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Overunity Machines Forum



A-King 21 - build discussion /investigation

Started by ramset, July 15, 2019, 09:13:34 PM

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partzman

Quote from: a.king21 on July 17, 2019, 11:41:59 AM

Or lets put it into joules.
Start 1 joule in  get one joule free.
So loop it and we get 1 joule in plus 1 joule in free - now we get 2 joules more free - minus system losses. and so on ......
So how did Benitez do it?  Take a look guys. Use your brains guys.
We can surely do it better with today's tech.  Can't we???

Let's see- I use 1 joule from my input to create .95 joule on the output (95% efficiency) and this is reasonable enough to complete the first cycle.  So next, I take the accumulated .95 joule and add it to the next 1 joule taken from the input for a total input of 1.95 joule which will then create a 1.85 joule output (again 95% efficiency).  And then I..... wait a minute!  I've used 2 joules to create 1.85 joules output in just two cycles so I'm in the hole already so no need to continue!  I must be doing something wrong.  Could you please explain my problem and why I don't see a gain?

Regards,
Pm

Void

I think you can blow an incandescent bulb due to internal arcing in the bulb caused by the high voltage destroying the filament,
but does that tell you anything meaningful about output power? I am inclined to think it doesn't.

They stated that "Ammeter shows current consumption as net 1.2A", and for the second bulb,
"The current in this lamp in this configuration was 1.8 - 2.3A", but the operating frequency was
around 1 MHz. You need a special RF ammeter or RF current probe of some sort to measure currents at
such a high frequency, but they made no mention of the exact measuring equipment they used in that document
that I saw. This was apparently done by a doctoral candidate. Who writes an engineering paper and doesn't list the exact
measurement equipment used? That seems kind of sketchy.



forest

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

TinselKoala

Quote from: a.king21 on July 17, 2019, 04:41:36 PM
Well it didn't do a bad job blowing those bulbs - so that's a good starting point...
Sure, blowing bulbs is fun.   ;D
But is it OU?    :o
https://www.youtube.com/watch?v=-aKuwNTiO7I

TinselKoala

Quote from: Void on July 17, 2019, 04:50:24 PM
I think you can blow an incandescent bulb due to internal arcing in the bulb caused by the high voltage destroying the filament,
but does that tell you anything meaningful about output power? I am inclined to think it doesn't.

They stated that "Ammeter shows current consumption as net 1.2A", and for the second bulb,
"The current in this lamp in this configuration was 1.8 - 2.3A", but the operating frequency was
around 1 MHz. You need a special RF ammeter or RF current probe of some sort to measure currents at
such a high frequency, but they made no mention of the exact measuring equipment they used in that document
that I saw. This was apparently done by a doctoral candidate. Who writes an engineering paper and doesn't list the exact
measurement equipment used? That seems kind of sketchy.
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?