Calculation
M₂ = M₁ · i · η
Reduction increases torque with ratio; losses reduce usable output.
M₂ = M₁·i·η
Gearbox Output Torque Calculator
Reduction increases torque with ratio; losses reduce usable output.
MINTSI
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Result
Select a target and calculate.
Understand the inputs
- Output torque M₂ — The torque usably available at the output shaft. As a result it follows from M₁·i·η.
- Input torque M₁ — The torque applied at the input shaft. Source: a torque sensor, or the drive motor's rated torque per its nameplate.
- Gear ratio i — The gear ratio i = n₁/n₂. Source: the gearbox nameplate, or computed from the tooth counts.
- Efficiency η — The gearbox efficiency – the fraction of input power that reaches the output after friction and meshing losses (0.95 = 95%). Source: the gearbox datasheet; typical value for one spur-gear stage about 0.96–0.99, markedly lower for a worm drive.
Example
100 N·m · 2.5 · 95% = 237.5 N·m.
Assumptions and limits
Constant ratio; load peaks and gear-face factors are excluded.
Understand Gearbox Output Torque Calculator
This calculator finds the missing quantity among output torque M₂, input torque M₁, gear ratio i and efficiency η from the other three. It shows why losses in multi-stage gearboxes multiply rather than add — and why several individually good stage efficiencies can still combine into a noticeably lower overall efficiency.
What does this quantity describe?
A gearbox's output torque follows from input torque multiplied by the ratio and the efficiency: M₂ = M₁·i·η. Efficiency η bundles all losses from gear-mesh friction, bearings and gear efficiency into a single figure.
For a two-stage planetary gearbox with stage efficiencies of 96% and 95%, overall efficiency is not their average but their product: 0.96 · 0.95 ≈ 91.2%. Although each stage individually counts as efficient, losses across multiple stages add up noticeably — an effect easy to underestimate when selecting multi-stage gearboxes.
Formula and variables
M₂ = M₁ · i · η
M₂ = M₁ · i · ηM₁ = M₂ / (i · η)η = M₂ / (M₁ · i)
| Symbol / input | Meaning |
|---|---|
| Output torque M₂ | The torque usably available at the output shaft. As a result it follows from M₁·i·η. |
| Input torque M₁ | The torque applied at the input shaft. Source: a torque sensor, or the drive motor's rated torque per its nameplate. |
| Gear ratio i | The gear ratio i = n₁/n₂. Source: the gearbox nameplate, or computed from the tooth counts. |
| Efficiency η | The gearbox efficiency – the fraction of input power that reaches the output after friction and meshing losses (0.95 = 95%). Source: the gearbox datasheet; typical value for one spur-gear stage about 0.96–0.99, markedly lower for a worm drive. |
Choose the inputs correctly
Enter three of the four quantities to find the fourth. Efficiency η is dimensionless and typically ranges 0.9-0.99 per gearbox stage. For multi-stage gearboxes, η is already the combined efficiency of all stages together, not a single stage.
How to use the calculator
Select the target quantity and enter the three known values with units. For multi-stage gearboxes, first find overall efficiency as the product of individual stage efficiencies before entering it into this calculator.
Worked example
Given M₁ = 100 N·m, i = 2.5 and η = 0.95, substitution gives M₂ = 100 · 2.5 · 0.95 = 237.5 N·m.
Understand the result and units
An output torque of 237.5 N·m sits below the ideal, loss-free value of 250 N·m (100 · 2.5), since 5% of transmitted power is dissipated as heat. Downstream component sizing must always use the real, efficiency-adjusted output torque.
Torque in newton-metres (N·m); ratio i and efficiency η are dimensionless ratios.
Why do losses multiply in multi-stage gearboxes?
Each gearbox stage passes on only a fraction η of the power arriving at it; the rest is lost as heat. With two stages in series, the second stage therefore doesn't receive the full input power, but only the share η1 of it — and the second stage applies its own efficiency η2 to this already-reduced power. The result is the product η1·η2, not their sum or average. With three or more stages, this effect continues and can push overall efficiency well below the seemingly good individual values.
Typical applications
The formula is used to size geared motors, conveyor drives, machine tools, and to estimate actually usable torque behind multi-stage gearboxes.
Assumptions, limits and common mistakes
The formula assumes efficiency constant at the operating point. In reality, efficiency depends on load, speed, temperature and lubrication, and can fall below the rated efficiency at partial load or unfavourable conditions.
Common mistake: For multi-stage gearboxes, do not add or average individual stage efficiencies — multiply them instead: two stages at 95% each combine to about 90.25%, not 95%. Also, do not use the ideal output torque (without η) as the real sizing basis for downstream components.
Frequently asked questions
How do I calculate the overall efficiency of a three-stage gearbox?
As the product of the three individual efficiencies, e.g. η_total = η1·η2·η3. At 97%, 96% and 95%, that gives about 88.5% instead of the seemingly uncritical individual values.
Why is real output torque smaller than M₁·i?
Because part of the transmitted power is lost as heat through friction in gear meshes and bearings. Efficiency η captures exactly this share.
Is a 95% efficiency per stage already very good?
For many gear stages, yes, especially spur gear stages, which often reach 97-99%. For multi-stage gearboxes, though, overall efficiency across all stages combined is what always counts.