# Formula for calculating output power after putting a rpm reduction gear

I'm thinking of starting of converting an old car to an electric car. I have a question, but google won't answer me my very long not very well explained question.

Take this scenario. I would take a small 2 HP electric motor that can run up to 5000 RPM, say I put a gear for reducing revolutions (e.g. 6:1 reduction gear). So the question would be, is there any formula that would give me the output HP at the output of the gear?

I hope you understood my not very well put question.

Cheers

tl;dr: Yes. The horsepower shouldn't change but the torque should go up by a factor of 6.

The thing to remember is that what you usually want is actually the torque of your motor. That's what actually does the useful work for you. In this case, though, you've been handed a horsepower number at a particular RPM.

We can get the equivalent torque at that RPM using the standard equation:

``````horsepower = (torque * rpm) / 5252
``````

So, if your cited horsepower is 2 HP at 5000 RPM, we expect that the torque output is 2.1 foot-pounds (yeah, ick, English measurements - blame the British empire).

At this point, we can introduce the reduction gear. The thing to remember is that we don't expect the output gear to affect the power of the motor at all. It will continue putting out its output torque at its previous angular velocity. The reduction gearing will then trade angular velocity for more torque at a ratio of six to one.

If you apply your reduction gear, you'll change the `rpm` in the equation. Plugging in 2 HP and 833 RPM (at a 6:1 reduction), we expect to see a torque of about 10.5 foot-pounds.

At the end of all this, you'll have a system with much more pulling power and a much lower top speed.

• Also of note, the 833 RPM Bob talks about is the output RPM ... the input RPM would logically remain the same (5k). – Pᴀᴜʟsᴛᴇʀ2 Mar 10 '14 at 1:21