I notice that motors are often described by their power rating. So, for example, you might have a motor that is described as a 1 HP motor.

However, that is actually not a complete statistic because the torque provided by the motor will vary by the RPM. For example, a 1 HP motor that runs at 1600 RPM will generate twice as much torque as a 1 HP motor that runs at 800 RPM (I think?)

So, given that, what is the right way to think about a motor's capability. Power x RPM? Does that have a measure?

  • $\begingroup$ Horsepower is torque (lb-ft) x rpm / 5,252, so twice the speed will only be capable of half the torque, if the speed/power curve is linear. Usually motors are designed to run at specific speeds, that is what the HP is based off, with significant power losses at different speeds. $\endgroup$
    – jko
    Aug 10, 2021 at 19:56
  • $\begingroup$ If you really want to quantify it, you have motor torque vs RPM curves. If just a number is given then it's just its rated performance which is a power level that is about the maximum continuous of what the motor was designed to run at. $\endgroup$
    – DKNguyen
    Aug 10, 2021 at 19:59
  • $\begingroup$ The type of the motor is just as important as its power rating. See en.wikipedia.org/wiki/Electric_motor#Types for a summary. Different types have very different power and torque curves versus RPM, and some types won't run at all except at their design RPM. $\endgroup$
    – alephzero
    Aug 10, 2021 at 20:03
  • $\begingroup$ I think you shall look into the "efficiency" of the motors. $\endgroup$
    – r13
    Aug 10, 2021 at 20:36

2 Answers 2


Horsepower is the single best measure of internal combustion engine performance.

Power = (2 * pi * N * T)/33,000 in imperial units. 33000 is the unit conversion of 33000 ft-lb/min/hp.

This means it is torque times RPM times a conversion factor. In your example, the amount of work you can get out of either engine is exactly the same, since you can put gears on the back end to make it go however fast or slow that you want. Power will be the same.

There are real world gains to be made in driveablility based on the shape of the torque curve. Cars and motorcycles are easier to drive/ride if the engine has better torque at low rpm. But the fastest engines won't behave this way because you can design an engine for high rpm power or for driveability. There are ways to try to get both (multi-vane turbochargers, Honda's VTEC, variable inlet and outlet geometry) but they are all compromises. F1 engines are not built for driveability.

Torque is so much trouble, because it is so misunderstood. The reality is that for a given horsepower, you can get whatever torque you want via gearing. What torque ought to be proportional to is acceleration (it's the F in F=ma). But torque is based on gears. The latest versions of US diesel pickups are in a "torque war," so one might think that having 1000 ft-pounds of torque in an engine means it's actually a really fast vehicle, but it isn't. These diesel engines run at a low rpm, so the gears required to run on the highway mean the trucks are slower than gas engines with less than half the torque. While the combination of power and torque and the rpm values of each peak can be useful to those who know how to interpret them well, my opinion is that torque is more of a marketing thing than an actual descriptor of engine performance.


sometimes you get the rating of the motor in Torque@rpm and power@rpm.

e.g. Mazda B6ZE(RS) 1.6 L European version (1994-1997) produced 66 kW (90 hp) at 6,500 rpm (sometimes briefly noted as 66kW@6500), and 129 N·m (95 lb·ft) at 5,500 rpm and (129Nm@5500rpm) (from Wikipedia).

You will notice that the maximum power is always at a greater rpm compared to the maximum torque. The difference between the two can be used to compare in a more meaningful manner different engines, in the absence of torque-rpm and power-rpm graphs.

  • $\begingroup$ Or you can say the motors that have different RPMs can produce the same power but the one with lower RPM requires more torque, which is a disadvantage. $\endgroup$
    – r13
    Aug 10, 2021 at 21:47

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