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The Young Effect, my gift to the free energy movement!

Started by captainpecan, November 16, 2008, 11:02:42 PM

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captainpecan

Quote from: capthook on November 17, 2008, 12:32:47 AM
Calculate out the total charges.
Like you properly gave: 1/2(.0047x V2)

Start cap C1: .5(.0047x19.35x19.35)= 0.8799

after run

C1= 10.35 : 1/2(.0047x10.35x10.35)= 0.2517
C2= 9.57 : 1/2(.0047x9.57x9.57)= 0.2152
C3= 4.62 : 1/2(.0047x4.62x4.62)= 0.0502

C1+C2+C3= 0.5171

Net loss: .8799 (start cap) - .5171 (total of C1+C2+C3) = 0.3628

So you have a net LOSS of .3628.  About a 40% loss.

Just adding the voltages up on the caps doesn't work...

Thanks for the vids though!

I understand exactly what you are saying... but I did not compare it to the energy of the first cap alone. That was the point of the first of the video. At the first of the video, I took C1, and discharged it into three caps parallel, with no leads or components.  That is the energy that I compared the end experiment to.

The question is, why is there more energy left in the 3 caps running it through the circuit, than there was without the circuit, directly draining them in parallel.

In the last video, I did not compare one cap to 3... I compared 3 caps to 3 caps.. all in parallel the entire time.

captainpecan

Quote from: nightlife on November 17, 2008, 12:28:26 AM
captainpecan, you are not recording the amps and voltage is not good with out amperage.

The first video you showed running the motor on 18 volts and then it stopped running but yet you showed have 19 volts still left. If it ran off 18 volts, it should have continued to run off the 19 that was still left. The reason why it didn't is because you used the amps and even though you ended with 19 volts, there was not enough amperage and therefore you really didn't gain anything and still had a loss.

The question is not amps. It's the total energy. And I know the equation says I have lost half my energy. The point is that I am getting the full amount of work on the motor, from each cap individually... After I already used the energy once to run the motor the first time.

By the way. It stopped running because it is a pulse motor and I only delivered 1 pulse. It was not supposed to keep running. The fact that the energy was left over for me to use again tells the tale. Equations true or false. This energy was recycled.

Michelinho


@captainpecan,

Nice work. I enjoyed the videos very much. Thanks.

Take care,

Michel

nightlife

 Voltage with amps = watts. Voltage without amps = no watts. Power is based on watts or joules per second. You must have a watt gain to have any gain at all.

capthook

Quote from: captainpecan on November 17, 2008, 12:53:55 AM

The question is, why is there more energy left in the 3 caps running it through the circuit, than there was without the circuit, directly draining them in parallel.


So you are thinking the 'circuit' is creating extra power you can harness?  Free energy?

Well - you commented on the (1) large spark when connecting the 3 capacitors in parrallel to "equalize" their charges.

There was (2) no such spark in the later test running it through the circuit.

So you lost more energy in the spark of (1) than you did running it throught the circuit (2).

Again - you need to measure (and fully calculate) energy in and then energy out.
No matter how you setup what you are doing - getting more out than in will never happen.