On 30/11/2020 09:30:42, The Natural Philosopher wrote:
On 30/11/2020 08:29, Jeff Layman wrote:
On 29/11/2020 16:53, Fredxx wrote:
On 29/11/2020 16:37:07, Andy Burns wrote:
Fredxx wrote:

Dave Plowman wrote:

What sort of resistance would you expect with an 18v cordless drill
motor?

Is this stripped-down and measuring just the windings?

I'm guessing but something in the region of 6ohmm -50/+100%

That'd only be 1.5 to 6A, so 27 to 108W, you can see 18V cordless
drills
specced as taking 30A so over 500W, some who have re-purposed drill
motors for use in robots claim nearer 900W

What is more important is that the resistance is independant of
position, subject to the segment to segment change under the
commutator. Assuming brushed of course.

Yes, my estimate is rather low.

This replacement motor has a surprising spec:
Â*Â*Â* http://www.robotcombat.com/products/BP393111-01.html

Perhaps the battery and motor control help to limit stall current rather
than just rotor resistance?

A few questions and comments...

What wire would be used in that motor, and what would be the length of
it?

According to
https://chemandy.com/calculators/round-wire-resistance-calculator.htm 10
metres of 2 mm diameter copper wire would have a DC resistance of
0.053 ohm. That motor has a resistance of 0.072 ohm, so we are in the
same ballpark. Wouldn't the motor have more than 10 m of wire? If so,
it would have to be at least 2mm diameter to keep the resistance low.
One other thing - the fusing current for 2 mm copper wire is about 220
amps, so once stalled that motor would be right on the edge of burning
out within a few seconds.

That table for the motor also says the no-load current is only 2.6
amps, so there is almost a factor of 100 between stall and no-load. Is
that usual?

The better the motor the bigger it is. The ideal motor has infinite
stall current and zero resistance.

And also zero no load current.

power is lostÂ* in electric motors in several ways.

(i) Through the rapid magnetisation and reverse magnetisation of the
armature. hysteresis means means you don't get back what goes in.

(ii) Eddy current losses. The radicallyÂ* changing magnetic fields
introduce currents in the iron which are subject to resistive losses.
That is why armatures are laminated.,

(iii) mechanical friction and bearing loss.

These three are all relatively constantÂ* at a given RPM and are
independent of the load, and are lumped into 'idle current' when
measuring the motor. Also known as 'iron losses'

(iv) heat lost due to the resistance of the motor windings (ideally
zero) known as 'copper loss'.

(v) energy lost as mechanical energy to drive the actual load.


It can be shown that peak efficiency occurs when the copper losses and
iron losses are equal.

I agree with you up to this point. Can you cite any reference as to why
this might be the case?

This model of an electric motor is not perfect, but works well enough to
predict performance quite accurately. Those of us who have measured
motor performance know that idle current varies with RPM quite a lot, so
it's important to measure it at the normal working voltage.

The theory of the motor is that It will spin up until a notional 'back
emf' marches the applied voltage.Â* The ratio of rotational speed to back
emf is expressed as the Kv or 'RPM/V'. The resistance of the motor is
deemed to appear 'in series' with the motor and therefore reduces the
applied voltage as 'seen' by the motor in proportion to the current
being drawn. Thus, in a less than perfect motor, the RPM will drop as
load is applied, and extra current is drawn in an attempt to match back
emf to applied emf.

Using longer thinner wires to wind the armature of a permanent magnet
motor simply means that its best efficiency working voltage is
increased. And the current is decreased in the same ratio.Â* In the end
the power is determined - as it is in a transformer - by the peak flux
density that can be sustained in the armature,Â* times the operating
frequency - or RPM.

So, the important thing to remember is that the resistance of the motor
has *almost no part to play* in determining what current it will draw.
That is far more determined by the *load that is applied*. In short the
motor will draw whatever current it needs to do, to supply the power to
the load, yea - even unto letting the magic smoke out in the attempt.

The only way to get more power out, is to increase the RPM. This we
found early in in making model planes fly better. Gearboxes to reduce
shaft RPM to sane levels allowed us to take armature speeds towards the
mechanical limits of the motorsÂ* = either brush bounce, or mechanical
disintegration. By increasing the applied voltage we gor more power
whilst keeping armature current - the killer - in reasonable limits. The
price was RPM up towards 100,000 , whereas a typical propellor operated
best below 10,000.

I'm surprised you haven't moved into solid state, wound stators and
spinning magnets with no wires to part with the armature?.