There are a few things where the units "drop out". The unit is "1".
Assignment: You live in an area with 50Hz AC current from the wall. How many cycles in a minute?
Answer: The Hz unit is actually 1/s. Maybe "cylces/s", but according to SI it is really 1/s. the cylce (just like turn) is dimensionless. 60 seconds per minute (min = 60s), N = F * t = 50Hz * 1min = 50 (1/s) * 1 min * 60 s/min = 3000 (cycles). That 3000 is dimensionless.
Another example is resistance of sheets of material. The resistance is expressed as "ohms / square". where the "square" is dimensionless (1m/1m).
Similarly the "turn" and "termination" parameters/factors are dimensionless quantities.
If you like the quantities in your formula better than those in the standard formula: Good for you!
The whole motorKVcalculation is a bit tricky: if you send an object moving at 10ms you can calculate how long it will take to travel 20m. But for a motor starting with say a CAD model of the motor, the materials used, go ahead and calculate the KV. Very difficult. I'm sure there are expensive motordesignsoftwarepackages that can do it, but there is probably just a numeric simulation involved to solve some equations that we can't analytically solve.
So you have this motor with 100kV and say 100 turns. Now I rewind that motor with aluminium wire because I heard aluminium is also a good conductor of electricity. I use the same thickness and manage to cram in the same 100 turns. So you predict my kV with your formula. The conductivity of aluminium is 61% of that of copper.
code to change the motor amp limit
Re: code to change the motor amp limit
rew wrote:There are a few things where the units "drop out". The unit is "1".
Assignment: You live in an area with 50Hz AC current from the wall. How many cycles in a minute?
Answer: The Hz unit is actually 1/s. Maybe "cylces/s", but according to SI it is really 1/s. the cylce (just like turn) is dimensionless. 60 seconds per minute (min = 60s), N = F * t = 50Hz * 1min = 50 (1/s) * 1 min * 60 s/min = 3000 (cycles). That 3000 is dimensionless.
Another example is resistance of sheets of material. The resistance is expressed as "ohms / square". where the "square" is dimensionless (1m/1m).
Similarly the "turn" and "termination" parameters/factors are dimensionless quantities.
If you like the quantities in your formula better than those in the standard formula: Good for you!
The whole motorKVcalculation is a bit tricky: if you send an object moving at 10ms you can calculate how long it will take to travel 20m. But for a motor starting with say a CAD model of the motor, the materials used, go ahead and calculate the KV. Very difficult. I'm sure there are expensive motordesignsoftwarepackages that can do it, but there is probably just a numeric simulation involved to solve some equations that we can't analytically solve.
So you have this motor with 100kV and say 100 turns. Now I rewind that motor with aluminium wire because I heard aluminium is also a good conductor of electricity. I use the same thickness and manage to cram in the same 100 turns. So you predict my kV with your formula. The conductivity of aluminium is 61% of that of copper.
@rew If you want to swap out the copper winding for aluminum and calculate the KV change while using only SI base unit derived variables in your formula, you can use these:
S=sqrt(C/Z)*sqrt(1/V)
C=S^2*V*Z
V=C/(S^2*Z)
Z=C/(S^2*V)
S = Change Factor of KV (rpm/v)
C = Change Factor of Winding Conductance (1/ohm)
V = Change Factor of Conductor Volume (meters^3)
Z = Change Factor of Conductor Conductivity (1/ohmmeters)
An aluminum wire will have 0.59574 as much conductance as an identical geometry copper wire.
A copper wire will have 1/0.59574=1.67858 as much conductance as an identical geometry aluminum wire.
Source: https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity
Re: code to change the motor amp limit
This formula may in the end be a bit more simple and practical because:
It's based on changes in resistance and resistivity rather than changes of conductance and conductivity.
It uses only one square root operation instead of 2
It still calculates KV changes using SI base unit derived variables, rather than legacy nonSI derived units "turns and termination"

D=sqrt(E/(V*N))
E=N*V*D^2
V=E/(N*D^2)
N=E/(V*D^2)
D = Change Factor of KV (rpm/v)
E = Change Factor of Conductor Resistivity (ohmmeters)
V = Change Factor of Conductor Volume (meters^3)
N = Change Factor of Conductor Resistance (ohm)

100kv 30 Turns 0.6ohm Wye Copper > 173.205kv 30 turns 1/2 Cross Section 0.671432ohm Delta Aluminum
D = 1.73205 = Change Factor of KV (rpm/v)
E = 1.67858 = Change Factor of Conductor Resistivity (ohmmeters) Copper to Aluminum
V = 0.5 = Change Factor of Conductor Volume (meters^3)
N = 1.1190533 = Change Factor of Conductor Resistance (ohm) = 0.671432 new ohm/0.6 original ohm
D=sqrt(E/(V*N))
D=sqrt(1.67858/(0.5*1.1190533))
1.73205=sqrt(1.67858/(0.5*1.1190533)) <—Correct
100kv 30 Turns Delta > 57.73kv 30 turns Wye
D = 0.5773 = Change Factor of KV (rpm/v)
E = 1 = Change Factor of Conductor Resistivity (ohmmeters)
V = 1 = Change Factor of Conductor Volume (meters^3)
N = 3 = Change Factor of Conductor Resistance (ohm)
D=sqrt(E/(V*N))
D=sqrt(1/(1*3))
0.5773=sqrt(1/(1*3)) <— Correct

^Verified
It's based on changes in resistance and resistivity rather than changes of conductance and conductivity.
It uses only one square root operation instead of 2
It still calculates KV changes using SI base unit derived variables, rather than legacy nonSI derived units "turns and termination"

D=sqrt(E/(V*N))
E=N*V*D^2
V=E/(N*D^2)
N=E/(V*D^2)
D = Change Factor of KV (rpm/v)
E = Change Factor of Conductor Resistivity (ohmmeters)
V = Change Factor of Conductor Volume (meters^3)
N = Change Factor of Conductor Resistance (ohm)

100kv 30 Turns 0.6ohm Wye Copper > 173.205kv 30 turns 1/2 Cross Section 0.671432ohm Delta Aluminum
D = 1.73205 = Change Factor of KV (rpm/v)
E = 1.67858 = Change Factor of Conductor Resistivity (ohmmeters) Copper to Aluminum
V = 0.5 = Change Factor of Conductor Volume (meters^3)
N = 1.1190533 = Change Factor of Conductor Resistance (ohm) = 0.671432 new ohm/0.6 original ohm
D=sqrt(E/(V*N))
D=sqrt(1.67858/(0.5*1.1190533))
1.73205=sqrt(1.67858/(0.5*1.1190533)) <—Correct
100kv 30 Turns Delta > 57.73kv 30 turns Wye
D = 0.5773 = Change Factor of KV (rpm/v)
E = 1 = Change Factor of Conductor Resistivity (ohmmeters)
V = 1 = Change Factor of Conductor Volume (meters^3)
N = 3 = Change Factor of Conductor Resistance (ohm)
D=sqrt(E/(V*N))
D=sqrt(1/(1*3))
0.5773=sqrt(1/(1*3)) <— Correct

^Verified
Re: code to change the motor amp limit
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...
devin wrote:The whole point of this is since midMay I've been trying to fully understand exactly why retermination from wye to delta results in an increase factor in rpm per volt of exactly sqrt(3) or 1.73205...
devin wrote:So... why exactly does retermination from wye to delta result in a change factor of kv equal to irrational number square root of 3?
Why does retermination from Wye to Delta increase the KV, also known as the RPM per Volt, by a factor of 1x1.73205?
First, both the Wye and Delta termination styles can have identical KV, KT, stall torque, stall wattage electrical, and copper volume when they have different numbers of “turns.”
The terminations between Delta & Wye phases have completely different geometries, and even when they have identical enclosed copper volume, the copper volume “energized” at stall differs between terminations— in the Delta version all the copper volume and all 3 phases are energized at stall and in the Wye version, 2/3rds of the copper volume and 2/3rds of the phases are energized at stall.
Simply the wye motor achieves its identical torque at stall with 2/3rds as much energized copper, 2/3rds as many energized teeth, and 2/3rds as much rotor magnetic moment in close proximity to the energized teeth.
When enclosed conductor volume and conductor resistivity stays constant, changes in electrical wattage at stall (100% duty cycle) result in linearly proportional changes in rotor kinetic energy at no load rpm, regardless of termination. For example, doubling the battery voltage will quadruple the electrical wattage at stall, and quadruple the rotor’s kinetic energy at no load rpm. Also halving the number of turns and doubling the cross section to achieve the same copper volume will quadruple the electrical wattage at stall and quadruple the rotor’s kinetic energy at no load rpm. Also reterminating a motor wye to delta will triple the electrical wattage at stall and triple the kinetic energy of the rotor at no load rpm.
With identical copper volume, resistivity, 0.6ohm resistance and identical performance, a Delta motor needs 100 “turns” in order to have the same KV as a Wye motor with 57.735 “turns.” In order to achieve the same copper volume, by a factor of 1x1.73205 the Wye wire has a larger cross section.
At stall, the wye motor has 1.5 times as much electrical wattage per energized tooth, but only 2/3rds as many energized teeth (& same copper volume per tooth).
1=1.5*2/3
In other words the 100kv Wye and 100kv Delta have identical performance because with same enclosed copper volume they have the same lead to lead resistance, and even though the wye motor has only 2/3rds as much energized copper and 2/3rds as many energized stator teeth, the wattage per energized per tooth is 1.5 times greater, therefore 1.5 times greater wattage per tooth * 2/3rds as many teeth = 1 change factor of performance — ie no net change in performance.
If the 1.5 times greater wattage per tooth combined with 2/3rds as many energized teeth explains why 100kv 100turns 1 copper volume delta and 100kv 57.735turns 1 copper volume wye have identical performance — then why does the same number of turns with a different termination result in a change factor in rpm per volt of 1.73205?
Changing the 100kv wye with 57.735 turns to 100 turns (same turns # as delta) with the same copper volume requires multiplying the number of turns times 1.73205 and reducing the wire cross section by a factor of 1/1.73205... — leading to a tripling of resistance and a reduction in wattage at stall by a factor of 1x(1/3). The 1x(1/3) reduction in wattage at stall results in a 1x(1/3) reduction in rotor kinetic energy at no load rpm. The 1x(1/3) rotor kinetic energy at no load rpm results in a reduction in rotor velocity at no load rpm of 1x(1/sqrt(3))=0.5773... change factor of rotor velocity, since kinetic energy changes at the square of velocity. In other words, since the wye is spinning slower than the delta by a factor of 1x(1/sqrt(3))=0.5773... with the same number of turns, the delta now spins relatively faster than the wye by a factor of 1xsqrt(3)=(1.73205...).
Simply the 100kv 100 turns delta and 100kv 57.735 turns wye have identical performance with identical enclosed copper volume due to 1.5 times as many watts per tooth at stall but 2/3rds as many energized teeth in the wye version (and same copper volume per tooth). Changing the 100kv 57.735 turns wye to 100 turns wye with the same copper volume decreases stall wattage electrical and no load rpm rotor kinetic energy by a factor of 1x(1/3)=0.333..., and therefore rotor velocity by a factor of 1x(1/sqrt(3))=0.5773... (because kinetic energy changes at the square of velocity and rotor no load rpm kinetic energy changes proportionally & linearly with respect to changes in stall wattage electrical @ 100% duty cycle when conductor volume and resistivity stay constant)  so the delta rotor with same turns now spins relatively 1xsqrt(3) or 1x1.73205... times faster than the wye motor when they have the same (100) number of turns and copper volume  thus changing the termination from wye to delta at the same number of turns produces a change factor of kv equal to the square root of 3, sqrt(3), or (1.73205...).
Therefore the change factor of the kinetic energy of the rotor at no load rpm is always equal to the change factor of electrical wattage at stall (100% duty cycle) when conductor volume and resistivity are constant, & the change factor of the KV is always the square root of the change factor of conductance when enclosed conductor volume and resistivity stays constant, as depicted:
Last edited by devin on 22 Aug 2017, 23:44, edited 35 times in total.
Re: code to change the motor amp limit
actually forget it I remember now
 Attachments

 Delta_diagram.gif (27.97 KiB) Viewed 1485 times

 Wye_diagram.gif (14.71 KiB) Viewed 1485 times
Re: code to change the motor amp limit
One could say the change factor of IKEACVE@CCV@S@FDC (individual kinetic energy of average copper valence electron @ constant copper volume @ stall @ 100% [full] duty cycle) = change factor of KER@NLR (kinetic energy of rotor at no load rpm)
CFOIKEACVE@CCV@S@FDC = CFOKER@NLR
For example at constant copper volume, if we double the voltage, the average copper valence electron has 4 times as much kinetic energy, and therefore the rotor has 4 times as much kinetic energy at no load rpm, & CFOIKEACVE@CCV@S@FDC = CFOKER@NLR.
It's a fancy way to say when the electrons go twice as fast, the rotor goes twice as fast.
Image Source: https://www.iggy.net/iggyexternal/media/external/periodical/img/coppervalence.png
Image Source: https://elementsunearthed.com/tag/forbital/
Caption: "The sorbital electrons create a spherical shape based on probability"
CFOIKEACVE@CCV@S@FDC = CFOKER@NLR
For example at constant copper volume, if we double the voltage, the average copper valence electron has 4 times as much kinetic energy, and therefore the rotor has 4 times as much kinetic energy at no load rpm, & CFOIKEACVE@CCV@S@FDC = CFOKER@NLR.
It's a fancy way to say when the electrons go twice as fast, the rotor goes twice as fast.
Image Source: https://www.iggy.net/iggyexternal/media/external/periodical/img/coppervalence.png
Image Source: https://elementsunearthed.com/tag/forbital/
Caption: "The sorbital electrons create a spherical shape based on probability"
Re: code to change the motor amp limit
If it turns out copper and aluminum wire at the same number of "turns" have different KV, in theory it's the result of different electron drift velocities in the copper vs aluminum windings with the same geometry. For example a copper wire with half the turns has double the electron drift velocity, and double the rotor rpm at no load speed.
Re: code to change the motor amp limit
The power of a good theory (formula) is that it can make accurate predictions about the real world.
So my theory/formula predicts that with aluminium wire and the same number of turns you get the same KV. Just a little bit less efficient motor overall.
So my theory/formula predicts that with aluminium wire and the same number of turns you get the same KV. Just a little bit less efficient motor overall.
Re: code to change the motor amp limit
hummie wrote:I rode many miles at 200 motor amps and am asking why it works
@hummie... i think this graph solves the riddle...
With a 200 motor amp limit setting & 48 battery amps limit setting, 0.0830ohm winding (0.0415ohm FOC detection * 2) & 35v pack, the motor amp limit would have been automatically reduced to a 120a motor amp limit due to the coded motor amp limit override in the vesc version that was used...
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