Shaft Torque from Power and Angular Speed

Power is the rate of doing work, and for a rotating shaft that is torque times angular speed:

\[ P = T\omega \qquad \Longrightarrow \qquad T = \frac{P}{\omega} \]

with $\omega$ in radians per second. The radian is not decorative: it is what makes the equation dimensionally clean, since $\omega$ is really an angle per unit time and the radian is the angle for which arc length equals radius. From rev/min it is $\omega = 2\pi n/60$.

The 5252

Every North American shop has the rule $T = 5252\,\text{HP}/\text{RPM}$ written on a wall somewhere, and it is exactly this equation with the unit conversions folded in. One horsepower is 33,000 ft·lbf per minute, so $T = 33000\,\text{HP}/(2\pi\,\text{RPM})$, and $33000/2\pi = 5252.11$. Nothing else is going on. A pleasant consequence: at 5,252 rpm a motor's torque in ft·lbf and its power in horsepower are numerically equal, which is why every dynamometer plot of torque and power crosses at that speed.

The design lesson: slow shafts are fat shafts

At constant power, torque is inversely proportional to speed. Put a 10:1 reduction after a motor and the output shaft carries ten times the torque. That is why the low-speed end of any gearbox is always the heavy end, and it is worth doing the arithmetic once to feel the scale of it. Ten kilowatts at 3,000 rpm is about 32 N·m — a 20 mm shaft handles it comfortably. The same 10 kW at 30 rpm is about 3,180 N·m, a hundred times the torque, and since shaft diameter grows as the cube root of torque, the shaft needs to be roughly 4.6 times thicker. Low-speed machinery is heavy not because it is old-fashioned but because torque is what steel has to resist and power is not.

It also explains the direction of most drivetrains. Electric motors and engines are cheap and light at high speed and expensive and heavy at low speed, so almost every machine puts a fast, small prime mover behind a reduction, and pays for the reduction rather than for a slow motor.

Two cautions

Nameplate power is output power at rated speed and rated load. A motor started under load, or stalled, or accelerating a large inertia, produces torque that has nothing to do with this equation — starting torque can be several times full-load torque, and the shaft and coupling have to survive it. That is what a service factor is for.

And this gives the torque, not the stress. What the shaft actually feels also includes bending from the pulleys, gears and couplings hung on it, and because the shaft rotates, that bending stress fully reverses every turn — which puts it into fatigue. See the shaft sizing page.