rew wrote:KV is essentially the voltage induced when using the motor as a generator. The voltage in each turn depends on the change in magnetic flux. This is measured at zero current, so there is no drop due to I*R. So the voltage over the motor is Nturns * <stuff that depends on the motor> * rotation speed.

Let's imagine a virtual motor where:

A = Mechanical Output Shaft

B = Rotor

C = Stator - lead-to-lead winding resistance 0.00353ohms

D = Ground

E = Rotor Kinetic Energy in Joules at no load RPM @ 4.2V DC (relative to C = Stator)

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.

When we measure the kinetic energy of the unpowered rotor at the same rpm as no load rpm 4.2V, we find the rotor's kinetic energy is 5000 joules.

Therefore:

Example #1

A = Mechanical Output Shaft

B = Rotor

C = Stator - lead-to-lead winding resistance 0.00353ohms

D = Ground

E = 5000 Joules = Rotor Kinetic Energy in Joules at no load RPM @ 4.2V DC (relative to C = Stator)

5000 Joules of kinetic energy is 5000 watt seconds of energy.

In other words, an average of 5000 watts of mechanical power could be extracted from the kinetic energy of the rotor for a duration of 1 second.

Assuming an air core motor, frictionless bearings, dielectric rotor and motor case construction and operation in a vacuum-- the only place the rotor CAN transfer its kinetic energy is to the mobile charge carriers in the copper windings. Therefore 100% of the rotor's kinetic energy relative to the stator WILL transfer first to the mobile charge carriers, then mechanically to the windings themselves through resistance, then to the ground.

Therefore if the transfer of 100 percent of the kinetic energy of the rotor to mobile charge carriers in the 0.00353ohm winding takes one second, the average wattage during that second would be 5000w and the average amperage in the windings during the second would be 1190.47amps.

5000 watts of power for 1 second from a 4.2V battery is also 1190.47 amps.

If we apply 4.2V to the 0.00353ohm winding with the motor stalled for 1 second, this is also 5000 watts of electrical power and 5000J of energy.

Simply at no load rpm 4.2V, the rotor has 5000j of kinetic energy it can transfer to the mobile charge carriers in the windings.

With motor stalled at 4.2V for 1 second, 5000j of energy will transfer to the mobile charge carriers in the windings from the battery. (ohm's law: 4.2V - 1190.12A - 0.00353ohm - 5000W)

Therefore in this case kinetic energy in the rotor at no load rpm is equal to kinetic energy applied to the mobile charge carriers in the winding in one second with the motor stalled.

Example #2

If we triple the cross section of the wire, we lower winding resistance to 0.00117 ohms. (ohm's law: 4.2V - 3589.74A - 0.00117ohm - 15076.92W)

In one second with motor stalled the energy transferred to the moving charge carriers in the windings is 15076.92 Joules.

The magnetic moment of the stator has tripled but the magnetic moment of the rotor has stayed the same.

At the same RPM which was no load rpm in the first example, the rotor still has 5000J of kinetic energy available to transfer to the moving charge carriers.

But in the new motor with 3 times the cross section winding there are 3 times as many moving charge carriers and they are moving at same voltage as in the first example, meaning the total cumulative power of the mobile charge carriers is 3 times larger than in first example.

Since at the same rpm which was no load rpm in the first example the kinetic energy of the rotor is still 5000J, only 5000J is available at this rpm to act in opposition to the now 3 times as many mobile charge carriers, each having the same voltage as before.

In order for the rotor to have the same completely-opposing effect on the total kinetic energy of the moving charge carriers in the windings as in the first example, since there are 3 times as many moving charge carriers in a 3 times thicker winding, and the 3 times larger number of mobile charge carriers has the same voltage as before, the rotor needs 3 times as much kinetic energy to completely counteract the 3 times greater kinetic energy of the mobile charge carriers in the new winding at the same voltage. (as in example 1 -- Back EMF V works in opposition to the battery voltage).

Simply since at stall there are 3 times as many moving charge carriers, each moving at the same voltage as in the first example, and cumulatively the charge carriers have 3 times the kinetic energy as the first example, the rotor would necessarily require 3 times as much kinetic energy @ the same total rotor magnetic moment to act in equal opposition to them (compared with example #1).

Since kinetic energy of the rotor increases proportionally to its velocity squared (V^2), only 1.73205 (sqrt(3)) times as much rotor velocity is needed to produce 3 times as much kinetic energy and counteract the tripled kinetic energy in the windings with tripled cross section and 1/3 resistance.

In a retermination of any motor from wye to delta, the resistance lessens by 1x(1/3), which leads to triple the number of moving charge carriers at the same applied voltage, which leads to triple the total kinetic energy of these moving charge carriers at the same applied voltage. To completely counteract the kinetic energy of this tripled quantity of moving charge carriers at the same applied voltage in the wye->delta retermination, the rotor necessarily requires triple the kinetic energy to act in equal opposition to the tripled total kinetic energy of the moving charge carriers. Triple the kinetic energy of the rotor only requires a 1x1.73205 [sqrt(3)] increase in rotor velocity since kinetic energy increases proportionally to velocity squared (V^2).

^This is why I THINK in any standard wye->delta re-termination the "Max RPM per volt" (KV) increases by a factor of 1x(1.73205) [sqrt(3)].