Calculate gear ratios, output RPM, torque multiplication, and power loss for single and two-stage gear trains with efficiency. Includes automotive gear reference.
A transmission converts rotational speed and torque between an input shaft and an output shaft through gear meshing. The fundamental trade-off: gears that reduce speed increase torque by the same ratio (minus friction losses). A 3:1 gear reduction triples torque while cutting RPM to one-third — this is why low gears feel so powerful in a car.
The gear ratio equals the number of teeth on the driven gear divided by the teeth on the driver gear. For compound gear trains with multiple stages, the overall ratio is the product of individual stage ratios. Real-world efficiency per stage ranges from 94-99% depending on gear type: spur gears achieve 97-99%, worm gears only 40-90% depending on lead angle.
This calculator handles single and two-stage gear trains, computing output RPM, torque, power, and efficiency losses. The ratio comparison table shows how different gear ratios redistribute speed and torque from the same input, which is essential for selecting the right gear for each operating condition.
Use this calculator when you need to compare how a gear ratio redistributes speed, torque, and power through a drivetrain or gearbox.
It is useful for vehicles, robotics, machine design, and any problem where the ratio itself is easy to see but the practical output values still need to be checked.
Gear ratio: R = N_driven/N_driver. Compound: R_total = R₁ × R₂. Output RPM = Input RPM / R. Output torque = Input torque × R × η. Power = Torque × RPM × 2π/60.
Result: Ratio: 5:1, Output: 600 RPM, 960 N·m, 60.3 kW
Stage 1: 36/12 = 3:1. Stage 2: 30/18 = 1.667:1. Total ratio ≈ 5:1. At 96% efficiency, output torque = 200 × 5 × 0.96 = 960 N·m at 600 RPM, which corresponds to about 60.3 kW.
Transmission calculations are most useful when you keep the application goal in mind: some systems need low-speed torque, others need high shaft speed, and many need a compromise across a range of operating points. Looking at ratio, efficiency, and total drivetrain reduction together gives a better answer than ratio alone.
The most common errors are mixing driver and driven tooth counts, forgetting the final drive, and assuming efficiency losses are negligible in multi-stage reductions. Backlash, bearing losses, and real load dynamics also matter, so the steady-state gearbox result should be treated as a baseline rather than the whole mechanical design.
The ratio of driven gear teeth to driver gear teeth: R = N_out/N_in. R > 1 is a speed reduction (torque increase). R < 1 is an overdrive (speed increase).
No — power in equals power out (minus friction losses). Gears trade speed for torque: P = T × ω is constant. What changes is the speed-torque split.
Engines produce peak power in a narrow RPM range. Multiple gears keep the engine near its power peak across a wide range of vehicle speeds.
The last gear reduction between the transmission output and the wheels (ring and pinion in the differential). Typically 3.0-4.5 for passenger cars.
Per mesh stage: spur/helical 97-99%, bevel 95-97%, worm 40-90%. Multiply efficiencies for compound trains: η_total = η₁ × η₂ × ... × ηₙ.
Continuously variable transmissions provide infinite ratios within a range. Their efficiency is typically 85-92%, lower than manual gearboxes but they keep the engine at optimal RPM.