Body
Hi Ole, and SvenIf you have a dc motor, perhaps you might have two? If so, two pulleys and a belt, and you have a genset, subject of course, to Lenz's law and then use your string of old fashioned incandescent light bulbs to adjust the load.It is a bit of nonsense, yet it gives a graphic indication of the energy that is of concern to us, and anyone else who may be interested.Those tungsten elements glowing away tell the story that digits on the meter leave abstract.Whether we heat a tank of hot water, or travel down a highway, it can be safely represented by those tungsten glowing elements.Have you noticed that leds are not much of an indication? They are already half way to dark energy, and quantum physics.In our new practical lives, we must use each and every tool in our toy box to use the free energy and not use the expensive grid power.For our testing purposes, those difficult to find, old fashioned bulbs work really well. I like the no nonsense clear bulbs, so the element is visible at most stages of glowing. Soon you will find no one sells them. I buy Chinese ones at the Dollar stores hear in Canada, enough to supply this need in the future.My Tesla spiral coil experiments popped quite a few with the induction stove top, a surprise, considering they were rated for 150 volts. A few can be destroyed without killing your bank account.Cheers Warren
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On Sun, Feb 11, 2018 at 5:30 PM, [email protected] [EVGRAY]<[email protected]> wrote:
Hi Sven,
A short doesn't dissipate the power. For greatest power transfer there has to be impedance match. To change the impedance of a transformer requires changing the number of turns on it. If matched for 24V load on the secondary the primary may be matched to some other impedance requiring some other voltage. Hector tells to use radio frequency rules. The mains grid would require dissipating MW power for an impedance match as it is of very low impedance. Working at high frequency is done by using perhaps 50 Ohm transmission lines. This requires 50 Ohm at each end of the line for an impedance match if using coaxial cables having a characteristic impedance of 50 Ohm.
I don't know if there has to be perfect impedance match for making free energy. But the load is probably critical. At least it is for the Jensen UDT. Perhaps the ideal load can be found by using the idea of Warren by connecting a motor and then braking the motor for using it as a variable load for best impedance match. The brake will have to be able to dissipate the power without setting anything on fire.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hi Ole, As I understand it, I can only lower the impedance of the primary winding in which I load the secondary winding best short circuit so that the capacitors can discharge quickly. Then the reactive circuit is apparently closed again. I would not know now how I can change the impedance of this transformer. I assume that at a very low resistance the voltage secondary collapses to a very low value. Maybe you could directly charge a battery via a FWBR. Greetings Sven
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Von: "onielsen@... [EVGRAY]"
Datum:11.02.2018 20:55 (GMT+01:00)
An: [email protected]
Betreff: Re: Aw: Re: Re: AW: Re: AW: Re: Re: AW: Re: Re: AW: Re: Re: Re: AW: [EVGRAY] Re: All information about Diode Plug.
Hi Sven,
This text from Hector tells that the impedance of eachpart (input and output side) of the circuit has to match the frequency.The charging of the capacitors must be done in no longer time than ahalf cycle which is 10ms at 50Hz. The output as well must be done inless time than a half cycle.
To do this the LC constant of theinput part formed by the output of the ferroresonant transformer and oneof the capacitors must be shorter than the time of half a period. Thesame applies for the output part of the circuit. As the capacitor is aconstant for both input and output only the transformers can be changed.Thus if the capacitors are chosen to match the ferroresonanttransformer the output transformer has to be chosen for the properoutput impedance.
The load has to discharge each capacitor in less than 10ms. The full period time is given as 2πf X square root(LC).Thus solve for L or C when choosing the other one of them. f is thefrequency in Hz L is the inductance in Henry and C is the capacitance inCoulomb. The same formula is used for the charging part of the circuitwhere L is the output inductance ot the ferroresonant transformer. Forthe output transformer the inductance is decreased by increasing theload as this cancels more of the primary winding's inductance. The finething about transformers is that they actually change the impedancebetween the primary and secondary windings. Thus choose thetransformation ratio to fit the wanted impedance which can go up or downor be one to one. This can be chosen for 24V at the output and thevoltage of the capacitors at the input. The voltage at the capacitors isthe peak voltage that must be transformed down to the peak of 24V. Thatpeak is different if the period time is different from a full wave whencombining the two semi-waves. A shorter pulse width at 50Hz requires ahigher peak value for delivering the same RMS voltage.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hello Ole, I found something here on evgray.
I quote:Transverter Secrets Revealed:
Data from 1999 series of Experiments
480/220VAC Universal transformer design :
The circuit here consists of 2 half LC tanks, AS know the RADIANT Energy
flashover occurs at TURN ON, one capacitor charges to maximal resonant value
then the cycle goes in opposed sinewave as the other charges in Resonant
condition the first one is DISCHARGED in the second transformer with a
nominal gain of 1.618 if properly tunned, this second circuit deplete C
Voltage value as near to 0 Volts as SCR switches off at a minimal remanent
voltage (minimal)before primary tank reverses sinewave to Reload half LC and
discharge the other loaded capacitor.....
The discharge does not affect input tunning as its at opposed non-coupled
relation this permits the full vector of the power components
to be decopled from the source (totaly non-reflective) as to permit a PERFECT
resonant tune of LC as to charge capacitor in RF radiant operation mode.
Now you have the BASIC and simple desing of the looped transverter circuit ,
just be Verry carefull with it ,use at minimal power and follow safety common
sense I dont want you to have any incident with this device ,as you validate
feel free to post (its quite demanding in its tuning) post this letter with
your results as EXTRA information for others to follow.
For basic triggering you can use multy-vibrator optoinsolated mode a simple
switching diode set up to triger the opossed capacitor when charged, Input ac
regulates the timing, LOAD regulates the discharge time so overal time must be
shorter (Lower impedance)to increase frequency (shorter pulse)(shorter
discharge time as to reduce capacitor Voltage value to near 0 .
Read ZPEV2.pdf ARK Files
Hector D Perez Torres
I understand this so that the load may only have a low impedance so that the energy remains reactive best discharged against a short circuit, so the power thyristors and diodes that can make these strong discharges at all possible, well a short circuit even hold the circuit not from. But where does this lead us back to reactive power?
I can not send such a potential directly through a 24Volt battery, it has only an internal resistance of 11 mOhm.
Or do you understand this text differently?
regards
Sven ...
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