Kingfish said:
John, I struggle to believe the stuff you write. You have mixed metaphors and facts and your explanations do not add up. Forget trying to explain this myth business… you are not conveying a message that is believable.
You appear to be unable to provide us with Math that proves your theory, yet you mock those of us that can. I asked you to clarify and you ran away. I want to see your math. Show us your evidence where you got this idea that all winds perform the same under the same power and load, and this relationship with copper-fill.
Be concise.
Where's the beef? KF
You guys who supposedly know the math should be embarrassed that you can't figure this one out, because it's actually quite simple. Miles has been telling people the same thing for years, which is how I learned to step back from the trees on this point to see the forest. He's smart enough stay out of any fracas. Me, I'm sick of seeing forum members ripped off by the idea that something called a high torque model of a given motor is possible.
I've proven that myth false with 6 years of real world use. I posted a chart shared by Astro that was compiled from the results of actual tests, which clearly demonstrates the myth to be false. Do you even understand the relationship of Kt and Kv how it applies to the number of turns on the teeth of a given stator? Can you at least agree that with a fixed wheel size and load, that the torque required from a motor to turn a fixed rpm (fixed speed) is a fixed amount of torque? Also that with a given rpm, the iron losses and parasitic losses of bearings, tires, etc. are the same, which leaves us only the copper losses to compare which motor does a better job of climbing the hill, ie producing the same torque at the same rpm with greater efficiency?
Let's examine the simplest case for comparison, which is 2 otherwise identical motors, one with double the turns on each stator tooth, ie double the Kt, and half the Kv, compared to the other. I'll refer to that one as the slow wind, and the other which has double the Kv as the fast wind. For the same copper fill (total copper on each tooth), the motor with twice as many turns can have only half the number of strands on each turn, so the resistance per unit of length is double. Twice as many turns means the length of wire on the windings is also twice as long, which brings the resistance in the copper up to 4X that of speed wind.
Now the simple formula you and the other myth promoters have obviously overlooked:
copper losses (heat created in the windings) = current squared X resistance.
Let's assign some amounts for the fast wind motor climbing the hill at the given speed. voltage is V, current is A, and phase to phase winding resistance is R. That puts power in at V*A and copper losses at A*A*R.
Now lets look at the slow wind motor. It has twice the turns, so it needs only half the current to make the same torque, but it needs double the voltage to get the same rpm. Power in is 2V*A/2, which is the same as the speed wind. We already know the iron losses are the same, so let's look at the copper losses, and don't forget resistance is 4 times higher with double the length and half the thickness to achieve double the turns. A/2*A/2*4R, and low and behold copper losses are the same. The math doesn't get any more simple.
Wow, speed and torque are the same, so power out is the same. Power in is the same. We already know the iron losses are the same, and now we know the copper losses in the motor are the same, so that means motor efficiency is the same. iow Performance is identical. With torque, rpm, power in, power out, efficiency all identical, please explain how the so called torque motor is better at climbing hills.
I can't make it any simpler for you to see KF. Either you get it or you don't.
