Plying CF as pseudo-inertia to scam N3

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MrVibrating
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Post by MrVibrating »

eccentrically1 wrote:After the actuator fires, the 1kg upper mass will have to dragged around like you say. That is the reaction, just delayed. The lower mass, or an angular replacement, will begin to lose momentum at the end of the actuator travel as the upper mass's reaction kicks in. Then, as always, the next question is how much does it cost to reset the actuator.
Looking forward to the results.
What happens when it's all mounted to a rotating wheel is getting a bit ahead for now, but on the face of it, i consider it a foregone conclusion that all GPE outputs and inputs sum to unity.

The form of gain here is very specific - with an effective N3 break, we can buy momentum at a fixed energy rate per unit from within the accelerating system. Each asymmetric inertial interaction leaves an accumulating excess of momentum, and so each subsequent interaction begins at a progressively-higher starting velocity. However the cost of operation of the asymmetric inertial interaction is speed-invariant, so it doesn't increase with rising velocity. From the external reference frame however (ie. us observers, outside the system) the value of the energy being expended on-board is being inflated by the ever-rising ambient momentum of the system inside which those interactions are being applied.

For example, suppose the internal acceleration applied by the jack is 1 kg by 1 meter / sec, then, per the standard KE term (KE = 1/2 mass times velocity squared), it is performing just 1/2 a Joule of work.

If however the system is already moving at some speed, say 1 meter / sec, then the on-board acceleration of 1 m/s brings its net momentum, relative to us stationary observers, up to 2 m/s.

Normally, due to N3, net system momentum is conserved, and the on-board forwards-acceleration will decelerate the net system by an equal amount, keeping net momentum constant. This is why you're not supposed to be able to change a system's net momentum from within that system - an external force must be applied.

And so while the on-board mass is accelerated by 1 m/s, the net system decelerates by the same amount, and only the relative velocity between the system's on-board components changes, ie. its internal distribution of momentum. Externally, the net momentum's constant, and us static observers see a net 1/2 m/s acceleration of the on-board mass, along with a 1/2 m/s deceleration of the net system.

On-board, the work performed was still a 1 kg-m/s acceleration, at a cost of 1/2 a Joule, but the net system momentum remains constant, and 1/4 of a Joule is spent accelerating some of its mass, while the other 1/4 of a Joule was spent decelerating the corresponding counter-momentum.

With an effective N3 break however, the net system is not decelerated - because we fiddled our way out of inducing the corresponding counter-momentum in the first place!

And so the static observer sees the on-board acceleration of 1 kg-m/s, at an on-board cost of 1/2 a Joule, as an acceleration from an initial 1 m/s, up to 2 m/s. And at 2 m/s, a 1 kg mass has 2 Joules - an increase of 1.5 J from the 0.5 J it began with..

And so while only 1/2 a Joule has been spent internally, externally, the system energy has risen by three times more than the internal work performed.

If we had some kind of accurate KE sensor sat on our desk monitoring the experiment, it would record an impact energy of 2 Joules. If we calculate the mass's energy as a function of its speed relative to us, we get 2 Joules. So the mass quite objectively has 2 Joules. Yet, the system only performed 0.5 Joules of work...

That is the gain principle, and yet to be demonstrated.. What we have here, so far, is an apparent N3 break - the secret sauce. The actual stir-fry comes next, but all the gravitational interactions will be a zero-sum game - the same amount of mass will rise and fall against the same gravity field, for zero net cost or benefit, besides skewing the distribution of momentum during the inertial interaction, and hence causing an N3 break and net momentum rise.

Likewise there's gonna be frictional losses etc. in a real-world build, but if you extrapolate the asymmetry i've described to higher velocities, you'll quickly see that the potential gain margins are many times greater than the potential entropic losses of even the shoddiest build..

Bottom line is that either this is an effective N3 violation or it isn't. If it is then we're in business. So far, i'm still confident.. moreso than ever in fact..
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

sleepy wrote:As I mentioned earlier in this thread,the energy it will cost to fire the jack will far outweigh any surplus energy gained in the reaction.I know that's a pesky detail,but it's not going away.It's like running up a series of falling rocks.The runner will stay still relative to the horizon but the rocks will accelerate a little as the runner steps on them.My legs are burning just thinking about it.
Essentially, the jack can be eliminated from the core concept - a force is applied between two masses. The work performed by that force is equal to the force magnitude, multiplied by the mass, multiplied by the resulting displacement or change in velocity.

In the scenario demonstrated above, an acceleration equal to twice that of gravity is applied between two masses. Due to the opposing counter-acceleration of gravity, this results in one mass remaining motionless, and the other accelerating downwards.

So this situation is very easy to delineate! Firstly, we can completely ignore the upper mass for now, since it has no momentum or energy to count. All the motion resides in the lower mass.

The lower mass has motion, and thus KE and momentum, from two sources - gravity, against which it has fallen, plus the inertial interaction - the applied acceleration.

We can easily disentangle to two by simply deducting the KE and momentum corresponding to the GPE of the fallen mass - its mass, times gravity, times the change in height.

This value is 9.81 Joules, corresponding to a 1 kg mass in freefall for one second.

Likewise, since P=mV, we know that the rise in momentum corresponding to that conversion of GPE to KE is precisely 9.81 kg-m/s. So, we can also deduct that away.

So that portion of the momentum and KE which was due to the initial GPE is conserved, and undone again when we re-lift the lower mass back to its starting height.

However, subtracting the momentum and KE corresponding to the initial GPE still leaves us with 50% of the momentum and KE remaining..

..and again, the maths are easy to understand why, because the force we applied in the inertial interaction was precisely 2 G. Half of that force was evidently occupied levitating the upper mass, which never budged an inch, and so the other half, a force of 1 G, applied to 1 kg for one second, accounts for the remaining 50% of momentum and energy.

In other words, we had an inertial interaction, and a gravitational interaction, both at the same time. Both interactions were equal in magnitude, for energy and momentum. If the fallen weight were relifted, against gravity, this would cost half the momemtum and energy we have. This leaves the other half accounted for by the inertial interaction...

...and all of that motion resides on just one of the two masses. The portion of the system momentum pertaining solely to the inertial interaction is fully, 100% asymmetrically distributed. It is a binary asymmetry - two masses pushed apart against each other, but all the motion resides on one of them, and the momentum of the other remains completely unaffected.

Hence if the upper and lower masses are suddenly locked together, the motion of the lower one will be shared evenly between both, per N3, and the net system of the inertial interaction - as distinct from the net GPE / GMH interaction - comprised of two masses bouncing back and forth off of one another, has gained momentum... specifically, 9.81 kg-m/s, at a cost of 96.17 Joules.

That particular investment is not wasted, but forms the foundation upon which we begin to pile successive layers of excess momentum, sourced purely from the asymmetric distribution of momentum during the interaction, and which forms the 'rolling platform' inflating the value of the on-board accelerations.

So to come back to your analogy, the net system of the runner plus rocks is gaining vertical momentum, whereas, without gravity's influence, the net momentum of two opposing inertias remains constant.

What the experiment above shows is that when we square up the input and output KE and momentum due to falling and then rising again against gravity, we're left with the energy and momentum pertaining exclusively to the inertial interaction - our legs pushing down on the rocks - and all of which has been apportioned entirely to the rocks, to the exclusion of any change in your own vertical momentum...

..hence, after subtracting or undoing the rise in momentum and KE caused by the drop, the net system of you plus the rocks has gained downwards momentum.

This rise in momentum between a pair of mutually-interacting inertias is the exploit. Unlike a conventional, N3-consistent mutual acceleration, wherein the energy cost of momentum rises sharply as it accumulates, the cost of momentum bought via a reactionless acceleration is constant, regardless of rising velocity. Meanwhile, the energy value of that momentum, externally to the system, continues to rise as a function of 1/2 the inertia times velocity squared, hence for example 10 Joules spent inside the accelerating system could cause a 100 Joule rise in momentum, as measured by the stationary observer.

In a nutshell, if you're floating though space at some ambient velocity, and you throw a mass forwards, in your direction of travel, but without incurring an equal opposing deceleration yourself, then a stationary observer would see the energy of that mass rise by more than you have spent - and substantially moreso, the faster your initial ambient velocity. So while a reactionless acceleration from 1 meter/sec might be 3x over-unity, the same reactionless acceleration from a starting speed of 100 meters/sec could be up to 10,000x OU.

I'm not suggesting we'll get anywhere near those efficiencies, certainly not to begin with anyway. Doubtless there's gonna be practical limits. But make no mistake, this is a prodigiously generous symmetry break, and inordinately more powerful than anything a prospective OB proponent could ever even dream of from their delicately-balanced wobbly springy things.. This is hardcore, kickass, keep-fingers-clear-of-moving-mechanism OU. From waving rocks around. While falling.

Or tell you what, how about waving rocks around, while falling, while being chased by a Rottweiler with a caged squirrel hanging in front of its face, off a rod taped to its back. That's gotta be OU?
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

ME wrote:
..so the sim begins with gravity enabled, and everything in freefall. Gravity is accelerating everything downwards, at 9.80665 m/s^2.
Immediately, a (massless) linear actuator applies an acceleration of twice gravity - pushing the two masses apart, by a rate of (2 * 9.80665 = 19.6133 m/s).
Hence, the upper mass hovers stationary in mid-air, while the lower mass plummets at twice its free-fall speed.
So, your CoM (Center of Mass) still drops in freefall, plus your setup expands... Here your drop is free, while this accelerated expansion is not. It reacts against the CoM of the system, or against each others weight (depends on how you look at it).
The same when turning your setup 90 degrees: Your CoM still drops in freefall (and now your two masses too), plus your setup expands.
It may even happily rotate while it drops: same difference on the expansion.
So no energy gains there, yet, by my reckoning... i'll test that shortly tho.
I agree with sleepy, it will cost energy.

...unless the illusion of reactionless motion gets you somewhere, I think I don't understand what you're trying.
OK, just check off the output KE and momentum due to the GPE interaction. We'll lose as much momentum re-lifting the mass, as it generated in falling, and likewise we'll re-convert all of the RKE sourced from GPE back into GPE..

...but after that zero-sum energy and momentum deal is accounted for, we still have that much energy and momentum again left over.

And as before, we have precisely the right amount of both - we have the right amount of momentum, and the right amount of KE, relative to the energy we spent firing the jack.

So everything sums to unity! Output = input, for both the inertial interaction, and the gravitational one!

The only thing out of the ordinary is that all of the motion - all of the KE and momentum - applied by the inertial interaction resides solely on just one of the interacting masses. The other has no motion.

This result is an N3 violation. A force was applied between two equal inertias, but only one of them moved. This means that the net momentum of the two masses has increased.

If both motions had changed by equal opposite amounts, the net momentum would remain constant, ie. zero change.

And although all of the motion between the two masses has been paid for during the N3 violation, so no excess work has been done between the two masses with regards to the relative change in motion between them, the net momentum of the system itself has risen, owing to the asymmetric distribution of momentum during the acceleration phase, and this new, ambient momentum of the net system has its own corresponding KE, of half its net mass times that net velocity squared.

This rise in energy of the net system has not been paid for in terms of the on-board accelerations, which cost the same, and perform the same amount of work, regardless of whether N3 is respected or not. Rather, this rise in energy of the net system is, effectively, free.


Perhaps a good exercise for you, and anyone else not quite seeing the light yet, would be to consider a pair of masses in free space, connected by a spring, bouncing back and forth off of one another. Consider what happens if N3 is violated every other stroke, as regards internal vs external energies and momenta..

..if frictional losses are disregarded, ie. the system's considered as fully-elastic, then the internal energy and momentum resonating between the two masses remains constant. However, despite this, due to the N3 break on alternate strokes, the absolute net system momentum and energy (as opposed to internal, relative momenta and energy) begin to rise and accumulate, independently of the conserved internal quantities.

In other words, the acceleration of the net system is not being paid for at the expense of the internal workload. As the net system accelerates, the energy and momentum being exchanged between the two masses never falters, neither rising nor falling. The system is not performing work on itself; ie. the internal momentum and energy is not converting to that of the net system.

Basically, when we have a reactionless acceleration, the net momentum of the system comprising the two inertias in question rises. A divergent reference frame is generated - the on-board measurements of energy and momentum can no longer be resolved equitably with those of observers in all other reference frames. N3 violations, just as hypothetical mathematical entities, generate energy and momentum from nowhere.

Here, we have a real N3 break - once we take away or undo all of the momentum and energy released by the input GPE, we have a substantial remnant momentum and energy residing on one mass only, which was caused by pushing against an identical mass, that nonetheless never moves.

Thus, although the net momentum between the two masses - relative to one another - is the same whether N3 is observed or not, the absolute net momentum of the masses themselves - as measured objectively from a stationary reference frame - and thus, that of the net system, rises when one mass can selectively ignore N3.

Obviously, the center of mass of the system gets lower as the lower mass descends. I'm not suggesting any kind of GPE asymmetry; input and output GPE's can only be equal. I'm not looking to re-lift any mass on the cheap.

The gain mechanism is simply the net rise in absolute momentum of the two interacting masses, due to the asymmetric distribution of momentum during the acceleration phase of the interaction, as demonstrated yesterday.
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Post by MrVibrating »

Before we even get to the matter of replacing the lower linear inertia with an angular equivalent, it seems worth considering that we may already have a sufficient advantage to progress...


Was it Wolff or Weiss who described their impression that the weights falling inside Bessler's wheels seemed to impact with excess momentum, over and above what would be expected from their fall alone..? Maybe they were simply thrust downwards, to generate a momentum asymmetry between them and whatever they were being accelerated against?

Then, upon colliding with the wheel, this excess directional momentum would be distributed evenly throughout the net system, providing an incremental skewing of the distributions of clockwise to counter-clockwise momenta, and thus a corresponding net acceleration..?
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re: Plying CF as pseudo-inertia to scam N3

Post by ME »

Sorry for your many words, but I will not consider the situation of an N3 break only to proof N3 broke: circular reasoning doesn't make motion perpetual.
Yes an N3 break should create momentum, energies etc... And I like the result probably about as much as many others here.

But in your accelerated frame of reference your N3-break is fictitious and not for free !
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re: Plying CF as pseudo-inertia to scam N3

Post by daanopperman »

Mr V

I think your concept is correct , but your implementation of it is not .

It can only work if the weights inside the wheel is not crucified to the wheel , which will render them useless .

Imaging the weights not being part of the wheel ( outside ) , but the effects manifested inside the wheel , then I think you will create the break .
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re: Plying CF as pseudo-inertia to scam N3

Post by sleepy »

Mr.V,
You have said a couple times,if you could create this reaction without incurring the penalty for this reaction,that you would have an N3 break and begin collecting surplus momentum.It's not just this concept,but essentially any concept that any of us has tried,that would work if only we could get a reaction without the penalty for it.But please don't think that I'm berating you.I have followed many of your "train of thought" posts and you have a wonderful sense of focus.Why not just have your jack fire against a stationary object? Then it would not have to retrieve the weight at the end of the jack.You only have to retrieve the jack.Just thinking out loud.
Trying to turn the spinning in my brain into something useful before moving on to the next life.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

ME wrote:Sorry for your many words, but I will not consider the situation of an N3 break only to proof N3 broke: circular reasoning doesn't make motion perpetual.
Yes an N3 break should create momentum, energies etc... And I like the result probably about as much as many others here.

But in your accelerated frame of reference your N3-break is fictitious and not for free !
Why is it fictitious?

Surely, "within the accelerated frame of reference" there is no symmetry break between the two masses, as motion is relative - ie. the relative speed between them is identical with or without N3. It is only relative to an external observer that the distribution of momentum is asymmetric.

And of course it isn't free, it's 96.17 J a pop, like i keep saying.

The only reason i've asked you (repeatedly, in multiple threads) to familiarise yourself with how reference frames diverge in an N3 break is that without this knowledge, nothing i'm doing here is ever going to make sense to you..

But it's cool, you know i appreciate your interest..

Just one step at a time though - why is it not a valid N3 break? I mean just the sim i've shown already, not extrapolating any further? The applied impulse pushes the two masses apart (using 96.17 J of input energy), against their mutual inertias, but accelerating only one mass..?

In addition to this asymmetric inertial interaction, there is also a perfectly-symmetrical gravitational interaction.

That is, the energy and momentum of the lower mass has two sources - half of it is from G*m*h, and the other half from the artificially-applied 1G down-force. It has precisely the right amount, and same amount, of both energy and momentum from both those sources.

And if we then subtract the momentum and energy corresponding to the GMH - ie. by relifting the mass, back to its initial height - we still have that other half of its momentum and energy that we bought with that 96.17 J impulse; so 96.17 Joules, and 9.81 kg-m/s of momentum. That hasn't gone anywhere. And all of it resides on one mass only. If the two masses were then reconnected, that momentum would be shared equally between them.

So it categorically is an N3 break by any meaningful definition. Well, the only meaningful definition will be if it's cyclable and gains momentum cumulatively, but it looks to me like this does just that.. We'll see over the weekend..

Still, if you can't or won't familiarise yourself with how momentum and energy evolve in an N3 break then you're never going to understand why i'm convinced 9.81 J per kg-m/s might just be the deal of the century..
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

daanopperman wrote:Mr V

I think your concept is correct , but your implementation of it is not .

It can only work if the weights inside the wheel is not crucified to the wheel , which will render them useless .

Imaging the weights not being part of the wheel ( outside ) , but the effects manifested inside the wheel , then I think you will create the break .
Honestly mate, any time i try and visualise how any of this might fit into the mechanics of rotation, i just kinda glaze over and drift off...

Seriously, it's taken me weeks - and i do mean weeks - of thinking about a vertical inertial interaction while in free-fall, to figure out a simple way of delineating the dynamics. And using two 1kg masses, driven apart by a 2G vertical repulsion, while everything falls at 1G, only occurred to me two days ago...

I think it's a beautifully simple demonstration... but with the emphasis on "simple"..

Still, just shooting from the hip here - suppose the two masses and scissorjack are hanging from a radial armature. As the armature reaches 9 o' clock, in free-fall, the hanging jack fires... again, the upper mass neither accelerates nor decelerates, but the lower mass races downwards at twice its free-fall speed, and slams into a rim-stop - sharing its momentum with the rest of the wheel...

...half of that momentum came from dropping the mass, and will be undone when relifting it. But the other half is unidirectional, and was induced with no corresponding counter-torque upon the net system..

Dunno, but if it's as simple as that then i expect to have a working sim before the weekend's out..
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Post by eccentrically1 »

The corresponding counter-torque is just delayed, the same as the N3 reaction. Once you run the sim, you'll see it.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

sleepy wrote:Mr.V,
You have said a couple times,if you could create this reaction without incurring the penalty for this reaction,that you would have an N3 break and begin collecting surplus momentum.It's not just this concept,but essentially any concept that any of us has tried,that would work if only we could get a reaction without the penalty for it.But please don't think that I'm berating you.I have followed many of your "train of thought" posts and you have a wonderful sense of focus.Why not just have your jack fire against a stationary object? Then it would not have to retrieve the weight at the end of the jack.You only have to retrieve the jack.Just thinking out loud.
LOL fantastic, and i don't mean that in any kind of condescending way.. it shows you're really trying to follow along, and if i can just walk you through the last few conceptual hurdles i promise you you're going to see something exciting...


OK, so taking your last point - why not just accelerate against a stator, ie. ultimately, against the Earth?

Answer - because we want the net system of interacting masses to be able to accelerate together.

If it were possible to perform an asymmetric inertial interaction against the Earth itself, then we'd be propelling it through space. However, we'd see no energy or momentum change, and no immediate benefit for the energy we're spending.

So in order to get any advantage from an N3 break, besides reactionless propulsion, we need a system in which both of the interacting masses are free to accelerate together, relative to us the observer.

Suppose we have two 1 kg masses in free space, connected by a spring. The spring unloads, accelerating each mass in opposite directions. Then the spring goes taut, and they bounce back together, ad infinitum..

Now, if either of those two collisions - when the masses bounce off one another, or else, when the spring reaches maximum stretch - affects only one of the masses, but not the other, then internally, there is no difference; the relative speeds between the masses is the same, regardless of whether all the motion resides on one mass, or else is equally shared between them.

But from an external observer's perspective, the net system would be accelerating.. both the masses, and the spring between them, would be gaining momentum in one direction, flying across the sky, accelerating each time an N3-violating but otherwise fully-elastic collision occurred.

Obviously, normally, with N3 respected, the two masses would just oscillate, wobbling back and forth, but never getting anywhere. The net system cannot self-accelerate. Some external force must be applied, in order to change its net momentum.


But with an N3 break, we're changing the net system momentum from inside the system. This costs energy - the energy of accelerating the masses, plus any energy required to negate N3.

Momentum bought from inside the accelerating system is cheaper than normal momentum, bought via an externally-applied force.

You'll be familiar with the standard kinetic energy term "KE = 1/2 mass * velocity squared" - well, this is also the usual list price for the energy cost of momentum.. It sets the objective energy value of momentum - how much work must be done to buy that momentum, and likewise, how much work that momentum can do.

Any gains we harvest, will only exist in the first place in relation to this standard energy term.

Inside the self-accelerating system however, the cost of momentum bought this way does not follow 1/2mV^2. And whatever it does cost, is a constant, regardless of rising momentum. Unlike momentum bought normally, via externally-applied forces, its cost does not rise the more we buy.

Outside the accelerating system however, the value of that momentum is still totally determined by the usual standard term, and so the more momentum we have, the more it is worth, and the more energy we can harvest from it.

So that's why we need both the interacting masses to be able to accelerate together, relative to us, external observers.

I do take your point, essentially that any other kind of symmetry break would be possible if only they were possible.. but they're not, are they? That's what makes this one special.

If you could drop a weight when its heavy, and pick it up when it's light, that too would create momentum and energy. Yet that isn't possible, because the components of GPE - G*m*h - are temporally-invariant - they don't change over time. GPE loads in a vertical wheel must be traveling in closed loops, with equal up and down distances, with static gravity and rest mass.. hence no gravitational asymmetry is possible.

Similarly, due to mass constancy, no N3 violation is possible either. At least, not through plying inertia alone.

But vertical inertial interactions, in a gravity field - ie., a uniform acceleration field - do present this possibility. For years now, i have intuited that it must be possible, and indeed, seems the single most self-consistent explanation for and interpretation of Bessler's success. But now i've actually demonstrated it, too..

The fallen mass here is accelerated downwards by two 1G forces - natural gravity, plus an identical 1G down-force, applied against the inertia of the upper mass.

The upper mass, in turn, is accelerating downwards at 1G, but also accelerating upwards at the same rate, hence no net acceleration occurs.

We could likewise consider the inverse interaction; the two masses are pulled together by a 2G attractive force, while both are in free-fall: this time, the lower mass's momentum remains constant, and the upper one races downwards at 2G to meet it..

Either way, same result - half the momentum generated corresponds to the conversion of GPE to KE, and will be lost again if the fallen mass were re-lifted. But the other half of its momentum was generated by pushing against another mass, which did not change its momentum in response.

And this is the important part.

That portion of the momentum corresponding to the drop against gravity is irrelevant. We'll lose it all again in the re-lifting. Its a zero-sum deal - this particular momentum is simply that of the GPE converted to RKE. It's a zero sum deal. The net momentum of a complete GPE cycle is zero, because all of that RKE gets reconverted back into GPE.

But the portion of the momentum belonging to the inertial interaction - bought by accelerating against the upper mass's inertia - is reactionless. It neither accelerated nor decelerated the other mass. If they were connected by a long piece of string, once all the slack's taken up and it goes taut, the momentum will be shared equally between both masses, and both will be moving in the same direction at half whatever speed the lower mass reached relative to the upper one. So their net momentum - ie., that of the closed system of interacting masses - is non-zero.

And unlike momentum bought via an externally-applied force, this reactionless rise in momentum does not get more expensive as the net velocity accumulates. A second such interaction will add the same amount of momentum again, at the same cost.


Don't worry for now about the practicalities of retrieving the accelerated lower weight, i'm sure a simple solution will become clear. I've already posited one option that seems interesting; of substituting the lower mass for an angular inertia instead, so that there is no change in GPE... for example, suppose the bottom of the jack connects, via a pivot, to the edge of a solid vertical wheel.. so as it expands, the upper weight is suspended as before, while the pivot connecting the lower end of the jack to the wheel gets lower, this time suspending the upper mass against the angular inertia of a wheel which does not itself get lower, merely rotating a solid, balanced disk..

So in that scenario, we've again generated twice as much momentum as we could've gained by dropping the mass over the same distance the lower pivot descended... yet nothing has descended.. so there's nothing to pick up again... net result, a reactionless rise in angular momentum. The two masses can be braked or collided against one another to share the unidirectional momentum gained, and the cycle repeated for the same input energy again..

Also, remember that the scissorjack itself is just a metaphor for the inertial interaction - which in essence is simply a mutual attraction or repulsion between two masses / inertias.

So think 'massless scissorjacks' for now - yet it's idealised, but that's the point - only the relative motions of the masses matters to our considerations, and the mechanical linkage could just as easily be lightweight pulleys and string, rather than a big hefty scissorjack. So look through the jack, see past it, and focus on the mass accelerations relative to one another and gravity / Earth.. The "special thing behind the stork's bills" is the inertias on either end of them, and their resulting distribution of momentum..

Don't worry, it'll come..
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Post by MrVibrating »

eccentrically1 wrote:The corresponding counter-torque is just delayed, the same as the N3 reaction. Once you run the sim, you'll see it.
Torque? No torque here yet.. it's just a linear acceleration.

When the masses are pulled back together - if they even need to be - then all we need during that interaction is a normal, symmetrical exchange of momentum.

The magic step has already transpired, during the initial acceleration.

Likewise, the momentum gained by dropping will be given up again when re-lifting. That's a symmetrical interaction.

The only special detail required is the one already demonstrated - a mutual repulsion or attraction, between two inertias, that only changes the momentum of one or other...

That in a nutshell is the exploit, and if it's cycled a few times, the energy value of the momentum generated - in terms of its collision energy or ability to raise a weight - will be greater than its cost of generation.

Again, when i say "effective N3 break", the only meaningful definition of "effective" is whether or not it's cyclable and the momentum gain cumulative. If this effect cannot do that, then i have nothing.

But i think it should, for two main reasons:

- Firstly, gravity is equivalent to an acceleration, and applies the same regardless of speed; that is, without air resistance, there'd be no terminal velocity, and rate of acceleration would remain constant up to impact.

Therefore, we gain the same amount of net momentum each time, regardless of how much speed we accumulate and thus, how fast the net system descends or rotates.. so long as we can keep inputting those 2G impulses, the upper mass will cease accelerating during its descent, while the lower mass will accelerate at 2G, hence gaining all of the change in momentum from that mutual acceleration against the upper mass.


- Secondly, inertia is not speed-dependent. So the reason the cost of momentum rises via 1/2mV^2 the more we buy is not because inertia itself increases with speed..!

Again, the actual reason for that is simply the practical constraints of reaction mass and N3.

But inertia itself is speed-agnostic. All it cares about is accelerations / decelerations. So if you have the same amount of mass, being accelerated by the same force over the same displacement, then the same work is being done, and the same energy spent, regardless of the rising velocity..

Hence, the conclusion of those two points is that the momentum should build up cumulatively, gaining by the same amount for each successive asymmetric inertial interaction, and at the same on-board cost, irrespective of rising net system velocity.

With both these conditions met, it is inevitable that beyond some low threshold of net speed, the system will have more kinetic energy than has been spent internally.


At any rate, a "delayed torque" could also be exploited to the same effect, if utilised in the right place..

For instance, if you could delay inducing a counter-momentum, angular or linear, then interleaving two such interactions allows you to launch a second reactionless acceleration off the back of the first. Then you could harvest the resulting KE gain, before releasing the counter-momentum. Net momentum would thus remain constant but energy out > energy in..

It seems to me certain, that in all of mechanics - gravity, inertia, springs and levers etc. - the only plausible route to OU is an effective N3 break.

And regardless, it remains the single most consistent explanation for Bessler's success, resolving many of the seeming contradictions.

But Bessler did say that there's more than one way to skin this cat - that earlier machines "worked on a completely different principle"... which must also have been effective N3 breaks... in other words, this may not be the only way to do it, much less the best way..

But for now, it's something, and the best i've found..
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re: Plying CF as pseudo-inertia to scam N3

Post by ME »

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

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
Their velocities at t=1 will be v&#8321;=v&#8322; = a&#8321;·t = 9.80665 m/s
Their distance from the CoM increases to d&#8321;=d&#8322; = ½a&#8321;·t² = 4.903325 m
Their energies are both ½m·v² = m·a·d = 48.0852 J (total=96.17038)
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.

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)
:-)
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
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Post by MrVibrating »

eccentrically1 wrote:The corresponding counter-torque is just delayed, the same as the N3 reaction. Once you run the sim, you'll see it.
...sorry to reply twice, just struggling to see things from your point of view - do you mean that the upper mass will experience a delayed upwards acceleration at some point? Cuz that would be a cool effect.. ;)

In reality, the momentum absent from the upper mass is not missing from the system, it's just all attached to the lower mass. Instead of each mass getting half the momentum, one of them won all of it. The amount of momentum caused by the acceleration is unchanged though, regardless of how it's distributed, equitably or not.

But without N3, its distribution has a net direction, instead of a cancelled net direction.

A delayed counter-momentum, under these circumstances, would itself be OU-momentum, since we already have all the momentum we've paid for, no more or less...

N3 breaks are nothing if not unintuitive.. but once momentum-symmetry's broken, all bets are off. There's simply no way to square everything off between other reference frames. The diverging system momentum becomes its own unique inertial frame, and the external universe the non-inertial frame, which, from the divergent frame's perspective, has an increasing net momentum, as well as directional inertia..

I began this thread musing about using CF as pseudo-inertia, but the winning twist i think is using gravity as pseudo-acceleration..

I keep wondering how this would play out in the cabin of a spaceship accelerating at a constant 1G... whether it would slow the ship or apply a load to the engines... and i suspect that actually, it would...

If that is correct, what would that tell us about the situation when the acceleration is gravitational, and so only equivalent to the ground accelerating upwards at 9.81 m/s?

It seems to me the implication is that 'the ground' would have to lose some of its 'upwards' momentum..

Yet of course, the ground isn't accelerating upwards, Earth isn't balooning outwards.. and so where is this momentum coming from, when the counter-acceleration it depends upon is only a pseudo-counter-acceleration, not a literal, classical one?

In short, whatever produces gravity, it looks like this system might present as some kind of exceptional 'load' upon it - ie. one that it does not usually support..

But sheet... i'm above my paygrade banging rocks together... Let's see what the weekend brings.. wanna try get some kind of decisive result instead of just extrapolations...

One thing that does seem clear however is that WM should be compatible with rendering this form of OU - ie. it can be reproduced in 2D and depends only on a transient balance of gravity and vertical acceleration, already demonstrated..

I'm also trying to ascertain the viability of daisy-chaining these interactions around the periphery of a vertical wheel, to circumvent the need to apply attractive interactions inbetween the repulsive ones.. ie. if all the masses are identical then there's no reason they need to act in monogamous pairs, and could instead alternate roles in successive interactions between multiple mechanisms..

But the net momentum rise is already proven, as far as i'm concerned - the fallen mass has the right amount of momentum for its GPE, but too much from the inertial interaction (all of it!). There is no question that the former is returned upon re-lifting. But the latter can only be undone by a second reactionless acceleration in the opposite direction...

Still, it's a fascinating subject i could rant on about all night, but i prefer to maintain the reserved, cautious dignity you've all grown to respect me for, ahem, so i'm goin' bed. 'Night.
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re: Plying CF as pseudo-inertia to scam N3

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