Plying CF as pseudo-inertia to scam N3

[MrVibrating]

To begin: your Energy values are useless... because energies only make sense when they relate to something, being relative to something.

LOL they're the energies of the masses relative to stationary (ground).

Your top weight may look stationary relative to the ground, but it's actually not and accelerates away from the CoM. It hoovers in that upright orientation.

Sorry, but non-sequitir? It hovers, relative to the ground. It does not change height. It has zero energy or momentum. It begins and ends with the same GPE. It has been used as reaction mass to raise a momentum.

Relative to the ground (the frame that counts, for now), it does not accelerate away from the CoM, or anything else; it does not move, period. The CoM accelerates down, away from it, however the CoM is not a particularly relevant factor here, since the jack is notionally massless (it could equally just be a pulley system), and only the lower mass is moving, and would thus need relifting..

Stationary "relative to the ground" is the only reference that matters, for all the motion here, or lack thereof..

When some bird thinks that top weight is a nice spot to land on, because from his perspective it's indistinguishable from any other pole, then it'll find itself a bit on unstable ground: the whole thing+bird will accelerate (<1G) downwards.

..pretty much.. i think. We could test this in sim tho - only, i'm thinking, won't the bird also have to accelerate the 'pole' system downwards against its inertia? To my addled thinking, it might be more akin to a falling mass that is also pulling another mass sideways, or spinning up a wheel or whatever, so while the bird will experience a 1G force, its acceleration will be determined by A=F/m..

Because things are in an accelerating Frame of reference, there's not a constant relation but a changing relation - hence that bird (while at rest after his super soft landing) shows a fictitious force (acceleration), which was already there but zero'd out.
At last that's my reasoning...

Only half-following this point - yes it's a non-inertial frame, albeit unconventionally. An accelerometer on-board will detect the net system accelerations. The sphere of fixed stars will turn into an accelerating tunnel. Maybe the "there but zero'd" bit will sink in after i go to bed, tired..

Your actuator accelerates two weights (m&#8321;=1 kg, m&#8322;=1 kg) with a&#8336;=19.6133 m/s² during &#8710;t=1 s.
Because m&#8321;=m&#8322; they'll split the available force equally and both accelerate from their common center (CoM) with a&#8321;=a&#8322;=a&#8336;/2 = 9.80665 m/s²

hmmm kinda - but would accelerometer readings support that interpretation? Not from the ground frame.. only the lower mass actually accelerates, and possesses all of the velocity.

Their velocities at t=1 will be v&#8321;=v&#8322; = a&#8321;·t = 9.80665 m/s

Again, this is not what happens, reposting the sim here for easy reference:



...all the momentum and KE's on the one mass, matey..

Their distance from the CoM increases to d&#8321;=d&#8322; = ½a&#8321;·t² = 4.903325 m

yes, but again, that's just an arbitrary point in empty space. It has no mass. We're not counterbalancing the system or anything - the CoM doesn't seem a useful reference for anything at this stage..?

Their energies are both ½m·v² = m·a·d = 48.0852 J (total=96.17038)

Again, that's not what happens with gravity in the mix is it?

That's relative to their common center, no matter the orientation in freefall or happening in freespace (there that "bird" may be some space debris, but that would likely be a collision).
Your freefall (in this case) just doubles the total energy amount, but isn't relevant.. or so I think.

..again, the only objective reference frame here is earth / gravity's / the static frame. The energy and momentum contributed by the freefall is indeed irrelevant, and a zero-sum deal. It was simply a necessary formality in order to facilitate the asymmetric inertial interaction, which, relative to the ground, given that neither it nor the upper mass are actually accelerating upwards, has resulted in a net rise in momentum, between those interacting masses, relative to the ground..

Better* to just have an actuator on the ground and shoot 1 kg up with 19.6133 m/s² (hopefully there's no bird around).
Or connect it to a bridge spanning a 10m deep valley and shoot 1kg down with 2G; you can always ditch the other 1 kg if you really want.
*) yeah well now you've made me doubt my own point... but it should be the same situation as your demo.

[...] nothing i'm doing here is ever going to make sense to you..
But it's cool, you know i appreciate your interest..

Maybe true... I'll try to shut up for now. And attempt to watch from a distance. (Can't promise)
:-)

Aha, you're wrestling the same issue as Sleepy - why not just accelerate one mass against Earth? Same answer i gave him - the closed system of interacting masses has to be free to accumulate net momentum from the N3 exploit in order to be useful... otherwise you'd just have a reactionless propulsion system. But if you wanna get energy and work out of one to drive an attached load etc., the system has to be free to accelerate. Basically, the same reason Bessler's wheels were all statorless, and why he insisted this was a necessary condition for any "true PMM"...

Evidently, the "true PM", to him, must refer to a runaway momentum / divergent non-inertial frame / bleedin' N3 break, innit.. :P

Whatever, just give it another few days - it's either gonna crash'n'burn this weekend, or else take flight.. If i do have a reactionless rise in momentum then i should be able to rectify, consolidate and accumulate it over multiple cycles. That's the acid test..

[eccentrically1]

The actuator in your sim is magical thinking. It doesn't matter if you run it in a linear mode or in an angular mode. Imagine trying to build it. At some point, the wheel is going to have to use some of its momentum to reset the actuator for the next firing, using whatever mechanism you think is best. A rim backstop, whatever. In the real world, there won't be enough momentum to fully activate it. The next firing will be less springy than the first one, etc.

[MrVibrating]

The "actuator" is purely a visual reference, at this stage.

I understand the conceptual hurdle. If i say "scissorjack" or whatever, it conjures an image of a literal scissorjack, a big heavy squeaky lump of high-friction mechanism, which obviously wastes energy, has a non-linear angular-to-linear acceleration, and has to be physically collapsed after extending, or vice versa. WD40 everywhere. Nightmare..

But all that really matters is the relative mass accelerations on either end of the jack; specifically with regards to the balance of momenta. Maybe instead of a jack we use pulleys, or rack and pinion or whatever - the practicalities of the transmission system is obviously tangential to whatever the distribution of momentum, so long as we're not attempting anything non-physical.

Or, perhaps that is exactly your point - that the accelerated masses must in turn be similarly decelerated when the actuator mechanism resets? Not necessarily.. i mean, not if they're not physically tied to the actuator mechanism, but only impelled by it via collisions, say.. alternatively, interactions might be daisy-chained around the perimeter, each mass in a pair alternating between roles in the reactionless acceleration, and swapping onto the next partner each successive interaction..

Whatever, that's jumping too far ahead. For now, all that matters is whether or not we can use gravity like this to generate reactionless accelerations which gain net momentum at a fixed, speed-invariant cost.

If so, then whatever the inefficiencies and losses, beyond some threshold speed the net output KE plus losses > input PE. It is simply a matter of fact that a reactionless acceleration launched from a prior such acceleration leapfrogs up the 1/2mV^2 multiplier, creating momentum, and from it, kinetic energy. If we can do that, we can afford all the losses we like, possibly even using an inferior brand of WD40.

[MrVibrating]

OK so i just knocked up the linear-to-angular version, here:



..on the left is the previous interaction, and on the right it is duplicated, but with the lower mass replaced with a massless ripchord, that spins up a flywheel.

The flywheel's angular inertia is identical in magnitude to the linear inertia of the upper mass (and likewise that of the lower linear mass it replaces).

So as before, the upper mass levitates, accelerating upwards and downwards at equal rates.

And the flywheel spins up, gaining juicy succulent momentum..

And sorry folks, but this is really the clincher:

Relative to the static frame (the only relevant one here for now), the flywheel is in motion, and the linear mass isn't... because it was held back by gravity, but the flywheel wasn't.

Net momentum has thus increased. The input energy is half what it was, because now we haven't dropped any mass downwards, so don't need to re-lift anything.

Yes, in this sim the lower end of the actuator still plunges down and would need relifting, and wouldn't be massless in real life.. but it's just a notional transmission system between two inertias, and could be replaced by anything you like, provided it simply commutes the required mutual acceleration.

So, anyway, again, the energy input, via the linear acceleration, is equal to the GPE the mass would've output, had it fallen that same height as the lower end of the jack...

[WaltzCee]

Is this the principle of this N3 scam?

[MrVibrating]

N3 = Newton's 3rd law.

An N3 scam would result in an asymmetric distribution of momentum between two interacting inertias / masses.

As a result, the 'net system' of those two interacting masses gains a net momentum from that asymmetric distribution.

Example: suppose i'm floating in space, holding a heavy lump of mass. If i throw it forwards, i'll also be propelling myself backwards. A stationary observer would see the thrown mass accelerate in one direction, and me accelerate in the opposite direction. The acceleration of each will be proportionate to the relative masses, so that the mass times change in velocity of each will be equal - the smaller mass will move faster, the larger mass slower, but the product of mass times velocity for each will be equal.

Because each mass is accelerated in opposing directions, the net momentum between them remains constant. If they were connected by a spring, they'd bounce back and forth indefinitely, without any net change in position.

If both masses were stationary relative to the observer when the impulse was applied, then each will be accelerated in opposite directions relative to the observer, so the two induced momentums are of opposite sign, positive vs negative, and thus their sum - and the net momentum of the system - is zero.

If on the other hand, both masses were initially moving at equal speed relative to the observer, then he sees one mass decelerate, and the other, accelerate by an equal opposite amount.. again conserving net momentum; what one gains, the other loses, so no net change in system velocity.

So N3 is why we can't have reactionless propulsion, and instead, need to carry, and eject, propellant, even if we had a free energy generator aboard our spaceship with which to accelerate that propellant.

You probably know all of this already, however there's another side to N3 that probably isn't so well-known, which concerns how the above principles affect the energy cost of momentum...

If you've ever calculated the energy of a moving mass, you'll've used the formula "kinetic energy = 1/2 mass times velocity squared". Well, what this expression is actually doing is setting the energy price of accumulating momentum.

Let's consider an example: suppose we have a 1 kg lump of mass, moving through space at a variety of different speeds; we'll use the KE=1/2mV^2 formula to calculate how much kinetic energy a 1 kg mass has different speeds, and thus, how much input or output energy is involved in the mass accelerating or decelerating between those speeds..

- at 1 meter / sec, a 1 kg mass has (1/2 * 1kg * 1m/s * 1m/s) = 1/2 a Joule

- at 2 meters/sec, 1 kg has (1/2 * 1 * 2 * 2) = 2 Joules of KE.

..so the difference in momentum between these two speeds is 1 kilogram-meter per second - and it costs 1/2 a Joule for the first kg-m/s, but then four times as much - 2 J - for the second identical rise in momentum.

Continuing on..

- at 3 m/s, 1kg has (1/2 * 1 * 3 * 3) = 4.5 J

- at 4 m/s, (0.5 * 1 * 4 * 4) = 8 J

..and so on. Fast forward to:

- 10 m/s - (0.5 * 1 * 10 * 10) = 50 J..!

..and so the energy required to increase the velocity of a given mass, by an equal rise in momentum each time, inflates by the 1/2 square of rising velocity. In short, it takes more work (input energy) to achieve the same increase in momentum, the more momentum we already have. The faster we go, the more expensive further acceleration becomes.

If however we could accelerate a mass, raising its momentum, without having to also pay for an equal opposite change in momentum, then two important things happen:

- the system's net momentum will change, purely from these internal interactions, and without the application of any external force

- the energy cost of momentum is slashed. If N3 symmetry can be entirely broken (as i believe i'm demonstrating here), then whatever the energy cost of the resulting acceleration, it remains constant, irrespective of rising speed, and likewise, every successive burst of acceleration has the same cost as the first, regardless of whatever velocity is reached.

In principle, then, the absolute minimum cost of momentum is simply 1/2 a Joule per kg-m/s - the initial rate set out by KE=1/2mV^2. So an ideal N3 break would fix that cost, no matter how much momentum we bought at that energy rate.

Hence, we could buy that 10 kg/m-s of momentum (1kg at 10 m/s), for the bargain price of just 10 * 1/2 a Joule = 5 Joules.

Yet, as we saw above, the normal energy cost of 1kg @ 10m/s is 50 J... 10x more than we've actually paid for it.

So at this point, we'd be 10x over-unity. If we'd input that 5J by dropping a mass over some height, we'd now have enough energy to re-lift ten times that much again.


When we compare the performance of a N3 symmetry break, to that of a notional gravitational asymmetry - suppose we could lift a weight for half its usual GPE, and drop it again at full value.. because GPE is purely a function of gravity times mass times height, irrespective of speed, such a system would have a fixed efficiency, outputting the same excess energy each cycle, regardless of speed. So, if we could raise say 10 J of GPE for just 5 J cost, we'd be 2x OU, and gain 5J excess energy each cycle.

Whereas, the efficiency of an N3 asymmetry increases the longer it goes on, as net velocity accumulates. Because inertia is not speed-dependent - a mass has the same resistance to a 1kg-m/s acceleration, regardless of how fast it's going - the input energy cost per cycle stays constant, just like the gravitational asymmetry. However, the output energy per cycle scales up with rising velocity, per KE=1/2mV^2, and so we'd have a fixed cost of purchase of momentum, but with an escalating real-world market value, when we came to cash it in..

The TL;DR in all this is:

- if you're marooned in space with nothing but a bat and ball, and the ball is attached to the bat by a long elastic chord, then you could maybe help ease your passing by burning off useless calories until your O2 expires, but you're never going to get anywhere no matter how much effort you put into your batting, because N3. Your net system of batter + ball will simply wobble back and forth pathetically.

- if however you'd mastered the art of zen tennis, and could somehow bat the ball without propelling yourself backwards, say, by controlling you chi or whatever, then your net system of batter + ball will start to gain momentum, accelerating a little each full cycle, you could then vector that thrust towards earth, button up your tunic and prepare for re-entry, frantically batting the ball downwards to brake your acceleration by gravity. Safely arriving back home at your desk, you'd calculate that the total energy you've expended was significantly greater than the calorific content of whatever you'd eaten for breakfast that morning... yet you wouldn't even feel particularly peckish.

In the case of the spinning basket ball, the dimensions become angular rather than linear, but the principle's the same - each time the guy flicks it with his finger, imparting torque, he has to flick it ever-harder the faster it spins in order to impart the same rise in angular momentum, precisely because the surface of the ball is gaining speed relative to his stationary body and finger. Obviously, he's also counter-accelerating the Earth in the opposite direction, by a tiny amount, but that equal opposing counter-momentum adds to the ball's momentum of opposite sign, and so the net change in momentum of the system is zero.

If however he could spin up the ball without applying counter-torque, then he'd be able to do so while standing on a turntable with perfect bearings, without spinning himself up in the opposite direction. Since the counter-momentum is thus zero, the net system momentum has risen - if he then grasps the ball to brake its spin, without cheating N3 anymore, then the ball's angular momentum would be shared / equalised with his and that of the freely-rotating platform he's standing on, and duly spin up a little..

..and just as in the linear case, the rotational kinetic energy value of angular momentum is set by the standard term "RKE = 1/2 angular inertia times angular velocity squared", hence the more we buy, and faster we wanna spin, the more input energy required to achieve a given rise in momentum, due to the ever-rising speed difference between the rotor and stator against which torque is being applied..

..again, the result of an angular N3 symmetry break is the same as for linear; reaction mass, ie. the stator, here, is able to be tethered to, and accelerate with, the accelerated mass of the rotor!

So in a nutshell, when we have an asymmetric distribution of momentum from an interaction between two free masses / inertias, the net momentum between them (ie. the net system momentum) can be increased or decreased from within the system, without the need for an externally-applied force, and furthermore the energy cost of momentum purchased via this method circumvents the practical causes of why that cost normally escalates by 1/2mV^2 the more momentum we buy, instead amounting to a fixed cost that does not change with rising net system velocity and thus momentum.

Since the energy value of momentum outside the accelerated system, where N3 still applies, remains a fixed function of 1/2mV^2, what we've effectively done is slash our input energy costs, while still benefiting from the normal, full energy value of the budget-priced momentum we've generated from thin air.

In all of mechanics and classical physics, this is the only possible consistent route to mechanical over-unity, and explanation for Bessler's success. Asymmetric gravitational interactions are impossible because the force fields involved, gravity and rest mass, are termporally-invariant, and so GPE is not a function of speed, and the integral (net product) of input force * distance will always be symmetrical to the output F*d integral when plotting a complete cycle. Leverage, by definition, merely trades 'F' for 'd', equitably and symmetrically. It is thus impossible to 'drop a weight when it's heavy, and pick it up when it's light'.

Over-unity is only possible for systems in which energy has a time component. Kinetic energy evolves by the half-square of inertia times velocity because input force has to rise to compensate the rising velocity difference between an accelerated mass and its reaction mass. In other words, standard inertial interactions, wherein two masses / inertias push or pull each other or collide with one another, are fertile ground for seeking an input / output asymmetry. The 'squaring' function suffixing the KE / RKE terms is the second time-derivative of motion, ie. acceleration; rate of change of position with respect to time. Conservation of mechanical energy (CoE) depends upon conservation of net momentum (CoM or CoAM for angular momentum) because the 'acceleration' component of force (F=m/A) - rate of change of position WRT time - subtracts from that of the accelerated mass in relation to that of its reaction mass, and so has to be compensated with an increasing force or displacement (distance or angle through which the force is applied), in order to maintain a constant change in acceleration.

And so, just as there is nothing you can do in a gravity field to create energy or momentum, likewise, there's nothing you can do with inertial interactions either. Mechanical over-unity is categorically impossible - written in stone into the very terms of the definitions of the respective field properties. You simply cannot make 5 from 2 & 2.

But....

...combine both sets of fields together in overlay, and 'gravity' is of course a downwards acceleration of 9.81 m/s^2 (again, the squaring function represents the 2nd time derivative; velocity in freefall increases by 9.81 meters per second of fallen height, for every second you continue falling).

Likewise, it is functionally equivalent, in many respects, to an upwards acceleration of the ground of 9.81 m/s^2.

Hence, everything around us is already subject to an effective 9.81 m/s^2 acceleration.

Hence, if we fire our reaction mass in the opposite direction, under the same acceleration, its net momentum, relative to us here on the ground, will remain constant, because there is no net acceleration being applied.

The opposing mass, which is being accelerated, is thus gaining all of the motion being imparted by pushing against the other mass, levitating it against further acceleration or deceleration.

If this opposing mass is being thrust downwards, as in my first example, then it is also using GPE (gravity times mass times height). This GPE converts to KE and momentum, and can be calculated by 1/2mV^2 for KE, and P=MV for momentum. Re-lifting the mass again, back to its initial height, will thus use up precisely that KE and momentum on the way back up.. leaving just the momentum and KE imparted by the applied acceleration.. all of which still resides on one mass only. If they were subsequently tethered together, much like the elastic between the bat and ball in the earlier example, then that momentum gets shared equally between the two masses, and thus their net momentum - that of the net system of interacting masses - has risen.

In my second example, the need to thrust an opposing mass downwards is obviated by pushing against the angular inertia of a wheel instead. This cleans up the picture by eliminating the GPE interaction and its associated momentum exchange, and we simply have a force being applied between two free inertias, that is accelerating only one of them; hence an asymmetric distribution of momentum, and net rise.

However, you may also notice that both examples push the world down...

..the whole planet is being pushed away from the upper mass.. a little more, by equal increments each successive cycle, as the net velocity, and thus OU efficiency, of our accelerating system increases...

You could maybe site generators centered at opposing points on different hemispheres... this would avoid changing Earth's net momentum, but while also applying a squeezing force.. that should keep Earth's momentum constant, and amount to full-blown classical OU - creation of momentum and energy ex-nihilo... though God know's what'd be happening WRT the quantum vacuum from whence these forces are borne..

But anyhoos, if you followed all that, then you now know as much as i do about the solution to Bessler's wheels; only an N3 break even stands a chance of replicating his success, the momentum gain is not environmentally-friendly, and definitely will destroy the Earth as we know it, if it was deployed en mass without controls. It would be impossible to buffer the Earth against the change in momentum from a single system by means such as gimbals, since the force and displacement concerned is vertical, and the system can't get progressively lower each cycle, and only carefully coordinated counter-balancing of multiple such systems would have a chance of cancelling their net influence, but which in turn, merely shifts the momentum and energy deficit out of the classical realm and into the quantum..

So basically, yes - just like spinning up a basket ball, give or take a few details.

[ME]

If you've ever calculated the energy of a moving mass, you'll've used the formula "kinetic energy = 1/2 mass times velocity squared". Well, what this expression is actually doing is setting the energy price of accumulating momentum.

Well, that expression can't be taken out of context of a moving system as some entity on its own. I suspect that's what you constantly try to do and get surprised when there may be some increase somewhere.
The PE=KE relationship is simply a consequence which emerges mathematically from some object going from one place to another under the influence of an acceleration.
We may multiply by mass to weigh the relationship during interactions. It's useless out of context (or frame of reference), and arguably an overrated quantity within the context, but just a nifty shortcut for calculating the transfer of momentum.

[MrVibrating]

...it's simply the product of force and displacement, and the singular quantity we wish to increase in any prospective OU system.

The "context" of a KE measurement is simply the rate or change in motion of a mass relative to a given reference frame, but which is also equitable with all other possible reference frames, due to the conservation of momentum and, ultimately, Newton's 3rd law..

When N3 is broken, the system's reference frame "diverges" from all other possible frames, gaining momentum and energy relative to them.

The question you need to challenge my friend is why does KE=1/2mV^2 - essentially the point i'm labouring is, why does momentum of a given inertia get more energy-expensive / valuable as velocity rises? Why doesn't energy scale linearly like momentum - basically, why do we have two quantities , P and KE, rather than just P as Newton originally contended?

Why does the potential to perform work, in terms of accruing Newton's vis viva, depend upon Leibniz's vector quantity?

The answer, as i've expounded and espoused ad nauseam, reduces to the practical mechanical implications of Newton's 3rd law.

If Newton's 3rd law is violated, then momentum can be raised from within a closed system of interacting masses, and the energy cost of purchase of that momentum is slashed, hence we reduce our input energy costs while maintaining the objective energy value of that raised momentum with respect to the static frame of Earth and gravity..

In essence, the momentum of a divergent frame is no longer a scalar with equal opposite components, but becomes vectored, just like output KE but without the 1/2mV^2 accumulator. Or, to put it another way, input KE, sans that 1/2V^2 exponent and measured from within the divergent frame, takes on the same basic dimensions as momentum, evolving linearly, while output KE, as measured from the external, static frame, remains bound to its usual dimensions, hence input energy < output energy.

There's nothing remotely mysterious about the maths. The goal remains to exploit an effective violation of N3 to raise reactionless momentum, thus avoiding the usual half-square accumulator on our input energy costs, while still cashing it in at full value in relation to the fixed reference frame of gravity and GPE.

In summary, the input and output energies are in separate reference frames, the former diverging from the latter.

Sure it's a bit of a head-fuck, but hey don't try understand, just calculate eh? Examine the maths, and then try to interpret and resolve what the results are saying..

But you're never going to succeed as a hefalump hunter if you steadfastly refuse to open your eyes and ears and really examine the life-cycle, habits and territory of this magnificent invisible species..

If you want OU then you're looking for a symmetry break, a system with more output than input energy / work.

Energy / work = force * displacement.

OU means excess energy / work from a closed-loop trajectory - ie. equal input and output displacements.

Hence if 'd' must be constant but E / W must not, then by simple elimination, F must be variable. F cannot be constant.

Furthermore, it has to vary passively, so that the cost of changing the force is less than, or decoupled from, any resulting excess output work that is so accomplished.

That's just the elementary facts of OU; I/O displacement must be a closed-loop, energy is simply force times displacement, hence OU implies a passively time-variant force; here, fulfilled by applying gravity as pseudo-inertia, or pseudo-counter-acceleration, depending on your point of view.

Because this does appear to interfere with Earth's momentum, it might be preferable to return to the original thread hypotheses, and see if i can replace gravity in the current examples with centrifugal force..

In the meantime though there's still one or two other things i wanna try with gravity..

[ME]

The question you need to challenge my friend is why does KE=1/2mV^2 - essentially the point i'm labouring is, why does momentum of a given inertia get more energy-expensive / valuable as velocity rises? Why doesn't energy scale linearly like momentum - basically, why do we have two quantities , P and KE, rather than just P as Newton originally contended?

I was actually counter challenging you. :-)
I'm basically saying that "energy" does not exist, but a simple and fundamental mathematical consequence.
The "removal" of energy gives rise to other fundamental questions, so that would likely become a slippery slope discussion...(let's not go there)

When you simply accept the existence of Force and Momentum then the whole concept of Kinetic and Potential Energy is basically unnecessary and questions like"what does KE" or "what does PE" stop making sense. PE/KE only makes life a bit easier.

I really like your brainstorms, but it's hard to read where your hypothesis ["IF N3 breaks then..."] starts and where your conclusion ["see here N3-broke"] is a either reminder of your hypothesis or some actual proof.

That's my reason I'd like to insert the notion that Energy is a consequence of displacement and not a malleable property. If things still happen, or could happen, as you claim, then it's all the better. That's all.

If Newton's 3rd law is violated, then momentum can be raised from within a closed system of interacting masses, and the energy cost of purchase of that momentum is slashed, hence we reduce our input energy costs while maintaining the objective energy value of that raised momentum with respect to the static frame of Earth and gravity..

I fully agree. For example: I'm able to recreate Silvertiger's simulation by doing just that. So it does have my interest.

[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.

[MrVibrating]

Ok here's an angle on it - simply Newton's laws:

- net momentum of a system can only be altered by intervention of an externally-applied force; otherwise, a force applied between two stationary masses can only cause equal opposing accelerations, keeping net momentum constant, and a force applied between two moving masses will decelerate one while accelerating the other, net momentum again remaining constant.

And so system momentum remains constant unless acted upon by an external force.

Here, gravity has counter-accelerated our reaction mass.

System momentum has thus risen.



- Second law: A=F/m - the upper mass is subject to equal 1G accelerations of opposite signs, so conforms to the first law, it's momentum remaining constant.

The lower mass is subject to 2 * 1G forces (1G from gravity plus the 1G internally-applied force), both vectored downwards, so accelerates downwards at 2G.

Alternatively, with the lower mass replaced by an angular inertia, it is now only subject to the 1G internally-applied force, albeit converted to torque, and so accelerates, spinning up. The upper mass, as before, remains subject to equal 1G forces of opposing signs, so still does not accelerate.

3rd law - the counter-momentum from our internally-applied force has been subjected to an equal opposing deceleration by gravity, so although the counterforce was manifested, the corresponding counter-acceleration and thus counter-momentum never transpired. Thus all of the momentum from the internally-applied force resides on just one of the interacting masses.

So on the one hand, the 3rd law has been entirely respected, however, in combination with the first two laws, and in juxtaposition against the effective ambient acceleration of gravity, the net effect is that the usual outcome of the third law is effectively circumvented - the externally-applied force of gravity does not need to perform net work - in terms of mass displacement - in order to skew the distribution of momentum and break N3... it doesn't even need to entirely brake or cancel the counter-momentum - it merely has to attenuate it, however slightly.. any vertical inertial interaction while falling, however slight the gravitational deceleration of the counter-momentum, will result in a corresponding momentum asymmetry and thus net rise.

So what i've shown here are idealised, extreme examples, to accentuate and simplify the underlying principle. In practice however, a clean and perfect 2G applied force is unnecessary, and the messy non-linear curve of a spring constant would suffice. The gain in net system momentum is simply equal to the effective deceleration of the counter-momentum, however shaky or incremental that margin; so long as friction's kept low enough, the gained momentum will accumulate over consecutive cycles..

But until i can demonstrate this it's all hypothetical.. i'm currently hashing together a rough'n'ready attempt at a closed-loop cycle, probably won't work, will share results here when done, but i'm thinking the real acid test will be showing that the input energy cost of the internally applied acceleration remains constant as the starting speed of the system is increased. If that checks out, then closed-loop gains are just a matter of engineering..

[MrVibrating]

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..

[eccentrically1]

You’ll get there eventually.
Would any other builders care to comment? Teachable moment!

[cloud camper]

Not trying to burst any bubbles here but we do introduce a couple of issues with this scheme.

First the upper mass that becomes stationary to the rotating wheel effectively disappears as a balancing mass on the descending side of rhe wheel. This unfortunately causes a countertorque to the wheel since the ascending side of the wheel now becomes "heavier".

Since a mass experiencing 0 G is not moving downward giving up GPE, it
has then "disappeared" as a balancing mass during the time it is stationary.

Second, the 2G downward accelerating mass has also disappeared as a balancing mass, producing a second countertorque. Since the downward accelerating mass is for all intents and purposes in free fall relative to the wheel, there is no balancing mass during this time interval either hence the wheel will try and reverse direction.

When the 2G weight does impact the wheel rim, momentum will be recovered but the wheel will have slowed considerably or even stop during the unweighted interval.

Since both the 0G stationary mass and the 2G reaction mass unweight the descending side of the wheel temporarily, we then temporarily induce a large countertorque on the wheel.

A possible way out of this dilemma would be to design a mechanism that is not a wheel but only drives a wheel as output.
This eliminates the requirement that the wheel be mass balanced at all times.

The devil always resides in the details!

[MrVibrating]

..The system i've been playing with tonight kinda fits your description:



..so a force is applied between two inertias, one angular the other linear, the former spins up against the weight of the latter, which itself is suspended against the angular inertia of the former.

What i was thinking was to basically replace the actuator with a spring that could be re-compressed on the ascending side, using the momentum from the flywheel.

Don't think it'll work though as it seems too simplistic..

Your concerns about underbalance are noted - however if the upper weight does need to eventually drop and ride back around the wheel (because it could, instead, just hang from a short chord fixed in place, waiting for something to come along and push up against it), then it'll be outputting GPE as it does so, and since we wouldn't expect to lift more weights than we're dropping, i believe all input and output GPE's will be equal, because we're not performing any net work from gravity...

..that is, the N3 break doesn't draw and energy itself, so all we pay for is the mass accelerations and GPE - both of which can't output any less than input, minus losses.. the only work gravity does is negative - limiting or preventing the acceleration of our counter-momentum during the inertial interaction.

Basically, since we're not drawing off GPE to actually accelerate any other mass inside the system, all of it remains available to the descending side of the wheel, in balance with its cost on the ascending side.

All of the input energy for the inertial interaction is instead provided by some form of pre-loaded PE, such as springs i guess.

The other point to stress is simply that the potential gains from an N3 break are far beyond the energies and momenta we're used to dealing with in GPE interactions, and so have the potential to easily handle a little underbalance here or there.. remember, at a fairly moderate edge-speed, for a large diameter wheel, of around 1 meter / second, a further internal reactionless acceleration of 1 meter / sec is potentially between 4 and 8 times OU.. whereas in GPE interactions we get excited about micro-torques... nah mate, if this works it is not going to pussyfoot. Think bigger!

Input and output GPE can only sum to unity here. The prospective gain mechanism is purely discount momentum from asymmetric mutual accelerations.

To understand it you have to think in two reference frames at once - the internal, divergent frame, relative to the external static one. Invest in the former, cash out on the latter.