A motor is a converter, and no converter is free. Some of what comes in the terminals leaves as heat in the copper, iron losses in the core, windage and friction at the bearings. What survives arrives at the shaft.

$\eta = \dfrac{100\, P_{out}}{P_{in}}$ — read aloud eta equals a hundred P-out over P-in, with $\eta$ the Greek letter eta. $P_{out}$ is the mechanical power at the shaft and $P_{in}$ is the electrical power drawn at the terminals, both in the same unit — watts, or kilowatts, as long as they match. $\eta$ comes out as a percentage: a pure ratio wearing a courtesy sign.

The electrical half of the pair is the plainest relation in the subject: $P = VI$, P equals V I, where $P$ is power in watts, $V$ is the voltage in volts and $I$ is the current in amperes. On a DC bus or a resistive single-phase load that is the whole story. Put the two relations together and you can walk from a shaft rating to the amps a feeder must carry — which is what the rest of this chapter keeps asking you to do.

One direction is worth burning in: the input is always the larger number. Solving for $P_{in}$ means DIVIDING by the efficiency, not multiplying. An answer where the shaft beats the terminals is not a remarkable motor; it is a rearrangement that went the wrong way, and the catalog says so in as many words.