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

--
*With her marriage she got a new name and a dress.*

Dave Plowman London SW
To e-mail, change noise into sound.

On 29/11/2020 15:20:48, Dave Plowman (News) wrote:
What sort of resistance would you expect with an 18v cordless drill motor?


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

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.

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.

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:
http://www.robotcombat.com/products/BP393111-01.html

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

On 29/11/2020 16:37, 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%

I can tell you with certainty it will be less than an ohm



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

Exactly,. at at least 60% efficiency so:

Input current = 500w/18v= 27A...40% losses 200W, so resistance
200/(27x27) = 0.25 ohm.

Even a poxy little 20W brushed motor us still under an ohm




--
"The great thing about Glasgow is that if there's a nuclear attack it'll
look exactly the same afterwards."

Billy Connolly

Its very hard on low voltage motors to tell if there are shorted turns as we
are talking fractions of an ohm here, I can tell you from my model boat
days, such a tiny short makes the motor unusable and eats the battery of
course. If we had an idea of your problem?
Brian

--

This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...

Blind user, so no pictures please
Note this Signature is meaningless.!
"Andy Burns" wrote in message
...
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.

They would do that, certainly some of the small model makers drill control
circuits stop stalling from frying the motor.
Brian

--

This newsgroup posting comes to you directly from...
The Sofa of Brian Gaff...

Blind user, so no pictures please
Note this Signature is meaningless.!
"Fredxx" wrote in message
...
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:
http://www.robotcombat.com/products/BP393111-01.html

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

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:
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?

--

Jeff

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 lost 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 radically 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 constant 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.

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. 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. 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, 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 motors = 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.


-

--
"The great thing about Glasgow is that if there's a nuclear attack it'll
look exactly the same afterwards."

Billy Connolly

On 30/11/2020 09:30, 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 lost 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 radically 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 constant 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.

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. 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. 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, 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 motors = 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.

Very interesting. Thanks for spending the time putting this together.
It's been saved for future reference.

--

Jeff

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?.

On 30/11/2020 08:29, Jeff Layman wrote:
On 29/11/2020 16:53, Fredxx wrote:

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,

Unless you are doing the adiabatic calculation (for a given speed of
trip), you can only speak of fusing current in the context of its heat
loss to the environment. Which is a lot less in the middle of a winding
than in free air.

On 30/11/2020 12:12:57, Fredxx wrote:
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?

So your post was unjustifiable twaddle, as per usual?