On 23/02/16 01:36, johnrhines@gmail.com [electricboats] wrote:
>
> Since electric motors have full torque at all rpms, what is the
> problem with spinning the motor at 650rpm and eliminating the gear
> reduction? Is it just a measure of efficiency? How would you
> calculate the load on a motor at 650rpm versus 1657rpm.
>
Assuming both ways require the same number of watts out in mechanical
effort, and watts in electrical equals watts out mechanical (ok, no such
thing, but...)
With a DC motor at least (perm magnet in my case), to run at half speed
but with the same power requires twice as much current. Since the
electrical losses are I Squared R (and R will constant as it's the same
motor), your losses will be greater at the lower speed.
By doubling the motor speed with a reduction gear you are significantly
reducing the I2R losses.
I'm not sure how this equation stacks up with PMAC or other multi-phase
motors, but I expect similar, if you want the same KW out but at half
the rotational speed, you are going to need more current, thus higher
I2R losses.
My E-Boat is direct drive and the motor gets quite warm (loss!, that
heat is battery power not being used to move the boat). While I have a
36v system I've discovered that the boat reaching full speed at about
60% throttle. When my lead acid batteries finally die, I'll probably
replace them with 48v of LiPo3 and fit a 2:1 reduction and run the motor
at 48v. The current will thus be lower so the I2R losses should be lower
and the motor will run cooler.
Batteries also tend to be able to deliver more power at lower rates of
discharge, so more volts and less current is probably a win there too.
Posted by: Matthew Geier <matthew@acfr.usyd.edu.au>
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