Decoupling Per-Cycle Momemtum Yields From RPM

[MrVibrating]

OK some updates & clarifications..

Here's an example of what (i think) i'm babbling about re. gravity basically reducing to an ambient time-constant rate of change of momentum:



..the net change in momentum / unit time depends only on the amount of gravitating mass - invariant of any proportion of non-gravitating system mass.

For cross-referencing consistency let's compare that back to a conventional 'rotating' wheel, with a varying net inertia:



'G' = a constant dp/dt, invariant of the gravitating to non-gravitating inerta ratio, or any spontaneous changes to it.


Now switching back to the linear analog, let's drop a second weight onto the first:



The mass of the 2nd weight being equal to the net of the other two, the collision thus between equal inertias.

So now let's look at rel vs abs momentum distributions; first, here's the system state if the frame immediately prior to impact:



..and here's the very next frame afterwards:



Upon collision:

- red + green gain 24.3352 p

- blue loses 24.2175 p

avg. ~ 24.27635 p

IOW conserving momentum.

However, pre-collision, the 2nd weight had 51.1319 p in the ground FoR:

- assuming N3, 51.1319 / 2 = 25.56595 p

- the collision actually exchanged -1.23075 p less than this

If instead we consider the relative momenta:

- impact speed = 4.2258 m/s * 11 kg = 46.4838 / 2 = 23.2419 p

- the collision actually exchanged +1.03445 p more than this

What's the nature of this apparent rel/abs discrepancy - presumably i've failed to account for something..?

[agor95]

I am talking to myself - this is off-the-cuff, seat-of-pants R&D.

And there is nothing wrong with some good R & D too inspire other members.

Both your R & D tests

1. Using a sliding mass 9.81 kg against a dropping mass of 1 kg.
2. Increasing MoI [Moment of Inertia] against a dropping mass of 1 kg.

Thought provoking.

All the Best

[MrVibrating]

..just to focus the attentions:

• if there's an effective N3 break to be had from inelastic collisions in a gravitating FoR, we must be looking at its potential components here

How might 'the riddle' find any interpretation here?

Should the mass of the 2nd weight be reduced to match the 1 kg 'OB' weight - ie. as would seem more natural in a simple wheel using equal mass weights?

[MrVibrating]

1. Using a sliding mass 9.81 kg against a dropping mass of 1 kg.

2. Increasing MoI [Moment of Inertia] against a dropping mass of 1 kg.

Thought provoking.

All the Best

Cheers, but isn't it? The basic dynamic being that 'momentum' is only substantiated in the instantaneous product of velocity and inertia, yet can only accumulate at a finite rate (G), whereas 'inertia' can vary faster or slower (or not at all)..

..if inertia's constant then all momentum is gained in the 'velocity' component, however any increase applied to the 'inertia' component thus absorbs the momentum gain instead of velocity / acceleration - so it's possible to entirely cancel gravity's acceleration..

..or even, in principle, effectively inverting it:

• you could apply sufficient negative inertial torque to cause a 9.81 rad/s² deceleration of a wheel while under acceleration by a falling weight!

Either way tho, the 'momentum yield' for a given mass * height is a function of the drop time; the more non-gravitating inertia, the longer the 'G-time', but with the increase in gain being in the 'inertia', rather than 'velocity' components of momentum.

If however we could somehow juggle up an asymmetric distribution of momentum and counter-momentum, any resulting non-zero energy sum might present another type of 'yield'.. dunno.. just digging bashically, 'scuse the mess..

[MrVibrating]

As mentioned, i equalised the inertia ratio of the collision to minimise the energy dissipated..

..however since it's probably more likely a simple wheel would use weights of equal mass for both over-balance and 'secondaries' for hitting 'em with, here both are now 1 kg:





Pre-collision detail:



..and post-collision:

[MrVibrating]

So in that last run, pre-collision there was:

• 4.6484 p on one side of the interaction, and

• 4.2258 + 0.42258 = 4.6484 p on the other

..so both 1 kg masses have induced exactly the same amount of momentum from gravity, over the same period..

..yet one at 11x the speed of the other..

..the latter having 11x the inertia of the former..


Post-collision, there's now

• 0.6039 p left on the secondary weight..

• ..and 0.79204 + 7.9204 = 8.71244 p on the other two masses..

• so we still have 0.6039 + 8.71244 = 9.31634 p

• remember we only had 2 * 4.6484 = 9.2968 p an instant before, so the slight net gain is only due to the continual acceleration of the system between these consecutive frames..

• ..crucially however it was symmetrically-distributed either side of the interaction pre-collision..

• ..post-collision, that distribution is now substantially asymmetric.. by a factor of 14.4

Is this an effective N3 break, then?

[agor95]

]Thought provoking.

Cheers, but isn't it? The basic dynamic being that 'momentum' is only substantiated in the instantaneous product of velocity and inertia, yet can only accumulate at a finite rate (G), whereas 'inertia' can vary faster or slower (or not at all).

I appreciate the examples you have created. Some readers may be invoked into thought.

Well will only know by them posting in your thread.

P.S. For those where English is not their first language "but isn't it?" means "It is thought provoking".

Cheers

[MrVibrating]

The asymmetry's clearly being caused by the 'relative d/p' - net momentum's conserved, but in an unusual, out of kilter way:

• 'relative d/p' = the 'effective' momentum developing between the two gravitating halves of the system

• although both halves gain equal p from G * t, the leaner one catches up with its inertially-laden partner

• it's this 'relative', internal momentum that's interacting and being redistributed in the collision

• precisely because N3 is still respected in the collision within the gravitating frame, the resulting momentum distribution in the ground frame is broken

So the hypothesis seems to have borne out; the 'efficacy' of the symmetry break now contingent on whether it can be made to accumulate momentum at profit, or else 'fling a heavy thing high', or something, i guess..


(@agor95 - ta mate)


ETA: the task, surely, is to consolidate an effective N3 break into an effective N1 break - ie. cyclical accumulation - then we can step into the 'energy' game..

[agor95]

One should not trust a computer on face value. It is best too check the output looks right.

In the presentation for my request item 1.

The green mass is 9.81 kg and the red weight is 1 kg supplying 9.81 newtons.

When the red weight pulls down. If we then apply the force by a robe, pulley setup.
That would help align the rope too pull along the horizontal.

The acceleration should be close to 1 m/s^2

The software calculations in the current setup might create inaccuracy.

So with this setup we are looking at the collision of a 1 kg mass accelerating at 9.81 m/s^2.

This is against another 1 kg + 9.81 kg mass accelerating at 1 m/s^2.

Now it's a task too validate the contact accuracy.

Cheers

[MrVibrating]

Nothing inaccurate in the sim, but also prolly nowt useful either..

Just because the momentum distribution was 50:50 prior to the collision, but not afterwards, with an end result of two equal momentas on the same vector catching up with one another and colliding, equalising their speed difference for an asymmetric distribution relative to the ground frame, doesn't mean we have any kind of 'effective' N3 break - we dropped both weights in the first instance, without applying a mutual force between them.. so, not really a closed-loop asymmetric inertial interaction, is it?

It's OK tho, as i think i've now found the solution.

Gonna pop down the offie for a wee dram.. gonna need a scotch & stogie for this..

[Fletcher]

LOL .. gotta love the highs and lows of this game :7)

[MrVibrating]

I'll spare the long & winding intro: TL;DR i think the riddle may be describing the Toys page interaction.

I think it might be a vertical inertial interaction in free-fall that pushes some weight upwards, whilst pushing more down..

Basically the exact same interaction i've been playing with the last ~5 years, only i wasn't applying enough internal force..

..if i'm right, the trick is collecting a PE discount on the resulting net of GPE + KE..

To recap, here's a 9.80665 N force applied vertically between two 1 kg weights in free-fall:




Now the exact same thing again, only using twice as much internal force, and with the lower weight quadrupled to 4 kg:




Finally, with the usual two alternate metrics on input work done:



• act P*t = 239.9453492

• act F*d = -239.9451895

• GPE out = 255.1359 - 14.71 = 240.4259

• net KE = 480.8519

• 480.8519 - 239.9453492 - 240.4259 = +0.4806508 J

..again, the riddle describing the gain conditions, rather than the gain or 'advantage' itself.

The advantage being a relative vs absolute FoR divergence..?


Is this anomaly real - can anyone else measure it independently?

[MrVibrating]

Can't see what might be causing error.. the gain's slight, for now, relative to the working load, but should be enough to close-loop, if it's real..

Not using any code, just the integer 'force' inputs you see there. Tried also controlling the actuator for 'acceleration', results identical..

Just a modest force applied between two modest weights in 1 G for 1 second.. how could that possibly break the sim..?

If real, how did i not notice this before? Ditto if in error?

Half a J takes 1 kg to 1 m/s - you wouldn't want it in the face, that's for sure..

[MrVibrating]

For another angle on it:

• consider opposing angular (or linear) accelerations - observing N3 - but in a rotating FoR

• thus one inertia is being accelerated in the direction of system rotation..

• ..the other being equally decelerated

So for example if the system was rotating at 2 RPM, inside which equal opposing 1 RPM accelerations are applied to a pair of angular inertias, then one's being accelerated to an absolute speed of 3 RPM, while the other's decelerated down to 1 RPM abs.

Because these onboard accelerations are equal and opposite, applied to equal inertias, they cancel each other's counter-torques, hence the system continues coasting at 2 RPM.

All three rotations remain in the ground FoR, and energy and momentum are conserved and constant - the two onboard rotors having spun up and slowed down, respectively, in equal measure; all we've thus done is pass some angular momentum from one hand to the other, but all that was lost with one hand gained by the other; all square, no anomaly.



Now just run that again, only this time, subject the system to some kind of reactionless acceleration - cause it to just hypothetically speed up, but without pushing or pulling against any external inertia, as would normally be unavoidable..

You could use inertial torque / the ice skater effect, but let's just assume it's gravity for now - the whole rotating FoR under a constant ambient acceleration, as from OB torque..

Now our little onboard mutual inertial interaction is no longer symmetrical! Just look:

• suppose this ambient acceleration was 1 RPM / sec - the same acceleration rate we apply between our angular inertias

• because angular momentum's absolute (either there's CF force or there isn't), our 'accelerating' rotor is thus still undergoing a 1 RPM acceleration relative to the system, but it's now a 2 RPM abs. acceleration from the ground FoR in which the whole system's accelerating at this uniform 1 RPM² constant..

• ..likewise, the 'decelerating' rotor still sheds 1 RPM relative, in response to the internally-applied torque, but instead of also decelerating down to 1 RPM in the ground frame, that 1 RPM relative deceleration is perfectly cancelled by the 1 RPM absolute acceleration of the net system, hence the 'decelerating' rotor remains at 2 RPM absolute

• net momentum's thus increased, and is asymmetrically-distributed between the two onboard interacting rotors - the 'accelerating' one has gained more absolute momentum than the 'decelerating' one has lost

..and so we come back to the linearisation - trying to use gravitation as the ambient acceleration to cause a momentum asymmetry.

The principle could form a third alternate solution for a 25% accumulator:

• reactionless accelerate-collide cycles between inertias in a 3:1 ratio are one potential route

• dumping half of per-cycle input energy to gravity along with the counter-momentum, then consolidating with an inelastic collision, also gets 25% better with every shot..

Neither of these 'mathematical possibles' having borne fruit thus far. Just can't mechanise 'dem maths..

..but now check out this one:

• suppose the internal relative acceleration is 4x the ambient / absolute acceleration (4G, bashically)..

• this causes a 50% imbalance of absolute momentum to counter-momentum in the ground FoR

• so for example instead of an internally-applied torque generating equal opposing momentum and counter-momentum, there'd be a -75% : +125% split - it's still a 1 RPM relative acceleration between free bodies, but one decelerates 'a quarter', the other accelerating by ¾..

• the point being that if we can induce momentum and CM asymmetrically then their sum is non-zero, and we have our momentum source; gravity's just the kickstart - this is making momentum from momentum itself..


• EG suppose you had 1 kg-m²-rad/s of angular momentum - send 60% clockwise and the other 40% CCW

• summing unequal positive and negative momenta back together leaves a remainder: +60% + -40% = +20%

• similarly, -75% + +125% = 50%

• if net mass / MoI is constant and half the momentum's gone.. so has 75% of the energy, too! Dissipated in the collision!

• hence a 25% efficient per-cycle momentum yield, consistent with the Toys page, on momentum made from the asymmetric distribution, and consequent gainful recombination, of other momentum.. in principle

This is what i think we might be looking at, anyway.

It's what the Toys page seems to be trying to convey - the lower hammer toy 'D' representing paired opposing angular inertias, the handles representing the internal input torque, in 90° phase with the GPE / OB cycle represented by the upper hammer toy 'C' and the two toys respective alignments to the 180° axle markings ('eyelets' on the axle 'A') - IOW the opposing angular accelerations are only applied in-between GPE output strokes / IE whilst the system's under OB torque, and, thus accelerating..

..so maybe the trick isn't so much to source or sink momentum or CM directly from or to gravity, so much as to use gravity to skew distributions of internally-applied +/- momenta, from which new momentum is then made from its subsequent non-zero cancellation..?

[MrVibrating]

Blabber blabber with the codeine & alcohol and clawed hand on throbbing spastic arm - you wouldn't believe the size of the box they sent me home with from the hopsical (hic), bleedin' two-weeks permanent blotto they're doling out there,gawd bless em.. let's see how much still makes sense tomorrow eh..