Real OU-Effect to Share with everyone!!!

[Frederic2k1]

Why the endless confusion with collecting a spark on a capacitor?
Wasn't captainpecan's 84 pages enough?

POWER IN MUST BE MEASURED!
And no where is that done!

Just because you are showing a larger voltage than the source means nothing.
What is a spark?  HIGH VOLTAGE!
So what do you get on a capacitor capturing that spark? HIGH VOLTAGE!

So let's see:

Power in:
12v source into a 4.7 ohm coil
I=V/R
12V/4.7Ohm = 2.553 amps
12V x 2.553A = 30.396 watts
30.396W x .1 seconds = 3.0396 wattseconds (W/S) or joules (J)

Even say you could pulse it in .01 seconds (1 hundredth of a second....doubtful... and a pulse of 1 microsecond - that's one millionth of a second - as listed on one of the diagrams is just silly thinking)

30.396W x .01 seconds = .30396 J or W/S

Power collected:
26V on a 30uF capacitor
.5(26x26x.000030) = 0.01014 J or W/S

So IF you could actually pulse it at .01 seconds (.1 would be more like it)

Power in:  .30396 J
Power out: .01014 J
Efficiency: 3.33%
Wasted power: 96.66 %

And that's being generous........
It's probably more like:

Power in (.1 second): 3.0396 J
Power out           : .01014 J
Efficiency:   .333%
Wasted power: 99.66%

You are not understanding the principle behind this experiment.
Wenn he first hit the two conductors without coil, then the source (the battery) was directly connected to capacitor, so there was low ohmic resistence, it was nothing more than a short circuit and the capacitors voltage climbed as high as the voltage of the battery:

So power collected on capacitor was

E = 0,5 x 12 V x 12V x 30 uF
= 0,00216 J or Ws

Then he connected the coil of 4,7 Ohm and hit again the two conductors.
In this case, because of inductive reactance and added 4,7 ohmic resistance the current must be lower (I = U / R).
The 12 Volt of the source stays the same and although the current is lower because of higher impedance, in the end the voltage on capacitor is higher and we have a gain in energy...


It is true overunity and it is exactly that what Tom Bearden said...

With the high inductive coil, he only wants to transfer the voltage without current. In every high inductance the current is nearly zero at the moment of switching. When you switch to slow, then the current will rise and the overunity effect is gone because the source will depleted.


kind regards

[capthook]

You are not understanding the principle behind this experiment.

The 'experiment' is hopelessly flawed, the observations and measurements are severely lacking and the conclusions drawn are completely wrong.
And it's comparing apples to oranges.

The first part has nothing to do with the second.  So to draw conclusion by comparing the two results is impossible and irrelevant.

It is as simple as this:
Take the second part of the - 'experiment'.

1. Power in: how many joules are drawn from the battery?
2. Power out: how many joules are collected on the capacitor?
3. Net loss/gain: 1 minus 2

And since no one ever bothered to measure/calculate the power in, I provided a reasonable calculation so that you might understand what was happening and how and where it all was wrong.....

[Tito L. Oracion]

@ captainhook

Sir mathematicaly you are correct but if we make sudden discharge of that cap to a coil, i think it will be converted into huge energy.

i think from that mathematic is not anymore a reality.

N.tesla, bearden.


Dream on...

o yeah
It is on dream why tesla made most of his inventions into reality. even bearden maybe is a dreamer next me ha ha ha;D

[Frederic2k1]

The 'experiment' is hopelessly flawed, the observations and measurements are severely lacking and the conclusions drawn are completely wrong.
And it's comparing apples to oranges.

The first part has nothing to do with the second.  So to draw conclusion by comparing the two results is impossible and irrelevant.

It is as simple as this:
Take the second part of the - 'experiment'.

1. Power in: how many joules are drawn from the battery?
2. Power out: how many joules are collected on the capacitor?
3. Net loss/gain: 1 minus 2

And since no one ever bothered to measure/calculate the power in, I provided a reasonable calculation so that you might understand what was happening and how and where it all was wrong.....

Do you then mean, that this is a rather more efficienter way to charge a cap than overunity ?

Nevertheless he has more energy in his cap, with same voltage reading of the source and lower current through ohmic resistance and induvtive reactance. ok agreed, it is still no proof for overunity...

[NRGFromTheVacuum]

Hello everyone,

Overunity is a word which is severely misunderstood.

When we talk about electrical overunity many people use to their common EE teachings say you need to measure watts in & watts out. However this thinking is flawed from the very beginning because their only taught to deal with electron current.

What must be understood in my particular experiment is the battery's energy and the energy which ends up in the capacitor, are completely different! The battery contains electrons which when caused to flow, generate heat, fall victim to entropy, and exhibit a positive curve in gravity. The capacitor contains, radiant energy, negative energy, Dirac-sea-holes, negative mass-energy electrons, dark energy, whatever you want to call it, and when caused to interact with matter, generates negative probability's, negative entropy, and exhibits a negative curve in gravity. The two types of energy must be understood before you can understand anything going on in the circuit.

If I charge one capacitor with electrons, and the other with Dirac-sea-holes. The one with Dirac-sea-holes will weigh less than the one with electrons.

In a future video, I will show this...


So if we want to properly measure the electrical overunity we must measure the input energy in watts, joules, or coulombs through the one load. Then instead of measuring the output watts, joules, or coulombs through one load. Use multiple loads in series with each another, each with its own capture circuit and  storage capacitor. Then take the total energy from each storage capacitor add it up and compare.