Plying CF as pseudo-inertia to scam N3

[eccentrically1]

If you were going to physically build this to show its repeatability and to prove the principle, (just the first example to keep things simple), please describe to everyone how you would go about it. This will perhaps illuminate some missing information.

On the descending side of a wheel, push a mass up, while pushing the edge of the wheel down.

The mass doesn't need to literally rise up, only its rate of descent needs to slow, briefly, while that corresponding counter-force is applied to the wheel.

This imparts some reactionless momentum to the wheel.

The wheel now has momentum from two sources - GPE, and the height the mass has fallen, plus the acceleration applied between the decelerated mass and the rest of the wheel.

Using the wheel to re-lift the fallen mass will only remove that portion of the wheel's momentum that came from GPE in the first place. That is, after relifting the mass, the wheel will still have some momentum, corresponding to the internally-applied acceleration.

Now apply a second such acceleration, identical to the first. This is where the magic happens...

..the second acceleration adds the same amount of excess momentum as the first, and costs the same energy to perform. The mass is re-lifted again, returning that portion of its momentum that came from GPE again, and again leaving only that momentum pertaining to the internal acceleration, which now sits on top of that momentum left over from the first acceleration. The KE value of the momentum now on the wheel at this stage is greater than the energy needed to reload the springs or whatever that caused the two internal accelerations.


That's using the second example.

Using the first, where a counter-inertia is thrust downwards, you could simply collide the lower mass with a rim-stop, transferring its momentum to the rest of the wheel...

..again, it doesn't have to be a 1 meter extension - a few centimeters will do. Whatever counter-momentum is decelerated by gravity will instead be present as forwards-momentum on the opposing mass - ie. the net change in momentum caused by the acceleration is equal regardless; gravity merely skews its distribution between those two masses.

So again, as we accumulate this residual, non-cancelled momentum, caused by gravity's passive deceleration of our reaction-momentum, over successive cycles, its cost of purchase remains constant, despite rising velocity, since it's just a mutual acceleration, a force, applied between two masses, and inertia is not velocity-dependent.

Yet the value of that momentum we've accrued is set by the standard KE terms, in which the energy cost of rising momentum is velocity dependent - the more momentum we have, the more energy it costs to add to, and likewise, the more our growing stockpile is worth, relative to the static reference frame of Earth and gravity.

So the energy asymmetry arises by buying reactionless momentum, on the cheap, from inside the system of interacting masses, by relying on the externally-applied static uniform gravity field to decelerate our reaction mass, thus skewing its distribution and causing a net increase, instead of equal opposing, and thus cancelling momenta.

That's the gain principle. It's that simple, if unintuitive, since we're only used to dealing with the usual, N3-determined cost of momentum.

It's no way near a build design yet. First i need to show that the input energy remains constant as system velocity rises. Then, if that's successful, i'll need to find the simplest, neatest mechanism possible for tapping off the KE gains and resetting the springs pushing the masses apart. If the prior test was successful then using the gain to re-compress the springs is simply a matter of engineering.

The deserving target of one's cognitive dissonance / incredulity is this notion that a given quantity of momentum can have two different, mutually-irreconcilable energy values at the same time, that we can conveniently treat as input and output, or source and sink - however this is simply an inescapable consequence of any effective exception to Newton's 3rd law - if momentum is induced without raising an equal opposing counter-momentum, then net momentum rises, and further, successive accelerations of the same kind begin at the remnant momentum & velocity of the preceding asymmetric exchange, piggybacking on that free surplus velocity, to effectively inflate the external, real-world value of the internally-applied momentum rise.

The simplest analogy i can make, to get your heads around this gain principle, is to picture a cannon mounted to the front of a sleigh, coasting at constant speed across a frozen lake:

- when the canon fires, the recoil will decelerate the sleigh, removing precisely as much momentum as has been imparted to the canonball.

- as a static observer, watching the sleigh travel from left to right across your field of vision, you see the system's net momentum, of sleigh plus canonball, as remaining constant.

Now however, suppose the cannon fired without any recoil - so the sleigh is not decelerated.

Alice, riding the sleigh, sees her speed relative to the fired canonball, as the same, regardless of whether there was recoil or not.

But Bob, watching from across the lake, sees the cannonball accelerate with more energy than has been supplied in its charge.

Furthermore, the excess energy Bob measures depends entirely on the ambient speed Alice was going when she fired the canon. Even though the charge contained a fixed amount of PE, and the relative acceleration between the sleigh and the canonball, as measured by Alice, is the same at whatever the initial velocity, and regardless of whether a recoil occurred or not, from Bob's point of view, the higher the sleigh's initial speed when the canon was fired, the greater the discrepancy between the measured rise in KE vs the charge supplied to the cannon.

A recoil-less 1 kg-m/s acceleration from an initial speed of 1 m/s is 4x OU; 0.5 J in, 2 J out.

But that identical 1 kg-m/s reactionless acceleration from a starting speed of 5 m/s is 10x OU. again, 0.5 J in, but this time, 5.5 J out..

From an initial speed of 10 m/s, that same 1 kg-m/s reactionless acceleration is worth 20x the 1/2 Joule we've paid for it.. the same 500 mJ in, for 10.5 J back out.

So it's very early days yet - i think we may have the principle by which to apply these reactionless accelerations, by letting gravity cancel out or just diminish counter-momenta, thus skewing the normal distribution of momentum and so causing a non-zero sum of momentum-plus-counter-momentum.

If this non-zero momentum sum can be accumulated though multiple such interactions, from forces applied solely between the two asymmetrically-endowed momenta, then after very few such interactions, the real-world (ie. GPE) value of this accrued, consolidated excess momentum should far exceed its costs of generation.

To familiarise yourself with the potential margins on offer here, have a play with the following KE calculator applet:

http://www.csgnetwork.com/kineticenergycalc.html

Simply enter 1 kg in the mass field, and 1 meter / sec in the velocity field, then click "calculate KE" to see the input energy cost of that acceleration.

Then increase the velocity in 1 m/s increments, checking the KE value each time.

You're adding the same amount of momentum each time - a succession of identical 1 kg-m/s accelerations, but the energy cost is spiraling - you're having to input ever-more energy to keep buying the same fixed quantity of momentum.

This is because of Newton's 3rd law! The initial kg-m/s only cost 500 mJ because the reaction mass was stationary relative to the accelerated mass! From thereon, the rising speed difference between the accelerated mass and the world it is being accelerated against, subtracts from the 'acceleration' component of the force (F=mA) that must be applied to achieve that consistent 1 kg-m/s acceleration, and the product of that ever-rising force over that constant displacement equals the escalating KE value you're calculating.

However, with a reactionless acceleration, in which the momentum induced into the accelerated mass can in turn be shared with the non-accelerated mass, thus accelerating the net system, the reaction mass is always stationary relative to the accelerated mass at the beginning of each successive cycle! Therefore the cost of that acceleration - and all subsequent such accelerations - is the nominal 500 mJ / kg-m/s we paid for the first 1 kg-m/s.

So for instance we could accelerate up to 10 m/s, with a value of 0.5 * 1 kg * 10 m/s * 10 m/s = 50 Joules, for an input energy cost of 10 discrete 500 mJ accelerations, equaling 5 Joules... et voila, we've just bought 50 Joules of KE with a 45 J discount...

Awesome or what? But this is the crazy world we live in.. there's nothing like this in inertial or gravitational interactions alone, but overlay one upon the other and this momentum gradient appears, and with it, a free KE escalator..

A leg up on Jacob's ladder..

The part I bolded is where your system analysis is flawed. Do you see it?

[sleepy]

A few pages back I made an attempt to bring a vague notion to light by using
the "donkey,carrot,cart" analogy. I have since thought more about it and I think I have more clarity now.What I was trying to say is this.Mr.V has 3 bodies in his reaction.1. the top weight which will appear to sort of hover as it is pushed upward.2.the bottom weight,which will gain momentum by combining the acceleration of gravity and the force of pushing against the top weight.3. the apparatus between the weights that will force the weights apart at the precise moment.In trying to imagine which one of these bodies would actually be connected to the wheel,I think I stumbled onto a problem.This configuration only works if all three bodies are in constant freefall through gravity.As soon as any one of them is connected to a wheel,it will cease to function as anticipated,because the 3 bodies combined will never have more mass than the total wheel (including those 3 bodies and maybe another set or 2). I know I'm not presenting this very well,so use your imagination and try to figure out which body will be connected to the wheel.

[eccentrically1]

sleepy, that's what I thought you meant, that was the magical thinking I referred to. The mechanism can't float above the wheel.

[cloud camper]

Thats definitely an issue that I have my doubts on.

The generated energy/momentum has to be relative to the wheel's rotating frame of reference not an outside stationary observer.

We're generating high energy values from a reaction occurring in free fall but from the point of view of an observer falling with the reaction there is no excess energy being generated.

And from the point of view of an observer rotating wirh the wheel there's no excess energy being generated as all he sees is a weight lagging and another leading for a short time then coming back together. Nothing unusual going on.

Per physics, all reactions are supposed to be invariant across different frames of reference. So a reaction occurring in one frame of reference should produce the same energies as viewed in any different frame of reference, accelerated or not.

When you push the accelerator to the floor of your car you feel a mysterious force pushing you back in your seat compressing the seat springs and you could say the springs are being compressed for free!

But when we decellerate back to a stop assuming our original non accelerated frame of reference we see that no overall work was done on or by the seat springs.

So we might be in a case here where we are only considering the positive effects of an accelerated frame and not considering the negative effects of bringing the reference frames back together as must happen with any conceivable wheel mechanism.

So we need to be super careful here.

We can appreciate Mr V's attempts to solve the problem in a most open source manner but all aspects of a theory must be considered especially without any physical proof.

[rlortie]

So we have to be super careful here:

When you push the accelerator to the floor of your car you feel a mysterious force pushing you back in your seat compressing the seat springs and you could say the springs are being compressed for free!

But when we decelerate back to a stop assuming our original non accelerated frame of reference we see that no overall work was done on or by the seat springs.

If you look out the window of your car after deceleration you will see that you, your car and seat have indeed changed in frame of reference. It is no longer where it was, therefore work was accomplished. You were not a stationary observer while the car was in motion. Inertia held you back compressing the springs upon acceleration and inertia forced you forward relieving the tension on the springs upon deceleration. Linear motion was achieved.

Per physics, all reactions are supposed to be invariant across different frames of reference. So a reaction occurring in one frame of reference should produce the same energies as viewed in any different frame of reference, accelerated or not.

[MrVibrating]

@CC

Yes, the gain is only manifested in the stationary frame, however it's a perfectly valid frame...

To put it another way, the stationary frame is the de-facto objective frame, from which a given symmetry is meaningfully broken. If you observe a symmetry break, your observation is valid, that symmetry was broken. For example if you're watching an object accelerate through space without ejecting reaction mass, then it's CoM that's broken, and not your reference frame.. which cannot be 'invalidated'.

But we also have an objective measure of that validity in the form of centrifugal / centripetal forces, which are RPM-dependent.

So it seems unassailable - we can generate the momentum asymmetry, we can hammer it into the wheel body, it thus starts to build up an excess of momentum in one direction, hence its ambient speed increases each time more momentum's added.. i call it "ambient speed" as it's always 'at rest'.. we're 'accelerating mass' indirectly, via CoM..

So normally, we generate momentum by accelerating mass, right? Performing work upon it. It costs precisely 1/2mV^2 this way.

But that's not what's we're doing here..

Here, we're creating momentum from nothing more than its asymmetric distribution and subsequent non-zero sum.

We knock this into the wheel body / main system via regular elastic collisions (for now; crude but foolproof).

So the net system's momentum must increase. Its momentum is a function of its angular inertia times RPM, and since the former's constant, the latter must rise to accommodate the gain.

We could measure the CF with a Newton meter, we could also measure the angular accelerations via accelerometers or cameras etc. etc., so there could be no room for doubt or subjectivity about whether the momentum rise, and its corresponding RKE value, was real..

Basically, in a nutshell, we don't need to prove 1/2mV^2 - that's a given.

The only bit that was really worth proving was the speed-invariance of the cost, although as i keep saying since inertia's not speed dependent, it couldn't have worked out any other way.

Provided that cost of momentum is constant across some useful RPM range, our investment regime is completely outside the bounds of the practical constraints normally enforcing the 1/2mV^2 standard.. which is basically N3 symmetry.

So the standard 1/2mV^2 value of momentum cannot fail us, and we don't need it to do anything it wouldn't normally do anyway.

The only real derring-do is the first part of the plan, generating the momentum in the first place, and i think we've covered all bases there..

This is the convergence point mate. We know his system was OU, that there had to be an OU exploit in basic mechanics, and so at some point we had to be ready to deal with that. Now is the time..

So, i can't believe it either, but it is what it is, and it appears to be telling us that if we just go through the motions and humour this concept, magic will happen, and we'll see something we have no intuitive point of reference for.. it's gonna go 'clack clack clack clack clack' and accelerate up to speed, not from a preponderance of weight, but momentum..

Apparently, anyway. Mate, you're looking at the same maths as me. I honestly do not know anything you don't here!

All we can be certain of, at this stage, is that we can definitely buy 9.81 kg-m/s of momentum, angular or linear, for precisely 96.23 J, and accumulate it, maintaining that price irregardless of the rising velocity.

For a single helping of momentum, it's a shite deal - 9.81 kg-m/s divided between two 1 kg masses means each is traveling at 4.905 meters/sec, so per 1/2mV^2 each has just 12.02 Joules of KE, so 24.04 J total output, from a 96.23 J input.

That's not just shite efficiency, it's massively under-unity! If we just spent that whole 96.23 J on a regular acceleration, we could get our 2 kg system straight up to 9.81 m/s without any effing about.

And likewise, the second and third interactions are also terrible deals, losing energy by buying much less momentum than could've been purchased via N3..

But by the fourth purchase, we're neck and neck with 1/2mV^2, and our disadvantage has dissolved.

By the fifth, that margin has reversed, and we now have the advantage over 1/2mV^2, at 1.24x OU relative to it. In other words we've performed 76% less work (in terms of F*d) in moving the internal masses, than the wheel itself will do if suddenly harnessed to a GPE load or whatever.

The weirdness resulting from a successful N3 break is precisely why it's considered inviolable. It's a runaway non-inertial frame, so all equivalencies with other frames become invalid..

Note, it's the 'equivalencies' that become invalid, not the reference frames themselves, which are all always equally valid in their own right. All would see the same energy and momentum rise in the anomolous system. And the equivalency between the internal cost vs external objective value of momentum is the specific one we're exploiting.. inequivalent I/O energies is the singular objective!

[MrVibrating]

LoL There's the rub - can you believe the first helping is under-unity?

What about the second, or third?

Do you believe that the fourth is precisely at unity? Why - because of the result, or because of the regime that caused it?

So then how can you deny that the fifth is OU? If that purchase is invalid, it's under precisely the same terms as the other four, so what's wrong with the first, massively-under-unity result?

Are we actually calculating F*d integrals, or just going by intuition and confirmation bias? Because that's the only way one could accept the 1:1 energy result of precisely four reactionless accelerations, but not three or five in the very same contiguous sequence..

Logic trap for any doubters right there, lol.. :P

[MrVibrating]

The part I bolded is where your system analysis is flawed. Do you see it?

Well no - on the contrary, it's already been tested in multiple iterations, and performs exactly as expected.

You seem to have highlighted what you think is the exploit, and claimed it's impossible without saying anything at all about it?

I'm going to take a stab and guess that you haven't quite grasped what's happening yet:

- the momentum gain principle is already proven

- it's speed-invariant; the cost of generating momentum doesn't rise with velocity

- the objective value of momentum is speed-dependent, per 1/2mV^2

Therefore, depending on the speed attained, output can be less than input, or equal to input, or more than input; they're completely thermodynamically decoupled because they have fundamentally different scaling properties WRT time.. input energy now simply has dimensions of inertia times velocity, whereas output energy has the additional time-derivative, squaring with velocity.

If you could simply articulate what you don't understand, i could maybe better help you with it..?

Do keep in mind however that i'm just trying to demonstrate the "fuel" for the fire, not how to build any specific engine using it. At this stage, all i can tell is that such engines are mechanically possible, and could take a considerable variety of forms, since the exploit is so general (again, the key ingredients are just two inertias, at least one of which is also subject to gravity at some point during its cycle).

[eccentrically1]

"So the energy asymmetry arises by buying reactionless momentum, on the cheap, from inside the system of interacting masses, by relying on the externally-applied static uniform gravity field to decelerate our reaction mass, thus skewing its distribution and causing a net increase, instead of equal opposing, and thus cancelling momenta. "

The 'external' gravity field is internal as well.
The momentum is asymmetric, but it isn't an increase.
The momentum wasn't bought on the cheap from inside the system, it has to begin from outside. Did the masses and jack throw themselves up in the air before they began freefall?


Try Fletcher's method, I think he uses spreadsheets to determine the "energy budget" or whatever he likes to call it to see if there is any gain.

[MrVibrating]

A few pages back I made an attempt to bring a vague notion to light by using
the "donkey,carrot,cart" analogy. I have since thought more about it and I think I have more clarity now.What I was trying to say is this.Mr.V has 3 bodies in his reaction.1. the top weight which will appear to sort of hover as it is pushed upward.2.the bottom weight,which will gain momentum by combining the acceleration of gravity and the force of pushing against the top weight.3. the apparatus between the weights that will force the weights apart at the precise moment.In trying to imagine which one of these bodies would actually be connected to the wheel,I think I stumbled onto a problem.This configuration only works if all three bodies are in constant freefall through gravity.As soon as any one of them is connected to a wheel,it will cease to function as anticipated,because the 3 bodies combined will never have more mass than the total wheel (including those 3 bodies and maybe another set or 2). I know I'm not presenting this very well,so use your imagination and try to figure out which body will be connected to the wheel.

LOL i've shown three simulated, idealised exagerated N3-breaking inertial interactions, just to highlight the principal and prove its practicality.

They were:

- a linear - linear interaction

- a linear- angular interaction

- and an angular - angular interaction

You've chosen to look at the first one, a linear - linear interaction.

On the lower end of page 8 i've shown one way we might incorporate this into a wheel - whatever the masses are attached to can momentarily fall relative to the rotating system, there's nothing mechanically contradictory about this.

For instance you could use bungees, knicker elastic, springs, hydraulics, pneumatics, a little CO2 or NO2 cylinder in a condom, immigrant labour, half a pound of Tannerite and a '22.. whatever. Just apply a force - any force at all, of any magnitude, of any form or origin - between two masses, at least one of which is inertially-suspended against gravity's acceleration, by the other inertia. That's the key condition. That's all.

So i dunno, just get one of those Lotus F1 carbon fibre scissorjacks off ebay and some R/C servos or something, aint they got them? They got everything else. What am i, Mr Fixer? Or else forget the linear - linear example and try something else, like angular - angular or angular - linear. BTW that's not necessarily an exhaustive list of options, either - maybe radial inertias could also work (not yet tested).

The exploit is simply augmenting "inertia" with more general sources of "resistance to acceleration", such as a static force field, purely for the purposes of achieving an effective break in momentum symmetry, and the examples i've given are simply proofs of principle, not necessarily the basis for any particular design, at this stage.

[cloud camper]

Mr V - was hoping you could answer a question on your mech below.

Are the input green and red output weights the same mass? If so you have shown how one weight can lift another of the same mass to a higher level.

That would be very convincing. Case closed!

Thank you.

Also could you post the WM file for download? I would like to experiment with it.

[MrVibrating]

So we have to be super careful here:

When you push the accelerator to the floor of your car you feel a mysterious force pushing you back in your seat compressing the seat springs and you could say the springs are being compressed for free!

But when we decelerate back to a stop assuming our original non accelerated frame of reference we see that no overall work was done on or by the seat springs.

If you look out the window of your car after deceleration you will see that you, your car and seat have indeed changed in frame of reference. It is no longer where it was, therefore work was accomplished. You were not a stationary observer while the car was in motion. Inertia held you back compressing the springs upon acceleration and inertia forced you forward relieving the tension on the springs upon deceleration. Linear motion was achieved.

Per physics, all reactions are supposed to be invariant across different frames of reference. So a reaction occurring in one frame of reference should produce the same energies as viewed in any different frame of reference, accelerated or not.

The easiest kinematic example to follow is a reactionless acceleration performed upon a frictionless rolling cart, or in micro-gravity etc.

So picture it - your cart's rolling across the screen from left to right, coasting at some ambient speed.

Upon the cart is a projectile of arbitrary relative mass to the cart.

And some N3-busting magic mechanism to fire it, in the direction of travel.

So when it fires, the cart experiences no recoil, and isn't decelerated in proportion to the momentum imparted to the projectile.

The cart's reference frame is now divergent.

If you then calculate the KE of the projectile as measured from aboard the cart, everything is normal between the projectile relative to the cart - it has the right amount of energy, the right difference in velocity, and the right amount of momentum.

But from the vantage of a stationary observer - or indeed, one in any other inertial frame - the net system of cart + projectile has gained momentum and KE.

You can calculate this on the back of a beer mat. Or with Notepad and calculator in Windows. It's piss-easy to do. Marcello won't, because everyone knows N3 breaks are impossible, and people might think he's gone soft. Besides, you could prove anything, hypothetically.

But yeah, that's the basic principle.

To really have fun with it, chop it up with a regular, N3-respecting reset stroke - so for instance the projectile could loop off the right side of the screen and re-appear back on the left, like Meteors or Pac Man, to re-collide with the cart... or else, the projectile could remain connected to the cart via a long, slack bungee. Either way, the point is to re-load and repeat the N3 break, to see how this affects the evolution of the resulting energy disparities.

Another way of considering the same issue is to consider multi-staging reactionless accelerations, the same way we stage rocket launches..

So the projectile could itself launch another - say, 50% of its mass or whatever - via a second or third-stage recoil-less interactions.. again, the point here is to plot the energy expended from within the accelerated, N3-fiddling reference frame, and also from the stationary external frame, to compare the two results, and thus get a feel for the energy gains that would be on offer if we could only actually cause an effective N3 break, somehow...

And what you'll see is monster gains. It quickly gets ludicrous, with outputs massively exceeding inputs.

And Marcello will be scoffing "meh - it's just hypothetical" but you'll be looking at these figures of dozens or even hundreds of times OU and thinking "so the only thing standing between me and these gains is N3?", hopefully followed by "Game on!", or something along those lines..

At least that's what happened in my case..

[MrVibrating]

Mr V - was hoping you could answer a question on your mech below.

Are the input green and red output weights the same mass? If so you have shown how one weight can lift another of the same mass to a higher level.

That would be very convincing. Case closed!

Thank you.

Also could you post the WM file for download? I would like to experiment with it.

LOL no of course not, tho i wouldn't have missed that (i'd hope!).

Like i say it's just a messy doodle - the green GPE weight bounces back out rather than being caught in the center, the only area of interest it demonstrates is during the first moments, when GPE is applying a mutual torque between the wheel and red mass, the latter also gravitating and so the greater proportion of momentum generated going into the wheel.

To be useful, the mechanism would need to coordinate the reset such that gravity did not interfere in the momentum distribution while resetting.

To be OU, that green mass would have to power five such interactions, all perfectly sequenced at the correct angles, one after another. If it could do that within 360° then it could re-lift the GPE, + 24% more.

Obviously, the more interactions you could squeeze into one full revolution, the larger that gain margin grows.

Hopefully, we can reduce a full cycle into something that covers a similar angle of displacement as this, though, and so then repeat it at say 45° increments for 8 bangs per cycle, OU after the first five, five-eighths into a complete revolution..

Or at the very least, at 5 * 72° intervals.

Or just one mechanism with a PE store that lasts at least 5 full cycles..

Whatever, it'll sort itself out, i'm sure. Just generate the gain, consolidate it, rinse and repeat, measuring and monitoring at each step, and the magic will happen.


...and remember, we don't need to wait til five interactions to simply see the magic - since the same magic makes us substantially under-unity initially. So if this is working, we'll see the first three lose precisely the energy predicted, gaining precisely the predicted momentum for just the right input energy, and on the fourth, we'll hit unity, with perfect agreement between input and output energies, despite arriving there from completely different scaling dimensions which only converge at precisely four interactions.. and so we'll know exactly what to expect from the fifth, by the time we've confirmed the result from the first one or two.. this is simply an inexorable path to success..!

[MrVibrating]

...so a quick followup for anyone who's been tempted to try following the maths of a divergent inertial frame, as above:


The only thing standing between us and these gains is an "effective" N3 break - rest mass is obviously constant so N3 is immutable, thus the only meaningful definition of the "effective" qualifier is whether or not the resulting momentum gain can be accumulated over successive cycles, either in series or parallel (or both!), and whether its cost of operation remains constant throughout these successive interactions or stages.. which basically boils down to not scaling up with velocity.

So these are the key tests i've performed with this candidate N3 break, and it has passed them, just as the (very basic) hypothesis predicted. It fulfills the necessary conditions for an input/output energy disunity. It's do-able.

You can have those gains you just calculated.

[MrVibrating]

Another quick recap of the maths so far demonstrated:

We can make 9.81 kg-m/s of momentum for the fixed, speed-invariant price of 96.23 Joules, which equivalently, works out to 9.81 J per kg-m/s.

Divided between two 1 kg masses, each has 4.905 kg-m/s and 48.11 J.

But per KE=1/2mV^2, at 4.905 m/s 1 kg has just 12.029 J...

So on the first cycle, our very first kg-m/s of manufactured momentum costs exactly four times more to make, than it is worth. To put it another way, it is 4x under-unity.

Everyone follow that so far?

So repeating that yield over successive interactions:

2 * 96.23 J = 192.46 J, and 2 * 9.81 kg-m/s = 19.62 kg-m/s

So there we doubled our input energy, and also our momentum.

Each 1 kg mass now has half that KE and momentum, so 96.23 J and 9.81 kg-m/s each.

And per KE=1/2mV^2, at 9.81 m/s a 1 kg mass has 48.11805 J precisely... which is also precisely half of our input energy!

So after two successive reactionless accelerations, using gravity to fully cancel the counter-momentum each time, we have spent exactly 192.46 J, to create momentum with a net value of precisely half that much.

So at two interactions in series, we're now 2x under-unity.

So now a third serving:

3 * 96.23 = 288.69 Joules in, for a momentum rise of 3 * 9.81 = 29.43 kg-m/s, divided between two 1 kg masses = 14.715 m/s of velocity each, and 144.345 J each.

But per KE=1/2mV^2, at 14.715 m/s 1 kg has just 108.265 J.

144.345 / 108.265 = 1.333, so we're now 30% or 1/3 under-unity after three consecutive cycles.

On to four, then - the equaliser:

4 * 9.81 = 39.24 kg-m/s, and 4 * 96.23 = 384.92 J

Divided by the two 1 kg masses, each has 19.62 kg-m/s and 192.46 J.

And, per KE=1/2mV^2, at 19.62 m/s a 1 kg mass has precisely 192.4722 J!

So we've hit unity after four successive generations of 9.81 kg-m/s each time, for a cost of 96.23 J each.

Our momentum is now worth precisely the energy we've paid for it.

Which, you may also notice, implies we can actually skip steps 1 - 4 entirely, and propel the system directly up to this threshold momentum and energy level by any other, conventional, means, ready for the unity-busting interaction:

So instead of simply considering this the fifth interaction in the series, we could manually crank the system up to 39.24 kg-m/s, using a 384.92 J PE store of any kind, and be in precisely the same condition in which we end up after four reactionless accelerations..

This would seem to massively simplify the requirements to manifest OU... all we need is that PE to produce that standard, N3-compliant amount of momentum, and then, with the system thus teetering on the brink of sanity, we generate that 'fifth' 9.81 kg-m/s of momentum using our super-secret special gravity trick, for a further investment of 96.23 Joules..

With me? OK let's tot that up then:

So we input 4 * 9.81 = 39.24 kg-m/s, and 4 * 96.23 = 384.92 J, either via these reactionless accelerations, or else, just by dropping a brick or summink, to produce the same momentum and KE from GPE. Whichever method's more convenient; both end results, identical.

So regardless of how we got the system to this state - 39.24 kg-m/s, and 384.92 J - whether we accomplished it with or without N3, makes zero difference.

So whether it's our fifth in an identical series, or our first, following an equivalent but entirely-conventional PE output, let's add another 9.81 kg-m/s of momentum to the system, for a further 96.23 J, and see how it ends up:

39.24 + 9.81 = 49.05 kg-m/s, and 384.92 + 96.23 = 481.15 J

So 5 * 9.81 = 49.05 kg-m/s, and 5 * 96.23 = 481.15 J, if we got here via five successive interactions, yet equally, we arrive at precisely the same outcome if we simply 'summarise' the net result of the first four with just a basic spring or GPE load.

Again, we divide the momentum and KE between the two 1 kg masses, so each has 240.57 J, and 24.525 kg-m/s.

But per KE=1/2mV^2, at 24.525 m/s, each 1 kg mass has 300.7378125 J !!!

Divide that my what we actually spent and we get 1:1.25 CoP - we're 125% OU, or "5 quarters" OU, or however one wishes to express it..


So one single asymmetric inertial interaction is sufficient to cause robust gains, provided the system is prepped to the threshold break-even point by any conventional means..


See what i mean about an inexorable conveyor belt of convergent simplicity? The most elegant solution to all this is right there in the maths, if we just follow them to their logical conclusions..

This would seem like a cool thing to double-check with a sim! Maybe over the coming weekend i'll try it out..

But in principle it could suggest that each individual mechanism we add to the wheel could perform one of these 'reduced' 5/4 gains, instead of climbing up to peak efficiency over a rising RPM range, from an initially under-unity output, we might be better off going for a fixed gain per interaction, with multiple but identical gains per revolution...

Such a machine would also want to accelerate straight up to its coasting speed, determined by that threshold-state pre-load condition.

We wouldn't have to tolerate being UU or even at unity, for a single interaction, let alone a single cycle...

Anyone still following..? Impeccable logic, or the rantings of a madman?