The 'free energy' spark

[PolaczekCebulaczek]

Interesting statement. Contradicts itself. Primary amps not affected, but energy comes from primary.

Because this is a matter of time frame, amps were affected during charging capacitance, they are not effected during discharge \phase, as i said, there is some efficiency there when max out but no OU.

[Zeitmaschine]

Take for instance the electrophorus https://en.wikipedia.org/wiki/Electrophorus , which is the first and oldest reliable "electricity generator". If done right, the electrophorus is a Hig Voltage AC Generator. And with this HV AC more can be done than with a spark which only discharges a positive or negative charge.

A mechanically driven device can generate electricity. That's not really new. The idea is, we do not use a generator but a resonator (which shakes the aether, as Kapanadze worded it).

Don't kill the dipole

Normally when we want to charge a capacitor we connect it to positive and negative of something like a battery. That's the dipole we are going to kill, because there is a flow of electrons (current) going through that battery while charging the capacitor. But if we connect one side of that capacitor to ground and the other side to high voltage, then this high voltage attracts electrons from ground and those electrons accumulate on the negative plate of the capacitor (charging it). The high voltage source (the dipole) will never see those electrons, hence that high voltage dipole is not going to be killed.

Because this is a matter of time frame, amps were affected during charging capacitance, they are not effected during discharge \phase, as i said, there is some efficiency there when max out but no OU.

Sounds almost as if you fear that one could find here OU nevertheless. At least it sounds not like someone that seeks OU.

[pomodoro]

Don't kill the dipole

Normally when we want to charge a capacitor we connect it to positive and negative of something like a battery. That's the dipole we are going to kill, because there is a flow of electrons (current) going through that battery while charging the capacitor. But if we connect one side of that capacitor to ground and the other side to high voltage, then this high voltage attracts electrons from ground and those electrons accumulate on the negative plate of the capacitor (charging it). The high voltage source (the dipole) will never see those electrons, hence that high voltage dipole is not going to be killed.

Try it out then , if you dare!

Charge a cap (10uF at least to avoid quick leakage) to 20kV and stick it up high above the ground. This is your stable high voltage supply which will give you free power apparently.
Now see if you can measure any voltage between ground and just one of those terminals of the charged up cap. This will be the voltage that can charge up another cap for free!

You will find that there is no voltage difference at all! No cap can charge for free when placed between earth and a high voltage battery with one terminal of the battery disconnected from earth.

I suggest you try this before assuming that Bearden was telling us some hidden secret.

You MUST use a stable HV sourve, and avoid corona discharge to ground from the unconnected terminal, stable like what you showed  in your diagram (battery)  ,hence the 10uF cap I suggested,  because any pulsating,DC or AC HV supply WILL give a voltage difference to earth due to the displacement current I mentioned earlier but  the displacement current  takes power out of the generator and its no free lunch!

Frolov and Bearden are a bunch of buttheads that have done OU a great disservice by disseminating pure BS.

[PolaczekCebulaczek]

Frolov and Bearden are a bunch of buttheads that have done OU a great disservice by disseminating pure BS.

absolutely true.

Sounds almost as if you fear that one could find here OU nevertheless. At least it sounds not like someone that seeks OU.

sounds almost as you don't know how electricity works. I've tried (and million of others since discovery of electricity) this configuration and my findings point to no OU and I know why.

killing dipole:
you can connect one side of cap to earth and other side to HV DC and electrons will deposit on one side of cap and other electrons will leave the other side of cap, where those electrons would go? to (now ionized) earth of course trough air, circuit is completed, dipole is killed and Tom Bearden can suck my nuts!

not much related to this topic but the cool thing about this system is that tesla coil is doing that on better scale, and if you remove electron from earth and put it back you create an electrical vibration in earth that you can detect at some distance.

[Zeitmaschine]

Frolov and Bearden are a bunch of buttheads that have done OU a great disservice by disseminating pure BS.

A great disservice? Then I'm sure you can point me to some posts of yours where you have done OU a great SERVICE. Because otherwise I can't see any reason why you are writing on this forum. If you think OU isn't possible (BS) then why you are interested in?

The same is true for PolaczekCebulaczek, you should trying to find a method how to do it, not to contradict each sentence an OU researcher writes. This would lead to a much more positive charisma of yours.

Bearden is BS in the eyes of vested interests? I indeed do believe this.

sounds almost as you don't know how electricity works. I've tried (and million of others since discovery of electricity) this configuration and my findings point to no OU and I know why.

OK, if you know how electricity works, then surely you can explain to me that one-sided ground current that we can see in the image below. I'm aware that it can be faked. Everything can be faked. But for the moment we assume that it is real. So how does it work?

The car battery here is used as metallic object. In the thick white wire the electrons (or whatever we should call an electric current) are oscillating (at 50Hz) without a closed circuit. Since this should not be possible according to the textbooks, it would make no sense at all to fake it, especially without any real need to, because Stepanov (and also Kapanadze) could claim OU without showing that odd ground current effect. So why is it that this clamp meter reads 8.2 amps? Any constructive ideas?

My guess: In the device we have high voltage without amperes (the open dipole) and - as a counter part - amperes without voltage (the ground current). Therefore Ohm's law does not apply. Now we should looking for the law that applies.