The Equations
\[ P = T\,\omega,\qquad \omega = \frac{2\pi n}{60} \]
\[ P_{\text{kW}} = \frac{T_{\text{N}\cdot \text{m}}\; n}{9549.3},\qquad P_{\text{hp}} = \frac{T_{\text{lbf}\cdot \text{ft}}\; n}{5252.1} \]
Power is torque times angular speed. With torque in newton-metres and angular speed in radians per second the answer is in watts. The two shortcuts fold the unit conversions into one constant.
9549.3 is 60,000 / 2π. 5252.1 is 33,000 / 2π. The second uses the mechanical horsepower, 33,000 ft·lbf per minute.
Constant power
\[ T = \frac{P}{\omega} = \frac{60\,P}{2\pi\,n} \qquad\Rightarrow\qquad \log T = \log\frac{60\,P}{2\pi} - \log n \]
Here n is the speed in rpm. At a fixed power the first term on the right is a constant, so torque is inversely proportional to speed. On linear axes that is a hyperbola. On log-log axes it is a straight line with slope −1, and higher power shifts the line up. So an engine’s power rating alone
does not tell you the pulling force at a given speed. It is also why gearing lets a small fast motor do the work of a big slow one.
Through a gearbox
\[ n_{\text{out}} = \frac{n}{i},\qquad T_{\text{out}} = T\,i\,\eta,\qquad P_{\text{out}} = P\,\eta \]
A reduction ratio i divides speed and multiplies torque. Losses (efficiency η) come off the power, so torque rises by a little less than the ratio.