How this instrument works
A motor's torque is the twisting force at its shaft — the quantity that actually turns a pump impeller, a conveyor roller, or a wheel against resistance. Power is the rate at which torque does work as the shaft spins: P = T × ω, where ω is angular speed in radians per second. Rearranged for torque, T = P ⁄ ω, and since ω = 2πN ⁄ 60 for a speed N given in revolutions per minute, the identity becomes T = P × 60 ⁄ (2πN).
The 2π turns revolutions into radians and the 60 turns minutes into seconds — both are unit bookkeeping, not physics, which is why the combined constant 60 ⁄ (2π) ≈ 9.5493 shows up as a shortcut in almost every motor-sizing handbook. An electrician checking whether a coupling can survive a 5 kW motor, or a pump engineer confirming a shaft can carry a load without shearing, reaches for exactly this relation: power and speed are what the nameplate and drive report, but torque is what the mechanical parts must actually withstand.
The result is the steady torque implied by rated power and speed, not the motor's starting or breakdown torque. Induction motors briefly produce several times that figure at switch-on — locked-rotor torque is a separate rating the manufacturer measures directly, not something this identity predicts. The common mistake is treating power alone as a proxy for pulling force: two motors can share an identical 5 kW rating and still differ enormously in torque, because for fixed power, torque falls as speed rises. A slow, heavy industrial motor and a small, fast one can output the same kilowatts while the slow one delivers far more twist at the shaft.
- Enter the motor's rated power in the Motor power field, switching its unit menu between W, kW, or hp to match the nameplate.
- Enter the shaft speed in the Motor speed, RPM field — the rated RPM figure stamped on the nameplate, not a load-dependent guess.
- Read the result in the Torque field; toggle its unit between Nm and ftlb depending on which standard your drawings use.
- Recheck the RPM figure before trusting the output — halving it doubles the torque, so a typo there moves the answer far more than one in the power field.
Worked example — a 5 kW motor at 1,750 RPM
A 5 kW motor — 5000 W — spinning at 1,750 RPM: T = 5000 × 60 ⁄ (2π × 1750) = 300,000 ⁄ 10,995.57 = 27.2837 N·m. That is the steady torque the shaft delivers while running at rated power and speed, the figure a coupling or gearbox downstream of the motor has to be sized against.
Double the speed to 3,500 RPM without changing the power and the torque is cut in half, not held steady: T = 5000 × 60 ⁄ (2π × 3500) = 13.6419 N·m. Same 5 kW, same motor family, torque nearly halved — a direct demonstration that RPM, not power alone, decides how hard a shaft can twist.
Questions
Why does doubling the RPM cut the torque in half at the same power?
Because power equals torque times angular speed, P = Tω, and here P is held fixed. If ω doubles, T must halve to keep the product constant — exactly why a 5 kW motor at 3,500 RPM produces 13.64 N·m while the same motor at 1,750 RPM produces 27.28 N·m. It is a trade-off built into the formula, not a quirk of any particular motor.
Is this the motor's starting torque or its running torque?
Running torque, calculated at rated power and speed. Starting, or locked-rotor, torque is a separate directly-measured rating that can run 150 to 300 percent higher for an instant at switch-on — check the motor's datasheet for that figure rather than deriving it here, since this formula assumes steady rotation.
My power rating is in horsepower — do I need to convert it first?
No. Switch the Motor power field's unit menu to hp and type the nameplate figure directly; the instrument converts to watts internally using 1 hp = 745.7 W before applying T = P × 60 ⁄ (2πN), so the torque reading is unaffected by which unit power was entered in.
Where do the 2π and 60 in the formula actually come from?
From converting RPM into radians per second. One revolution equals 2π radians, and one minute equals 60 seconds, so a speed of N RPM equals angular speed ω = 2πN ⁄ 60 rad/s. Substituting that into P = Tω and solving for T gives T = P × 60 ⁄ (2πN) — the constants are unit conversions, not separate physics.
Why doesn't my calculated torque match the number on the motor's nameplate?
Nameplate torque is usually quoted at full-load power and full-load RPM together. Enter the rated power but the no-load, synchronous speed instead of the loaded speed and the result runs a few percent high, because induction motors slip and turn slightly slower under load than at no load.
Does this formula work for DC motors as well as AC motors?
Yes. T = P × 60 ⁄ (2πN) follows from P = Tω, which holds for any rotating shaft regardless of what drives it. Enter the mechanical power actually delivered at the shaft, not the electrical power drawn from the supply, which is larger by whatever the motor's efficiency loses to heat.