Plying CF as pseudo-inertia to scam N3

[MrVibrating]

Had to leap out the bath for this one, gonna be freezing in a mo, so quickly, here's the solutions to the AP wheel, and "a great craftsman" clues:


The AP wheel depicts three 'quarters'.

3/4 = (3 * 25%) = 75% unity.

Bear with me on this.

5/4 = (5 * 25%) = 125% unity.

And of course, 4/4 = (4 * 25%) = unity.


Hopefully now you can already see where this is going...

4/4 = unity in the reference frame of the ground - the meeting place or boundary line between what lies above the earth, and what lies beneath.

At 5/4 we open a portal to the heavens, whose bounty flows down from above.

But at 3/4 we open a pit, into which the vis viva - the 'living', motive essence - is sucked down, irreversibly lost..

Thus the AP wheel motiff is a curse, the meaning of which would only be obvious to someone else who has discovered the means by which to climb higher upon Jacob's ladder...

He's basically attributing the work and motives of his detractors to unholy influences.

Which segues us neatly into the next realisation:

“a great craftsman would be he who, as one pound falls a quarter, causes four pounds to shoot upwards four quarters.�

..cut-n-pasted from a recent post on JC's blog...

...do you see it, then?

4/4 = unity, remember...

In other words, a weight dropping off the fifth rung of the ladder has sufficient advantage to cause all 4 weights on the rungs below it to ascend one rung upwards, thus the weight on the fourth, 'unity' rung in turn ascends to the fifth rung, therein to be endowed with the divine '5/4', 25% bonus, and so primed to perpetuate the cycle.



In case anyone missed it, this is also basically the solution to the Toys page - it's a depiction of the ladder, and the N^2 / N kg-m/s input-efficiency constant is represented by the hammer toys. The staff and chain (items A and B) represent the 5/4 ladder, and the upturned top that RKE / CF is now an output rather than an input.


If anyone still doubts we're there, i salute your pragmatism..




ETA; On reflection, i suppose another good candidate is this squared vs linear relationship of (e^2 / P)...

..since it's ((N^2) / (N)), a '4' on the right corresponds to 16 on the left.

Since N = Force / mass, it could also be equivalent to "weight" and thus "pounds" in Bessler's conceptual framework.

And since it is the same numerical value on both sides, his mental image could've been:

((pounds^2) / (pounds))

Whereas, with our modern view of gravity as a universal effective acceleration, we've initially derived it as ((9.81^2) / (9.81)), and then generalised it to any distribution of force & mass, ((F/m^2) / (F/m)), then substituting (F/m) for the arbitrary constant N, hence ((N^2) / (N)) and ultimately ((e^2) / (P)).

Either way, these are the proportions and ratios that any such clues must be alluding to..

[eccentrically1]

Ok I see how you get 'Bessler's constant' now. That's going to be another issue related to the one below.

The trouble I'm having now is, in your example, the two masses, in the first purchase, got 4.905 kg-m/s each of momentum.

So let's stick with my special rate, and see how it pans out. The only condition is, you have to use a pair of masses. The deal's only valid for couples.

So we make our first purchase. 96.23 J for 9.81 kg-m/s. Divide that momentum between our two 1 kg masses, so now they have 4.905 kg-m/s each, and thus also 4.905 m/s each. At 4.905 m/s 1 kg has 12.02 J, per 1/2mV^2.

So two positives, not one positive and one zero as in the original concept of the 'not-so-mutual-acceleration phase' ? I'm taking it you are still talking about that? A force is applied between the two masses in freefall and the lower mass gets all the momentum because the upper mass is at rest from the cancellation of gravity by the applied force pushing up against it?
If on the other hand both masses each get 4.905 kg-m/s then, they are both traveling the same speed, in the same direction, at half the acceleration of gravity, in apparent freefall?
Wut?

You can also see though how all conservation laws are met during the exploit - if you're seeing any magic then just ask.

I'm seeing magic...

[sleepy]

eccentrically1,
I've been banging on about how this concept only works in freefall,and now I see you mention weights and a mech in freefall. Did you and I have a meeting of the minds? Or did Mr.V mention this and I missed it?

[MrVibrating]

@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

[MrVibrating]

eccentrically1,
I've been banging on about how this concept only works in freefall,and now I see you mention weights and a mech in freefall. Did you and I have a meeting of the minds? Or did Mr.V mention this and I missed it?

I've told you repeatedly that this is wrong, and provided multiple examples refuting your conclusion.

Please address those given answers directly if you wish to keep pressing this argument.

[sleepy]

What are talking about freefall? Are you kidding? Have you changed your mind now and realize that all three bodies must be in freefall to accumulate any accleration? I feel like I'm in the twilight zone. Could you please clarify your current position?

[sleepy]

Ok. We're obviously posting simultaneously. Let's try this. It seems you have changed your mind about just the top weight alone being able to accomplish anything. Correct?

[MrVibrating]

@sleepy

The gravitating inertia is the reaction mass for the non-gravitating inertia, and vice versa.

There is no more complete or concise answer to your question.

In the current general discussion, we should assume that both inertias are angular, not linear.

No free-fall is required.

Rather, a mass which would be able to accelerate ahead of the wheel's rotation, due to its gravitation, instead has that acceleration arrested, and countered by an opposing force (ie. a counter-torque) which in turn applies positive momentum to the wheel.

If this is the first cycle, then, the upper mass initially hovers in mid-air, because it is not subject to any net force and so not accelerating.

In subsequent cycles, it is already moving, but again, upon its descent, the gravitational acceleration it would've undergone is instead countered, again with a corresponding force of twice that magnitude applied to the wheel. So again there is no net force acting on the gravitating inertia, and so its motion (speed) remains constant whilst that of the wheel is further accelerated.

Rectifying the resulting uni-directional momentum rise between the two inertias yields a 25% ratio of output energy to input energy, an initial 75% loss, however since the 25% remainder is conserved in terms of CoE and CoM, the next such cycle only loses 50% of its input energy. The third only loses 25% - as you can see, each successive interaction inheriting the 25% momentum and velocity left over from each preceding one - hence after four cycles our output energy is finally equal to 4 * 25% = 100% of our input energy.

A fifth such cycle thus takes us into OU territory, at 5 * 25% = 125%.

And so our input / output efficiency continues to diverge another 25% per cycle up to whatever speed we can mechanically maintain synchronisation of the inertial interaction's acceleration / deceleration phases relative to gravity's vector.

This should be a complete answer, so please try to have at it before popping the next question.. ;)

[MrVibrating]

here's a comically-crude graphic of how i think Bessler's internal cosmogony basically mapped out:




...pretty basic, and self-explanatory..


As outlined already, 'Bessler's constant' resolves with the 'worldly' energy value of momentum at precisely (4 * (N^2 J / N kg-m/s)), where it is equal to 1/2mV^2 and thus unity.

This motive force being harnessed - the divergence between BC and 1/2mV^2 - is descended from above, to where it mathematically diverges to infinity.

Below that 'unity' plane of the Earth, the motive force is sucked down, thermodynamically, irreversibly lost to the second law of thermodynamics (2LoT) and the entropies of whatever was included in Bessler's definition of "friction" (which i think basically means any form of "negative torque", whatever its provenance).

The first rung down from the 'unity plane' of Earth takes us through that portal to what lies beneath, the "three quarters" of the AP wheel denoting the first rung down into the descent, into perdition, destruction, and 'loss' per se...

If this chimes with anyone then hopefully better, more fuller revelations might come in time...

[MrVibrating]

..going back to using (N^2 J / N kg-m/s) since (e^2 / P) only represents the fields and dimensions, and doesn't preserve the equality of N on either side of the term, which is of course the defining property of the constant - ie. the unit energy cost of momentum.

It seems to me that (e^2 / P) is insufficient unless the numerical equality of 'e' and 'P' is also explicit, otherwise the /4 axis is lost, which seems an integral feature of the fully-optimised symmetry break, ie. operating at peak input efficiency.

[MrVibrating]

...this is how input and output energies evolve as a function of RPM:






...this just gives some indication of the kinds of gains on offer. Obviously this shows the optimum ratio with no regards to the kinds of practical speed limits an actual mechanism is likely to be subject to.

But you can appreciate what i mean by "monster gains"..

Provided the mechanical principle behind a prospective mechanism is sound, robust OU should be within relatively easy reach...

[Grimer]

Ok I see how you get 'Bessler's constant' now. That's going to be another issue related to the one below.

The trouble I'm having now is, in your example, the two masses, in the first purchase, got 4.905 kg-m/s each of momentum.

So let's stick with my special rate, and see how it pans out. The only condition is, you have to use a pair of masses. The deal's only valid for couples.

So we make our first purchase. 96.23 J for 9.81 kg-m/s. Divide that momentum between our two 1 kg masses, so now they have 4.905 kg-m/s each, and thus also 4.905 m/s each. At 4.905 m/s 1 kg has 12.02 J, per 1/2mV^2.

So two positives, not one positive and one zero as in the original concept of the 'not-so-mutual-acceleration phase' ? I'm taking it you are still talking about that? A force is applied between the two masses in freefall and the lower mass gets all the momentum because the upper mass is at rest from the cancellation of gravity by the applied force pushing up against it? ...or in the case of the yoyo, the applied force, the string, pulling up against it.
If on the other hand both masses each get 4.905 kg-m/s then, they are both traveling the same speed, in the same direction, at half the acceleration of gravity, in apparent freefall?
Wut?

You can also see though how all conservation laws are met during the exploit - if you're seeing any magic then just ask.

I'm seeing magic...

With the mention of free-fall this thing is beginning to make sense to me at last.

With the yo-yo the free energy arises between the difference between free fall of the yoyo and restricted fall on a string which induces rotation (RKE - 2nd derivative) and increase in rotation, (PKE - third derivative).

The smaller the axle the slower the fall and the more the energy induced before it hits the ground. Different axle sizes generate different total energies.
Also, the smaller the axle the greater the time from start to finish and the greater the energy induced. By coupling two yoyos of different parameters, different energies, one should be able to get the combination to rise up against the force of Newtonian Gravity (rise up into the gravitational wind in other words) in an analogous manner in which Ventomobils can sail into the atmospheric wind.

Ventomobils rely on their wheels being able to grip on the earths surface so one will have to provide a vertical rack and pinion system for one of the yoyo's.

You have obviously understood this stuff better than me eccentrically one - but I am beginning to catch up. :-)

[Grimer]

here's a comically-crude graphic of how i think Bessler's internal cosmogony basically mapped out:




...pretty basic, and self-explanatory..


As outlined already, 'Bessler's constant' resolves with the 'worldly' energy value of momentum at precisely (4 * (N^2 J / N kg-m/s)), where it is equal to 1/2mV^2 and thus unity.

This motive force being harnessed - the divergence between BC and 1/2mV^2 - is descended from above, to where it mathematically diverges to infinity.

Below that 'unity' plane of the Earth, the motive force is sucked down, thermodynamically, irreversibly lost to the second law of thermodynamics (2LoT) and the entropies of whatever was included in Bessler's definition of "friction" (which i think basically means any form of "negative torque", whatever its provenance).

The first rung down from the 'unity plane' of Earth takes us through that portal to what lies beneath, the "three quarters" of the AP wheel denoting the first rung down into the descent, into perdition, destruction, and 'loss' per se...

If this chimes with anyone then hopefully better, more fuller revelations might come in time...

That's a nice diagram MrV. What drawing program are you using for that?

[MrVibrating]

Just MS Paint off Windows XP!

Zero graphics skills here mate...

[Grimer]

To my disgust I have discovered that I was close to the answer 7 years ago.
It has taken me all that time to fill in the last piece of the puzzle.

On the 26th November 2010, 5:32 am Post subject: Conservative Flaw
I wrote:

I have just realised that there's a big flaw in the argument that gravity is conservative in that it fails to take something vital into account.

Further down the page I wrote:

The familiar yo-yo children's toy manifests a smaller Counter-Gravito-Motive-Force (CGMF). It is only partially grounded by the stationary child's hand at the top of the string. It attenuates the acceleration due to gravity by inducing Newtonian gravity to Ersatz gravity. In other words it transduces most of the Newtonian acceleration towards the earth into acceleration towards its centre.

It's only now that I realise the significance of Precession Kinetic Energy (PKE) that all the pieces have fallen into place.

Well I suppose 7 years is the appropriate time for an apprenticeship and also the time that it takes travellers in the fairy tales to fulfil their mission. :-)

It's all over chaps. Time for the builders. I wonder who will be first. It would be appropriate for it to be John but I'm afraid he's a lost cause. Still, he did write the book and without it I doubt if many of us would be here.