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

[Groundloop]

@captainpecan,

Attached is my first idea for an implementation of your system.
The idea is to run the energy into capacitors is steps and then
discharge the last capacitor into the load for a new cycle.

Very good work done, you have showed something that I have suspected for a long time,
that the capacitor loss is only occurring for a parallel draining scheme. Great work.

Groundloop.

[4Tesla]

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.

This was very interesting, but nightlife is correct.  Think of a dead car battery.. remove the cables and the voltmeter will still show 12+ volts even though there isn't any juice left.  I bet if you measure the current after the caps are hooked up that you will find very little there.

Great work though!

Jason

[captainpecan]

@captainpecan,

Attached is my first idea for an implementation of your system.
The idea is to run the energy into capacitors is steps and then
discharge the last capacitor into the load for a new cycle.

Very good work done, you have showed something that I have suspected for a long time,
that the capacitor loss is only occurring for a parallel draining scheme. Great work.

Groundloop.

I'm glad you see what I'm trying to get across. And you took the thoughts right out of my head with your schematics... Good work.

[capthook]

In vid 1:  you claim a gain of 'free energy'
Again - adding cap voltages doesn't equal total energy.

Plug the voltages into the formula to get joules: watt/sec
.5(.0047xV²)

C1 start: 18.22 =0.7801
after run
C1: 9.7 = 0.2211
C2: 9.29 = 0.2028

C1+C2 = 0.4239

System loss: .7801 - .4239 = 0.4239

No energy gain.  Do you see??

- -

4Tesla - a capacitor is joules - total amount of energy.  The amps will depend on the resistance (ohms) of the circuit you are discharging it into.

[captainpecan]

This was very interesting, but nightlife is correct.  Think of a dead car battery.. remove the cables and the voltmeter will still show 12+ volts even though there isn't any juice left.  I bet if you measure the current after the caps are hooked up that you will find very little there.

Great work though!

Jason

Thanks for checking it out. And yes, I do talk voltage, without talking amps. This is only because we are dealing with capacitors and not batteries, and because I really have no better way to make comparisons of the work I am seeing, compared to the energy the formulas say is in those caps after they are split. I am seeing the same work done with the caps individually, "AS IF" the actual energy in the capacitors follows the same trend that the voltage does. The point is, work is done... energy is recycled, and then work is done once again with each of the caps. Then adding the work done from each cap individually, shows that the first discharge, gave me free work. The second and third discharge gave me the work I paid for.. in a sense...