SDT engineering selection guide · 02

Motor Torque, Speed and Power

Use torque and speed together to define mechanical output, compare operating points and avoid selecting a motor from nominal power alone.

Engineering purpose

Use the sequence below to establish a defensible starting point, then confirm the decision with model-specific performance, tolerances and test evidence.

Rotational powerP = TωP in W, T in N·m, ω in rad/s
Speed conversionω = 2πn / 60n in revolutions per minute
Practical formT = 9550P / nP in kW, n in rpm, T in N·m
01

Power Is Not a Standalone Requirement

A power value can represent many torque-speed combinations. A 1 kW requirement at 1,000 rpm needs roughly three times the torque required at 3,000 rpm. The motor, transmission and drive must therefore be evaluated at every important operating point.

  • State continuous and peak torque
  • State the corresponding speed for each torque
  • Separate shaft output power from electrical input power
02

Define the Complete Torque–Speed Envelope

Real machines accelerate, run, decelerate, reverse and hold. Plotting these states creates the required operating envelope and exposes peak events that a single rated point cannot show.

  • Include breakaway and friction torque
  • Add acceleration torque and external forces
  • Identify overspeed and field-weakening operation
03

Account for Transmission Efficiency

Gearboxes, belts, couplings and bearings change both speed and available output torque. Use realistic efficiency at the selected ratio and load rather than assuming an ideal transmission.

  • Output speed is approximately motor speed divided by ratio
  • Output torque is motor torque multiplied by ratio and efficiency
  • Check gearbox continuous and intermittent limits separately
04

Separate Mechanical and Electrical Power

Mechanical output power is lower than electrical input power because of motor, drive and transmission losses. The supply and thermal design must be based on input current and total losses at the real operating point.

Worked example

Calculate motor torque for a 750 W load

Given

  • Required load output: 0.75 kW
  • Motor and load speed: 1,500 rpm (1:1 transmission)
  • Transmission efficiency: 90%

Method

  1. Load torque = 9550 × 0.75 / 1500 = 4.78 N·m.
  2. Required motor mechanical power = 0.75 / 0.90 = 0.833 kW.
  3. Required motor torque at 1,500 rpm = 9550 × 0.833 / 1500 = 5.30 N·m.

Engineering result: The motor must provide at least 5.30 N·m at 1,500 rpm before service margin, acceleration torque and thermal duty are considered.

Common mistakes

What to Avoid

  • Selecting from kW alone
  • Using no-load speed as the loaded operating speed
  • Ignoring gearbox efficiency
  • Confusing electrical input power with shaft output power

Selection check

Confirm Before Selection

  • Torque and speed at every operating state
  • Acceleration time and load inertia
  • Transmission ratio and efficiency
  • Continuous, peak and regenerative conditions

SDT engineering note

Apply General Knowledge with Product-Specific Evidence

This guide supports preliminary engineering. Final selection must be confirmed against the complete application, selected motor and drive data, declared operating conditions, applicable standards and agreed validation criteria.

Application-specific support

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