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Hi Ole, I can connect the Dc engine with a generator that acts like a brake at last. I still have a Ferroresonanzrafo if you can test it as a load, but I think that will be the same problem. maybe we do not even know how we have to deal with this energy. Regards Sven
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Von: "[email protected] [EVGRAY]" <[email protected]>
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An: [email protected]
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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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Hi Sven,
This text from Hector tells that the impedance of each part (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 a half cycle which is 10ms at 50Hz. The output as well must be done in less time than a half cycle.
To do this the LC constant of the input part formed by the output of the ferroresonant transformer and one of the capacitors must be shorter than the time of half a period. The same applies for the output part of the circuit. As the capacitor is a constant for both input and output only the transformers can be changed. Thus if the capacitors are chosen to match the ferroresonant transformer the output transformer has to be chosen for the proper output 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 the frequency in Hz L is the inductance in Henry and C is the capacitance in Coulomb. The same formula is used for the charging part of the circuit where L is the output inductance ot the ferroresonant transformer. For the output transformer the inductance is decreased by increasing the load as this cancels more of the primary winding's inductance. The fine thing about transformers is that they actually change the impedance between the primary and secondary windings. Thus choose the transformation ratio to fit the wanted impedance which can go up or down or be one to one. This can be chosen for 24V at the output and the voltage of the capacitors at the input. The voltage at the capacitors is the peak voltage that must be transformed down to the peak of 24V. That peak is different if the period time is different from a full wave when combining the two semi-waves. A shorter pulse width at 50Hz requires a higher 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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