Delayed Lenz or not?... post your explaination!

[synchro1]

Synchro1 you can wonder about all manner of silly ideas.  The problem is that any idea is only as good as the evidence that supports it.

@MarkE,

You maintain that "Magnet Wave" is a made up term. It takes time for magnetizem to travel from point A to point B through a core. How do you propose we describe this event?

[MileHigh]

Luc:

You saw an increased power draw when the generator coil is loaded.  Before you were insisting that there was no increased power draw.  You should try to account for your past errors and try to square them away with your readers.

Input to Dremel on load is 60vdc @ 0.500ma = 30W
Input to Dremel off load is 60.3vdc @ 0.475ma = 28.64W

Power difference is 1.36W of extra power consumption by prime mover when coil is on load

Output is a 25 Ohm load @ 7.84vrms = 2.46W
then we subtract - 1.36W =  1.1W of gained power output without affecting prime mover.

I hate to say it again but you are leading yourself down a garden path.  Also, you are making real measurements here, you should not be ignoring the resistance of the coil.  The coil capacitance measurement is also meaningless and has no affect and should not be considered.

Here are some issues that have to be factored in:  1) You have no idea what the efficiency of the Dremel is.  2) You have no idea if the efficiency of the Dremel will change under different supply voltages, loads and RPM.  3)  You are drawing a conclusion without having enough data to support the conclusion.  4)  You are not correctly relating the waste heat with the "payload" power that goes into the generator.

Here is your real data:

UNDER LOAD:

Input:  30 watts electrical

Generator output:  2.46 watts
Motor output:  27.54 watts heat
Total output:  30 watts

Note:  The motor heat output power is derived by subtracting the generator output power from the input electrical power.  That is how the power is split in your setup.

NO LOAD:

Input:  28.64 watts electrical

Motor output:  28.64 watts heat

The above is the real analysis of your data.  That's all that you can conclude from your experiment with the caveat that the heat dissipation in the coil windings was not accounted for.

Output is a 25 Ohm load @ 7.84vrms = 2.46W
then we subtract - 1.36W =  1.1W of gained power output without affecting prime mover.

The above calculation is an invalid calculation.  If you review what I wrote above this should become clear to you.

MileHigh

[synchro1]

Luc:

You saw an increased power draw when the generator coil is loaded.  Before you were insisting that there was no increased power draw.  You should try to account for your past errors and try to square them away with your readers.

I hate to say it again but you are leading yourself down a garden path.  Also, you are making real measurements here, you should not be ignoring the resistance of the coil.  The coil capacitance measurement is also meaningless and has no affect and should not be considered.

Here are some issues that have to be factored in:  1) You have no idea what the efficiency of the Dremel is.  2) You have no idea if the efficiency of the Dremel will change under different supply voltages, loads and RPM.  3)  You are drawing a conclusion without having enough data to support the conclusion.  4)  You are not correctly relating the waste heat with the "payload" power that goes into the generator.

Here is your real data:

UNDER LOAD:

Input:  30 watts electrical

Generator output:  2.46 watts
Motor output:  27.54 watts heat
Total output:  30 watts

Note:  The motor heat output power is derived by subtracting the input electrical power from the generator output power.  That is how the power is split in your setup.

NO LOAD:

Input:  28.64 watts electrical

Motor output:  28.64 watts heat

The above is the real analysis of your data.  That's all that you can conclude from your experiment with the caveat that the heat dissipation in the coil windings was not accounted for.

The above calculation is an invalid calculation.  If review what I wrote above this should become clear to you.

MileHigh

The math you're applying here is backwards. The heat loss formula measures a negative value, not a positive one.

[MileHigh]

Corrected the original posting with this new sentence:

Note:  The motor heat output power is derived by subtracting the generator output power from the input electrical power.

[synchro1]

Corrected the original posting with this new sentence:

Note:  The motor heat output power is derived by subtracting the generator output power from the input electrical power.

That only Works if the generator output is less then the input.