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Explaining electric OU devices (?)
Posted on Wednesday, March 09, 2005 @ 23:04:24 UTC by vlad
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In the free-energy yahoo group, Koen van Vlaenderen writes: Hi,
I am new to this group, and I would like to offer a practical theory that explains right away the extra energy in the 'output' with respect to the 'input'. My theory confirms almost all of John Bedini's thoughts/comments on electrical OU devices.
My theory can be found at: www.truth.myweb.nl
Most of you have heard of the electric power law:
P = V*I where P is electrical power (energy conversion), V is voltage and I is current.
For instance, a resistor converts electrical power into heat, where V is voltage across the resistor and I is the resistor current. The transmitted heat is radiated away in the form of Transversal ElectroMagnetic radiation (TEM waves). By the way, a truly negative resistor reverses this conversion: it assimilates heat energy and converts this into electricity.
Now, P = V*I = V* dQ/dt where dQ/dt means the change of charge per second (current in unit Ampère).
I suppose the big secret of OU engineering is making use of power factor P = dV/dt * Q, so the only thing you do is move the d/dt operator from Q to V. It's as simple as that. This does not change the unit of P, it is still Watt or Joule/sec. Now, the scalar field that John Bedini already mentioned years ago on his site, is simply S = -dV/dt which is the time-differential of electric potential. The electric field is defined as E = - grad(V), which is the space-differential of the electric potential, so the definition of the scalar field S is very similar to the definition of the electric field.
How come electrical engineers simply failed to exploit the scalar-electrostatic power P = dV/dt Q = -SQ ??? This is a long story which has something to do with "gauge conditions" (the Coulomb- or Lorentz gauge condition). Because of gauge conditions, that are in itself a totally illogical and an unphysical theoretical assumptions, electrical engineers missed the whole point of scalar field engineering, since these gauge conditions are simply the condition S=0 (no scalar field !). At my website you can find how scalar field S is defined more fully (involving also the magnetic potential A), and how Tesla's longitudinal electrical waves are in fact Longitudinal Electro-Scalar waves, and how scalar field S can explain longitudinal Ampère forces. My theory describes a generalised electrodynamics with broken (and restored) gauge symmetry, based on an extra scalar field S. It does not contradict all the verified and accepted laws of classical electrodynamics, so with respect to Tom Bearden and the AIAS group, I am in a much more confortable position. Thus far my theory has not been refuted, and it has been published in the scientific press, in the form of two peer reviewed papers.
Lets go back to the engineering principle: power in the form P = Q dV/dt. So we need a large Q (lots of STATIC charge) and a large dV/dt (a very abruptly changing-in-time voltage). John Bedini's circuit designs contain capacitors with very big C that can store a lot of (static) charge, and John is making use of the very abrupt back EMF pulses of fairly big coils. The milli-farad capacitors are such that all the capacitor charge can 'feel' the back EMF pulse, so basically the capacitor will ALSO store energy (power) in the form P = Q dV/dt beside storing energy (power) in the form of P = IV. Probably the capacitor converts this stored energy into an extra charging current ( a trade-off between extra voltage with extra charge). Even better would be if all the charge in battery#2 (the one that is charged) could 'feel' the back-EMF pulse, since there is much more charge (Q) in the battery than in the capacitor. But I suppose not all the battery charge can be potentialized, only a small portion of the total charge, unless the battery has a certain optimized geometry. John's site contains a picture of such a geometry.
I proved mathematically that scalar-field static charge power Q dV/dt is stronly related to longitudinal electro-scalar waves (also known as Tesla waves that are not Hertz waves). I can prove also that Whittaker's decomposition of the Coulomb potential (that falls of as 1/r), into a set of electric potential waves implies a powerflow in the form of longitudinal electro-scalar waves (and NOT powerflow in the form of bidirectional TEM waves, as suggested by Tom Bearden).
So the "negative resistor" function is not a matter of converting heat (TEM waves) back into electricity, it is a matter of converting Tesla waves (longitudinal electro-scalar waves, or LES waves) into electricity. Gabriel Cron made a statement about the capacitor as "negative resistor", and this can be understood by involving the scalar field S and extra assimilated energy in the form of longitudinal Tesla waves.
I would like to thank John Bedini for his great work, and the fact that he shared his findings with us. It shall not be forgotten, John, as long as we live. I would like to thank also Dr. Lindemann, and ask him to study my theory and give as much comments as possible, especially the most sceptical remarks from which we can learn the most. I suppose my theory also applies to the Testatika and the EvGray Motor. Hal Puthoff already knows about scalar field S, and his collegue Michael Ibison confronted me with fine remarks, such as: how can scalar field S be sourced? Michael is not convinced that S exists, but Hal likes my theory very much (his actual words). Russian scholars know about it too (dr. Chubykalo, V. Onoochin, etc...), and also Jack Sarfatti commented: "well done".
Much more important to me than acceptance of the theory is its practicle value and its support for the free energy community. Let me know what you think of P = Q dV/dt as a design principle.
The great Tesla said that the longitudinal electric waves carries much more energy than the Hertz wave. So be it.
Koen
http://home.wanadoo.nl/raccoon/
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Re: Explaining electric OU devices (?) (Score: 1)
by mojo on Thursday, March 10, 2005 @ 12:42:49 UTC
(User Info | Send a Message) | Hi,
It can be argued that the idea of S=dV/dt is equivalent to changing the formulation of the A 3-vector potential into a 4-vector potential with the additional component being the same dV/dt term.
mojo |
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Re: Explaining electric OU devices (?) (Score: 1)
by ElectroDynaCat on Thursday, March 10, 2005 @ 21:20:12 UTC
(User Info | Send a Message) | | Its always a pleasure to have someone interested in this field, BUT please, learn the difference between energy and power before you go any further. Otherwise, have fun! |
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