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ENERGY AMPLIFICATION

Started by Tito L. Oracion, February 06, 2009, 01:45:08 AM

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Magluvin

Well as we have gone over earlier in the thread, discharging a cap directly into another cap, we lose 50% of our initial energy. That loss is avoided by using inductors in the path from cap to cap. In the sim of the circuit above, I even used very low ohm switching and no resistance added to the circuit. Still lossy.  Cap to cap is like a full air tank into an empty. Once they equal out, 50-50 in each tank, the total amount of energy left in both tanks total is 50% of what was in the first full tank.  :o

Now, if we insert a turbine in the path of air pressure pipe that was connected to a fly wheel, then as we release the pressure from the full tank, it begins to fill the empty tank, but, as the pressure flows, the turbine spins up the flywheel(inductor) and once the empty tanks pressure reaches the same pressure as the first tank, the fly wheel is near max rpm, and continues to feed the previously empty tank from the tank that was full, and in the end, using check valves(diode), the empty tank could be the full tank and the first empty, if it were not for inefficiencies of the total system. ;)

So I have a valid argument with the circuit presented, as there are losses in dumping caps into caps. ;)   

Mags

totoalas


stupify12

Hello everyone.
8) what you have is the proof of the what actually the true nature of capacitor and a cap discharge. Okay i give you this hint. All the necessary parts and set up to have free energy has already been posted here and well understood by our friend a.king21.

I will explain it in words. We need a pre charge capacitor lets say 1.5v in a 1uf to 10uf 250v to 400v cap ( the purpose of smaller capacity is to fill it with charge faster) , then we discharge this cap to a coil which has very low resistance-we also have a many turns of secondary coil which is induced by the capacitor 400v discharge which make will result into multiplication of electrostatic energy. The output of the secondary is to charge a HV cap lets say .50uf 2000v to 4000v, this 4000v energy is to create a HF transformer( this is the part which create the extra energy=frequency)- the primary is also a very low resistance which also has a Secondary- capture only the charge which is more POSITIVE=Spike/Surge of CEMF/BEMF to charge another BANK of capacitor it depends how many capacity you need- it is still be charged faster because you have here a HF charger.Again capture only the POSITIVE and connect it  on the POsitive of the BANK, and for the negative- I already give you all the hint of it to where you will get it. We call it charge funnel of energy from the environment.

The capacitor Bank should be timely discharge back to the first capacitor when your switch of the cap 400v is OPEN Circuit. So you can charge it fully. At the same time the path of your discharge is true a transformer, so clock your discharge with 50/60hz.

P.S. with capacitor discharge you can also fast charge a battery it depends upon your frequency of choice. The pre charge could be from 9volt battery, electrostatic machine, Aerial capacity.

Quote from: DilJalaay on February 02, 2014, 01:34:45 PM
:) ;) :D ;D ::) :-* :-[

Cadman

From Tito's posts

QuoteSuppose two capacitors of C value initially charged at Q charges each one,(that is at V potential).

When capacitors are in parallel, the battery of 2*V waste Q charges to fill one capacitor, and other Q charges to fill the other at the same time. The capacitors rise to the same voltage as the battery (2*V).

Then you've got the capacitors charged at 2*V volts.

Now you disconnect the parallel.

Now you've got two C capacitors, each one charged at 2*Q. This means that each one of the disconnected capacitors have the same 2*V volts as the battery.

Now connect them in series. Now you've got 2*V volts of one 2*Q/C capacitor plus 2*V volts of the other 2*Q/C capacitor (series connection). That is 4*V.

They say that two series C capacitor are equal to C/2 because (in terms of voltage):

4*V = 2*Q/C + 2*Q/C = 4*Q/C = 2*Q/(C/2)

Now, it's clear that two C series capacitors at 4*V are the same as a unique C/2 capacitor at 4*V (but charged at half the charge!).

Why free energy?

Now, connect the series capacitors to the battery. The 4*V volts will be "dumped" into the battery. Then, the charges stored in the C capacitors will go to the battery, until the potential level of the series capacitors rise 2*V volts of the battery again.

Now, if you've got 2*V in the series connection, you must have V on each capacitor when you disconnect them. That means that each one of the capacitors is charged at V so the relation of charges is Q/C that is initial conditions.

SO YOU HAVE THE SAME CHARGES ON CAPACITORS THAN WHEN YOU STARTED THE CYCLE, SO THE CHARGES THE BATTERY GAVE ON PARALLEL CHARGING PROCCESS RETURN TO THE BATTERY ON SERIES DISCHARGING PROCCESS, so THE BATTERY DOES NEVER DISCHARGE.

Now only one method is missing

As it goes back and forth then it should be doing some magnification there, and we have to step that up first ok.


"for us to use the pure high voltage, we have to add a small caps to make or take effect in a coil ok "
That's it you have it.
The only secret now is the design.


i have gave you all the tools you need.

Quoteok here is a simple toy to scrutinize : let say for example we can make a make and brake right?, and since we know that a diode can permit a current and voltage to pass then why not after the diode lets make a lot of coils but not permitting bemf to go back ;D is that clear? oh boy my ape English really makes secrets. ;D



http://www.energeticforum.com/renewable-energy/16496-self-induction.html#post248775
QuoteOne more thing, about Tesla's bifilar coil. I made an interesting setup once, basically a solid state Bedini type oscillator, but instead of having a cap receiving the coil collapse I connected a bifilar coil. Meaning the output of the power coil was connected to a bifilar coil with a diode (like in normal Bedini setup). The wires on the other side of the bifilar were left open. Running the circuit for even just a fraction of time would charge the bifilar coil like a cap, and the neat thing was that it could sit there for hours after the circuit had been shut off, when I would short its open wires it would discharge with a bang just like a cap, well, because it is

Think about it

Regards