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For any rotating machine, power equals torque times angular speed: P = T·ω, where ω = 2πN/60 in radians per second. Working in the practical units of kilowatts and rpm, this reduces to the well-known motor formula T = 9550·P/N (the 9550 comes from 60000/2π). It tells you the shaft torque a motor produces at a given power and speed — the key to matching a motor to a load.
| Quantity | Formula |
|---|---|
| Torque (N·m) | T = 9550 × P(kW) / N(rpm) |
| Power (kW) | P = T(N·m) × N(rpm) / 9550 |
| Angular speed | ω = 2πN/60 (rad/s) |
| Power (SI) | P(W) = T(N·m) × ω(rad/s) |
A key insight: for a fixed power, torque is inversely proportional to speed. A low-speed motor delivers far more torque than a high-speed one of the same power, which is why gearboxes trade speed for torque. To convert, 1 N·m = 0.7376 lb-ft and 1 kW = 1.341 hp.
The practical formula is T = 9550 × P/N, where T is torque in newton-metres, P is power in kilowatts, and N is speed in rpm. The constant 9550 equals 60000 divided by 2π.
Power equals torque times angular speed (ω = 2πN/60). Rearranging for torque in N·m with power in watts gives T = 60000·P(kW)/(2πN) = 9549.3·P/N, rounded to 9550.
Use P(kW) = T(N·m) × N(rpm) / 9550. For example, 100 N·m at 1500 rpm gives 15.7 kW.
For a fixed power, torque and speed are inversely related (P = T·ω). So a high-speed motor produces less torque than a low-speed motor of the same power, which is why gearing is used to gain torque.
Multiply newton-metres by 0.7376 to get pound-feet, or divide pound-feet by 0.7376 to get newton-metres. For example, 49.4 N·m is about 36.4 lb-ft.
The formula gives the torque at the power and speed you enter, which is normally the rated (full-load) operating point. Starting torque and peak torque are different and are quoted separately as multiples of rated torque.
Yes. The relationship P = T·ω is universal for any rotating machine, so the formula works for AC induction, synchronous, DC and servo motors alike, as long as you use the mechanical (shaft) power.
Use the actual shaft speed at the load point. For an induction motor that is the full-load speed (slightly below synchronous due to slip), which is what the nameplate rpm gives.
It is ω = 2πN/60, converting rpm to radians per second. A 1500 rpm motor turns at about 157 rad/s. Angular speed is needed for the SI power formula P = T·ω.
Multiply kilowatts by 1.341 to get mechanical horsepower, or divide horsepower by 1.341 to get kilowatts. So 7.5 kW is about 10.06 hp.
Ideally no — a gearbox conserves power (minus small losses) while trading speed for torque. A 10:1 reduction cuts speed to one-tenth and multiplies torque roughly ten times at the same power.
Estimate the load torque from the force and radius (T = F×r) or from the load power and its speed. The motor must supply at least that running torque, plus extra for acceleration and starting.
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