Plying CF as pseudo-inertia to scam N3

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MrVibrating
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Post by MrVibrating »

@CC

Not sure what you think you're looking at, but i'm brainstorming potential means of applying momentum without incurring equal opposite counter-momentum.

That is the gain mechanism of interest here.

MT 6 has balls riding curve slats nest to a ladder. I've never been able to work out what it's supposed to mean - are the balls riding up or down the ladder? Where's the transmission between the ladder and wheel? I don't see what MT 6 is trying to do, but neither do i see anything to do with momentum asymmetry.

Besides, i'm a long way from putting a complete design together. Again, all i'm looking at here is momentum distributions.

In the animation you're looking at above, the wheel is driven by a constant-velocity motor, and the armature by another such motor turning twice as fast in the opposite direction.

Although it is powered tho, the only mass changing height is that of the armature, not the mass suspended on the end of it, so the net change in GPE per cycle is minimal... at least compared to what it would be if the bobs were also changing height..
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re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

...I expect to go thru lots of prospective mechanisms yet, this is just purely a quick examination of one simple example of interacting angular-angular inertias, one of which is also subject to gravity.

Again, if the GPE is not doing any work on the way down besides converting to KE and back, then i do not care about it. GPE-in = GPE-out.

That would only change if the GPE did some other work on the way down, such that it landed with less KE, so was no longer able to restore its own GPE.

But even if it fell some distance in free-fall, it's still converting its PE to KE, it's still conserved, and when it lands again all of that KE can be converted back to GPE, minus any losses of course.

If you just look at the scale of the potential I/O disunity, we could arguably even afford to waste GPE, and still keep replenishing it.

However nothing here is actually in "free-fall" - the reason the bob isn't getting lower is because a counter-torque is being applied between it and the wheel, applying momentum to the wheel instead of letting the weight drop.

As such, there simply are no "inadvertent underbalancing moments" to worry about.
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re: Plying CF as pseudo-inertia to scam N3

Post by cloud camper »

No, the bob weight is not the issue as it never changes height, it is the lever which cannot be massless.

When the lever breaks free at the hub, it temporarily "floats" in space as it
disconnects from the hub.

During this interval, the descending side becomes unloaded and the wheel
reverses (or at least suffers a large temporary countertorque which you have not addressed in any of your calculations).

So far, every solution you have proposed is a variation of MT6, with of course
the same fatal flaw.

This is not to say I believe your theory is without possible merit, but we need to try something different than variations of MT6.
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re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

OK here's the same thing again, except this time the motor powering the wheel has been removed.

The motor controlling the angle of the armature is still present, but now, instead of applying a constant counter-torque, it is 'geared' to the angle of the wheel, via this function:

(-2*(body[1].p.r))

..where "body[1]" is the wheel.

So the angle of the armature is still a direct inverse function of the wheel angle, but neither is powered by a constant-velocity motor any more.

Obviously, this is not particularly realistic - a real transmission system like this would probably need a stator, or a least some kind of coaxial gear train, all of which adds friction.

The reason i find the motion in the above animations interesting is because it at least makes a start on replicating the maths; an inertial interaction is initiated whilst one of the angular inertias involved is gravitating.

As demonstrated, this causes an asymmetric distribution of momentum.

Obviously however, if we then mutually decelerate the two inertias whilst the gravitating one is now rising, we perform an inverse asymmetric momentum exchange, undoing the one generated during the acceleration phase.

So maybe now you might see why i found that motion interesting - parking the bob in the middle negates having to perform that inverse-asymmetric interaction.

So in principle we can mutually-decelerate the inertias while neither are gravitating, hence yielding a regular, symmetrical distribution of the asymmetrically-apportioned momentum we just generated when accelerating them.

So the sequence is; generate an asymmetric momentum distribution whilst gravitating, then mutually decelerate the inertias whilst unaffected by gravity to consolidate that asymmetric distribution to a non-zero sum, and then repeat the interaction, carrying off from the preceding interaction's remnant velocity.


So in the sim below, the only gravitating mass is the blue armature.

The bob, again, doesn't make any significant change in height, so its only real contribution is as an angular inertia, given by its mass times the arm length squared.

The system is given a small push-start to knock the blue armature at 12 o' clock out of balance, precipitating its fall, driving the wheel which in turn drives the armature to keep the bob level at all times:

Image


...so as you can see, the bob is currently not parking in the middle, the armature simply continuing its rotation, and so converting the RKE generated from dropping the blue armature, back into GPE.

So it accelerates on descent, whilst applying an asymmetric distribution of momentum to the wheel.

But from 6 o' clock back up to 12 o' clock, that interaction is perfectly reversed, and whatever momentum and KE that was borrowed from GPE on the way down is repaid on the way back up.

So as ever, GPE-in = GPE-out, but also momentum too.

So next i need to do another test, this time parking the bob in the center, as shown in the first animations..

At the moment i don't think this could generate gains, since the bob isn't strictly rotating with the system. As explained ad-nauseum already, i expect that we'll want a system in which the bob can get lower in the field, and consequently would accelerate, but for the fact we're going to arrest that acceleration and instead apply it to the wheel.

If that works then we'll need to figure some way of harnessing the gains, either through CF or GPE workloads, with the former presumably being the more efficient option..

For now tho i'm stuck playing with this thing, until anything more promising shows up..

Again, i'm just kicking out ideas and seeing where they lead. Predict, test, record, repeat. If i already knew exactly what i was doing, what would be the point? It's R&D, not trying to prove anything... i'm just trying to falsify the hypothesis - or not - as honestly and accurately as i can.

You just can't please some folks - you come up with an apparently physical momentum asymmetry and everyone assumes it can only be applied using an exploding scissorjack in free-fall using an umbrella as a parachute or something..

..come up with the same thing using two moving parts and they complain it's too simple, or underbalanced, or not finished yet etc.

I'm just playing with an apparently-promising discovery, making small predictions, experimenting and revising as necessary. Same way i always have.

The same way most of us use this forum..

Yes it's nuts, and yes it'll probably fizzle out like all the rest. But for now it's still very early days and i've barely made a dent in some of the predictions kicked about on just the first few pages.. so i've potentially got years of work to keep me busy if needed. Hopefully it won't be, and i still hope i can crack this definitively before the year's out..
Last edited by MrVibrating on Wed Nov 22, 2017 11:20 pm, edited 1 time in total.
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

cloud camper wrote:No, the bob weight is not the issue as it never changes height, it is the lever which cannot be massless.

When the lever breaks free at the hub, it temporarily "floats" in space as it
disconnects from the hub.

During this interval, the descending side becomes unloaded and the wheel
reverses (or at least suffers a large temporary countertorque which you have not addressed in any of your calculations).

So far, every solution you have proposed is a variation of MT6, with of course
the same fatal flaw.

This is not to say I believe your theory is without possible merit, but we need to try something different than variations of MT6.
Mate i appreciate your reserve, no escaping it i've been getting a bit manic since finding this. I just don't see your objection tho, and feel like you're not really following my rebuttal - look at the last sim above, unpowered but for the GPE it begins with.

It converts that GPE of the blue arm into KE and momentum on the way down..

..and then converts that momentum and KE back into GPE on the way up.

As you can plainly see, it is a zero-sum deal. It's just coasting, sinusoidally against gravity, neither gaining nor losing anything!

My point is that here, on the rising side, we're explicitly undoing any momentum asymmetry we generated on the descending side.

This is why i've repeatedly stressed that the mutual deceleration phase has to be performed whilst horizontal to gravity, so that we get just a regular, symmetrical redistribution of the asymmetrically-distributed momentum generated during the acceleration phase.

Does the above sim answer your contention here? The descending side does not become unloaded because the arm's being torqued upwards, against gravity and towards the wheel's direction of rotation..



PS. above sim attached here if anyone wants it..
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re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

OK here's that last sim again, everything's absolutely identical and untouched, except this time the armature parks the bob in the center before rising back up:

Image


...so just as before, given exactly the same push-start, the blue arm gravitates downwards outputting its GPE, but now locks into the center before coasting back up.


Evidently this is totes batshit so take this with a hefty grain of meth, if this is a real gain (?) i'm not sure it's even the one i was expecting, so hopefully we can make it go away again.

Need to try knock up a different variation on it, see if the effect persists..


:?
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re: Plying CF as pseudo-inertia to scam N3

Post by MrVibrating »

OK here's the same thing again, everything as before, except this time i've replaced the purely mathematical transmission system with a gear train, via a stator wheel.

I think this 'stator wheel' could be made co-axial to the wheel shaft (i'm sure i've done it before, can't remember how), however all three gear wheels here are virtually massless, at 10 mg each, so hopefully won't interfere too much..


As before, it is coasting with the same push-start as the others, and the GPE of the blue bar is the only form of PE:

Image


...so we have the same sinusoidal behaviour from the sim before last.

So now i need to try it again whilst parking the bob at the center as the blue bar coasts back up..
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Re: re: Plying CF as pseudo-inertia to scam N3

Post by ME »

MrVibrating wrote:
MrVibrating wrote:
ME wrote:And Grimer, did you also check the SI units of e²/p ? Does that match another theory?
A key detail in this is the numerical base-rate equality between the respective values of e^2 and P (a vector here, not scalar), as a function of F/m.

It is true, we could assign any arbitrary base-rate, without altering the respective curve shapes or thus their inevitable intersection, however this would transpose the vertical position of the input line integral relative to the KE line.

For example we could assume any initial ratio, such as 1:6, and extrapolate the evolution from there. However this would raise the unity intersection accordingly, and we would lose this defining /4 axis.

The 9.81 value of e^2 and P on both sides is obviously not trivial, being this specific constant of F/m.

Hence any expression reliant on e^2 / P has to first define their equality in terms of F/m.

Using 'A' for 'acceleration', the most meaningful expression would thus be equivalent to A^2 J / A kg-m/s.

However the concepts of "acceleration squared times Joules" or "acceleration times kg-m/s" seem unwieldy... hence why i just define "N" for "number" as F/m and use that.

I've had absolutely zero constructive feedback on any of this remember so i'm just doing as best i can, trying to find the minimal accurate expression.

Doubltess there'll be more succinct ways of formalising ((A^2 J) / (A kg-m/s)), but it's accurate for now. (e^2 / P) only conveys the fields and dimensions, and not their numerical equality or its F/m derivation.

..further to this, A^2 / A is just too misleading - acceleration of what, with respect to what? Acceleration per se?

"Acceleration squared times energy per acceleration times momentum" is just too discombobulating, and needlessly so.

The only logical thing to do is assign some proxy parameter predefined from F/m, exactly as i've done. If anyone can come up with anything better, please share your thoughts..?

Not that this issue's particularly important at this stage - so long as we know what we mean, we can just dub it Bessler's Constant. That's what i'm doing anyway.
You are not kidding, do you.
mrVibrating wrote:Again, i'm just kicking out ideas and seeing where they lead. Predict, test, record, repeat. If i already knew exactly what i was doing, what would be the point? It's R&D, not trying to prove anything... i'm just trying to falsify the hypothesis - or not - as honestly and accurately as i can.
No kiddo, it's random nonsense and kicking people.
You do not predict, test... or show any interest in honesty and accuracy at all.
mrVibrating wrote:[...]However the concepts of "acceleration squared times Joules" or "acceleration times kg-m/s" seem unwieldy... hence why i just define "N" for "number" as F/m and use that.
I've had absolutely zero constructive feedback on any of this remember so i'm just doing as best i can, trying to find the minimal accurate expression.
What are your talking about, at least you confirm this "N" of yours is just arbitrary..
And please reread your own thread/t.
Plenty of attempted feedback of multiple people, but you simply (let's say) "refused".
mrVibrating wrote:please share your thoughts..?
Besides this post, here's an undeserved gift.
Already told you, besides what you hope, want or aim it to be, in physics the expression Energy per Momentum, or f(v)=E(v)/p(v), leads to v/2.
We can do the dimensional analysis, just without the Unicode. Now it's ugly. Just as we seem to like it.
(kg m^2 / s^2) / (kg m / s) = (kg m^2 / s^2) * (s / (kg m)) = (kg/kg) * (m^2/m) * (s/s^2) = m/s

We could do the same thing for any variant of your "N" versus "N²" - examples.
And then you'll see that all those variations you tried neither make any sense from a physics point of view, nor are as compatible with each other as you made it look like: in my opinion all variants are more like 'rubbish' than "unwieldy". It all just shows you have zero insights.

I expect your "troll"- or "copy"- card again: make it a good one.
Marchello E.
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Post by MrVibrating »

Marcello everyone can see you're hopelessly confused.

You evidently don't know what an "arbitrary constant" means.

We have a conserved numerical relationship between the amount of momentum a given amount of energy buys.

It is enumerated by force over mass, F/m.

For gravity, that value is 9.81^2 / 9.81.

If a different force were involved, then we'd need to apply F/m to find that number.

That is not 'arbitrary' or random.

It is not "energy per momentum" as a function of V^2. That only gives the regular value of momentum per 1/2mV^2. It is frankly amazing that you could still not understand this. Everything you're desperately looking up is telling you the same thing; exactly what i'm telling you...

A speed-invariant energy per momentum of 9.81^2 J / 9.81 P is thermodynamically decoupled from 1/2mV^2.

In other words you cannot even derive (9.81^2 / 9.81 P) for two 1 kg masses from 1/2mV^2, for any multiples of that rate apart from precisely 4.

The expression you're trying to find has to be equivalent to:

4 * (F/m^2 J / F/m P) = 1/2mV^2

There is no 'problem' to solve here, there is no need for a shorter expression purely to grasp the relation. Reformulating it to a shorter expression doesn't change what it is or why we're using it, which is precisely because of the fact that it only intersects with 1/2mV^2 at an exact multiple of four, and for self-evident reasons (the V^2 e/p is not of course constant, and dividing 'p' into doubled 'm' halves 'V', quartering 'e').

So whether you were able to further reduce (or indeed, expand upon) it, or not, it makes no difference - we don't need that to proceed, and it has no bearing on the validity of results thus far demonstrated.

I've admitted that despite being accurate and adequately-short, (F/m^2 e / F/m p) seems unwieldy, and unintuitive outside of this specific context of an effective N3 break. But in response to this, the best you've come up with is:

(kg m^2 / s^2) / (kg m / s) = (kg m^2 / s^2) * (s / (kg m)) = (kg/kg) * (m^2/m) * (s/s^2) = m/s

...which is quite a bit longer, and wrong; remember, we've spent 96.23 J to buy 9.81 P divided between two 1 kg masses, so resulting in a 4.905 P net rise per iteration, and you're looking for a shorter, more 'concise' expression for this relationship than say (i = (F/m^2 J / F/m P), where 'i' = elapsed iterations.

Since you don't seem to know how to do this, and since it makes no difference to anything anyway, and especially since your motivations are exclusively egotistical, your best line of attack would be to try and prove that we cannot produce an effective reactionless rise in momentum in the first place - not for 96 J or any sum.

If you can't do that, then maybe try prove that multiple iterations are impossible, that the cost does rise with velocity, maintaining full symmetry with 1/2mV^2. Then there wouldn't be a 'constant' to describe in the first place.

As things stand, your thoughts are all over the place, your 'position' in tatters, arguing incoherently just for the sake of it, clearly not understanding what you're arguing or why, just some disembodied stream of unfocused contrarianism.. One way or another, you need to get a grip.

Use an online graph plotter to look at the divergence of x^2 vs (x^2 / x).

Are you trying to argue that they're equal? Or that (F/m^2 J / F/m kg-m/s) is impossible?

Is (F/m^2 J / F/m kg-m/s) even possible, Marchello?

Everything here hinges on (4 * (F/m^2 J / F/m kg-m/s)) = 1/2mV^2.

Nothing you say has any bearing on its demonstrated validity. If you can contract it, great, if not it makes absolutely diddly-squat difference to anything. I'm not expecting anyone here to be resolving it with the field equations anytime soon, so chill, this is happening... come to peace with it.
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Post by eccentrically1 »

MrVibrating wrote:@eccentrically1

LOL don't worry you're there:

- one mass gets 9.81 kg-m/s of reactionless momentum... ie. its counterpart wasn't accelerated at all, its motion remaining constant.


- critically however, the motion that was input into the lower inertia was nonetheless leveraged against the inertia of the upper mass. We couldn't have accelerated the lower inertia otherwise, without having something to push against..

- so if we now decelerate the two inertias against one another, scrubbing off that 9.81 m/s speed difference, but without interference from gravity in the resulting momentum distribution, then we just get a bog-standard N3 result, and the momentum difference is shared evenly between them.

Hence we're dividing 9.81 kg-m/s of reactionless momentum, from a single 1 kg mass, into the second 1 kg 'reaction mass', that was 'left behind' (but still sitting right next to the wheel, so hasn't run off anywhere), and so 9.81 kg-m/s divided between two 1 kg masses leaves 4.905 kg-m/s on each.

So if you recall, Dwayne got most upset because he found that simply dropping two 1 kg masses at 1 G for 1 second causes the same rise in momentum as dropping 1 kg at 2 G for 1 second - and the same amount we get after also inputting this 96.23 Joules.

And on that count, he's right of course - we end up with no more momentum at the end of each drop, in all three cases.

The only difference is its distribution with regards to signage. Its polarity, if you will. In the first two examples of inert mass drops, the net momenta all cancel, so the system's net momentum at any given moment basically has 'scalar' dimensions - that is, magnitude, but no direction or orientation in space or time.

Whereas, in our third example, when we input this 96 J of work, the net result is not an increase in absolute momentum, but rather in the form of its asymmetric distribution, and thus non-zero sum.

Hence we've generated momentum with 'vector' properties, of magnitude and direction.

And this is what makes our momentum special.

Its why we can accumulate it over successive cycles.

And where our KE gain appears to arise from.


Given that we can't of course remain in free-fall all day, it's worth stressing again that GPE-in = GPE-out.

So we apply a small mutual acceleration while one or both inertias are descending, followed by a mutual deceleration while horizontal to gravity, then coast around and repeat the cycle.

No work is done by gravity upon the inertial interaction, and when re-lifted, the weight returns all KE and momentum lent by gravity, leaving only our 9.81 P reactionless rise, distributed into two 1 kg masses at 4.9 P each and 12 J each, so 24 J total, 1/4 of our 96 J input energy.

So yeah, that's basically it... output KE rises another 25% each successive cycle thereafter.

The only magic should be the resulting divergence between the way the input and output energy integrals evolve, the latter inevitably intersecting the former and leaving it fiddling with its laces on the starting blocks..

I totally appreciate that grasping the maths and mechanics here is like watching two completely different movies only to be told they were different angles on the same story..

You have it tho - the sequence of mechanical actions is the story of the maths, and vice versa.


Again, i suspect there's a good deal more to be fleshed out here, and all i've done is lay a few stones across the quagmire, just to provide the easiest possible path to navigate across.

But if you consider that using equal inertias probably isn't necessary (just making it easier by keeping the divisors symmetrical), we're currently seeing it through a kind of Minecraft filter - all our basic elements crudely-hewn from arbitrarily-proportioned blocks, purely for ease of comprehension, rather than practical mechanical expedience...

So thus far, we've barely scratched the surface.

We definitely seem to have the keys to the safe though...

:O
Don't worry, I've been there and done that, and I'm already at the finish line, lol.

This is not a reactionless acceleration-asymmetrically distributed momentum-nonzero-polarity-magic sum you're seeing here. That is not what is happening. I've tried to say this as best I can, but no one else seems to see it, maybe it got buried in the thread-torrent.

Simply, the reaction force is gravity. That should be the first dealbreaker.
The acceleration is not "reactionless". It's not even "effectively" reactionless. If anyone needs further explanation, it is in an earlier post of mine.

The second dealbreaker is the lower inertial mass (or the wheel) losing angular momentum when the upper inertial "push-off" mass must be accelerated (or as you say, decelerated against the lower mass).

Even if this is timed to be horizontal to gravity and we get a standard N3 result (*which we got in the beginning phase, no non-standard N3 result there - see first dealbreaker*), there won't be enough momentum leftover for the wheel, and the spring/lever/storkbill/whatever-connection dujour, and the upper inertial mass and for the spring/ whatever-connection dujour - using some of the wheel's momentum - (using some unidentified mech!) to reset and, and finally coast around with 25% leftover to begin the next acceleration. In the real world, Bessler's constant does not compute.

All we've done up to this point is seen simulations can prove F(net)=ma and F(a)= -F(b) and KE=1/2 m(v^2). They can prove these equations when the conditions are ideal. Massless rods and springs, frictionless joints, vacuums. We have also seen here on the forum they can show overunity if the program is set up for those conditions. WM2d is great if you don't misinterpret it.
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Post by MrVibrating »

@eccentrically1

Cheers mate, you're wrong on all counts, but i'm not going to press you on it further. Thanks for your thoughts anyway.
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Post by MrVibrating »

..it is funny tho how everyone thinks it's wrong, all for completely different reasons, with no one agreeing with anyone...!

I guess i should think about elevating this somehow..
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re: Plying CF as pseudo-inertia to scam N3

Post by cloud camper »

I am still stuck thoroughly on the fence as I don't see any way to implement other than MT6 variations which do not work no matter how much you try and sex them up!

I have thought of a different way of applying the idea which I am attempting to test in simulation.
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Post by eccentrically1 »

MrVibrating wrote:..it is funny tho how everyone thinks it's wrong, all for completely different reasons, with no one agreeing with anyone...!
You’re welcome.

We’ll see who’s right.
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Post by Grimer »

eccentrically1 wrote:Don't worry, I've been there and done that, and I'm already at the finish line, lol.
Ooo! I am glad to hear that. I have the finishing line in sight but I'm only at it in theory..... and nobody believes theory apart from the originator. :-)
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