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Shaft power and torque calculator

Calculate steady power, torque or rotational speed from the other two quantities at the same shaft section; inspect units and zero limits.

MECHANICAL POWER · 1.0.0

Shaft power, torque and rotational speed

Find the third quantity from two values at the same shaft section. Explore unit conversion, inverse calculations and zero boundaries step by step.

P = Tω ω = 2πn/60 P(kW) = 2π × n(rpm) × T(N·m) / 60 000

Steady rotation with torque and speed in the same power-transmitting sense. Both refer to the same shaft section. Negative quantities, braking and reverse power flow are excluded.

Inputs and units

Use a decimal point or comma. Do not use thousands separators. Scientific notation is accepted.

This quantity is calculated; choose its result unit.
0 or 10⁻⁶–10¹² N·m equivalent. Steady torque magnitude at the same section.
0 or 10⁻⁶–10⁶ rpm equivalent. Rotation direction is outside this model.

Calculation result

Calculator loading. The method and worked example remain readable without interactive controls.

This calculator processes its inputs, calculation and CSV in your browser. The calculator does not send or store them.

From formula to result

Rotational work is dW = T dθ. For constant torque and speed, dividing by time gives P = Tω. Torque in N·m and angular speed in rad/s give power in W. This tool calculates the magnitude of positive power transfer; direction and four-quadrant drive behavior are excluded.

One revolution is 2π radians and one minute is 60 seconds. For n in rpm, use ω = 2πn/60. Divide W by 1,000 for kW or by 1,000,000 for MW. One kN·m is 1,000 N·m. Inputs are first converted to these common units.

For torque, T = 60,000 P(kW)/(2πn); for speed, n = 60,000 P(kW)/(2πT). The familiar factor 9,550 is rounded. This tool evaluates 60,000/(2π) ≈ 9,549.296586 instead.

An inverse calculation requires a nonzero divisor. With P = 0 and a zero divisor, infinitely many values are possible. With positive P and a zero divisor, no finite solution exists. Finite torque at zero speed can give P = 0; thermal or mechanical acceptability cannot be inferred from this identity.

Worked example

For an invented auxiliary machine, take P = 75 kW and n = 1,500 rpm at the same shaft section. Find torque, then distinguish fixed-torque and fixed-power speed changes.

  1. Convert to common units: 75 kW = 75,000 W.
  2. Angular speed: ω = 2π × 1,500 / 60 = 157.079633 rad/s.
  3. Torque: T = 75,000 / 157.079633 = 477.464829 N·m.
  4. At fixed T and n = 1,350 rpm, P becomes 67.5 kW. At fixed P and n = 1,350 rpm, T must be about 530.516477 N·m. The quantity held constant changes the answer.
Worked example
QuantityValue
75 kW, 1,500 rpm477.464829 N·m
Same torque, 1,350 rpm67.5 kW
Fixed 75 kW, 1,350 rpm530.516477 N·m

Limits of the result

  • The result is mechanical power at one section. Crankshaft, gearbox output, propeller shaft and useful thrust power are not interchangeable. No transmission, motor, drive or propeller efficiency is assumed.
  • Mean torque multiplied by mean speed need not equal mean power in a varying process. Vibration, torque pulsation, acceleration, inertia and transients require time-dependent T(t)ω(t).
  • The tool does not select shaft diameter, strength, torsional vibration limits, couplings, bearings, motor loadability, service factors or class compliance. Numerical bounds are not equipment limits.
  • A result cannot replace manufacturer data, measurement-uncertainty analysis or qualified engineering review. Display rounding does not improve input accuracy.

Method references

Sources support the underlying relationships. Numerical examples and software bounds were chosen for this educational tool. Links checked 8 October 2026.

Read the method and its limits

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