Home/Mechanical Engineering/Power-Torque-Speed Relationship

Formula & Calculator

Power-Torque-Speed Relationship

Relates the power transmitted by a rotating shaft to the torque it carries and its angular velocity.

Written by the Vorzep Team

Mechanical EngineeringMachine DesignCalculator

Power–Torque–Speed CalculatorRotational Mechanics

P = T · ω
P = power (W)  ·  T = torque (N·m)  ·  ω = angular velocity (rad/s)
⟹ Solve⟷P, T, ω
W
N·m
rad/s
Please fix the errors above.
Solve for:
Presets:
Power—
—
P: —T: —ω: —
ω in rad/s. To convert from RPM: rad/s = RPM × 2π / 60
✓ Copied!
Power Gauge—
Low (< 100 W) Medium (100–1000 W) High (1000–5000 W) Very High (> 5000 W)
P = T · ω  ·  Units: W, N·m, rad/s

Variables

SymbolQuantityUnit
PTransmitted powerW
TTorque on the shaftN.m
omegaAngular velocityrad/s

What it means

In rotational systems, the mechanical power developed is given by P = T ω, where P is power (watts), T is torque (N·m), and ω is angular velocity (rad/s). This equation is the rotational analogue of P = F·v. It is essential for sizing motors and engines: a motor with high torque at low speed can provide high power, and vice versa. The relationship is used in gearboxes, where gear ratios adjust torque and speed. In vehicles, engine power is often rated at a certain RPM. For electric motors, torque‑speed curves are crucial for control. The equation also applies to generators and turbines. In practice, efficiency reduces the output power, so P_output = η·P_input. This relationship helps engineers balance torque and speed requirements in design. It is also used in calculating the power of wind turbines and hydraulic systems. Understanding this equation is vital for powertrain design and energy conversion efficiency.

Worked example

Power‑Torque‑Speed – Two Examples

Real‑World
Scenario 1 – Electric Motor: Torque 100 N·m at 10 rad/s. Find power.
ParameterValue
T100 N·m
ω10 rad/s
1P = T·ω = 100×10 = 1000 W = 1 kW
ResultP = 1 kW✓
Scenario 2 – Engine: A car engine produces 200 N·m at 3000 RPM. Convert RPM to rad/s: ω=3000×2π/60≈314.16 rad/s. Find power.
ParameterValue
T200 N·m
ω314.16 rad/s
1P = 200 × 314.16 = 62,832 W ≈ 62.8 kW (≈ 84 HP)
ResultP ≈ 62.8 kW✓
Key insight: Power = torque × angular speed – higher speed or torque increases power.

Common mistakes

  • Torque T: In N·m.
  • Angular velocity ω: In rad/s – convert from rpm if needed (ω = 2πN/60).
  • Units: T·ω → W (if T in N·m, ω in rad/s).
  • Power vs. torque: This relationship is instantaneous power; for varying loads, use average.
  • Efficiency: If the machine has losses, output power = T·ω × efficiency.

Applications

Power in rotating machinery is the product of torque and angular velocity, P = T ω. This relationship is fundamental in the design of electric motors, engines, and turbines. It allows engineers to determine the required torque for a given power output and speed, or vice versa. In automotive engineering, it is used to match engines to transmissions and to calculate the power at the wheels. In industrial drives, it helps size motors for conveyors, pumps, and fans. The formula also applies to wind turbines, where power generated depends on rotor torque and rotational speed. By using this relationship, engineers can optimise the performance and efficiency of mechanical drive systems.

  • Electric motor and generator sizing
  • Internal combustion engine power and torque curves
  • Transmission and drivetrain design
  • Wind turbine and hydro turbine power calculation
  • Industrial machinery drive system design