Decoupling Per-Cycle Momemtum Yields From RPM

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Post by MrVibrating »

I said:
"..this 0.25 : 1 thing.."
- but that's not right is it? It misses that quadratic aspect that the careful repetition preserves; ie. he doesn't say in the reiteration, '1 pound falls as 4 pounds rise' but rather specifically, '16 ounces fall..' - implying perhaps that if the first example relates to one mechanical action, the second describes four such actions in parallel, but either way, quite deliberately framing the key mathematical relationship as:

• (4 * 0.0625) : 1

Is that right? Dunno, work now, back later..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Fletcher wrote:Bessler said he used different "principles" in his different wheels. So don't get hung up on which OB system he used, imo.

Take a look at MT20 and MT's 44 and 48. Completely different operational OB systems.

The common factor is the Prime Mover.

Where we agree is that the Prime Mover costs less than it produces and that which is deployed to the various OB systems to create surplus torque, imo.

Regardless, if we each pull it all apart (in agreement or not on what those significant parts are) eventually someone will put it back together again like Humpty Dumpty on the wall. So discussion is always worth having.
MT 48 is interesting in that it perfectly encapsulates the physical constraints that this thread is aiming to circumvent - the cost of the input GPE's is constant, but the amount of momentum they can induce from over-balancing torque is decreasing with RPM, each one causing slightly less acceleration than the one before.. yet always costing the same to lift.

In order to make an OU 'MT 48' you'd want each ball drop to cause an equal rise in RPM, at least for a few drops, in spite of rising RPM.

Just another way to help visualise the very specific constraint we're up against.

FWIW, you can, easily, stabilise momentum yields by steadily increasing the MoI, gaining momentum in that form rather than acceleration - in this case each drop has equal G-time and momentum yield, however it's increasingly of the 'MoI' component, rather than the 'velocity' component, which is where the KE gain would be.. and stabilising momentum yields at the cost of getting slower and heavier is never gonna break unity.

The wheel has to gain momentum in the 'velocity' component..

..yet, in spite of this, the per-cycle momentum yield has to be constant across some (if only small) range of RPM.

It thus seems to me we have little option but to find some way to add momentum to the weights landing in the OB position, which in turn can only be sourced from G*t. Beyond that, i've nothing, nada, zilch mate.

Just kinda hoping this riddle might inspire a way forwards..

4 ounces of weight vs 1 pound of inertia? Some kind of vMoI interaction involving an MoI change of a factor of four? I've literally no clue what it means for now.. just that it seems to involve a 'squaring' function, perhaps, and likewise, 'quartering', where 'quarters' are fixed-cost momentum yields, each 75% lossy, but with a 25% per-cycle efficiency accumulator.

You know i'm gonna come up with something here, dog with a bone and all that, just a matter of brew time.. day off tomoz, hot tub & bottle of scotch might do it..
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Post by MrVibrating »

• (4 * 0.0625) : 1
OK how about this - pretty obvs. - maybe it describes the proportions of an effective inertial asymmetry; ie. the ratio of momentum to counter-momentum that can be achieved, or else, perhaps, the amount by which a given momentum can somehow be traded up for more?

Both these hypotheses would have corresponding hypothetical energy solutions, so perhaps these are worth plotting out to see if any match the key '25% efficient' net accelerations indicated by the Toys page..


The way the riddle carefully distinguishes "16 ounces v 4 pounds" rather than simply "1 pound v 4 pounds" strongly implies that the exploitable property is dependent upon the cooperation of discrete properties of the four-fold input components, whatever they are..

..ie. that 4 small 'things', in concert, have 4x the effective potential as 1 bigger thing of equal net worth.

Mind, that's just from a 30 min dry soak in the bath - spitting feathers here, lemme go whet me whistle and i'll come back to this refreshed..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

I found curious the use of 16 ounces and not 1 pound too.
A while back i tried using the movement of 4 small weights that move together applying their accumulated force to move one big weight. The plan was to free the weight at the bottom and replace it with a weight at the top, to again have 4 weights to apply force to the next big weight. It was not an easy task in the real world, and i didn't get the impression it was very promising, so i gave up.
A small weight at 2-3-4-5 oclock all apply force by leverage to a big weight at 6.
The weights then find themselves at 3-4-5 and the one at 6 continues to rotate with the wheel. The weight, that just come over the top , then adds itself to the other 3 weights to apply the weight of 4 small weights to the big weight which was at 5 and is now at 6.
It wasn't actually 3-4-5 oclock, because the wheel was divided into 8 sections and not 12, but i think you get the gist of what i mean.
This did have the need to lift the weight in a flash, because of the need to get the weight up and out of the way before the following weights started applying force to the following big weight.
Maybe i'll have a go at doing it with the simulator.

Edit; I think i have learned a lot since then and my biggest mistake was being too greedy.
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Fletcher »

MrVibrating wrote:
But I would just like to add this friendly little note of caution:-

A great craftsman would be that man who can "lightly" cause a heavy weight to fly upwards!
Who can make a pound-weight rise as 4 ounces fall, or 4 pounds rise as 16 ounces fall.


If he can sort that out, the motion will perpetuate itself. But if he can't, then his hard work shall be all in vain. He can rack his brains and work his fingers to the bones with all sorts of ingenious ideas about adding extra weights here and there. The only result will be that his wheel will get heavier and heavier - it would run longer if it were empty! Have you ever seen a crowd of starlings squabbling angrily over the crumbs on a stationary mill-wheel? That's what it would be like for such a fellow and his invention, as I know only too well from my own recent experience!
- JC's AP

He refers back to this riddle later ("even Wagner, wherever he is, will have heard that one pound can cause the raising of more than one pound").
https://www.besslerwheel.com/forum/view ... rter#44346
Tinhead wrote:Stewart wrote:
Bessler actually says (from AP part 1 chapter 43):

Der wird ein grosser Künstler heissen/
Wer ein schwer Ding leicht hoch kan schmeissen/
Und wenn ein Pfund ein Viertel fällt/
Es vier Pfund hoch vier Viertel schnellt. &c.

He will be called a great craftsman,
who can easily/lightly throw a heavy thing high,
and if one pound falls a quarter,
it shoots four pounds four quarters high. &c.

Hi Stewart, love your translation, thought I might add my thoughts to it.

Künstler = craftsman, artist, skilled person
Easily/lightly = in the German context it is along the lines of "without much effort"
I also agree with the use of "throw" and "shoot" (as in shoot an arrow).
Good work mate :)

Just one thing I would like to highlight, in regard to the 1 quarter down and 4 quarters up, I think it is quite important that he is not using any units ...

Could be a 'thinking trap', 1st thing coming up to mind is the vertical up/down, but it could mean anything. Maybe the circumference of a wheel?

Just my 2 cents.
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Post by MrVibrating »

I think i might be starting to see something, a path forwards perhaps..

There's a potential momentum source that i haven't yet managed to test:

• suppose we could apply a linear force, downwards, upon the axle / support posts, in propelling a mass upwards

So we've raised some 'upwards' linear momentum, against the inertia of the Earth.

• now suppose it gets up so high, before gravity pulls it back down again, landing on the OB side of the wheel..

The key point is that the linear-to-angular conversion is accomplished by gravity, without incurring a counter-torque component within the wheel system.

Thus an otherwise inertially 'closed' rotating system is continually being fed with more fresh angular momentum, converted by gravity and raised as linear momentum against Earth.

I don't know yet a) whether this is actually possible (the linear-to-angular conversion), nor b) if it would furnish fixed-cost momenta in spite of rising RPM.


However, now that you have the general picture of what i'm considering, it seems to follow as a consistent train of logic in the latter pages of MT (even assuming they're in preserved order, always in question); beginning with MT 132:

Image

..linear inertial interactions against the axis / earth, featuring a square hub.

Now 133:

Image

..main feature: dropping weights in the rotating frame in order to pump some fluid against the terrestrial frame..

..insert: that 'fluid' now a 'pound-weight' being accelerated & raised against the axis - ie. levering the pound-weight upwards by pushing the earth downwards..

..on to 134:

Image

- main feature = same deal; angular displacements in the rotating FoR leveraging linear displacements against the axis / earth, now depicting the inputs divided into 'four quarters'

- insert = a kind of Roberval planar linkage for doing the same thing

135:

Image

..any linear displacement thru the fixed axis is still a linear inertial interaction with the earth - if a pole is thrust upwards, the earth is thrust downwards via the axis..

136:

Image

..but how to convert these linear momenta into angular..?

Now, note here in this scan of MT that '143' actually precedes '137':

Image

..so first we see "pounds in equilibrium" (hence a potential visualisation for "pounds" in the 'part 1 riddle' - and again using a Roberval-type planar linkage..

..and then, 'MT137' - the most immediate impression being that is represents a multiplied schematic of this same Roberval-type angular-linear linkage..

Culminating in the Toys page, MT 138-141:

Image

..so if that is a thread of continuity, the lower hammer toy 'D' is raising linear momentum against the 'square' hub (ie. the earth via the fixed axis), and affirmed via the alignment of the lower toy mid-way between the left and right notches on 'B', representing the system axis - ie. this initial linear acceleration does not apply torque or counter-torque to the axis..

The scissorjack, culminating in 'one quarter', represents the linear-to-angular conversion of that momentum as it is given to the OB weights represented by the upper toy 'C', and resulting in the accumulating clockwise v counter-clockwise momentum asymmetry represented by 'A'.

The upturned whistling top - by dint of being upside-down - implying a linear, as well as angular, interaction with the earth..


So to briefly resummarise: besides kiiking and 'classic OB', perhaps a 3rd way to gain 'statorless' momentum is by converting linear momentum applied against the fixed axis to earth, into angular momentum, via gravity. I don't yet know however if this is possible, much less if it offers opportunity for trapping momenta at RPM-invariant efficiency..

However with regards to 'the riddle', suppose it did allude to an effective energy asymmetry..

• to "lightly" cause a heavy weight to fly upwards

..puts 'lightly' in scare-quotes; there's a reservation involved, contingent upon something..

If 4 'ounces' fall while 1 lb rises equal height - an effective GPE asymmetry but stay with me - then a) we're talking about an input energy discount, rather than excess output - so, logically consistent with the actual implicit physics of OU - and b) that discount can only be due to the 'ounces' being in an anomalously-accelerated reference frame; that is, one that is not inertially interacting with its environment, or, thus, exchanging counter-momentum with it; furthermore if the output value of that input workload is being raised by a factor of 'four', then the relative velocity delta between I/O FoR's must be a factor of two - that is, the absolute accelerations of the 'ounces' is twice as high, in the ground FoR, as the actual work being performed in the anomalously-accelerating rotating FoR.

Hence a pound weight might be accelerated upwards against a full compliment of linear counter-momentum transferred to earth via the support posts, yet "lightly" with respect to the axis of rotation, since no counter-torque has been applied..!

An effective reactionless angular acceleration of four 'ounces' of GPE in a rotating FoR might, thus, conceivably, raise 'one pound' of GPE in the ground FoR..

Little sketchy perhaps, but does frame a specific type of interaction that can be further investigated; namely, trying to get gravity and rotation to stably convert linear into angular momenta, and then, without the input work each cycle having to increase with rising RPM..
Last edited by MrVibrating on Tue Nov 10, 2020 8:51 pm, edited 1 time in total.
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Robinhood46 wrote:I found curious the use of 16 ounces and not 1 pound too.
A while back i tried using the movement of 4 small weights that move together applying their accumulated force to move one big weight. The plan was to free the weight at the bottom and replace it with a weight at the top, to again have 4 weights to apply force to the next big weight. It was not an easy task in the real world, and i didn't get the impression it was very promising, so i gave up.
A small weight at 2-3-4-5 oclock all apply force by leverage to a big weight at 6.
The weights then find themselves at 3-4-5 and the one at 6 continues to rotate with the wheel. The weight, that just come over the top , then adds itself to the other 3 weights to apply the weight of 4 small weights to the big weight which was at 5 and is now at 6.
It wasn't actually 3-4-5 oclock, because the wheel was divided into 8 sections and not 12, but i think you get the gist of what i mean.
This did have the need to lift the weight in a flash, because of the need to get the weight up and out of the way before the following weights started applying force to the following big weight.
Maybe i'll have a go at doing it with the simulator.

Edit; I think i have learned a lot since then and my biggest mistake was being too greedy.
Yes, i think only a diverging input FoR can actually embody a PE:KE asymmetry, with the work being done in the accelerating FoR always at unity efficiency, minus losses.

The mathematical OU route the Toys page depicts ideally dissipates fully 75% of all input energy each cycle - yet, perversely, this affords the optimal OU efficiency, with a 25% per cycle accumulator as net momentum builds up in the ground FoR; the net input PE simply summing with RPM while net output KE squares.

TL;DR we can't create or destroy energy... we have to trick the laws of physics into doing so, on their own terms.. 'OU' being an anomalously-open system; open to what? The terms substantiating and underwriting 'energy' itself. Mechanical PE:KE symmetry is enforced by nature's own 'connectedness principle' of N3 forming an 'inescapable circle' of mutual inertial interactions regulating the relative metric of momentum's 'velocity' component against an assumed net-zero (or simply, time-constant) value.

If the laws of physics must hold true in each FoR then we can only hope to gain energy by cultivating effective FoR anomalies..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Fletcher wrote:
MrVibrating wrote:
But I would just like to add this friendly little note of caution:-

A great craftsman would be that man who can "lightly" cause a heavy weight to fly upwards!
Who can make a pound-weight rise as 4 ounces fall, or 4 pounds rise as 16 ounces fall.


If he can sort that out, the motion will perpetuate itself. But if he can't, then his hard work shall be all in vain. He can rack his brains and work his fingers to the bones with all sorts of ingenious ideas about adding extra weights here and there. The only result will be that his wheel will get heavier and heavier - it would run longer if it were empty! Have you ever seen a crowd of starlings squabbling angrily over the crumbs on a stationary mill-wheel? That's what it would be like for such a fellow and his invention, as I know only too well from my own recent experience!
- JC's AP

He refers back to this riddle later ("even Wagner, wherever he is, will have heard that one pound can cause the raising of more than one pound").
https://www.besslerwheel.com/forum/view ... rter#44346
Tinhead wrote:Stewart wrote:
Bessler actually says (from AP part 1 chapter 43):

Der wird ein grosser Künstler heissen/
Wer ein schwer Ding leicht hoch kan schmeissen/
Und wenn ein Pfund ein Viertel fällt/
Es vier Pfund hoch vier Viertel schnellt. &c.

He will be called a great craftsman,
who can easily/lightly throw a heavy thing high,
and if one pound falls a quarter,
it shoots four pounds four quarters high. &c.

Hi Stewart, love your translation, thought I might add my thoughts to it.

Künstler = craftsman, artist, skilled person
Easily/lightly = in the German context it is along the lines of "without much effort"
I also agree with the use of "throw" and "shoot" (as in shoot an arrow).
Good work mate :)

Just one thing I would like to highlight, in regard to the 1 quarter down and 4 quarters up, I think it is quite important that he is not using any units ...

Could be a 'thinking trap', 1st thing coming up to mind is the vertical up/down, but it could mean anything. Maybe the circumference of a wheel?

Just my 2 cents.
Rainer
Ah wow thank you Fletch for that timely intervention - on seeing it i think i had read this previously, but totally forgotten about it, and this really does re-frame the possible interpretations doesn't it?

I've been robotically mumbling the wrong riddle to myself the last week. doh!

Code: Select all

"He will be called a great craftsman, who can easily/lightly throw a heavy thing high, and if one pound falls a quarter, it shoots four pounds four quarters high."
OK gimme another week to re-think everything..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

To put it simply.
Two closed circuits are rotating around the same axis.
They both respect all the laws of physics.
One is trying to catch up with the other, and in doing so it causes the other to accelerate.
The connectedness, or whatever he called it, is the connection between the two, which can not be anything other than progressive/evoluative.
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Post by MrVibrating »

Remember, fellow sleuths, that 'quarters' are actually the units of momentum gain accumulating on an OU-efficiency trajectory:

• each 'quarter' is a fixed rise in net system momentum

It makes zero effective difference how much momentum this is - just that it's the same amount each 'cycle'. Generally, smaller bites of the apple are going to keep input costs lower, as energy squares with velocity remember..

• each 'quarter' of this statorless angular momentum gain has the same energy cost of production / acquisition

The same internal work done causing the same net rise in system momentum (specifically, velocity) each cycle.

• each 'quarter' has a unitary efficiency of just 25% - whatever it is we have to do to bag these momenta necessitates dissipating fully 75% of all input work / energy every cycle

This might be a single loss event, or compounding from two or more in sequence (ie. inelastic collisions in a 3:1 inertia ratio, or else perhaps sinking all counter-momentum in a 1:1 inertial interaction - thus sacrificing half the input energy along with the CM to G, while consolidating and accumulating the remainder with an inelastic collision.

Additionally is the latest purely-hypothetical process of acquiring linear momentum against earth via the wheel's fixed axis - pushing a weight upwards against the earth, then relying on gravity to invert and re-apply it as angular momentum along with OB torque; haven't yet progressed to modelling its efficiency, but you can already see it still involves an inelastic collision.

Have to say tho that upon reconsideration of the clarified wording my thoughts keep tending back to possibilities for raising a GPE while counter-balanced in some respect (ie. 'lightly / easily'), while again thinking about those Roberval type linkages, introducing, as they would, a linear component into an otherwise-angular rotation cycle. Similarly, perhaps, the Toys page scissorjack might convey a linear transfer or transmission between the angular actions of the hammer toys..

We need to come up with prospective mechs for using gravity to convert or transfer grounded linear momenta into a floating rotating frame.. literally 'shooting' projectiles upwards, perhaps at a slight angle, to land in an OB position.. a game of ballistics, then, if it's even viable.. IOW what simple sim might the proposal be reduced to? Some kind of radial 'pin ball' firing mechanism, as from an unloading spring? Chimes with that Weissenstein demo anecdote, right?

We could program a linear actuator to do that while measuring its work done; the only outstanding reservation i have right now being step 2: 'then a miracle occurs' - that is, expecting gravity to somehow convert linear into angular momentum.. is such a proposition actually even tenable in the first place? Is it really just a simple case of letting it follow a ballistic arc?
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Robinhood46 wrote:To put it simply.
Two closed circuits are rotating around the same axis.
They both respect all the laws of physics.
One is trying to catch up with the other, and in doing so it causes the other to accelerate.
The connectedness, or whatever he called it, is the connection between the two, which can not be anything other than progressive/evoluative.
Yes, it's a massive over-simplification, yet really does convey the raw mechanics of what's going on in an OU interaction:

• the KE equation fixes the PE cost (as well as value) of momentum as ½mV², or in angular terms, ½Iw²

• because it squares with speed, accumulating it in smaller purchases is cheaper than buying it all at once - however keeping the unit energy costs down despite rising net speed means keeping the internal relative velocities down; ie. basically EMGAT / dragging the 'stator' around with the rotor; effectively, requiring reactionless angular momentum rises

• thus a 1 rad/s acceleration of 1 I of inertia resulting in a 1 L rise in momentum costs ½ J; 10 such cycles have a net cost of 5 J, but a KE value in the ground FoR of 50 J, hence 10x OU

The energy gain factor is, directly, the accumulated velocity difference between internal and ground FoR's - a half-Joule workload in the internal frame worth 10 J or 50 in the ground frame, depending only on the relative / absolute speed difference.
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

Yes, it's a massive over-simplification, yet really does convey the raw mechanics of what's going on in an OU interaction:
Some bloke once said "If you can't explain it simply, you don't understand it well enough".
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Hi MrVibrating

I appreciate the efforts you are making in working out a way forwards.

It appears you know the route in theory and you have not reached an implementation method.

I know it's like walking through mud. However if it was easy then it would not be worth doing.

All the Best
[MP] Mobiles that perpetuate - external energy allowed
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Post by MrVibrating »

Yes, making energy is as simple as applying a reactionless rise in momentum; eg. two 1 kg masses both at 1 m/s = 1 J of KE, now apply a 100% asymmetric 1 m/s inertial interaction between them, thus only accelerating one, so now one mass is at 2 m/s, the other still at 1 m/s.. hence 2 J + 0.5 J = 2.5 J, a rise of 1.5 J.. that internal acceleration only cost ½ J, however.

So i've tried every way i can think of to cause effectively-asymmetric inertial interactions, the most obvious way being to skew the momentum distributions by applying some other force or acceleration such as the ice-skater effect or gravitating weight.. In the former case, CF workloads square with RPM, and in the latter, per-cycle G-times, and thus momentum yields, decrease by the inverse square of RPM, either way still enforcing PE:KE symmetry.

The other approach is to forget about asymmetric inertial interactions, so much as relative speed in relation to the gravitational constant - ie. kiiking, effectively gaining or losing momentum unilaterally directly to the I/O G*t asymmetry, but again, dependent upon the same input energy and output momentum constraints of rising CF force and/or reducing momentum yields.

Frustrating, since each requisite step can trivially be demonstrated in isolation, yet a full mechanism that can enact a complete cycle of the only possible general mathematical solutions remains elusive, and for the same persistent reasons.

We know there's a way outa the loop tho..
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