MTM5

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MrVibrating
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Re: MTM5

Post by MrVibrating »

Logically the zero-momentum frame must be that of the rotor angle, as opposed to any of the inertias.

Setting the problem out from this perspective should help solve it. We know net dp should be zero around that FoR at all times, and when the two satellite discs are braked the freely-turning wheel base does halt; in conjunction with the working 2-body formulas as shown below, it shouldn't be that difficult to solve the 3-body version when the grey disc is unlocked:

Image

Throwing it out there if anyone's up for the challenge.. I'll make a stab over the w/e, no more time for now..
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Re: MTM5

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..will spend some time on this from tomorrow, but just two little ideas for now; make the dp exactly 1 kg-m²-rad/s, as this will make it easier to recognise numbers that sum to '1', and maybe if i get really desperate, try putting the question to chatGPT or similar generative AI..

Another thought that came to me earlier was to remember to treat translationals as relative, rotationals as absolute.. so it's more likely the former need adding and subtracting from one another than the latter..
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Re: MTM5

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Spent a few hours on this today and don't seem to have got anywhere.. it's seriously doing my nut in..

I did discover that a body's orbital L - usually calculated as the point-mass times radius-squared times angular velocity - can be equivalently formulated as 0.5 * body[n].mass * sqrt(body[n].v.x^2 + body[n].v.y^2) for the blue disc and 2 * body[n].mass * sqrt(body[n].v.x^2 + body[n].v.y^2) for the green one.. not much use for it here but equivalent formulations are just interesting in their own right..

Made no headway on the 3-body solution tho. The 2-body version's simple enough:

Image

..the blue rotational momentum on the left is balanced by the blue orbital moment about green plus green intrinsic on the right, very straightforward.

When the grey axis is also free however it all goes out the window:

Image

..remember, if we drop the anchors on the blue and green axes, leaving the central one free, all motion ceases, proving the net rotational momentum is a constant nil.

You'd think it was just a simple case of adding and subtracting the angular motions of the base wheel from the two satellites, yet i'm sure i've tried every permutation without success.. A problem that should have a simple empirical solution seems to turn into some kind of Medusa - i can get the delta low, but can't make it disappear, yet we know it's not real so a 'solution' is all or nothing..
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Re: MTM5

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..just realised - there may be a use for that reformulation of the orbital momentum after all:

• WM can output rot and trans KE for any body

• the sum of these discrete rot and trans KE's equals kinetic() - WM's internally-calculated net KE total

• i could previously invert the rot KE's by their velocities to derive the MoI's and thus momenta, but not the trans KE's..

..obviously you can't just add the translational momenta to the rotationals and expect a constant sum from the mixed dimensions; clearly there are translational momenta being exchanged with the world via the fixed axis, and summing those to zero is a matter of inference and irrelevant to the objective of determining the net rotational momentum, but the standard I=(KE/w²)*2 inversion cannot be applied to the translationals since they're based on mass rather than angular inertia, so re-multiplying that derived 'I' by the velocity would give an excess value, unrepresentative of the true orbital momentum..

But since i can now also derive orbital AM from trans KE, that sum should now be valid.

Furthermore in principle i don't see that i should be adding and subtracting velocities to curve-fit momentum any moreso than one would for the KE's - angular motion's absolute, and indeed KE is only meaningful in relation to the absolute FoR - so the sum of products of derived MoI's and metered velocities must represent the net system momentum just as surely as the sum of rot and trans KE's equals net system KE.

That seems like pretty infallible logic to me. This method of deriving net system momentum from KE's - having this baked-in cross-check with kinetic() - has to likewise peg the net angular momentum. Similarly, if net KE is constant whilst coasting, the net momentum derived from it must also be constant.

If i can just plot a constant value, it should already come out split down the middle, either side of the motor.. basic logic right?

If it still comes out non-constant, then there would seem little option but to try factoring in some kind of rot-trans hybridisation of momenta going on - for instance maybe the momentum's non-constant because what we're measuring as 'net system AM' isn't actually closed, but includes a reactive translational component being exchanged with the world via the fixed axis, and it is this rot-trans chimera that is being cancelled when braking the satellite moments, halting the system, rather than purely-internal angular momenta.. But that's premature speculation for now; hopefully this new approach will show the zero-sum..
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Re: MTM5

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Turns out that deriving orbital momentum from trans KE produces an absolute number, which thus still needs appropriate signing, which thus means using operator logic, basically devolving into something that looks more of a kludge than an elegant function..

I did realise however that controlling the motor for 'zero acceleration' wasn't actually coasting, since the green disc naturally wants to vary its speed independently of the blue one owing the the latter's MoI variation relative to grey. I switched to controlling for torque, and thus 'zero torque' when coasting, which also allowed me to fine-tune the imparted momentum as a function of N2 over time, so that exactly 1 kg-m²-rad/s is applied in each direction:

Image

..should've done that a week ago, but now all i need to do is shuffle the right-side equations to reach that target figure of '-1'.. and then test to ensure the solution's impervious to the masses or MoI's of the discs.. Finally seem to be getting a handle on it tho..
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Re: MTM5

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Something curious here, that seems to go to the heart of the issue:

In the following test run, the motor progressively torques the blue disc's 0.25 I up to 4 rad/s = 1 L over 1 second, and thus presumably imparting the corresponding -1 L - that we're trying to measure - to the green and grey discs beneath.

The expectation that it's really there is three-fold; the basic assumptions of N3 and N1 (ie. assuming we have an inertially-closed system), and also the brake test in which braking blue and green axes cancels all motion.

So the test below spins up the system for the first second, then, at 1.5 seconds, the green disc's axis is locked against the grey disc; grey and blue remaining free and coasting. Half a second later at the 2 second mark, the blue disc is likewise locked against green, halting all motion:

Image

Ignoring all the other right-side meters, pay attention to the provisional "Net Grey Rotational" in the lower right corner; this is the grey disc's own moment, plus the point-mass mr² of the green and blue discs relative to its center, then multiplied by its angular velocity:

• after locking the green axis, during the period between 1.5 and 2 seconds, the 'net grey rotational' should represent all of the momentum on the 'bottom side' of the motor; all of the counter-momentum - at least by my flawed reckoning - and yet it shows a value of '-0.964'.

• when the blue axis also locks at 2 secs however, net system momentum nonetheless goes to zero..

Where's that missing -0.036 L hiding during this period?

Somehow, things aren't quite as they seem..

Is the missing counter-momentum still there, hidden within the system or some lame accounting oversight? Or else, is it outside the system, beyond the fixed axis - an internal angular momentum that is only balanced by an external translational counter-momentum? (woo-wavy hands)
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Re: MTM5

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..perhaps that hypothesis can be tested by anchoring the system to pseudo-planet - a mass great enough to be all but unmoved by these momenta, yet light enough to embody them within the 16 digits discriminable range we have here..
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Re: MTM5

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..but before drawing any conclusions about the missing -0.036 L, i should probably keep on trying to calculate the -0.964 L that's apparently the real net counter-momentum from the instant the spin-up's finished.

I can't for now see any combination of those components that makes that value, so something's obviously wrong with how i'm trying to formulate it.

Yet all i'm doing is the same technique that works for the two-body system - and again, angular momentas must be absolute, no? The instantaneous angular momentum of a rotating body is its MoI times its absolute angular velocity. Or should i be adding and subtracting relative velocities - such as subtracting the grey disc's motions from green and blue, or green from blue?

If i can just get it to plot the evolution of that -0.964 L counter-momentum i'll at least be able to see where it starts to diverge from the top-side 1 L of momentum.. For now i'm not even sure i can trust the latter; is there even a deficit at all?

Updated version attached (corrected translational sum, added pauses for both brake points for comparing before / after frames)
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Re: MTM5

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Getting there - seems that adding and subtracting velocities is the key after all; the trans momenta thus cancel (so the system is inertially closed), and the 'missing' counter-momentum is restored by subtracting the grey disc velocity from the blue one's orbital velocity.

This now results in the correct counter-momentum being calculated when the green disc's braked, now showing '-1' instead of '-0.964', but it's still calculating gibberish all the way up to that point:

Image

Looks like it fully-conserves momentum just fine - as you'd expect - so must be a matter of correctly compensating the velocities to show the constant zero-sum..
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Re: MTM5

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Rabbit hole eh?

I've come to conclude that all the compensations i'm trying to apply by adding and subtracting velocities are misguided; the correct momenta are functions of the absolute velocities, just like the KE's (the rot + trans net of which sum to kinetic() remember), and there's obviously a rotational-translational exchange arising between the system and world, as previously noted.

To highlight this i've unpinned the central fixed axis from the background - which was just grounding unmetered momenta - and instead pinned it to a floating planetesimal (1e9 kg and 1e9 kg-m²) so that these counter-momenta can also be metered.

Here's the result with the grey axis locked:

Image

..and here it is again, unlocked:

Image


Bearing in mind that the objective of tracking the momentum zero-sum was simply calibration of the metrics - leaning on N3 as a sanity-check by cancelling momentum and counter-momentum - the result of this little detour is that i have less confidence in any of the compensations i've been trying to apply than i do in the untouched metrics based on absolute velocities.

The sum of rot and trans KE's based on absolute velocities matches kinetic(), so this should be adequate confirmation that the momenta based on the same inertia and velocity values are equally sound.

As such, i'm satisfied that the sum of the rotational and translational momenta represents all of the primary system momentum. That is to say:

• blue.moment * blue.v.r

• green.moment * green.v.r

• grey.moment * grey.v.r

And:

• blue.mass * |blue.v|

• green.mass * |green.v|

..those five values describing the full system momentum state and its components at any given instant, no matter which axes are torqued, locked or coasting. Again, the KE's based on those same components neatly sum to kinetic(), so the corresponding momenta must be equally valid..

Trying to plot a zero-sum between rot and trans momenta is misguided - the rots should be cancelled or not by counter-rots, the trans by counter-trans. If there's transfer of momentum between dimensions then net dL over a cycle will be proportionate to net dP when considering the net system including planet.

Anyone disagree with these conclusions or have any thoughts to add?

I'm gonna float the planet in the energy-gain sim, monitoring net system momentum as above, to see how counter-rot and counter-trans momenta play out over eg. a full cycle, quarter-cycle etc.
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Re: MTM5

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Sounds good ..
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Re: MTM5

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OK so now the complete closed system is comprised of four discs, the base one being a free-floating 1e13 kg 'planet', also having an MoI of 1e13.

The momentum meter sums all four rotationals on one plot and all four translationals on another - the 'net' is obviously mixing units and dimensions, but then so is the interaction; any of the components can always be plotted out individually as required, for now though it's covering the net closed-system momentum..

It is increasing over successive cycles however, so a brake test will be in order to see if that amounts to a real change in resting momentum state or position..

For a quick preview here's the first quarter-turn:

Image

..you can already see there that rot and trans momenta are directly inversely related - note that the sign of the rotational momentum's negative, even though all the rotations we can actually discern are in the positive direction; which can only mean that the counter-angular momentum being induced to the planet, slight as its velocity component may be, is still greater than the positive angular momentum developing on the wheel..

The next key objective however has to be the search for any force or torque asymmetries between input and output workloads that would reveal the presence of an effective N3 break, which remains the working hypothesis on the source of the energy disunity..

For that, we can compare the forces calculated by the sim to those derived from the discs' inertias and accelerations per N2. May be busy this weekend but will start working in this direction ASAP..
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Re: MTM5

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Coming back to measuring the disunity, it becomes apparent that simply breaking down the net KE into its component KE's reveals further context in which to interpret the momentum deltas:

Image

For instance note that the wheel's KE isn't increasing even though the radius of the blue disc is increasing; implying that the wheel motor sees a constant MoI, as if the blue disc were remaining at fixed radius, so long as the green one it's mounted to is rotating.

This is thus one area where my momentum calcs were derailing, since i was including the blue disc's point-mass mr² in the grey wheel's net MoI calc. I was also however including its mr² as relative to and part of the green disc's MoI..

..i could see however that the net KE components only count the blue trans KE once of course, hence trying to calculate two different dL's for it was obviously a mis-step.

But besides clearing that up, it's also revealing to see the blue and green dKE's separately, since blue rotational is a load on the k-motor, whereas its translational dKE is being powered by CF-PE as it's pulled outwards, hence a load on the w-motor.

Hence this sharpens up our view of the gain mechanism - of what the exploit is, and where it's arising.

Here's the max-freq results from the above Q1:

kTa = 12.8330934
wTa = 4.704379623
CFPE = -0.999937481
net in = 18.537410504

weight dKE = 23.50718262
rotor dKE = 1.79549284
net dKE = 25.30267546
diff = +6.765264956
CoP = 1.36

For context, here's the contents of the KE meter:

Weight = 0.5*body[21].moment*sqr(body[21].v.r)
Rotor = (0.5*body[5].moment*sqr(body[5].v.r))+(0.5*body[21].mass*sqr(body[21].v))
Wheel = (0.5*body[1].moment*sqr(body[1].v.r))+(0.5*body[5].mass*sqr(body[5].v))

..these summing to meet kinetic().

So the most-striking detail we notice is that the blue rotational's 183% of the work done by the k-motor - the spin-up evidently having been subsidised by something.

The rotor dKE is obviously the blue translational KE as it's reluctantly pulled outwards by CF force, against its efforts counter-torquing back in the opposite direction.

The output CF-PE is 1 J, the rotor dKE is 1.79 J, and the total load on the w-motor is 4.7 J. Subtracting the former from latter leaves 2.9 J of work done by the w-motor that must be contributing to the blue rotational dKE.

Again subtracting the former from latter leaves 20.59 J of dKE attributable only to the kTa of 12.83 J, so an excess of +7.76 J and a k-motor CoP of 160%.

Looking at the velocities, the rotor actually decelerated slightly over the drop, from its initial 2 rad/s at TDC, down to 1.85 rad/s at 9 o' clock, hence the KE gain isn't attributable to a reactionless acceleration of the green rot / blue trans component.

That only leaves the inter-reaction - or lack thereof - between the two motors; the system's OU because the k-motor isn't applying a reciprocal compliment of counter-torque upon the w-motor when spinning up the weight.

The wheel motor's basically somewhat agnostic about the rotational acceleration of the weight..

..seemingly consistent with the hypothesis that counter-torque from the spin-up is effectively being sunk to CF force and time..
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Re: MTM5

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Finally moved on to Q2.

It's a bit of a data-dump, but the results are interesting:

Q2 - from 90° to 180° BDC:

Weight dKE = -20.23901484
Rotor dKE = +15.4965887
net dKE = -4.74242614

kTa = -5.253228188
wTa = 5.965301163

net in = 10.707727303
net out = 5.253228188
diff = -5.454499115
CoP = 0.509

So after netting a CoP of 1.36 from Q1, we've just sacrificed almost half our system energy in Q2, with a big non-dissipative loss.

To put that in perspective, here's the results for the full 180°:

Q1 + Q2 - from 0° TDC to 180° BDC:

dKE = 20.56085981
kTa = 7.580634691
wTa = 10.67061019

net in = 18.251244881
net out = 20.56085981
diff = +2.309614929
CoP = 1.126

..so we still come away with some gain, but clearly the only real benefit of performing Q2 was any practical expediency it affords us in setting us up for the larger gain arising in Q3, and in cycling the interaction. Q4, presumably, will also be another loss phase.

Tentatively, this may suggest that there's a more-optimal cycle possible.. for instance, what if we only spin up for the entire drop, then only spinning down during the lift? Will have to try this at some point..

Of course, there's no shortage of options for de-spinning without incurring a whacking-great non-dissipative loss - it'd be daft to design-in any unnecessary losses - so continuing with the current regime without exploring alternatives would be cargo-cult engineering. More to the point, a better, more-optimal application of the effect is evidently possible, once again reminding us that a practical embodiment is unlikely to closely resemble the ones we're currently looking at..

Will break out Q's 3 & 4 next.. feat. that 'big dipper' in the wheel motor plots..
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Re: MTM5

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Q3 - from 180° to 270°:

Weight iKE = 3.76869954
Weight fKE = 1.67375899
Weight dKE = -2.09494055

Rotor iKE = 18.79232299
Rotor fKE = 28.07744186
Rotor dKE = +9.28511887

net dKE = +7.19017832

kTa = 7.231189465
wTa = -7.78963977

net in = 7.231189465
net out = 14.97981809
diff = +7.748628625
CoP = 2.07


Q1 -> Q3 - from 0° TDC to 270°:

dKE = 27.75103813
kTa = 14.81182142
wTa = 2.882274697

net in = 17.694096117
net out = 27.75103813
diff = +10.056942013
CoP = 1.56


Discussion:

As anticipated, Q3 has the highest single-stroke yield. The full 360° cycle is rolling over gains and losses, three steps forwards, two steps back.. but none of the four strokes actually require any of the preceding ones, and will achieve the same CoP's no matter how they arrived in the strokes' starting configurations..

Regarding the 'spin-up' of the weight; this in reality amounts to a spin-down in Q3, the weight spin decelerating and the motor's acceleration instead being applied to the rotor at close to unity efficiency (~7 J of work by the k-motor causing ~7 J KE rise). The gain is thus almost entirely PE on the w-motor, the k-motor having performed an extra 7 J of work against it.

This apparent mirroring of values makes Q3 particularly interesting, suggesting it's somehow straddling some kind of axis of symmetry in the gain conditions. I'll presumably be looking at this quadrant in much more detail as it's the best candidate for singling out thus far.. 200% in one fell swoop, and we're still only playing with arbitrarily-chosen weights and MoI's etc..

Q4 is presumably only going to further dilute the current Q1-3 CoP of 1.56, but i'll plot that out too tomorrow, and likewise for the second 180° from Q3-4.

Then i'll look at halving the stroke count, see what happens when we spin up for the entire drop, and de-spin for the entire lift..
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