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Author Topic: Капанадзе  (Read 7315 times)

Offline NickZ

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Re: Капанадзе
« Reply #105 on: Today at 04:59:25 PM »
   If you do go, bring your tea and cups, while watching the device warm the water up.
    IF I COULD GO, BELIEVE ME,  I WOULD.

    NZ   

Free Energy | searching for free energy and discussing free energy

Re: Капанадзе
« Reply #105 on: Today at 04:59:25 PM »

Offline AlienGrey

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Re: Капанадзе
« Reply #106 on: Today at 05:10:07 PM »
are you talking about a so called Капанадзе device? To start with you don#t have much if any build information
like coil winding details and the alarming thing is it has no way of generating accelerated electrons or ions, 
or any other published extra collection method. I saying that we don't know what the the modified CFL
device does nor are we likely to, fun ain it.

AG

Offline r2fpl

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Re: Капанадзе
« Reply #107 on: Today at 05:37:57 PM »
I would if I could. But, I don't see any free electrons, nor free energy of any kind by the use of a Tesla coil/Kacher circuit, by itself.   Nor have I seen a Kacher self running. Which makes me wonder why after all these years, if they are the source of free electrons.   We have been using the reverse windings on the HV Kacher's balun, and also on the grenade coils, for years. Nothing to show for it, so far. 

FE is closer to you than you think. I have done FE many times to see that it exists, possibly you did it too just didn't understand it. Everyone has it before their eyes. Let me say more, you think that's not it! I have been dealing with the subject for several years to understand what I saw in the first year.
It is so trivial that I'm afraid if I told you you wouldn't be taking me seriously. That's why I prefer to wait. I know you are nervous because time passes and nobody wants to say anything.
« Last Edit: Today at 08:59:40 PM by r2fpl »

Free Energy | searching for free energy and discussing free energy

Re: Капанадзе
« Reply #107 on: Today at 05:37:57 PM »
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Offline Void

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Re: Капанадзе
« Reply #108 on: Today at 07:04:45 PM »
Hi Void
When i had first looked at it i wondered the same. But then, i realized that they don't back at all.
The reason is that they step on different nodal points over the long tesla coils and when current is
induced to one of the thick coils, the other is inducing voltage. They would be bucking coils only if
they were inducing current the same time at the opposite direction.

Hello Jeg. Those are interesting ideas. If Bunk and Guglodrom don't give any further details, then
all a person can do is try some different approaches and see what the result is. :)


Void,
1. If secondary is partnered coil, it will yield always zero, no matter how you drive it.
2. If that driving module is 12V input, in whole device there are no circuitry which
steps down voltage to 12 V for driving module to operate.
It can work only if driving module is 220V input.
Those are 2 things which bothers me around that device I saw in clips.
??

Hello v8karlo. In the 'all in one' circuit, I don't think it is self-looped, so the battery
is not being charged. In the 'Monster' circuit, I am not certain if they are self-looping it,
but they have added some other circuitry, so it may possibly be stepping down the output voltage
down to the required DC voltage for charging the drill battery they are using.


Have you any idea what the voltage output would be (open circuit/ unloaded) at amplifiers like in the attached image?
https://www.k-po.com/RM-KL-405-V.2.html?last=10

Hi Jeg.  Not sure about the open circuit/unloaded voltage, but amplifiers like that may have problems if you don't have a load connected,
due to the reflected power back to the transistors. You could blow the output transisitors, if the amplifier doesn't have protection circuitry.

For a 50 ohm resistive load:
If you take the peak output power of that amplifier as 200W (when amplifying a complex waveform, like for AM/FM)),
and use a resistive load of 50 ohms (equipment for ham radio use is typically designed to work into a 50 ohm load),
then you would have:
P = V^2/R  (assuming a purely resistive load)
200 = V^2 / 50
10000 = V^2
V = sqrt(10000)
V = 100V RMS (when driving a 50 ohm resistive load, and assuming a sinewave)

If you can run 400W into a 50 ohm resistive load (that little HF amplifier will probably overheat),
then the output voltage on a 50 ohm resistive load would be: 141.42V RMS.
 

So guys, call Bunk, and get some help with this. Or don't, it's up to you.
But, that's what I would do. Maybe go and meet with him, if possible. Or watch this device fade out, like all the others have.
    NickZ

That would be great if they are willing to cooperate with other people that way.


R2fpl: I wish you good luck!


Offline r2fpl

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Re: Капанадзе
« Reply #109 on: Today at 08:38:54 PM »
R2fpl: I wish you good luck!

Void - you don't have to wish me luck because the results largely take time for testing, but luck would also come in handy :)


Free Energy | searching for free energy and discussing free energy

Re: Капанадзе
« Reply #109 on: Today at 08:38:54 PM »
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