Selfrunning Free Energy devices up to 5 KW from Tariel Kapanadze

[jbignes5]

I think you got the wrong impression. Think about the Tesla coil. At one end the field is weak and at the other end it is very strong. This would make the same setup as the air gun vortex. The ferrite is the focus. The copper is merely the injector of the pulse and also the squeezing element. Once the vortex forms it wants to stay formed. So it rolls just about where it was formed. Periodic blasts of one way current tap the vortex and coax it to speed up. And the vortex should interact with the Tesla coil raising the low voltage end to a high voltage end and the other end becomes super high voltage.


There is something to this that makes sense.

[wattsup]

Hey wattsup,

Concerning your idea of the parallel stripes I would expect them to destroy themselfs if you do not have them strongly glued on a epoxy-sheet. There might show up heavy current-density
http://www.youtube.com/watch?v=6uSKFmgmQvs
Regards Kator01

@Kator01

I don't think the video you referred to would properly apply to this at all. There they are showing the effect on a wire with high amperage. If you look at the size of that transformer. Here we are dealing with high voltage. The voltage will only follow along all those lines just like it would if it was a coil or many coils in parallel.

The capacitor on 7 and 8 is a small bias to the coil. There will be a voltage difference between the two winds and they create a small pool of energy there to bias the coil.

How does that sound? Bounce a drum stick on a loose drum skin and it doesn't bounce back in return as much as if the drum skin was tightly stretched. This is the same premise I believe. But there are more elements to this equation. Surface area and even geometric shape of the fields surface determine the output.

@jbignes5 and @All

The FBT is the central part of all this action. The HV positive alone is energizing the caduceus coil via the pulses coming from the HV positive spark gap, while the HV negative is going through this rather long coil located inside the FBT. The HV is not really being solicited in a classical load loop since only the HV positive is being used to generate the primary spark gap. The output spark gap is being energized via the HV negative side. Both the negative and positive sides of the HV output never touch each other. The interface of these two is the Caduceus Coil to Output Bucking Coil. In this manner the HV is not really being loaded in a conventional manner.

Thanks to @Osiakosia's post here;
http://www.overunity.com/7679/selfrunning-free-energy-devices-up-to-5-kw-from-tariel-kapanadze/msg307246/#msg307246
you can see the TBC-110ЛА is listed there and very interesting information we never had before.

The secondary (not the HV) goes from pins 4 to 9 so you have;

4 to 5 is one coil of 80 turns.
5 to 6 is one coil of 80 turns but from the other values it seems this is a thicker wire so less resistance.
6 to 7 is one coil of, well as Murphy's Law would have it, they forgot to put that one in the table but I would safely put minimum 80 turns but if I look a the others, 6-7 is about 33% of 7-8. 7-8 being 610 turns, I would say 6 to 7 is 200 turns.
7 to 8 is one coil of 610 turns with a 2.2pF - 10kv capacitor.
8 to 9 is one coil of 190 turns before this wire goes to the Caduceus Coil Center Tap (CCCT) output.

So the flyback secondary coil is around 1160 turns with a cap tap at around 360 and 970 turns.

In normal flyback usage this secondary coil will supply various output voltage levels taken from the 4 to 9 pins. So when the primary pulses, the secondary provides output. The difference here is now this coil is not only doing what is usually does, but now it also is riding on the HV negative going right through to the CCCT. So here we have a very new effect to look at.

9 to 10 is one coil of 1200 turns producing the high voltage output.
1 to 2 is one coil of 48 turns. This primary coil is left open.
2 to 3 is one coil of 48 turns. This primary coil is the only drive coil.

Notice it is not a 5-6 turns primary hammering and hammering on the HV coil. It has 48 turns, obviously much thinner wire but even so, the primary manages to pulse the HV and secondary. But most importantly, it manages to survive the unusual HV negative (HVN) feeding back into the FBT. So what could that effect be? Maybe the HVN is neutralizing the secondary flyback and produces what I am tempted to call a "negative diode" that prevents any CCCT harmonics from returning and potentially harming the FBT. Who really knows but at least now I know how the TBC-110ЛА is precisely built. Most all FBTs have a secondary output coil (otherwise you would probably call then a Tesla coil) so it just remains to be seen how we can emulate the original or close enough to have the effect. A low level investigation of this effect may be the best way to further understand it.

wattsup

PS: Sorry for long post.

[verpies]

Below are my interpretations and rephrasings of the tuning instructions pertaining to a device using  ferrimagnetic core from a TV deflection Yoke.
I would appreciate any verifications of the issues and additions marked in blue color and clarifications of questions marked in red color:


***********************************************************************
Do not concentrate on particular frequencies but on the tuning process itself.
The same frequencies will not work in all devices since every ferrite and every winding has slightly different properties . 

To tune the Yoke device use the following steps:

1. Connect the output of a frequency sweeping Signal Generator to the 15-turn winding (the generator can be connected to the 50-turn winding as well, but connecting it to the 15-turn winding makes the tuning process easier)

2. Connect the transverse 1-turn "U" shaped winding to the input of a Spectrum Analyzer (A computer based spectrum analyzer is sufficient)

3. Find a frequency peak above 1MHz  of the largest magnitude on the Spectrum Analyzer. Let’s denote this frequency as F1. What is the recommended frequency range of the Signal Generator's sweep?  From 0 to 50Mhz (see pt.17a) or a narrower range ?

4. Disconnect the Signal Generator and the Spectrum analyzer (connected in pt.1 & pt.2 above)

5. Connect the output of a Signal Generator set to frequency F1 (sine, square or rectangular/duty?) to the transverse 1-turn "U" shaped winding.

6. Connect the 50-turn winding to the input of a Spectrum Analyzer and look for a frequency peak with the largest magnitude,  that is not caused by an exponentially decaying ringing waveform and is not the frequency F1.  Let’s denote this other frequency as F2.

7. Disconnect the generator from the transverse 1-turn "U" shaped winding (connected in pt.5) leaving this winding unconnected to anything.

8.  Connect the output of a Signal Generator set to frequency F2 (sine, square or rectangular/duty?) to the 50-turn winding.
     Does the input of the Spectrum Analyzer remain connected to the 50-turn winding ALSO (together with the output of the Signal Generator), see pt.6 where it was connected and never disconnected ?
  a) If the output of a Signal Generator is connected to the 15-turn winding instead, the frequency/spectral response of the signal induced in the bifilar winding does not change much, but its amplitude becomes larger.  The latter may be helpful with some measurements.

  9. Connect the bifilar winding to an oscilloscope and a small incandescent light bulb in parallel. Be careful as the high voltage induced in the bifilar winding may damage the test equipment (if the voltage is too high then use a voltage divider or a scope probe of an appropriate voltage rating). The light bulb also helps to lower this voltage. 
Are there any capacitors connected to the bifilar winding in parallel, at this stage?

10. Slightly vary the frequency of the Signal Generator connected to the 50-turn winding in order to find the maximum voltage amplitude induced in the bifilar winding, as observed on the oscilloscope. Make sure that the signal induced in the bifilar winding does not represent an exponentially decaying ringing waveform and resembles a constant amplitude sine wave, as closely as possible.

11. Repeat pt.10 with the output of the Signal Generator connected to the 15-turn winding.

12. Connect a second Signal Generator set to ~50 Hz sine wave without disconnecting the first generator connected in pt.8 and adjusted in pt.10 & pt.11. The output of this ~50Hz generator will be denoted as the modulating frequency F3.

13. Vary the F3 frequency in order to obtain the highest amplitude of the signal induced in the bifilar winding and have the waveform induced by the first high-frequency generator (F2) ride-on-top of the ~50Hz sine wave (F3) from the second generator(see: waveform addition).

14. Connect a more powerful incandescent light bulb to the bifilar winding (or a light bulb if no light bulb was connected before in pt.9 ). This will also lower the induced voltage in the bifilar winding.  Are there any capacitors connected to the bifilar winding in parallel, at this stage?

15. Fine tune F2 and F3 in order to obtain the brightest light from the light bulb. At this point the bulb should be dissipating approximately 150W of power.

16. Apply high voltage sparks from a flyback transformer to the transverse 1-turn  "U"-shaped winding via a spark-gap. Was the 1-turn "U"-shaped winding left unconnected between pt.7 and pt.16 or was there a capacitor connected to it in parallel?
   a) Keep in mind that applying sparks from flyback to the 1-turn U-shaped winding causes much higher voltages to be induced in all windings, including the 15-turn and 50-turn windings where the Signal Generators are connected, thus those generators are in danger of being damaged. Protective buffers are highly recommended in order to protect the Signal Generators.
   b) With the HV sparks at the 1-turn U-shaped winding the bifilar winding should be able to output approximately 1KW, which can be verified with a chain of stronger incandescent light bulbs in series.

17. The ferrite material of the yoke core may change its properties during the operation of the device affecting the F1, F2 and F3 frequencies, thus the first Signal Generator should be capable of outputting signals above the common 2MHz limit of many signal generators.
   a. It might be interesting to sweep the signal generators connected in pt.1 from 0 to 50MHz in order to probe the frequency characteristics of the core and windings.
   b. Note that not every ferrite material and shape may function well.
   c. Keep in mind that strange effects might happen including extreme, unknown and enormously strong fields and radiations, that might or might not affect the health of the experimenter/operator, albeit adverse effects have never been scientifically confirmed.

[semenihin-77]

As told, the bulb method is totally in-conclusive in estimating real power... but purelly from personal experience in the field, i have the impression that Kacher's output combined with inverter's, is more than what one should expected but seeing individual bulbs brightness at each mode (plain kacher mode or plain inverter mode).
Just interesting...  nothing more

Note, the sparking kacher is a sign of losses. Kacher's input can be as much as 20 watts or so, but its capacitative coupling and lamp power transfer could be below 50%.
Semenhih, says that his inverter takes 15 watts (i would swear it takes more than that). So if, say, it takes 15 watts Kacher alone should output below 10w (pretty obvious in comparison)

Anyway, i find it amusing!

Shaw will not hold, I can not for the entertainment of the crowd put the experience if you so blind that would see the truth, I wash my hands. Look for the energy of ferrite, and so on.
It is unfortunate that attitude.
Delamorto

[xenomorphlabs]

Shaw will not hold, I can not for the entertainment of the crowd put the experience if you so blind that would see the truth, I wash my hands. Look for the energy of ferrite, and so on.
It is unfortunate that attitude.
Delamorto

Don't be discouraged. It is understandable that people are struggling to see power measurements and sooner or later someone will do them.
The last diagram you posted is really interesting as it reflects what Kapanadze is doing in the newest device.
Instead of the slitted copper tube he just seems to be using the copper of the output wire itself going through the core of the crossed coil. That wire is most likely being fed with a 50 Hz sine signal (from the PVC box) just like you supposed to attach to both sides of your copper tube.
The power of the sine signal is maybe only 10 Watt. So the current amplification might take place
inside the crossed coil due to a property of that crossed coil in conjunction with the HV excitation.
I speculate that effect to be more related to the electric field of the coil thus increasing the current  in the wire. Such element is explicitely mentioned in TK's patent and referred to as "current amplifier".