Poss. Symmetry Break?

[ME]

A possible explanation could be, -as you introduce a sudden change at some point-:

In the last case Angular Momentum is conserved because you engage on Motion, and Energy follows.
L=I*w[0] == (2*I)*(w[1]) --> w[1]=0.5*w[0]

In the Spring-case the Spring has maximum potential which will be divided, and Angular momentum follows.
E[rke]=0.5*I*w[0]² == 2*0.5*I*w[1]² --> w[1]=0.707*w[0]

I guess there's no mechanism operating between spring and gearing?

[MrVibrating]

OK so obvious idea - i gave an inital velocity of 60°/s (10 RPM) to each disc, spinning in opposite directions, CW & CCW.

Then i placed a rotary spring on the blue disc. The blue disc thus unloaded its energy into winding up the spring.

The precise frame the blue disc ground to a halt, i paused the sim and simply transferred the spring over to the red disc, still coasting in the opposite direction.

Then i let the spring unload into the already-spinning red disc..

This is as close as i can currently concieve of to a direct reversal of the previous test.

Results:

Blue disc left coasting with nanojoules, pretty much all of its energy transferred to the red disc.

As you'll likely have anticipated already, the red disc ends up with precisely twice its original energy - the two energies summed linearly. Each began with 1116.226 J, for a 2232.452 J net total, all of which now resides on the red disc.

No quadrupling here. Yet.

So not quite the inverse of the previous test. Will have to ascertain how and why it differs, and whether it can be corrected...

But since we don't have the quadrupling, we can't have held on to all of our momentum either... So let's take a look at that:

- we began with 244200 kg-m^2-°/s, on each wheel, so 488400 in total.

- the blue disc ends with peanuts - virtually all the system momentum that could be transferred, now resides on the red disc

- but it only has 345400 kg-m^2-°/s

- so the net system momentum has been reduced, from 488400 to 345400...

- 488400 / 345400 = 1.41401273-etc.





Is it stalking me or something? Should i go a use this in a lotto ticket, or do i only have 1.4 days left to live? WTF does it want from me?

[MrVibrating]

A possible explanation could be, -as you introduce a sudden change at some point-:

In the last case Angular Momentum is conserved because you engage on Motion, and Energy follows.
L=I*w[0] == (2*I)*(w[1]) --> w[1]=0.5*w[0]

In the Spring-case the Spring has maximum potential which will be divided, and Angular momentum follows.
E[rke]=0.5*I*w[0]² == 2*0.5*I*w[1]² --> w[1]=0.707*w[0]

Yeah maybe something like this, like i say i'm sure it's not anomalous because i've checked it bottom up and top down... it's an unintuitive result tho, and still doing my nut in (see above)..

I guess there's no mechanism operating between spring and gearing?

Trying to swap between them for now, but this weird ratio keeps getting in the way. I've tried ignoring it but it doesn't seem to be going away...

[MrVibrating]

Here's an anim of that last test:



..first pausing at min energy on the blue disc, then transferring the wound spring, and pausing again at max KE on the red disc.

anyway, it's getting on for 1:41, so i'm off to 141 myself a nice hot mug of 1414, before 14140.

141 ever4b1dy!

[Fletcher]

Here's some alternate methods of coupling you can try.

Note that one uses a rod connection and the other a rigid joint that suddenly comes into existence.

In both cases the rpm drop from 20 rpm in blue disk to 10 rpm each. Accordingly the RKE quarters but AM is conserved.

N.B. I often make latches out of rods and stops that I activate to hold something - sometimes energy disappears there also because its not exactly realistic interaction i.e. a proper latch is a 'softer' engagement.

[MrVibrating]

Last post this eve, this is all still sinking in...

But clearly, we have a symmetrical consistency here - when we divide one disc's energy between two, net momentum rises by 1:1.414, and when we attempt the opposite, adding two discs' energies to one, our net momentum is reduced by 1.414:1.

Understanding what this ratio means and why it arises must be a valuable lesson, whatever its implications..

[ME]

You manually manipulating the conditions, that's what's happening.

I think (not sure) it's non-intuitive because they look the same but are not.

The spring-disk is hold by some peg -or so I imagine a practical implementation-. Then some peg breaks on spring stress, now that spring-stress-potential sets two disks in motion. --> conservation of energy

The geared disk is missing half its teeth, until it engages by rotation. Because of gearing at least the velocity will become equal. --> conservation of momentum.
There's no quadrupling as the sum of energy is just half of what it was before.
E[rke, final] = I[left]*(0.5*w[0])²+I[Right]*(0.5*w[0])²

For the last one I guess you load a flywheel, and then transfer that flywheel. No wonder you transfer energy.

----
I found an interesting (middle?) thingy

Take the two wheels with gears, body[1] and body[2]
Set the gear-ratio (going from [1] to [2]) to a formula:

min(1, 0.0001 + Body[1].p.r/90 )

Meaning: Body[2] gears with a very small amount, going up and until 1:1 depending on the current angle of Body[1] and being reached within 90°.

When Body[1] start with 120°/s then Body[2] remains almost stationary.
After 90° The energy is shared (and halved), and the velocity of both will be 84.68°/s (0.707 * 120°/s)
Angular Momentum is not conserved (and the absolute sum is about 1.414 times the initial)

Now this sliding-gear-case acts like the spring-case.

With this example I contradict my previous thought, and enhance your dilemma... sorry for that. :-)

[MrVibrating]

Here's some alternate methods of coupling you can try.

Note that one uses a rod connection and the other a rigid joint that suddenly comes into existence.

In both cases the rpm drop from 20 rpm in blue disk to 10 rpm each. Accordingly the RKE quarters but AM is conserved.

N.B. I often make latches out of rods and stops that I activate to hold something - sometimes energy disappears there also because its not exactly realistic interaction i.e. a proper latch is a 'softer' engagement.

Cheers mate, i also tried the magic rigid joint, same result.

Sooo... energy per disc 1/4 the orignal, 50% destruction of net energy...

This is a non-dissipative loss, my friends.

The missing energy hasn't simply been dissipated to low-grade heat.

It's been written out of existence by a spontaneous change in inertia.

This is precisely the same animal as an OU gain. What we have here, is bona fide 'under-unity'.

This is a symmetry break. Even if we never manage to invert it, we've finally earned the thread title.

And should it turn out to be irreversible, we'll at least have gained a valuable insight on our mission... and put this damn 1.414 thing to bed. I'm sure the sqrt 2 relationship can't be coincidence, but for now i just can't see how it all fits together. All we want is 1:1 momentum transfer. No more, no less. I need to run the linear tests Marcello mentioned, see if the same ratio shows up.. back tomorrow night..

[MrVibrating]

You manually manipulating the conditions, that's what's happening.

I think (not sure) it's non-intuitive because they look the same but are not.

The spring-disk is hold by some peg -or so I imagine a practical implementation-. Then some peg breaks on spring stress, now that spring-stress-potential sets two disks in motion. --> conservation of energy

The geared disk is missing half its teeth, until it engages by rotation. Because of gearing at least the velocity will become equal. --> conservation of momentum.
There's no quadrupling as the sum of energy is just half of what it was before.
E[rke, final] = I[left]*(0.5*w[0])²+I[Right]*(0.5*w[0])²

For the last one I guess you load a flywheel, and then transfer that flywheel. No wonder you transfer energy.

----
I found an interesting (middle?) thingy

Take the two wheels with gears, body[1] and body[2]
Set the gear-ratio (going from [1] to [2]) to a formula:

min(1, 0.0001 + Body[1].p.r/90 )

Meaning: Body[2] gears with a very small amount, going up and until 1:1 depending on the current angle of Body[1] and being reached within 90°.

When Body[1] start with 120°/s then Body[2] remains almost stationary.
After 90° The energy is shared (and halved), and the velocity of both will be 84.68°/s (0.707 * 120°/s)
Angular Momentum is not conserved (and the absolute sum is about 1.414 times the initial)

Now this sliding-gear-case acts like the spring-case.

With this example I contradict my previous thought, and enhance your dilemma... sorry for that. :-)

Sorry didn't quite follow your initial contention there, but it's late, i'm frazzled, been banging the metal all day.. you seem to have tripped into the same rabbit hole tho. Nightmare eh? I actually did consider the CVT transmission but unlike you it takes me half an hour formulate basic concepts.. i did try alternate ratios but saw significant immediate losses, which i attributed to the braking torque caused by the effective rise in MoI (gear-up a given MoI and you have an effecively higher MoI per input torque) - IIRC a 10:1 ratio causing a tenfold drop in net energy, with a 10:1 division of the remainder.. (or something like that, didn't take notes)..

[Fletcher]

Mr V .. I changed out the connection between the side-by-side two identical disks in my earlier sim and replaced the rod with a rope, then a spring, then a dampened spring element.

The sim went crazy with huge energy and AM increases. Obviously this is not correct.

Yet my sim of using a 'hard' connection (the rigid joint being turned on) of two identical disks also doesn't appear right in some respects. If the connecting method were perfect I would expect the rpm to be shared equally between the two disks.

That should see a drop in system Kinetic Energy [ formula is Kinetic () ] to that representative of the new rpms, so a quarter. The AM should be shared equally between tthe two disks, and it is.

The lost energy must be attributed to deformation of the pin as I think ME suggested. A non perfect elastic collision in effect. But the sim can't show this detail. I would expect the first disk to slow quickly and the second disk to speed up quickly (some lag).

What I see is in one frame (set at 1000) the rpm changes from 20 to 10 i.e. instantaneous change with no lag.

Just something to think about when using these methods.

[Trevor Lyn Whatford]

Hi Mr V,

a few years back I built a three wheel build, wherein there was 3 separate ratchet bike wheels in line linked together with elastic cord , each being a 360 pendulum, the weight spread was similar to the apologia wheel on Bills site, only at the outer rim. http://www.orffyre.com/mt.html
The idea was to have the falling weight on one wheel (pendulum) pull the next wheel (pendulum) to the fall position and so on, each pendulum wheel, tripping the next.

It work for a while until my input energy ran out, it was a bit chaotic at best, but needless to say there was nothing there to drive a wheel, still glad I built it though, because my question was answered.

Edit, I even tried it with a fly wheel but it was no better.

[agor95]

It is a good thing; you tried and let us know.

So you had a main wheel; hopefully going clockwise.
3 ratcheted wheels with a weight; hopefully going counter CW.

The main wheel was fix to rotate at a set speed?
You stored CF using spring along the bottom arc?

Then released it near the top section?

You are so near.

Good Luck.

[MrVibrating]

Mr V .. I changed out the connection between the side-by-side two identical disks in my earlier sim and replaced the rod with a rope, then a spring, then a dampened spring element.

The sim went crazy with huge energy and AM increases. Obviously this is not correct.

Yet my sim of using a 'hard' connection (the rigid joint being turned on) of two identical disks also doesn't appear right in some respects. If the connecting method were perfect I would expect the rpm to be shared equally between the two disks.

That should see a drop in system Kinetic Energy [ formula is Kinetic () ] to that representative of the new rpms, so a quarter. The AM should be shared equally between tthe two disks, and it is.

The lost energy must be attributed to deformation of the pin as I think ME suggested. A non perfect elastic collision in effect. But the sim can't show this detail. I would expect the first disk to slow quickly and the second disk to speed up quickly (some lag).

What I see is in one frame (set at 1000) the rpm changes from 20 to 10 i.e. instantaneous change with no lag.

Just something to think about when using these methods.

WM's sprung dampers do have a habit of creating energy, but generally any kind of damping (a dissipative loss) is gonna muddy the waters here.

I also noted problems with causing a rigid joint to appear - most tries, it just pinned both discs to the background for some reason. Obviously, if they're also subject to dissipative losses then that's another hindrance.. It also seemed a flaky technique in the first test, Discs1, where the pin kept causing the blue disc's 20 RPM preset to be copied over to the red disc's field, and also kept glitching during the run, causing the sim to calculate infinitesimals for the red disc before the rigid joint had even released.

So i'll keep in mind the risks in drawing conclusions from any particular test technique, and continue manual cross-checking of results, esp. dubious ones..

But despite WM's vagaries, i hadn't anticipated this 1:141 thing, so my own predictions of results were also off - i'd assumed 1:1 momentum transfers should be trivial, so if i can understand why this pesky ratio keeps spoiling the party i'll at least have found higher ground..


The lesson so far seems to be this striking asymmetry in nature - if we divide the energy of one rotor with a second identical static one, net momentum rises by 1.414:1, and net energy halves.

But when we try to reverse that sequence, combining the final energy of each rotor back into one, net momentum drops 1:1.141, and net energy is conserved.

Since yesterday i've been thinking that we must be coming up against 2LoT, somehow - because we're trying to reduce the system's entropy for free (getting the motion of two discs back into one). This could mean that the halving of energy is actually dissipative, yet this seems utterly paradoxical as net momentum actually rises, so since we don't appear to have lost any, you'd think we could just trade it back...


Later i'll try the linear tests mentioned yesterday. For now tho, i'm just trying to run the one-into-two test backwards in a mental replay.. When we share the energy of one disc with a second, net energy halves, but when we try the reverse, it doesn't double.

So what could be different about the reverse test (two-into-one), compared to a video of the forward test played backwards?

When we play that one-into-two video backwards, something non-physical must be happening, that doesn't arise when we attempt the sequence manually...

I wonder what that could be?

[ME]

This is what I think:
... All manipulations are Conservation of (spring potential) energy, and there was never a transfer in momentum.
This Energy is transferred or activated to get two discs in motion. This results in a 2 times √½ for velocity.
The angular momentum formula will probably become something like L=I·(&#969;+&#945;·&#8710;t) <-- I don't yet know if I can/may write it like this, but it's not a constant.
Only in the test where you suddenly engaged the gearing (in &#8710;t=0) the angular momentum halved.
When doing this gradually with CVT (&#8710;t>0) it generated an angular acceleration, torque and a transfer in energy again.

[MrVibrating]

Obviously momentum cannot literally be 'transferred', but merely induced by an exchange of energy - a point i've been making since the concept of momentum transfer first arose in another thread.

The energy halves (in the 1-into-2 test) because RKE is a function of MoI, which doubles when the second disc is engaged, hence our given RKE is divided between a second MoI, and half the original system energy is thus written out of existence - it no longer applies as it did previously - and the remaining half is evenly divided between the two discs, 2 * 0.25 the initial energy.

At each stage, we can cross-reference the figures by dividing a body's KE by half the square of its velocity, to derive its instantaneous inertia (usually constant), then multiply that by its velocity again to cross-check momentum... So far i've seen no mathematical inconsistencies, the sim appears accurate. It's too premature to be considering differentials (for me anyway) at this stage; the sim's based on quasi-static integrals and i'm used to thinking in similar terms, and in this case at least, i'm fairly sure that the sudden division of avaliable momentum into double the inertia causes an instantaneous thermodynamic disappearance of half the system energy, so it's a kind of binary change rather than an 'analog' variable - the conditions and corresponding integrals change completely between consecutive simulation frames..

But i suspect you're onto the correct line of reasoning as to why we're so far unable to reverse the process. Like i say, i think we may be inadvertently butting up against the 2nd law.. certainly on thin ice, in that respect - system's want to minimise their energy, so we're likely gonna hit a wall trying to push this one back into a higher-energy state..

But ain't taking my foot off the accelerator till i can see the writing on the wall... the fine print, hopefully explaining the root of this 1414:1 enigma. Understanding how and why this arises seems to be a pivotal aspect of what we're up against..