Body
So I would say that depends on the impedance of the resonance coil I have the ability to adapt by different windings on each leg, the impedance. If I increase the inductance on the resonance leg, the voltage will also increase, the current will be lower with a smaller capacity of the capacitors. Maybe I should test that again dealing with higher voltages makes everything more difficult. I am now at 450 volts max. Now only the current continues to increase with higher capacity. Greetings Sven
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-------- Ursprüngliche Nachricht --------
Von: "Norman Wootan [email protected] [EVGRAY]" <[email protected]>
Datum:14.11.2017 20:59 (GMT+01:00)
An: [email protected]
Betreff: Re: [EVGRAY] Re: Ferroresonanz transformer Three - phase transformer control
As Hector and I both warned, use capacitors that have a rating 10X greater than voltages of circuit. at ferroresonance, both voltage & current see a 10X multiplication! In my "Ferroresonance Capture" trials, I collected the HV caps needed to safely play with the multiplied voltages.
Warren, thanks to "Ole" explanation, "You got it"!
On 11/14/2017 11:39 AM, Warren Keillor [email protected] [EVGRAY] wrote:
Norm
My feeble brain is starting to get the ferroresonant phenomena. It is super saturation transitions that we might want to set up?
OK , we saturate a coil, then zap it with an improbable big zap driving it over the top, using a capacitor.
I have some 100mf 4000 volt capacitors.
Can they be in the useful range of what is needed?
Cheers Warren
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On Tue, Nov 14, 2017 at 10:27 AM, Norman Wootan [email protected] [EVGRAY]
<[email protected]> wrote:
Excellent explanation "Ole" of conditions of iron at "Super saturation"! A big capacitor discharge through a "saturated flux iron core" drives it into a "super saturated condition" thus you see current go "exponentional"! Can we say "EMP"? "Neutral Spike" same thing! Thanks Ole!
On 10/29/2017 5:47 PM, [email protected] [EVGRAY] wrote:
Hi Sven,
The text is Hector explaining the parametric change of an iron core from being a good magnetic conductor to being a poor magnetic conductor. When saturated it's hard to put more magnetic flux into the core. Its magnetic conductance becomes like that of vacuum (thousands of times less than that of iron). If inputting a little more current (magnetic flux) at this stage it doesn't take much energy. This is a little more charge in the first or primary capacitor. But as the inductance drops off (nearly vanishes) the current becomes big.
"Bringing a transformer into saturation with a strong short impulse and reaping the energy + extra energy on the secondary side would make me understand or grasp that."
I just did part of the experiment. That is connecting a charged capacitor to an inductor making it go into saturation. The attached PDF-documents show the voltage across the capacitor being discharged by connecting it to the coil. The current running between the capacitor and inductor is also shown as well as the power (energy flow) between the components. The current from the discharging capacitor should then be stopped at its first peak.
Just added an extra curve to also include the power integrated which is the energy. The next part of the experiment would be to add a second coil around the core. This coil is then to be calculated for for the same inductance as the primary coil has when the core isn't saturated. But the secondary coil has to have this value when the core is saturated and when using the same value of capacitance to store the output energy if I comprehended correct what Hector tells? Comparing the voltages of the similar input and output capacitors then tell if more energy was output than being fed to the transformer.
Regards
Ole
---In [email protected], <s.friedrich@...> wrote :
Hello Ole,
I think so, too, I would like to transfer the energy into the resonant circuit and not extract it directly from the primary coil, of course, protective measures must be taken at idle.
I read again and again when I have time the pdf RE-Various-v1 through and there is on page 41 Proof OU Property:
1.27. Proof OU property
you need 2 IDENTICAL capacitors and a dual coil wound core. First coil is SATURATE the core by
ONG PULSE .... at X Volts N farads. Second coil is TAILORED to RECOVER
the collapse of a capacitor in a logarithmic semi-resonant path ...
The resultant VOLT FARAD relation is Computed ... If SECOND equiva- lent capacitor has a HIGHER charge than the
IDENTICAL prime pulse one at SAME farads you have OU ... as simple as that ... CAPACITORS MUST BE OIL (NO
LITICS) (Tuning is a bitch!) Learn to TUNE ELECTROMAGNUM and you can make the REACTOR CORE WORK ...
This simple TEST is what you need. Sooner or later someone has to replicate this as is DAM simple as the battery
SHUNT to test DC OU ... (Used By Konzen) Only here you measure, Voltage potential in input capacitor - Discharge -
voltage potential in OUTPUT capacitor ...
2 CAPACITORS .... the one in. Sorry but no one on the BOOK can tell me
the OUTPUT has 1,618 Joules potential more than the initial charge .. Get that you get OU ... HOW? "TUNING"
How is done? Discharge capacitor into coil as Current is maximum disconnect input capacitor at O V Connect Output
capacitor to SECOND coil, as current LOGISTIC CURRENT DROP Occurs wile
LOGARITMIC VOLTAGE INCREASE OCCURS Cut of COIL as VOLTAGE is MAXIMUM and CURRENT = 0
THERE you have ZPE ZERO "POINT" ENERGY but in reality is 1-0: 1-0 10-10 one Hi zero low ... You have 2 Zeros and 2
ONES!
I just Explain what Tesla did not say Aether String to a galactic cluster or KITCHEN sink you will get OU, what
changes is the gain or "Q" of object you use ...
First like RV "Refresh yourself with the reading" you need to GET OU, Next is what you are going to do with it
"applications".
Bringing a transformer into saturation with a strong short impulse and reaping the energy + extra energy on the secondary side would make me understand or grasp that.
Best regards
Sven