I guess i'm saying KV is inversely proportional to the number of turns. You're saying that resistance is related to the square of the number of turns if you keep the copper volume constant. True. So now KV becomes inversely proportional to the resistance. Call the inverse of the resistance conductance, and you get KV = C(motor) * C(termination) * sqrt(conductance). Sure.

But... does this formula help you in some insight or quick calculations if you make an easy change on your motor?

The formula for the period of a swinging weight is does not include the weight. This leads to the insight that the weight does not matter.

## code to change the motor amp limit

### Re: code to change the motor amp limit

rew wrote:I guess i'm saying KV is inversely proportional to the number of turns. You're saying that resistance is related to the square of the number of turns if you keep the copper volume constant. True. So now KV becomes inversely proportional to the resistance. Call the inverse of the resistance conductance, and you get KV = C(motor) * C(termination) * sqrt(conductance). Sure.

But... does this formula help you in some insight or quick calculations if you make an easy change on your motor?

The formula for the period of a swinging weight is does not include the weight. This leads to the insight that the weight does not matter.

@rew so it seems we agree now that at constant copper volume, the change factor of kv is always the square root of the change factor of conductance, regardless of wye or delta termination, but assuming identical rotors, stators and winding quality, exactly as depicted in this drawing:

rew wrote:Call the inverse of the resistance conductance, and you get KV = C(motor) * C(termination) * sqrt(conductance).

Actually the formula for constant copper volume winding changes would be much simpler:

sqrt(conductance change factor) = kv change factor

^no termination factor, motor constant, or number of turns is necessary at all to calculate and predict changes to kv at constant volume copper.

devin wrote:this formula appears to confirm that in any winding-only change when copper volume stays constant, the factor of change of kv is always square root of the factor of change of winding conductance.

If we look at the formula for the last transformation from wye @ 100kv to wye @ 150kv at constant copper volume:

150 = 7.79625 x 1 x 19.24

-conductance change factor: x 2.25054

-kv change factor: x sqrt(2.25054) = 1.50017

^if we switch this copper winding for an equal volume & geometry & number of turns of tungsten wire, it seems obvious to me this would change the motor constant number (7.79625).

let's imagine the motor constant number changes from 7.79625 to an even 5 in the switch to an equal volume and turns from copper to tungsten. (different resistive power loss-- different km)

150kv = 7.79625 x 1 x 19.24 turns

X = 5 x 1 x 19.24

96.2kv = 5 x 1 x 19.24 turns

^ @ rew we see from your formula a different motor constant number at the same winding volume and number of turns (less conductive material) has lowered the kv.

^so here is an example with same number of turns, same volume, same geometry but different material, different conductance, therefore different KM, therefore different KV.

^it seems obvious that similarly, a thinner wire of copper (less total copper volume, higher resistive loss, less conductance, same turns) would have a different resistive loss at the same number of turns from the original copper volume, which similarly would change KM, & therefore similarly change KV despite the same copper material & same length wire & same number of turns.

### Re: code to change the motor amp limit

I agree that the resistance is proportional to kv. Yes. I haven't disagreed on that. I still don't believe it causes it. Magnetic inductance I believe decides kv/kt (which are inverse). The conversion of turns physically of the wire ,with a core or not, is Linearly related I think into infinity, u can float a frog, and the voltage produced by that field is proportional to the field, and that field and voltage are MORE SO the DETERMINING factor in why the rotor gives out its torque and speed curve of a motor. Is my best basic math hunch.

Wattage passing a core in a wire of any size must have its magnetic field concentrated, overlain, to make a strong enough field to strengthen it. Ideally there would be no back emf that limits speed and people do field-weakening to skirt around the back-emf limit. The weakening of the magnetic field not the changing of the wire resistance is my guess at the method but I don't know how it really works .

Wattage passing a core in a wire of any size must have its magnetic field concentrated, overlain, to make a strong enough field to strengthen it. Ideally there would be no back emf that limits speed and people do field-weakening to skirt around the back-emf limit. The weakening of the magnetic field not the changing of the wire resistance is my guess at the method but I don't know how it really works .

### Re: code to change the motor amp limit

OK. Now I get it.

Suppose I have two Y wound motors. One I rewire with thicker wire to make the end-to-end resistance 3x smaller. The second I rewire to delta. You will not be able to measure the difference between the two changed motors. Terminal resistance and KV will be identical.

A useful insight after all!

Suppose I have two Y wound motors. One I rewire with thicker wire to make the end-to-end resistance 3x smaller. The second I rewire to delta. You will not be able to measure the difference between the two changed motors. Terminal resistance and KV will be identical.

A useful insight after all!

### Re: code to change the motor amp limit

rew wrote:OK. Now I get it.

Suppose I have two Y wound motors. One I rewire with thicker wire to make the end-to-end resistance 3x smaller. The second I rewire to delta. You will not be able to measure the difference between the two changed motors. Terminal resistance and KV will be identical.

A useful insight after all!

[20 May 2017, 05:50]

devin wrote:4. Now we shorten the original wye winding to 1/3rd the original number of turns as delta retermination, while keeping the same crossection as original wye winding -- this is 1/3rd the original copper volume

devin wrote:^this wye is not the same number of turns as the wye in example 3, but is it the same kv anyway because of the same conductance?

@rew so about 8 days ago I asked if you could take 2/3rds of the turns out of a delta and retermimate to wye and get identical performance to the delta. (wye now has same conductance and resistance as the original delta but 1/3rd the turns and 1/3rd the copper volume.) my theory is the delta will have the same kv as the wye with 1/3rd as much copper and turns & would have identical performance to the delta. what's your opinion?

### Re: code to change the motor amp limit

With 1/3rd the copper, you cannot get "identical performance".

Because I think you can substitute thinner wire (but same number of turns) at no damage to the KV, it is easy to create a configuration with the same KV, but less performance. Thinner wire, or "less copper" as you say, will increase the losses.

Because I think you can substitute thinner wire (but same number of turns) at no damage to the KV, it is easy to create a configuration with the same KV, but less performance. Thinner wire, or "less copper" as you say, will increase the losses.

### Re: code to change the motor amp limit

rew wrote:With 1/3rd the copper, you cannot get "identical performance".

Because I think you can substitute thinner wire (but same number of turns) at no damage to the KV, it is easy to create a configuration with the same KV, but less performance. Thinner wire, or "less copper" as you say, will increase the losses.

devin wrote:Re-Winding the 100 KV Wye to 150 KV Wye Keeping Same Copper Volume:

-Wye

-150kv = 100kv original wye x 1.5 [kv change factor 1.5] <- more kv

-19.24 turns = 28.86 original turns / 1.5 <- less turns

-2.59807 cross section = 1.73205 original cross section x 1.5 [keeps copper volume constant after turns reduction] <- more cross section

-1 copper volume <- same copper volume

-0.03688 ohm resistance lead-to-lead = (((original 0.0830 ohm)/1.5 [less turns])/1.5 [more cross section]) (0.01844 ohm VESC detection) [resistance change factor 0.44433...] <- 0.44433 times as much resistance

-27.11496 siemens conductance = original 12.04819 siemens x [2.25054 conductance change factor] = 1 / 0.03688 ohm <- 2.25054 times as much conductance

...

150kv = 7.79625 [A] x 1 [B -- Wye] x 19.24 [C]

150 = 7.79625 x 1 x 19.24

150 = (2.25 x 3.46500 original Constant(motor) value) x 1 x 19.24

...

-conductance change factor: 2.25054 | 12.04819 original siemens x 2.25054 conductance change factor = 27.11496 new siemens

-kv change factor: sqrt(2.25054) = 1.50017 | 100 original kv x 1.50017 kv change factor = 150.0 new kv

-notice motor(constant) number has necessarily increased by the same factor [2.25] as the conductance change factor [2.25] for the equation to provide a valid answer

@Rew-- looking at the equation for the rewinding wye 100kv to wye 150kv at same copper volume:

150kv = 7.79625 x 1 x 19.24 turns <-- 150kv wye formula, 1 copper volume

@rew What will happen if we reduce this wye 150kv copper volume precisely in half while keeping the same resistance by reducing wire length by a factor of exactly 1 / sqrt(2) (less turns and less resistance) and decrease wire cross section by a factor of exactly 1 / sqrt(2) (less cross section and more resistance)? This should be precisely identical resistance & conductance to the original 150kv wye wire at 1/2 the original copper volume.

Starting with equation from 150kv wye example:

150kv = 7.79625 x 1 x 19.24 turns <-- 150 kv wye at 1 copper volume (original copper volume)

K kv = (7.79625 x X) x 1 x (19.24 turns / sqrt(2)) <-- reducing the number of turns by a factor of 1/sqrt(2)

K kv = (7.79625 x X) x 1 x (13.604 turns)

^According to my theory, since the conductance is identical to the original winding, the kv should be identical, regardless of the wire volume, so these are my predictions:

150 kv = (7.79625 x X) x 1 x (13.604 turns)

150 kv = (7.79625 x sqrt(2)) x 1 x (13.604 turns)

150 kv = (7.79625 x 1.41421) x 1 x (13.604 turns)

150 kv = (11.02556) x 1 x (13.604 turns) <-same kv & resistance wye? 0.5 copper volume?

^So according to my theory this is a wye 150kv with half the copper volume of the original wye 150kv, with identical resistance, conductance, kv and performance as the original wye 150kv -- i think.

### Re: code to change the motor amp limit

Look you can draw circles and squares all day, but I don't understand what you're getting at.

You claim to have found a relationship between the conductance and the KV. If now you're saying that based on that formula you can make a better motor with less copper, then I'd say that your formula is not correct after all.

I have studied your post a bit better now. Your formula was deduced under the assumption that the copper volume was kept the same. Now you're changing the copper volume and you start getting funny results. The good news: your formula still stands. The bad news: You are using it wrong.

So as to your question: When you reduce the copper volume by a factor of two, and reduce the number of turns, I would say the KV increases. In your example by a factor of sqrt(2).

You claim to have found a relationship between the conductance and the KV. If now you're saying that based on that formula you can make a better motor with less copper, then I'd say that your formula is not correct after all.

I have studied your post a bit better now. Your formula was deduced under the assumption that the copper volume was kept the same. Now you're changing the copper volume and you start getting funny results. The good news: your formula still stands. The bad news: You are using it wrong.

So as to your question: When you reduce the copper volume by a factor of two, and reduce the number of turns, I would say the KV increases. In your example by a factor of sqrt(2).

### Re: code to change the motor amp limit

rew wrote:Look you can draw circles and squares all day, but I don't understand what you're getting at.

@rew If one wants to represent the square root of 3 with 100% accuracy in decimal form this takes infinite paper volume (more paper mass than the entire solar system including the sun and jupiter... in fact more paper mass than the entire milky way galaxy and by extension more paper mass than the amount of mass which is visible in the entire observable universe). It isn't merely my opinion that it takes less paper to represent "irrational" numbers like the square root of 3 with 100% accuracy through geometric representations rather than decimals. Also, for me, sometimes complex ratios are easier to understand visually through pictures, charts or graphs -- & professionally I am a visual artist & yet also a software engineer which may be another reason I'd whimsically prefer an equally valid visual geometric proof over an equally valid decimal numeric proof.

rew wrote:You claim to have found a relationship between the conductance and the KV. If now you're saying that based on that formula you can make a better motor with less copper, then I'd say that your formula is not correct after all.

@rew Less copper at same torque per amp, kv, wattage and conversion efficiency would improve the power to weight ratio, producing a slightly "better" motor with less copper if one is considering power to weight ratio (or if a motor manufacturer's accountant is considering the copper budget).

rew wrote:I have studied your post a bit better now. Your formula was deduced under the assumption that the copper volume was kept the same. Now you're changing the copper volume and you start getting funny results. The good news: your formula still stands. The bad news: You are using it wrong.

^mathematical evaluation of this in the next posting....

rew wrote:So as to your question: When you reduce the copper volume by a factor of two, and reduce the number of turns, I would say the KV increases. In your example by a factor of sqrt(2).

^mathematical evaluation of this in the next posting....

### Re: code to change the motor amp limit

Your pencil has finite thickness. Your compass is not perfect, you have to visually set it to the right width. My calculator can EASILY outperform your not 100% accurate drawings.devin wrote:rew wrote:Look you can draw circles and squares all day, but I don't understand what you're getting at.

@rew If one wants to represent the square root of 3 with 100% accuracy in decimal form this takes infinite paper volume (more paper mass than the entire solar system including the sun and jupiter...

But I prefer to do the sensible thing with decimals and cut them off somewhere in the neighborhood of where the accuracy stops. (In college I had to do a lab assignment. I jotted down numbers with one extra digit than the rest. when the instrument was about a third between two marks, I'd jot down say 1.23 when the needle was about a third along 1.2 to 1.3. "You can't see if it's 1.22 or 1.24, so that last digit is not valid!" they said. True. But I have a +/- 1 or 2 in that last digit, the others have a +/- 5 because they are rounding the whole digit away. So even if I'm not 10x more accurate, I'm still more accurate. )

Yes, but your simple: less copper, same KV motor will not have the same wattage. It can handle less current. In the extreme case we used 100x thinner wire, but the same number of turns. So now you have a wimpy motor that can handle 100x less current (at best) (but the same RPM/V, same maximum voltage, but 100x less wattage (formally called power) ).@rew [b]Less copper at same torque per amp, kv, wattage and conversion efficiency would improve the power to weight ratio, producing a slightly "better" motor with less copper if one is considering power to weight ratio (or if a motor manufacturer's accountant is considering the copper budget).

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