I'm planning to compare a couple different windings. ...prepping the experiment...

## code to change the motor amp limit

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

Last edited by devin on 01 Aug 2017, 00:02, edited 1 time in total.

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

test winding 1: 18 turns delta

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

FYI: Even though I'm not posting, doesn't mean I'm not anxiously awaiting the results....

If you're doing KV testing and are considering testing both delta and Y winding, then you should bring out both ends of each winding.

For AC induction motors they terminate all six wires as follows onto a connection block:

Now you connect all A-B-C- together and power from C+ A+ B+ (Y configuration), or connect A-C+ to first ESC phase, B-A+ to the second etc for Delta.

Now you can, without having to rewind or take t he motor apart test both the Y and D configuration!

If you're doing KV testing and are considering testing both delta and Y winding, then you should bring out both ends of each winding.

For AC induction motors they terminate all six wires as follows onto a connection block:

Code: Select all

`A- B- C-`

C+ A+ B+

Now you can, without having to rewind or take t he motor apart test both the Y and D configuration!

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

@rew On a semi-related side note I was thinking about the interesting relationship between kinetic energy of the electricity in the windings at stall compared to the kinetic energy of the unloaded rotor at no load rpm...

For example with a 100kv motor and 25v battery & 0.05ohm winding:

Battery Voltage: 25v

Winding Resistance: 0.05ohm

Stall current: 500a

Stall wattage: 12500w electrical

No Load RPM: 2500rpm

No Load RPM Rotor Kinetic Energy: 1 unit

Now we double the voltage of the battery:

For example with a 100kv motor and 50v battery & 0.05ohm winding:

Battery Voltage: 50v <- x2 increase

Winding Resistance: 0.05ohm <- no increase

Stall current: 1000a <- x2 increase

Stall wattage: 50000w electrical <- x4 increase

No Load RPM: 5000rpm <- x2 increase

No Load RPM Rotor Kinetic Energy: 4 units <- x4 increase

Simply doubling the battery voltage quadruples the electrical kinetic energy (electrical wattage) at stall, and also quadruples the kinetic energy of the rotor at no load rpm (since kinetic energy increases at the square of velocity).

In other words quadrupling the electrical kinetic energy at stall quadruples the kinetic energy of the rotor at no load rpm.

Halving the number of turns of a wye motor and doubling the thickness of the wire to retain same copper fill also causes the same quadrupling of the electrical kinetic energy at stall and quadrupling of rotor kinetic energy at no load rpm.

For example with a same size 200kv motor and 25v battery & 0.0125ohm winding:

Battery Voltage: 25v <- no increase

Winding Resistance: 0.0125ohm <- 1x(1/4) decrease

Stall current: 2000a <- x4 increase

Stall wattage: 50000w electrical <- x4 increase

No Load RPM: 5000rpm <- x2 increase

No Load RPM Rotor Kinetic Energy: 4 units <- x4 increase

In my theory, halving the number of turns of a wye motor, but not increasing the wire thickness to retain the same copper fill will only result in a doubling of electrical kinetic energy at stall (not quadrupling as in the last example) -- and this doubling of electrical kinetic energy at stall should only result in a doubling of rotor kinetic energy at no load rpm (compared to quadrupling in the previous example).

For example with a same size wye, half the turns of the 100kv (same number of turns as the 200kv), but same wire thickness as 100kv motor and 25v battery & 0.025ohm winding:

Battery Voltage: 25v <- no increase

Winding Resistance: 0.025ohm <- 1x(1/2) decrease

Stall current: 1000a <- x2 increase

Stall wattage: 25000w electrical <- x2 increase

No Load RPM: 3535rpm ??? <- 1xsqrt(2) [1x1.41421] increase ???

No Load RPM Rotor Kinetic Energy: 2 units ??? <- x2 increase ???

KV: 141.21kv instead of 200kv??? (3535rpm / 25v)

@rew ^ ...by extension the theory implies changing the wire thickness with the same number of turns changes the kv because it changes the stall electrical wattage and stall electrical kinetic energy, therefore it would according to the theory, proportionally change the rotor's acquired kinetic energy at no load rpm.

For example with a 100kv motor and 25v battery & 0.05ohm winding:

Battery Voltage: 25v

Winding Resistance: 0.05ohm

Stall current: 500a

Stall wattage: 12500w electrical

No Load RPM: 2500rpm

No Load RPM Rotor Kinetic Energy: 1 unit

Now we double the voltage of the battery:

For example with a 100kv motor and 50v battery & 0.05ohm winding:

Battery Voltage: 50v <- x2 increase

Winding Resistance: 0.05ohm <- no increase

Stall current: 1000a <- x2 increase

Stall wattage: 50000w electrical <- x4 increase

No Load RPM: 5000rpm <- x2 increase

No Load RPM Rotor Kinetic Energy: 4 units <- x4 increase

Simply doubling the battery voltage quadruples the electrical kinetic energy (electrical wattage) at stall, and also quadruples the kinetic energy of the rotor at no load rpm (since kinetic energy increases at the square of velocity).

In other words quadrupling the electrical kinetic energy at stall quadruples the kinetic energy of the rotor at no load rpm.

Halving the number of turns of a wye motor and doubling the thickness of the wire to retain same copper fill also causes the same quadrupling of the electrical kinetic energy at stall and quadrupling of rotor kinetic energy at no load rpm.

For example with a same size 200kv motor and 25v battery & 0.0125ohm winding:

Battery Voltage: 25v <- no increase

Winding Resistance: 0.0125ohm <- 1x(1/4) decrease

Stall current: 2000a <- x4 increase

Stall wattage: 50000w electrical <- x4 increase

No Load RPM: 5000rpm <- x2 increase

No Load RPM Rotor Kinetic Energy: 4 units <- x4 increase

In my theory, halving the number of turns of a wye motor, but not increasing the wire thickness to retain the same copper fill will only result in a doubling of electrical kinetic energy at stall (not quadrupling as in the last example) -- and this doubling of electrical kinetic energy at stall should only result in a doubling of rotor kinetic energy at no load rpm (compared to quadrupling in the previous example).

For example with a same size wye, half the turns of the 100kv (same number of turns as the 200kv), but same wire thickness as 100kv motor and 25v battery & 0.025ohm winding:

Battery Voltage: 25v <- no increase

Winding Resistance: 0.025ohm <- 1x(1/2) decrease

Stall current: 1000a <- x2 increase

Stall wattage: 25000w electrical <- x2 increase

No Load RPM: 3535rpm ??? <- 1xsqrt(2) [1x1.41421] increase ???

No Load RPM Rotor Kinetic Energy: 2 units ??? <- x2 increase ???

KV: 141.21kv instead of 200kv??? (3535rpm / 25v)

@rew ^ ...by extension the theory implies changing the wire thickness with the same number of turns changes the kv because it changes the stall electrical wattage and stall electrical kinetic energy, therefore it would according to the theory, proportionally change the rotor's acquired kinetic energy at no load rpm.

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

the motor spun, but then a winding stuck out & got caught on a spinning magnet so will have to re-wind... wasn't able to check the kv...

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

I was informed today of the experimental results of a test by an anonymous researcher...

15 turns vs 30 turns using same thickness wire & same termination (the 15 turn motor had half as much copper fill as the 30 turn motor).

The findings were that despite the 1/2 copper volume of the 15 turn motor, its KV was nearly exactly double that of the 30 turn motor, which strongly suggests to me that within unknown constraints, for a given termination and number of turns, kv remains constant regardless of wire thickness or % copper fill.

Regarding this theory:

^It appears to remain valid as long as copper volume stays constant to say that when conductance changes by a certain factor, for example x4, then the kv changes by the square root of the conductance change factor, for example x2, including when re-terminating between Y <-> Delta.

The experimental results do suggest however that when the copper volume in the motor changes, the square relationship between conductance and kv changes is no longer quite so simple...

15 turns vs 30 turns using same thickness wire & same termination (the 15 turn motor had half as much copper fill as the 30 turn motor).

The findings were that despite the 1/2 copper volume of the 15 turn motor, its KV was nearly exactly double that of the 30 turn motor, which strongly suggests to me that within unknown constraints, for a given termination and number of turns, kv remains constant regardless of wire thickness or % copper fill.

Regarding this theory:

^It appears to remain valid as long as copper volume stays constant to say that when conductance changes by a certain factor, for example x4, then the kv changes by the square root of the conductance change factor, for example x2, including when re-terminating between Y <-> Delta.

The experimental results do suggest however that when the copper volume in the motor changes, the square relationship between conductance and kv changes is no longer quite so simple...

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

Based on the experimental results I think these are the correct KV formulas for all winding changes regardless of copper volume or termination.

-----------------------

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/V)

C=S^2V

V=C/S^2

-----------------------

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/v)

100kv 30 Turns Delta -> 200kv 15 turns Delta (0.5 change copper & 2 change conductance)

2=sqrt(2)*sqrt(1/0.5) <-correct

100kv 30 Turns Delta -> 200kv 15 turns Delta (1 change copper & 4 change conductance)

2=sqrt(4)*sqrt(1/1) <-correct

100kv 30 Turns Delta -> 100kv 30 turns Delta (0.5 change copper & 0.5 change conductance)

1=sqrt(0.5)*sqrt(1/.5) <-correct

100kv 30 Turns Delta -> 100kv 30 turns Delta (2 change copper & 2 change conductance)

1=sqrt(2)*sqrt(1/2) <-correct

100kv 30 Turns Delta -> 50kv 60 turns Delta (1 change copper & 0.25 change conductance)

0.5 = sqrt(0.25)*sqrt(1/1) <-correct

100kv 30 Turns Delta -> 57.73kv 51.96 turns Delta (1 change copper & 0.333... change conductance)

0.57735=sqrt(0.333)*sqrt(1/1) <-correct

100kv 30 Turns Delta -> 57.73kv 30 turns Wye (1 change copper & 0.333... change conductance)

0.57735=sqrt(0.333)*sqrt(1/1) <-correct

100kv 30 Turns Wye -> 173.205kv 30 turns Delta (1 change copper & 3 change conductance)

1.73205=sqrt(3)*sqrt(1/1) <-correct

100kv 30 Turns Wye -> 173.205kv 30 turns Delta (0.5 change copper & 1.5 change conductance)

1.73205=sqrt(1.5)*sqrt(1/0.5) <-correct

^Simply, "turns" & "termination" aren't part of this KV equation, only conductance and copper volume...

-----------------------

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/V)

C=S^2V

V=C/S^2

-----------------------

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/v)

100kv 30 Turns Delta -> 200kv 15 turns Delta (0.5 change copper & 2 change conductance)

2=sqrt(2)*sqrt(1/0.5) <-correct

100kv 30 Turns Delta -> 200kv 15 turns Delta (1 change copper & 4 change conductance)

2=sqrt(4)*sqrt(1/1) <-correct

100kv 30 Turns Delta -> 100kv 30 turns Delta (0.5 change copper & 0.5 change conductance)

1=sqrt(0.5)*sqrt(1/.5) <-correct

100kv 30 Turns Delta -> 100kv 30 turns Delta (2 change copper & 2 change conductance)

1=sqrt(2)*sqrt(1/2) <-correct

100kv 30 Turns Delta -> 50kv 60 turns Delta (1 change copper & 0.25 change conductance)

0.5 = sqrt(0.25)*sqrt(1/1) <-correct

100kv 30 Turns Delta -> 57.73kv 51.96 turns Delta (1 change copper & 0.333... change conductance)

0.57735=sqrt(0.333)*sqrt(1/1) <-correct

100kv 30 Turns Delta -> 57.73kv 30 turns Wye (1 change copper & 0.333... change conductance)

0.57735=sqrt(0.333)*sqrt(1/1) <-correct

100kv 30 Turns Wye -> 173.205kv 30 turns Delta (1 change copper & 3 change conductance)

1.73205=sqrt(3)*sqrt(1/1) <-correct

100kv 30 Turns Wye -> 173.205kv 30 turns Delta (0.5 change copper & 1.5 change conductance)

1.73205=sqrt(1.5)*sqrt(1/0.5) <-correct

^Simply, "turns" & "termination" aren't part of this KV equation, only conductance and copper volume...

Last edited by devin on 14 Aug 2017, 22:52, edited 11 times in total.

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

The clear implication is if you know how much the kv has changed and how much the conductance has changed, you can determine what percentage the copper volume has changed. In other words there is a direct correlation between kv, conductance & copper mass.

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/V)

C=S^2V

V=C/S^2

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/V)

C=S^2V

V=C/S^2

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

That's what I've been saying all along.devin wrote:which strongly suggests to me that within unknown constraints, for a given termination and number of turns, kv remains constant regardless of wire thickness or % copper fill.

In practice, copper volume does not change. When you rewind you always try to cram in as much copper as you can, resulting in, hopefully, the same copper fill (about 100%: as much as will fit).you can determine what percentage the copper volume has changed.

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

rew wrote:That's what I've been saying all along.

In practice, copper volume does not change. When you rewind you always try to cram in as much copper as you can, resulting in, hopefully, the same copper fill (about 100%: as much as will fit).

^Using half the turns with the same thickness wire, the copper volume is 1/2. If wire with twice the girth was placed within the motor on the right, KV [rpm per volt] and KT [torque per amp] would stay the same as when the thinner, more modest wire was used.

Left Motor: 30 turns, delta, 47.9V, 23856erpm, 3408rpm, 71.14kv | Right Motor: 15 turns, delta, 47.6V, 47444erpm, 6777rpm, 142.3kv | 71.14kv Left Motor x 2 = 142.3kv Right Motor

@rew I agree you’ve been saying KV changes in proportion to turns and termination only and not wire thickness, which is technically true.

But it would also be true to say KV changes in proportion to conductance and copper volume only (and not termination), both of which are SI base unit derived variables in contrast to “turns and termination.” (https://en.wikipedia.org/wiki/SI_base_unit)

My aim since May has been to uncover why re-termination from wye to delta changes kv by a factor of exactly 1.73205 or the square root of 3... which I don't think anyone has yet fully described the correct answer to this but I intend to in my next posting...

...But for now there's 2 accurate formulas to calculate kv from winding changes... a new one and an old one... let's compare their differences:

-------------------

Formula 1:

KN=CT

K=(CT)/N

K = kv = max rpm per volt no load

N = turns = # of wire turns per tooth

C = constant = original kv x original # turns

T = termination = 1 for wye | 1.73205 for delta

kv x turns = constant x termination

original winding:

100kv x 30 turns = 3000 constant x 1 termination

now we change the number of turns:

50kv x 60 turns = 3000.00 constant x 1 wye termination

now we change the termination:

86.62kv x 60 turns = 3000.00 constant x 1.73205 delta termination

-------------------

Formula 1:

-Accurately predicts KV changes from winding “turns,” “constant,” & “termination” changes.

-Unfortunately, variables “Turns,” “Constant,” & “Termination” are not derived from internationally recognized SI base units such as “Kilogram,” “Second,” “Meter,” etc.

-------------------

Formula 2:

S=sqrt(C)*sqrt(1/V)

C=S^2V

V=C/S^2

S = Change Factor of KV

C = Change Factor of Conductance

V = Change Factor of Copper Volume

original winding:

100kv, 30 turns, 0.05ohm, 1 copper volume

1=sqrt(1)*sqrt(1/1)

now we change the number of turns (same wire thickness, assuming there was extra space):

50kv, 60 turns, 0.1ohm, 2 copper volume

C = (1/0.10ohm new resistance)/(1/0.05 original resistance) = 0.5 = Change Factor of Conductance

V = 2.0 = Change Factor of Copper Volume

S=sqrt(C)*sqrt(1/v)

0.5=sqrt(.5)*sqrt(1/2)

Therefore:

S = 0.5 = Change Factor of KV

Therefore:

New 50kv = Original 100kv x (S = 0.5 = Change Factor of KV)

-------------------

Formula 2:

-Accurately predicts “KV” changes from winding “conductance” and “copper volume” changes

-Accurately predicts “Conductance” changes from winding “kv” and “copper volume” changes

-Accurately predicts “Copper Volume” changes from winding “kv” and “conductance” changes

-Formula #2 has one less variable than Formula #1 -- works for both delta & wye without explicitly specifying which termination style is used

-In contrast to Formula #1, variables “Conductance,” & “Copper Volume” in Formula #2 are indeed derived from internationally recognized SI base units such as “Kilogram,” “Second,” “Meter,” “Ampere,” etc.

-------------------

Conclusion

-Formula 1 highlights the correlation between # of turns, termination, and KV, regardless of wire thickness, but requires one additional variable & doesn’t utilize exclusively SI base unit derived variables.

-Formula 2 highlights the correlation between conductance, copper volume and KV, regardless of termination -- only requires 3 instead of 4 variables (termination style delta/wye is not needed to be specified for the formula to work), and unlike Formula 1, Formula 2 utilizes exclusively SI base unit derived variables.

-------------------

devin 14 May 2017, 23:54 wrote:i'm trying to conceptualize in my mind how sqrt(3) enters into the equation of the difference in kv between wye and delta...

The whole point of this is since mid-May I've been trying to fully understand exactly why re-termination from wye to delta results in an increase factor in rpm per volt of exactly sqrt(3) or 1.73205...

So... why exactly does re-termination from wye to delta result in a change factor of kv equal to irrational number square root of 3?

I’m pretty sure I’ve solved it, & will answer this in my next post…

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