How to Calculate Gearbox Output Torque: Formula, Worked Example and Common Mistakes

Output torque is the single most checked number on a gearbox selection. It is also one of the easiest to miscalculate, because the formula has three terms that all matter and two of them are often estimated rather than measured.

The standard formula

For an input power P (in kW), input speed n1 (in rpm), ratio i, and gearbox efficiency eta, the output speed n2 is n1 divided by i, and the output torque T2 in Newton-metres is: T2 = 9550 × P × eta / n2.

Equivalently, T2 = 9550 × P × eta × i / n1. Both forms give the same answer; the second is more convenient when ratio is known and output speed is not.

Why the efficiency term is not optional

On helical inline units (R series), efficiency at full load is typically 0.95 to 0.98 per stage. On parallel-shaft (F series) and bevel-helical (K series), full-load efficiency is typically 0.94 to 0.97 per stage. On helical-worm (S series), full-load efficiency is typically 0.50 to 0.85 depending on ratio and lead angle. Skipping the efficiency term on worm units produces a calculated torque that is 15 to 50 percent higher than the gearbox can actually deliver at the output shaft — which then looks fine in the rating table and fails in service.

Worked example: 15 kW motor, 1460 rpm input, target 28 rpm output

Step 1: Ratio. i = 1460 / 28 = 52.14. Step 2: Output speed n2 = 28 rpm. Step 3: Pick efficiency for the chosen family. For a K-series helical-bevel, two-stage efficiency at full load ≈ 0.94. Step 4: Compute. T2 = 9550 × 15 × 0.94 / 28 = 4810 Newton-metres. The 9550 constant converts kW and rpm to Nm correctly; do not use a different conversion factor.

The reverse calculation: torque required by the driven machine

Often the input is the process torque required at the driven shaft, not the motor power. In that case, the calculation inverts to required motor power. P = T2 × n2 / (9550 × eta × i). For an agitator needing 19000 Nm at 28 rpm through the same K-series, P = 19000 × 28 / (9550 × 0.94 × 52.14) = 1.13 kW mechanical at the output. Add 15 to 20 percent for mechanical losses and start-up, and you arrive at a 1.5 to 2 kW motor. The selection team should confirm the actual driven-machine torque and start-up profile before finalising the motor size.

Common mistakes

1. Skipping efficiency on worm units: the single most common error. Always include eta for S-series and confirm the eta value with the factory for ratios above 50.
2. Using motor nameplate power without derating: motor nameplate power is the maximum continuous rating, not the load. If the driven machine draws 80 percent of motor nameplate, base the torque calculation on that 80 percent value.
3. Confusing output speed with output shaft speed after the gearbox: these are the same number in a single-stage unit, but in a two-stage arrangement they differ by the second-stage ratio. Always compute n2 from the total ratio, not the first-stage ratio alone.
4. Ignoring the service factor on the input power: if the service factor requires 1.5, base the input power on the worst-case load × 1.5, not the average load. Underrated gearboxes fail thermally before mechanically.
5. Treating the gearbox as 100 percent efficient: even helical-in-line units are not 100 percent efficient. The 2 to 5 percent loss becomes significant over time on continuous-duty applications.

When to ask the factory

If the calculated torque is within 10 percent of the next-size-up gearbox in the rating table, ask the factory for the thermal derating curve and the actual measured efficiency at your ratio. Selecting the smaller unit and relying on the catalogue figure is the second most common cause of warranty disputes.

For project-specific torque calculation review, send the motor power, input speed, target output speed, mounting position and driven-machine description to sales@chinfar.com.