http://spectrum.ieee.org/green-tech/advanced-cars/the-charge-of-the-ultra-capacitors/1

Quote from: Groundloop on November 17, 2010, 01:56:26 PM
@Sprocket,

If there is volt and you short that volt into a load then there is current.
(This current may be very small but it is still current.)
You are correct when you say that it takes substantial current to charge a battery.
But that is not the point. The point is that if I can light ONE led directly (or via
a JT) and the light NEVER goes out, then the capacitor self charge effect is a free
energy device.

@e2matrix,

You need to connect a high Ohm resistor across you wires to measure the constant
current flow out of a self charging capacitor. Try a 10 Mega Ohm. Then calculate
the current flowing. If there was no current then how is it possible to blink a LEDfrom the effect.[/]
Groundloop.

That thing is called Radiant Energy. As you may know, radiant energy is a wattless manifestation of electrical energy. It's also called "displacement current" and that kind of energy doesn't involves electron flow. Just pure voltage. And that pure voltage can perform work but it cannot charge a battery.

Quote from: Magnethos on December 02, 2010, 06:10:46 PM
That thing is called Radiant Energy. As you may know, radiant energy is a wattless manifestation of electrical energy. It's also called "displacement current" and that kind of energy doesn't involves electron flow. Just pure voltage. And that pure voltage can perform work but it cannot charge a battery.

Hi there! 

ELECTRON has the nature effect of repel and attraction.

A radiant energy is a balance thing, then therefore anything that is moving is seeking for balance because of the effect of attraction acting as a vacuum, and as it flows then it performs energy, and therefore as the facts says that we should make a strong disturbance or make a strong rung for us to disturb the radiant energy so that we can use it and that is AC of tesla.

i think in order for a lamp to produce light, then there should be an electron flow, cause its the flow of electron that makes friction in the conductor for it to glow ok! 'Collision'

It is considered wattless because our meter cannot anymore measure or reach that energy, but still it is a combination of current and voltage, and sometimes because of very very high voltage, the led or lamp lights without ground because the high voltage goes back and forth very fast and make use of the electron of the conductor from start to end of the light bulb's tail to move very very little and that produce lights.

Yes! it cannot charge a battery because it has only very little current.
To charge a battery we have to induce electron in the battery and pulling some protons, so therefore as electrons grows in numbers then attraction, voltage, currents also grows, so therefore protons are being suck from the other side and the source is empty space or wire connected to it just to balance the sucking effect of the growing electrons. and that makes an unbalance source of energy and really really want to seek balance. 

Sorry sir it is not my intention to block you but that is what i  understand.   


noicaro L. Otits reporting 

Don't worry about blocking me, we're just only discussing some interesting things. I will explain my opinion also. 

After reading some different things about this field, I cannot fully understand what an electron is. Ok, I've read the classic books about physics and electronics but it's not a fully or enought detailed explanation that I can understand.
As you know, there is a thing called Radiant energy. And that radiant energy is only composed of current-free energy. That means, it only flows the electric field (the magnetic field is hypothetically absent).
So, E = M x C² is not correct for the radiant part, since M = 0 (there is no amperage flow).
Radiant energy is expressed as  E = ∆T x C², where T means Time. Radiant flow is a flow of SUB-electronic (smaller than electron) electric particles.

I think the electron is a vortex flow. Radiant energy flow is not measurable with classic meters, but it's claimed that there are some special meters that are capable of measuring it.

Attraction means vortexes rotating in the same direction. Repulsion is a phenomenon that involves that one vortex is rotating in the opositte direction of the other.

Quote from: Groundloop on November 16, 2010, 11:18:28 PM
@Sprocket,

>>They all at most increase to 20-30mV

And that is the point, they ALL self charge. The next test is to try to get rid
of the 20mV. How many times do you need to short out the capacitor to get
it to zero volt. One time? Ten times? Hundreds of times? Is it possible
at all to get to zero volt? So after N times of discharge into a load, you capacitor
still wants to climb to 20mV. Every time you discharge you get Volt times current (V*I).

What I want to find out with my tests is how BIG that number "N" is.

Groundloop.

Hi Groundloop,
I just also tested it with a 2700 uF 35 Volt electrolytic cap.
The cap charges up to around 20 to 50 millivolts, depending
on if I touch it or not.
Then when shorted only very quickly, it recharges fast.
But when you short it out for longer time, it need also much longer to recharge again.
So the trick is to extract only short portion of it , so it can  recover faster.

Also in my case only the case and the negative pole of the capacitor are the real
contacts that have the effect.

I just also made another test
like this one:

http://www.youtube.com/watch?v=oPIHaXey1DQ

I tested this just with my 2700 uF 35 Volts electrolytic cap with my 3 Watts 12 Volts LED lamp.
I fed this from a 10 Volts DC Nokia mobile Phone charger. The Captret drops the voltage from 10 Volts to around 8 Volts at the LED and the LED lamp is much darker then. You can also connect the plus of the power supply also directly to the plus of the LED lamp and not connect it to the cap plus pole->same effect.
Then the cap works as an additional load resistor..

BUT: the Wattmeter went this way from 3Watts input and cos phi 0.48 to
0 Watts and cos phi=1 !!!
So I don´t draw any power from the grid and still get light output....

But the light output is much lower this way.
I will try it also with a 47 uF cap as Ibpointless2 did and report back..
Many thanks.

Regards, Stefan.