Decoupling Per-Cycle Momemtum Yields From RPM

[MrVibrating]

In a nutshell, the issue is:

• what do the maths of OU do, that the interaction above doesn't?

To see an answer, we can only refer back to that perennial question:

• what's the final 'action' in an OU interaction?

..so a system or process is at unity efficiency, then something happens.. and now net of input < net output; what just happened? What is the 'X' event?

?


Marcello, or any other dabblers in the dark arts got thoughts here?

[MrVibrating]

..one point of reference re. that last run:

• note that initially, absolute angle of green makes a clean break on red..

• ..its rate of progression then tails off, barely nudging past 45° total relative displacement after 10 secs..

• ..the relative acceleration per-cycle being an inverse function of RPM..

Gravity's acceleration is a constant rate of change of momentum / unit time, but per-cycle G-time as a function of RPM is a variable...

..and so back to square one; the route to the implicit solution framed right there in the thread title..

[MrVibrating]

Once again, the only sane way forwards seems to be to retreat to the center of rotation; to radial GPE's.

• if the drop-weight from which momentum is being gained each cycle isn't on the 'descending' or 'ascending' sides, then its effective G-time per cycle seems somewhat more insulated against RPM's - it's no longer being thrust downwards with the descending side of the wheel, so, isolated from that acceleration and inertia etc.; i thus can't shake the thought that a radial drop might offer a much more consistent drop time per cycle, and thus wedging open that window for exchanging momentum with gravity and time..

There would presumably still need to be an OB system, but that in itself cannot establish the kinds of coupling relationships between fields and machine that would substantiate OU efficiency; or, more generally, decouple the input / PE FoR from that of the KE / ground.

It all reduces down to removing the momentum source from the rotating FoR - or at least, from its inherent speed / time constraints.

We have a closed system of masses interacting about a fixed axis; momentum thus only has one way in or out - gravitation as a function of exposure time. If our collective neocortices are struggling to integrate that physical concept right now, remember that our cerebellums learned and mastered it circa age ~5, down the park, on the swings..

The only place we're getting fresh momentum each cycle is from G*t.

G*t shrinks with RPM. This is what usually causes and substantiates PE:KE unity - KE squares with RPM, and accordingly, our per-cycle momentum yields - how much momentum our PE buys - is reduced by the inverse-square of RPM. This simple fact should lift the veil of fog from the solid brick wall we've all committed to perpetually fling ourselves against..

..if not illuminating, by implicit reason, any potential path around it..


On this point, another question of translation (or transliteration) comes to mind:

"At present, as far as I'm concerned, anyone who wants can go on about the wonderful doings of these weights, alternately gravitating to the centre and climbing back up again, for I can't put the matter more clearly."

- JC's AP

The original text is:

"Zur Zeit mag noch ein jedes rahten,
Durch was für wunderbare Thaten
Dies’ Schwere nach dem Centro kehrt,
Und jene in die höhe fährt. x
Denn deutscher darff ich hier nicht reden,
Noch öffnen alle Fenster-Laden;"

..for which prof. Google translates:

"At the moment each may still guess
What wonderful deeds
This ’heaviness returns to the Centro,
And that goes up. x
Because I can't speak German here,
Still open all window shutters;"

..he obvs. considers this a juicy morsel, so interpretation's everything, but could this actually be an allusion to the solution to our central underlying physics problem here?

IE. something to do with radial GPE's, as opposed to ones occurring at any radius?

Or perhaps something to do with returning momentum to the axis of rotation?

IE. these types of interpretation would thus be speaking - or at least alluding - to the actual physics of the problem.. whereas, perhaps, Mike Senior's transliteration was unconsciously biased in expectation of a context of a prospective GPE asymmetry, but thus inadvertently flipping the signpost for all those to follow, with a misleading, misconstrued interpretive context..?

[ME]

what's the final 'action' in an OU interaction?

The universe that's correcting a mistake... but can't.

We are so occupied in finding a solution that we overlook that a 'solution' actually means 'balance'.
It is much, much harder to create a serious mistake on purpose.
Our brains usually do not operate that way, so we have to name it differently:
We need a magic trick. Not necessarily for our entertainment, but for physical reality.

Marchello, or any other dabblers in the dark arts got thoughts here?

We only know how the math would look after we have a verified perpetual motion machine.
But when the situation of a PMM is, most likely by definition, not balanced, then the math should be neither.
So we'll not find a closed-form equation for this non-closed-path situation, but something that's only solvable via numerical integration.

An example is the chaotic function: x &#8594; 4x-x². This is related to the logistics map, circle squaring and the Mandelbrot function.
Another example is the 3-body problem. Where, although we use the Newton's equations of motion, it is not likely that three orbiting objects follow a predicable path.

Nevertheless, I still think a perpetual motion machine will find balance at some maximum rotational velocity where the centrifugal force on the mechanism drives out all gravitational influence.
Thus the familiar thing that pops up again in one of the OU-wheel formulae is that PE:KE relation. This time it related Mass, Initial acceleration, Wheel size, Mass displacement and final rotational velocity.

-I know, it does not help much-

[Leafy]

I have a question about the ice skater effect. Does it matter if we pull the mass in or out quickly or slowly? Are there cases which speed remain constant when mass is being moved inward or outward?

[agor95]

Hi Leafy

If you think on it then you will appreciate when your arms are out stretched they are moving at some speed as you rotate.

When you do work too pull them in your rotation rate increases.

Your arms want to continue at the same speed. So the distance your arms travel should not change. However the distance in a circle around the body shortens as they get smaller.

So half the circumference twice the rotation rate. This keep the speed the same.

It's believed pulling in your arm's fast increases the work you do, but the principle stays the same.

Regards

[MrVibrating]

For completeness i dotted the i's and crossed the t's on that last cofig:



• braking torque now a function of rel. speed

• 'spin' phase added - applying constant torque when not braking

..so the wheels - and thus red/greed OB weights - are being pushed apart as they descend, prior to being braked against one another - just like the linear version, pushing falling weights apart vertically.. so now performing 'full' closed-loop inertial interactions - mutually accelerating and braking - whilst gravitating..

..and of course it's still unity.


In conclusion, what really becomes apparent here is that the design and implementation of the OB system is potentially a critical bottleneck condition standing in the way of a momentum divergence / energy asymmetry:

• consider for a moment ANY prospective over-balancing system you like - NOT a GPE asymmetry or energy gain; just continual OB and corresponding acceleration, but running at unity efficiency, KE = PE..


• ..accepting that GPE = GMH, all three of which are invariant, the very notion of a GPE:KE asymmetry basically a logical non-sequitur - a continually-OB system can only run at unity minus losses; neither non-dissipative losses or energy gains are mathematically possible..

• ..but no worries: i'm gonna gift you 100 J, gratis - you're welcome - just tell me in what form you need it so it can be added to your wheel - a spring, a battery, whatevs - and i'll get straight onto Amazon..

• ..so your free box of 100 J arrives on the porch tomorrow afternoon, pristine and the right size and colour etc.; now add it to your OB wheel: et voila - a continually-OB wheel with a 100 J 'gain'!

Q: How is that gain manifested? What's it doing, or done?

IOW if I/O GPE can only ever be symmetrical, how can a continually-OB wheel even physically express an energy gain from any other source?


The question's obvs. of central importance if we're pitting an otherwise-viable sequence of asymmetric inertial interactions against a GPE system that's always going to preclude realisation of the desired resulting gain conditions.

One obvious further optimisation to the above rig would be to re-sequence the actuators to be a function of the relative internal angle, rather than absolute angle in the ground frame..

..or perhaps also synced to the clock, like the motor & brake..

..but would that free up enough 'displacement potential' with which to embody an energy gain? Not sure i'm quite framing the central problem there..

Remember, KE itself can't be 'OU' or 'excessive' in any way - by definition, a system only ever has precisely the right amount of KE for its given inertia and velocity distributions in the ground frame..

So, on the one hand:

• 'OU' thus can only be KE we've paid less PE for; ie. an effective PE discount

..while on the other, PE can also only ever be precisely the right amount of energy for its given inertia and velocity distributions in its frame.. tho this is not necessarily the ground frame! At the same time tho:

• this KE-that-we've-paid-less-PE-for nonetheless needs reinvesting pronto in more PE with which to perpetuate the loop - presumably of the 'gravitational' kind, or else elastic PE..

And to top it all off, the last thing that happens in an OU interaction - the final act that tips it from unity to gain - only finds solution in an effective asymmetric inertial interaction, whether during the acceleration or deceleration / collision phases.

You don't need to accumulate reactionless momenta if the gain-interaction begins at sufficient speed - ie. any reactionless acceleration at any speed is OU - the core dynamic and central exploit is simply that, freed from the usual necessity of gaining momentum relative to a stator, its energy cost of accumulation no longer needs square up with torque and angle to match the resulting KE; instead summing linearly with RPM, while KE of course still squares.

So the central exploit is simply that with reactionless momentum, doubling the system RPM only needs doubling the input energy / work done, even tho it results in four times the rotational KE.

These are the basic foundational stepping stones we need to negotiate some route across..

..but if mech OU's possible, there can't be any conflicting steps..

[MrVibrating]

what's the final 'action' in an OU interaction?

The universe that's correcting a mistake... but can't.

We are so occupied in finding a solution that we overlook that a 'solution' actually means 'balance'.
It is much, much harder to create a serious mistake on purpose.
Our brains usually do not operate that way, so we have to name it differently:
We need a magic trick. Not necessarily for our entertainment, but for physical reality.

Marchello, or any other dabblers in the dark arts got thoughts here?

We only know how the math would look after we have a verified perpetual motion machine.
But when the situation of a PMM is, most likely by definition, not balanced, then the math should be neither.
So we'll not find a closed-form equation for this non-closed-path situation, but something that's only solvable via numerical integration.

An example is the chaotic function: x &#8594; 4x-x². This is related to the logistics map, circle squaring and the Mandelbrot function.
Another example is the 3-body problem. Where, although we use the Newton's equations of motion, it is not likely that three orbiting objects follow a predicable path.

Nevertheless, I still think a perpetual motion machine will find balance at some maximum rotational velocity where the centrifugal force on the mechanism drives out all gravitational influence.
Thus the familiar thing that pops up again in one of the OU-wheel formulae is that PE:KE relation. This time it related Mass, Initial acceleration, Wheel size, Mass displacement and final rotational velocity.

-I know, it does not help much-

Hmm i meant with regards to mathematically formulating an energy gain; IOW overcoming the cognitive dissonance and recognising that if 1 + 1 can't equal 3, then it can't be framing our problem properly either; ie. if CoM and CoE must hold at every step in an OU interaction, the gain arises because, not in spite, of them.. and respecting their application as far as possible in the conversion of PE to KE, the one, singular concession that does break PE:KE symmetry is N3 and its effects on N1 in locking I/O energies into the same FoR.

The last thing that happens in a gain interaction - the only indulgence that delivers the goods, as well as being consistent will all B's clues - is an effective N3 break: a reactionless 1 kg * 1 m/s momentum rise whilst already at 1 m/s has a nominal cost of ½ J, for a 1.5 J KE rise. N3 is enforced by two things - mass constancy, and I/O time symmetry of fundamental force interactions between free bodies; the former's non-negotiable, but time symmetry of closed loops thru gravity are amenable to speed, and thus, force * time, manipulations, per kiiking or classic OB - we can gain momentum without applying counter-torque at the system axis.. just as B's wheels did, and as he explicitly stated the energy gain was dependent upon..

IOW in summary, pretty much all B's clues - and the documentary evidence - unambiguously frame auto-acceleration of the wheel's PE reference frame; inside the wheel, the cyclical workload generating the momentum gain from gravity and time each cycle was impervious to / agnostic of the system RPM - the same internal work done each cycle, raising the same amount of momentum, for the same PE cost. Any system can only ever have the 'right amount' of KE relative to the ground FoR, but equally, every input workload cost and spent the 'right amount' of PE in its, anomalously-accelerated, reference frame.


If we can't even formulate a prospective gain in the first place we've nothing to even aim for.. but nebulous, conflicted hand-wavium.. an operating OU wheel is obviously not in thermodynamic equilibrium - a kind of dynamic disequilibrium, or 'homeostasis, perhaps, but one explicitly substantiated by the anomalous 'divergence' of the input FoR - that is, the FoR of the workload causing the momentum gain is accelerating without pushing / pulling or torquing against some external body or inertia, IOW in a state of isolation from inertial interaction with its wider environment, where the 'equilibrium' balancing the diverging PE and KE values is simply the 'velocity' component of the anomalous momentum gain, transposing the KE value of the PE by precisely the square of that difference.

I ain't gonna condescend with repetition of the scientific method, but formulating and testing 'hypotheses' is kind of the whole gig.. and 'testability' is kind of crucial to that,so.. i think it's simpler than you're making out, and pretty much garden-pathed by what we know of the physics and evidence.. Poirot i am not, sir; this is all rather self-evident (esp. that last point)..

[MrVibrating]

I have a question about the ice skater effect. Does it matter if we pull the mass in or out quickly or slowly? Are there cases which speed remain constant when mass is being moved inward or outward?

In a non-gravitating system, no:

• the same radial displacement from the same initial RPM will cause the same change in rotational KE, instantaneously, and so regardless of radial speed

• the total centrifugal PE is exactly equal to (and is, physically) the rotational KE

• the radial velocity does still determine the radial KE of course, so slamming into the rim or axis inelastically will waste energy.. but that's just dissipative losses and kind of extrinsic to the conservation dynamics you're asking about

Whereas in a gravitating system of course, the accelerations / decelerations caused by these inertial torques can affect rising vs falling period lengths under gravity, and thus the ability to rotate or accelerate at all is critically dependent upon radial translation speeds relative to RPM and I/O 'G-times'.

Re. your second question: yes - with two or more masses changing radius at the same time - albeit, at different, carefully-controlled radial speeds - it is possible to change radius without changing MoI, or, thus, incurring accelerations or decelerations that CoAM would otherwise cause. In the system below, this principle has been leveraged to isolate 'pure' over-balancing torque:



..tho in retrospect that's a little over-engineered; the trick's only really useful if applied to a single pair of radially-moving masses - as soon as you double that up, to two pairs in alternate phase, then their +/- MoI fluctuations cancel out anyway, with no need to control their radial speeds independently..

..but yes, short answer - it's possible.

[agor95]

As you are looking at this aspect. You may look into this concept.

https://steampunks.ddns.net/thecake.html

The rotation rate has to be high enough to flail out the masses.

All the Best

[Leafy]

Thanks

What I don’t understand is when we pull the mass in, we do work against the centrifugal force so energy increased, but when the mass move outward it does work and loose energy. Where does this energy go?

[agor95]

Hi Leafy

Of cause your arms want to fly out tangentially all by themselves.

Think above moving a toy car forward, which takes work.
Then move it back and that takes work.

Where does all the work go?

All the Best

[MrVibrating]

Thanks

What I don’t understand is when we pull the mass in, we do work against the centrifugal force so energy increased, but when the mass move outward it does work and loose energy. Where does this energy go?

Whether the mass is moving in or out, the positive or negative work done against or by CF force is precisely equal to the change in rotKE..

In practice this means that if not otherwise harnessed, all of the rotKE will convert to radial KE..

..if the masses collide fully-inelastically at some finite radius, then the rotKE loss is dissipated..

..and if the collision's 100% elastic, they bounce right back in, performing equal work against CF and the wheel's acceleration that it and deceleration transferred to it on the way out, thus restoring all rotKE upon arriving back at the initial radius!

Just as force * distance or torque * angle satisfy the work-energy equivalence principle, so too does CF force * radial displacement, with the work in/out instantaneously equal and inverse to the change in rotKE - plotting CF force on one axis and radius on the other, the resulting area under that curve is the total work in / out, and, assuming CoE, the total change in rotKE too.

[MrVibrating]

In consideration of the above points re. conflicting means and ends:

• there's an evident juncture in the potential routes across here:

• if the asymmetric inertial interaction were performed between the inertias of an internal weight vs that of the wheel / net system, then we're aspiring to manifest the gain, in the first instance, as rotKE on the wheel.. which then needs harvesting - without recourse to a stator - and re-investing as GPE or sprung PE

And this is where that last rig is pitting itself against conflicting, mutually-exclusive requirements..

..thus by elimination, the only other - and thus, best - path forwards, in terms of prioritising interaction objectives, is:

• an asymmetric inertial interaction between internal inertias only - not between them and the wheel / net system

IOW, within the rotating FoR, a reactionless acceleration and corresponding KE gain is applied to a mass / weight, immediately re-invested in GPE, such that the wheel itself never 'sees' or partakes in the momentum asymmetry, only benefiting from its returns via passive overbalance (ie. w/o torquing the main axis, either from within or without), via the resulting subsidised GPE lifts.

Basically, the OB system is, most likely, also the same interaction applying the reactionless momentum rise paying for its GPE.. suggesting that the gain arises at some point in the very act, and by the very means, of raising the GPE / 'flinging a heavy thing upwards'..

TL;DR -'the riddle' seems to most consistently align with the implicit accommodations above, of an asymmetric inertial interaction raising GPE instead of KE gains.. IOW whereas a reactionless 1 m/s acceleration of 1 kg whilst already at 1 m/s would raise 1.5 J of KE gain / PE discount, vector that 1 kg upwards against gravity and instead of converting to velocity, you'll get height..

The solution has to be an effectively-reactionless acceleration upwards, rather than 'around' - a linear, rather than angular, asymmetric inertial interaction..?

[ME]

Hmm i meant with regards to mathematically formulating an energy gain; IOW overcoming the cognitive dissonance and recognising that if 1 + 1 can't equal 3, then it can't be framing our problem properly either;

I wrote about a PE:KE closed-form symmetry on a systems level: from start-up to full-speed.
I think this remains fixed, and thus limited.
Hence Bessler needed a larger wheel with heavier weights for more power.
When there would actually be a "1+1=> 3"-situation then that wheel would have flown apart in no time or burst into flames.

What you want is the energy-formulation of what the mechanisms are doing in between. And that's what I wrote about too.

i think it's simpler than you're making out, and pretty much garden-pathed by what we know of the physics and evidence..

You asked me, and I think it's a bit more complex... Let's find out :-)

What I don’t understand is when we pull the mass in, we do work against the centrifugal force so energy increased, but when the mass move outward it does work and loose energy. Where does this energy go?

You could see the wheel's rim as the ground and the wheel's axle as the sky.
You add potential energy when you go to the center, you get that back when it 'drops' back down to the rim... when it moves along a radial pole.
Attache a spring in and you can make a weight bounce between rim and somewhat closer to the rim while the whole systems speeds-up and slows down again.

On a space-station you'll get things like this: https://www.besslerwheel.com/forum/viewtopic.php?t=6639