A-King 21 - build discussion /investigation

[TinselKoala]

Use 1 joule to create .95 joule and recover 0.95 from original joule

Eat your cake and have it, too?

[Void]

Which echoes my opinion from when I first worked on this. The paper is incomplete in its description of the data and how it was collected. It might pass for a 10th grade science fair project in the USA (and many things like it probably have, already). A graduate EE thesis.... well, it leaves one wanting. If we can't reproduce overunity measurements, and we don't know how the authors made their OU measurements, but we are familiar with artefacts in HVRF  measurement that they may not be... what are we to conclude?

We can only conclude that the results are questionable with the information we have
in that document. For all we know based on that document, they could have placed a potato
in series with the bulb across the output and estimated the bulb current based on the amount of steam
coming off the potato. The results are interesting if the stated measured bulb current was measured accurately though,
and if the average input power really did remain at around 4.2W while the bulbs were lighting.
There is no way to know if this might really be the case with the information provided however.
You just have to take their word for it.

[partzman]

Use 1 joule to create .95 joule and recover 0.95 from original joule

I don't understand.  Could you be more specific as to how I recover .95 joule from the original joule?

Regards,
Pm

[a.king21]

I don't understand.  Could you be more specific as to how I recover .95 joule from the original joule?

Regards,
Pm

You don't use the output.  You feed it back into the input. Feeding back loses the 0.05%  You are assuming that the output is used. It is not.  Benitez explains it better than me. In fact why don't you read Benitez. You only use the output when you are happy with the gain in the circuit.   
Then there is a specific way to use the output.  Rick outed the process in his video when discharging one capacitor into another. So I  thought "Ah well, Rick's  let the cat out of the bag,  may as well join the party".
But this stuff is only 100 years old guys.

[Void]

You don't use the output.  You feed it back into the input. Feeding back loses the 0.05%  You are assuming that the output is used. It is not.  Benitez explains it better than me. In fact why don't you read Benitez. You only use the output when you are happy with the gain in the circuit.   
Then there is a specific way to use the output.  Rick outed the process in his video when discharging one capacitor into another. So I  thought "Ah well, Rick's  let the cat out of the bag,  may as well join the party".
But this stuff is only 100 years old guys.

I have experimented with the general approach. The problem I have found is it all starts
to fall apart when you try to power a real load. The supply battery runs down just as it would
if you were powering the load directly with no attempt to feed back. Power consumption is power
consumption after all. That energy has to come from somewhere. If the energy is coming from the battery,
then the battery  will run down. If someone can demonstrate a way to pull in extra energy externally from
the environment or wherever else and use that to power a real load, so the supply battery doesn't run down,
then I would be happy to see that demonstration. It might all sound good in theory, but unless you can
pull in extra energy somehow from external to your setup, your source battery is going to run down. I hope we can
all at least agree on that.