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    Calculates locally in your browser · two-stage compound train
    Power transmission / gear trains

    Gear ratio

    Work out the overall ratio of a two-stage gear train from tooth counts, then follow speed and torque through each mesh. Mesh efficiency is applied per stage, not once at the end.

    DirectionReversedMeshes2Train efficiency94.1%
    Total ratio
    9.000:1
    3.00 × 3.00
    Output speed
    161.1 rpm
    From 1450 rpm input
    Output torque
    203.23 N·m
    Includes 94.1% mesh efficiency
    Output power
    3.429 kW
    0.215 kW lost in the train
    Stage-by-stage ratios
    Tooth counts, ratio and the speed and torque after each mesh.
    9.00:1 total
    Stage-by-stage ratios
    StageTeethRatioSpeed afterTorque after
    Stage 118 → 543.000:1483.3 rpm69.84 N·m
    Stage 220 → 603.000:1161.1 rpm203.23 N·m
    Total—9.000:1161.1 rpm203.23 N·m
    Power——3.644 kW in3.429 kW out

    Ratios multiply across stages; torques do too, but each mesh takes its own efficiency bite.

    Inspect the model
    Choose a quick result view, a scaled comparison, or the complete diagnostic set.
    Speed and torque after each stage
    Speed falls by the ratio while torque rises by the same factor.
    Speed and torque after each stage0.000342.20684.401026.601368.801711.0024.00InputInput483.3369.84After stage 1After stage 1161.11203.23After stage 2After stage 2Speed (rpm)Torque (N·m)
    Torque multiplication
    Output torque against total ratio at this input torque.
    Torque multiplication0.00060.70121.40182.10242.80303.500.5:10.5:11:11:12:12:13:13:15:15:18:18:112:112:1Torque (N·m)
    Canonical equations
    The exact relationships implemented by this engine.

    i = N_driven/N_driver · teeth count sets the ratio

    i_total = i₁·i₂ · multiply across stages

    ω_out = ω_in/i_total · inverse of the ratio

    T_out = T_in·i_total·ηⁿ · n meshes, each with its own loss

    engineering methodology

    A clear path from inputs to results.

    A gear pair trades speed for torque. The ratio is set entirely by tooth counts, and it multiplies across stages — two 3:1 stages give 9:1, not 6:1. This engine follows speed, torque and power through a two-stage train, applying efficiency at each mesh.

    Step-by-step methodWorked exampleAssumptions explained

    How to use

    1. Enter the tooth counts for the first stage: driver and driven.
    2. Add the second stage tooth counts, or set them equal for a single-stage train.
    3. Enter the input speed and torque.
    4. Set the efficiency per mesh. Spur gears are typically 97–98%.
    5. Read the total ratio, output speed and output torque.
    6. Check the power row: the difference between input and output is the heat the gearbox must dissipate.

    The formula

    i=NdrivenNdriver,itotal=i1i2,ωout=ωinitotal,Tout=Tin itotal ηni = \frac{N_{\text{driven}}}{N_{\text{driver}}}, \qquad i_{\text{total}} = i_1 i_2, \qquad \omega_{\text{out}} = \frac{\omega_{\text{in}}}{i_{\text{total}}}, \qquad T_{\text{out}} = T_{\text{in}}\, i_{\text{total}}\, \eta^{n}

    Reference: Standard involute spur-gear kinematics; AGMA 2001 for load ratings.

    Worked example

    18:54 followed by 20:60 at 1,450 rpm (default inputs)

    Inputs
    • Stage 1: 18 driver teeth, 54 driven.
    • Stage 2: 20 driver teeth, 60 driven.
    • Input 1,450 rpm at 24 N·m.
    • Efficiency 97% per mesh.
    Calculation path
    1. 1

      Stage 1 ratio: 54/18 = 3.0:1. Stage 2 ratio: 60/20 = 3.0:1.

    2. 2

      Total ratio: 3.0 × 3.0 = 9.0:1.

    3. 3

      Output speed: 1,450 / 9 = 161.1 rpm.

    4. 4

      Output torque: 24 × 9 × 0.97² = 203.2 N·m.

    5. 5

      Power: 24 × 1,450 × 2π/60 = 3.64 kW in; about 3.43 kW out.

    Result

    9:1 total · 161.1 rpm · 203.2 N·m · 3.64 kW in and 3.43 kW out, with 0.21 kW lost to mesh friction.

    Interpreting the result

    Torque rises ninefold and speed falls ninefold, which is the whole purpose of the train. But efficiency is applied per mesh, so two meshes at 97% give 94.1% overall — the loss compounds. That 0.21 kW becomes heat in the oil, and on a continuously running gearbox it sets the cooling requirement.

    Assumptions & limitations

    • Ideal involute geometry with no backlash, deflection or manufacturing error.
    • Efficiency is a constant per mesh, independent of load and speed. Real efficiency falls at light load.
    • The gears are external, so each mesh reverses rotation direction.
    • No idler gears are included; an idler changes direction but not ratio.
    • Shaft, bearing and seal losses are folded into the mesh efficiency.

    Frequently asked questions

    Why do ratios multiply across stages?

    Because each stage’s output speed becomes the next stage’s input. Speed is divided by i₁ then by i₂, so the overall division is i₁ × i₂.

    Why does torque rise by the ratio?

    Power is conserved (less losses), and power is torque times speed. If speed falls by a factor of nine, torque must rise by roughly nine to keep power constant.

    How much efficiency should I assume?

    97–98% per spur mesh, 95–97% for helical, 40–90% for worm depending on lead angle. Add bearing losses of about 1% per shaft.

    What does an idler gear do?

    It reverses the output direction without changing the ratio, because it acts as both driven and driver with the same tooth count.

    Does a worm drive behave the same way?

    The ratio is the number of worm-wheel teeth per worm start, which can be very high in one stage. Efficiency is much lower — typically 40–90% — and some worm drives are self-locking.

    What about backlash?

    Backlash does not change the ratio but it does add positioning error in reversing applications. It is a manufacturing tolerance, not a kinematic parameter.

    Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.