Flippin' Flywheels

[MrVibrating]

Consider a blank wheel, with just a single radially-sliding mass.

It slides from the center, out to the rim, under CF only (no gravity for now).

This induces a negative torque on the wheel, slowing its rotation.

But suppose instead, that as it is dragged outwards by CF, it pulls on a ripchord, which spins up a central flywheel.

The objective is to convert the CF workload directly back into angular momentum, to try and offset the losses normally incured. So when the sliding mass reaches the rim, the central flywheel has been spun up, in the same direction of rotation as the main wheel, such that when they're clutched together the momentum is returned from the flywheel back to the main system, and we've extended a mass, raising our MoI, without having to sacrifice RPM.


That would be an N3 violation... except it doesn't work, because spinning up the flywheel induces a counter-torque of its own upon the net system, such that when they're clutched together, the flywheel merely returns the momentum it sapped from the main wheel when being spun up.

It makes no difference if the direction of spin of the flywheel is inverted, relative to the main wheel - the net momentum is always split 50/50, and remains constant.

Everyone happy? So now, consider this minor alteration:

- instead of spinning up one flywheel, do two - both coaxial (tri-axial?) to the main axis of rotation (ie. in the center, on independent bearings) - but spin them in opposite directions.. so one CW, the other CCW.


- So we've converted our linear excursion into two, equal and opposite angular momentums, on-board our net system angular momentum.

- This hasn't applied a net counter-torque on the system - the two induced counter-torques were equal and opposite, cancelling completely..

However, now we're no better off than we started - when we clutch the flywheels to the main system, one will accelerate it, the other will decelerate it by an equal amount.. and we still incur the net deceleration from the radial excursion.

- But here's the trick; what if we flip one of the flywheels, once it's spun up, before clutching them back onto the main wheel?

We could thus arrange both to add positive torque...


So a mass slides outwards under CF. In so doing, it spins up two flywheels in opposite directions. One of them then flips, rotated thru the Z plane like a pancake toss, the sign of its angular momentum flipped with it..

By my (simple) reckoning, this gets us a free radial excursion - ie. we can 'freely' raise our MoI without decelerating our RPM...

If this works, then we're OU. The cost of retracting the mass inwards rises as a linear function of velocity, while net RKE scales as the half square of MoI times rising velocity, hence there's a breakeven, unity, threshold RPM, below which the interaction's under-unity, and above which it is over.

So for example it might cost 2 J to induce 1 J worth of anomolous momentum, to begin with... but as RPM builds, another 2 J in buys 2 J out, but the next 2 J in gets 4.5 J out, then 7, 10, and so on - the peak interaction efficiency limited only by practical constraints on RPM.



S0 without getting too sidetracked on how and why CoE depends upon N3, it just does, the proposition here is simply that we might gain net system momentum by spinning up two flywheels and then flipping one of these momentums from negative to positive, before recombining them with a base angular momentum.

Flipping a flywheel in a tangential plane induces precessional torque, presumably accelerating or decelerating the flywheel depending on the direction of rotation..? But does such a rotation induce a negative torque on the net system? And besides which, surely a stator could be used to ground any such counterforces - ie. pass the flywheel to the stator support, perform the z-rotation, then pass it back to the main rotor, now inverted..


Again, for clarity, the objective is not to cause a net acceleration when the two positive momentums are clutched back onto the main wheel - although that would be a double win.. All we need tho is to re-extend a radially-sliding mass back out to the rim, from the center, without incurring the usual slowdown. More accurately, we're gonna initially take the RPM hit, but use that CF workload to spin up two opposing flywheels, both dead center sharing the one axle. Then we flip one of those flywheels - the one with the negative momentum - and now we have two positive momentums. These are both braked against the wheel, transferring their welly to the net system's oomph.

Even if the system still slows down a little... even if the amount of momentum saved is absolutely teeny, it's still free momentum and energy..

What thinks ye?

[MrVibrating]

...lookin' for weaknesses here, but suppose the negatively-spun momentum brings that flywheel to a full stop, relative to the static external frame? For example the system began at 3 RPM, the mass slides out, decelerating the main rotor down to 1 RPM, while accelerating both flywheels by 1 RPM in opposite directions, which, relative to the decelerated main system frame, results in one flywheel at 2 RPM positive, 1 RPM positive on the main wheel, and the other flywheel's 'negative 1' RPM, relative the main wheel, fully stationary relative to us. We could just pick up that static wheel, flip it around and sit it right back where it was, with no benefit, as it has zero actual momentum in the first place..

But that'd be a failure to test the hypothesis, not a negative result.. so long as the flywheels are small and light enough, they can be spun up to much higher actual, intrinsic, velocities than the main wheel - maybe the radial excursion brakes the main wheel to a virtual standstill... no problemo, so long as the two flwheels are spun up as fast as possible, with real momentum not just relative velocity, we can flip the sign of the negative portion, and retrun it as postive momentum, yeilding a non-zero net momentum.

Obvioushly, it would then be a simple matter of retracting the mass inwards while the flywheels are locked stationary relative to the main wheel, so accelerating the net system by the applied inertial torque, before repeating the outbound momentum recovery and reversal.. It costs energy to pull the mass inwards, equal to the RKE gained, and initially, we lose it all again each time the mass falls back out. But by offloading it onto equal opposite flywheels, and then inverting one, we can effectively rectify a counter-torque as positive torque.


Seems like a clumsy build, but it's conceptually simple enough it need never get that far.. i must be talking utter shite, no? Can't be this simple..

[ME]

First thought (the rest I don't know yet):

instead of spinning up one flywheel, do two - both coaxial (tri-axial?) to the main axis of rotation (ie. in the center, on independent bearings) - but spin them in opposite directions.. so one CW, the other CCW.

Because they are on independent bearings (basically independent flywheels), why not just keep them rotation in one direction (perhaps I missed the exact 'why') it saves the flip. Spinning up N flywheels (no matter the orientation) takes (1/N) of the available centrifugal-potential-energy.

Anyway, as long as your net energy >100% it's still OU.

[Trevor Lyn Whatford]

Hi Mr V,

I think that the effort to flip a spinning flywheel has been ignored, if you have a bike wheel handy, give it a spin then try to flip it, it as not as easy as you may think.

With that said I love your inventive thinking, so keep it up until you have the ultimate brain wave, that can set us all free.

Giving it more thought as I was writing the above, if you go for a small tilt on a horizontal wheel instead of a flip you maybe able to play with CF and Gravity once you get the timing sorted out, at least two weights would be need though to play that game.

[sleepy]

It seems to me that the weight will require an input of energy three times.
Once to move the weight to the rim.Again to keep it rotating against what is now massive counter torque once it's out there.And one more time to move it back to the center against CF.I don't think the center flywheels could ever provide enough energy simply from the force of the weight revving them up.
It's like asking a yo-yo to return all the way to the top without assistance from your hand.It'll come close the first time,and less each time after that.

[MrVibrating]

@Marcello

The energy asymmetry here varies as a function of RPM - it's principally an inertial asymmetry, an effective N3 violation, and so the energy efficiency follows the same pattern as any other putative N3 break - it has an arbitrarilly-low threshold RPM at which it is at unity, below this speed it is under-unity (so will resist gaining momentum, but the energy being lost isn't being dissipated as heat - it's not a frictional loss).. and above that threshold RPM it is progressively more OU the faster it rotates.


So the energy efficiency is almost a sideline, a consequence of the real game, which is momentum symmetry. Understand the break in momentum symmetry, and the energy efficiency explains itself..

I'll try to summarise the core concept:


- we want to rectify an asymmetric net sum of momentum from pumping a radially-sliding mass in and out

- sliding a mass outwards under CF raises the system's MoI, inducing negative torque

- pulling it back in does the opposite

- if we can slide out without sacrificing RPM, we gain momentum from the completed interaction


So perhaps we can somehow convert the PE of the CF workload, back into RKE on the main wheel?

- if the mass slides out, spinning up a central flywheel, that angular acceleration exerts an equal opposite counter-torque upon the rest of the system

- so extending the mass has caused the usual deceleration due to the rising MoI, but also, now, this other, additional deceleration, caused by the counter-torque from spinning up the flywheel

- we could still dump that KE on the flywheel back into the main system, but now we're no better off than we started, having recovered the momentum lost to the flywheel, but while still suffering the MoI-induced deceleration.

- it makes no difference which way we spin up the flywheel relative to the net system's direction of rotation - the net momentum remains unaffected by the flywheel, and we still incur the MoI-induced deceleration


And so this is where we consider adding a second, opposing flywheel, spinning them both up equally, in opposite directions relative to the wheel's rest frame.

- now, we still incur the usual MoI-induced deceleration. No getting away from that. The extended mass is orbiting too slowly for its rising radius, and its inertial resistance to that acceleration is braking the whole system.

- but the work done in spinning up the two flywheels has NOT incurred a corresponding counter-torque. So the net system has ONLY been decelerated by the MoI rise, and not by converting that CF PE back into RKE, as otherwise occurs when spinning up a single flywheel. We've dodged a bullet, at this point..


- If we add the positve momentum from one flywheel back to the net system, our net momentum has now increased!

- however when we add back the negative momentum, too, our net system momentum balances back to zero

..which brings us to the coup de grace;

- flip that last flywheel's momentum, reversing its sign, before adding it back to the net system.

- now we have even more anomolous momentum..

- clutch the whole stack back together and draw the radially-sliding mass back inwards again, lowering the MoI and raising the net system velocity

- now just rinse and repeat, each time the mass slides out under CF, it accelerates both flywheels equally in opposing directions, incurring no counter-torque from them, yet still adding these momentum back into the system, having inverted the negative portion.



The net momentum rises further each cycle. Don't worry about energy efficiency for now - hell, just assume it's a great big inefficient loss, that's fine for now.. just follow the momentum; it's principally a momentum asymmetry, so the energy balance depends entirely on the validity of that premise..

[MrVibrating]

...just wanna really spell out this point - this is a momentum asymmetry in the first instance, so the energy efficiency is not the principle point of interest - an energy asymmetry will inevitably fall out of an effective N3 break, trust me on this, but i'm not asking anyone to consider energy symmetry here, just yet.

The question is purely whether or not net system momentum is constant.. energy balance, or lack thereof, is only relevant in consequence of this principal objective. If it costs a megajoule to steal a half gram meter second, that's still a miracle, if bittersweet..

[MrVibrating]

Hi Mr V,

I think that the effort to flip a spinning flywheel has been ignored, if you have a bike wheel handy, give it a spin then try to flip it, it as not as easy as you may think.

With that said I love your inventive thinking, so keep it up until you have the ultimate brain wave, that can set us all free.

Giving it more thought as I was writing the above, if you go for a small tilt on a horizontal wheel instead of a flip you maybe able to play with CF and Gravity once you get the timing sorted out, at least two weights would be need though to play that game.

Cheers mate, i did consider this 'precessional torque' and whether or not it will induce a counter-torque on the net system, concluding that in principle, any such counter-forces could be grounded via a stator.

However we could circumvent inducing precessional torque in the first place by simply using a single gearwheel to invert the torque from the momentum, instead of the sign of the actual momentum itself.

Physically lifting one flywheel off, while still spinning, flipping it round to face the other way, then re-attaching it, is obviously impractically awkward.. yet still a perfectly valid thought experiment, to get to grips with the concept. But a stator-mounted gearwheel would make it easier to actually implement the same effect.

I like this so far.. it's a two-stage inertial interaction, "motion from motion" as Jim calls it..


In its simplest form, the prospective symmetry break is simply being able to extend a mass outwards radially, without incurring the corresponding RPM drop. We still slow down at first, but while adding (free?) momentum to the flywheels, which can then be added back to the main system.



The decisive experiment would be as follows:

Baseline:

- let the mass slide out while the flywheels are both locked to the main wheel, and note the starting and ending momentum and energy


Test

- repeat, letting both flywheels spin up, again noting final momentum and energy

- then brake the positve flywheel to the main system, adding back its momentum

- and finally, add back the negative momentum from the other flywheel, but with its sign flipped to positive.


If the final momentum and energy is more than the baseline, it's also more than we began with.



So what i need to do now is work out how to sim this. Converting the linear translation into torque on a flywheel is going to be the tricky part..

A spindle and ripchord seems the simplest option, but WM2D doesn't have a widget for this, so i'd have to improvise a maths function..

I'm thinking maybe add a motor between the flywheel and main wheel, applying 'acceleration', with an 'active when' condition taking the acceleration of the radially-sliding mass as its input.. this should offer versatile gearing options while still repsecting CoM, i think... (would appreciate any 2nd opinions here)..

[ME]

Interesting theory, thanks for sharing !

When a weight travels from a low radius to a large radius because of a centrifugal force effect, then it's basically a change in position under influence of acceleration resulting in some predetermined velocity at such distance: Conservation of energy is still a mathematical consequence (at least for this part).
What happens with this rotating Atwood mechanism is that you actually make use of a desired effect which is: this weight should end up with some small velocity (preferably zero) at the rim, which also helps against structural damage by redirecting most if this hit momentum.

When I take Sleepy's yo-yo example and use the flywheel's velocity to bring this weight back to its smallest radius then the end result will also be that the flywheel's velocity becomes 0 (despite some losses).

if we can slide out without sacrificing RPM, we gain momentum from the completed interaction

Yes, but keeping RPM will also require the weight to gain in tangential velocity (and momentum)... That's why the wheel slows down in the first place: it tries to keep its momentum. So it's reversed: you need to add momentum to keep the RPM's.
According to my unsubstantiated guess it could require the same amount of momentum as now stored in the flywheel, or the actual end-velocity of the weight at hit-point only directed inline with rotation to keep its RPM's (perhaps answers this: the brachistochrone)

The energy asymmetry here varies as a function of RPM - it's principally an inertial asymmetry, an effective N3 violation, and so the energy efficiency follows the same pattern as any other putative N3 break - it has an arbitrarilly-low threshold RPM at which it is at unity, below this speed it is under-unity (so will resist gaining momentum, but the energy being lost isn't being dissipated as heat - it's not a frictional loss).. and above that threshold RPM it is progressively more OU the faster it rotates.

I need to chew on this a bit more, it tastes either bitter or sweet.


Edit to reply on upcoming post (I guess there's no need to add a new post):

For starters though, just work out how you could gain energy from a violation of Newton's 3rd law; the game is principally momentum gain, and the energy gain falls out as an almost incidental consequence of that.

True.
Energy as a number does some tricky stuff when brought out of frame, I'll only believe in a CoE violation when some mass gains in height (or something equivalent). But inside a frame a resulting acceleration over distance just leads to some new velocity, by definition.

[MrVibrating]

It seems to me that the weight will require an input of energy three times.
Once to move the weight to the rim.Again to keep it rotating against what is now massive counter torque once it's out there.And one more time to move it back to the center against CF.I don't think the center flywheels could ever provide enough energy simply from the force of the weight revving them up.
It's like asking a yo-yo to return all the way to the top without assistance from your hand.It'll come close the first time,and less each time after that.

The weight wants to move out to the rim, under CF.

Letting it do so, raises the MoI, decelerating the system. For now, we let this happpen, and take the RPM hit.

But on the way out, we let it spin up a pair of opposing flywheels.

Because they're opposing, the net counter-torque from these accelerations is zero.

Therefore, i'm thinking that when the mass arrives out at the rim, the work it has done in accelerating these flywheels is basically free - the deceleration due to the rise in MoI is the same as if the flywheels were locked to the main wheel throughout, except they weren't, and so the PE available from the CF output workload hasn't come at the expense of net RKE..


In other words, i'm assuming that the angular deceleration caused by extending a mass outwards radially is the only deceleration so incurred, and that if we use that PE to load up a spring, say, the PE in the spring hasn't come from RKE...


So that's an easy assumption to test, if i haven't done so and forgotten it already.. does loading CF PE into a spring reveal it to be RKE after all?

Will try this out later..

[MrVibrating]

@Marcello - cheers, back in a mo, gotta go eat...

For starters though, just work out how you could gain energy from a violation of Newton's 3rd law; the game is principally momentum gain, and the energy gain falls out as an almost incidental consequence of that.

It's basically the same reason why reactionless thrust would be a CoE violation - energies as measured from the internal and external reference frames diverge with the rising net momentum, following a distinctive pattern (caused by circumventing the usual half-square accumulator on KE)..

Back later with more thorough explanations, or retractions, as the case may be ..

[MrVibrating]

Quick update (couldn't sleep)..

I did the test mentioned above - two identical flywheels with radially-sliding masses - one loads a spring on its way out, the other doesn't, and both systems end at the same velocity.


radius here is 3 meters, with a spring constant of 1 N/m - a half-decent output workload by any measure

Thus the CF PE is available to perform work - it doesn't come at the expense of RKE, and if we don't tap it, it just gets dissipated away.

So, waste not, want not..

---



This seems to clear the only potential for doubt i've had so far. The implication is, spinning up a pair of opposing flywheels will not cause any deceleration of the net system - it's a free ride - and converting that PE straight back into RKE and angular momentum looks like a doddle; braking the positive momentum boosts the net system velocity to higher than it would otherwise have been after the same radial translation with the flywheel locked to the main system, and then braking the second one after flipping it, or else just inverting its torque, adds yet more additional momentum.

I suspect that, in principle, we may be able to reset a mass out to the rim without sacrificing any drop in velocity, gaining net momentum each complete cycle. Again, though, even a partial effect will be a miracle, and OU seems guaranteed..

[MrVibrating]

How far this can go, performance-wise, seems to be limited by how small and fast the flywheels are, relative to the main system. This is because the negatively-spun flywheel begins each acceleration by shedding actual momentum, only gaining it again after passing through stationary, and accelerating back up in the opposite (real) direction.


Relative to the rest frame of the main wheel, both flywheel masses accelerate equally in opposite directions. But from the external frame, the positive flywheel accelerates ahead of the main rotation, gaining momentum, while the negative one first decelerates down to stationary, then accelerates back up the other way...

So this 'crossing of the curve' of actual, objective momentum is a limiting factor.

Worst-case extreme = light, fast main system with high MoI flywheels at low velocity = the negative flywheel will just lose real momentum, being braked towards stationary and never breaking through and back into real-world velocity territory. At best here, it'll have no momentum to contribute back, and at worst it will have sapped some net momentum

Best-case extreme = heavy, slow main system with high-geared low-MoI flywheels spinning much faster that the main system RPM. This will minimise the foray into dead or negative real work intergal, and the maximal share of negatively-spun momentum will remain real and recoverable.

Bottom line is that a practical build will have a self-limiting optimal RPM, rather than a runaway feedback loop; Efficiency drops as main system velocity approaches the actual (real) velocity of the negatively-spun flywheel, rendering its velocity difference progressively more relative (and thus impotent) than actual.

Basically, so long as the momentum is real, and not just relative velocity, the proposed interactions seem viable..

[eccentrically1]

Wouldn't this have the same problem any wheel has? The spring loading mass has to be returned to center? Otherwise you just have a build of 3 flywheels?

[MrVibrating]

So here's the same deal, but this time the spring has been replaced with the two opposing flywheels.



They're driven by a rack and pinion i found online (credit wherever it's due) - a ripchord would be ideal but can't see a safe way to implement this in WM2D without bypassing the sim with a parallel (and likely erroneous) set of home-made formulas, defeating the point of simulation - it'd just be an animated model that did whatever i told it to. So a mathematical transmission, using pure algegra, is a bad idea - better to implement a physical mechanism that relies only on whatever values the simulation computes.

The rack and pinion, and the two gear wheels via which they drive the flywheels, are almost massless, so have negligible momentum and GPE (i haven't even bothered balancing them).

The whole system is free to rotate in response to any induced counter-torques from spinning up the flywheels, but as you can see, there aren't any.

Gravity is currently being used to drive the drop weight, just so you can clearly see what's happening.

So the next step will be to turn gravity off, and spin it up, applying CF instead.

Then we can compare the final momentums against a baseline in which the flywheels remain fixed to the main wheel.

If we're over the baseline, then we have effectively sent a mass back out to the rim without incurring the full compliment of deceleration that would normally apply, and we're ready to create momentum and energy..