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 <> 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.

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