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What's_nitro?
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03.25.2008, 11:27 PM

Have you considered doing it yourself? I imagine the stator is pretty big inside that motor so it may not be that hard to do.

Now, would it be half as many turns for twice the kV, and just use a larger wire? Or is it not that simple?
   
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Takedown
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03.25.2008, 11:33 PM

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Originally Posted by What's_nitro? View Post
Have you considered doing it yourself? I imagine the stator is pretty big inside that motor so it may not be that hard to do.

Now, would it be half as many turns for twice the kV, and just use a larger wire? Or is it not that simple?
Ill be using 22awg wire found in nice 1/10 540 racing motors like sike did. It really doesnt matter what kv range Ill be getting after its rewound as long as its better than what it is right now. If thats the case then Ill try my hardest and maybe Ill have you guys guide me threw this step by step. I really want to pull some power out of my e-warhead conversion and this looks like the perfect motor. Just to low of kv... I also have to grind both ends of the shaft to 8mm to fit the couplers.
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johnrobholmes
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03.25.2008, 11:59 PM

Winding isn't too hard, but speeding that outrunner up might cause problems. What target RPM are you trying to get it to hit, and what is the magnet/stator count?

For small wind changes, the kV is tied to winding in a semi linear fashion. It is really complicated to actually calculate them off the cuff though.

Taken from Reliance.com, but there are other forces to be accounted for.

This equation can be further developed to obtain the voltage for movement of a conductor at constant velocity through a uniform magnetic field:
E = NBLv sin P x 10^-8

where:

E = voltage
N = number of turns
B = flux density in lines per square inch
L = length of the conductor in inches
v = velocity in inches per second
P = the angle between the conductor and flux field



You do realize you just ordered a 2 pound motor right? :D

Last edited by johnrobholmes; 03.26.2008 at 12:00 AM.
   
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Takedown
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03.26.2008, 12:04 AM

Quote:
Originally Posted by johnrobholmes View Post
Winding isn't too hard, but speeding that outrunner up might cause problems. What target RPM are you trying to get it to hit, and what is the magnet/stator count?

For small wind changes, the kV is tied to winding in a semi linear fashion. It is really complicated to actually calculate them off the cuff though.

Taken from Reliance.com, but there are other forces to be accounted for.

This equation can be further developed to obtain the voltage for movement of a conductor at constant velocity through a uniform magnetic field:
E = NBLv sin P x 10^-8

where:

E = voltage
N = number of turns
B = flux density in lines per square inch
L = length of the conductor in inches
v = velocity in inches per second
P = the angle between the conductor and flux field



You do realize you just ordered a 2 pound motor right? :D

This is why I want someone else to rewind it for me. Im horrible at math and im scared I'll screw up. Oh and no I didnt order it yet. Most likely will though. The motor rpm I want is around 600kv or so... Its a 14 pole, 52/40 size stator.

Last edited by Takedown; 03.26.2008 at 12:10 AM.
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johnrobholmes
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03.26.2008, 12:21 AM

If you want 500 to 600KV, just get a motor that is in that target range. A 16 ounce motor would be plenty, even a 10 ounce outrunner is a lot for a vehicle.

What voltage do you plan to use?
   
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