Decoupling Per-Cycle Momemtum Yields From RPM

[Leafy]

You guys are the best. I think I can say eureka. Can the gravity wheel solution this simple...again.lol

[Leafy]

MrV and ME,

Please check this calculation. On the left is the initial position. We have a ball connecting to a disc and can rotate about its joint.

Ball mass = 1kg
Disc inertia = I
R= 1m

Now let it fall clockwise and reach the position like the right picture. The total energy of the right figure should be mg2R, right?

If the disk energy + ball mass (should be 0) > mg2R we have OU. Please check.

Edit: the right picture is a bit misleading. I drew the little block on the wrong side.

[Leafy]

Never mind, there is no gain. My thought was the weight going to the center works against the CF to give energy. It just leech off energy from the imbalance.

[MrVibrating]

Guys, i think i've sussed the riddle - it's bleedin' obvious:


What if 'one quarter' relates to the work potential of a drop, rather than the height distance itself?

For "one pound falls one quarter" let's assume 1 kg over 1 meter of height.

"Quarters" thus have implicit dimensions of 'force' and 'displacement' or work / energy.

Hence the riddle - obvs. describing the efficiency of the exploit - is simply framing how much output work a given GPE - or simply, perhaps, a given input energy - can perform, when using the secret exploit.

Of course tho, it also thus frames the nature of the exploit itself..

So if the 'input' side of the riddle is simply a given GPE:

1 kg * 1 meter = 9.81 J..

..by simply inverting the KE equation, e = ½mV², we can solve for V = (2e/m)^½ and find for how many 'quarters' of output energy a GPE may be worth in terms of a reactionless acceleration involving four times the mass:

• assuming N3 (no exploit), 9.81 J would be enough energy to accelerate 4 kg to 2.21 m/s

• suppose instead that the 9.81 J of output GPE is split four ways: 9.81 / 4 = 2.45

• 2.45 J is enough to accelerate 4 kg by 1.1 m/s

• four such reactionless accelerations in succession, thus having the same fixed cost, would thus accelerate 4 kg to 4.4 m/s, for that 9.81 J of output GPE

• 4 kg @ 4.4 m/s = 39.23 J of KE

• 39.23 / 9.81 = 4


The units are irrelevant of course, only the dimensions matter.

The 'quarters' riddle is describing the efficiency of accumulating reactionless momentum.. just like the Toys page.

Think about it - we know 'quarters' can't be 'height' for the input dimensions since that'd be framing an I/O GPE asymmetry which we know he knows is irrational..

..yet 'height' in the context of a GPE output isn't merely displacement but also acceleration or 'force' - so for a given mass, a given height is also an energy potential..

..yet if the 'output' side of the riddle - '4 lbs by 4 quarters' - cannot pertain to a GPE input, and by default (sheer paucity of alternate possibilities) must instead relate to some kind of inertial interaction; 'mass * distance' is obviously not an energy term, so the likeness is not transferable, and basically meaningless in that context: force * displacement is thus the dimensions of a 'quarter'..

..the riddle thus impervious to cultural or historical differences in interpretation, a purely abstract yet mathematically-precise generalisation of the I/O efficiency of raising momentum sans counter-momentum.

Note also that the maths resolve so neatly as to frame this 'quadratic' relationship very precisely - 'mechanics' basically reducing to gravitational interactions, and inertial interactions, all else being but variations - if "4 lbs * 4 quarters" ain't also GPE, it can only be an inertial interaction, and only asymmetric inertial interactions can produce mechanical OU anyway, so the riddle's perfect fit for the efficiency of such work conversion cannot be coincidence, and once again i claim my £5 etc. etc..

[Robinhood46]

I think that that is well worth 5 quid.
But i'm taking 5 quid back for that nonsense you shared last month, so we'll call it quits.
Keep up the good work and i hope your hand is recovering well.
Best regards

[MrVibrating]

Never mind, there is no gain. My thought was the weight going to the center works against the CF to give energy. It just leech off energy from the imbalance.

Don't be too despondent; something like this has to be on the right track.

I've tried so hard to break this symmetry.. we know it can be done tho, so keep at it.

For another angle on the problem, consider the following system:

• you could apply + inertial torque to counter-balance gravity..

• ..causing a weight thus rising / the system turning, at constant speed - without being decelerated by gravity..

• ..yet you'd find that reducing MoI while preventing the corresponding acceleration causes exactly the same 9.81 kg-m/s per kg of weight as before - IOW keeping your velocity component, but shedding the inertia component of momentum back to gravity..


The point for contemplation being that if all gravity wants is its due share of momentum, what might be the possibilities for decoupling its GPE value from its roKE value..?

CF force = mass * RPM² * radius; so, doubling radius doubles CF force, but doubling RPM quadruples it. So the value of a given quantity of angular momentum is related between rotating and ground FoR's via 'absolute speed' in that ground FoR..

Obvs., to decouple the two we can only have any hope of manipulating / exploiting an RPM-invariant workload, such as torque * angle of an oscillating MoI or something... something that'd cost the same each time, raising the same momentum for the same input work done, ie. reactionlessly..

Just food for thought..

[MrVibrating]

I think that that is well worth 5 quid.
But i'm taking 5 quid back for that nonsense you shared last month, so we'll call it quits.
Keep up the good work and i hope your hand is recovering well.
Best regards

Ah rats.. still, fair cop. I did basically predict tho that the solution would involve various words, spelled out with letters and spaces, so '2 out of 3' and all that..

[Leafy]

Force stacking theory:

Suppose we induce a force over time on a mass. The equation is

F x t = mv

Suppose there is a stacking force always acting on the mass like gravity or CF. The equation becomes

(F+f) x t = mv

The higher the apply force, the greater energy return from the stacking force.[/code]

[MrVibrating]

..so if the 'stacking force' is an ambient time rate of change of system momentum (mv), a prospective PE:KE asymmetry requires an external stacking force with an energy cost / momentum yield invariant of system velocity.. IOW, in an alternate FoR to that of the KE / ground..

..again, it's an effective momentum asymmetry we're really talking about.. the cost / benefit asymmetry usually precluded by N3..

[MrVibrating]

BTW, for what little it may be worth:

• for 'shoots' or 'jumps' - how about "launches"..?

Thus '1 kg falling 1 meter launches 4 kg to 4 m/s'..

Or perhaps better yet, note that the word 'accelerate' never appears in MS's translation of AP - did Bessler even use the word in German? Perhaps '1 J of GPE accelerates 4 J of KE' is the more concise intention..

[Robinhood46]

What about "advances"?

[agor95]

BTW, for what little it may be worth:

• for 'shoots' or 'jumps' - how about "launches"..?

Thus '1 kg falling 1 meter launches 4 kg to 4 m/s'..

Or perhaps better yet, note that the word 'accelerate' never appears in MS's translation of AP - did Bessler even use the word in German? Perhaps '1 J of GPE accelerates 4 J of KE' is the more concise intention..

So you have dropped 1 kg down 1 metre then accelerated four 1 kg masses along a horizontal plain at 1.1 m/s each.

I suppose you can launch a rocket car.

Cheers

[MrVibrating]

What about "advances"?

I find it curious that 'acceleration' didn't appear part of his parlance..

Haven't checked all his works yet but it doesn't appear in AP..

[MrVibrating]

Once again, a clear certainty is that whatever inspirations lead up to the 1712 breakthrough, by some point prior to the publication of AP in 1717 Bessler had, incontrovertibly - and not least, independently - resolved the vis viva problem; discerning not just the distinction and relationship between momentum and energy, but furthermore the interdependence of their conservation laws upon N3 symmetry - and thus that the collection cost of piecemeal momenta could be less than their sum value.

A further point of note is that, unlike the Toys page interaction which depends upon dissipative collisions for the deceleration phase of the asymmetric inertial interactions, the 'quarters' riddle ascribes useful work potential to all of the input energy - it doesn't include dissipative losses..


Does this perhaps allude to internal MoI's oscillating without collisions? As perhaps driven by a crank and conrod - thus alternating in directions but without impacts..?

Or perhaps neither example necessarily represents an actual implementation of the principal, so much as the general principle itself..? IE. 'the efficiency of accumulating unilateral momenta'..

Note also that the solution works out identically for angular or linear dimensions - so substitute 'kg' for kg-m² and 'meters' for rad/s, 'force' with torque and you get exactly the same 400% KE of PE..

..obvs., linear implementations seem less useful to our ends than angular..

..however there's the rub - the external 'stacking force' skewing momentum distributions from internal inertial interactions seemingly can only be gravity and time, yet if the system is gaining 4x the RPM then per-cycle G-times would usually be reducing accordingly - down to 25% of their initial value for the first 'quarter' acceleration of 4 kg-m² by 1.1 rad/s..

..except, this is one GPE output driving four reactionless accelerations, four times in a row.. from a single G*t yield..

IOW, if each of the four asymmetric inertial interactions sinks 100% of counter-momentum to G*t, then the momentum yield of the GPE output must be constant.. geddit?

It isn't decreasing with RPM; apparently, because that one, singular GPE output is driving all four reactionless accelerations..!

So instead of one GPE cycle per inertial interaction, the riddle is obvs. framing a 1:4 ratio.. one individual weight's G-time is divided into four simultaneous reactionless accelerations, four times in succession.. for a single drop!

MT 133-134 are thus evidently related to this theme:

• radial GPE driving opposing angular accelerations

Perhaps suggesting radial GPE's still provide some relief for per-cycle G-times from rising RPM..

• each apparently has an angular work potential of 'four quarters'

It's not clear why or how a momentum asymmetry might be implemented here, to me anyway; i've been doodling with that basic 'CND-symbol' mechanism for years - drop'n'clap - but never sussed how to rectify consistent angular momenta from it.. however if the illustrative theme is more general, depicting the physics of the exploit rather than implementations per se.. as B. said; 'no individual images present a working mechanism, but between them lay out all the ingredients of a solution..'

Likewise, when he mentioned employing 'different principals', there's a variety of different ways of leveraging unilateral momentum from G*t - perhaps even inclusive of non-dissipative, as well as dissipative, solutions..

Also, note that whereas the Toys page solution may or may not factor in a cost for sustaining the external stacking force (depending on whether it relates to a 1:1 or else 1:3 inertia ratio) - but either way definitely includes collision losses - the 'quarters' riddle seems to provide for neither - again, perhaps this is to focus upon or highlight the physical form of the PE:KE asymmetry, rather than its practical efficiency in actually harnessing it, but equally, at this stage, may be hinting at a more-optimal, non-dissipative solution..

The only transiently-'reactionless' stacking forces available seem to be gravity and inertial torque (ice skater effect), tho in both cases thus far i've been unable to break I/O energy symmetry, and obvs., only G*t can actually alter net system momentum, inertial torques just moving it around.

Still, because it features this principal of isolating per-cycle momentum yields from RPM - by quartering each one into four separate momenta from every drop (obvs. solution, no? doy..) - the solution to the riddle does seem to speak directly to a resolution of the thread topic, as expected (!), via sheer exhaustion of alternate possibilities; confirming that we're basically there, at the threshold, knocking at the right door..

.. get yer specs on peeps.. should be some keys lying around somewhere..


ETA: NB - merely rendering a force, in principal, costs no work, just an arbitrarily-low 'switching' energy to set up or align the force field; it is only if +/- work is performed, in the form of displacement by or along its gradient, that the associated energy cost presents as a load upon the field's energy supply - usually, via the inevitable intercession of N3; in short, 'forces' per se are basically 'free', but +/- F*d is always 'work' and thus I/O unity is normally expected.. unless N3 is violated / circumvented, breaking that feedback between load and supply: thus the 'quarters' riddle is framing the same basic gain dynamic as an effectively-Lenz-less motor or transformer, for the same basic decoupling of I/O energy FoR's - IE. reactionless acceleration of the input FoR relative to that of the ground and the rest of the universe..

[MrVibrating]

So you have dropped 1 kg down 1 metre then accelerated four 1 kg masses along a horizontal plain at 1.1 m/s each.

Angular or linear accels, or any combo, in terms of the dimensional maths it works out identically.

The key point i think is that one, unitary G*t yield is divvied-up to sink 4 parallel counter-momenta, four times in a row.

Instead of a 1:1 ratio of GPE to asymmetric inertial interactions, or an asynchronous ratio, as i've previously tested; you avoid diminishing G*t / momentum yields by grabbing them all from the same GPE, at whatever current RPM and thus G*t as a function of it.. rather than spreading the accumulation of 'reactionless' momentum to OU efficiencies over the course of multiple GPE cycles - and thus, constraining the resulting yields to the inevitably-diminishing G*t returns of rising RPM's and shortening cycle periods, defeating the objective..

'Grab it while it's hot', basically..

This, manifestly, is directly framing a solution for the entire raison d'etre and very title of this so-called thread topic.

On a scale of 1-10, where 1 = a gift horse and 10 = a silver platter, this has gotta score at least a '7.5'.. i mean we gotta be well past the halfway point - how many variations can there be on using single G*t yields for multiple asymmetric inertial interactions?