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

[Groundloop]

@4Tesla,

You relly can't compare a capacitor with a battery! The VOLTAGE over a capacitor terminals is directly related to the total WORK a capacitor
can do when discharged into a load.

Work must be done by an external influence to move charge between the plates in a capacitor. When the external influence is removed, the charge separation persists and energy is stored in the electric field. If charge is later allowed to return to its equilibrium position, the energy is released. The work done in establishing the electric field, and hence the amount of energy stored, is given by (See image attachment).

Groundloop.

[4Tesla]

Okay, but he isn't measuring joules.. he is measuring volts.. if the energy is still there when the caps are disconnected from the circuit.. then why did the motor stop?

Jason 

[captainpecan]

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.

Yes... I understand EXACTLY what you are saying.  I have been through this many times, believe me. Just for 1 second, forget what that equation is saying to you. Compare the work that is done. See if it matches what the equation says. I'm not supposed to be able to get that much work out of the caps after they are split.

You know what, it really doesn't matter. Thank you for your comments. I'm not saying your figures are wrong. As you are figuring it EXACTLY like it has been taught forever. I'm questioning what has been taught, not your figures. What is taught does not seem to match the work I'm seeing.

And the spark... You may be right, or you may be wrong.  Honestly, after doing the exact experiment over a 100 times and still getting the same results, I know there is energy entering the system.  The only question is where from. Possibly RF, in that case, any gains I receive would not be there if the project moved... lol...  I simply showed an interesting effect I noticed.

Like I said, there is no point in arguing this.

[Groundloop]

@4Tesla,

The motor did stop because it is NOT a complete motor. He has no switching to let the motor run.

What he is doing is charging up ONE capacitor, run it through the coils and INTO a second capacitor.
The energy is moving the magnets away from the coils (thus creating work). Now the first capacitor
is discharged until the voltage in the second capacitor reach the 1/2 of the start voltage. The math
says that we should lose 50% in the process. Now he measure both capacitors and find that the
total voltages combined is greater than the start voltage. Voltage over a capacitor is in direct relationship
with the total work a capacitor can do. So if you discharge both capacitor into a load, one at the time,
then you will find that the work those two capacitors can do is greater than the work the first capacitor can do.

The 50% loss formula only applies when discharging a capacitor into a parallel load. When we discharge a capacitor
through a load and into another capacitor then it seems that the formula is wrong.

This is how I understand what he is doing.

Groundloop.

[captainpecan]

Okay, but he isn't measuring joules.. he is measuring volts.. if the energy is still there when the caps are disconnected from the circuit.. then why did the motor stop?

Jason 

And this is the point of these videos.  This is where you need to try and put what you have learned about how a motor uses up the energy fed to it.  And for just a minute, think what would happen if the energy did not get used up.  What if it just passed right through to the other side and you caught it?  Then you could use it again.  The true meaning of "Conservation of energy".

Two capacitors in paralel... 1 at 18v, and 1 at 0v.  A capacitor will simply try to balance the energy between the two. Hence, that is why half the energy flowed through the motor and landed in the second cap, and the other half, never left the first cap.  When the capacitors are balanced, they will not do any more work until there is a place of lower voltage to flow to.

I hope I explained that well enough.