BREAKING : **~Imhotep~**'s Free Energy Radiant Oscillator Lite

[Peterae]

Hi Chaps
In the UK you can buy a power factor correction box just plugs into the mains near ther meter.
http://cgi.ebay.co.uk/Power-Electricity-Energy-Saver-Save-5-to-30-UK-Pins_W0QQitemZ330278143356QQcmdZViewItem?hash=item330278143356&_trkparms=72%3A1301|39%3A1|66%3A2|65%3A12|240%3A1318&_trksid=p3286.c0.m14


Peter

[sulake]

Generally "volts" is not a technically acceptable measurement for the state of battery charge.

Generally, state of a battery charge is useless when trying to find out, if your circuit is OU or not. Electrical energy is stored in a chemical form and that causes problems. When you charge the battery and then stop charging, the voltage starts to go down for a while. What is the current charge of your battery right now? Are you sure? No you are not!
Using battery will cause huge errors in your energy calculations all the time. You can not ever accurately state how much energy is in your battery. Also, all cheap normal meters give huge errors when you are using pulsed DC or just any other shape of AC than pure sine wave. You can not state how much energy went in to the battery.
Instead of batteries, you can use capacitors. Again, electrolytic capacitors are not the best choice because they attend to develop charge after you have emptied them. Again because of the chemicals.
If you want to measure the energy production of your circuit very accurately, here is what you can do.
And this is very accurate and easy!

Just use two capacitors, another one is the charged power source and the other is the one that you charge. Even easier if the capacitors are identical. Run your circuit for a while and if you end up with more volts than you started, you have OU!
Say, you have 0.5 Farads capacitor charged to 12 Volts. You then run your circuit for a while and shut it down. Then you have 9Volts left in the source and 3Volts charged to the empty one. Congratulations, your circuit is 100% efficient. There is just no chance of error!
If you end up with 9Volts in the source and 9Volts also in the empty one then there just is no possibilities, you have Free Energy! You have 150% efficient circuit. But you can not make a statement like this if you use batteries. The error marginal is just so huge when using batteries that you are just wasting your time.
When you use capacitors as the source and to store the output, all you need is a cheap ass voltage meter and you can have very accurate results.



It is very likely, that when one of us makes a circuit that is over unity, the free energy is only a small amount, say, like 5% gain in total energy. This is likely to happen like this, because we don’t yet have a law of free energy, so we will bump into it by accident. And if your system has a 0…50% error factor in it all the time, then you will miss this valuable free energy that you just made.
Greatest invention of a human history, that you made, but missed it just because your experiment was so poorly made.
After you have found a circuit that produces just a little amount of free energy, we can all replicate your circuit and start to fine tune the circuit to maximize the free part of it!

[Kator01]

Dear one,

you mix up Charge ( Q ) with W ( Energy ) in your formula.

Energy W for 12 Volt in a Capacitor with 0.5 Farad = 12 V exp2 * 1/2 + 0.5 F = 36 Joule ( Watt sec )


Two caps with 9 Volt will have : W = 2 * ( 9 V exp2 *1/2 * 0.5 F ) = 2 * 20.25 J = 40.5 Joule

which gives you 107,5 % more energy in both caps related to 36 Joule - not 150 %

Please be more accurate.

Kator

[sulake]

Oh! How stupid of me! 
If the energy stored in capacitor (12V 0.5F) is divided in two capacitors the resulting voltages would be 8.5V and 8.5V not 6V!
My bad! Well, it was two a clock at night in my time zone.

The point was, using only capacitors, gives a firm proof of the efficiency of your circuit, without the need of a fancy and expensive measurement instruments.
Anybody can do this and the results are there for others to replicate. This is not the case when the system involves unknown amount of energy stored in a chemical form. You just can not replicate something that includes an “almost dead battery that is almost empty”.
Voltage of a battery does not tell accurately about the energy stored in it. Even if it where brand new off the shelve one. Voltage of a capacitor does tell you that.

[dodoshlodo]

Hi Everyone,

I've been following this forum and the Imhotep Radiant Oscillator for a while now but this is my first post...

I just wanted to share my simple replication with everyone.

This video is great for someone starting out at this who wants to get a radiant oscillator working. Instead of a relay I'm using the Earth Ground sort of as an antenna/oscillator:
http://au.youtube.com/watch?v=wXKR7rJdze8

This video goes more in depth into the radiant oscillator, now using a PWM circuit. I can achieve some 'Tesla Like' Transmission of energy and one wire transmission. I can also light up 100V neons on the low voltage side and create one wire spark gaps that spark into thin air. Anyway its al explained in the video, check it out.
http://au.youtube.com/watch?v=1_AGbBsa45E

I have been noticing some very strange things happening with this circuit and am haing loads of fun experimenting with it. If someone can tell me how i could implement secondary battery charging into this it would be much appreciated

-shlodo
I haven't had much success