Gear Ratios: Trading Speed for Torque

The Constraint at the Teeth

Two spur gears mesh only if their teeth are the same size — same module m (in millimetres, the pitch diameter divided by the tooth count) or, in imperial terms, the same diametral pitch. Where the teeth touch, they cannot slide through one another, so the pitch-line velocity at the contact point must be identical for both wheels:

• v = ω₁·r₁ = ω₂·r₂, where r is the pitch radius and ω the angular velocity in rad/s.
• Because pitch radius is proportional to tooth number, r ∝ N_teeth, this rearranges to the defining kinematic law: ω₁/ω₂ = r₂/r₁ = N₂/N₁.
• The gear ratio is that tooth-count ratio: N = N_driven/N_driver = N_out/N_in.

A pinion of 12 teeth driving a gear of 48 teeth gives N = 48/12 = 4. The big wheel turns a quarter as fast as the little one. Nothing about the material, the power, or the torque enters yet — this is pure geometry, a rolling-without-slipping constraint identical in spirit to the wheel-and-axle. Every additional mesh in a train multiplies these ratios: a two-stage reducer with stages of 4 and 5 delivers an overall N = 20, so a 1800 rpm motor emerges at 90 rpm.

Why Torque Goes Up When Speed Goes Down

The speed law is kinematics; the torque law is a conservation statement. In an ideal, lossless gear pair the mechanical power delivered to the input shaft equals the power taken from the output shaft. Rotational power is the product of torque and angular velocity:

• P = τ·ω (units: N·m × rad/s = W).
• Setting P_in = P_out gives τ_in·ω_in = τ_out·ω_out.
• Rearranging: τ_out/τ_in = ω_in/ω_out = N.

So the output torque is multiplied by exactly the same factor N that the speed is divided by. This is the mechanical-advantage heart of the machine. Equivalently, you can get it from statics alone: the tangential force F transmitted through the mesh is common to both gears (Newton's third law at the tooth face), so τ = F·r means the torque on each shaft scales with its own radius — again the ratio N. A 2 N·m motor through a 4:1 reduction yields 8 N·m at a quarter of the speed. The gearbox is not an engine; it is an impedance-matching device for rotational power, letting a fast, weak, high-efficiency prime mover drive a slow, stubborn load.

A Worked Example: Sizing a Winch

Suppose you must lift a 500 kg load at 0.10 m/s using a drum of radius 0.15 m, driven by an electric motor that runs most efficiently at 1500 rpm. Work backward through the chain:

• Force and torque at the drum: F = mg = 500 × 9.81 = 4905 N. Torque τ_drum = F·r = 4905 × 0.15 = 736 N·m.
• Drum speed: the rope pays off at v = ω·r, so ω_drum = 0.10/0.15 = 0.667 rad/s ≈ 6.4 rpm.
• Required ratio: N = ω_motor/ω_drum = 1500/6.4 ≈ 235:1 — a job for a worm drive or a multi-stage planetary reducer.
• Motor torque (ideal): τ_motor = τ_drum/N = 736/235 ≈ 3.1 N·m.
• Power check: lifting power = F·v = 4905 × 0.10 = 490 W; motor power = τ_motor·ω_motor = 3.1 × 157 ≈ 490 W. Consistent — power is conserved.

Real hardware adds losses. At 90% overall efficiency the motor must supply 490/0.90 ≈ 545 W and about 3.4 N·m. The gearbox has let a modest 0.7 kW motor lift half a tonne — the entire point of trading speed for torque.

The Controlling Variables and Their Scales

Three quantities govern what a gear train can do, and each has a characteristic scale:

• Ratio N — set by tooth counts. A single spur mesh is practically limited to about 6:1 or 7:1 before the size disparity becomes awkward; big reductions are stacked in stages or done with worms (up to ~300:1 in one pass) or planetary sets (typically 3:1–12:1 per stage).
• Reflected inertia — a load inertia I_load seen through a ratio N appears at the motor as I_load/N². Kinetic energy ½Iω² is frame-invariant, and since ω scales by N, inertia must scale by N² to keep it consistent. This is why a 100:1 gearbox makes a heavy load feel 10 000× lighter to the motor — and why backlash and torsional wind-up matter so much in precision servos.
• Efficiency η — friction at the tooth flanks and bearings. Well-cut spur or helical gears run at 97–99% per mesh, so even a five-stage train keeps ~90%. A worm drive, which slides rather than rolls, may fall to 50–70% and can be self-locking (the load cannot back-drive the input), a feature exploited in hoists and machine-tool feeds.

The module and face width, not the ratio, set the torque capacity: tooth bending stress follows the Lewis equation σ = F/(m·b·Y), so a bigger module or wider tooth carries more load before it snaps.

Planetary Gears and Continuous Ranges

The most versatile arrangement is the epicyclic (planetary) gearset: a central sun gear, an outer ring (annulus), and a set of planets on a common carrier. Its kinematics obey the Willis equation:

• ω_sun + N_r/N_s · ω_ring − (1 + N_r/N_s)·ω_carrier = 0, where N_r and N_s are ring and sun tooth counts.
• Hold the ring fixed and drive the sun: the carrier output ratio is (1 + N_r/N_s):1 — e.g. a 20-tooth sun and 80-tooth ring give 5:1 in a package that is compact, coaxial, and load-sharing across several planets.

Because a planetary set has two degrees of freedom, choosing which member is fixed, driven, or output yields several distinct ratios from one gearset — the basis of the classic automatic transmission, where clutches and brakes reconfigure the same planetary stack into different speeds. Automotive continuously variable transmissions (CVTs) take this further, using a belt on variable-diameter pulleys to sweep the ratio smoothly, keeping an engine near its most efficient rpm regardless of road speed.

Where the Trade Shows Up in the Real World

Once you see the speed-for-torque exchange, it appears everywhere:

• Bicycles: a 53-tooth chainring driving an 11-tooth sprocket gives a chain ratio of 4.8:1 — the rear wheel turns nearly five times per crank revolution for high-speed cruising. Shift to a 34/32 combination (≈1.06:1) and the wheel barely outpaces the cranks, but torque at the rear axle jumps by more than 4× for climbing.
• Automobiles: first gear near 3.5:1 combined with a 3.7:1 final drive gives ~13:1 total, multiplying a 250 N·m engine to over 3 kN·m at the wheels for launch; overdrive top gear below 0.8:1 drops the engine to a fuel-sipping 2000 rpm at 110 km/h.
• Wind turbines: a three-stage gearbox steps a 15 rpm rotor up to ~1500 rpm for the generator — a ~100:1 increase that trades the rotor's enormous torque for the speed a synchronous generator needs.
• Clocks and instruments: tiny reduction trains turn a fast escapement into the once-per-12-hours crawl of an hour hand — an overall ratio in the thousands, achieved with modules under 0.3 mm.

Subtleties, Limits, and Misconceptions

The clean N-in, N-out picture hides several traps:

• Gears don't create energy. The torque multiplication is real, but power out never exceeds power in — indeed it is always slightly less. A 'torque multiplier' wrench and a lever obey the same conservation law; there is no free lunch, only rebalanced coordinates.
• Direction reverses. Two externally meshing gears spin in opposite senses. An idler gear between them restores the original direction without changing the ratio — its tooth count cancels out.
• Backlash and elasticity. Real teeth have clearance (typically 0.02–0.2 mm) and finite stiffness, so a train winds up under load and lashes on reversal — a dominant error source in robotics and CNC positioning, mitigated by anti-backlash or harmonic (strain-wave) drives.
• Speed increase costs stability. Running a train backward (overdrive) is possible only if it isn't self-locking; worm drives generally cannot be back-driven at all.
• Ratio ≠ diameter ratio exactly. It is the tooth-count ratio that is exact; pitch diameters follow only if the module matches, which is why you can never mesh gears of different modules.

The deepest point: a gearbox is an impedance transformer, the mechanical analogue of an electrical transformer where turns ratio n plays the role of N. Voltage↔torque and current↔angular velocity map cleanly, and just as a transformer conserves VI, the gearbox conserves τω.

Speed reduction vs. speed increase: what a gear ratio does to each quantity

Quantity	Reduction (N > 1)	Overdrive (N < 1)	Why
Output speed ω_out	÷ N (slower)	× 1/N (faster)	ω_out = ω_in / N
Output torque τ_out	× N (higher)	÷ N (lower)	τ_out = N·τ_in (ideal)
Power P	unchanged	unchanged	τω conserved, minus losses
Reflected inertia	÷ N² (lower)	× N² (higher)	energy ½Iω² invariance
Typical use	cranes, winches, 1st gear	highway top gear	match load to prime mover