Re: comparative efficiency test

On Nov 10, 6:21 pm, Camilo <camilo...@xxxxxxxxxxxxxx> wrote:
On 9 nov, 23:33, nottooo...@xxxxxxxxxxx wrote:

On Nov 10, 3:29 pm, Camilo <camilo...@xxxxxxxxxxxxxx> wrote:

Power out = Torque * angular velocity
(Torque in Nm and angular velocity in rad/s)

well but without using a torque meter ¿how would you measure motor´s
torque?

You can wind a string on the shaft and use it to lift a weight.

You can use F=mg and torque = F*r to find the torque from that.
(assuming steady state)

This is handy because you can also use it to measure speed at the same
time.

that ´s the right answer... I really had thought about that method
but I was unsure of it, but you confirmed to me. Obviously isn´t a
professional solution but for an initial test is okay I guess.
Well, so when I perform the test, ¿could I affirm (by common sense)
that the motor what lift faster the load is the most powerful motor of
both?

The method is actually pretty accurate, as long as the string on the
shaft doesn't pile up into multiple layers (use a long shaft or feed
it back out the other side to another, smaller weight). But yes, more
suitable for slow or geared motors. And no you can't assume the faster
lifter is more powerful. You'd need to repeat it with different masses
to get a proper picture of performance. Different motors will have
their peak power output at different loadings and speeds - and their
peak efficiency elsewhere again.

If you have a way to measure RPM, Helmut's flywheel method might be
even easier.

In both cases, if it's accelerating, don't forget to take into account
the rotational intertia of the armature.


.

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