Re: Declining motor torque with lower voltage.

--- In CAD_CAM_EDM_DRO@yahoogroups.com, "cnc_4_me" <cnc4me@g...>
wrote:

>
> I found this statment googling around today.
>
> "It is common motor knowledge that motor torque declines by the
> square of the change in supply voltage."
>
> First...Is this true.
>
> And second if true would this be the proper way to appy this.
>
> 1) You have a treadmill motor rated 130VDC, 2.5HP 6700RPM, and
> 373 oz-in continuous.
>
> 2) The motor is run from a 80VDC power supply.
>
> Well, the change in voltage is 130VDC - 80VDC = 50VDC.
> The square root of 50VDC is aprox 7.
> So the motor torque declines by 373 0z-in / 7 = 53 oz-in
>
> ???? Is this right ????
>
> Wally

I have a similar dead spot in my thinking.

It would seem a motor as above at 24VDC spinning at the percentage
speed. (ie: 50% voltage and 50% of max speed) that the current would
drop and you would get half the power. but then at 1/100th power,
or a very slow feed, there would be no power at all. Obviously, the
driver increases the amps to compensate. Or we would use much
simplier drivers. Since we use these in CNC applications, there is
something going on besides simple arithimitic.

It may be that the reference above is for AC motors ? or motors
without the driver circuit ? I seem to remember that AC motors have
a 4:1 (or there about) volts to hertz ratio. that is key in the
variable speed drives operation.

Dave