Working Attraction Magnet Motor on Youtube!?

[nightlife]

Can anyone tell me what is the best thing to use for magnetic shielding and how good it actually is?

[Pirate88179]

@ nightlife:

Hey, we all do.  I think I speak for a lot of us, if not all, on this forum when I say that we want someone (even if it isn't you or me) to make a successful device.  If we didn't think it possible, we would not be here. I still have hope that it will happen.  I think we have some of the most creative minds in the world on Stefan's site and if it can't be done here, then, who will do it?  Thanks.

Bill

[shruggedatlas]

@ Shruggedatlas:

So, you are saying that nasty word "equilibrium" due to the fact that any "excess" energy can be explained by the potential energy stored in the ball by the researcher lifting it to the starting point?  I don't want to put words in your mouth so correct me if I am wrong.  So, this would be like saying that "hey, once I lift this weight inside this clock, it runs for a week on free energy."  I see your point and don't disagree.  I am not a physics expert and don't claim to be.  I think you have raised a valid question that should be answered, or at least, looked in to. I am still experimenting with the SMOT as it relates to a rotating wheel but I have not progressed far enough to voice an opinion either way.

Bill

(For nightlife's benefit, my analysis refers to Omnibus's SMOT model, where point A is the bottom of the dish receptacle that catches the ball, point B is the bottom of the SMOT ramp, and point C is the top of the SMOT ramp, where the ball falls through the ramp and to the bottom of the receptacle, point A.  Refer to http://gggttt.host.sk/sketch.pdf for a diagram of what Omnibus is talking about.  The motion of the ball is discontinuous, as the reasearcher manually lifts it from A to B, then the ball proceeds spontaneously from B to C, and lastly the ball falls down at C and ends up at rest back at A.)

I am no physics expert either, and I had to spend a few hours getting to understand this.  Fortunately, it is just the basics, high school stuff at the worst.  No calculus is involved.

I do not think I am saying exactly what you are saying.  I am not so much focused on what the researcher does.  Remember, Omnibus focuses on the ball rising spontaneously from B to C, thus spanning h2.  The energy expended in this fashion, as I understand it, is mg(h2) - (Mb - Mc).  Omnibus claims the energy mg(h2), or at least a portion of it, is never properly "returned" by the time the ball returns to A, and is therefore in excess (or "overunity").  The point of my analysis is to show that this spontaneous energy gain is in fact fully accounted for and does not remain "in excess" by the time the ball returns to A.

Omnibus is more knowledgable than I am in physics and spent much longer working on his theory, so it is quite possible I missed something.

[Omnibus]

@shruggedatlass,

Please, understand that ball at C gains, I repeat, gains, energy mgh2 (height h2), as well as kinetic energy plus other kinds of energy, at the expense of the energy (Mb - Mc), where Mc = 0. Again, energy (Mb - Mc) is spent energy, while energy mgh2 plus kinetic and other energies are gained energies. Energy (Mb - Mc) is spent in order to gain energy mgh2 plus kinetic and other energies. This is the CoE in its "transformation" part which firmly holds and which is the part almost always considered in scientific analyses--the energy (Mb - Mc) is transformed in equivalent quantity into mgh2 plus kinetic and other energies. In SMOT as in any experiment whatsoever, there is no violation of CoE in its "transformation" aspect, formulated still in Lomonosov's works. Lomonosov isn't known in the West at all but in fact, as far as I know, he was the first to formulate that aspect of the CoE principle. When people say that the principle of CoE has never been proven to have been violated they mostly mean exactly this "transformation" part of the CoE principle which, indeed, has never been violated. As I said, SMOT doesn't violate it as well. What is new in SMOT is that it demonstrates the production of the energy mgh2 + kinetic etc. from no source. The magnetic energy (Mb - Mc) spent to produce said mgh2 + kinetic etc. energies upon closing the A-B-C-A loop is fully compensated when that A-B-C-A loop closes. There's no depletion of the magnetic potential energy reservoir when the A-B-C-A loop closes. Thus, upon closing the A-B-C-A loop, although there's no loss or gain of magnetic potential energy, the ball loses more energy (mgh1 + mgh2 + kinetic etc.) than the energy that was imparted to it (mgh1).

[Omnibus]

Compare what I said with this statement from your post: "The energy expended in this fashion, as I understand it, is mg(h2) - (Mb - Mc)." In fact, what you should have said is not that that quantity is energy spent but that it's a balance of the spent and gained energy as point C. That balance at point C is zero. At C magnetic potential energy (Mb - Mc) is lost, while gravitational potential energy plus kinetic plus other kinds of energy are gained at the expense of losing that magnetic energy. That is, mgh2 + [kinetic +] - (Mb - Mc) = 0.