Plying CF as pseudo-inertia to scam N3

[MrVibrating]

LOL so again, how can we accept the result for four successive interactions - unity, so no cognitive dissonance induced - while rejecting the results of the very same interactions repeated three times, or five?

The UU and OU results at 3 and 5 interactions must be as valid as the unity result at 4 interactions, since all lie in a contiguous sequence, depending on the identical predicates and conditions..

Logic trap...

[ME]

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.
...
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..

Is that my caricature? (honest question)
I don't feel myself familiar with scoffing at hypothetical stuff. Only at contradicting stuff like hypothetical explanations making things more complex and explaining nothing.
(So perhaps it's indeed true, sorry bout that)

To be honest, I have no clue what you're actually showing, and I still wait for my coin drop.
Still stuck on that linear example, it's the same conclusion as I had at page 2:
(The circular-one seems similar but then with RKE)

1. You drop some (complex) object.
2. It accelerates 1G due to Earth.
3. While it drops this object is forced to mechanically expand outwards with an acceleration of a=9.81m/s2.
4. A). one part shoots up, and B) one part shoots down.
5. Part A appears to be motionless(+a-G), Part B accelerates with 2G (-a-G)
6. A). The drop [2] costs PE[G, height], and B) the expansion [3] costs PE[a, distance]

7. A). As far as I'm concerned it [4] could also be a horizontal action
--. B). and the total cost [6] will be the same per clock tick.
--. C). and the kinetic energy will be the same: two dropping down and sideways (v²=v[x]²+v[y]²)
--. D). and the momentum downwards will, now accumulated, be the same.
--. E.) The extra momentum this setup has by expanding horizontally [7A], is compensated once the hovering [5A] weight detaches and drops down to Earth.
--. F). ... but then this 'new' GPE-->KE of the hovering weight (height [5A]) needs to be compensated as it's now extra. But for that we can still drop the whole construction [7A] because it didn't drop (per clock tick [7B]) as much as that 2G weight [5B]... that's the imbalance I see.
--. G). From this [7E] it may be recognized as a height for width issue, but without a wheel.

8. My found imbalance [7F] is a non-issue when you consider this as two separate actions where the whole system [1] is dropping in freefall with its own GPE [6A] and with added expansion energy [6B], and no matter in which direction it expands. Hence the issue is misrepresentation.

9. I haven't really checked, but it's save to say that in the end (at some depth) you need the same amount of energy to bring it back to where it started [1], no matter in what orientation it expanded.


Obviously I missed something, and you probably tried to explain it already (I already hear the "doh!" echoing).
In case you want to try again, I numbered the best I could for pinpointing my error.

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...

I don't get it, perhaps you could show the maths.

Also I'm a bit confused (I blame my non-native-Englishness) as it seemed like you thought it was already a done deal, and now you still looking for an "effective" N3 break. But then again your test proved your hypothesis...
Confused :-)

Perhaps I simply never get it Game on halt.

ps.
For your latest maths-example: you lost me at the first bolded sentence... how do yo get there? Ok, I'll try to muddle through it.
And stop typing! I'm still trying to decipher your first few pages :-)

[MrVibrating]

A quickie for any of the maths heads:

- what's the formula describing this efficiency gradient?

At one interaction, we're 4x UU.

At two, 2x UU

At three, 0.66x UU

At four, unity

Five = 1.25x OU

Six = 1.5x OU

Seven = 1.75 OU

Eight = 2x OU

Etc.

Presumably it's somehow the reciprocal of the obviated V^2 multiplier? Or something else?

What's the most succinct formalisation here?

[eccentrically1]

On advantages and disadvantages of... thread

So wassit look like, how'd i build it, what's a bill of materials look like etc. etc.? How the flip would i know, you're the mechanical geniuses, you work it out. Half of you are bloody retirees, no excuses lol..

Funny!

[ME]

OU=(interactions/4)

[Furcurequs]

MrVibrating,

Pushing upward against a hovering mass doesn't get you anything that you couldn't have from just pushing off of the earth - or pushing off of something rigidly affixed to the earth, of course. If the mass you are pushing off of is itself descending in the earth's gravitational field, however, then obviously that is a different situation, but there is a cost.

You have already done an accounting of the energy involved. The lower mass will end up with its initial kinetic energy plus the energy you've input to push the masses apart plus the kinetic energy due to the conversion of the gravitational potential energy of both masses' travel downward in the earth's gravitational field.

The upper mass, though, will have descended in the gravitational field at a constant speed while simply maintaining its initial kinetic energy. Maybe this is what you are overlooking. To move it back upward without it losing its speed, you must give it back the kinetic energy you swiped from it as it was descending.

I believe this is a zero sum game.

...and, btw, you are not the first to have thought about such things. Do you remember the slinky drop? The lower end of the spring hovers in place because the upward force from the spring above it is equal in magnitude to that portion's weight until the spring has fully collapsed.

It's just a propulsion problem where mass is being propelled downward to make the lower portion "hover." That mass just happens to be above the hovering portion.



Gif is from here: http://www.gifbin.com/985815

I think this was the original youtube video:

https://www.youtube.com/watch?v=wGIZKETKKdw

[MrVibrating]

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.
...
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..

Is that my caricature? (honest question)
I don't feel myself familiar with scoffing at hypothetical stuff. Only at contradicting stuff like hypothetical explanations making things more complex and explaining nothing.
(So perhaps it's indeed true, sorry bout that)

To be honest, I have no clue what you're actually showing, and I still wait for my coin drop.
Still stuck on that linear example, it's the same conclusion as I had at page 2:
(The circular-one seems similar but then with RKE)

1. You drop some (complex) object.
2. It accelerates 1G due to Earth.
3. While it drops this object is forced to mechanically expand outwards with an acceleration of a=9.81m/s2.
4. A). one part shoots up, and B) one part shoots down.
5. Part A appears to be motionless(+a-G), Part B accelerates with 2G (-a-G)
6. A). The drop [2] costs PE[G, height], and B) the expansion [3] costs PE[a, distance]

7. A). As far as I'm concerned it [4] could also be a horizontal action
--. B). and the total cost [6] will be the same per clock tick.
--. C). and the kinetic energy will be the same: two dropping down and sideways (v²=v[x]²+v[y]²)
--. D). and the momentum downwards will, now accumulated, be the same.
--. E.) The extra momentum this setup has by expanding horizontally [7A], is compensated once the hovering [5A] weight detaches and drops down to Earth.
--. F). ... but then this 'new' GPE-->KE of the hovering weight (height [5A]) needs to be compensated as it's now extra. But for that we can still drop the whole construction [7A] because it didn't drop (per clock tick [7B]) as much as that 2G weight [5B]... that's the imbalance I see.
--. G). From this [7E] it may be recognized as a height for width issue, but without a wheel.

8. My found imbalance [7F] is a non-issue when you consider this as two separate actions where the whole system [1] is dropping in freefall with its own GPE [6A] and with added expansion energy [6B], and no matter in which direction it expands. Hence the issue is misrepresentation.

9. I haven't really checked, but it's save to say that in the end (at some depth) you need the same amount of energy to bring it back to where it started [1], no matter in what orientation it expanded.


Obviously I missed something, and you probably tried to explain it already (I already hear the "doh!" echoing).
In case you want to try again, I numbered the best I could for pinpointing my error.

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...

I don't get it, perhaps you could show the maths.

Also I'm a bit confused (I blame my non-native-Englishness) as it seemed like you thought it was already a done deal, and now you still looking for an "effective" N3 break. But then again your test proved your hypothesis...
Confused :-)

Perhaps I simply never get it Game on halt.

ps.
For your latest maths-example: you lost me at the first bolded sentence... how do yo get there? Ok, I'll try to muddle through it.
And stop typing! I'm still trying to decipher your first few pages :-)

Sorry, should have clarified - 9.81 Joules per 1 kg-m/s of momentum is how much energy it costs to create that much momentum, by fully-cancelling counter-momentum via gravity.

And yes, it is a done deal, and i'm sorry if i'm confusing people with silly rhetoric - just trying to methodically address any potential points of doubt.

Might need a little polish and detail on the maths, but the broad strokes are there. I've already sent the obligatory crank email to the UK DoE, so i'm that convinced or deluded, whichever it is.

But the take-home point in that first example - the linear inertial interaction in free-fall - is that all of the input momentum and KE from the inertial interaction was manifest on one mass only.

Yes, equal momentum and KE was also provided by GPE.

When we repay that GPE, we give back all the KE and momentum it lent us.

And we're left with just the momentum and KE we input by pushing the masses apart, or pulling them back together, whichever way up we fried it. Just the momentum and KE from the applied inertial interaction.

All of which lies on one mass only, and is of one sign.

So it's not a GPE exploit - we haven't gained anything by lifting anything. That's not the objective.

The purpose of the applied interaction was not to cause the upper mass to levitate; that's incidental.

The purpose was simply to prevent it from counter-accelerating against the lower mass's inertia!

But, unequally, the lower mass was accelerated against the upper mass's inertia!

Sure, both were uniformly subject to gravity at the same time, which is the only reason we got this result... yet, gravity performed no net work in generating this momentum asymmetry! GPE-in = GPE out, regardless of whether we generate the momentum asymmetry or not.

So the resulting asymmetric distribution from an inertial interaction is not a 'load' upon the GPE, and gravity's role in skewing the momentum distribution is entirely passive (ie. as opposed to 'active', ie. performing work).

In a conservative vertical rotation, sometimes the two masses are parallel to the gravity vector (ie. at 3 o' clock and 9 o' clock), and other times, anti-parallel (at 12 and 6 o' clock), so we have a passive time-variation in the resultant momentum distributions between a pair of interacting linear inertias constrained to a vertical orbital trajectory.

As you suggest, perform the mutual acceleration without gravity, or just horizontally to gravity, and each 1 kg mass ends up with 9.81 kg-m/s and 48.11805 Joules, oriented in opposite directions, after 1 second.

Because the two 9.81 kg-m/s momenta are equal but of opposite sign, their sum is zero, and the system's net momentum is unchanged.

Equivalently, if both masses were initially moving in uniform motion, in the same plane as the mutual acceleration, then one accelerates while the other decelerates by an equal amount, and again, net system momentum remains constant, net change in momentum is zero.

The first law states that only an externally-applied force can cause a net change in momentum of an otherwise-closed system!

But it doesn't say that this externally-applied force has to perform any net work upon the system!!!

Of course, if it did, then it's momentum would no longer be constant, but then, it would not longer be a closed system, would it? The new "net system" would encompass the applied force as internal to the system, rather than external, and so would be expected to present as a load upon that force's energy source.

Yet here, gravity (or whatever the type of force used) is not performing any net work in simply rendering the momentum asymmetry.

So the system remains effectively thermodynamically closed.

Like i say, bit of an edge-case scenario, but still fully Newtonian.


And so, with a perfect 2 G mutual force applied between the masses, no matter how fast they're already moving, when aligned to gravity on the descending side, we only input momentum of one sign, and conversely, when ascending on the opposite side, if we pull them back together at 2 G, only the upper mass will accelerate, the lower one's speed remaining constant regardless of both the applied force and gravity, and so only inputting momentum of the opposite sign..

Since we don't want that opposing quotient, all we need to do is pull the masses back together while they're at a horizontal phase in their orbit, so at 90° right angles to the first stroke. Thus we reset them in full accordance with N3, and simply dividing that rise in unbalanced momentum between the two masses, thus accelerating that closed system via internal expenditure of work only..

And so we generate an asymmetric momentum distribution on the first stroke, and then consolidate / share it / equalise it / bring the net system up to a new equilibrium coasting momentum / velocity on the second, reciprocal stroke, 90° later... gaining - or rather, generating - 9.81 kg-m/s each full cycle of this inertial interaction.



Alternatively, we could simply frame it this way:

- suppose we have PE sufficient to induce 10 kg-m/s of momentum between two free masses

- per N3, we get +5 kg-m/s and -5 kg-m/s change in momentum, net zero

- but sans N3, we could get a 8 / 2, or even a 10 / 0 distribution, as here..

- obviously, 8 - 2 = 6, divided between the two masses = a 3 kg-m/s acceleration of the net system, from an internally-applied interaction

- or, as here, we could have a fully asymmetric interaction, hence 10 - 0 = 10 / 2 = 5 kg-m/s acceleration of net system inertia.



So we can effectively create momentum, from nothing, by summing an asymmetric distribution of the stuff.

Creating momentum this way has a constant energy cost, irrespective of the velocity at which the net system is already moving.

Whereas, conversely, the energy cost of raising momentum by accelerating an inertia - ie. the conventional method - squares with rising velocity, per KE=1/2mV^2.

So the input energy (internal mass accelerations) / RPM is a flat-line plot, whereas the output energy (RKE) / RPM is a diagonal line.

So there's some (low) theshold RPM beyond which the line integrals for input vs output energy fields intersect.

That intersection arises at precisely four consecutive fully-asymmetric gravitationally-assisted reactionless accelerations.

From thereon, it seems, efficiency climbs by a steady rate (25%?) per successive reactionless acceleration.

In short, it's cheaper to create momentum as you go along, than to physically accelerate mass up to speed across some speed range. Well, not so much necessarily 'cheaper', but fixed - so its relative value depends upon how much you accumulate, and its objective value per the 1/2mV^2 standard.

As demonstrated, this differential can be positive, negative or equal.

Rate of change of energy vs rate of change of momentum is different with and without N3. One's temporally-variant, the other, constant

So this appears to be fully consistent with Newton and Noether, CoM and CoE at every stage... except for the net result, basically. Bona fide OU.

Or else just inveterate lunacy...

I'm easy, tho, either way its a helluva trip..

[MrVibrating]

OU=(interactions/4)

Thanks...


So, umm... why, then? I mean from the physics POV...?

[MrVibrating]

mis-post

[MrVibrating]

MrVibrating,

Pushing upward against a hovering mass doesn't get you anything that you couldn't have from just pushing off of the earth - or pushing off of something rigidly affixed to the earth, of course. If the mass you are pushing off of is itself descending in the earth's gravitational field, however, then obviously that is a different situation, but there is a cost.

You have already done an accounting of the energy involved. The lower mass will end up with its initial kinetic energy plus the energy you've input to push the masses apart plus the kinetic energy due to the conversion of the gravitational potential energy of both masses' travel downward in the earth's gravitational field.

The upper mass, though, will have descended in the gravitational field at a constant speed while simply maintaining its initial kinetic energy. Maybe this is what you are overlooking. To move it back upward without it losing its speed, you must give it back the kinetic energy you swiped from it as it was descending.

I believe this is a zero sum game.

...and, btw, you are not the first to have thought about such things. Do you remember the slinky drop? The lower end of the spring hovers in place because the upward force from the spring above it is equal in magnitude to that portion's weight until the spring has fully collapsed.

It's just a propulsion problem where mass is being propelled downward to make the lower portion "hover." That mass just happens to be above the hovering portion.



Gif is from here: http://www.gifbin.com/985815

I think this was the original youtube video:

https://www.youtube.com/watch?v=wGIZKETKKdw

Yes i was reminded of the slinky drop phenomenon!

You raise the most interesting objection so far, however...

- it's already accounted for.

Dammit, mate - honestly, you had me shuddering with that impending sense of dreaded clarity, embarrassment and final relief... but no! Denied. I've checked and re-checked...

It pans out like this:

- the energy that would've been output, had the inertially-suspended mass been allowed to accelerate under gravity, is instead manifested on the lower mass, and has been input via the internally-applied force instead. That portion of the work is done by our input energy, not gravity.

- if we consider just the angular - linear example, then there is no change in GPE throughout the initial interaction. Again, the rotor receives exactly the same momentum and KE that the mass would've generated, had it fallen, even though it hasn't, again all provided internally.

It could then fall and rise again for a symmetrical gravitational interaction.

If the system's already rotating, then the upper mass is moving through the gravity field, and so work is being done, regardless of whether that mass is accelerating, or else the acceleration being applied elsewhere, such as to the wheel.. so again, i see input / output work from gravity as zero, but input inertial work rectifying to one sign of momentum only.

I'll think more on this, but does that answer your objection satisfactorily? If not, please persist, you have to be right, cos one way or another, this is batshit..

[MrVibrating]

LOL, the thing is, the slinky's really is doing this too, no? It's gaining momentum downwards, not just from the descent of its center of mass, but from the asymmetric accelerations of its opposing ends relative to us stationary observers!

The question is, does it strike the ground with only the momentum corresponding to the GMH of its center of mass - with the internally-applied momentum summing to zero relative to us / the ground...

...or else, does it land with more momentum, owing to the apparent asymmetric distribution corresponding to the internal acceleration relative to us / ground?

If the latter is correct, then has this asymmetry been caused by an output of additional work by gravity?

If the spring is re-stretched while horizontal, lying along the ground, before being rotated and lifted back to its initial height - is the net momentum (spring + earth) balanced? Has more work been performed against gravity than when it dropped?

[MrVibrating]

..mate, hopefully you're right, and my above retorts are full of holes and have deviously dodged the thrust of your objection...

...but at the same time, dammit:

- JB defo had a symmetry break.

- There's only a finite number of hiding places for such a beast within classical mechanics.

- His systems were vertical and statorless. Both apparently necessary conditions, despite his apparent denial of the possibility of a gravitational asymmetry (which i maintain we must regard as axiomatic anyway). Yet his systems apparently required vertical rotation regardless.

- Similarly, statorless requirement implies a runaway momentum / N3 break

- So logic dictates that somewhere in classical mechanics there's an effective N3 break that depends upon variable orientation WRT gravity

- Here is one. A seemingly perfect-fit candidate. Nothing like it in either gravity or inertial interactions alone, but bring 'em together and this possibility arises...

- The point i'm making is that there is, obviously, a solution to be found... but that it's invariably going to end up looking quite a lot like this. Almost by definition, it's going to concern the same issues, and tick all the same boxes...


Now that's all as circumstantial as it gets, of course... but gentlemen, we surely couldn't be any warmer..?

We have the correct spatio-temporal dimensions for our input / output energy asymmetry, a fully causal effective N3 break, speed invariant input efficiency, speed-dependent output efficiency...

So far it seems clear that if real, it can match the performance characteristics of JB's wheels - ie. as regards preferential coasting velocity, load-matching abilities, very slow but high energy operation, etc..

(using lots of "etc's" only cos other consistencies seem bound to fall into place)

So yeah.. if this ain't it, it's probably still a good dry run.

Either way, you're welcome! ;P

[Silvertiger]

Mr.V. I must admit your posts are far too long to read. But, in its most simplistic description, it will take more energy to hold one object up while the other accelerates away than it would to simply let it drop. There essentially needs to be an acceleration greater than that of gravity to drive them apart, which then becomes the problem. You are starting out with having a primed mechanism of stored energy. Once this energy is used up, how will you get it back? How will you recompress the spring so-to-speak? The UKDOE will take one look at it and not reply because of the same problem that has always existed.

As far as the slinky goes, what "asymmetry" are you talking about? You mention asymmetry a great deal. It's just acceleration. The slinky is just a spring contracting at a rate relative to its K constant.

[MrVibrating]

..OK here's a min-bending answer

- if Wayne's contention is valid, how would we explain the unity result at 4 interactions? Surely all interactions should be equally gainful, corresponding to the neglected negative torque while re-raising the non-accelerated mass? Yet that's not how the gain evolves, instead centering around this four-fold axis.


Another, side issue - if this madness does pan out, we're gonna need a snappier term for "reactionless acceleration"... or a word for the momentum produced this way..lol

[Silvertiger]

I don't know who Wayne is nor do I know his contention and so cannot communicate on this frame of reference lol.