## code to change the motor amp limit

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

I could reason that out with about 50% chance of getting it right. Or you can look it up on the internet. There is a topic here about someone with a 7500W motor with 540 (he says) or 580 KV (the page he links says). That motor is sold explicitly in Y and D configuration, and I'm guessing with the same wire.

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

a source i found claims:

if 100 turns = delta = 100kv

100 turns = wye = 57.735kv

& that this is because:

57.735kv wye x sqrt(3) = 100kv delta

&

if 100 turns = delta = 100kv

57.735 turns = wye = 100kv

& that is because 100 turns delta / sqrt(3) = 57.735 turns wye

i can't claim to know if it's right, and i don't yet understand the reasoning.

http://www.bavaria-direct.co.za/info/

if 100 turns = delta = 100kv

100 turns = wye = 57.735kv

& that this is because:

57.735kv wye x sqrt(3) = 100kv delta

&

if 100 turns = delta = 100kv

57.735 turns = wye = 100kv

& that is because 100 turns delta / sqrt(3) = 57.735 turns wye

i can't claim to know if it's right, and i don't yet understand the reasoning.

http://www.bavaria-direct.co.za/info/

Last edited by devin on 14 May 2017, 11:32, edited 1 time in total.

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

Sounds about right.

But the important question is: why would you want a high or a low RPM/V?

A low RPM/V means that low amps cause lots of torque. Yay low RPM/V!

IF you are limited in the voltage you can handle (e.g. because you want to use "safe" voltages, < 60V, or just because your VESC can only go up to 60V) you might not be able to reach your target RPM (=speed in vehicles) with the limitation in the voltage. Yay high RPM/V!

But the important question is: why would you want a high or a low RPM/V?

A low RPM/V means that low amps cause lots of torque. Yay low RPM/V!

IF you are limited in the voltage you can handle (e.g. because you want to use "safe" voltages, < 60V, or just because your VESC can only go up to 60V) you might not be able to reach your target RPM (=speed in vehicles) with the limitation in the voltage. Yay high RPM/V!

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

using this simulator and comparing the same system but replacing the 36 volt pack at 100% throttle with a 72volt pack at about 50% throttle and getting the same speed the high voltage system is shown to lose one percent efficiency. To me that seems an insignificant loss for double the potential speed. Can I assume about the same loss of efficiency with the vesc and a similar voltage and throttle comparison.

http://www.ebikes.ca/tools/simulator.html

the loss of one percent is consistent with the 72 volt battery vs the 36 volt battery regardless of the "grade" and load when staying the same speed.

this "efficiency", is it both the motor and esc equally?

http://www.ebikes.ca/tools/simulator.html

the loss of one percent is consistent with the 72 volt battery vs the 36 volt battery regardless of the "grade" and load when staying the same speed.

this "efficiency", is it both the motor and esc equally?

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

i'm trying to conceptualize in my mind how sqrt(3) enters into the equation of the difference in kv between wye and delta...

this is what flashed into my mind (from geometry) but i still haven't completely solved how it relates:

i think the following true, but if my math turns out wrong please let me know, just trying to learn.

if a circle has a diameter of 2 meters...

an equilateteral triangle inscribed in the circle in such a way that all three corners touch the circuference of the circle has a side length = sqrt(3) meters (1.73205... meters)

57.735kv wye x sqrt(3) = 100kv delta

at 100kv:

100 turns delta / sqrt(3) = 57.735 turns wye

this is what flashed into my mind (from geometry) but i still haven't completely solved how it relates:

i think the following true, but if my math turns out wrong please let me know, just trying to learn.

if a circle has a diameter of 2 meters...

an equilateteral triangle inscribed in the circle in such a way that all three corners touch the circuference of the circle has a side length = sqrt(3) meters (1.73205... meters)

57.735kv wye x sqrt(3) = 100kv delta

at 100kv:

100 turns delta / sqrt(3) = 57.735 turns wye

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

Firist, IF you can fit 100 turns, then it makes no sense to go to "57 turns". Especially with the low-turns-numbers that our motors have. IF you do that, you need to increase the wire thickness. But more common is to simply have two motors with different KV one in Y and one in D configuration.

@hummie Yes, a very small hit in performance is expected.

@hummie Yes, a very small hit in performance is expected.

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

hummie wrote:36 volt pack at 100% throttle with a 72volt pack at about 50% throttle and getting the same speed the high voltage system is shown to lose one percent efficiency.

@Hummie, please correct me if I am wrong about this, but it seems to me:

72V @ 50% THROTTLE is equivalent to 36V @ 100% THROTTLE in wattage for apples to apples "efficiency" comparison only if:

% THROTTLE = % PWM DUTY CYCLE

with the 72V system, i'm certain if you check the simulator at certain rpms (low rpms)... 50% throttle DOES NOT equal 50% pwm duty cycle.

at all the rpms at which 50% throttle does not equal 50% pwm duty cycle, 72v at 50% THROTTLE is not necessarily the same WATTAGE (electrical) as 36V at 100% THROTTLE, and therefore not an apples-to-apples "efficiency comparison."

the only RPMs at which this efficiency comparison might be valid (thanks to varying back emf with varying rpms) is within the range of rpms with the 72V system where 50% THROTTLE = 50% PWM DUTY CYCLE

72V @ 50% THROTTLE @ 25% PWM duty cycle = 18V Effective PWM Voltage applied to motor windings

^not equivalent to 36 effective volts applied to windings

72V @ 50% THROTTLE @ 50% PWM duty cycle = 36V Effective PWM Voltage applied to motor windings

^equivalent to 36 effective volts applied to windings

36V @ 100% THROTTLE @ 100% PWM duty cycle = 36V Effective Voltage applied to motor windings

^equivalent to 36 effective volts applied to windings, and therefore a proper apples-to-apples efficiency comparison compared to 72V @ 50% Throttle @ 50% PWM Duty Cycle

36V @ 100% Throttle @ 50% PWM duty cycle = 18V Effective PWM Voltage applied to motor windings

^36V @ 100% throttle isn't necessarily equivalent to 36v effective pwm voltage applied to the motor windings

72V @ 100% throttle @ 100% PWM duty cycle = 72V Effective PWM Voltage applied to motor windings

^with a 72v supply @ 100% throttle @ 100% pwm duty it is possible to apply 200% of the max effective voltage to the motor windings when compared to a 36v supply @ 100% throttle @ 100% duty

In simple terms, effective voltage applied to the motor windings is proportional to %pwm duty cycle and pack voltage, and not necessarily proportional to "throttle."

Effective voltage applied to windings is also inversely proportional with RPM, meaning at greater rpms, effective voltage is lower with the same % duty cycle because of increasing back emf voltage, in opposition to pack voltage, with additional rpms. Back EMF V = Pack Voltage at 100% duty @ "no load rpm", which is why the motor won't go any faster than no load rpm under its own power, and why watts electrical and amps are both close to ~0w and ~0a @ 100% throttle @ 100% duty cycle at no load rpm.

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

thats' way too much math. Assuming 50% throttle is 50% duty is what I did.

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

That was way too much. But Hummie's assumption is just about right.

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

I'm trying to conceptualize the following:

1) I start with a wye terminated motor.

2) I set the VESC to 120a motor amp limit and 20a battery limit with a 50V pack and 0.06944 ohm winding (0.03472 ohm vesc detected lead to virtual ground point)

3. Full throttle close to standstill, I expect:

1000w = 20A battery amps x 50V pack v = 120A motor amps x 8.333...v effective pwm voltage = 120A motor amps x 120A motor amps x 0.06944 winding = ((8.333...v effective pwm voltage / 50v pack voltage) x 50V pack voltage) x 120a motor amps = ((%16.666... duty cycle / 100) x 50V pack voltage) x 120A motor amps

-1000w

-120a motor amps

-20a battery amps

-8.333...v effective pwm voltage

-16.666...% duty cycle

4. Since at full throttle close to standstill I expect 120A motor amps through a wye winding, i expect 120a motor amps through phase "A" as well as 120a motor amps through phase "B" full throttle close to standstill

Simply 120a motor amps energizing phase A + 120a motor amps energizing phase B = 2 x Phases energized each with 120a Motor amps = "240" Variable

5. So now i take the same motor and all I do is reterminate to delta. I still face the 120A motor amp limit of the VESC.

6. There are 2 conductive pathways lead to lead through a delta termination, and one pathway has twice the resistance of the other pathway, so 2/3rds of the 120a amps will take the low resistance pathway, and 1/3rd of the 120a amps will take the "high resistance" pathway through the other 2 windings.

7. If we add up the number of amps in each phase, it is as follows: 120A x (2/3rd) = 80 A through phase A -- then: 120A x (1/3rd) = 40A through phase B and 40A through Phase C since these 2 phases are connected in series for the second conductive pathway.

Simply 80a motor amps energizing phase A + 40a motor amps energizing phase B + 40a motor amps energizing phase C = (2 x Phases energized each with 40a Motor amps) + (1 x phases energized with 80a motor amps) = "160" Variable

8. So basically considering the 120a motor amp limit with VESC, the wye would effectively have 240 amps "in the phases" if you add up how many amps would be in each separate phase at full throttle accelerating close to standstill, while delta would effectively have 160 amps "in the phases" if you add up how many amps are in each phase at a given time.

9. With the resistance of the Delta termination dropping to 1/3rd of the original resistance ohms value of the wye termination, the calculation i expect for full throttle delta close to standstill would be:

333.333...w = 6.666...A battery amps x 50V pack v = 120A motor amps x 2.777...v effective pwm voltage = 120A motor amps x 120A motor amps x 0.023148 winding = ((2.777...v effective pwm voltage / 50v pack voltage) x 50V pack voltage) x 120a motor amps = ((%5.555... duty cycle / 100) x 50V pack voltage) x 120A motor amps

-333.333...w

-120a motor amps

-6.666...a battery amps

-2.777...v effective pwm voltage

-5.555...% duty cycle

10. Comparison with full throttle close to standstill values for wye termination (same motor):

1000w = 20A battery amps x 50V pack v = 120A motor amps x 8.333...v effective pwm voltage = 120A motor amps x 120A motor amps x 0.06944 winding = ((8.333...v effective pwm voltage / 50v pack voltage) x 50V pack voltage) x 120a motor amps = ((%16.666... duty cycle / 100) x 50V pack voltage) x 120A motor amps

-1000w

-120a motor amps

-20a battery amps

-8.333...v effective pwm voltage

-16.666...% duty cycle

11. Retermination of wye motor to delta appears to drop the available electrical wattage for full throttle close to standstill acceleration by 2/3rds considering 120A motor amp limit of the VESC?

wye: 1000w (in example)

delta: 333.333w (in example)

This is as far as i've gotten so far with the logic of re-termination of wye to delta... to be continued...

1) I start with a wye terminated motor.

2) I set the VESC to 120a motor amp limit and 20a battery limit with a 50V pack and 0.06944 ohm winding (0.03472 ohm vesc detected lead to virtual ground point)

3. Full throttle close to standstill, I expect:

1000w = 20A battery amps x 50V pack v = 120A motor amps x 8.333...v effective pwm voltage = 120A motor amps x 120A motor amps x 0.06944 winding = ((8.333...v effective pwm voltage / 50v pack voltage) x 50V pack voltage) x 120a motor amps = ((%16.666... duty cycle / 100) x 50V pack voltage) x 120A motor amps

-1000w

-120a motor amps

-20a battery amps

-8.333...v effective pwm voltage

-16.666...% duty cycle

4. Since at full throttle close to standstill I expect 120A motor amps through a wye winding, i expect 120a motor amps through phase "A" as well as 120a motor amps through phase "B" full throttle close to standstill

Simply 120a motor amps energizing phase A + 120a motor amps energizing phase B = 2 x Phases energized each with 120a Motor amps = "240" Variable

5. So now i take the same motor and all I do is reterminate to delta. I still face the 120A motor amp limit of the VESC.

6. There are 2 conductive pathways lead to lead through a delta termination, and one pathway has twice the resistance of the other pathway, so 2/3rds of the 120a amps will take the low resistance pathway, and 1/3rd of the 120a amps will take the "high resistance" pathway through the other 2 windings.

7. If we add up the number of amps in each phase, it is as follows: 120A x (2/3rd) = 80 A through phase A -- then: 120A x (1/3rd) = 40A through phase B and 40A through Phase C since these 2 phases are connected in series for the second conductive pathway.

Simply 80a motor amps energizing phase A + 40a motor amps energizing phase B + 40a motor amps energizing phase C = (2 x Phases energized each with 40a Motor amps) + (1 x phases energized with 80a motor amps) = "160" Variable

8. So basically considering the 120a motor amp limit with VESC, the wye would effectively have 240 amps "in the phases" if you add up how many amps would be in each separate phase at full throttle accelerating close to standstill, while delta would effectively have 160 amps "in the phases" if you add up how many amps are in each phase at a given time.

9. With the resistance of the Delta termination dropping to 1/3rd of the original resistance ohms value of the wye termination, the calculation i expect for full throttle delta close to standstill would be:

333.333...w = 6.666...A battery amps x 50V pack v = 120A motor amps x 2.777...v effective pwm voltage = 120A motor amps x 120A motor amps x 0.023148 winding = ((2.777...v effective pwm voltage / 50v pack voltage) x 50V pack voltage) x 120a motor amps = ((%5.555... duty cycle / 100) x 50V pack voltage) x 120A motor amps

-333.333...w

-120a motor amps

-6.666...a battery amps

-2.777...v effective pwm voltage

-5.555...% duty cycle

10. Comparison with full throttle close to standstill values for wye termination (same motor):

1000w = 20A battery amps x 50V pack v = 120A motor amps x 8.333...v effective pwm voltage = 120A motor amps x 120A motor amps x 0.06944 winding = ((8.333...v effective pwm voltage / 50v pack voltage) x 50V pack voltage) x 120a motor amps = ((%16.666... duty cycle / 100) x 50V pack voltage) x 120A motor amps

-1000w

-120a motor amps

-20a battery amps

-8.333...v effective pwm voltage

-16.666...% duty cycle

11. Retermination of wye motor to delta appears to drop the available electrical wattage for full throttle close to standstill acceleration by 2/3rds considering 120A motor amp limit of the VESC?

wye: 1000w (in example)

delta: 333.333w (in example)

This is as far as i've gotten so far with the logic of re-termination of wye to delta... to be continued...

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