code to change the motor amp limit

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Re: code to change the motor amp limit
As I've said before: with the same termination the KV only depends on the number of turns. Turns times KV is constant.

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Re: code to change the motor amp limit
You quoted but forgot to read "with the same termination".

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Re: code to change the motor amp limit
Hi,
You are now progressing to valid scientific methods. Good.
Many people do not understand when a formula is applicable. So when I give them that the voltage across red led is 2V at 20mA, and I ask them to calculate the current limiting resistor for a 5V system, they will plug the given numbers into Ohms law and calculate R= 2V/.02A = 100 Ohms. Or 5V/0.02A = 250 Ohms.
What I THINK is happening is that in the article you quoted this guy fell into this trap and confused A cross section of the wire with A cross section of the LOOP that the wire makes.
Case in point, the flux depends on the number of electrons wizzing around. If it is 1A going around 10 times or 10A going around 1 time does not matter. Especially in an aircoil, the magnetic flux spreads out over the available area. So the flux would be inversely proportional to the area enclosed by the loops of the coil.
He still has a good grasp of the material at hand, because when asked to increase the KV he immediately draws the conclusion that the number of turns must be reduced, implying the validity of the proposition that you're trying to disprove.
You are now progressing to valid scientific methods. Good.
Many people do not understand when a formula is applicable. So when I give them that the voltage across red led is 2V at 20mA, and I ask them to calculate the current limiting resistor for a 5V system, they will plug the given numbers into Ohms law and calculate R= 2V/.02A = 100 Ohms. Or 5V/0.02A = 250 Ohms.
What I THINK is happening is that in the article you quoted this guy fell into this trap and confused A cross section of the wire with A cross section of the LOOP that the wire makes.
Case in point, the flux depends on the number of electrons wizzing around. If it is 1A going around 10 times or 10A going around 1 time does not matter. Especially in an aircoil, the magnetic flux spreads out over the available area. So the flux would be inversely proportional to the area enclosed by the loops of the coil.
He still has a good grasp of the material at hand, because when asked to increase the KV he immediately draws the conclusion that the number of turns must be reduced, implying the validity of the proposition that you're trying to disprove.

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Re: code to change the motor amp limit
A 3 tooth wye air core motor with optical sensors has 3 solenoids.
KV testing is performed with a 4.2V DC source, commutation timing is achieved via optical sensors, and effective voltage to motor is 4.2V directly from the DC source with no PWM, revealing a 100kv motor constant.
Each solenoid has 50 turns, a lead to virtual ground point resistance of 0.0415ohm and a solenoid magnetic moment vector area of 0.00064516 m^2, and a solenoid length of 0.01 meters.
Since this is a wye motor and when voltage is applied to any 2 leads, two of the 3 solenoids are wired in series, the lead to lead resistance of this motor measured at stall is 0.0830ohm.

The magnetic moment of a single layer winding solenoid, measured in newton meters per tesla, is given by the following formula:
u=NIS
Where:
u = Magnetic Moment of Solenoid in newton meters per tesla
N = Number of identical turns
I = Current in Amperes
S = Vector Area (Solenoid Cross Section Area in Square Meters)
Source: https://en.wikipedia.org/wiki/Magnetic_moment

The flux density of a single layer winding solenoid, measured in tesla, is given by the following formula:
B = ((u0)*N*I)/L
Where:
B = Flux Density in Tesla
u0 = 4pi(10^7) = Absolute Vacuum Permeability
N = Number of identical turns
I = Current in Amperes
L = Solenoid Length in Meters

The Auxilliary H Magnetic Field Strength, measured in newtons per weber, is given by the following formula:
H = (NI)/L
Where:
H = Auxilliary H Magnetic Field Strength in Newtons per Weber
N = Number of identical turns
I = Current in Amperes
L = Solenoid Length in Meters

As per Ohm's law:
I = V/R
W = IV
Where:
I = Current in Amperes
V = Volts DC
R = Resistance in Ohms
W = Power in Watts
https://en.wikipedia.org/wiki/Ohm%27s_law

Therefore:
Magnetic Moment of A Single Solenoid (Rotor Removed) (Example #1)
u=N(V/R)S
3.26466=50*(4.2/0.0415)*0.00064516
u=NIS
3.26466=50*101.20481*0.00064516
Flux Density of A Single Solenoid
B = ((u0)*N*I)/L
0.63588 = ((4pi(10^(7)))*50*101.20481)/0.01
HField of A Single Solenoid
H = (NI)/L
506024.05 = (50*101.20481)/0.01
Where:
N = 50 = Number of identical turns
I = 101.20481 Current in Amperes
V = 4.2 = Volts DC
R = 0.0415 = Resistance in Ohms
S = 0.00064516 = Vector Area (Solenoid Cross Section Area in Square Meters) = 1 square inch
W = 425.06020 = Electrical Power in Watts
u = 3.26466 = Magnetic Moment of Solenoid in newton meters per tesla
B = 0.63588 = Flux Density in Tesla
H = 506024.05 = Auxilliary H Magnetic Field Strength in Newtons Per Weber
u0 = 4pi(10^7) = Absolute Vacuum Permeability
L = 0.01 = Solenoid Length in Meters

Now we increase the wire cross section to 3 times the original size, while keeping the number of turns identical.

This gives:
Magnetic Moment of A Single Solenoid (Rotor Removed) (Example #2)
u=N(V/R)S
9.79635=50(4.2/0.01383)0.00064516
u=NIS
9.79635=50(303.68764)0.00064516
Flux Density of A Single Solenoid at Stall
B = ((u0)*N*I)/L
1.90812 = ((4pi(10^(7)))*50*303.68764)/0.01
N = 50 = Number of identical turns
I = 303.68764 Current in Amperes < 1x(3) more current than Example #1
V = 4.2 = Volts DC
R = 0.01383 = Resistance in Ohms < 1x(1/3) less resistance than Example #1
S = 0.00064516 = Vector Area (Solenoid Cross Section Area in Square Meters) = 1 square inch
W = 1275.48807 = Electrical Power in Watts < 1x(3) more power than Example #1
u = 9.79635 = Magnetic Moment of Solenoid in newton meters per tesla < 1x(3) more Magnetic Moment than Example #1
B = 1.90812 = Flux Density in Tesla < 1x(3) more Flux Density than Example #1
u0 = 4pi(10^7) = Absolute Vacuum Permeability
L = 0.01 = Solenoid Length in Meters

^Tripling the cross section of the wire at the same number of turns has tripled the magnetic moment of the solenoid. Despite the tripling of the magnetic moment of the solenoids with tripled cross section, the magnetic moment of the permanent magnets in the rotor remained constant.
Since the magnetic moment of the permanent magnets is fixed in magnitude, the gyromagnetic ratio of the rotor magnets increases proportionally to rotor angular momentum.
In rotational systems, power is the product of the torque τ and angular velocity ω. (Source: )
Since the gyromagnetic ratio of the rotor magnets at a particular rpm has both a mechanical torque component acting on nearby magnetic objects in proportion to their BField flux density, the rotor itself has a kinetic energy, and an angular momentum measured in rpm, and since power is the product of torque and angular velocity, and since the torque of a spinning rotor is proportional to its kinetic energy...
...from the magnetic moment of the permanent magnets, the magnetic properties of nearby stator and windings, the kinetic energy of the rotor and its angular velocity, we can derive the power in watts of the torque times the angular momentum of the rotating magnetic field produced by the spinning rotor magnets which is acting on the mobile charge carriers in the windings.
Since the rotating magnetic field generated by the spinning rotor magnets has a definite "available wattage" to do work on nearby magnetized objects (limited by the kinetic energy of the rotor at a particular rpm), the effect on a current in a winding caused by the spinning magnets is limited to the "available wattage" present in the spinning rotor. (Rotor torque acting on charge carriers in windings proportional to rotor kinetic energy relative to stator, angular momentum, and total magnetic moment of the permanent magnets).
If at a certain rpm the available wattage present in the rotating magnetic field of the rotor magnets' influence on mobile charge carriers in the windings is less than the opposing electrical wattage of those same mobile charge carriers in the same windings, then current present in the nearby wires cannot be reduced all the way to 0 as would be the case at no load rpm for a given applied voltage, since there is not enough available rotor wattage to counteract the winding wattage.
If in Example 1, at a particular rpm, the spinning magnetic field of the rotor has sufficient wattage to completely counteract the electrical wattage in the nearby wires, bringing current to 0 despite the 4.2V closed circuit, then this particular rpm is "no load rpm" and the rpm per applied volt is the KV.
In Example 2, when the winding cross section and electrical wattage found in the windings triples, then at the same rotor rpm which was no load rpm in Example 1, since there is no additional kinetic energy in the rotor at the same rpm as before, there is insufficient additional wattage provided by the rotating magnetic field of the rotor magnets to completely counteract the tripled electrical wattage found in the windings.
Further complicating matters is the fact that while the rotor's magnetic field is rotating, the stator's magnetic field is also rotating due to the commutations.
In example 1, at no load rpm winding current is brought down to 0 as a result of the rotational wattage of the rotating magnetic field of the rotor magnets.
In example 2, at the same rpm as was no load in example 1, the wattage in the windings is 3 times larger in magnitude than the wattage that the rotor's rotating magnetic field is exerting upon the mobile charge carriers in the windings, therefore the rotor's effect on the windings is insufficient to bring winding current to 0. Since the same RPM in example 2 is insufficient to bring current to 0 despite the 4.2V dc applied voltage, it is not no load rpm, and therefore no load RPM in example 1 and 2 must be different. Since no load rpm in example 1 and 2 must be different, and only the cross section of the wire changed, it is implied that a change in cross section of a winding wire must cause a change in KV.
KV testing is performed with a 4.2V DC source, commutation timing is achieved via optical sensors, and effective voltage to motor is 4.2V directly from the DC source with no PWM, revealing a 100kv motor constant.
Each solenoid has 50 turns, a lead to virtual ground point resistance of 0.0415ohm and a solenoid magnetic moment vector area of 0.00064516 m^2, and a solenoid length of 0.01 meters.
Since this is a wye motor and when voltage is applied to any 2 leads, two of the 3 solenoids are wired in series, the lead to lead resistance of this motor measured at stall is 0.0830ohm.

The magnetic moment of a single layer winding solenoid, measured in newton meters per tesla, is given by the following formula:
u=NIS
Where:
u = Magnetic Moment of Solenoid in newton meters per tesla
N = Number of identical turns
I = Current in Amperes
S = Vector Area (Solenoid Cross Section Area in Square Meters)
Source: https://en.wikipedia.org/wiki/Magnetic_moment

The flux density of a single layer winding solenoid, measured in tesla, is given by the following formula:
B = ((u0)*N*I)/L
Where:
B = Flux Density in Tesla
u0 = 4pi(10^7) = Absolute Vacuum Permeability
N = Number of identical turns
I = Current in Amperes
L = Solenoid Length in Meters

The Auxilliary H Magnetic Field Strength, measured in newtons per weber, is given by the following formula:
H = (NI)/L
Where:
H = Auxilliary H Magnetic Field Strength in Newtons per Weber
N = Number of identical turns
I = Current in Amperes
L = Solenoid Length in Meters

As per Ohm's law:
I = V/R
W = IV
Where:
I = Current in Amperes
V = Volts DC
R = Resistance in Ohms
W = Power in Watts
https://en.wikipedia.org/wiki/Ohm%27s_law

Therefore:
Magnetic Moment of A Single Solenoid (Rotor Removed) (Example #1)
u=N(V/R)S
3.26466=50*(4.2/0.0415)*0.00064516
u=NIS
3.26466=50*101.20481*0.00064516
Flux Density of A Single Solenoid
B = ((u0)*N*I)/L
0.63588 = ((4pi(10^(7)))*50*101.20481)/0.01
HField of A Single Solenoid
H = (NI)/L
506024.05 = (50*101.20481)/0.01
Where:
N = 50 = Number of identical turns
I = 101.20481 Current in Amperes
V = 4.2 = Volts DC
R = 0.0415 = Resistance in Ohms
S = 0.00064516 = Vector Area (Solenoid Cross Section Area in Square Meters) = 1 square inch
W = 425.06020 = Electrical Power in Watts
u = 3.26466 = Magnetic Moment of Solenoid in newton meters per tesla
B = 0.63588 = Flux Density in Tesla
H = 506024.05 = Auxilliary H Magnetic Field Strength in Newtons Per Weber
u0 = 4pi(10^7) = Absolute Vacuum Permeability
L = 0.01 = Solenoid Length in Meters

Now we increase the wire cross section to 3 times the original size, while keeping the number of turns identical.

This gives:
Magnetic Moment of A Single Solenoid (Rotor Removed) (Example #2)
u=N(V/R)S
9.79635=50(4.2/0.01383)0.00064516
u=NIS
9.79635=50(303.68764)0.00064516
Flux Density of A Single Solenoid at Stall
B = ((u0)*N*I)/L
1.90812 = ((4pi(10^(7)))*50*303.68764)/0.01
N = 50 = Number of identical turns
I = 303.68764 Current in Amperes < 1x(3) more current than Example #1
V = 4.2 = Volts DC
R = 0.01383 = Resistance in Ohms < 1x(1/3) less resistance than Example #1
S = 0.00064516 = Vector Area (Solenoid Cross Section Area in Square Meters) = 1 square inch
W = 1275.48807 = Electrical Power in Watts < 1x(3) more power than Example #1
u = 9.79635 = Magnetic Moment of Solenoid in newton meters per tesla < 1x(3) more Magnetic Moment than Example #1
B = 1.90812 = Flux Density in Tesla < 1x(3) more Flux Density than Example #1
u0 = 4pi(10^7) = Absolute Vacuum Permeability
L = 0.01 = Solenoid Length in Meters

^Tripling the cross section of the wire at the same number of turns has tripled the magnetic moment of the solenoid. Despite the tripling of the magnetic moment of the solenoids with tripled cross section, the magnetic moment of the permanent magnets in the rotor remained constant.
Since the magnetic moment of the permanent magnets is fixed in magnitude, the gyromagnetic ratio of the rotor magnets increases proportionally to rotor angular momentum.
In rotational systems, power is the product of the torque τ and angular velocity ω. (Source: )
Since the gyromagnetic ratio of the rotor magnets at a particular rpm has both a mechanical torque component acting on nearby magnetic objects in proportion to their BField flux density, the rotor itself has a kinetic energy, and an angular momentum measured in rpm, and since power is the product of torque and angular velocity, and since the torque of a spinning rotor is proportional to its kinetic energy...
...from the magnetic moment of the permanent magnets, the magnetic properties of nearby stator and windings, the kinetic energy of the rotor and its angular velocity, we can derive the power in watts of the torque times the angular momentum of the rotating magnetic field produced by the spinning rotor magnets which is acting on the mobile charge carriers in the windings.
Since the rotating magnetic field generated by the spinning rotor magnets has a definite "available wattage" to do work on nearby magnetized objects (limited by the kinetic energy of the rotor at a particular rpm), the effect on a current in a winding caused by the spinning magnets is limited to the "available wattage" present in the spinning rotor. (Rotor torque acting on charge carriers in windings proportional to rotor kinetic energy relative to stator, angular momentum, and total magnetic moment of the permanent magnets).
If at a certain rpm the available wattage present in the rotating magnetic field of the rotor magnets' influence on mobile charge carriers in the windings is less than the opposing electrical wattage of those same mobile charge carriers in the same windings, then current present in the nearby wires cannot be reduced all the way to 0 as would be the case at no load rpm for a given applied voltage, since there is not enough available rotor wattage to counteract the winding wattage.
If in Example 1, at a particular rpm, the spinning magnetic field of the rotor has sufficient wattage to completely counteract the electrical wattage in the nearby wires, bringing current to 0 despite the 4.2V closed circuit, then this particular rpm is "no load rpm" and the rpm per applied volt is the KV.
In Example 2, when the winding cross section and electrical wattage found in the windings triples, then at the same rotor rpm which was no load rpm in Example 1, since there is no additional kinetic energy in the rotor at the same rpm as before, there is insufficient additional wattage provided by the rotating magnetic field of the rotor magnets to completely counteract the tripled electrical wattage found in the windings.
Further complicating matters is the fact that while the rotor's magnetic field is rotating, the stator's magnetic field is also rotating due to the commutations.
In example 1, at no load rpm winding current is brought down to 0 as a result of the rotational wattage of the rotating magnetic field of the rotor magnets.
In example 2, at the same rpm as was no load in example 1, the wattage in the windings is 3 times larger in magnitude than the wattage that the rotor's rotating magnetic field is exerting upon the mobile charge carriers in the windings, therefore the rotor's effect on the windings is insufficient to bring winding current to 0. Since the same RPM in example 2 is insufficient to bring current to 0 despite the 4.2V dc applied voltage, it is not no load rpm, and therefore no load RPM in example 1 and 2 must be different. Since no load rpm in example 1 and 2 must be different, and only the cross section of the wire changed, it is implied that a change in cross section of a winding wire must cause a change in KV.
Last edited by devin on 10 Jun 2017, 07:30, edited 7 times in total.

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Re: code to change the motor amp limit
Who is paying me to check your homework?
Your end result is wrong, there must be something wrong with your reasoning.
I still think you're confusing the electrical power that goes into the motor, at stall being 100% losses, with something going on with the magnetic fields. You are throwing with formulas that I am unfamiliar with. I would have to study to get to know them and their limitations.
Your end result is wrong, there must be something wrong with your reasoning.
I still think you're confusing the electrical power that goes into the motor, at stall being 100% losses, with something going on with the magnetic fields. You are throwing with formulas that I am unfamiliar with. I would have to study to get to know them and their limitations.
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