I'm planning to compare a couple different windings. ...prepping the experiment...
code to change the motor amp limit

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Re: code to change the motor amp limit
Last edited by devin on 01 Aug 2017, 00:02, edited 1 time in total.

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Re: code to change the motor amp limit
test winding 1: 18 turns delta

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Re: code to change the motor amp limit
@rew On a semirelated 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.

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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 rewind... wasn't able to check the kv...

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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 reterminating 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 reterminating 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...

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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.

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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
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