁/N₂ = T₂/T₁ = d₂/d₁.
• Speed (velocity) ratio = N_first/N_last = (product of teeth on driven gears)/(product of teeth on drivers). Train value is its reciprocal, N_last/N_first.
• A simple train of gears A-B-C-D gives N_A/N_D = T_D/T_A: the intermediate gears cancel out of the magnitude.
• Direction: an even number of external meshes gives the same sense as the input; an odd number reverses it. Mark directions on the sketch, or carry signs (− for each external mesh, + for internal).
• Compound train example form: N_A/N_D = (T_B·T_D)/(T_A·T_C) for A-B compounded with C-D.
• Reverted-train condition (coaxial input and output, standard gears): r₁ + r₂ = r₃ + r₄. With equal modules this becomes T₁ + T₂ = T₃ + T₄; with different modules, m₁(T₁+T₂) = m₃(T₃+T₄).
• Practical reduction per external spur mesh is kept to about 1:6-1:8 (centre distance and tooth strength), which is why multi-stage trains exist.

Epicyclic trains: why they differ and how to solve them

• An epicyclic train has two degrees of freedom: fix or drive *two* members and the third is determined. Hence a planetary set gives several ratios simply by changing which member is held - the basis of automatic gearboxes.
• Geometry (coaxial sun-planet-annulus): r_A = r_S + 2r_P, so with one module T_A = T_S + 2T_P.
• Assembly condition for n equally spaced planets: (T_S + T_A)/n must be a whole number. Adjacent planets must also clear each other: (T_S + T_P)·sin(180°/n) > T_P + 2 (approximate, in teeth).
• Method 1 - tabular (superposition): (i) imagine the arm fixed and give the first gear +1 revolution, writing every member's revolutions from the ordinary train ratios; (ii) multiply the whole row by y (so the first gear turns y); (iii) add x to every member (the arm's actual rotation, since adding a rigid-body rotation changes nothing geometrically). Apply the two given conditions, solve for x and y, read off all speeds.
• Method 2 - relative velocity / algebraic: motion relative to the arm is an ordinary train, so

(N_last − N_arm)/(N_first − N_arm) = train value with the arm fixed (signed). One equation, two unknowns - plus the given data.

• Method 3 - Willis equation for the standard sun-planet-annulus set (all speeds in the same sense):

N_S·T_S + N_A·T_A = N_C·(T_S + T_A), i.e. ω_S + k·ω_A = (1+k)ω_C with k = T_A/T_S.

• Sign convention is everything: choose clockwise positive, keep it for every member, and let the algebra return negative speeds for reversed members.

The five ratios of one simple planetary set

With k = T_A/T_S (sun S, annulus A, carrier C):

• Hold nothing and the set free-wheels (neutral). Real automatic transmissions (Simpson and Ravigneaux sets) stack two or more planetary sets with clutches and brakes to get 3-10 forward ratios plus reverse from the same gears.

Torque in epicyclic trains

• In steady state (no angular acceleration) the algebraic sum of external torques on the train is zero: T_input + T_output + T_holding = 0. The torque on the fixed member is the holding / fixing / reaction torque, taken by the casing or brake band.
• The fixed member carries no power (zero speed), so for an ideal train T_in·ω_in + T_out·ω_out = 0; output torque magnitude = input torque × speed ratio.
• Useful shortcut: holding torque = input torque × (ratio − 1) when input and output rotate in the same sense.
• With efficiency η, output power = η × input power, so output torque = η × T_in × ratio, and the holding torque adjusts to keep the torque sum zero.
• Gearbox casing reaction and foundation bolt loads are computed from this balance - a standard design-office calculation.

Advantages, limitations and special trains

• Advantages of epicyclic gearing: coaxial input and output; very high ratio in small volume; load shared by 3-5 planets so torque density is high; balanced radial loads on the central shafts (low bearing load); ratio changes without moving gears into and out of mesh.
• Limitations: more parts and tighter tolerances; planet bearings are highly loaded and run at high relative speed; needs brakes/clutches for ratio changes; certain configurations recirculate power and have poor efficiency (an epicyclic reduction can be far less efficient than its "arm-fixed" basic efficiency); harder to inspect and repair.
• Ferguson's paradox: an arm carries one planet meshing with three coaxial gears of nearly equal tooth numbers; as the arm turns, one gear creeps forward, one stands still and one creeps backward - enormous ratios with few teeth, used in instruments and counters.
• Humpage's reduction gear: a compound bevel epicyclic giving a large coaxial reduction, long used in machine tools and hoists.
• Differential: a bevel epicyclic with the crown wheel as carrier - it splits drive torque equally between the two half-shafts while allowing different wheel speeds on a turn (2N_crown = N_left + N_right). A limited-slip or locking differential is added when one wheel can spin on a loose surface.
• Multi-stage industrial reducers combine a planetary stage with helical stages - the planetary sitting at the low-speed rotor end in wind turbines and at the output in mill drives - to reach 1:100 and beyond.

Indian standards and practice

• IS 2535 - basic rack and preferred modules for cylindrical gears (module series such as 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8, 10 mm); Indian design uses the metric module, not diametral pitch.
• IS 4460 (parts) - calculation of load capacity of spur and helical gears (bending and surface durability); ISO 6336 and AGMA 2001 are used alongside it for industrial and wind-turbine gearboxes.
• Tooth proportions follow the 20° full-depth involute system; accuracy grades follow the IS/ISO 1328 family. IS numbers are revised periodically - confirm the current edition on the BIS catalogue before quoting one in a design document.
• Automotive gearboxes and differentials are type-approved under the CMVR / AIS framework; gearbox noise, efficiency and oil specification (IS/API GL-4, GL-5 grades) are part of that documentation.
• Industrial gearbox ratings are quoted with a service factor; Indian suppliers (Elecon, Shanthi Gears, Premium Transmission, Bharat Gears) publish catalogues on this basis.

Key points

• Gear train - two or more meshing gears transmitting motion between shafts with a defined speed ratio.
• N₁T₁ = N₂T₂ - the single-mesh relation from which every train value is built.
• Train value - N_last/N_first = product of driver teeth / product of driven teeth; speed ratio is its reciprocal.
• Idler - changes direction and bridges distance in a simple train, but not the ratio magnitude.
• Compound train - gears paired on a common shaft; ratios multiply, giving large reductions compactly.
• Reverted train - input and output coaxial; needs r₁+r₂ = r₃+r₄ (T₁+T₂ = T₃+T₄ at equal module).
• Epicyclic train - at least one gear axis revolves; two degrees of freedom, so two conditions are needed.
• T_A = T_S + 2T_P - tooth-count condition for a coaxial sun-planet-annulus set.
• Tabular method - arm fixed +1 rev, multiply by y, add x; solve the two given conditions.
• Relative-velocity rule - (N_L − N_arm)/(N_F − N_arm) equals the arm-fixed train value.
• Willis equation - N_S T_S + N_A T_A = N_C (T_S + T_A), signs in one sense.
• Annulus fixed, sun driving - ratio 1 + T_A/T_S, output in the same sense; the most used planetary reduction.
• Carrier fixed - the set becomes an ordinary train and the output reverses.
• Lock any two members - the whole set rotates as one: 1:1 direct drive.
• Holding torque - T_in + T_out + T_hold = 0; the fixed member takes reaction but transmits no power.
• Differential - bevel epicyclic; 2N_crown = N_outer + N_inner, torque split equally.
• Assembly condition - (T_S + T_A)/n integer for n equally spaced planets.
• Efficiency - each spur mesh is roughly 97-99%, but a recirculating-power epicyclic layout can fall far below the product of its mesh efficiencies.

Formulas and examples

Example 1 - compound train, speed and direction. Gear A (20 T) on the motor shaft runs at 1,200 rpm clockwise and meshes with B (60 T); C (25 T) is keyed to B's shaft and meshes with D (75 T).

• Speed ratio = (T_B·T_D)/(T_A·T_C) = (60 × 75)/(20 × 25) = 4,500/500 = 9.
• N_D = 1,200/9 = 133.33 rpm. Two external meshes (even) → D turns clockwise, same as A.
• Ideal torque gain on D = 9 × input torque; with 98% per mesh, η = 0.98² = 0.9604, so output torque = 8.64 × input torque.

Example 2 - designing a reverted train for a ratio of 12. Input and output must be coaxial, module 3 mm for all gears, minimum 24 teeth.

• Split the ratio: A-B gives 4, C-D gives 3 (4 × 3 = 12).
• Take T_A = 24 → T_B = 96, so T_A + T_B = 120. The coaxial condition needs T_C + T_D = 120 with T_D/T_C = 3 → T_C = 30, T_D = 90 (30 + 90 = 120 ✓).
• Check speed ratio: (96 × 90)/(24 × 30) = 8,640/720 = 12 ✓ (train value = 1/12).
• Centre distance: both pairs = 3 × 120/2 = 180 mm - identical, which is exactly what the condition guarantees.

Example 3 - simple epicyclic by the tabular method. Sun S = 30 T, planets P = 20 T, annulus A = 70 T (check: 30 + 2×20 = 70 ✓). The annulus is fixed; the sun turns at 300 rpm clockwise. Find the arm and planet speeds. (Clockwise positive.)

• Annulus fixed: x − 3y/7 = 0 → x = 3y/7. Sun = 300: x + y = 300 → 3y/7 + y = 10y/7 = 300 → y = 210, x = 90.
• Arm (output) = 90 rpm clockwise. Check by formula: ratio = 1 + 70/30 = 3.3333; 300/3.3333 = 90 rpm ✓.
• Planet absolute speed = x − 1.5y = 90 − 315 = −225 rpm, i.e. 225 rpm anticlockwise; relative to the arm it spins at 315 rpm.
• Four planets can be fitted: (30 + 70)/4 = 25, a whole number ✓.

Example 4 - torque and holding torque for Example 3. The sun receives 5 kW at 300 rpm; the arm drives the load; the annulus is bolted to the casing. Assume no loss.

• ω_S = 2π × 300/60 = 31.416 rad/s → T_S = 5,000/31.416 = 159.15 N·m.
• ω_arm = 2π × 90/60 = 9.4248 rad/s → T_arm = 5,000/9.4248 = 530.5 N·m (= 159.15 × 3.3333 ✓).
• Holding torque on the annulus = 530.5 − 159.15 = 371.4 N·m, in the same sense as the input torque; the casing bolts must carry this.
• At 96% efficiency the output torque becomes 0.96 × 530.5 = 509.3 N·m and the reaction becomes 509.3 − 159.15 = 350.2 N·m.

Example 5 - compound epicyclic by the relative-velocity rule. Gear A (40 T) is fixed to the casing and coaxial with output gear D (36 T). The arm, turning at 120 rpm clockwise, carries a compound planet B (20 T, meshing A) and C (24 T, meshing D). All gears have the same module (40 + 20 = 60 = 36 + 24 ✓, so the layout closes).

• Arm-fixed train value from A to D = (T_A/T_B)(T_C/T_D) = (40/20)(24/36) = 2 × 0.6667 = +1.3333 (two external meshes → positive).
• (N_D − 120)/(N_A − 120) = 1.3333 with N_A = 0 → N_D − 120 = 1.3333 × (−120) = −160 → N_D = −40 rpm, i.e. 40 rpm anticlockwise, opposite to the arm.

Example 6 - automobile differential on a turn. Propeller shaft 1,200 rpm, final-drive pinion 15 T, crown wheel 60 T, rear track 1.5 m, and the outer wheel measures 320 rpm.

• Crown wheel (carrier) speed = 1,200 × 15/60 = 300 rpm.
• Inner wheel = 2 × 300 − 320 = 280 rpm.
• Turn radii: R_out/R_in = 320/280 = 8/7 and R_out − R_in = 1.5 m → R_in = 10.5 m, R_out = 12.0 m (check 12/10.5 = 1.1429 ✓).
• Both wheels still receive equal torque - which is why a single spinning wheel robs drive, and why locking or limited-slip differentials are fitted to tractors and off-road vehicles.

Where it is used

• Automotive driveline: manual gearboxes (compound trains with an idler for reverse), differentials in every car, truck and tractor (Tata Motors, Ashok Leyland, Mahindra), and planetary automatic transmissions in Maruti Suzuki, Hyundai, Toyota and Kia models assembled in India.
• Agricultural and construction machinery: tractor final drives and hub reduction epicyclics (Mahindra, TAFE, Sonalika), planetary travel and swing drives on excavators and cranes, winch and hoist gearboxes.
• Wind and power: multi-megawatt wind-turbine gearboxes (one planetary input stage plus two helical stages, roughly 1:100) supplied to Suzlon-class turbines; mill, kiln and conveyor reducers from Elecon, Shanthi Gears and Premium Transmission.
• Machine tools and metrology: lathe headstock and back-gear trains, feed gearboxes, indexing heads, dividing heads; clock and counter trains (hour-to-minute 12:1 reverted trains).
• Railways, aerospace and robotics: traction motor pinion-and-gear drives, turboprop and helicopter planetary reduction gearboxes (HAL), harmonic and planetary gearheads in robot joints and ISRO actuators.
• Job roles: transmission design and validation engineer, gear manufacturing/quality engineer (hobbing, shaping, grinding, over-pin and span measurement), NVH engineer, service and warranty engineer on driveline failures.

Interview questions

• Q: Does an idler gear change the speed ratio of a simple gear train? A: No - it cancels out of the magnitude; it only reverses the direction of rotation and bridges the distance between distant shafts.
• Q: What is the difference between an ordinary and an epicyclic gear train? A: In an ordinary train every gear axis is fixed in the frame; in an epicyclic train at least one gear axis revolves about another axis on a carrier, giving the train two degrees of freedom.
• Q: Why does an epicyclic train need two inputs or conditions? A: Because it has two degrees of freedom - the sun, annulus and carrier speeds are linked by one equation, so two of the three must be specified.
• Q: State the condition for a reverted gear train. A: The sum of the pitch radii of each mating pair must be equal (r₁+r₂ = r₃+r₄), which reduces to equal tooth-count sums when the modules are the same.
• Q: For a sun of 30 teeth and annulus of 90, what reduction do you get with the annulus fixed? A: 1 + 90/30 = 4, with the carrier output turning in the same direction as the sun.
• Q: How do you get reverse from a planetary set? A: Hold the carrier and take the output from the annulus (or the sun) - the set then behaves as an ordinary train and reverses direction.
• Q: What is holding torque and who carries it? A: The reaction torque on the fixed member, equal to input torque × (ratio − 1) for a same-sense drive; it is carried by the casing, brake band or clutch, and the fixed member transmits no power.
• Q: Why is torque equal on both wheels of an open differential even when their speeds differ? A: The bevel planets are free to spin, so they can only balance equal torques on the two sun gears; the speeds adjust to satisfy 2N_crown = N_left + N_right.
• Q: What happens if one wheel is on ice with an open differential? A: It spins at nearly twice the crown-wheel speed while the gripping wheel stops, and since the torques are equal the vehicle gets almost no tractive effort - the reason for limited-slip or locking differentials.
• Q: Why are planetary gearboxes preferred in wind turbines and hoists? A: Coaxial layout, three or more planets sharing the load, balanced radial forces and high torque density for the volume and weight.
• Q: Can an epicyclic train be less efficient than the product of its mesh efficiencies? A: Yes - in layouts where power recirculates internally, the circulating power multiplies the losses, so overall efficiency can be far below the basic arm-fixed value.
• Q: How are several ratios obtained from one planetary set in an automatic gearbox? A: By holding or driving different members with brakes and clutches - and locking any two members together gives direct 1:1 drive.

Common mistakes

• Inverting train value and speed ratio, then reporting the output speed as 9 times the input instead of one-ninth.
• Treating an idler as a reduction stage, or forgetting that it flips the direction of rotation.
• Solving an epicyclic train with ordinary train equations - the ratios apply only *relative to the arm*.
• Dropping or mixing signs in the tabular method, especially at the internal (planet-annulus) mesh, which does not reverse direction.
• Forgetting the arm's own rotation when converting planet speed relative to the arm into absolute speed.
• Omitting the holding torque, so the torque sum does not close and the casing reaction is understated in design calculations.
• Choosing tooth numbers that break T_A = T_S + 2T_P or the (T_S+T_A)/n integer condition, giving a set that cannot be assembled with equally spaced planets.
• Assuming the differential makes the two wheels turn at the same speed - it averages them, it does not equalise them.

Related topics

Theory of machines: mechanisms and gears · Kinematic analysis: velocity and acceleration · Belt, rope and chain drives · Flywheels and turning moment diagrams · Design of machine elements: gear design and shafts · Automobile transmission systems