Motor Torque⚠ unverified
Electrical / Motors · Mechanical torque from output power and angular speed
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| P | P | W | 1000.0 | Output power |
| omega | ω | rad/s | 100.0 | Angular speed |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| torque | T | N*m | Torque |
The science & history
Understanding the Parameters
- Power $P$ — the mechanical output in watts.
-
Angular speed $\omega$ — in rad/s ($\omega = 2\pi n/60$ from rpm $n$). As $\omega \to 0$ the same power implies very large torque (why gearing down trades speed for torque).
-
$T$ — the resulting torque in newton‑metres.
Derivation (Approaching a Proof)
Starting from the rotational power relation $P = T\omega$ (derived from $dW = T\,d\theta$; see Motor Power), simply solve for torque:
$$T = \frac{P}{\omega}.$$
The inverse dependence on $\omega$ is the mechanical statement of "gear reduction": a gearbox that lowers output speed by a factor $n$ raises available torque by the same factor (minus losses), because power is conserved.
History
Like $P = T\omega$, this is a direct consequence of classical rotational mechanics. It is the everyday tool for reading a motor's torque from its rated power and speed, and for sizing gear ratios in drives.
Related Concepts: Motor Power, Torque Constant, Maximum Torque, Starting Torque
Notes: Registry calculator motor-torque (unverified). Requires $\omega$ in rad/s; this is the shaft
(mechanical) torque, distinct from the electromagnetic torque before friction/windage losses.