Do bigger motors provide more power if everything else in the setup stays exactly the same?

I see a lot of people putting bigger motors on their boards for more power

I’m not extremely well versed in the science behind this, but my understanding is that a bigger motor will actually require more power put in to achieve the same performance as a smaller motor on lower settings. Is this correct?

I understand that a larger stator should be capable of accepting more power, and more than likely has better thermal capabilities so that it can perform for longer…

But my question to you fine folk is that if you only replace the smaller motors for bigger motors, and leave everything else (including your current settings) the same, will performance be worse, same or better?


bigger motor needs bigger power. e=mcfuck if I know though.


You consumption will go up, but if the motor KV is the same the only difference will be consumption and cooler running temps…I think


In every way that actually matters, this is the case.


This is an interesting hypothesis. Is it because more wattage needs to go into the bigger motor to achieve the same current?

If there is more wattage going in, then surely there is more power right?

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The motor will have more winding resistance because of the bigger stator, so it will actually in theory pull more current to get you the same performance as the smaller motor


Riiiight, so more current out of the battery for the same current into the motor due to higher resistance.


i would say yeeesss, cause the more stator volume you have the more torque and power you can draw

it’ll be at the cost of efficiency of course blah blah

i would definitely be able to notice a difference between say a 2004 motor and 2306 of the same manufacturer, same kv, same everything, same props, and build on my quads

i dont no the science but in practice i think there’s some difference

maybe smaller motors get saturated much more, meaning it’s power is overall less than a larger motor at the same current and viltage idk

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It wasn’t a yes or no question :rofl:

I can’t see how you would have more power delivered to the wheels when everything else apart from the size of the motor stays the same. It does make sense to me that the bigger motor would draw more power for the same performance and would likely perform better from a thermal perspective, but what I really wanna know about is the power to the ground.

It’s not a defining contribution to this thread, but i have never seen an esk8 calc that asks you for your stator size. It only ever asks for power values and kv.

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Woulda thought a larger stator diameter could maybe provide more torque. Not sure that equates to more power tho. Windings at a larger distance from shaft is like a longer lever arm…?


More torque would equal more power to the definition i have constructed in my brain for sure.

The length of the stator affecting power is an interesting proposition… say you had a motor with exactly the same copper mass in the stator but one was short and girthy, but one was long and skinny, you reckon the longer one would have more power?


Preface: I know fuck all about technical aspects of motors and their performance :joy:

Expand the example, one motor with very wide but short stator. Other with very narrow but super long.

My gut feel would be longer but narrower stator could potentially spin to higher rpm’s with less power input?

Wider but super short stator potentially more torque but might need more power input to spin at the same speed as the skinny one?

Like a torquey big block vs high revving race engine…?


I am also in this boat :rofl: that’s why we need big brains :grin:

I can see your reasoning with the short and girthy motor providing more torque tho.


We’ve already had nearly the same question asked:

I’m basing my answer off the agreed upon facts in that thread, because my memory is poopoo.

As for your question: when you increase motor size, if the winding thickness stays the same your KV will change, or you have to change the winding thickness to have the KV stay the same. So you can’t actually keep all other factors equal.

In any case, data shows that larger motors have a smaller internal resistance. This makes them more efficient at converting whatever electrical power you do put through them into mechanical power rather than heat.

(In practice since our ESCs drive the motor with a specific amount of current, this would mean less voltage is required to achieve that amount of current, i.e. less battery power (thus battery current) is needed to drive bigger motors at the same current level.)

So yes, bigger motors do actually provide more power (either in top speed or torque or both, depends on the new KV) at the same settings, although that amount is likely negligible. The biggest difference is that now you can actually push the motors harder before overheating, i.e. raise your ESC settings.


I thought it allowed a specific amount of current but output a specific wattage? P=V*I and all that

Only if you’re already saturating them or hitting temp cutoffs :wink:

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Thanks, i guess i would have found that, had i used the search button. But i don’t think its unhealthy for the topic to be expanded upon and with your help they have now been inextricably linked together.


I’m afraid I don’t understand the question. Could you rephrase it please?

it certainly doesn’t hurt to revisit topics after some time has passed. If anything starting a new thread that ultimately points to an older one will help somebody find what they’re looking for.

I just went from 6374s to 6396s without changing the settings and on the street I can’t feel the difference, other than the motors are slightly cooler, which is the main reason I went bigger.

I haven’t fully read the older thread but I’m kind of interested in the amount of braking current generated by the same KV motor of a different size. Is there any difference there?


It depends. There are two main kind of losses in the motors, copper losses and iron losses. Copper losses = I^2*Rphase*constant

If my understanding is correct, this constant is basically the integral of the three sine waves…
In FOC the graph of the phase currents looks the same as the graph of a three phase AC. The integral of a sine wave under the period of pi is 2. As the period of a full sine wave is 2*pi, the absolute value of the area under a full sine wave is 4. We have 3 of these sine waves under the period of 2*pi, so this constant should equal 3*4/(2*pi) which equals 6/pi

So we can get the copper losses by I^2*Rphase*6/pi.
If we move to a bigger motor while keeping the same KV, the resistance goes down, therefore you will have less copper losses. With the same input power, lower losses mean a higher output power due to a higher efficiency. Note: the difference in output power is nowhere near enough that you can feel it.

The other type of losses is iron losses. Iron losses equal the idle current of motor (I0) times the BEMF which it is spinning at, which is nearly the same as the voltage applied to the coils. This BEMF roughly equals Vbattery*duty_cycle. When current flows through the motor, the actual voltage varies a bit, usually not more than by 2V, so lets just leave this out not to overcomplicate the equations.

So iron losses equal I0*Vbattery*duty_cycle. The idle current depends on the quality of the motors construction, better magnetic properties leading to a lower idle current. It also depends on the motor’s size, higher size means higher idle currents. And it’s also influenced by the rotational speed of the motor, by how much, I don’t exactly know right now.

However, what we can say for sure, is that if we take the exact same motor, and just scale it up in size, keeping the quality of the construction as a constant, then the idle current and thus the iron losses increase. Meaning, that a bigger motor has higher iron losses all else being equal. Which means for the same input power, we get less output power. But again, the difference is so small, that you won’t feel the difference.

This was my take on it, if there are some electrical engineers here, feel free to correct me if i messed something up, I am not an electrical engineer yet. (although I am currently enrolled in a university course as a first semester EE bachelor student)