Decoupling Per-Cycle Momemtum Yields From RPM

[Johndoe2]

Mrvibrating I think your last post is going in the right direction.
Your initial design was way to complex and chaotic imo.

[agor95]

Keep on with the good work.

The use of springs that have varying 'K' value as they distort
and the MOI, on average, increases with rotation rate.

All the Best

[MrVibrating]

..just got to thinking: instead of lifting a pre-loaded spring, maybe it could be loaded after, by borrowing a trick from MT 26:



..and MT 27:




You can see what i'm getting at - in principle, we could load a leaf spring simply by pressing at its center, with a linear action..

..this obviates having to move an axis or bob - only the central spring section needs bending - hence circumventing any need to apply torques / counter-torques when loading, whilst still collecting 'em on output by releasing the spring to swing the bob..


Also thought of another way of reversing spring torque, tho it needs a third dimension; use a rotary spring, load it by counter-rotating two masses / inertias, then lock 'em together, remove the locked spring, flip it face over, heads for tails, and re-insert - the two parts will thus 'unwind' in the same directions they were wound..

Still not sure if any of this is right, wrong, or just utterly useless and trivial.. need to run some basic tests to see if pre-loading springs per MT 27 results in a net torque or momentum change, as improbable as it undoubtedly is..

Given that it follows MT 24 and MT 25, which also appear to use leaf springs (somehow apparently translated into 'poles with threads', go figure), the latter hinting that they need to curl..




..and also, recall that MT 18 tells us that curling leaf springs are applicable in a working design, promising that he "will show more than speak of it at the appropriate place."



..so if they're part of the long-lost big reveal ('lost', or just obfuscated in arcane glyphs, per MT 134?), maybe MT 27 really is intended to intimate side-loading of leaf-springs, in logical sequence with the preceding diagrams..

Either way, 'something you can do with leaf springs' has to somehow square up with fixing p-c momentum yields.. Right track? Looks right. But still at the sandbox level, looking for inspiration / useful experimentation..

[MrVibrating]

Things are livening up a bit:



The levers are now hinged, and the two bobs are interconnected by a rigid rod, restraining one another, and thus allowing the central GPE / vMoI to side-load the springs.

Not often you see a truly chaotic system in WM, but this phasing never repeats exactly - there's no harmonic behaviour at all:



Tantalisingly, there appears to be certain randomly-arising modes that cause the wheel to accelerate, or decelerate.. but it looks like net momentum's conserved, and the net KE never goes out of range..

However this is still just mucking around, an incremental variation on the former sim. What i really want to try is just two levers in 90° cross-config, in place of the black masses here, with a bob on each end that alternates as a hinge, passing the springs just an inch thru the axle to attach to the opposite rim before unloading in the same direction they compressed..

[MrVibrating]

Bessler was heard to press down on a spring when reloading the weights after translocation - something apparently went 'boing'!


Was he side-loading a leaf spring with PE?

Wolff got the impression the weights were attached to moveable levers or springs..

MT 18 says leaf springs are part of the solution..

They, or something they do, are almost certainly represented by the lower hammer toy 'D' on the Toys page..

..and it's 90° out of phase with the interaction represented by the upper toy 'C'..

Might both hammer toys represent the same weighted spring mechs, undergoing spring loading in the lower toy, and 90° later, passing across the axis in the upper toy? The scissorjack showing that leverage can be used, since the objective is not to raise GPE but simply to invert the unloading spring's torque.. and since the levers are diametric already they only need to move a small radial distance to effectively cross the wheel's diameter.. switching sides, and thus torque signs..

We'll see..

[MrVibrating]

Ooh or maybe it's t' other way round, like - 'C' is a radial drop of the vertical mech thru the axis, in the process preloading the horizontal mech 'D'?

[MrVibrating]

The perennial question is, what caused the OB in the one-way wheels; if OU depends on leaf-springs, it doesn't necessarily follow that OB does too - maybe the OB mech is intrinsic to the energy-gain mech (ie. using the same masses / components), or else maybe the OB mech is slave to the master energy gain mech (ie. separate 'prime mover' driving a classic-OB loop)?

Simplicity favours the former - the GPE system integrated into the same components, and presumably, actions, responsible for generating the PE discount / fixed momentum yield..

Spring sag with fixed axes doesn't generate an OB moment - ie. simply allowing the bobs to droop on one side, does not introduce under-balancing torque to that side (as far as i've been able to determine).

Perhaps hinged axes will be more accommodating to an OB scheme, tho this is yet to be determined..

[MrVibrating]

Maybe the hammer toys represent vMoI's / inertial torques, rather than leaf-springs?

IE. consider the armatures in MT 133 / 134 - they may imply directionality of inertial torques?

Maybe inertial torques lift a central radial weight?

Maybe it's the other way around - a central radial weight drives the inertial torques??

It's essentially still the same cross-shaped interaction - i've already tried moving opposing masses in and out at the same time, with no momentum gain, but when phased in alternate sync like this:



..you get an angular momentum gain. It's a little shaky there - if the central 'beam' weight isn't heavy enough, no momentum gain. The FoR of the mechanism is the wheel itself, not external coordinate space, so it finds its own groove instead of following a rigid sync as a function of angle. It's obviously dissipating copious energy via the 'stopper' ropes, and can't be very efficient..

But it is gaining momentum, which can only come from a G*t asymmetry, so closer study of exactly how it's doing so may pave a path forwards..

It's basically 'pumping' angular momentum back and forth between CW and CCW signs, no? Via the intermediary of the central radial GPE / MoI variation (the beam causes its own positive inertial torque as it moves into the center, and a negative one when moving back out).

This is pretty much what i've been trying to deduce from MT's 40 thru 133/134, the Toys page and the Kassel prints - ie. it looks like the right principle, the right orientation / profile in terms of depending on being cross-shaped, it gains momentum, asynchronously to its rotation in the static FoR, everything goes around together with no counter-rotating parts..

Can it be better regulated / optimised to work more consistently?

It obviously needs metering up to the nines.. but if we can get it working on a reliable and robust per-cycle basis (ie. equal yield per stroke / cycle), how does its efficiency develop as a function of speed - is it constant or variable? It already appears to have an inherent speed limit, which can be varied as a function of the central beam's mass / weight.

Maybe a kind of sprung-vMoI / pulley system could harvest the KE gain as CFPE, thus perpetuating the radial translations causing the G*t asymmetry?

Should prolly knock up a scratch-built high-detail model with variable inputs for everything..


Works better than leaf-springs anyhoo..

[Georg Künstler]

MrVibrating wrote:

This is pretty much what i've been trying to deduce from MT's 40 thru 133/134, the Toys page and the Kassel prints - ie. it looks like the right principle, the right orientation / profile in terms of depending on being cross-shaped, it gains momentum, asynchronously to its rotation in the static FoR, everything goes around together with no counter-rotating parts..

Can it be better regulated / optimised to work more consistently?

It can, you only must eleminate the backfall to a second system, technically you waste a part of the energy.
This will result in an asymmetric energy distribution.

But we can do is also better, a permanent fall over.
Maybe one of the most interesting part is, as more you waste as more you get.

Nature doesn't waste anything and will react.
So Nature will automatically fill the gap and refill the losses.

Maybe this sentences are hard to understand for an technician, but they are universal. Nature will balance the thing out.
Technically you have to disturb this balance.
But we have to disturb the energy balance, left and right in the wheel.

[MrVibrating]

The losses on that rig are simply far too high; the hope was to generate a gain in momentum, from gravity, by pumping the MoI's in and out on alternate strokes... but this isn't happening. Instead, it's just gaining momentum from brief, random moments of overbalancing torque.

It doesn't even have an inherent preferential direction - tuning the mass distributions it'll just as happily accumulate overbalancing momentum in the opposite direction, just as haphazardly..

It's obvious just from the animation that the energy being dissipated by the inelastic stopper-ropes is much greater than the rotational KE gain from that unstable overbalancing regime.



It gets some things right - in terms of resolving Bessler clues with the implicit ingredients of an OU interaction - but you can't be a little bit pregnant, and this is obviously not 'it', yet.


Inelastic collisions are an essential ingredient - you cannot cross the threshold into OU by dropping a weight, or performing work against CF force, for instance; your net system energy following either such inputs will be precisely equal to its previous value plus whatever GPE or CF work that last stroke adds, but no more..

..the final action in an OU interaction - the last thing that happens before PE / KE symmetry is broken - is that inelastic collision. That's what consolidates per-cycle momentum gains into successive accelerations of the net system, and then that growing net system momentum - specifically its velocity component - becomes the 'travelator' inflating the KE value of continuing PE inputs upon it.

Each inelastic collision will necessarily waste 50% of the per-cycle input PE.

That's a best-case scenario; assuming a 1:1 MoI ratio between the colliding bodies.

The Toys page shows a 5-cycles-to-OU sequence; a 25% per-cycle efficiency accumulator.

This could only arise in one of two ways; either:

• from an interaction with a 3:1 MoI ratio between interacting bodies

For example suppose a minimum system requires four discrete masses / weights, and involves somehow applying a unilateral momentum gain to either just one of them at a time, which then collides its momentum gain back into the other three (ie. such as re-locking back to the wheel along with them, perhaps as by landing on rim-stop), or else the other way around - unilaterally accelerating all but one, and again consolidating with a collision to share back the momentum gain and so accelerating the system FoR, inflating subsequent internal accelerations..

Inelastic collisions with a 3:1 MoI ratio dissipate 75% of input energy per cycle, keeping all of the momentum along with 25% of the remaining input energy. If the accelerations are unilateral then the system reaches unity at the fourth such cycle, and 125% at the fifth, continually increasing in efficiency by 25% per cycle thereafter.


• Or else, the '5-cycs-to-OU' regime could also arise if the MoI ratio is 1:1, thus wasting 50% of input energy per cycle in the inelastic collision, but also wasting 50% of the momentum generated each cycle; ie. so you pay each cycle to generate some momentum, but only half of it is unilateral, say (ie. a 50% inertial asymmetry, instead of 100%), prior to each collision.. thus you only get to collide & consolidate half the momentum you're paying for each cycle, still losing another 50% of that in the collision, for the same 25% per-cycle efficiency accumulator / 75% p-c loss profile.


Either regime will reach 125% in 5 consecutive cycles, per the Toys page.

Both are entirely abstract mathematical propositions though, with no actual design instructions - it's up to us to work out how to express and embody them with real solutions.. "to design mechanisms that do what the maths do.."

The collision sequencing in a working gain config will be surgical, given that it'll necessarily be wasting 50% of per-cycle input energy at best; you obviously don't chop out 50% of your input energy unless you know you're gonna get more back..

Flailing around randomly like that ain't gonna cut it..

[MrVibrating]

This is a weighted vMoI (green = weight, red = vMoI).

Additionally, a blank disc (blue) sits coaxially, which can either be braked against the former, or else left to coast on its own.

Everything begins in equal motion at 1 rad/s.

The weight is allowed to drop under OB torque.

As it does so, the vMoI opens. This slows the drop, increasing its momentum yield.

We complete the GPE interaction with more momentum that we would've got from dropping with a constant MoI.

However, we've gained the excess in the 'MoI' component of momentum, whereas it's the velocity component that we need in order to break energy symmetry.

If we were to simply retract the MoI, the resulting rotational KE gain will be precisely equal to the input CF-PE workload.

So we'd have come away from the GPE interaction with more momentum than it would've paid out at constant MoI, but our final KE would be equal to the input GPE plus input CF-PE.


This rig attempts to work around that issue, by first colliding the momentum gain with another, equal MoI, thus halving its speed, and thus the amount of CF-PE required to retract the MoI.

To that end, the blue disc is given an MoI of 2.5, to match that of the weighted vMoI at full extension, thus they're in a 1:1 ratio the instant the brake bites, initiating the inelastic collision while reducing the required 'reset' workload:




..the technique is found to be ineffective, and PE-KE symmetry remains intact.

An extended run of 5 full turns makes little difference:



..it's a nice, logical-feeling process, yet has certain inherent constraints, such as relying on external coordinate space to sync the actions as a function of absolute wheel angle..

..but more problematically, the 'final stroke' here in this interaction is closing the vMoI; and thus inevitably, the rotational KE is only ever going to rise by whatever that final CF-PE input workload - if we had 10 J of rotKE and do 1 J of work against CF force, we'll end with 11 J of rotKE, no more, no less.. ie. it's impossible to envisage a situation in which a final CF workload will accelerate a mass or inertia to OU efficiencies..

The only way to cause such accelerations, is via successive inelastic collisions. Sheer conservation of momentum has to be the cause of the punch-through..

[MrVibrating]

Think about that, cus it seems significant - a 'red line' to ponder:


• we can't punch thru with a GPE output

Dropping a weight cannot be the 'final stroke' in a gain interaction.

But also:

• we can't punch thru with a CF workload either

Same deal; final KE will be whatever it was, plus whatever we've just added, but never any more.


The coup de grace sealing the OU deal can only be an inelastic collision..

..specifically, an ongoing sequence of them, consolidating consistent momentum gains for consistent PE expenditure.

That's what 'mechanical OU' is.

[silent]

I think you're spot-on Mr.V and no doubt the impacts that people heard is no doubt for precisely the reason that you've deduced.

silent

[Georg Künstler]

MrVibrating wrote:

The coup de grace sealing the OU deal can only be an inelastic collision..

..specifically, an ongoing sequence of them, consolidating consistent momentum gains for consistent PE expenditure.

That's what 'mechanical OU' is.

In my construction I have an elastic collision, this is much more effective.
The other sentences I can agree on.

[MrVibrating]

Perhaps, but now you have bounce and presumably escalating internal speeds, and input energy squares with speed if given the slightest chance, and i'm all about trying to prevent that..

Look mate, take two 1 kg masses. Propel one to 1 m/s, without bothering with reaction mass. Collide it with the other inelastically, so they're now at equal speed:

• the moving one sped up the stationary one

• conversely the stationary one slowed the moving one

Because their masses are equal, so is their resulting velocity distributions.

Because KE squares with velocity, each has equal KE.

Inelastic collisions with a 1:1 mass / inertia ratio dissipate the minimum waste heat, while rendering the maximum remaining KE from the interaction. More crucially, they reset the velocity difference between the two interacting bodies, thus resetting the 'V²' multiplier on the input energy cost of raising further momentum; it is this constant resetting of the relative speed difference that pegs the input energy cost of momentum, in spite of the ever-rising system velocity, allowing us to buy, for example, a 10 m/s acceleration of a 1 kg system in ten discrete accelerations costing ½ J each (per KE=½mV²), so 5 J total input energy, yet for a KE value of 1 kg * 10 m/s = 50 J, hence 10x unity.

If the collisions were inelastic, or the mass / inertia ratios were not 1:1, the result would be less efficient; peak OU efficacy converges towards these conditions: it's what we need to observe to design a system that will take the shortest, fastest and most direct route up the OU ladder.


I have already long-since generalised these principles. To make a shortcut calculation of how efficient a given inertia ratio will be, simply add both sides of the ratio together, to find the number of consecutive cycles that must be repeated to reach unity efficiency, and thus, adding one more, OU, as well as the per-cycle efficiency accumulator, hence:

• for a 1:1 ratio of interacting bodies, 1 + 1 = 2, hence the system will reach unity in 2 cycles, thus having a 50% per-cycle accumulator, wasting 50% of input energy per cycle, and reaching 150% on conclusion of the third cycle.

• for a 2:1 inertia ratio, 2 + 1 makes '3', so we have a 33.33% per-cycle efficiency accumulator, making unity in three cycles and 133% on the fourth

• for a 3:1 ratio, 3 + 1 = 4 cycles to unity, 75% loss p-c, hence a 25% p-c accumulator and 125% at the fifth

• etc.

The only other factor to consider is whether we get to keep all, or just some, of the momentum we pay to generate; for example if we had to pay equally for momentum and counter-momentum, but then sink the latter to gravity to keep the now-unilateral remainder, we've also sacrificed half our input energy before we even get to the collision, which wipes another half, thus again giving us a 75% p-c loss / 25% p-c efficiency accumulator / 5 cycs to 125% envelope.


That's pretty much it bruv - the implicit principles of mechanical OU, read between the lines of the standard KE and momentum equations; p=mV and ke=½mV².

All we need to do is think / work thru the only possible permutations of the above dynamics, one of which must be viable in practicable form..