## code to change the motor amp limit

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

simpler to compare results with different windings.

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

By rewiring a motor from Y to Delta, you are trading amps for volts. So if you have motor that is exactly in the corner with 120A (you say max current of VESC) and 50V (max voltage of VESC) then you can only lose by rewiring it.

If we take your numbers, 120A, 50V as the maximums that your motor controller can handle, then if your motor can take 70A, 50V in Y, you can rewire it to delta: 120A, 29V In that case it doesn't matter.

I THINK I see you doing calculations wrong.

Lets take a simple numbers example: I have a motor that runs 5000 RPM at 50V (KV = 100), can take 100A and It has 0.01 ohm resistance.

To push 100A through it at standstill requires 0.01 Ohm * 100A = 1V. 1V * 100A = 100W of losses: these end up as heaqt in the motor, 100W from the battery. By the time you're doing 900 RPM, the motor requires 9V * 100A = 900W of current to produce the energy on the wheels. You still need 1V, 100W to make that current flow through the motor. You are now drawing 20A, 1000W from the battery. If that's the current limit for the battery, you start dropping current if you continue to go faster.

If we take your numbers, 120A, 50V as the maximums that your motor controller can handle, then if your motor can take 70A, 50V in Y, you can rewire it to delta: 120A, 29V In that case it doesn't matter.

I THINK I see you doing calculations wrong.

Lets take a simple numbers example: I have a motor that runs 5000 RPM at 50V (KV = 100), can take 100A and It has 0.01 ohm resistance.

To push 100A through it at standstill requires 0.01 Ohm * 100A = 1V. 1V * 100A = 100W of losses: these end up as heaqt in the motor, 100W from the battery. By the time you're doing 900 RPM, the motor requires 9V * 100A = 900W of current to produce the energy on the wheels. You still need 1V, 100W to make that current flow through the motor. You are now drawing 20A, 1000W from the battery. If that's the current limit for the battery, you start dropping current if you continue to go faster.

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

rew wrote:To push 100A through it at standstill requires 0.01 Ohm * 100A = 1V. 1V * 100A = 100W of losses: these end up as heaqt in the motor

wouldn't these 100A only end up entirely as heat if the motor were locked in a clamp at full throttle standstill? my understanding is if the mechanical load is below the threshold that entirely locks up the motor, then some of this 100a is channeled into acceleration rather than heat? doesn't more motor amps and more effective volts and more duty cycle at standstill result in more torque, quicker acceleration (not only heat)? not only wasted amps?

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

The 100A will produce torque. That torque will induce a force on the vehicle and result in acceleration.

If you're good at calculus you can calculate the complete curve. I'm lazy, haven't practiced enough calculus the last two decades so I'm rusty.

Suppose my motor produces 5N for each A of current. My motor can do 20A, and the total weight of my bike is 100kg.

So at full throttle I start out with 5N/A * 20A = 100N. That means I accellerate at 100kg/100N = 1m/s.

You can now think of the motor voltage split into two parts. One part is the 1V to generate the current, and the other part is there to produce actual work at the wheel. So that first part is constant as long as the current remains 20A. And the second part will linearly increase from zero to whatever the speed dictates.

If you're good at calculus you can calculate the complete curve. I'm lazy, haven't practiced enough calculus the last two decades so I'm rusty.

Suppose my motor produces 5N for each A of current. My motor can do 20A, and the total weight of my bike is 100kg.

So at full throttle I start out with 5N/A * 20A = 100N. That means I accellerate at 100kg/100N = 1m/s.

You can now think of the motor voltage split into two parts. One part is the 1V to generate the current, and the other part is there to produce actual work at the wheel. So that first part is constant as long as the current remains 20A. And the second part will linearly increase from zero to whatever the speed dictates.

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

seems to me if considering 2 motors of the same physical dimensions, same winding resistance, same kv, same applied motor amps close to standstill, the one with higher measureable torque per amp is more "efficient."

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

Same KV means same torque per amp.

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

rew wrote:Same KV means same torque per amp.

then it would seem

if you start with wye 100kv

reterminate to delta 173.205kv (33.3333% original resistance)

or rewind wye with shorter thicker wires (same copper volume) to 173.205kv (resistance and length and turns divided by square root of 3 because shorter wires, then divided by the square root of 3 again because of thicker wire = 33.333% original resistance).

these should have exactly the same winding resistance, torque, performance, acceleration and efficiency?

the wye must be closer to saturation since this equal performance is achieved through 2/3rds as many phases and 2/3rds as many close proximity magnets in use at a given time when compared to 3/3rds phases and 3/3rds "close proximity to energized phase" magnets used at all times by delta?

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

in retermination of wye to delta, resistance drops to 1/3rd of original, while kv increases to 1xsqrt(3) of original

when resistance drops to 1/3rd of original, "conductance" increases to 3 times the original

when conductance increases to 3 times the original, kv increases by sqrt(3) times the original

if winding conductance increases to 2 times the original (for example, wye wind cross sectional area doubles, same number of turns), does this imply kv will increase by sqrt(2) times the original?

apparently:

%173.205 of original kv = 10 x sqrt(%300 of original conductance) in the example of 100kv wye retermination to delta

&

%173.205 of original kv = 10 x sqrt(%300 of original conductance) in the example of 100kv wye rewinding to 173.205kv wye with same volume copper fill (shorter, thicker wires wye)

for windings does:

% of original kv = 10 x sqrt(% of original conductance) ?

for example does:

%141.421 of original kv wye = 10 x sqrt(%200 of original conductance wye -- only change wye winding cross section doubled -- same # turns) ?

when resistance drops to 1/3rd of original, "conductance" increases to 3 times the original

when conductance increases to 3 times the original, kv increases by sqrt(3) times the original

if winding conductance increases to 2 times the original (for example, wye wind cross sectional area doubles, same number of turns), does this imply kv will increase by sqrt(2) times the original?

apparently:

%173.205 of original kv = 10 x sqrt(%300 of original conductance) in the example of 100kv wye retermination to delta

&

%173.205 of original kv = 10 x sqrt(%300 of original conductance) in the example of 100kv wye rewinding to 173.205kv wye with same volume copper fill (shorter, thicker wires wye)

for windings does:

% of original kv = 10 x sqrt(% of original conductance) ?

for example does:

%141.421 of original kv wye = 10 x sqrt(%200 of original conductance wye -- only change wye winding cross section doubled -- same # turns) ?

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

yesdevin wrote:these should have exactly the same winding resistance, torque, performance, acceleration and efficiency?

No. When you properly drive a three-phase motor like this, you always drive the motor by spinning a wheel.devin wrote:the wye must be closer to saturation since this equal performance is achieved through 2/3rds as many phases and 2/3rds as many close proximity magnets in use at a given time when compared to 3/3rds phases and 3/3rds "close proximity to energized phase" magnets used at all times by delta?

You've drawn a circle with points A D E on it. So we rotate that circle and apply the X-coordinate of each point to each of the phase-termination wires. the Y-coordinate is ignored. If you draw in the Delta configuration between the ADE terminals, you'll see the voltage across each of the windings visualized as the difference in X-coordinates. At some points in the rotation a phase may be vertical so the voltage differential across it will be zero. But you'll see all three windings energized all the time.

Draw in the Y termination scheme and exactly the same happens. Only when the phase is exactly vertical is there no voltage and no current running through a phase winding.

For further visualization, draw a timeline from the center of the circle downwards. Make 36 time-ticks along this line. Now for each 10 degree increment, bring down the X-coordinate of each of the phases down to it's timeline. Use three different colors for the three terminations.

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

1. We start with a wye winding

wye

100 turns

100 length

1 crossection

1 conductance

1 resistance

1 copper volume

100kv

2. Now we simply reterminate to delta

retermination to delta

100 turns

100 length

1 crossection

3 conductance

1/3 resistance

1 copper volume

173.205kv <--- what exactly caused 100kv x sqrt(3) -- [100kv x 1.73205] kv transformation?

was it simply:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^when making any change to conductance of a winding, is this the proper formula to calculate the expected new kv?

3. Now we shorten and thicken the original wye winding to the same kv as delta retermination, while keeping the same copper volume

wye

57.735 turns <- 100 original turns / 1.73205 [sqrt(3)]

57.735 length <- 100 length / 1.73205 [sqrt(3)]

1.73205 crossection <- 1 x 1.73205 [sqrt(3)]

3 conductance <- 1 x 1.73205 x 1.73205 [sqrt(3)]

1/3 resistance <- 1 / 1.73205 / 1.73205 [sqrt(3)]

1 copper volume

173.205kv <- is this correct

is this correct ?:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^apparently this wye winding has the same electrical resistance, torque, kv, performance and efficiency as the wye to delta retermination

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

wye

33.333 turns <- 100 original turns / 3

33.333 length <- 100 length / 3

1 crossection <- same as original wye

3 conductance <- winding 1/3rd length so inverse 3/1 conductance, same conductance as examples 2. and 3.

1/3 resistance <- winding 1/3rd length so 1/3rd resistance

1/3 copper volume

173.205kv <- is this correct?

is this correct ?:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^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?

5. Now we triple the crossection of the original wye winding (3x copper volume) but with the original number of turns, how does this change kv? (assume original winding had spacers allowing this)

wye

100 turns <- original turns

100 length <- original length

3 crossection <- 3x original crossection

3 conductance <- winding 3x crossection so 3x conductance, same conductance as examples 2. 3. & 4.

1/3 resistance <- winding 3x crossection so 1/3rd resistance

3 copper volume

173.205kv <- is this correct?

is this correct ?:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^has tripling the cross section of the original wye but with the same number of turns achieved 100kv x sqrt(3) = 173.205kv??

6. Now we double the crossection of the original wye winding (2x copper volume) but with the original number of turns, how does this change kv? (assume original winding had spacers allowing this)

wye

100 turns <- original turns

100 lengh <- original length

2 crossection <- 2x original crossection

2 conductance <- winding 2x crossection so 2x conductance, more than originally but less than delta retermination

1/2 resistance <- winding 2x crossection so 1/2 resistance

2 copper volume

141.421kv <- is this correct ( 100kv x 1.41421 [sqrt(2)]

is this correct ?:

sqrt(2 new conductance / 1 original conductance) = 141.421kv new / 100kv original kv

^has doubling the conductance by doubling the crossection (rather than tripling) increased kv to 100kv x sqrt(2) = 141.421kv ?

In simple terms is the change in kv caused by a change in a winding geometry always proportional to the change in conductance of that winding?

What exactly puts kv x sqrt(3) into the equation of wye -> delta retermination?

wye

100 turns

100 length

1 crossection

1 conductance

1 resistance

1 copper volume

100kv

2. Now we simply reterminate to delta

retermination to delta

100 turns

100 length

1 crossection

3 conductance

1/3 resistance

1 copper volume

173.205kv <--- what exactly caused 100kv x sqrt(3) -- [100kv x 1.73205] kv transformation?

was it simply:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^when making any change to conductance of a winding, is this the proper formula to calculate the expected new kv?

3. Now we shorten and thicken the original wye winding to the same kv as delta retermination, while keeping the same copper volume

wye

57.735 turns <- 100 original turns / 1.73205 [sqrt(3)]

57.735 length <- 100 length / 1.73205 [sqrt(3)]

1.73205 crossection <- 1 x 1.73205 [sqrt(3)]

3 conductance <- 1 x 1.73205 x 1.73205 [sqrt(3)]

1/3 resistance <- 1 / 1.73205 / 1.73205 [sqrt(3)]

1 copper volume

173.205kv <- is this correct

is this correct ?:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^apparently this wye winding has the same electrical resistance, torque, kv, performance and efficiency as the wye to delta retermination

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

wye

33.333 turns <- 100 original turns / 3

33.333 length <- 100 length / 3

1 crossection <- same as original wye

3 conductance <- winding 1/3rd length so inverse 3/1 conductance, same conductance as examples 2. and 3.

1/3 resistance <- winding 1/3rd length so 1/3rd resistance

1/3 copper volume

173.205kv <- is this correct?

is this correct ?:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^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?

5. Now we triple the crossection of the original wye winding (3x copper volume) but with the original number of turns, how does this change kv? (assume original winding had spacers allowing this)

wye

100 turns <- original turns

100 length <- original length

3 crossection <- 3x original crossection

3 conductance <- winding 3x crossection so 3x conductance, same conductance as examples 2. 3. & 4.

1/3 resistance <- winding 3x crossection so 1/3rd resistance

3 copper volume

173.205kv <- is this correct?

is this correct ?:

sqrt(3 new conductance / 1 original conductance) = 173.205kv new / 100kv original kv

^has tripling the cross section of the original wye but with the same number of turns achieved 100kv x sqrt(3) = 173.205kv??

6. Now we double the crossection of the original wye winding (2x copper volume) but with the original number of turns, how does this change kv? (assume original winding had spacers allowing this)

wye

100 turns <- original turns

100 lengh <- original length

2 crossection <- 2x original crossection

2 conductance <- winding 2x crossection so 2x conductance, more than originally but less than delta retermination

1/2 resistance <- winding 2x crossection so 1/2 resistance

2 copper volume

141.421kv <- is this correct ( 100kv x 1.41421 [sqrt(2)]

is this correct ?:

sqrt(2 new conductance / 1 original conductance) = 141.421kv new / 100kv original kv

^has doubling the conductance by doubling the crossection (rather than tripling) increased kv to 100kv x sqrt(2) = 141.421kv ?

In simple terms is the change in kv caused by a change in a winding geometry always proportional to the change in conductance of that winding?

What exactly puts kv x sqrt(3) into the equation of wye -> delta retermination?

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