Plying CF as pseudo-inertia to scam N3

[Furcurequs]

MrVibrating,

Maybe Marchello hasn't yet looked closely enough at what you are claiming to be doing to know for sure yet, but unfortunately I have. ...and so I must say that you have simply been talking out of your ass, if you don't mind my own jargon. ...lol

I'll give you this. Although you are spouting a lot of nonsense and gibberish, there is a certain consistency to it all. That would seem to indicate that you are making up your own terminology based upon your own confused misconceptions of the real physics involved here.

To those uneducated in physics like (I have great reason to assume) yourself, maybe your pseudo-scientific mumbo jumbo sounds "sciencey" enough that they think you actually know what you are talking about, but fortunately or not for me, I have the education and knowledge to actually know better.

Hopefully for those who aren't just plain clueless, the calculations I've already shown make sense and will stand on their own merit.

Of course, though, the results of your own simulations show that there is no excess energy. The simulation output shows the speeds and positions of the masses and from that information it is quite easy to see that there is no energy that is unaccounted for considering the initial kinetic energy, the energy you input during the interaction, and the gain in kinetic energy due to the conversion of gravitational potential energy of the two masses during their descent in the gravitational field of the earth.

The only screw up seems to be when you apply your own unique understanding and unique math.

In the situation where you claim you end up with 125% of the energy you input, in the type of interaction you actually described you would instead end up with only about 40% of your input energy for a LOSS of 60% - and that is truly AT BEST.

In testing the spirit of the OU messiah, I must conclude that he has a devil - an evil spirit - for he doesn't seem to care much about the truth.

[MrVibrating]

LOL wow fantastic that's great Dwayne and i truly wish you the best of luck with all of that. You've certainly given me much to think about, and i want to thank you again for all your efforts, really pulling your finger out and giving it your best shot, sterling work all round, if only we had a dozen more like you.

And for chissakes no, i am not the messiah - this isn't even a proper messiah's hat, it's just a crash helmet... look, it comes right off. Forgot i was even wearing it.

[MrVibrating]

OK i think we're making this harder than it is:

- Marchello presumably agrees that per 1/2mV^2, 1 kg at 4.905 m/s has 12.029 J (ie. not 48J!).

- He presumably also accepts that a regular N3-consistent mutual deceleration would produce such an equitable distribution of the 9.81 kg-m/s generated.

- Therefore he tacitly accepts that N^2 J/kg-m/s is valid for at least one full cycle.

- Since this in turn validates that counter-momentum can be inverted / cancelled / rectified (whatever it is we're doing here), then it is pretty clear what N^2 actually is:

- It's equal to the force of the static uniform field divided by the mass, ie. so having dimensions of acceleration, per F/m=A

- It is thus patently obvious that it is the inverted counter-momentum component of our input workload! Specifically it is the fixed cost of a 9.81 kg-m/s acceleration nested within an effective (not physical) 9.81 kg-m/s accelerating reference frame...!

- So this seems incredible - i'd only previously hypothesised that something like this might be possible - but it appears that by briefly adopting the free-fall reference frame, we gain the benefits of its 'effective acceleration' as if it were a physical upwards acceleration of the ground itself... as if the whole world were plunging upwards or outwards at 9.81 m/s^2.. in short, we're accelerating an acceleration with a virtual acceleration. That old 'negative GPE' cliché again.. (yeah me neither)

- This doesn't exactly allay my concerns we're potentially performing N^2 J-kg/ms net work upon Earth's resting state..

- There is no such thing as terminal velocity in vacuum.

- inertia is velocity-agnostic.

- Therefore N^2 J/kg-m/s is constant, ie. invariant of speed.

- If KE=1/2mV^2 = 1/2MoI*RPM^2 = 4(N^2J/kg-m/s) = unity

- Then 3(N^2J/kg-m/s) = (0.75 * (1/2mV^2)) = 3/8mV^2 = 75% unity

- And 5(N^2J/kg-m/s) = (1.25 * (1/2mV^2)) = 5/8mV^2 = 125% unity



Ecce homo, ergo elk.



The £5 is all mine and i'm splurging the lot on Irn Bru and curly wurlys!



Now let's go build perpetual motion machines..

[cloud camper]

OK folks, turns out I’m a battle tested fence sitter from the Steorn days myself and my current magnet motor project is a spinoff from Alsetalokin and Overconfident’s ideas from 2008.

So I’m not challenging the math here and rather hope it works out.

But just looking at the main issues involving testing and it would seem that this scheme is essentially an enhanced version of MT6, 7, or 8 with some extra sauce.





The idea behind these MT’s is that weights are raised by the ascending side of the wheel, then separated from the wheel as they are translated horizontally to the rim where they are then dropped, hopefully powering the wheel as the weights impact the lower rim.

All looks great except for one ugly factor.

And that is while the weights are translating horizontally they are not creating any positive torque.

Neither do they create any positive torque while falling.

The only time positive torque is created is when the falling weights impact the lower rim.

What happens to the wheel in the meantime?



Sadly the wheel is unloaded on the downward cycle and the wheel then instantly reverses direction.

And even if the jack spring assembly remains attached to the wheel, we have the same problem as the two reaction weights are offloaded.

So the elephant in the room on MT6, 7, and 8 has now conveniently taken up residence with our new Reactionless Acceleration process and has not been informed that the wheel reversing direction is not playing nicely with the new theory.

This does not argue that there is not a genuine N3 break with this scheme, only that in order to utilize it we need to solve the MT6, 7 and 8 problem before any physical results can manifest.

This is the devil in the details and why MT6, 7 and 8 do not work.

So while I am hopeful the math confirms the possibility, the physical application appears just as problematic as a solution to MT6, 7 and 8.

[MrVibrating]

Hi CC

The principle's so general it could be applied in just about any way you can envisage.

Balls following radial curved trajectories are exchanging angular momentum with the wheel body.. all we're talking about here is taking control of a dynamic that occurs anyway - momentum production and management. Gravity-assisted momentum control.

As for your reservations, i take it you're concerned about the practicalities of an underbalancing moment caused by dropping mass into free-fall.

Surprisingly, although this certainly seems to make for an inelegant implementation, it is, nonetheless, mathematically sound. GPE-in = GPE out, so that underbalancing moment is actually no insurmountable sticky-spot.

A far more difficult sticky spot to wade through is simply the startup sequence through the first three cycles, which is effectively destroying most of the energy you're trying to start it with. Without that bump-start, PE store or else the "4-in-1" trick, this is a non-conservative sticky-spot!

My own thoughts on potentially promising approaches haven't really changed since i laid them out on the bottom of page 10.

Those examples should help clarify that no actual free-fall phase is even necessary; we do not have to drop any weights or apparatus from the wheel to land below, this is entirely unnecessary... it is sufficient merely to apply a counter-torque to a descending mass which arrests its gravitational acceleration, while applying it vicariously to the angular inertia of the net system instead, for example.

Basically if you're envisioning something necessarily clunky, there's probably a simpler way. The range of possible designs is limitless. All we're doing is using gravity to invert counter-momentum from an applied inertial interaction, then summing that skewed distribution to a net rise, accumulating it and repeating.

No reason an elegant design couldn't be whisper-quiet.

The only situation in which a requirement to drop a weight would be a concern (and we have no such requirement, but just hypotheticaly) would be if it were caused to perform some other work on its way down, other than simply converting its GPE to its own KE and back.

In our case, gravity performs no net work - all of the momentum we accumulate has been generated ourselves, by a purely internal expenditure of work. Each cycle that we buy more momentum, we've generated it by spending energy. However because it's unidirectional instead of equal opposing momenta, it can only be cancelled by an opposing asymmetric inertial deceleration, ie. while while rising. So provided we perform that deceleration while not gravitating, we get a regular N3 result and so get to keep it, and build on it.

After years of playing with gravity-type wheels, we get a pretty good feel for the perils and pitfalls, and inadvertent underbalancing moments are absolutely one of the main ones. Watch what's in this hand, not what's not in that hand..

But also we develop progressively more conservative expectations of just how much energy an effective gravitational asymmetry would even be able to produce, were it even possible. I mean, even if we could achieve full cancellation of input GPE, output GPE is just that, a linear, fixed quantity..

And of course, in reality, since gravitational asymmetries appear intrinsically impossible due to constancy of G and rest mass, our hopes can soar on the slightest suggestion of any advantage... the tiniest torques..

Mate, seeing a working Bessler wheel is gonna kick you square in the nuts.. It's gonna come alive, and send shudders down your spine.. ;)

[Grimer]

I have written down a handful of calculation-attempts all with the intent to replicate your page-11 math. So I'll not waste any time and space showing that boring result. They all fail in one way or the other. That's at least 100 lines (in compacted notation) of useless math...

As a last attempt I tried something weird, and I guess I know how you got your epiphany... It needs some cleaning up, and of yet it doesn't completely match your math: I guess I'll have to redo it a second time. But it does indicate a similar increase like you found. But suspect of being based on something 'unusual'.

So at this moment it appears I'm closing in on either confirming or debunking that theory of yours. I don't know how long that will take.

That sounds encouraging. You (and my grandson) were a great help to me on the pendulum maths. Be assured you will get full acknowledgement when I collect my Nobel prize. ;-)

[ME]

Glad you wrote it like that, made me delete a lot of my already written reply.

Therefore he tacitly accepts that N^2 J/kg-m/s is valid for at least one full cycle.

Actually I do not.
This factor is bugging me the whole time. What is it?
As far as I can determine you take it as:
E/p=(½mv²) / (mv) = ½v [J / (kg·m/s)] = ½v [m/s]
But what is its validity?

[MrVibrating]

..Haven't i just explained this?

It's the energy cost of the momentum we're buying. 9.81 kg-m/s for 96.23 J, from a 1 G force times 1 second.

Since 96.23 = (9.81 * 9.81) = 9.81^2, the work performed by that 96.23 Joules evidently accelerated a 9.81 m/s acceleration by 9.81 m/s.

One of those 9.81 m/s accelerations was the one we physically applied ourselves between the two inertias.

The other was gravity's effective acceleration.

However remember that the lower inertia in this linear case was subject to a net 2G acceleration..?

You've just jogged my memory on something important - the lower inertia could be angular, so not itself gravitating...

So surely in that case the input constant would have to be simply N J / kg-m/s, rather than N^2..?

I need to go back and check the angular-linear example... since i now suspect it's N J if the lower inertia's angular, and N^2 only if it's linear..


Either way tho, N is evidently equal to our acceleration in m/s^2.

Its 9.81 value is obviously no coincidence.

Which surely means its dimensions must be squared?

In other words, i'm correct that if the lower inertia is angular and thus not itself gravitating, our optimum momentum yield would be N J / kg-m/s rather than N^2, however this is simply because the acceleration isn't nested within another 9.81 m/s acceleration.. so we don't obtain a squared result.

However we're still dealing with a quantity of energy with dimensions of acceleration, so the baseline rate must indeed be N^2 J per kg-m/s of reactionless momentum.

As for the validity of its constancy (ie. its speed-invariance), i've performed three tests of this:

- it is valid regardless of whether initiated from a standing start, or else with any arbitrary initial system velocity

This simply follows from first principles; that inertia is speed-invariant, and there is no such limit as 'terminal velocity' in a vacuum.

- it can thus be accumulated and repeated over successive cycles, at constant yield

- we can circumvent the first four cycles entirely, and leap straight to a 5/8mV^2 result

Therefore we can maintain the N^2 J / kg-m/s purchase rate regardless of velocity.

Obviously we don't need a lot of velocity to reach absurdly-high efficiencies if we aim for multiple cycles per rotation of a wheel. Conversely with few per revolution, velocity will have to be able to rise to reach higher efficiencies - the key factor determining peak efficiency being simply the number of preceding cycles and their net efficiency.

For example, consider the machine Bessler claimed he could build, if God were to grant him sufficient time, which would turn very slowly, yet with very high power, all the while emitting a steady chatter...

...this would be consistent with very many, very small reactionless accelerations per revolution. The value of N^2 per cycle could thus be extremely low - a very small amount of momentum, over a very small angle of rotation of the net system.

Since N^2 is not time-dependent, it doesn't depend on the 1 second impulse used in my examples - it scales to whatever the applied period.

But because its resulting inertial frame diverges from Earths, and thus 1/2mV^2, as a function of # of prior elapsed cycles, there's no practical reason why we couldn't squeeze say 64 very small reactionless accelerations into a one full system rotation, thus attaining an 8mV^2 (1600% unity) outcome after just one revolution.

If you think about it, the number of cycles we could squeeze into a given angle of system rotation is really just an engineering challenge - with modern fabrication and automation, we could have one reactionless acceleration per degree * 360 or whatever.. it's basically arbitrary.

But that's getting a little ahead..

Assuming for now we just have 1 reactionless acceleration per full system revolution, then efficiency is a direct linear function of # of elapsed prior cycles, and thus RPM.

N^2 J / kg-m/s is the input efficiency, in terms of the energy cost of reactionless momentum.

1/2mV^2 is the output efficiency, in terms of the value of the momentum accrued over # cycles.

The equality of 4(N^2 J/kg-m/s) = 1/2mV^2 is objectively proven because N = F/m, and so 9.81 N / 1 kg = 9.81 m/s^2.

Thus 4 * ((N^2 J/kg-m/s) * 1 second) = 4 * (96.23 J / 9.81 kg-m/s) = 384.92 J / 39.24 kg-m/s.

39 kg-m/s divided between the two 1 kg masses leaves 19.62 m/s velocity on each.

Per 1/2mV^2, ((1/2 * 1 kg) * 19.62^2) = 192.47 J, on each mass, so 384.94 total.

Thus we have proven irrefutably from first principles that 4 * ((N^2 J/kg-m/s) * 1 second) = 1/2mV^2.

Therefore we have also proven that 3 *((N^2 J/kg-m/s) * 1 second) = (0.75 * (1/2mV^2)) = 3/8mV^2 = 75% unity.

And equally, that 5 * ((N^2 J/kg-m/s) * 1 second) = (1.25 * (1/2mV^2)) = 5/8mV^2 = 125% unity.


If anything at all needs further clarification, just ask... this is about as far as i've got tho..

Obviously this is just the coarsest tip of the iceberg, and the underlying Lorentz transformations, Lagrangian rotations etc. would be all but impenetrable to me. All i can do right now is measure, record, predict and test etc.


ETA: presumably the 2 mJ discrepancy between 4 cycs / 1/2mV^2 is just rounding error.

I've only included the (* 1 sec) qualifier in the term above since i've only deduced that the period has no effect on the 1/2mV^2 equality, rather than having demonstrated it explicitly - not that i'm in any doubt, just a concession to accuracy / honesty.

[MrVibrating]

Here's that linear-angular example again, side-by-side with the linear-linear equivalent:




So i'm out of practice with angular systems right now - is N squared on both sides, or just the left?

[ME]

..Haven't i just explained this?
It's the energy cost of the momentum we're buying. 9.81 kg-m/s for 96.23 J, from a 1 G force times 1 second.
Since 96.23 = (9.81 * 9.81) = 9.81^2, the work performed by that 96.23 Joules evidently accelerated a 9.81 m/s acceleration by 9.81 m/s.
One of those 9.81 m/s accelerations was the one we physically applied ourselves between the two inertias.
The other was gravity's effective acceleration.

Yes you mentioned before, I must have missed the explanation.
Please mind my slowness, it feels like I'm constantly missing some crucial information.

Could you repeat the same calculation for when that duration is 3 seconds (actually "t" seconds)?

[MrVibrating]

OK, just had a nice long bath and realised a full solution to the Toys page:

- two pairs of input / output integrals are represented in the horizontal plane

- differentiation between 5 successive cycles of these integrals are represented in the vertical plane

So integration in the horizontal axis, and differentiation in the vertical axis.

The shape of the anvil on the lower hammer toy signifies that this pair of integrals represent a quantity that has squared dimensions. The handles obviously signify that it is an input integral.

The upper hammer toy has a circular anvil, likely representing a pair of integrals dealing with a linear quantity.

So in short, the lower toy represents input energy, and the upper toy represents output momentum.

The scissorjack represents the axis of differentiation, applied by the uniform acceleration of gravity.

And upon that axis, 5 consecutive cycles of the above energy-for-momentum exchange yields something extraordinary.

The upturned whistling top, apparently penned in afterthought, is spinning upside down, upon its handle - a physically-impossible situation, but so representing an output of RKE or angular momentum or CF, instead of an input. This is how the gain is harvested, but also, its alignment in the vertical plane beneath everything else marks it out as the result of the differentiation in that axis.

So the solution to the Toys page is simply:

5 * ((N^2 J/kg-m/s) * 1 second) = (1.25 * (1/2mV^2)) = 5/8mV^2 = 125% unity.

It's not a clue intended to help find the solution, since it requires foreknowledge of the solution.

Rather, it is Bessler's way of saying "..Look.. I've been here before you!"

[MrVibrating]

Yes you mentioned before, I must have missed the explanation.
Please mind my slowness, it feels like I'm constantly missing some crucial information.

Could you repeat the same calculation for when that duration is 3 seconds (actually "t" seconds)?

Yes i'm curious too. Provisionally however, if N is always squared, then the reason is because it encompasses nested accelerations, and so already includes the relevant time-derivatives.. hence further extrapolation in the time domain has no effect on its ratio.

But yes, let's double-check this too, defo on the list of to-do's.

I also have to get up to speed with the equivalent angular quantities to ensure consistency.

For a first cycle however i think you're there - first you need to be satisfied that after KE and momentum from GPE is notionally repaid, we're left with a 96 J / 9.81 kg-m/s acceleration applied to only one of the two inertias.

If and only if you're satisfied with that result, then dividing the resulting momentum to produce the net rise in system velocity depends on nothing more than N3, and so can be taken for granted as perfectly viable.

At which point, by extension, since there is no dispute that 1/2mV^2 remains valid in our external, static reference frame, you can with certainty confirm that after one such cycle, each mass has 4.905 m/s velocity and thus 12.02 Joules.

Likewise, there is no dispute that 2 * 12.02 = 24.04.

We definitely spent 96.23 J to begin with, we have 24.04 J remaining, so our ratio of input energy to output KE was 96.23 / 24.04 = 4; we have precisely 1/4 of our input energy remaining, per (1 * (N^2 J / N kg-m/s)) = 1/8mV^2 = 25% unity.

Hence if you can just get all your ducks lined up in a row, you should see that the equality is already proven for one complete cycle, without trying to convince or garden-path you into anything controversial or improbable.

Likewise, all relevant conditions determining the speed-invariance and thus constancy of that transaction value are already established or implicit, hence you're only a penny-drop away from the validity of this predicate, and hence the 1/4mV^2 value of (2 * (N^2 J / N kg-m/s)), for 50% unity and beyond.


So in summary, we have two self-consistent means of confirming the validity:

- 'Backwards', from a presumption of (4 * (N^2 J / N kg-m/s)) = 1/2mV^2 and solving for N as a function of F/m (per F=mA)

- Or 'fowards' by running the full-cycle interaction in step-wise solutions.


I will get on with testing the time integration and angular equivalencies, but take your time, assume nothing, question everything, will try to clarify anything i can..

[MrVibrating]

LOL if this pans out then we need to find a suitable Greek symbol or something for "Bessler's constant"..

How about a little unicode AP wheel lol?

[MrVibrating]

Here's a 3 second burn of both linear-linear and linear-angular:




..i'll work up a breakdown as before, same drill: deduct KE and corresponding momentum (P) sourced from GPE, to leave only input J / P applied between the two inertias.

This should result in a value of N consistent between both sides, and presumably equal to ((1/2mV^2) / 4)...

[MrVibrating]

Linear-linear case:

For accuracy i will use the same five significant digits as 1 G = 9.80665 m/s^2.

The lower 1 kg descended by 88.2598 meters, producing KE equal to GMH = (9.80665 * 1 * 88.2598) = 865.53296 J.

The upper 1 kg has not changed height, so still has all of its GPE and cannot have contributed any KE or momentum.. or else we'd effectively be able to drop it twice.

So subtracting the 865.53296 J of GMH applied by gravity's 1 G acceleration, from the 1731.06692 J total, leaves 865.53296 J - basically half as much energy again - input internally between the two inertias, and corresponding to the additional 1 G acceleration it applied to the lower inertia.

In summary, we applied two 1 G forces, both performed an equal acceleration of the lower inertia only, and both spent the same energy, having performed equal amounts of work.

Thus half the final momentum and velocity was due to GPE, and the other half due to the 3 second 2 G mutual impulse.

58.83990 / 2 = 29.41995 m/s of velocity from each source.

So 29.41995 P from GMH, plus 29.41995 P applied internally.

Since only the lower inertia has been accelerated by either force, we only need to repay KE and P to GMH from its motion.

Leaving it with 29.41995 P and corresponding 432.76672 J (per 1/2 * 1 kg * 29.41995^2) from internal work.

So to recap, we ended up with 1731.06692 J of KE.

Half that - 865.53296 J - came from GPE.

We also ended with 58.83990 m/s of velocity and equal 58.83990 kg-m/s momentum.

Repaying half the 58.83990 m/s that came from GPE, we're left with 29.41995 in m/s and kg-m/s, with an energy value of 432.76672 J, which is precisely half of the 865.53296 J that didn't come from gravity, but was instead applied internally.

The reason half our input energy has disappeared is because KE squares with velocity, and we've just halved velocity, so we're left with a quarter of our net total energy, and half our internally-applied input energy.


That remaining 29.41995 P divided into the upper inertia per N3 leaves each at 14.70997 m/s.

1/2 * 1 kg * 14.709975^2 = 108.19168 J on each, so 216.38336 J total KE.

So over a 3 second impulse we input 865.53296 J of internally-applied work, yielding 216.38336 J of output KE, and 865.53296 / 216.38336 = 4.

Therefore ((N^2 J/kg-m/s) * 1) is fully time-symmetrical to ((1/2mV^2) /4).

Thus ((N^2 J/kg-m/s) * 4) = 1/2mV^2, and ((N^2 J/kg-m/s) * 5) = 5/8mV^2, etc.


Next we need to confirm consistency with the angular example...