Automotive & Engines
Intermediate

Power from Torque & RPM

Engine power in kW from torque and rotational speed.

Formula

P=τ×N9549P = \frac{\tau \times N}{9549}

Variables

PPower (kW)
τTorque (N·m)
NSpeed (rpm)

Example

300 N·m at 5000 rpm: P = 1.5M/9549 ≈ 157 kW

How it works

This formula converts how hard an engine is twisting the crankshaft (torque) and how fast it is spinning (rpm) into output power. It comes from the rotational power relationship P = torque × angular speed, with the rpm value converted into radians per second and the result expressed in kilowatts, which is why the constant 9549 appears. In practice, higher torque or higher engine speed both increase power, but only if the engine can maintain that torque at that speed.

When to use it

Use this when you know an engine’s torque at a particular rpm and want to estimate its power output in kilowatts. It is common in automotive engineering, dyno testing, and comparing engine performance across different speed ranges.

Worked example, step by step

Given: torque τ = 420 N·m, speed N = 3200 rpm Step 1: Substitute into the formula P = (τ × N) / 9549 P = (420 × 3200) / 9549 Step 2: Multiply torque by speed 420 × 3200 = 1,344,000 Step 3: Divide by 9549 P = 1,344,000 / 9549 ≈ 140.8 Final answer: P ≈ 141 kW

Common mistakes to avoid

  • Using torque values in lb·ft without converting them to N·m first
  • Putting the rpm value into the formula without using the constant 9549 or an equivalent unit conversion
  • Assuming more rpm always means more power, even when torque drops sharply

Did You Know?

Power is torque times speed — which is why high-revving engines make big horsepower from modest torque.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026