Kapanadze Cousin - DALLY FREE ENERGY

[Grumage]

     I say "someone" because I'm not an electronics expert as is Grum, or Akula, or so many others that are working on this.


   

Dear Nick.

I am no expert!! Hoppy can run rings round me!! But I do know the feeling of frustration! My field was heavy electric but having had a couple of years at night school with Radio and Television repair has helped. But Valves (Tubes) aren't used much these days. 

Now there is something rather special about GeoFusion's original Royer circuit and split core. It is to do with resonance!! Just get yourself a couple more TIP3055's and if you have another spare Yoke just have another go. PM me if you need any help. 

Cheers Grum.

[Khwartz]

That's always true only for DC.
For AC that is not generally true because:
average Amps * average Volts <> average Watts.

Usually the more high frequency content, the less true Amps*Volts=Watts is.

You are true the conditions in AC are not the same, even "True RMS" meters are not accurated for too high frequencies.

Nevertheless, read my post again, it was essentially about the input measurement, so DC.

Now, on the output side, a pure linear load as a resistor, will EXACTLY produce a proportional consumption power in any case, of the "real effective" current going through, and drop of voltage will be the EXACT indicator of the lost of energy of the electrons flowing through it.

Then the problem is not at all in the Ohm's "law", but is the accuraty of the instrument to registrer the drop of voltage, e.i. the energetic potentials difference.

So, if we are in the range of the frequencies the meter calibrated for, all is fine and you have your true voltage, so true amps by the I = U/R formulas.

If not, a very professional way is to eat a closed volume of something (gaz, liquide, etc), and by the gradient of temperatures and few calculations (through the "thermic capacity" of the mater chosen) you will know the dissipated wattage, and knowing your resistance, you have your amps and volts by: I = P/R2 and U = scareroot(P/R).

For knowing exact power dissipated by the resistor, you can heat water and you will need 1 cal of energy, 4.18 J, to increase of 1 °C 1 liter of water at 15°C. So of course if the gradient is 1°C per seconde, so you have 4.18 J/s, so 4.18 W. If the resistor is 1 ohm, you get 4.18 W / 1 ohm2 = 4.18 amps and s.root(4.18 W / 1 ohm) =~ 2.04 V of drop of voltage. (Dear Woopy, did I made any error? ).

Could look a little complicated but UNATTACABLE as a mater of proof and results., and need only CHEAP instruments and materials.

Cheers.

[Khwartz]

Hi All Nickz, Hoppy, Khwartz

.../...

HI Khwartz

I Understand. thank you for your time and explinations.
Will try to do so what you have pointed out for measurements on the device and devices coming up soon.
Hmmm Luxmeter. Interesting saw it on google. might have to see how I get one. Might be handy
That should be a method of showing it too, Showing bulb from mains and one from the device to see differences.
Well I do see the circuit consuming small amounts of amps because removing the leads from the battery does not Spark that much at all or make drastic spark noise.
Thank you again , Yes I think I will take these steps.

Cheerz~

You are mostly welcome, dear GeoFusion! Your progresses are ours too!

[Grumage]

HI Khwartz

I Understand. thank you for your time and explinations.
Will try to do so what you have pointed out for measurements on the device and devices coming up soon.
Hmmm Luxmeter. Interesting saw it on google. might have to see how I get one. Might be handy
That should be a method of showing it too, Showing bulb from mains and one from the device to see differences.
Well I do see the circuit consuming small amounts of amps because removing the leads from the battery does not Spark that much at all or make drastic spark noise.
Thank you again , Yes I think I will take these steps.

Cheerz~

Dear GeoFusion.

Or you could just make this..........

http://www.overunity.com/12736/kapanadze-cousin-dally-free-energy/msg363918/#msg363918

It is really cheap to make and although it has some minor drawbacks it does work!!

Cheers Grum.

PS. Please note Verpies comments below that post!!

[totoalas]

Dear Nick.

I can understand your dilemma with this new circuit. I have been in contact with T-1000 today and he seems to think it is another way of mixing frequencies... He keeps on saying to me that there are "many roads to Rome"!!

I am very sure that clarification will be available in the not too distant future   So for now my suggestion to you is to keep tinkering with what you have. You have all elements in place, I think. Do you have a choke within the main coil? This needs to be in parallel with a capacitor to create a freewheeling secondary frequency to that of the primary drive oscillations.

Dear MenofFather.

I think that choke you have shown as an error is for HF filtering. However I may be wrong, but that's what it looks like to me.

Dear Verpies and Hoppy.

Attached is a picture of my "Watt Box". A simple device made from a Poundland Solar PV cell. Small micro ammeter. 60 Watt lamp and a 10k ohm pot for calibration. I wanted something that did not need to have  batteries  replaced....This is an OU site afterall!! I hope you approve??

Cheers Grum.

Another way to Rome  ( in reference to your picture)
LDR   exposure  by jonnydavro    resistance measurement
and side by side light comparison by XEE2  with normal lighting against test bulb
totoalas