More RPM = more windings or stronger magnets ?

I don't understand the theory well enough to answer your question, but I suspect the question isn't valid - I think it might be too simplistic to have an answer. There's no such thing as "torque not being an issue". PM DC electric motors have a torque/RPM curve. You trade off one for the other as a function of the load you have placed on the motor. The speed the motor will actually turn in any given application is a function of the load you put on the motor. At low speed, the motor will produce more torque, and at high speed, the motor will produce less torque. How fast it ends up spinning is a function of where the application will stabilize along the torque/RPM curve of the motor.
The two commonly specified points on the curve are the stall torque (the point where the speed is zero and the torque is at it's max), and the no load RPM (the point where the torque is zero, and the RPM is at a max.
Changing the voltage, or the magnets, or the windings, will change the curve. Increasing the voltage, I believe, will move the curve up so that both torque and max RPM will increase.
But changing the windings, or the magnets, might change the shape of the curve, causing one end to increase while decreasing the other (just a guess on my part). So it might increase the RPM in the low torque range of the curve while decreasing the RPM in the high torque side of the curve. If that was so, the change might both increase and decrease the RPM (if you try to pretend the torque doesn't matter).
So, I wonder for example, if increasing the magnet strength might increase the stall torque, and increase the RPM when under high torque load, but could actually decrease the no load RPM rating of the motor because you have increased the back EMF as the motor is spinning.
If this is the type of dynamics that actually happen in motor design, your question has no simple answer because it would depend on the specifics of the motor in question and the application it was used for (how much load it was placing on the motor at different RPM levels).
Then of course the winding question is also ignoring wire resistance. If you change the windings, do you also get to change the wire so the total winding resistance stays the same? Or are you talking about adding or removing windings without changing the gauge of the wire (which of course isn't for the most part practical because a given motor only has so much space for windings so you can't just add more without reducing the wire size which is yet another unspecified part of your question).
And then there's the 12V issue. Real 12 V sources like batteries have power limits and their voltage will drop as you draw more current from them. Are you assuming infinite power with a fixed 12 volt supply?
I would guess that in real PM motors, the resistance of the winding will play a very important role in the no load RPM speed of the motor if you have a fixed voltage infinite current supply. So again, you probably need to be more specific about whether you question is purely theoretical and applies to an example where the we pretend the wires have no resistance and the armature has no friction or whether you question is more real-world in nature.
Or maybe, none of this is important because like I said, I don't understand motor theory well enough to answer the question. :) Maybe John, who knows everything, and knows a lot about electric motors, can shed more light on it. :)
My guess however, is if there is one simple correct answer, is that if you increase windings while keeping total winding resistance the same you will get more torque and higher RPMs. And if you use a stronger magnet, you will likewise get more torque and higher RPMs. But this would be contrast to what the other person posted who thought you needed to reduce windings to get higher RPMs.
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Curt Welch http://CurtWelch.Com /
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Weaker magnets and less windings. If you remove magnets from running motor RPM will increase so high that motor will explode (if your power supply can give enough power).
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So, you are saying the magnets are not needed to make a PM DC motor spin? That's absurd.
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Curt Welch http://CurtWelch.Com /
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No, this is not what I as saying.
This is simplified formula:
Voltage of power supply = Number of windings * RPM * Strength of magnetic field + current * ohm resistance
Removing the magnet is a process. During this process strength of magnetic field goes to zero. If power supply can produce enough power, according to the formula, RPM goes very high (if resistance is low). After you removed the magnet, strength of magnetic field is zero, formula is U=I*R (Ohm's law), torque is zero and motor will (if still in one piece) eventually stop running.
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On Sep 10, 5:24 am, snipped-for-privacy@kcwc.com (Curt Welch) wrote:

I showed the formula earlier. Any electrical engineer will verify this. If you reduce the field so much, the torque will eb so weka that the motor will not spin. It is possible to explode a motor. A series connected one is the classic on no-load or a differential compound one.
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On Sep 10, 5:24 am, snipped-for-privacy@kcwc.com (Curt Welch) wrote:

I showed the formula earlier. Any electrical engineer will verify this. If you reduce the field so much, the torque will be so weak that the motor will not spin. It is possible to explode a motor. A series connected one is the classic on no-load or a differential compound one.
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Do you mean that you could actually remove the magnets to achieve higher RPM ? If so, what would the emf (on the armature) push against ?
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This is correct. Field wekening speed up a motor but r educes torque since Torque proportional flux X armature current.
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pogo wrote:

Much confusion here. Start by reading this DC motor tutorial from MIT:
http://lancet.mit.edu/motors
DC motor behavior is relatively simple. DC motors are also generators, and if rotated, will generate a voltage. The unloaded voltage from the motor being rotated at a given RPM is the "back EMF" some have mentioned. When running as a motor, the motor still generates a "back EMF", and as the back EMF approaches the input voltage, the motor goes no faster. The speed a motor will achieve with no mechanical load is the "no load speed", and output torque is zero at that speed, because, of course, there's no load.
If the motor is not permitted to rotate at all, it will generate some torque at zero RPM. That's the "stall torque".
On a graph of mechanical load vs RPM, the two points defined by stall torque (max torque, zero RPM) and no load speed (zero torque, max RPM) define motor behavior. You can draw a straight line between those two points, and for a simple DC motor, that's its torque curve. The MIT site has a great little lab apparatus for demonstrating this and a demo video.
Driven from a constant voltage, current consumption will be maximum at stall, and near zero at no load max RPM. Maximum output mechanical (torque * RPM) is obtained at half the no load speed.
So those are motor basics.
The original poster wants their motor to go faster. The question is whether their motor has any mechanical load on it. If it does, it won't max out at no-load speed; it will max out at some lower speed where the torque needed to move the load is equal to the motor's output torque.
For the no-load case, going to fewer turns on the windings will indeed increase the motor's top speed. But it will also increase the motor's current consumption at low speeds. This may burn out the windings. Large DC motors often have a resistance in series with the motor during startup for this reason.
Note that DC motors are generally heat-limited. The limitation is watts dissipated, not input voltage.
If the original poster wants their motor to go faster with some mechanical load on it, putting fewer turns on the armature may not be the answer. The motor may just overheat.
Realistically, nobody rewinds small DC motors. They're so cheap it's better to find a motor properly rated for the job.
Or crank up the input voltage. A DC-DC power supply with current limiting would be a good way to step up the voltage without burning out the motor. A 12V motor can probably tolerate 50VDC provided the current at startup is limited. Heat is the problem, not insulation voltage limits.
John Nagle
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In hindsight, I can see where I should have specified no load. It might have removed a good bit of confusion.

Yep - right again! I was just trying to establish a relationship between motor speed vs. things you *could* change assuming a constant voltage.

And this is what I will probably do in the real world application I have in mind. I was trying to minimize the problem but I think instead I just made it more confusing. Whoops!

Anyway ... thanks as usual!
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