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 Subject: RE: Field-free A potential in MEG Date: Sat, 12 Jan 2002 10:48:57 -0600   Paul,   Just how we do it is one of the things I'm not yet at liberty to release.  But any good electrodynamicist these days, can calculate the normal A-potential and also the field-free A-potential.  The field-free A-potential is still a vector potential, hence it has direction and motion as well as magnitude.  The B-field is made from the A-potential, which is more primary (the B-field, contrary to electrical engineering, does not and cannot even exist as such in mass-free space, since it is defined in terms of magnetically charged mass --- mass with magnetic poles).  So when you reverse the magnet in the normal polarity sense, yes you have reversed the A-potential, or the "made-from-it" B-field would not have reversed.  That's just B = del cross A.  And for the E-field, E is just E = - dA/dt.   I will admit that the chief scientist of an important experimental group in a large company was rather stunned at the type of output we were able to obtain.  The MEG may look like just a transformer, but it is not.  It is a completely different breed of cat.   To find how a force-free A-potential actually behaves, one does lots of little experiments with it.  Thereby one discovers some little "tricks" and novel effects that allow one to use it.  As I said, I'm not at liberty yet to reveal those little tricks; for one thing, they are not my personal discovery, but that of the bench guys on the MEG project.   From any nonzero potential, no matter how small, you can collect as much energy as you have intercepting charges available to collect.  This is easily seen with the electrostatic scalar potential (voltage) V.  the amount of energy W in joules, collected from a single potential V, is just W = Vq, where q is the amount of charges (in coulombs) that are present to intercept and diverge the energy flows comprising the V.  The trick is to make a little dipole (this is true either magnetically or electrically) which has a static potential between its ends.  Electrically that is V, magnetically that is magnetostatic scalar potential.  This dipole, once made, then must not be destroyed.  In short, one must figure out how to collect lots of energy from the potential so freely furnished by that dipole, then dissipate that collected energy in the load and circuit losses, without using half the collected energy to destroy the dipole.  That's about as plainly as I can say it.   We moved into magnetics experiments about a decade ago because of a compelling feature: Running the return flux back through the source dipole of a permanent magnet does not destroy the dipole, as contrasted to the return current in an electrical circuit which is forcibly rammed back through the source dipole in the generator and does destroy that dipole (hence the potential that is potentializing the external circuit).  In the magnet, the materials "lock in" the charges (the poles) so that they cannot physically disperse.   Notice that what I said is NOT adequately covered in conventional circuit theory, no matter at what level applied.  The standard circuit theory prohibits and excludes all COP>1.0, because it prescribes circuits which always use half their collected energy to kill that source dipole furnishing the potentialization (the potential energy) to the external circuit for it to intercept, catch, and use.  Unless this "use half the collected energy to destroy the source dipole" function of the standard closed-current-loop circuit can be violated, all one's efforts will fail and his circuits will never produce overunity performance, no matter how much energy he collects in his external circuit.   That's about as straightforward as I can say it.  As inventors and in a --- hopefully --- commercial enterprise eventually, we cannot release the "kit of parts" and actual "kit assembly and tuning instructions" ahead of securing patent rights,  if we wish to retain our patent rights.  I do get a few tirades from misunderstanding folks who insist we should just "give it away".  My response to them is simple:  They ask that the five of us who invented the MEG just "give away" the results of about 10 years very hard work of the group.  I ask them to show me that they practice what they preach; we will be happy to receive their own wages for the last 10 to 50 manyears, as a "gift" to the project to show they do what they prescribe.   Haven't had a single taker on that suggestion yet!  Apparently lots of folks advocate one thing, while actually practicing another.   So one must keep one's sense of humor. In the patenting etc. business, there is a prescribed procedure which any person or group has to comply with.  We have already released far more information -- real, hard, technical information -- than most other inventors have ever released on their overunity systems.  There are two very rigorous papers in Foundations of Physics Letters, e.g., showing exactly how the energy is received from the active vacuum.  And the rigorous theory for it is there in those two papers.   Hang in there and best wishes,   Tom Bearden   Subject: Field-free A potential in MEG Date: Fri, 11 Jan 2002 00:18:19 -0600 Dear Tom, Does the "field-free A potential" that you write about in your original 69 page paper on the MEG have a, I don't know the correct term, polarity of some kind? What I mean is, would the A potential be the same if I turned the permanent magnet around?  Or would it switch "direction" like the B field would? If it would stay the same, then how does one determine which way the current flows if the coil interacted ONLY with the A potential (as I understand it, it is theoretically possible to induce a voltage in a coil by interacting it with the A potential alone). Figure 12 and 13 in the paper would have me believe that the A potential is not affected by the orientation of the magnet.  If that is the case, then again: how does one determine the sign of the voltage in the coil? Please excuse my ignorance on the matter, but clearing up these elementary concepts would really help my visualization of the model. As always, thank you. Regards, Paul