Short answer - i'm explicitly claiming an effective CoE violation. Your incredulity is entirely appropriate. It sounds like complete heresy. I'm saying it's meticulously measured and a direct consequence of CoM and CoE holding precisely as they're supposed to, beyond any possibility of error. I am absolutely susceptible TO error, but because of that i've done my due diligence, to eliminate my own stupidity as a factor.
Dancing around this issue point-by-point when i haven't presented you with evidence of the claim is probably redundant.. like i say if i can't enlist any help with it by the w/e i'll post it up here, though i'm setting my expectations low, just as you are.. On Tue, Jun 5, 2018 at 5:20 AM, John Shop <quack...@outlook.com> wrote: > On 5/06/2018 2:40 AM, Vibrator ! wrote: > > Your view of what is conserved and why is too simple, and essentially > incomplete. > > All force interactions perform work against the vacuum activity > manifesting that force - the discrete, quantised energy exchanges between > the respective force carriers in question, traded in units of h-bar - > essentially, 'ambient' quantum momentum. > > When we input mechanical energy to a such field, there is no number > scribbled down in a book somewhere - rather, it's an emergent calculation > determined by the application of the relevant F*d integrals being mediated > at lightspeed - ie, essentially instantaneously, as they pertain to the > respective dimensions of the given energy terms. > > Thus if output and input energy terms are in different respective > dimensions, any equivalence between net energies as a function of changes > in time and space is dependent upon further conditions with regards to how > each term scales in the other's domain. > > If both input and output energy terms are in the same fields and domains, > then their equality is a given. And yet, it would be a step too far to > conclude that the Joule we get back out was 'the same' Joule we put it. > When we spend 1 J lifting a weight, so having performed work against > gravity, there isn't a tab somewhere saying "gravity owes Bob 1 J". The > fact that we only get 1 J back out from the drop is simply an incidental > consequence of the invariant input vs output conditions. But it's not > manifestly 'the same' Joule you put in - just the same amount of energy / > work. > > I agree with you. It is not manifestly the same joule. So depositing > money in the bank may be a better illustration (or pumping electrical power > into the electricity grid). I can deposit $1000 in one city in $20 bills > and pull the same amount out in another city in $50 bills. It is not > manifestly the same cash that I have taken back out, but the bank makes > sure that the amounts always balance! So Nature does the same job as the > bank tellers and accountants. Whenever you do the calculation correctly, > after allowing for incomings and outgoings, the overall energy balance > sheet always balances perfectly - which is almost the same as saying that > gravity owes Bob 1 J! > > You might wonder who the tellers and accountants are that work for mother > Nature. The simple answer is that they are Newton's equations. When > applied correctly the spreadsheet always ends up balanced because the > equations themselves are balanced. I believe that you can achieve an > imbalance, but not by operating in accord with Newton's equations. You > have to do something a lot more subtle and sneaky and discover an effect > that has not been noticed and a term that has not been included in the > equations. And it is bound to be a small effect (eg < 1% of energy being > exchanged) or it would have been noticed a long time ago. > > With the right change in those determinant conditions, we can get more > out, or less. An under-unity, or over-unity result. > > > Consider the case for so-called 'non-dissipative' loss mechanisms, in > which the energy in question has NOT simply been radiated away to low-grade > heat. I'm talking about 'non-thermodynamic' losses, in the literal sense. > For example: > > - Due to Sv (entropy viscosity - the subject of Rutherford's first paper > in 1886), a small NdFeB magnet will rapidly leap across a small airgap to > latch onto a lump of 'pig iron', in less time than is required for the > iron's subsequent induced magnetisation ('B', in Maxwell's terms) to reach > its corresponding threshold (Bmax, or even saturation density - Bmax - if > its coercivity is low enough). > > So the iron's level of induced B, from the neo, continues increasing long > after the mechanical action's all over. > > We could monitor this changing internal state, using a simple coil and > audio amplifier, tuning in to the so-called Barkhausen jumps, as > progressively harder-pinned domains succumb to the growing influence of > their lower-coercivity neighbors. After some time, the clicking noise > abates, and so we know the sample's at Bmax. > > We now prise them apart again, however because B has risen, so has the > mechanical force and thus work involved in separating them. > > Quite simply, due to the time-dependent change in force, which did not > occur instantaneously at lightspeed, the system is mechanically under-unity > - it outputs less energy during the inbound integral, than must be input > during the outbound integral over the same distance. > > So we could input 2 J, but only get 1 J back out. > > By my calculation you have got nothing out. You let the magnet fly and > collide into the pig-iron so that the 1 J you might have recovered from its > kinetic energy ended up as heat during the collision. > > Following this the permanent magnet slowly magnetises the pig-iron. To > the extent that this is slow (due to magnetic viscosity) and occurs in > jumps (generating Barkhausen noise), this process is lossy and generates > heat by jiggling the domains. The fact that you have forced pinned sites > to become magnetized means that some of the induced magnetization will be > retained. So that now when you try to prize them apart you are also > working against some permanent magnetism. So the energy required to force > pinned sites to switch magnetization (some of which was dissipated as heat) > now has to be put back into the system as the force required to return the > permanent magnet back to its initial position. So you have to put in both > the kinetic energy (1 J) that you failed to recover and the energy (1 J) > that resulted in the pig-iron becoming magnetized and warmer. > > Yet this 'loss' has not been dissipated as heat - it's simply energy that > never existed, never came to be, in the first place. Energy that could've > been collected, had we constrained the neo's approach speed, to allow > induced B to keep up... but which wasn't, because we didn't. > > Thus the extra Joule we had to input has performed more work against the > virtual-photon-spehere (being the EM mediator), than it in turn has output > back into the mechanical realm. Assuming ultimate conservation - as you > would seem to - we've raised the vacuum energy by 1 J, with a 50% > under-unity EM-mechanical interaction. > > I agree that the energy can be stored as "vacuum energy" but I disagree > that any *text book process* can create or destroy energy. If you think so > then you have not fully read the small print of the text book! > > Yet we don't need such exotica as obscure magnetic effects to achieve this > feat... simply consider a moving mass, colliding inelastically with an > equal, static one: > > - so we could have 1 kg flying into a static 1 kg > > - or equally, a rotating 1 kg-m^2 angular inertia being instantly braked > against an identical static one > > Since spontaneously doubling the amount of inertia that a given conserved > momentum is divided into accordingly halves its speed, we end up with half > the kinetic energy. > > "Ah", but you say, "the collision converted the other half of the KE into > heat!" > > That is correct. That is how Newton's equations are correctly applied. > > But is that actually what happens? If we began with say 1 kg * 1 m/s > linear momentum, so half a Joule, which then inelastically scoops up > another, static 1 kg, we now have 1 kg-m/s divided into two 1 kg masses, > hence a net system velocity of 0.5 m/s, and 125 mJ on each, for a 250 mJ > net KE. > > Notice that we've necessarily assumed full conservation of our velocity > component, simply sharing it evenly between the two masses, in order to > conserve net momentum. > > Given that the original KE value of 500 mJ was a function of that > conserved velocity, and that the final KE of 2 * 125 mJ is also dependent > upon the equitable distribution of that same conserved quantity.. where > does the velocity and thus momentum that could constitute mechanical heat > come from? How could we have accelerated the air and molecules around the > system, if not by transferring momentum and thus velocity to them? Which > would mean we'd have to have LESS than 0.5 m/s of velocity and thus less > than 0.5 kg-m/s of momentum and so less than 125 mJ on each 1 kg mass! > > Sorry I don't understand your argument. An experiment of allowing a 1kg > lead mass travelling at 1m/s to collide with a similar stationary one so > that they both travel away with half the velocity does not need any "air > and molecules" for the interaction. But in order to bring them to the same > velocity a force between them does need to be applied. If this force is > frictional, then the energy obviously turns into heat. Similarly if the > force results in the lead being forced to change shape, then the energy > appears as heat (try breaking a reasonable diameter steel wire by flexing > it back and forth with a pair of pliers until it fatigues and fractures - > then touch the flexed section to see how hot it has become!) > > There can be no paradoxes.. > > In short, elastic collisions conserve net energy, but not net momentum - > try calculating the same interactions fully elastically and you'll > necessarily be invoking a rise in momentum. > > The same interaction (one moving mass attaching to a stationary one so > that they both move away joined) *cannot* occur elastically unless you can > think of some mechanism to absorb the kinetic energy - such as a spring > acting between them and a ratchet to stop the spring from pushing them > apart again afterwards. Then when you do the calculation you discover > that the kinetic energy loss has become potential energy stored in the > ratcheted spring. The energy spread-sheet always balances perfectly or you > have made a mistake. > > Conversely, inelastic ones conserve net momentum, but not energy. This > loss, by the very nature of its constituent terms and conserved quantities, > is non-dissipative. Only its non-reversibility with respect to time > prevents easy access to energy gains. This is entropy, albeit acting on a > level beyond strict 'thermodynamics'. > > Your language here seems a bit unusual. "Inelastic" is usually understood > to be "dissipative" almost by definition - heat generation being the result > of inelastic and dissipative mechanisms. Mainstream physics still regards > heat energy to be unrecoverable although there is no good reason except > statistical ones why this should not be possible. > > Like i've always said, the explicit instructions on how thwart CoE and CoM > are implicit within their terms of enforcement. Read between the lines, > they tell you precisely what not to do if you don't want to get a unity > result. > > > Without this kung fu, i would never have been so stupid as to take a > second look at Bessler's claim, let alone tackle it with confidence. But > with it, the evidence of Leibniz et al meant that i couldn't fail. Success > was guaranteed. There had to be an unnoticed symmetry break riding through > the middle of classical mechanics, an elephant in the custard, that with a > little determination could be tracked and cornered... and now i've bagged > it. > > Not just wounded it. Not "close, but i'm running out of hamsters". There > was a fully-grown African bull elephant perfectly concealed in the custard > bowl, and i've totally harnessed it, by "accelerating without > accelerating", and now nobody will believe me and it's so unfair etc. > > Sorry but since you are talking of *textbook physics*, all physicists will > be absolutely certain that you have made a simple mistake based on some > conceptual misunderstanding. Given the opportunity, (and if they are not > fed up with fielding crackpot questions) they will be happy to point it out > to you to save you from further embarrassment. >