Decoupling Per-Cycle Momemtum Yields From RPM

[MrVibrating]

Look people, this is obviously some kind of torque sequence:



There's twice as much going left-to-right, as back the other way.

Each of the vertical bars on the chain represents one torque.

So there's two torques driving left-to-right for 180° of rotation..

..followed by a single right-to-left torque for the final 180° of each cycle.

The only two torque sources possible in a statorless wheel are weights, and CF workloads:


• weights directly change the system momentum

• CF workloads alone can only cause torques that depend upon conservation of momentum (the ice skater effect), however in varying the weights' rising vs falling period they can further manipulate momentum from or to gravity


Each full cycle of the three torques above, the axis must gain momentum, from gravity and time.

It has to gain the same amount each cycle, for the same input work.

Each cycle has to end with a collision, which consolidates the per-cycle momentum rises.


So what are the correct torque sources to assign to that sequence - is it a pair of gravitational torques, followed by an inertial torque, or else, an inertial torque coupled with a gravitational torque both going one way, with just one or the other coming back?

What about the 'V' and 'T' shapes atop each item?

The 'V' seems to lean towards the right side, as if reaffirming the direction of the torque asymmetry?

If the 'T' is the initial startup torque, are these single torques thus over-balancing torques? Or perhaps, it represents any ambient torque being applied to the axis, such as from overbalance or a push-start?

Expanding slightly on that line of thought, is that startup torque represented by the single 'T' bar simply propogated on down through the sequence, such that the double-bar links are that persistent torque, plus another, pulsed every 180°, such as a counter-balancing torque? Thus what we'd be looking at is a system that's only under a net torque every half-cycle?


Does the upper hammer toy 'C' represent a 'stamper' - basically a mass or pair of masses moving radially through the axis?

Angular lift with radial drop, or other way around?

All the clues seem to intimate that a radially-moving mass is propelled by a scissorjack / leverage operated by a pair of long lever weights, whether radial or diametric...

WTF is that interaction, and how's it map up to that torque sequence, to cause a constant per-cycle momentum-from-gravity gain?

Does each left-to-right or right-to-left 'chain link' have to culminate in a collision, or just each successive pair? Because that could double the KE dissipated..


..i mean for instance suppose two torques go left-to-right, then there's a collision, before the one torque comes back right-to-left, followed by another collision, hence each per-cycle momentum yield has its energy reduced twice; assuming best case 1:1 ratios between the colliding masses / inertias, each wastes 50% of whatever KE remains on that momentum, thus after the first collision we still have all our momentum but at only half the energy, so 50% dissipated, and then a second 1:1 collision coming back the other way would reduce the remaining KE by a further 50%, hence losing a total of 75% of per-cycle input energy, for the 25% per-cycle accumulator leading to 125% at the fifth cycle?


Or rather than torques, could they represent actual momenta?

In which case, we're looking directly at a momentum asymmetry, with twice as much in one direction over the other.

Then again, maybe rather than 'twice as much momentum' going one way over the other, it's a case of paired reciprocal momenta - obeying N3 - followed by a lone momentum..? So, an interaction that results in a unilateral momentum, but beginning with equal opposing momenta?


Again, note the offset of the lower toy 'D', relative to these left / right axle points, compared to the alignment of the upper toy 'C' - indicating 90° of separating angle..

This implies that the lever-weights it represents are acting 90° out of phase to the radial load represented by the upper toy.

This is why a minimum mechanism has a cross-shaped profile.

The most obvious and apparent solution to the above jigsaw pieces is that a pair of lever weights lift the central radial weight - whether for purposes of over-balance, inertial torque from the ensuing MoI variation, or both - or else, vice-versa, and that, rather than being an asymmetric GPE interaction (impossible), this somehow rectifies a consistent unilateral momentum rise.. yet what variations on this haven't i tried already?

Does the upturned spinning top represent a conversion of CFPE into GPE, as the means by which to harvest rotKE gains? Ie. to power radial lifts, rather than angular ones (which would directly convert rotKE into GPE)? For a momentum benefit?

Why are the kids twisted and armless? Somehow implying 'reactionless' momenta / torques, per the ice-skater effect? Or perhaps indicating springs, such as leaf springs, which seem to automate much of the process of changing radius at certain angles?

Does their square anvil refer to the 'stamper' / radially-sliding mass / the upper toy 'C', whose round anvil signifies the stampers transition through the axis?


These are the jigsaw pieces that form the reveal.

There's OB torques, and inertial torques. Nowt else without a stator.

The only possible source / sink for momentum is a +/- G*t asymmetry.


Whatever the torque sequence, it has to produce equal momentum rises, for equal input work done.

It has to waste 75% of per-cycle input energy on one or more collisions.

WTF is the interaction???

Finite permutations! This is not a plethora of possibilities, but converges to an irreducible nucleus of paying in at 'V1' whilst cashing out at 'V2'.. we need to instigate a divergent inertial frame, ie. the FoR of the input energy workload needs to be accelerating around with the system, EMGAT, like the 'chicken run' demo, which means subverting N3 by ultimately buying momentum from gravity and time, instead of from a stator / external FoR, relative to which speed, and thus further input energy costs, inevitably increases, so that our net input energy sums linearly with rising RPM, while output KE continues to square..

We have the general theoretical solution.. we just need to work out how to apply it to a physical, mechanical system..

[Georg Künstler]

The chain on the right is presenting a timing chain, as you use it today in your car to time the opening and closing of the valves.

In the toy page you have an upper and a lower parallelogam.
The system itself is presenting a top heavy pendulum, look at the figures.
The upper figures are bigger.
A top heavy pendulum will fall over, and therefore the game will be over.

But look at Besslers construction again, the parallelograms are moving between two borders.
We have a catch the fall function.
An oscillator which will swing between this borders, shifting the weights during the fall.
A mechanical flip flop.

The top heavy function will stay as top heavy function because you waste some energy.
A very clever construction, but not driving the wheel, it only shows you how a top heavy pendulum stays top heavy, of course with an collision.
A double collision in one go.

The only possible source / sink for momentum is a +/- G*t asymmetry.

I agree on that, what I try in my construction is to extend t on one side of the wheel.
I use the collision in that way that gravity has to do the acceleration again and again.
Like a repelling function, catch the fall.
This is generating an asymmetric energy distribution in the wheel.
You read it right, asymmetric energy distribution, not asymmetric torque.

The asymmetric torque follows the asymmetric energy distribution.

[MrVibrating]

Asymmetric energy distributions aren't gains - your net total's still the same - and there's no combination of elastic (ie. conservative) collisions that will yield a non-conservative result.

The figures shown on the toys page aren't literal mechanisms, like i say it's showing a torque sequence that produces a momentum gain every full cycle, five of which sum to 125% of unity.

Your 'top-heavy pendulum' idea would be invoking a GPE asymmetry.

I don't understand why you're still throwing out guesses after i've already shown you exactly how to calculate gain from first principles, using only the momentum and KE equations?


Look, sir, one more time, just for Herr Künstler:

• p=mV

• KE=½mV²

• take a 1 kg mass, and accelerate it reactionlessly to 1 m/s

• you now have 1 kg-m/s of momentum

• you've spent ½ * 1 * 1² = ½ J of PE

• collide it, inelastically, with a stationary 1 kg mass

• each now has a speed of ½ m/s, and resolving the KE equation:

½ * 1 * ½² = 0.125 J

• so we had ½ J and now we have 2 * 0.125 J = 0.25 J, ie. ¼ J

We've dissipated the other ¼ J, a 50% loss..


Now, with both 1 kg masses moving at ½ m/s, repeat the cycle - add another 1 m/s, ex nihilo, to one mass, for another outlay of ½ J, and splat it into the other, non-accelerated mass, again redistributing the reactionless momentum rise:

• we end with two 1 kg masses moving at 1 m/s, so checking the KE formula again, each mass now has:

½ * 1 * 1² = ½ J

..there's two of them, so we have 1 J in total, and we've also spent 1 J in total PE.



1 + 1 = 2, therefore a series of asymmetric inertial interactions with inertias in a 1:1 ratio will hit unity at the second cycle, with a 50% dissipated loss per cycle, but also a 50% per-cycle efficiency accumulator; that is, each cycle is growing more and more efficient, by a whole 50% every successive cycle!

So let's punch through to OU efficiency; watch closely, it's just the exact same cycle one more time:

• we have two 1 kg masses moving at 1 m/s; apply one more reactionless 1 m/s acceleration to either one, then collide it into the other..

• we finish with both 1 kg masses moving at 1.5 m/s

• we've spent 1.5 J of input PE

• how much KE do we have?

½ * 1 kg * 1.5² m/s = 1.125 J on each mass, thus 2.25 J

2.25 J of output KE, divided by 1.5 J of input PE, equals 1.5.

So in three cycles we went from a 50% loss, to unity, then 150%.

How much KE will we have after a 4th cycle? Each 1 kg mass will be at 2 m/s, so has ½*1*2² = 2 J each, for 4 J of total KE, for which we'll have spent just 4 * ½ J = 2 J, so 200% efficiency at four cycles.

You can surely see the rest - 250% at the fifth, 500% at the tenth, 1,000% at the 20th and so on, ie.:

• 1 kg at 10 m/s has 50 J, we have a pair of 'em so 100 J, for which we'll've invested the princely sum of 20 * ½ J = 10 J, the former being fully 1,000% of the latter

If that was a 1 kg-m² wheel it'd be 1,000% efficient by the time it hit 20 rad/s from a standing start..


Calculate these things! See them for yourself!

Momentum equals inertia times velocity.

Kinetic energy equal half the inertia times the velocity squared.

These two equations automatically, naturally break input/output energy symmetry when reactionless momentum rises are consolidated with inelastic collisions.

Conversely, if you drop a weight, your new KE is the old KE, plus the GPE you've just added, the momentum yield from which inevitably decreases with RPM, perfectly observing PE/KE unity. Alternatively, perform some work against CF force, and your new KE will be the old value, plus the integral of CF force times radius that you just input; in either case, you end with no more or less energy than you've paid for.

In a statorless wheel, CF workloads and GPE workloads are the only means to affect any acceleration, much less net momentum change. Just those two physical principles. Nothing else available. Any interaction culminating in either is inevitably bound to unity. You're literally enforcing PE:KE symmetry with these actions.

But the instant that 1 kg mass, moving at 2 m/s, hits that other 1 kg mass moving at only 1 m/s, conservation of momentum and Newton's 3rd law will naturally leave each with 1.5 m/s, and thus a 2.25 J total KE, yet for only 1.5 J expended..



There's no magic, no mystery, no need to guess if elastic or inelastic collisions are better - just run the calcs on the back of a fag packet, you get unity, 'fully elastic' = fully conservative, the clue's in the name; if you want a non-conservative outcome you need to orchestrate non-conservative interactions: so you should WANT to calculate them - it's just two formulas, they're incredibly simple to solve, and i have no time for any wild guesses or hunches that aren't laser-focused on gaining a fixed momentum yield for a fixed cost five times in a row or bust.

The Toys page is definitely portraying an energy gain cycle with a 25% per-cycle accumulator, so each pair of single and double bars of the 'chain' represent a cycle that's 75% inefficient, but which gains a consistent rise in momentum, moving down the chain-sequence, hence 25% efficiency at the first cycle, then 50% at the second, then 75% at the third, 100% / unity at the fourth, and 125% at the fifth. A sixth would take us to 150%, then 175% and so on.

I've already explained the only possible reasons why it could be five-cycles-to-OU instead of the shortest-conceivable route above of just three cycles; either half the momentum gets sunk to gravity * time, or else one mass collides with another three (or vice versa) - either regime produces a 25% per-cycle efficiency accumulator.


We're currently looking for a way of doing this using a combination of radial loads and angular ones (various flappy arms and 'stampers', bashically) or whatever else anyone can read into the clues. The Toys page clues are all about the maths of OU however - they're not literal mechanical diagrams, and OU is a mathematical abstraction, not something our 'mechanical mind', for all its perspicuity, has experience enough to model. Cerebellums like single reference frames. Stationary ones especially, though they'll take on moving ones at a stretch (YMMV). We don't instinctively brace against lifting when tugging on our bootstraps.. studies have shown we correctly anticipate the outcomes of inertial and gravitational interactions before we can even walk..

We're idiots at this - left to the visual imagination, the mind almost inevitably drifts back to trying to invoke an effective GPE asymmetry.. that is literally all it can think of! The first place we go. I mean come on, we suck at this.

Calculate.

Mass.

Velocity.

Momentum.

N1, 2 & 3.

PE and KE.

Especially, the role of FoR's in the above terms.

Ask, and then understand why, mechanical energy squares with velocity, instead of just summing like momentum?

You've had a fairly comprehensive set of answers here - i suspect i may be wasting my time - but you can easily calculate OU in the bath or on the bus. You could do it eating peas and rice, or drinking Malibu on ice. You could quantify it in a queue, or generalise it on the loo. Tot it up in bangs and blows, count 'em on your pinky toes, it's a simple, causal tally-up, not a guessing game, christ in f#*£...

[MrVibrating]

I'm getting pretty tired of feeling like i'm reading out of a freaking crystal ball here - to calculate OU use the momentum equation, and the energy equation, and calculate the input and output energy of accumulating reactionless momenta. That's it.

That's your "42".

It's how sufficiently-large values of 1 + 1 can equal 3.

It's how a 'perpetual motion machine' could actually operate.

It's the only way Bessler's wheel's could've worked.

He pretty much expressly states it's how they worked, to the positive exclusion of a putative GPE asymmetry.

EMGAT.

The bangs were not prima facie dressing, but entirely causal and functional.

Why else would you deliberately waste copious KE on ~8 collisions p-c, when the objective is to conserve more than 100% of input energy?

In a statorless wheel, you can only source momentum from gravity and time (gravity alone just gives you a pendulum).

The amount of momentum a weight can add to a wheel is physically, spatially and temporally, constrained by RPM; gravity's a constant acceleration, and the faster a wheel spins, the less time the weight spends exchanging momentum with gravity each cycle; this momentum drop-off with RPM gives us the standard rotKE efficiency of ½Iw² - ie. the energy cost of lifting the weight never changes, regardless of RPM, but the amount of output work it can perform, in terms of adding further momentum to the wheel, is ever-diminishing.

There's no alternative way forwards here, so we must find a way to add more momentum, from G*t, than is otherwise obtained by over-balancing torque; weights have to land in an overbalancing position, but also imparting significant reactionless momentum as they land. Again, where'd'ya get reactionless momentum from in a statorless wheel? A +/-G*t asymmetry. Nowhere else to look.

Wolff cogently described how the weights heard landing on the descending side seemed to be imparting more momentum than could've been obtained by their fall.

He was convinced this was the instant of energy gain.

The implicit 'maths of OU' agree with him.



Again, the singular objective is making the weights land on the descending side with additional momentum, which ultimately can only have come from an effective +/-G*t delta, and by definition, the input energy cost / workload involved in generating that momentum gain is unaffected by RPM, at least across some practical RPM range.

Familiar examples of a +/-G*t asymmetry are classic overbalance (radial lifts w/ angular drops), and kiiking / swinging.

And that's your lot, bottom of the barrel. Those are all the pieces of the physical solution, if not the actual parts list.

It's just accumulating momentum at fixed energy cost, invariant of RPM. That's it. That's what the solution looks like. That's how you make energy. It's how Bessler's wheel's worked. It's basically how he and everyone else said they worked.

Momentum's just mass times velocity. You just multiply them together, and that's your 'momentum'.

'Energy' just multiplies mass by half and velocity by itself, and then multiplies the two together.

Everyone can understand 'recoil', and that N1 and N3 are really about conserving the net mV product.

Everyone can thus understand inertial FoR's and why energy squares with velocity.

Why, if speed is relative, then so is KE.

That symmetry of PE and KE is dependent upon them sharing the same FoR.

That an open system, by definition, has differing I/O FoR's.

How, with EMGAT satisfied, the 'V²' value of each input stroke could begin at a relative 'zero'..

..while the absolute 'V²' value of the net system relative to the ground could rise to many RPM's..

..so input energy would thus simply sum, the total being the fixed per-cycle cost multiplied by the number of elapsed cycles..

..while output KE squares with RPM..


All my efforts at demistifying this are farcical if this ain't clicking anywhere.. may as well be reading the tea leaves..

[Fletcher]

I'm getting pretty tired of feeling like i'm reading out of a freaking crystal ball here - to calculate OU use the momentum equation, and the energy equation, and calculate the input and output energy of accumulating reactionless momenta. That's it.

That's your "42".

It's how sufficiently-large values of 1 + 1 can equal 3.

It's how a 'perpetual motion machine' could actually operate.

It's the only way Bessler's wheel's could've worked.

He pretty much expressly states it's how they worked, to the positive exclusion of a putative GPE asymmetry.


There's no alternative way forwards here, so we must find a way to add more momentum, from G*t, than is otherwise obtained by over-balancing torque; weights have to land in an overbalancing position, but also imparting significant reactionless momentum as they land. Again, where'd'ya get reactionless momentum from in a statorless wheel? A +/-G*t asymmetry. Nowhere else to look.


Again, the singular objective is making the weights land on the descending side with additional momentum, which ultimately can only have come from an effective +/-G*t delta, and by definition, the input energy cost / workload involved in generating that momentum gain is unaffected by RPM, at least across some practical RPM range.

Familiar examples of a +/-G*t asymmetry are classic overbalance (radial lifts w/ angular drops), and kiiking / swinging.

And that's your lot, bottom of the barrel. Those are all the pieces of the physical solution, if not the actual parts list.


All my efforts at demistifying this are farcical if this ain't clicking anywhere.. may as well be reading the tea leaves..

Just solve the mechanics for reactionless momenta and you're done and dusted mate, job done. And no rewrite of the Classical Laws required to boot !

No more crystal ball gazing and looking into the tea leaves in yer lunch break. Few have put such a consistent effort into digging their way to the bottom of the well. You deserve to find your pieces of eight.

[Georg Künstler]

Hi MrVibrating,
I am fully aware of the formulas which you presented.
The insights from you get better and better but have a gap.
I am here on the Besslerwheel board to help you, and not to offend.

There's no alternative way forwards here, so we must find a way to add more momentum, from G*t, than is otherwise obtained by over-balancing torque; weights have to land in an overbalancing position, but also imparting significant reactionless momentum as they land. Again, where'd'ya get reactionless momentum from in a statorless wheel? A +/-G*t asymmetry. Nowhere else to look.

You describe an collision between two objects with an hard impact, such an collision is wasting energy, in addition it is slowing down the object which is the driver of the wheel.
An impact(s) does not drive the wheel, something like this you can read in Bessers writings.
So during the impact something is happen to create the overbalance.
An additional move of masses occurs because you made this impact.
I have described this as an wrong fall, a catch of the fall.
It is a tilting mechanism.

The reactionless momentum will also not occure, there is an reaction between the masses and the turning wheel.
The reaction is made in that way that gravity has to work several times on one side of the wheel.
Described as parasitic oscillation.
Without action there is no reaction.
Without action no turning of the wheel.
Both, the driver and the outer wheel run in the same direction.

weights have to land in an overbalancing position,

true, it happens also in my wheel.
Here you describe a positive feedback loop.

The problem with the mathematic is that it can't see how the masses are interacting.
Mathematics will not tell you how to build the connectedness principle.

From a model you can now make your mathematical equations.

[MrVibrating]

Maybe MT 133 holds the key here: the lower variation using a smaller lever weight to operate the diametric lever only works that lever, not both at once, implying a directionality..

Whereas MT 134, with its radial lever weights, could only seem to facillitate directional asymmetries that were incidental to its relative vs absolute speeds - ie. if already rotating in one direction then operation of each pair of radial levers would slow one whilst accelerating the other, and since angular momentum is absolute not relative, one would have more than the other during those equal opposing angular accelerations in a rotating FoR. Subsequent inelastic collisions during those states might thus shuttle momentum around, 'as poltergiests through walls' etc., dunno..

The scissorjack, too, on the Toys page, seems to encapsualte the same core interaction, of two opposing angular displacements (the handles) converting to a linear displacement, and culminating in "one quarter" (ie. 25%), signified by the otherwise mechanically non-viable 'arrowhead' atop it, albeit one with a 90° aspect (ie. mechanically, either it freely rotates about its own axis, or else it's fixed to one or other side of the terminal scissor pair and thus rotates as it extends and retracts.. thus merely a 'motiff', signifying something abstract)..

'Quarters' being the currency of a gain interaction (ie. four to unity, five to 125%).. AP wheel = ¾ = 75% loss (a hex upon Wagner!) and each 'hammer strike' of MT 133 or the Toys page interaction is applying / imparting one of these 75% loss / 25% efficiency accumulator strokes..

So what of MT 134 then, using radial instead of diametric levers, and with 'four quarters' per interaction?

Previously i'd interpreted this as implying each interaction is primed with PE to its unity threshold - the speed it would reach after four consecutive 75% loss cycles - since at that point it makes no difference how the system reached that state, and any conventional acceleration will do exactly the same job.. thus the magical 'fifth' cycle that tips the system efficiency over unity can actually be the first cycle, provided you have that startup PE on board and can keep repleneshing it..

But it's just struck me that perhaps MT 134 is a 'null hypothesis', intended to show symmetry, and thus the comparitive merits / asymmetry of the preceding design? IE. 'four quarters' equalling unity, perhaps the point is not that the actions's primed to a threshold velocity, but simply that it's a non-viable prinicple; that diametric levers offer something that radial ones do not (presumably, 4x the MoI, per mr² (ie. a diametric lever can be twice as long as a radial one, hence having up to four times the MoI / one quarter the speed for the same momentum distribution))?

I'm gonna spend more time focusing on really elementary systems - a single diametric lever weight, with a raidal GPE / vMoI, and really careful telemetry, especially wrt axial / orbital MoI and momentum distributions.. (a diametric lever basically being an orbiting axis with an oscillating rather than rotating action).

The objective of course being to try to 'scoop up' momentum-from-gravity with the big lever on that high-MoI orbiting axis, and then 'hammering' it into the central axis using brakes etc., whilst making various MoI changes to force an effective momentum asymmetry / per cycle momentum gain yadda yadda..



Hope B's pissing his pants at us fumbling around - we have all the components - implicitly the gain interaction must be fashioned from angular displacements paired to radial ones - yet i feel like a monkey trying to work a can opener.. pure hit'n'miss, stumbling forward by sheer exhaustion of alternate possibilities.. depsite having a map and compass..

Upon waking every morning i go through the same 'boot up' routine over coffee - "where do you get statorless momentum? From gravity and time". "How do you fix its energy cost?".. tumbleweed for the rest of the day.. fleeting what if's? - often not even including gravity - followed sooner or later by the inevitable nah, stoopid's..

The same amount of input work has to cause the same amount of net system momentum rise each cycle, the net input energy thus summing with velocity, while the net output KE squares with velocity, as KE is wont to do.. nowt unnatural about it, just investing frugally, like..

..tho not too frugal, is the key - 25% efficient, for each and every cycle, being the magical per-cycle efficiency to aim for, wasting fully 75% of all input energy every single cycle..


Can this momentum gain envelope be accomplished with diametric levers and a radial GPE / vMoI?

Previously i was trying to operate both at once. That was stupid. I should focus on a more directional interaction, that operates the levers individually in relation to whatever the radial GPE / vMoI is doing.. just let the other hang, inactive, while trying to nab a consistent per-cycle momentum yield from one at a time.. that's the lesson of MT 133, no?


I have to start bagging good-looking momentum yields at some point, just thru sheer elimination..

[MrVibrating]

OK got an idea.. it's ambitious, gonna be tricky, don't know how i'm gonna do it, but:

• thought of an interpretation of MT 137 that seems to fit here


MT's 133, 134, 135 and 136 are all angular-linear conversions, then there's the mysterious stellation of MT 137, before the final angular-linear interaction of the Toys page..

As yesterday's fubar with diametric levers shows only too well, you're never gonna get anywhere trying to move opposing pairs of levers..

..but furthermore, you can only get 180° around even if using single levers, because by that angle an initially clockwise lever action is going to need resetting with an ACW action.


As discussed previously, it seems the only way to be able to generate torques consistently in one direction only - ie. so that the 'reset' stroke is also pushing in the same angular direction as the first stroke - is if a diametric-type lever moves a short distance through or across the axis, switching its pivot point to the opposite rim.

In this way it 'steps' around the rim, each end alternating as the pivot as it swaps sides each stroke.


You can follow this sequence around in MT 137:



• select any one line

• let the nearest corner be a pivot

• it folds down from there, the other line from that corner showing the new position

• now staying with that line in its new position, make its other end the imaginary pivot, and the line off that corner again becomes the new resting position

And so you just repeat that sequence around, forever..


• only adds unidirectional torque

• equal angle of displacement each stroke (ie. RPM invariant)

• equal torque * angle workload each stroke (RPM invariant)


Alternating the pivot points is gonna be hard enough; if the lever needs a radial / linear GPE to drive it too then i've little idea how to incorporate that yet.. really just thinking about the purely physical viability of 'walking the rim' with a diametric lever like this.. only producing positive torques from both strokes, type situation..

[MrVibrating]

..there's no sugar-coating it - yes, i'm partly inspired here by yesterdays idiocy; ie. just lose the opposing lever entirely, because then you at least get something rather than nowt. Then it's a case of working how to get more of it, but at least you've made that start..

Torquing an inner, gravitating lever is a great way to add angular momentum from gravity.

The only problem is, to keep doing so you need to reset, and hence equally counter-torque.

This solves that.. no?

[MrVibrating]

..cos why else has 137 got such a fat 'axle', right?

It's much wider than needs be otherwise.

But the inner circle's not so much about the axle itself, as the displacement angles of the lever rotations, right?

Yeah i know there's 1e6+1 alternate interpretations, i'm just feelin' out this one..

[MrVibrating]

..tracing me finger around MT 137 there, i make 17 steps around to arrive back at the starting position.. is that right?

[Robinhood46]

MrVibrating,
I've been following your posts very closely, although i must admit to not being able to entirely follow all of your thoughts.
I don't think the 17 steps is right, i find 24, unless you are counting something different to what i think you are counting. ??
I agree that the center of mt137 probably doesn't represent a axle.
If the points at the rim represent the swivel points, then the two different lines could represent the scope of the swing. This was covered by PathFinder a while back, but only supperficially, in my opinion.

In this way it 'steps' around the rim, each end alternating as the pivot as it swaps sides each stroke.

I'm all for the stepping around the rim, and i don't think it is that difficult to achieve the mechanism needed to do this either. I think this may be why Bessler covered the ends of the weights with a cloth when allowing people to see them. Maybe they had spring loaded pegs on the ends (retractable), maybe they had some other give away sign?
I have given this a lot of thought for the last 6 months and the best i can come up with to achieve the stepping is this.
https://www.youtube.com/watch?v=_gmSCeQIhl4
You need to right click on the video and play on loop.

[Wubbly]

Path_finder should have a similar animation of walking MT137.

Maybe something like this?:

https://www.besslerwheel.com/forum/view ... 9257#59257

[MrVibrating]

MrVibrating,
I've been following your posts very closely, although i must admit to not being able to entirely follow all of your thoughts.
I don't think the 17 steps is right, i find 24, unless you are counting something different to what i think you are counting. ??
I agree that the center of mt137 probably doesn't represent a axle.
If the points at the rim represent the swivel points, then the two different lines could represent the scope of the swing. This was covered by PathFinder a while back, but only supperficially, in my opinion.

In this way it 'steps' around the rim, each end alternating as the pivot as it swaps sides each stroke.

I'm all for the stepping around the rim, and i don't think it is that difficult to achieve the mechanism needed to do this either. I think this may be why Bessler covered the ends of the weights with a cloth when allowing people to see them. Maybe they had spring loaded pegs on the ends (retractable), maybe they had some other give away sign?
I have given this a lot of thought for the last 6 months and the best i can come up with to achieve the stepping is this.
https://www.youtube.com/watch?v=_gmSCeQIhl4
You need to right click on the video and play on loop.

Very good mate, yes, that's pretty much exactly what i had in mind, with only these finer point worth clarifying:

• diametric levers = longer = higher MoI = more torque / momentum gain per stroke

• this is the bit that makes it slightly more complicated - the lever has to be torqued between positions, ie. a turning force has to be applied to the lever, against the wheel body..

..it can't just passively step around under gravity, because each stepping action has to be an asymmetric inertial interaction, ie. the wheel gaining more momentum than the lever does counter-momentum, because of its gravitation; something like this:



Important to stress the the whole point of the walk is that each stride forms one of these asymmetric momentum distributions, with only the wheel OR lever gaining most of the applied acceleration each stroke, before each ends in an inelastic collision, sharing back the momentum gain to the other, non-accelerated body.

So it's still very much a 'spin & brake' scheme, attempting to apply a torque between two inertias such that only one accelerates, then colliding and repeating.

I'm sure you're right it's 24 steps not 17 (weird odd number, i was bleary eyed before bed).

I'm glad you agree there's nothing intrinsically non-physical about it (ie. just the action, not the hoped-for results), but i'm still no clearer on how to make a good (buildable) sim of it..

All i've thought up thus far is that i could begin with just one lever (obviously, complex enough already!) that lands in little 'pac-man' rotor slots built onto the rim at the appropriate points / angles..

..but i'm only considering doing it really roughly by triggering rigid joints and pin joints to become active / inactive at just the right key-frames..

..in turn this means the sim will only run at its preset frequency, since changing it will screw up all the 'active when' frame-counts.

You can also cause dodgy / non-physical collisions when suddenly activating rigid joints, so it is a quick'n'dirty technique..

But initially of course i just wanna measure it - get a reliable feel of its accumulating efficiency, even if it's just on paper.. ain't even got to be a sim at first, strictly i shouldn't need one if i were any good at physics (!)..

eta: just reading that back it ain't all that clear - i meant to say that each 'pac-man' rotor-receptacle thingy, at its respective loci, will be driven by its own 'motor', obviously logging the torque * angle of each.

But on that point, of how to supply the torque - if we're to salvage anything from this 'MT 40-ish' / 'scissorjack' interaction he kept drawing, of levers connected to a radial linear load, that latter item would seem the most likely torque source..

This is where my visual imagination just turns to a pitch-black room.. since a radial linear displacement presumably needs to remain at right-angles to the lever its operating, no?

Perhaps that's not necessarily true, but either way i've been trying to envision that radial linear load (surely an output GPE) walking around with the diametric lever/s..

Pondering that some more, each radial-GPE-plus-diametric-lever would form a 'cross piece', and it'd be those cross-pieces walking around the perimeter..

If these driving GPE's require a radial / linear drop, then presumably that implies an angular lift.. ie. lifting the over-balanced side of the wheel. Again, nothing impossible in sim-land / on paper, but means it had better make enough energy to nonetheless still raise the even-heavier weights that would be needed to cause it to overbalance in the opposite, desired, direction and so keep re-lifting those radial GPE's..



Just for neatness and simplicity, i've tried and tried to find ways to use the radial load as an OB load, because wouldn't that tie up nicely?

So instead we'd somehow lift the radial load up through the axis, popping up into a neat OB position and that OB torque being the momentum source we accumulate..


It just doesn't begin to add up tho - passive OB torque cannot provide OU momentum levels, due to diminishing yields with RPM (which seems utterly intrinsic and intractable) - to get to OU levels of efficiency, we need - consistently, in spite of some range of RPM - more momentum than could be gained by a passive weight drop alone.

This means we must instead initiate active inertial interactions / reactionless accelerations, which only seems possible by gravitationally sinking counter momenta.

Furthermore if the radial linear load is lifted up rather than dropped down through the axle, then it must be lifted by the diametric lever falling. That would become the output GPE. Hmm could that still apply a unlilateral torque to the wheel in the process? Should i try lifting a radial GPE with clockwise or ACW levers, while trying to get an angular momentum rise from both sunk counter-momentum and OB torque? But then how's it reset every 180°? You still have to walk the lever around in order for each and every stroke to apply torque in the same direction, right?

Dunno, gonna give it a few more days to consider before going off half-cocked again.. 'walking the lever' / alternating pivoting ends seems like a potentially promising avenue to pursue - especially getting some 'quality' (ahem) metering on it eh? Robin above has shown it's do-able, so let's try do it with active torques and see if we can't whip up a nice 'staircase plot' of consistent momentum rises..


That's what the final, working OU interaction's gonna look like y'all, one of them staircase plots, with equal step heights across some RPM range, and equal input work per step. That's all mech OU is. Piece of piss right?

[MrVibrating]

Path_finder should have a similar animation of walking MT137.

Maybe something like this?:

https://www.besslerwheel.com/forum/view ... 9257#59257

Yes, that sir is it!

Just with actively torquing between positions - each input torque causing the wheel to accelerate more / gain more momentum than the lever.

Each has to be a near-as identical per-stroke momentum rise.

The very specific beasty we're trying to nail here being the thread title.. overbalancing momentum yields (specifically +G time per cycle) necessarily diminish with RPM.

But if each stride were a perfectly-asymmetric interaction - only accelerating the wheel, while the lever's motion remained constant (because gravity) - we'd presumably be paying the same T*a each stride, for an equal-sized step up the staircase every stroke..

Then it's just a case of matching the lever's MoI to that of the wheel - another thing diametric levers can do that radial ones can't - to get the shortest number of steps to OU, be that 3 or 5 or whatevs..

Basically it'd be a working version of the 'chicken run' spoof from earlier in this thread.. essentially using a 'walking' diametric lever as an internal 'carry-along', transient, stator.

Sounds grand as usual, just watch me eff it right up..