Vibrator, do you have a machine that generates energy, a device that powers
itself?

If so, then yes it is beyond question that you have done it, call me
captain obvious.

Then it is a question of if you are honest, personally I would be willing
to consider that is possible as I believe that CoE and CoM have been
violated in the past by other devices.

But really, you need to say that yes, the device powers itself if you want
to be beyond any possibility you are wrong.
If it does, then assuming you are acting in good faith, you need someone
else to replicate it.
It might be a good idea to provide a video with as much transparency as
possible to ensure people are willing to construct replicate your device.

If you DON'T have a device that can run continuously, then you really need
to disclose all the details so people can understand the principle, and
help you work out how to build out.

So really, you either should have a device that can power itself...  In
which case you should video it and help someone replicate it, have that
person sign an NDA if you wish.

Or you should be seeking help designing and building a device that can
power itself (and ideally a load).
Anything else is vanity, a waste of time etc...

So, which is it, do you want help to replicate something you have already
done?  Or to build something that you believe you have proven but not yet
demonstrated?



On Wed, Jun 6, 2018 at 12:43 AM, Vibrator ! <mrvibrat...@gmail.com> wrote:

> 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.
>>
>
>

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