Rotating Magnetic Field's and Inductors.

[MileHigh]

Do I have it wrong then that it is the area under the spike?  I thought it was the area under the spike that was what is supposed to be compared.

So the marginally smaller area under the one with rotor is more power?  the height of the voltage with rotor is less than without.

It is the area under the spike.  The average current reported by the scope for both of the current pulse traces is effectively the area under the spike.

The smaller area under the one with rotor represents less power.  The area for the one without the rotor in place is the larger area.  And that translates into a higher input power.

[verpies]

Why is the Voltage on the yellow Trace got this funny squiggly bit from one to the other Capture, what could this funny squiggly bit be I wonder? ?

A "squiggly bit" like this can mean:
1) An LC oscillation
2) The response of the coil to a changing external flux (from the approaching and departing magnet)
3) Superposition of the following 2 currents:
    a) When a resistive coil is powered from a constant voltage source, the current through it has the shape of an inverse exponential curve (see the 1^st plot).
    b) When a coil is subjected to a varying external flux from the approaching and departing magnet, the coil responds by inducing current in its windings to keep the magnetic flux through itself - constant  (see the 2^nd plot).

I don't think it's Case #1 because the capacitor is separated by a diode that would allow it to make only ½ of the LC oscillation.

I wish I had scopeshots with properly connected scope probes so I could make sure of that (with a moving and stationary rotor, of course)

[EMJunkie]

No, the only thing that has been proven is that adding the rotor increased the overall impedance of the coil.  Why is that?  Where is the power going without the rotor vs. with the rotor?  What is transpiring inside the coil when there is no rotor in place vs. when there is a rotor in place?

I have already stated that the coil will be more efficient without a rotor in place.  Brad can prove that for himself if he wants to by doing the test.  The lower the final current in the coil (shorter pulse) before the transistor switches off the more efficient the coil will be.

Please stop the Straw Man argument where you state that I am denying Faraday, it's laughable.

But the waveforms MileHigh, the Waveforms...

Coils, Magnets, Time Rate of Change (Velocity of the Permanent Magnets) - Faradays Law of Induction...

   Chris Sykes
       hyiq.org

[MileHigh]

But the waveforms MileHigh, the Waveforms...

Coils, Magnets, Time Rate of Change (Velocity of the Permanent Magnets) - Faradays Law of Induction...

If you can make a cogent technical argument about the test being discussed, please feel free to do so.  If you can't add any value to the technical discussion then please just read.

[verpies]

Understanding that the approaching magnet induces a voltage which raises the impedance of the coil thus reducing the input power.

Generally yes, but because the "impedance" is a different concept, to be absolutely correct you'd have to state that:

the approaching magnet induces a voltage across the coil, which subtracts from the power supply voltage, thus reducing the overall voltage available to push the current into the coil and the overall current flowing in the coil decreases.
...or
the approaching magnet induces a current in the coil, which subtracts from the current pushed by the power supply, thus reducing the overall current flowing in the coil


However, a departing magnet does the opposite.
Furthermore, an approaching magnet of a different polarity, does the opposite, too...