Plying CF as pseudo-inertia to scam N3

[ME]

I don't see any other forum member trolling with relevant shit. There's only a handful capable of doing that sort of "trolling" here.

MrVibrating, I found this on the internet. Just as I seem to find all kinds of prepared answers to you nonsense:

• Ad hominem (Latin for "to the man" or "to the person"), short for argumentum ad hominem, is an argumentative strategy whereby an argument is rebutted by attacking the character, motive, or other attribute of the person making the argument, or persons associated with the argument, rather than attacking the substance of the argument itself.

Nice going V.

Then i showed that the ratio of input energy to single-signed momentum was equal to N^2 J/kg-m/s, regardless of the impulse period.

And then I found this nice piece of text somewhere. I thought... hmm pretty. Let's copy, add some colors and show it to mister V. Perhaps he likes it too...?

This is somewhat what I had initially intended, but removed because I thought you moved on... (so yes, a copy/paste, but from the archives). It's a waste anyway, one way or the other.

Linear-linear case:

For accuracy i will use the same five significant digits as 1 G = 9.80665 m/s^2.

The lower 1 kg descended by 88.2598 meters, producing KE equal to GMH = (9.80665 * 1 * 88.2598) = 865.53296 J.

The upper 1 kg has not changed height, so still has all of its GPE and cannot have contributed any KE or momentum.. or else we'd effectively be able to drop it twice.

So subtracting the 865.53296 J of GMH applied by gravity's 1 G acceleration, from the 1731.06692 J total, leaves 865.53296 J - basically half as much energy again - input internally between the two inertias, and corresponding to the additional 1 G acceleration it applied to the lower inertia.

In summary, we applied two 1 G forces, both performed an equal acceleration of the lower inertia only, and both spent the same energy, having performed equal amounts of work.

Thus half the final momentum and velocity was due to GPE, and the other half due to the 3 second 2 G mutual impulse.

58.83990 / 2 = 29.41995 m/s of velocity from each source.

So 29.41995 P from GMH, plus 29.41995 P applied internally.

Since only the lower inertia has been accelerated by either force, we only need to repay KE and P to GMH from its motion.

Leaving it with 29.41995 P and corresponding 432.76672 J (per 1/2 * 1 kg * 29.41995^2) from internal work.

So to recap, we ended up with 1731.06692 J of KE.

Half that - 865.53296 J - came from GPE.

We also ended with 58.83990 m/s of velocity and equal 58.83990 kg-m/s momentum.

Repaying half the 58.83990 m/s that came from GPE, we're left with 29.41995 in m/s and kg-m/s, with an energy value of 432.76672 J, which is precisely half of the 865.53296 J that didn't come from gravity, but was instead applied internally.

The reason half our input energy has disappeared is because KE squares with velocity, and we've just halved velocity, so we're left with a quarter of our net total energy, and half our internally-applied input energy.


That remaining 29.41995 P divided into the upper inertia per N3 leaves each at 14.70997 m/s.

1/2 * 1 kg * 14.709975^2 = 108.19168 J on each, so 216.38336 J total KE.

So over a 3 second impulse we input 865.53296 J of internally-applied work, yielding 216.38336 J of output KE, and 865.53296 / 216.38336 = 4.

Therefore ((N^2 J/kg-m/s) * 1) is fully time-symmetrical to ((1/2mV^2) /4).

Thus ((N^2 J/kg-m/s) * 4) = 1/2mV^2, and ((N^2 J/kg-m/s) * 5) = 5/8mV^2, etc.


Next we need to confirm consistency with the angular example...

Green: inserted
Red: corrected

Linear-linear case:

• For accuracy i will use the same five significant digits as 1 G = 9.80665 m/s^2.
  
  Not that more accuracy helps here!
  The lower 1 kg accelerates with A=2G, for t=3 seconds. It descends: h=½a·t² = 88.25985 m
  
  The lower 1 kg descended by 88.2598 meters, producing KE equal to AMH = (2*9.80665 * 1 * 88.2598) = 1731.06692 J.
  The upper 1 kg has not changed height, so still has all of its GPE and cannot have contributed any KE or momentum.. or else we'd effectively be able to drop it twice.
  So subtracting the ...<snip> WHY? It is 1731.06692 Joules. done.

From the Center of Mass perspective:

• The complete system (2 kg) drops with 1G due to gravity for 3 seconds.
  v[t] =g·t = 29.41995 m/s dropping speed of the CoM
  E[drop] = ½m·v² = ½m·(g·t)² = 865.533458 J
  
  The system expands with 2G, or accelerates each 1 kg with a=1G away from the Center of Mass for 3 seconds.
  v[t] =a·t = 29.41995 m/s expanding speed away from the CoM
  E[expand] = 2·½m·(a·t)² = 865.533458 J
  E[total] = E[drop]+E[expand] = 1731.06692 J
  
  For the momentum stuff (note: g=a):
  The top has a momentum of p[top]=m·(+g-a)·t = 0
  The bottom has a momentum of p[bottom]=m·(+g+a)·t = 58.8399 kg·m/s

But you continue:

So subtracting the 865.53296 J of GMH applied by gravity's 1 G acceleration, from the 1731.06692 J total, leaves 865.53296 J - basically half as much energy again - input internally between the two inertias, and corresponding to the additional 1 G acceleration it applied to the lower inertia.

WHY do you need to subtract things: all energies are accounted for.
As you said many, many, many times yourself: KE=½m·v² ... done.

since there is no dispute that 1/2mV^2 remains valid in our external, static reference frame,...

So what were you doing, and what are you doing next?

In summary, we applied two 1 G forces, both performed an equal acceleration of the lower inertia only, and both spent the same energy, having performed equal amounts of work.
Thus half the final momentum and velocity was due to GPE, and the other half due to the 3 second 2 G mutual impulse.
58.83990 / 2 = 29.41995 m/s of velocity from each source.
So 29.41995 P from GMH, plus 29.41995 P applied internally.
Since only the lower inertia has been accelerated by either force, we only need to repay KE and P to GMH from its motion.
Leaving it with 29.41995 P and corresponding 432.76672 J (per 1/2 * 1 kg * 29.41995^2) from internal work.
So to recap, we ended up with 1731.06692 J of KE.
Half that - 865.53296 J - came from GPE.
We also ended with 58.83990 m/s of velocity and equal 58.83990 kg-m/s momentum.
Repaying half the 58.83990 m/s that came from GPE, we're left with 29.41995 in m/s and kg-m/s, with an energy value of 432.76672 J, which is precisely half of the 865.53296 J that didn't come from gravity, but was instead applied internally.
The reason half our input energy has disappeared is because KE squares with velocity, and we've just halved velocity, so we're left with a quarter of our net total energy, and half our internally-applied input energy.

.
WTF?
WHY complicate things, and what are you doing anyway? All velocities (in progress) are already accounted for.
You keep combining things from different frames of references, and recombine them again... At least that's what I suspect.

Two things: There's energy in the drop, there's energy in the expansion....
You may look at it from one way, or the other. Combine the shit out of those frames and energy will never be conserved.

So here we are:

If you can't or won't get a grip on basic CoM and CoE you have nothing to add here but noise

I mean if you can't understand how N3 enforces CoE then you can't even measure or calculate anything - none of your velocity, acceleration, force, displacement or momentum or KE maths has any context without an inertial frame, which depends entirely upon N3.

Do you even have any clue why "KE squares with V" at all, or is it just some given fact you found somewhere? I seriously doubt you understand.

I'm not trolling even when I do what you claim I do. Was I wrong with my information of angular velocity? Where should I have found that?
The pattern: I am able to show you the complete algebraic proof: so you can fill in the numbers. I usually do, but I short-cut this time because otherwise I flaunt my "math-wizz" or something according to you...

You found a brain fart, i've said repeatedly they're probably there, get over it, i haven't come here with a fully-fledged theory to expound, i've simply laid down some BS to kick-start myself into some research, followed up on some of it and found this. It's a process of discovery, not divine inspiration. If you can find any errors that actually alter the result, then fix them instead of just gloating over it..?

In all your confusions, please point out, in the tsunami of bla bla, which are to good words, and which should be ignored: links to posts, quotes to texts....

You are barely able to some some proper calculus where you drop your system in 1G=9.81 m/s² giving a velocity of 9.81 m/s in 1 second... while expanding that system with 1G=9.81 m/s giving a momentum of 9.81 kg m/s but because it drops the top one is 0 kg m/s and the other mysteriously 2*9.81 m/s....
Even in that extremely unlikely case where you have some hidden validity in your talk (it's random so there must be something): this use of numerology is confusing by definition. We have algebra to avoid this exact same issue.
Hence: There is a difference between a=9.81 m/s² and g=9.81 m/s², even though they look the same and have the same units. Maybe I looked it up, what's your excuse?
Hence my 3 seconds proposal. Do you think I looked that one up too? And even when I did: What ABOUT that? Still a good idea, right?

This "Cost of energy": N^2 J/kg-m/s = (1/2mV^2) /4)
I showed way earlier: N=E/p=v/2 (did I look that up?) But duh! N depends on that velocity... just as I show, and not mystically assumed, hinted or hidden in drunken texts.
And re-insert your N-factor: E=½m·(N)²=½m·(v/2)²=m·v²/8 ... compared with its original ½m·v²: And what factor will we find when comparing these?
You take a slightly different route (so I couldn;t have looked it up), but with the same result: and it is 4 = E / E[E/p].
You will ALWAYS find 4 this way !!

And then there is time, this value N depends on time. Time per (what you call) reactionless interaction, or something.
Time gives nice intervals, but I don't think you have a clock on board your mechanism or otherwise some working pendulum doing it's timed latching during freefall.
Not that even an off-site timer solves your (math) issue, but still.
I'm not talking about "typos" , or even about the units of this introduced N^2. I'm most of the time even overlooking the physical validity of this factor!

Could it be that you realised you'd inadvertently confirmed its value for one cycle already, and now you're trying to back away from that because if it's true for one cycle, it must be so for any number?

NO ! I told you I already found your factor 4 way before.

I want you to co-discover the reason of this N-factor. You use it, you describe it a vaguely. It has and I see this mathematical thing.
Yet it has no grounds and validity in physics !
But 'm not a physics guy. I simply don't know. I see you use it here, there and everywhere... but why: that remains a mystery?
If it is THAT important: find the reason. The rest you already figured out, including the possible planetary side effects...
By the way, you also need at least two or more cycles before even considering a constant. And 1 second, as told, is treacherous and so is one cycle, especially when combined and so is all kinds of values designed to be the equal but are actually not: as explained.
Why do I have to defend myself against your insane assumptions?

At least I don't try to attack you on some personal level, or use all kinds of mathematical distractions which looks and sounds almost right or otherwise complicate the shit out of things.
And you show your 'decency' by pointing me out as some sort of troll?
Why? It's simple: you don't want me here and thus trying to flush me out.
I'm likely one of the few able, willing and trying to understand your thing, while also able to do some math on iit anyway while not acting like some fanboy: aka, an annoyance.

True, I admit that I'm not making your live easy. Wait until you find yourself some real physicist.

Sorry for adding noise relative to your level of understanding.
I was under the impression you could handle some, my bad.

I'm going now, bye.
Have a confusing life.

[MrVibrating]

I thought you asked for that linear to angular stuff somewhere.... Don't know which post, don't care to look it up (it seems I'm good at it).

I use Unicode characters to make them more readable... I like it that way.
Those are formula's I copy pasted from my own list !
I have more.... likely website use LaTeX, or Sub/superscript, or images or I don;t know... who cares any way.
Where do you show the speed of radians per second, or even the notion that linear gravity is converted into angular acceleration ? Where?
I show

Yes this troll is attacking you personally... sure that my MO here.
Assumptions!
Why do I need to defend myself against this... hmm name it.

BS, no one thinks in terms of S.I. symbols, we process units mentally in their respective dimensions - i've only resorted to using "P" in place of kg-m/s after ensuring to clearly define it and in situations where i'm calculating momentum repeatedly. Similarly, if i used "l" for angular momentum i'd be defining it as i introduced it, and using ASCII not exotic character sets - it's a bleedin' discussion forum for kicking ideas about, not a bleedin' journal publication.

It looks like you're trying to paint a veneer of plausibility or 'authoritative legitimacy' you simply do not deliver on, while trying to exploit that ruse to mis-represent and undermine what i'm doing.

You've contributed SFA so far, i don't need your assistance or blessings and seem to be doing just fine on my own for now, thanks.

Anyone unable to follow what i'm doing can just sit back and wait; either this work will result in successful builds or it won't. I'm convinced it's now a foregone conclusion, but don't expect anyone to be holding their breath, much less understanding the gain principle.

Post any BS you like, just don't expect any more engagement from me.

[MrVibrating]

Green: inserted
Red: corrected

Linear-linear case:

• For accuracy i will use the same five significant digits as 1 G = 9.80665 m/s^2.
  
  Not that more accuracy helps here!
  The lower 1 kg accelerates with A=2G, for t=3 seconds. It descends: h=½a·t² = 88.25985 m
  
  The lower 1 kg descended by 88.2598 meters, producing KE equal to AMH = (2*9.80665 * 1 * 88.2598) = 1731.06692 J.
  The upper 1 kg has not changed height, so still has all of its GPE and cannot have contributed any KE or momentum.. or else we'd effectively be able to drop it twice.
  So subtracting the ...<snip> WHY? It is 1731.06692 Joules. done.

The upper mass is stationary, Marcello. It has no energy.

WHY do you need to subtract things: all energies are accounted for.

To determine the input energy cost of using gravity to cancel counter-momentum.

That's the whole brief, repeated yet again in the prior post, after you asked me to verify its constancy over three seconds.

So you know perfectly well why i'm delineating the input energy and momentum sources.

You were the one who asked me to do it.

As you said many, many, many times yourself: KE=½m·v² ... done.

since there is no dispute that 1/2mV^2 remains valid in our external, static reference frame,...

So what were you doing, and what are you doing next?

Measuring input energy, not output energy, like you asked me to, and which the whole gain principle depends upon.

You know this already, i'm entirely depending on output energy conforming to its standard terms, all results are based upon them, the sims the calcs and everything.

Yet you just copy out my own results and claim to be 'correcting' them, while adamantly ignoring the input efficiency you just had me cross-check again.

In summary, we applied two 1 G forces, both performed an equal acceleration of the lower inertia only, and both spent the same energy, having performed equal amounts of work.
Thus half the final momentum and velocity was due to GPE, and the other half due to the 3 second 2 G mutual impulse.
58.83990 / 2 = 29.41995 m/s of velocity from each source.
So 29.41995 P from GMH, plus 29.41995 P applied internally.
Since only the lower inertia has been accelerated by either force, we only need to repay KE and P to GMH from its motion.
Leaving it with 29.41995 P and corresponding 432.76672 J (per 1/2 * 1 kg * 29.41995^2) from internal work.
So to recap, we ended up with 1731.06692 J of KE.
Half that - 865.53296 J - came from GPE.
We also ended with 58.83990 m/s of velocity and equal 58.83990 kg-m/s momentum.
Repaying half the 58.83990 m/s that came from GPE, we're left with 29.41995 in m/s and kg-m/s, with an energy value of 432.76672 J, which is precisely half of the 865.53296 J that didn't come from gravity, but was instead applied internally.
The reason half our input energy has disappeared is because KE squares with velocity, and we've just halved velocity, so we're left with a quarter of our net total energy, and half our internally-applied input energy.

.
WTF?
WHY complicate things, and what are you doing anyway? All velocities (in progress) are already accounted for.

? I performed the calculation of input efficiency over 3 seconds as you requested.

You have neither performed the same calculation yourself, nor checked my solution.

So if you think you know a simpler way, we're yet to see any sign of it.

You keep combining things from different frames of references, and recombine them again... At least that's what I suspect.

LOL you 'suspect'?

If the standard KE terms are producing the correct results then by definition all such calcs must be in the same reverence frame, duh?

If we were applying alternative reference frames, then all KE calcs should be equivalent between them!

That equivalency would only be broken by exploiting an effective violation of N3 - yet you freely admit you don't even grasp the concept of inertial frames and thus CoE's dependent on N3 in the first place!

Without that understanding, all of your velocity, momentum and energy calcs are built on a bed of marshmellow. You don't know why KE squares with velocity... it just does!

Two things: There's energy in the drop, there's energy in the expansion....
You may look at it from one way, or the other. Combine the shit out of those frames and energy will never be conserved.

LOL what?

Where are you getting this alternate FoR from, and why is its equivalence from any other frame broken? Where are you getting this from - the sims and calcs were all based in Earth's FoR, ie. the standard static frame.

At no point in any of the calcs you asked me to perform was any other FoR invoked in any way whatsoever.

All GPE and additional input energy, and output energy, and all velocity and momentum, is measured relative to the static frame.

Again, you're throwing faeces, making up BS that hasn't happened and which you thus cannot reference. It's all insinuation and facile attempts at misrepresentation.

So here we are:

If you can't or won't get a grip on basic CoM and CoE you have nothing to add here but noise

I mean if you can't understand how N3 enforces CoE then you can't even measure or calculate anything - none of your velocity, acceleration, force, displacement or momentum or KE maths has any context without an inertial frame, which depends entirely upon N3.

Do you even have any clue why "KE squares with V" at all, or is it just some given fact you found somewhere? I seriously doubt you understand.

Because of the practicalities of N3, like i've patiently explained to you a thousand times.

It is you who have freely admitted you don't understand this, a few days back:

I see CoE as a byproduct of N2, not N3.

...so may we assume it's finally sunk home for you then?

I'm not trolling even when I do what you claim I do. Was I wrong with my information of angular velocity?

It's just irrelevant disingenuity isn't it? What's it got to do with the ratio of input energy to momentum - ie. the specific subject of enquiry of the calcs you asked me to run?

Where should I have found that?

I explicitly gave the MoI as 0.12500 kg-m^2, so RKE / 1/2MoI = RPM! Why would you need to go trawling to a physics site to copy-paste the derivation if you already knew all this? I've repeatedly given RKE as half angular inertia times angular velocity squared, in case anyone didn't know it already. The objective (set by yourself) was to assess the ratio of input energy to momentum generated, which i have done, and you have not.

The pattern: I am able to show you the complete algebraic proof: so you can fill in the numbers. I usually do, but I short-cut this time because otherwise I flaunt my "math-wizz" or something according to you...

LOL no you've copy-pasted formulas you've had to go look up and which you evidently don't even know how to use off the top of your head.

But you haven't once performed a single calculation of the ratio of input energy / momentum, which is the whole basis of the exploit here.

Nor have you checked any of my results either. Instead you found a mis-calculated discarded quantity not used in the results, without even bothering to calculate the correct value which replaced it in the subsequent line, derived accurately from 1/2mV^2. Had you actually done so, you'd've confirmed that the output KE was precisely 1/4 the input energy. But that was not your interest or motivation. It never was in the first place, because you're not actually here for the same reasons as the rest of us...

You found a brain fart, i've said repeatedly they're probably there, get over it, i haven't come here with a fully-fledged theory to expound, i've simply laid down some BS to kick-start myself into some research, followed up on some of it and found this. It's a process of discovery, not divine inspiration. If you can find any errors that actually alter the result, then fix them instead of just gloating over it..?

In all your confusions, please point out, in the tsunami of bla bla, which are to good words, and which should be ignored: links to posts, quotes to texts....

Input energy / input momentum = N^2 J/kg-m/s.

(N^2 J/kg-m/s) = ((1/2mV^2) /4) = ((N^2 J/kg-m/s) *4)

((N^2 J/kg-m/s) *5) = ((1/2mV^2) *0.25), etc.

You are barely able to some some proper calculus where you drop your system in 1G=9.81 m/s² giving a velocity of 9.81 m/s in 1 second... while expanding that system with 1G=9.81 m/s giving a momentum of 9.81 kg m/s but because it drops the top one is 0 kg m/s and the other mysteriously 2*9.81 m/s....
Even in that extremely unlikely case where you have some hidden validity in your talk (it's random so there must be something): this use of numerology is confusing by definition. We have algebra to avoid this exact same issue.
Hence: There is a difference between a=9.81 m/s² and g=9.81 m/s², even though they look the same and have the same units. Maybe I looked it up, what's your excuse?

I've said repeatedly that gravity is an effective acceleration, and the ground isn't literally accelerating upwards at 9.81 G, as if that needs clarifying, duh.

Glad you finally looked it up tho, don't take my word on it eh.. much less basic logic.

Hence my 3 seconds proposal. Do you think I looked that one up too? And even when I did: What ABOUT that? Still a good idea, right?

No, as i patiently explained repeatedly already, since N is equal in magnitude to F/m it implicitly has squared dimensions, hence the input ratio of energy to momentum already includes all the relevant time derivatives and so is invariant to time translation - we'll always derive the same proportion of energy to momentum irrespective of the impulse period.

This is patently obvious from first principles, as i've already pointed out, but i gave you the benefit of the doubt that you were just struggling to follow basic maths, so proved it anyway with a sim and fully-annotated breakdown, which you've duly ignored, surprise.

This "Cost of energy": N^2 J/kg-m/s = (1/2mV^2) /4)
I showed way earlier: N=E/p=v/2 (did I look that up?) But duh! N depends on that velocity... just as I show, and not mystically assumed, hinted or hidden in drunken texts.
And re-insert your N-factor: E=½m·(N)²=½m·(v/2)²=m·v²/8 ... compared with its original ½m·v²: And what factor will we find when comparing these?

The effective value of input energy per cycle, ie. the output energy from one full cycle is equal to 1/4 the input energy.

The output energy from a second cycle is equal to 1/2 its input energy.

The output energy from a third is 3/4 the input energy.

A fourth is fully-conservative.

A fifth is 125% OU, equal to (1/2mV^2 *.025).

Can you not see that we have two different metrics for the energy cost / value of momentum?

How can anyone not see this?

It converges with 1/2mV^2 at precisely 4 cycles, for perfectly straightforward reasons (halving V quarters KE), is under-unity below that threshold number of cycles and over-unity above it.

As repeatedly demonstrated eight ways from Sunday already...

Marchello, you're evidently incapable of following a simple premise, just quit already, your performance is farcical.

You take a slightly different route (so I couldn;t have looked it up), but with the same result: and it is 4 = E / E[E/p].
You will ALWAYS find 4 this way !!

Precisely, and so ((N^2 J/kg-m/s) *4) = unity, and ((N^2 J/kg-m/s) *5) = 125% unity.

Again, you're struggling and failing to completely ignore this without tacitly admitting it left right and center..

And then there is time, this value N depends on time.

No it does not you blithering buffoon, i've repeatedly explained it already includes its dependent time derivatives, and furthermore just demonstrated its constancy: N^ J / N kg-m/s is intrinsically, self-evidently, obviously time-invariant.

I've proven it is, and you've completely ignored that, much less made any attempt to test it yourself.

So you're resorting back to guessing... and guess what? You're wrong. Mistaken. In error. Factually at odds with the mathematical results, and willfully ignorant about it.

Time per (what you call) reactionless interaction, or something.
Time gives nice intervals, but I don't think you have a clock on board your mechanism or otherwise some working pendulum doing it's timed latching during freefall.
Not that even an off-site timer solves your (math) issue, but still.

There's a clock in the upper left corner of the sim there.

You're not suggesting a clock attached to the wheel would show some kind of relativistic effects, presumably?

I'm not talking about "typos" , or even about the units of this introduced N^2. I'm most of the time even overlooking the physical validity of this factor!

LOL you can say that again.

Could it be that you realised you'd inadvertently confirmed its value for one cycle already, and now you're trying to back away from that because if it's true for one cycle, it must be so for any number?

NO ! I told you I already found your factor 4 way before.

Er no, i found it, demonstrated the /4 series, but could not understand where it came from, which i've since resolved myself.

You still don't know where it comes from, and can still only comprehend values of N^2 J/kg-m/s = (1/4 * 1/2mV^2), though even there you're refusing to even admit that much, let alone the corollary condition that ((N^2 J/kg-m/s) *4) = 1/2mV^2... much less that ((N^2 J/kg-m/s) *3) = ((1/2mV^2) * 0.75) or that ((N^2 J/kg-m/s) *5) = ((1/2mV^2) * 1.25).

It's all right here in the thread Marchello.. you're 20 pages behind what's happening. Events have hopelessly overtaken you. Just do the dignified thing and bow out. So long and thanks for all the fish.

I want you to co-discover the reason of this N-factor. You use it, you describe it a vaguely. It has and I see this mathematical thing.
Yet it has no grounds and validity in physics !

What on earth are you blathering about? It's the amount of momentum we buy with a given amount of input energy?

Momentum scales linearly, while KE squares with velocity. If 1 kg is moving at 10 m/s it has 50 J, and 10 kg-m/s. If we share that momentum into another stationary 1 kg mass, each is left with 5 kg-m/s and 12.5 J, one quarter of our initial energy.

If you think you've overturned the founding tenets of classical physics, why not go and start your own thread?

The value of N is equal to the effective acceleration of the static uniform field, equal to F/m, and it squares because we're applying a 9.81 m/s acceleration nested within another 9.81 m/s acceleration.

All of this has come from me, none of it from you. On the contrary, i've patiently explained these points to you repeatedly, to no avail. You will not learn, you will not think. You're a lost cause and contributing only noise. Give up already!

But 'm not a physics guy. I simply don't know. I see you use it here, there and everywhere... but why: that remains a mystery?
If it is THAT important: find the reason. The rest you already figured out, including the possible planetary side effects...
By the way, you also need at least two or more cycles before even considering a constant. And 1 second, as told, is treacherous and so is one cycle, especially when combined and so is all kinds of values designed to be the equal but are actually not: as explained.
Why do I have to defend myself against your insane assumptions?

All of this is demonstrated and proven already, inside out and top to bottom.

All of it whistled straight over your head mate. Forget it, just quit.

At least I don't try to attack you on some personal level, or use all kinds of mathematical distractions which looks and sounds almost right or otherwise complicate the shit out of things.
And you show your 'decency' by pointing me out as some sort of troll?
Why? It's simple: you don't want me here and thus trying to flush me out.
I'm likely one of the few able, willing and trying to understand your thing, while also able to do some math on iit anyway while not acting like some fanboy: aka, an annoyance.

No you've understood nothing, you're hopelessly confused on the basics of classical physics, you admit you can't even understand why CoE depends upon N3, you cannot understand the N^2/N relationship between a squared vs linear quantity, you cannot understand that inertia is speed-invariant, you can barely sum one cycle of N^2/N, let alone 2, 4 or 5 etc., you're insinuating errors that don't exist, you've not performed a single measure of an interaction's resulting momentum distribution, you ask me to perform self-evidently redundant cross-checks then totally ignore them and misrepresent their results...

True, I admit that I'm not making your live easy. Wait until you find yourself some real physicist.

Sorry for adding noise relative to your level of understanding.
I was under the impression you could handle some, my bad.

I'm going now, bye.
Have a confusing life.

It's just a shame, that's all. Given the singular purpose of this place, you might think a fully-consistent OU energy gradient would garner a little objective appraisal, galvanise some degree of 'productive collaboration'..

But instead it's just a one-way street with me doing all the slog, having to repeat everything ad nauseem while deflecting specious accusations of erroneous conclusions and absolutely zero corroboration of the detailed symmetry break between I/O energies.

I provided all the information as accurately and honestly as i could, and this is all the response i've got.

So yeah, proper physics-talking dudes are probably gonna have some fun with this, one way or another.

If it's wrong and i've made some crass mistake, i'll finally find out and no harm's been done. At least, asides from prickling my already-rock-bottom sense of dignity.

If it's right - and obviously i'm convinced it's incontrovertible - then we all benefit, even if the forum missed out on its own red-letter day...

[MrVibrating]

Okay if anyone doesn't like usage of the letter "N" as an arbitrary constant describing the input energy cost of momentum, since its value is derived from F/m we could simply write it as:

((F/m^2 J) / (F/m) kg-m/s)) = (1/2mV^2) /4))

Obviously, replacing it with "G" or "9.81" would make it exclusive to gravity, and incorrect in all other circumstances.

Since the same value is used on both sides of the expression, and it should hold true no matter the force field involved, the F/m value is the only correct definition, and so i chose "N" for "number", made it a capital cos it's a vector quantity (with dimensions equal to acceleration)... and that's all there is to it.

However you choose to represent it, its point is that it's a speed-invariant cost of momentum purchase, so doesn't rise with RPM like RKE does.

So please just ignore Marchello's bone-headed whimperings, represent the value any way you like, it's just the optimum cost of momentum obtained by full-cancellation of counter-momentum, using any static uniform force field you like. It's speed-invariant, time-invariant, and if your wheel design exploits 'artifical-G' from a horizontal centrifuge, driving the centrifuge from the Bessler wheel will supercharge the CF and thus the momentum yield per cycle... But no matter, N^2 J always buys N kg-m/s of momentum.

Even if its corresponding 1/2mV^2 value is four times less... or 125% more.. or thousands of times more.

So if that doesn't warrant having its own parameter then i don't know what does, but use ((F/m^2 J) / (F/m) kg-m/s)) = (1/2mV^2) /4)) by all means if you find it more comfortable..

I'm just trying to keep things as simple and comprehensible as possible.

[eccentrically1]

Edit - My responses to MrV are in blue...

On page 2, I replied that once the system is started, the upper mass that is acting as a push-off for the wheel will be the negative angular momentum as soon as the mechanism reaches its end of travel. The wheel won't feel its braking effect until then. The upper mass is going to be at rest, and when the delayed reaction kicks in it will have to be accelerated by the wheel. I don't see that this issue has been addressed but maybe you did, sorry in advance if I missed it, in which case you can ignore me too.

Perfectly reasonable request for clarification.

(And thanks for being so civil! Sorry if i didn't address this issue sufficiently.)

Again, the masses are mutually accelerated against one another's inertia, whilst one or both of them are gravitating.

Then they're mutually-decelerated against one another, when neither are gravitating.

This is the magic step/mechanism , no one has found a way around it. Everyone's design stalls as this point. Why would a preloaded force (spring, jack, etc) between the upper mass and the wheel all of a sudden work now?

This rectifies momentum from gravity's acceleration.

Because we're applying a mutual force between two inertias, but only one of them is actually accelerating, we're inputting momentum of one sign only.

This is the reactionless acceleration, but the reaction force for the acceleration is hiding there in plain sight- Earth's gravity! Similarly, the reaction force for Cp acceleration is inertial.

So when we then come to mutually-decelerate them, instead of having induced positive and negative momenta that sum back to zero, we end up with two positive momenta of equal value.

That's not the way you described it earlier or the way I see it. The preloaded force is divided into a positive momentum (the wheel) and a zero momentum {the upper mass providing the inertial (fictitious) force}.

All of this momentum has been paid for - none of it is 'excess'. At no point do we depend upon 'extra' momentum that hasn't been produced by an internal expenditure of work.

Yes, the momentum is preloaded in the mech for startup. No one has been able to design a mech that internally produces that needed extra momentum.

So it's not "free momentum from gravity" - rather, gravity passively assists in causing a momentum asymmetry between the two inertias, all of which has always been paid for, but which sums to a step-wise net increase instead of mutually cancelling.

The momentum is asymmetrically distributed, but not increased. Lots of interactions have asymmetric results, but in the final analysis, all of the different system properties (different types of energies, linear and angular momentums, different types of forces,) sum to zero. If they don't, we haven't included all of the contributing properties in that final analysis and our results are skewed.

As we repeat that cycle, each new one begins at the net ambient velocity the last one left off at, but the ever-rising speed has no effect on the energy cost of the momentum we're adding each cycle - it's a constant 9.81 J/kg-m/s.

We haven't repeated a cycle yet, or I missed that page. If I didn't, I'd like to see you (or anyone with 2d!) simulate a design, the simpler the better. Just a 1G preloaded spring (not a massless spring please) with a 1 kg mass on one end, one mech on each side for balance, or whatever the force/mass match needs to be for the concept to work since the wheel and balancing mech will also have mass and some startup friction, attached to the wheel spokes, turn on air resistance, fire it off, and see if it gains speed, even after the opposing mech is fired. Did I miss that?

Thus because KE=(1/2mV^2) or (1/2MoI*RPM^2), input and output energies are completely thermodynamically decoupled, only converging - almost incidentally - at 4 interactions.

The input energy is GPE and elastic potential, of external origin to the wheel system. The springs didn't load themselves, or attach their selves up on the wheel spokes. When you include the wheel environment, the input and output are thermodynamically coupled.

Thus an OU result is just as easily obtained as a unity result from the same process... it's just one more cycle of precisely the same interaction.

As i say, if you can reach the second rung on the ladder, you can surely reach the fifth..



PS. on page 6, look again at the green disc demonstration; this is the basic effect, albeit exaggerated to a functionally-useless degree since no further such accelerations are possible after the first.

On page 8 i made a small start on considering mechanisms that might be applied to produce and control the process.

Obviously, all earlier findings are subject to later correction - this is research & discovery not gnosis - but the core exploit is simply a mutual deceleration, following a not-so-mutual acceleration. Any system that can accumulate the resulting gain over four consecutive cycles has no more energy or momentum than it should have, therefore we're all saved and/or doomed.

Well, I agree the green disc on page 6 is useless.

The mech on page 8 is a small start, it seems to have fallen by the wayside.
Looking forward to any results from your concept, from anyone.

[sleepy]

Mr. V wrote
[/quote]In summary, we applied two 1 G forces, both performed an equal acceleration of the lower inertia only.

This is the "impossible part" that is taken for granted in the equation.Please think this through.Both weights are being accelerated evenly.There is no way to borrow force from one interaction and apply it to the other,unless the mech between the weights is free to move toward the lower weight.Then you must consider "where will this contraption be hooked to the wheel". If the upper weight must be free to move,and the lower weight must be free to move,and the firing mech must be free to move,then there is no way to harness the reactions.This entire concept hinges on this point,but it is being disregarded.

[MrVibrating]

Well, I agree the green disc on page 6 is useless.

That tone suggests you see no merit in it whatsoever... but its purpose is absolutely 100% identical to all the other tests. It performs precisely the same function.

It's only impractical to use because it's a single protracted acceleration, and once the velocity is equalised with the radial weight lever, no further inputs are practical, since the weight lever's orientation to gravity is changing too fast.

However it also very neatly demonstrates the effect in an entirely different setting to a falling scissorjack, and makes the principle much more intuitive to follow, since everyone can instantly grasp the key points:

- The weight lever wants to gravitate downwards

- However a torque is being applied between it and the wheel, preventing its downwards acceleration whilst accelerating the wheel

- At the end, when the weight lever is locked onto the wheel, because all of the acceleration was applied to only one of these two interacting angular inertias (the wheel), when it is locked against its 'reaction mass' (the weight lever), both take off together instead of cancelling or bouncing off one another in opposite directions. So we've input momentum of one sign only, not equal opposing momentas. They thus sum linearly, instead of mutually-cancelling.

The ratio of input energy (e) to momentum (P) obtained is equal to (e^2 / P), or equivalently F/m^2 J/kg-m/s.

So instead of taking one big bite of momentum like this, if we had instead taken 5 much smaller bites, we'd have generated 125% more RKE than we'd spent on those 5 blips of counter-torque.

The effect is this flexible because it's time-invariant - the ratio of (e^2 / P) is the same, no matter how small the quantities, and so 5 bites of momentum bitten off this way are 125% OU, no matter how big or small they are.

It's some relief that the mechanical principle looks so simple and obvious to you... because that is really all there is to it; use gravity to generate a momentum asymmetry between two inertias, then equalise their velocity to a new net increase and repeat 5 times to hit OU.

Alternatively, we can input (5 * (e^2 / P)) and immediately achieve 125% unity in one single interaction.

But all it boils down to is accelerating an inertia, so generating momentum, whilst using gravity or some other external force to avoid inducing a corresponding counter-momentum in the other mass you're pushing against.

As a result, you only generate momentum of one sign, and since there's no counter-momentum to cancel it out with, sharing / dividing that momentum with the reaction-mass-that-never-moved accelerates the net system, just a lil' bit.

The KE you're left with after doing that once is only 25% what you began with.

But if you can repeat it any number of successive times, that efficiency rises another 25% each time.

Thus after 4 cycles your output energy finally equals your input energy.

And after 5 you have 5/4 your input energy, a 125% gain.

After eight, 200%, etc. etc.

[MrVibrating]

This is the "impossible part" that is taken for granted in the equation.Please think this through.Both weights are being accelerated evenly.There is no way to borrow force from one interaction and apply it to the other,unless the mech between the weights is free to move toward the lower weight.Then you must consider "where will this contraption be hooked to the wheel". If the upper weight must be free to move,and the lower weight must be free to move,and the firing mech must be free to move,then there is no way to harness the reactions.This entire concept hinges on this point,but it is being disregarded.

Mate i've addressed this repeatedly.

The linear-linear example is used as a simplified, reduced test-bed for examining the relationship between input energy and resulting momentum sign.

According to the maths, it could be built into a working PMM. The physical practicalities of such a contraption notwithstanding, there's nothing physically impossible about it.

But why you would want to attempt such a design is the question? Why not just obviate thrusting a linear mass downwards, and apply torque to a wheel instead?

As far as i can see there'd be nothing gained from a linear-linear design, besides the challenge, noise and spectacle of it running. Or even trying to run. The angular-angular option seems much more elegant, and since power = energy / time, and OU efficiency is a function of accumulated momentum, the higher RPM's possible will significantly boost performance.

Just forget about linear-linear as a build option for now, it would only be worth trying for fun, & horribly clunky unless designed to a high standard.. but it's totally unnecessary to plunge a mass downwards, and torquing a wheel that has equivalent angular inertia has precisely the same net effect.

[eccentrically1]

That tone suggests you see no merit in it whatsoever... but its purpose is absolutely 100% identical to all the other tests. It performs precisely the same function.

I see the same merit as you:

PS. on page 6, look again at the green disc demonstration; this is the basic effect, albeit exaggerated to a functionally-useless degree since no further such accelerations are possible after the first.

However a torque is being applied between it and the wheel, preventing its downwards acceleration whilst accelerating the wheel

And the torque is :

The situation is still idealised however, as a motor is being used to apply a constant angular acceleration.

Where is the merit?

I can't respond to the rest of your reply since you haven't responded to the remarks I made in blue?
Looking forward to anyone's results with your concept.

[cloud camper]

Since we are talking about a Reactionless process, by definition this means there can be no reaction on the wheel either.

This leads us directly to the MT6 problem for which there is no solution.

[eccentrically1]

CC, it isn't reaction less, earth's gravity provides the reaction force. See my post above the last one.

[MrVibrating]

And the torque is :

An angular force applied between the two angular inertias.

Where is the merit?

The energy cost of purchase of reactionless momentum is speed-invariant.

The energy value of momentum is speed-dependent.

That is the only merit of any of this.

I can't respond to the rest of your reply since you haven't responded to the remarks I made in blue?
Looking forward to anyone's results with your concept.

Yeah i'm not holding my breath either..

[MrVibrating]

CC, it isn't reaction less, earth's gravity provides the reaction force. See my post above the last one.

Precisely.

In essence, the reference frame of the on-board inertia the wheel is being torqued against is that of gravity's free-fall reference frame.

As far as a Bessler's wheel is concerned, it's literally as if the ground beneath it were rushing upwards at 9.81 m/s^2.

In other words the internal inertial frame isn't simply diverging to infinity, but specifically to the 'infinite displacement' of the effective (ie. 'virtual') acceleration of gravity... in which there is no actual net displacement.

Everyone needs to remember tho that the whole Earth is still accelerating upwards towards a weight that isn't accelerating back down...

..so even though a 9.81 kg-m/s momentum divided by the mass of Earth produces a vanishingly small net acceleration, pretty much any acceleration is non-trivial, especially when extrapolated en masse as a global alternative to conventionals.


We're bickering here like savages who've just dug up a radium crystal, arguing over who saw the shiny first and what it all means.. all i know is, i don't wanna be stashing it under my pillow..

Anyone seen the news reports saying we're in for a quake-tastic 2018? Because our rotation undergoes periodic variations? How in the hell that happens in the first place (surely our angular inertia would have to change?) i don't know, but the point is they're predicting hella lotta rumbles from a shift in day-length of microseconds... tiny change to our resting momentum, big knock-on consequences down the line...


It is of course alarmist and completely unwarranted to draw any conclusion between the unprecedented event of Bessler's longest and largest, most powerful wheel demonstration in Kassel, and the unprecedented flooding that ensued two weeks after it ended.

Correlation is not causation. So even though we now realise that system was likely extracting megajoules of negative GPE by propelling the earth upwards, and that this would in turn cause an excess of gas and liquid to slosh 'downwards' towards the southern hemisphere, rebounding with unprecedented tidal surges and storms a few weeks later, the fact that this actually happened does not imply any direct link. The word "coincidence" exists for a reason, and this is one.

Besides, the same prediction would likewise apply to the fluid components of the Earth's outer core - just like this day-shortening apocalypse we're apparently in for - causing equally dramatic effects on the opposite side of the world.

Which would take us all the way down to New Zealand, which in 1717 still lay undiscovered by Europeans and so no such records even exist.

The only long-term records we do have for the area are geological. Which actually do show that an 8.2 mega-quake did strike the area at around that time, although the exact date remains undetermined. But even if it was sorta mid-November-ish, two swallows do not a summer make.

So the direct line of cause and effect we could draw between these events is purely hypothetical, and unprovable.

Not unfalsifiable... just, not in any good way (ie. if there's no one left to say "told you so!" then it's all kinda moot anyway)..

Wotta lotta rhubarb eh..? Still, i just hope no one's still expecting this to be free energy from nowhere...

[eccentrically1]

And the torque is :

An angular force applied between the two angular inertias.

Which is supplied by the motor, external to the disc.

Where is the merit?

The energy cost of purchase of reactionless momentum is speed-invariant.

The energy value of momentum is speed-dependent.

That is the only merit of any of this.

That applies to any possible interaction. We seem to be in agreement here.

I can't respond to the rest of your reply since you haven't responded to the remarks I made in blue?
Looking forward to anyone's results with your concept.

Yeah i'm not holding my breath either..

Sorry to hear you've thrown in the towel. It was a nice try.

[MrVibrating]

So hang on a minute - given gravity's distinction as a curvature of space-time, does this scupper our chances of replacing it with magnetism to co-opt that net linear force for intentional propulsion?

Throughout, i've simply assumed force is force, F=mA.

In my head i'm imagining a box containing a pair of contra-rotating Bessler wheels, employing lumps of iron instead of just inert masses.

At one end of the box is an electromagnet we can switch in and off at will.

Together, the clockwise and counter-clockwise magnetic angular inertias are drawn towards the EM, both "downstrokes" in-sync at the same time, whilst their corresponding accelerations are arrested and applied instead to their flywheels via internally-applied forces, as here.

It seems to me that the EM, and thus box + net system, is accelerating towards a pair of masses that are not accelerating back..? If we then switch off the EM as the masses reach BDC and coast around for another identical cycle, is this displacing the net system a little each cycle?

I'm thinking the excess force here is from the magnetic field's asymmetric acceleration of the box + net system, relative to the decelerated attraction of the iron masses? Again, we'd be gaining net momentum by burning off (dissipating) the KE corresponding to the cancelled counter-momentum...

Not asking or expecting clear answers of course, just kicking it out there.. it's all angels on pinheads till the proof's physical..