function of VIC: rethinking signals...

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actually this question is to get dynodons response:

is it possible that the secondary was less turns than the primary?

Steve meyer says in a radio interview that it only takes 11 volts to pull an electron from hydrogen.

if the primary has 60 volts at 600 turns, and the secondary has 20 volts at 200 turns, then 1 amp at the primary should equal 10 amps at the secondary

now with such a strong 20 volt signal and such a high amperage on the secondary, the magnetic flux is greatly increased in the core and the amp restriction is very high in the chokes at resonance...with the 20 volts passing through to the wfc waveguide..and a very high magnetic field at the chokes... a very strong opposition to current flow past the chokes...

thinking outside the box... were the drawings misleading?

i really dont think so, but its worth a try.
any comments from others ?

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Good questions again.

I suggest you read up on Rutherford–Bohr. Losing electrons happens by use of adding energy.
Normally, that would be photonic energy and you measure that in eV like with LEDS.
The ground state energy level of the electron in a hydrogen atom is −13.6 eV
Hydrogen's ionisation energy is 1312 kJ mol-1
I am pretty sure that Steve Meyer ment this and not normal 11V stuff......

http://en.wikipedia.org/wiki/Bohr_model

http://en.wikipedia.org/wiki/Hydrogen

http://www.chemguide.co.uk/atoms/properties/hspectrum.html

Steve

 

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Ali,as you can clearly see in the pictures of the coil packs,the secondary coil is alot larger than the primary.It is also evident in the readings taken of the coils as well.All of the coils were traced out to their connections.
Don

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@All

Hey guys 

@Ali

The main point at Stans 3-23 was that he was using relatively small AWG 29 copper wire.
AWG 29 = no more than 0.5-2 amps... My primary AWG 22 heats up a lot from 2 amps so AWG 29 should pass even less...

"Stanley:  ....so voltage can take over and perform in dead short conditions...."  How can voltage can perform in dead short conditions - wich is what we get in water capacitor - almost dead short. Then how can voltage perform in dead short with no amps... should be dissipated immediately in other words overhelmed by the losses if you will.  Mystery to me.

Minde

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What if one of the tubes were insulated? By some kind of surface oxidization or other materials. In this way we really have a water capacitor, but we need strong electrostatic field between the tubes, this means high voltage, to can pull away the molecule.

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Hi Ali
interesting rethinking.
why does meyer chooses the MUR1560 diode (600v 15A) .Why such high current rating ?
However then as Minde said why does he uses awg29 wire that is good for a maximum of an amp or so?
These 2 question contradict each self.

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Stephen Meyer is referring to 11eV, not 11v

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Yes! Let the voltage do the work! Not the potential voltage between the exciters, but the eV(electron volts).

In physics, the electron volt (symbol eV; also written electronvolt[1][2]) is a unit of energy equal to approximately 1.602×10−19 joule (symbol J). By definition, it is the amount of energy gained by the charge of a single electron moved across an electric potential difference of one volt. Thus it is 1 volt (1 joule per coulomb, 1 J/C) multiplied by the electron charge (1 e, or 1.602176565(35)×10−19 C). Therefore, one electron volt is equal to 1.602176565(35)×10−19 J.[3] Historically, the electron volt was devised as a standard unit of measure through its usefulness in electrostatic particle accelerator sciences because a particle with charge q has an energy E=qV after passing through the potential V; if q is quoted in integer units of the elementary charge and the terminal bias in volts, one gets an energy in eV.

The electron volt is not an SI unit and its value must be obtained experimentally.[4] Like the elementary charge on which it is based, it is not an independent quantity but is equal to (1 J/C)(2 h α / μ0 c0)0.5 It is a common unit of energy within physics, widely used in solid state, atomic, nuclear, and particle physics. It is commonly used with the SI prefixes milli-, kilo-, mega-, giga-, tera-, or peta- (meV, keV, MeV, GeV, TeV and PeV respectively). Thus meV stands for milli-electron volt.

Atomic properties like the ionization energy are often quoted in electron volts.

In chemistry, it is often useful to have the molar equivalent, that is the energy that would be produced by one mole of charge (6.02214129(27)×1023) passing through a potential difference of one volt. This is equal to 96.4853365(21) kJ/mol.[3]

13.6 eV: The energy required to ionize atomic hydrogen. Molecular bond energies are on the order of one eV per molecule.