Plying CF as pseudo-inertia to scam N3

[MrVibrating]

Here's that final angular momentum again in alternative units:




...so, i want to know if (N^2 J / kg-m/s) remains valid in the equivalent angular dimensions.

Like i say, bit rusty with the conversion process, so need to chew this over, duh.. yes i know there's obviously a simple way involving Pi or something, just.. need.. Chinese take-away first. Low glucose eh..

[MrVibrating]

..still munching, and binging on Twin Peaks, but just realised, that last set of calcs finally laid clear where this /4 divisor arises:

- for perfect cancellation of our applied counter-momentum, we need to apply a force equal to twice that of the static uniform field

- thus for any fully-asymmetric distribution of momentum obtained via this method from any static uniform force field, we'll need to input precisely equal work as output by that field

- since repaying that contribution from the field for our zero-sum net deal with it eliminates half the velocity corresponding to the total net energy (the PE borrowed from the static field plus our equal contribution applied between the inertias), the value of our remaining input energy is quartered, per 1/2mV^2

Thus we get this consistent *25% or 1/4 axis of multiplication at peak efficiency, regardless of the static force field involved.

So that's the 'magic 4' demystified.

Plus magic five, and all other multiples of the constant... from 1 to whatever.. each successive cycle inheriting the remnant velocities of all preceding ones in a kind of *25% factorial series, type stuff..

[MrVibrating]

..Googling for inspiration for how to interpret the ratio of input energy to momentum in the linear-angular case, i happened across this surprisingly-prescient post from JC:

https://johncollinsnews.blogspot.co.uk/ ... m-and.html

..had i read that in 2011 i'd've probably scoffed at the rationale..! Pretty bang-on the money tho eh?

[MrVibrating]

..phew what a lot of bother... all the boss number dudes seem to think converting between angular and linear momentas is invalid - found one or two engineers offering formulas i couldn't read, then spent ages trying to work out how to calculate the angular equality from first principles - the sim there used a disc with properties 0.5 kg @ 0.5 m radius, MoI = mr^2 = 0.12500 kg-m^2, a magnitude of inertia precisely equal to that of the 1 kg linear inertia. Therefore RPM = 8 * l, where 'l' is angular momentum in rpm-kg-m^2 and so we can divide a given change in RPM by 8 and.. pffft... there's gotta be a simpler way, no?

Notice that in the angular case, the net KE is precisely half of in the linear case, since we have zero GPE output. All of the momentum there on the flywheel has been input internally.

So all we really need to show is that a four-fold rise in angular momentum - however its units or dimensions are represented, so long as they're consistent - corresponds to a 4^2 = 16-fold rise in RKE, right? Ie. use inductive reasoning, assume the predicate's valid and test the resulting prediction, which will either conform, or not..!

Accordingly, at 16 J RKE we have 19.1 rpm-kg-m^2..

...and at 256 J RKE (16^2 J) we have 76.4 rpm-kg-m^2


76.4 / 19.1 = 4, and 4^2 = 16, therefore the angular equivalencies of N^2 J/kg-m/s are also valid.


So it's speed-invariant, time-invariant, applies to all possible force fields and all possible inertias in any plane or axis.


It's actually even simpler than we're making it out, i think - if momentum doubles then so does velocity, assuming inertia's constant, therefore KE quadruples per 1/2mV^2, and 2^2 of course = 4.

So N^2 J/kg-m/s - whatever the relevant dimensions - because a 4-fold increase in momentum means a 16-fold rise in KE... purely because KE squares with velocity, whereas momentum scales linearly.

Therefore i think we can consider the validity of the constant as axiomatic, and perfectly inevitable from first principles.

Hence there's really nothing remotely surprising that (4 * (N^2 J/kg-m/s)) = 1/2mV^2, or equally, that N^2 J/kg-m/s = (1/2mV^2) /4) = 1/8mV^2...

...and thus that (5 * (N^2 J/kg-m/s)) = 5/8mV^2, or that (8 * (N^2 J/kg-m/s)) = mV^2.

It's all the same maths, the same process, same ratios. We can argue no special case for any multiple of cycles. The results of one cycle, with a 75% loss, are as valid at that of four cycles for unity, five for 125% or eight for 200% unity.

It thus seems indisputable that over-unity efficiency as as trivially obtained as unity or under-unity...


If anyone has any questions, anything at all i've omitted or insufficiently justified, with regards to this attempt at mathematical proof, please fire away.

Obviously i accept that i would not be able to read a rigorous "mathematical proof" of anything worthy of that official designation, but as far as the basic mechanics is concerned, i think i've dotted all the i's and crossed the t's..

[ME]

I don't know what you did with your linear stuff, but it's just 1731.06692 J in total.

Attached to your disc you have something going at a 2G acceleration.
With E= ½I·ω² and L=I·ω then in that example it gets an angular acceleration of (4G) rad/s² thus after 3 seconds an angular velocity of (12G) rad/s or (360G/Pi) rpm.
(note: 'G' is here just the number 9.80665)

[MrVibrating]

..so below are a couple of examples of weight levers:




...could be any kind of weighted armatures, attached to the rim or the axle.

And we could be looking at the ascending, or descending side of the wheel.

For sheer practical purposes however, let's suppose we're looking at the descending side of the wheel:

- We need to co-opt a situation in which the weight levers want to drop away from the wheel, out-accelerating it due to their gravitation. In short we want levers that want to flop downwards, and which will overtake the wheel's speed of descent as they drop.

- However we're not going to let them do so - instead, propping them up against that downwards acceleration, with an opposing force twice as strong, applied between these levers, and the wheel.

- So during their descent phase, counter-torque or linear counter-force is applied to the levers or the weights on their ends, preventing their gravitational acceleration, and instead commuting twice that acceleration to the wheel.

- At the end of their drop phase, they've done their job and it really doesn't matter too much what they do next, until it's their turn to descend again and be torqued against.. so they could just hang limp from the rim, or if they're co-axial with the wheel they could be towed around by it, like a 360° pendulum... whatever. GPE-in = GPE out. We're making our money by buying cut-price momentum.

This is not intended as the basis for a design, it's purely to provide some kind of visual reference for how an asymmetric inertial interaction might be applied between two inertias, one of which is gravitating.

As for how to apply these torques, and how to power them via CF workloads from the resulting RKE gain, i couldn't make any better guesses than you... the image above is only meant to convey a simple means of causing and perhaps rectifying an asymmetric distribution of momentum, which is the prime mover here. The potential range of possibilities for doing this are practically unlimited, but i'm too frazzled to think of anything more elaborate right now... just trying to offer a desperate anything to jog the visual imagination, rather than a big empty nothing.. or a damned falling scissorjack.

So yeah, the theory looks great, can't knock it. Solid.

I just have no sorry idea how to build it.

Yet.

[MrVibrating]

I don't know what you did with your linear stuff, but it's just 1731.06692 J in total.

Yes, that's what the KE meter's there for!

With the "linear stuff" i deducted the KE and momentum that came from GPE, leaving only the internally-input single-signed momentum.

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.

Just like you asked..

Attached to your disc you have something going at a 2G acceleration.
With E= ½I·ω² and L=I·ω then in that example it gets an angular acceleration of (4G) rad/s² thus after 3 seconds an angular velocity of (12G) rad/s or (360G/Pi) rpm.
(note: 'G' is here just the number 9.80665)

There is nothing anomalous about either case - the exercise was purely to test the consistency of the N^2 J/kg-m/s input efficiency between angular and linear cases, purely for completeness.

You questioned it's validity, so i've now proven it for all conceivable scenarios - any inertias in any plane or axis, angular, linear or any combination, for any static uniform force field, from gravity to 'artificial gravity' / CF to magnetic force etc.

So it's generalised, cross-referenced with standard KE terms, solved backwards and forwards from first principles.. At this point, i don't see what else there is i can do or need to do with the theory. It's apparently rock-solid. We can definitely build perpetual motion machines.

[Grimer]

..Googling for inspiration for how to interpret the ratio of input energy to momentum in the linear-angular case, i happened across this surprisingly-prescient post from JC:

https://johncollinsnews.blogspot.co.uk/ ... m-and.html

..had i read that in 2011 i'd've probably scoffed at the rationale..! Pretty bang-on the money tho eh?

Here is the "surprisingly-prescient post from JC:" in full.

Angular Momentum to Linear Momentum - and back again?
Sometimes when looking for a solution to a mechanical problem it helps to reverse the sequence of events required - or as Professor Eric Laithwaite liked to do, consider an analogy.

What we seek is a way of converting the downward linear force of gravity into rotational motion. As a means of seeking a solution for those of us who believe Bessler's wheel operated purely by gravitational force, perhap it might help to study the work of those whose ambition is to reverse the process. They wish to generate a unidirectional or linear force from a rapidly rotating weighted lever. There are a number of sites devoted to the experiments http://jnaudin.free.fr/html/IPEmain.htm for instance.

Now whether or not you believe it can be achieved, and I do, it is a fact is it not, that if Bessler's wheel is successfully built and is proved to be driven by gravity, and a working model demonstrated, then it follows that an engine that converts angular momentum to linear momentum can also be developed, since the action or process of one must be the reverse of the other.

Such an engine would have its own extraordinary abilities. It could move over land and water and rise upwards against gravity. Space drive with no emissions.

JC

The mistake John and some other members of the forum make is to imagine that Bessler's Wheel is driven by gravity, implicitly by Newtonian gravity (NG).

It isn't.

NG only acts as a catalyst for the generation of energy by Ersatz Gravity (EG).
anyone who cavils at the term Ersatz Gravity can substitute centripetal/centrifugal force.

The generation of EG is clearly shown by the behaviour of the yo-yo.
In falling its angular momentum increases, in other words its acceleration towards the centre increases.

What is change in acceleration towards the centre, change in the 2nd derivative, change in Rotation Kinetic Energy (RKE)?

It is the 3rd derivative, Precession Kinetic Energy (PKE).

The energy of a yo-yo when it make contact with the ground is far more than the Newtonian Gravity potential energy that has been destroyed. This can easily be proved experimentally by letting the rotational energy of the yo-yo drive it up a hill. It will be found that the height to which the yo-yo rises is greater than the distance fallen.

Another more spectacular demonstration of third derivative energy is to let the yoyo fall far enough so that it is torn apart by the hoop strain energy generated.

Newtonian gravity can't do that can it! The only way NG destroys anything is by impact with the groud.

There's a marvelous example of the destructive force of EG in an accident on the German high speed railways. I'll see if I can dig out a link.

Found it.
https://www.youtube.com/watch?v=0fMWlSjeMnM
I'd hate to have been the passenger who had the unravelled steel tyre come up through the floor right beside his seat.

[MrVibrating]

I don't know what you did with your linear stuff, but it's just 1731.06692 J in total.

Attached to your disc you have something going at a 2G acceleration.
With E= ½I·ω² and L=I·ω then in that example it gets an angular acceleration of (4G) rad/s² thus after 3 seconds an angular velocity of (12G) rad/s or (360G/Pi) rpm.
(note: 'G' is here just the number 9.80665)

Marchello i don't know what you're doing mate but it just looks like you go fishing for math-y type stuff and then come and copy-paste it here to try look smart. Is that unfair? It's a consistent pattern.

Why ask me to go perform some tests of predictions, and then ignore the results and copy-paste the sim outputs without any analysis of the test's objective - specifically, to test the constancy of the input efficiency?

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?

Why do i feel like i'm being trolled?

Intellectual dishonesty = pathoskeptic behaviour.

This looks like textbook trolling to me..

[ME]

??

[ovyyus]

Exactly the sort of textbook response you'd expect from a pathoskeptic troll :D

[MrVibrating]

Dude he isn't doing any maths, can't you see he's just copied the outputs from the pictures i've uploaded, and then copy-pasted the standard RKE formula from some other website without even bothering to replace the unicode characters with ASCII... Duh. As if he just reflexively thinks in those terms and is so familiar using them he knows all the unicode key codes off the top of his head...

He hasn't calculated anything mate. He asked me to integrate the input efficiency over three seconds. I did it, then he just copies the RKE values i posted, copy-pasted the RKE formula he found online and hasn't even pretended to check the constancy of the input energy to output momentum.

He can't even seem to understand the divergence between a linear and squared function, despite pages and pages of simple demonstrations.

He's not 'helping', has contributed nothing but noise, i don't need him for anything and have answered all his questions repeatedly in detail.

If you think he's genuine you're being had.


ETA: Ovyyus - you can't be expected to know what's happening here, not having followed the thread.

And it probably looks like i'm just having a bit of a melt-down.

And in all fairness, i expect i probably am.

But the exploit i've discovered here has absolutely no effect on the standard KE outcomes for a single full cycle; the masses always end up with precisely the right amount of KE.

So the standard KE terms produce the correct results. The theory is not only fully consistent with 1/2mV^2 and 1/2MoI*RPM^2, it explicitly depends upon them holding precisely as they're supposed to.

The whole advantage here is a discount on input energy...

...i've found an input energy cost of momentum that is linear WRT speed.

Whereas under the standard KE terms, energy squares with speed.

So copy-pasting my own KE outputs with copy-pasted formulas from another website has absolutely nothing to do with analysis of the results, and everything to do with a mindless attempt to impress casual observers.

If he ain't outright baiting me, he's still adding nothing but noise.

[Furcurequs]

...and we have some new contestants in this, the 2017 year, edition of "Who's the Bully?"! ...lol

[ME]

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.

[MrVibrating]

I don't know what you did with your linear stuff, but it's just 1731.06692 J in total.

Reads as:
"I can't even follow how you've messed up so badly but you've over-estimated the energy and there's only 1731 J"

..precisely the amount i posted a bleedin' picture of.

What i calculated was the ratio of input energy to output momentum over three seconds, like he requested.

So he said:

"Could you repeat the same calculation for when that duration is 3 seconds (actually "t" seconds)?"

And i said "OK" and promptly did so.

I posted the full annotated results, showing that (N^2J/kg-m/s) is fully time-symmetrical to ((1/2mV^2) /4).

Followed by a solution of where that /4 divisor arises (an outstanding question thus far).

He doesn't acknowledge any of it and instead insinuates i've just claimed an excess of energy, pretending to 'correct' it, simply copying out the same energy i've just posted sims and calcs of...

...and this:

Attached to your disc you have something going at a 2G acceleration.
With E= ½I·ω² and L=I·ω then in that example it gets an angular acceleration of (4G) rad/s² thus after 3 seconds an angular velocity of (12G) rad/s or (360G/Pi) rpm.
(note: 'G' is here just the number 9.80665)

...simply deriving the RPM from RKE as a function of MoI and acceleration / time, and obviously copied wholesale complete in its original unicode characters.

Nothing to do with anything, nothing useful, no rational response to having his requests met, and trying to pretend i've made errors that do not exist (cos if they did we can be sure he'd be all over them).

He's either genuinely thick as two short planks, or just trolling.

I'm wasting no more time with him either way.