Dooh! You're right. Though I suspect a too small motor would have higher magnetic losses too, but I haven't got a clue how you would measure those so that's kinda irrelevant.
FWIW: from looking at motor charts it's obvious that all the factors that affect motors do so in a non-linear(?) fashion. The efficiency curve is non symetrical and drops off much steeper on the "too little" current side then it does on the "too much" side. It's concievable that a too small motor would be MORE efficient at lower speeds than it's larger cousins. If you aren't running into overheating problems it might be better to stick with the small motor. Joe Smalley wrote: >This was a estimate of the losses from using a TOO SMALL motor. > >The windage losses, magnetic losses, and brush losses would also be >applicable to a larger motor. The main difference between a too small motor >and a larger motor loss is in the internal resistance losses which this test >was designed to measure in his application. > >Joe Smalley >Rural Kitsap County WA >Fiesta 48 volts >NEDRA 48 volt street conversion record holder >[EMAIL PROTECTED] > >----- Original Message ----- >From: "Peter VanDerWal" <[EMAIL PROTECTED]> >To: <[EMAIL PROTECTED]> >Sent: Friday, September 27, 2002 11:41 PM >Subject: Re: How much power am I losing? > > >>There is also friction and windage losses, not to mention magnetic >>losses. And then there is the fact that the motors resistance changes >>as it heats up and that the resistance through the brushes on a moving >>comutator is different than on a stationary one. >> >>All you can be sure of with the below method is that the losses will be >>MORE than you calculated. >> >>You can get a bit closer by measuring the amount of power the motor >>takes to spin at the same RPM without a load and factor out the >>resistance losses. The remainder will be 'some' of the unaccounted for >>losses (friction and windage) but wont account for the extra losses due >>to high current (magnetic, etc.) >> > > >
