Real OU-Effect to Share with everyone!!!

[CTG Labs]

Hi all,

Figures look good!  But they are quite close together and we have not accounted for the energy he uses from his hand to do the manual switching, hopefully this can be done with mosfets and a digital switching circuit, but that will need current to run.  But I suspect a spark is required and so a more elaborate and power hungry switching circuit will be required...


D.

[PaulLowrance]

IMO the 123% example clearer because the blue cap voltage changes by a lot more, from 1.84 volts, rather than just 0.01 volts. 0.01V might be too difficult because of cap absorption. Although the example where the blue cap changes by 1.84V for 123% eff. is a lot clearer.

PL

[PaulLowrance]

Figures look good!  But they are quite close together and we have not accounted for the energy he uses from his hand to do the manual switching, hopefully this can be done with mosfets and a digital switching circuit, but that will need current to run.

Using a MOSFET is a good idea! Although I'd imagine such hand energy is on the order of micro joules at most. The spark lasts about 0.5ms, the electric force between the plate and lead in terms of Newtons is extreme low, amounting to almost nothing. From what I see, the only down fall could be dielectric absorption. We can't tell dielectric absorption in the videos because his DMM has too low of input resistance. You can see the cap voltage drops over time, mostly because of the DMM. If the DMM was removed, then there's a real chance that the blue caps DC voltage could rise a lot in about one hours time. That's the only way to tell what the caps *real* voltage is; i.e., let the cap sit disconnected and undisturbed for awhile.

PL

[capthook]

The whole 'experment' is bunk.

Consider the time constant while discharging a capacitor:

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/capdis.html
(Capacitor Discharge Calculation)

Consider a resistance of 4 ohms:

The time constant is .03 seconds.  This means in .03 seconds the capacitor will lose about 2/3 of its charge.
26V x 66% = 17 volts consumed
And that's with a pulse of .03 seconds

[PaulLowrance]

The whole 'experment' is bunk.

Consider the time constant while discharging a capacitor:

http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/capdis.html
(Capacitor Discharge Calculation)

Consider a resistance of 4 ohms:

The time constant is .03 seconds.  This means in .03 seconds the capacitor will lose about 2/3 of its charge.
26V x 66% = 17 volts consumed
And that's with a pulse of .03 seconds

You forgot to consider "Inductance."  He's momentarily touching the plate, which connects the battery to a ferrite core coil.

PL