Re: volts/rpm

Chris,

In a perfect world, yes. What makes it imperfect is the motor's
armature resistance. The current (in response to torque load) has to
pass thru this resistance, which produces a voltage drop. This
voltage subracts from the power supply voltage, leaving the
difference to turn the motor. The part you are looking for is this
probably:

(1) If you know the armature resistance accurately, what is the peak
mechanical power (Watts) the motor can develop at a given power
supply voltage?

W = V squared / 4 times R

(2) What torque will the motor generate at peak power if the no-load
speed is known for the given supply voltage?

in-oz = ( W times 1351 ) / RPM RPM is 1/2 of no-load here

(3) What will the stall torque be?

Twice what is calculated in (2)

(4) What will the stall current be?

I = V / R

(5) What will be the maximum continuous torque and current?

About 10% to 20% of what was calculated in (3) and (4)

(6) What will be the max continuous power?

Between 1/3 and 2/3 of (1)

I hope that helps.

Mariss

--- In CAD_CAM_EDM_DRO@y..., ccs@m... wrote:
>
> Is the voltage I measure on a dc servomotor being rotated by a lathe
> spindle at a known rpm essentially the same voltage I would need to
> apply to make it turn at that rate, provided I can source and (it
can
> handle) sufficient current for the torque load?
>
> Is there any easy way to estimate the current handling
> of a motor with no specs whatsoever?
>
> I know someone - I think Mariss - posted an excellent description of
> how to determine servomotors specs this way, but I have not been
able to
> find it in my search of the list archives.
>
> Thanks,
>
> Chris Stratton
>
> -------------------------------------------------
> Christopher C. Stratton
> Engineer, Instrument Maker, and Horn Player
> ccs@m... 617 628 1062
> http://web.mit.edu/~stratton/www/brassbuild.html