Into the Vanishing Point..

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Gregory
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re: Into the Vanishing Point..

Post by Gregory »

Here is a sim attached.
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Post by MrVibrating »

Cheers guys, sussed it now, Greg's nailed it... i keep meaning to find out if it's possible to make a net momentum meter in WM by summing inputs from other meters, i expect it should be... but in the meantime i've been summing manually, such as to find the peak momentum point last night - must've done around 20 sums before converging on ~2.3 meters radius... a 'net momentum' meter would've shown this at a glance..

But yep, i was measuring their momenta relative to the static frame without noting the signs. Duh. Lack of sleep, that's my excuse and i'm sticking to it. Got in from work last night, and, after thinking about it all day, just interrupted the sim at 2.3 meters and sat a 1 kg 'target' mass in the path of the rotor at peak momentum, got pretty much the energy result i'd calculated, minus losses which i'd neglected, but then got stumped trying to square the net momentum.. so went and did some linear tests and suddenly it seemed like creating momentum was trivial, and i even ended up Googling CoM to re-check what i thought i knew!

I spent a few weeks trying to create momentum from relative motion (my 'discs' experiments last year), so it's not as if this is all new to me..

Yet i was staring at this and scratching my head:

Image

Dum, dumdum dumB... LOL it would've clicked eventually. :|

And thanks again for the handy reference sheet Fletch - i think i already had it, but damn i shouldn't need it - i knew i was obviously doing something wrong, but thought it was on the sim-side..
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Re: re: Into the Vanishing Point..

Post by MrVibrating »

Gregory wrote:Here is a sim attached.
Ahh excellent, cheers mate - so how did you pull up those meters? I can edit their equations to use my own masses but if there's an easier way i could sure use it..

For instance net X + Y would be useful...
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Post by MrVibrating »

Fortunately all the angular momentum so far is of the same sign... Ahem.

(probably why i'd got used to neglecting it)
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re: Into the Vanishing Point..

Post by Gregory »

MrVibrating wrote:
Gregory wrote:Here is a sim attached.
Ahh excellent, cheers mate - so how did you pull up those meters? I can edit their equations to use my own masses but if there's an easier way i could sure use it..

For instance net X + Y would be useful...
Simple. Just start with a meter which is the closest to what you want to measure. In your case you can start with a angular momentum graph.

Double click on the meter or Ctrl+I. Inside the meter's properties you can manually edit/rewrite everything you want. For example you can name the X,Y axes, you can add as many Y graps as you want, and most importantly you can edit/rewrite the formula/equation for all Ys.

For angular momentum of a body you will get something like this:
9.54929659e+004*(Body[2].moment *1.00000000e-004* Body[2].v.r*1.04719755e-001)

The multiplying parameters are some kind of conversion values I believe, depends on what SI/Units settings u use, etc.
But basically it is: Body[2].moment*Body[2].v.r

.v.r stands for velocity.rotational
.v.x is velocity.xdirection
.v.y... you know...

Also:
.a.r acceleration.rotational
.a.y acceleration.ydirection
.p.x position.xdirection
.p.r position.rotational

So, you can study the equation for all of your graphs, you can combine them, you can use the constant "pi" and math functions like abs, and sin/cos. All in all you can create a new meter and totally rewrite it as you want. Check what I did with the two momentum meter...

Attached the formula language reference guide.

Cheers!
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Post by MrVibrating »

Amazing, cheers for typing all that out! I've been working my way into using formulas but this really helps clarify what i need to know, will be making good use of it so thanks again..!
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Post by MrVibrating »

There's a potential exploit i mentioned a while back, that i can't seem to refute.. might be worth further investigation.

As usual, it's quite unintuitive - basically, generating momentum, by paying to destroy counter momentum.

So suppose you have to equal angular inertias, both stationary. A momentary torque impulse is applied between them, setting them in opposite motion, CW & CCW, and having equal opposite momentum.

The net momentum at this stage is obviously zero. Nonetheless. the two opposing momenta we have are real enough - rotation is intrinsic, absolute motion, not just relative - a system is rotating or not regardless of whether it's moving relative to something else or not. The inertial effects aboard each rotor are real..

..And so what if, on one of these rotors, we then pull half of its orbiting mass into the center, paying energy to perform work against CF, but which ultimately destroys angular momentum (by pulling into the exact center where orbital motion goes to zero)...?

So suppose we initially had 10 kg-m/s clockwise, and the same -10 kg-m/s going anticlockwise. And then on one or other rotor, we destroy half of its angular momentum, by de-orbiting it, while also doubling its RKE.

So now we've paid some energy to destroy some momentum, and one rotor now only has 5 kg-m/s, while the other, untouched one still has 10 kg-m/s.

Then we simply collide them again.

Doesn't this leave a net unbalanced momentum? In other words we've generated a non-zero net system momentum, by the act of destroying some counter-momentum...

We've paid energy for it, yet gained momentum in an otherwise closed system, and it also seems that this residual momentum is, by definition, reactionless in nature..

I'd initially had doubts about whether it can be cycled, however bear with me slightly further:

- when starting with both rotors stationary, as above, the initial impulse brings both into real motion, however, suppose instead that we gained some net momentum successfully as predicted, but now when we come to attempt a second cycle, both masses are already in uniform motion...

- suppose both are at 1 RPM in the same direction, and we apply a 1 RPM acceleration:

- this leaves one rotor with 2 RPM, while the other is braked to a standstill

- if we destroy some angular momentum on the moving rotor, say if it had 10 kg-m/s but we reduce that down to 5, while again doubling its RKE, since the other rotor is stationary, the net system momentum is down to 5 kg-m/s.

- if however we pull the mass inwards on the stationary rotor instead, leaving the moving one with its 10 kg-m/s or whatever, then we pay nothing to make that translation (cos there's no CF), while also halving its MoI. When we subsequently collide them again, this re-equalises the momentum distribution, so we'll have 5 kg-m/s on each rotor, but the halved-MoI rotor will need to spin twice as fast in order to embody that momentum at its reduced mass radius... which seems to suggest a conflict between CoM and CoE - if CoM dominates then everything equalises, but if CoE takes precedence then there can't be an even distribution of that momentum, and again a net momentum would result...


So the long and short of the proposition is that for a single cycle at least, we can effectively buy momentum from within a closed system, by paying to dispose of counter-momentum, leaving an unbalanced remainder of either CW or CCW sign.

That in itself seems like it could be an eminently useful trick... if only we can work out how to apply it..

The manner in which free energy from an N3 break is usually encountered is when considering why a reactionless thrust system is overunity - and it's ultimately because it pays a fixed price for momentum, not scaling via the half-square of rising velocity. So the energy benefits are something emergent - buy enough momentum and eventually you exceed breakeven, in terms of the on-board energy expenditure compared to your KE as seen from the static frame.

Yet Bessler's wheels seem to gain energy multiple times per cycle - there's good reason to suppose from Bessler's own comments and clues that one mechanism causes two power pulses - so two complete OU energy interactions - per 360° cycle.

The runaway, unbalanced momentum may simply be that of the main wheel body and axle (the external 'peritrochium' structure), and the KE gains those of the weight interactions benefiting from the free velocity increase, thus boosting the value of accelerations relative to gravity's static reference frame..

However the other possibility is the the excess momenta are embodied in the first instance not induced directly within the wheel body, but only within the flight of masses rotating internally relative to the wheel body, and then transferring their momentum gains upon colliding with it - more consistently with Wolff's impressions.

So we have this potential dichotomy - is the excess momentum induced in the first instance directly within the wheel body, or rather within the flight of other internal parts, and then transferred from them to the wheel body via a collision...?

And so the answers to this question come to bear on the potential utility of the aforementioned 'buy momentum by paying to get rid of counter-momentum' lark.. - which, if it turns out to depend upon both rotors possessing absolute motion, probably leans more towards the latter possibility. Whereas, if we still generate a momentum asymmetry even when the initial impulse is decelerating the reaction mass, rather than accelerating it, then we can proceed accumulating momentum gains over successive cycles by inducing it directly within the mass of the wheel housing and axle.

So, it looks like we may be able to apply PE to destroy counter-momentum while converting to KE - as a random example, start with 10 kg-m/s and 10 Joules, add another 10 J for a result of 20 J KE and 5 kg/m/s... then re-collide elastically with the other, 'primary' momentum, still at 10 kg-m/s, leaving a 5 kg-m/s net system momentum - a full-cycle momentum gain, at the very least.

And so even if we can't accumulate the same benefit over successive cycles with rising velocity (and i'm doubtful, but let's not dismiss the possibility entirely just yet), perhaps a limited single cycle closed-loop momentum gain is still a sufficient advantage to leverage an exploit..

What keeps coming back to me is the purpose of a GPE interaction in all this? The thing is, if you already had an angular momentum asymmetry (a lá N3-dodging), then surely you could apply that directly to the rotating masses, with no need for GPE interactions...

Yet Bessler assures us that no aspect of his designs are 'pro forma', or for appearances' sake - every detail is a functional, necessary part of the mechanism..

And suppose instead of a momentum asymmetry, what you had in the first instance was momentum unity (consistent with N3), but a directional KE asymmetry - so equal positive and negative angular momentum, but with a free change in MoI (ie. radially translating while not rotating), and hence unequal CW vs CCW RKE? If you simply collide the two rotating masses, the KE difference is presumably squandered, and no net momentum remains...

But suppose instead of simply colliding the masses, we rather use the greater KE to raise a GPE, while dropping and thus accelerating a GPE against the lower-KE rotor - so one rotor gains angular momentum, while another loses it...!

And so perhaps the ultimate purpose of the GPE interaction is to convert a non-N3-bothering RKE sign asymmetry into an angular momentum sign asymmetry, which does challenge CoM, and thus by extension CoE?

Sorry, i'm rambling i know... it's cos i have what looks like a possible niche, the merest toehold on a potential leg-up.. but it really all depends upon the route ahead.. we can see a summit, but whether it's the summit, or just another dead-end, is the issue i'm wrestling with. Is this anything of value, and if so how to proceed with it? Paying to destroy counter-momentum, that is..?
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Post by MrVibrating »

...it's counter-intuitive cos really, it's spending energy to buy reactionless momentum..

But what we actually spend on is sinking counter-momentum; getting rid of it, so with regards to the actual form of input work, we're not directly spending energy to raise more momentum.. we're not accelerating something, but decelerating it.

But because it's counter-momentum, this raises the net remainder...

So we'd ultimately be 'purchasing momentum', by not purchasing momentum..

Or to put it another way, buying negative counter-momentum.

"Cheap, at half the price"..
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Post by MrVibrating »

..struggling to get the formula right for this net momentum meter..

All i need is to sum the X + Y momenta for two bodies orbiting a common axis - however i want to keep using these linear dimensions, rather than switching to angular dimensions. Summing two angular momenta i think i can manage, but as linear X + Y components i can't find the right expressions..

Been trying stuff like this:

(Body[5].mass * Body[5].v.x) + (Body[5].mass * Body[5].v.y)*(Body[31].mass * Body[31].v.x) + (Body[31].mass * Body[31].v.y)

..or...

Abs(Body[5].mass * (Body[5].v.x + Body[5].v.y)) + Abs(Body[31].mass * (Body[31].v.x + Body[31].v.y))

..but i'm still seeing sinusoids (ie. X-axis only) instead of neatly summed integrals..
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Post by Gregory »

Hi MrV,

Try this, summing the x and y components for both body separately, then adding those together:
(Body[5].mass * Body[5].v.x + Body[5].mass * Body[5].v.y) + (Body[31].mass * Body[31].v.x + Body[31].mass * Body[31].v.y)
Shorter version:
Body[5].mass * (Body[5].v.x + Body[5].v.y) + Body[31].mass * (Body[31].v.x + Body[31].v.y)

Or this one for the Abs() version:
(Abs(Body[5].mass * Body[5].v.x) + Abs(Body[5].mass * Body[5].v.y)) + (Abs(Body[31].mass * Body[31].v.x) + Abs(Body[31].mass * Body[31].v.y))
Shorter version:
Body[5].mass * (Abs(Body[5].v.x) + Abs(Body[5].v.y)) + Body[31].mass * (Abs(Body[31].v.x) + Abs(Body[31].v.y))

Summing angular momentum will work similar, like:
Body[5].moment * Body[5].v.r + Body[31].moment * Body[31].v.r
Abs(Body[5].moment * Body[5].v.r) + Abs(Body[31].moment * Body[31].v.r)
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re: Into the Vanishing Point..

Post by Fletcher »

Yeah .. create individual OUTPUTS for each mass (linear or rotational) from Measure menu.

Then rename One OUTPUT as the Master OUTPUT (use Window > Appearance to overwrite the description).

Then in the Master enter stuff into y2, y3 field etc (they stack) - copy the other OUTPUT fields into the y2, y3 fields for Equation so they stack in one Master Output.

Now you need to sum them - to do that look at the Master OUTPUT reference in Properties (it might be say *Output[2] - Sum of MV, for example).

In the y3 or y4 field you need to create a summation Label like SUM and Equation in the Equation Filed like Output[2].y1 + Output[2].y2 + Output[2].y3

Now you have all stuff handily together doing what you want it to do in terms of display etc.
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Post by Gregory »

I just realized that if you want to measure momenta of weights rotating with a wheel moving in a circle, then you will always get some kind of sinusoid result, because the y and x values are changing according to sin/cos function and they will not cancel each other perfectly as they are not changing the same way with respect to position around the circle. Even with the absolute formulas you will find an annnoying sinusoid vibration at the top and bottom lines of the graph. (I tried this myself with a morning coffee, haha :)

But there is a simple workaround. Try this instead, it should work fine:
Body[5].mass * |Body[5].v| + Body[31].mass * |Body[31].v|

|Body[5].v| gives back the speed of Body[5] without a direction component, so it is always positive/absolute.

Working with the above formula you will always measure "total quantity of motion" aka net momentum from an outside and absolute perspective without any directions involved. So this way you can forget that annoying sinusoid component... And you can add this together for as many bodies as you like.

And finally if you want to extend it with a "direction of rotation" +/- component, then this is a way to go:
IF(Body[5].v.r > 0, 1, -1) * Body[5].mass * |Body[5].v| + IF(Body[31].v.r > 0, 1, -1) * Body[31].mass * |Body[31].v|

This works fine as long as the bodies are rigidly pinned(rigid joint) to something which rotates around. Otherwise it is better to use Body["main wheel"] inside the IF function, and that still assumes that your bodies are fixed to a single point/radius on the wheel and not moving forward/backward related to the wheel.

Hope this helps. Cheers!
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Post by MrVibrating »

Brill, thanks so much for this, will get to it again this evening...


All i want to do for now is map out this interaction accurately, to clarify in precise figures what the conditions are for non-constant momentum.

We can already see it's sharply non-linear, with obvious boundary conditions:

- if the fixed-radius mass is zero, then the sliding mass represents all of the system mass, and net momentum remains constant all the way in and out, with a high variation in forces and KE.

To put a finer point on it, the interaction is time-invariant under these conditions - there's no 'inertial drag'.

- if, OTOH, the fixed mass is infinite (or just much higher than the sliding mass), then this inertia limits the angular acceleration and thus forces and KE that the inbound mass can attain - and due to its narrowing radius, the amount of space it's accelerating through is dropping at constant rate all the way in, even though RPMs are climbing all the way in. Under this condition, all of the momentum being shed by the inbound sliding mass is being neatly transferred over to the fixed-radius mass, in real-time, without delay.

For instance, at an initial speed of 1 RPM, pulling a 1 kg mass inward over 10 meter radius at 1 mm/s while a 100 kg mass remains at 10 M radius, the sliding 1 kg mass begins with around 1.047 kg/m-s of momentum, which then decreases linearly to down to zero at the center, and during this transition the momentum of the fixed-radius 100 kg mass is rising, in step, by the exact amount, milligram per millimeter per millisecond. IOW the momentum transfer is instantaneous, and so net momentum again remains constant, because the 'inertial drag' is maximal.

Yet between these extremes lies this open range in which inertial drag is partial, limiting angular acceleration without precluding it - the fixed-radius mass can limit the accelerations and thus forces and KE's attained by the inbound mass, while at the same time the inbound mass can significantly accelerate the fixed-radius mass, and so raising its momentum.

And so this non-linear curve emerges - typically the inbound mass will still be able to undergo angular acceleration initially, also accelerating the fixed mass, too... but then invariably, as its radius further declines, so does its intrinsic velocity, and thus forces and momentum, as the fixed mass increasingly represents all of the moving system mass.

So if you're able to swing hefty-enough masses around at sufficient radius, and have enough power, you can cause the system momentum to vary by hundreds of kg-m/s, in a statorless system without incurring any immediate counter-momentum.

Incidentally, the system also underlines a point i've laboured previously, which is that there's actually no such thing as "momentum transfer", as we normally consider it. It's a false premise, albeit one that is usually consistent enough to suffice. For example, as explained above, if the fixed mass is much greater than the sliding mass, we see nothing to challenge our notions of momentum transfer - we see an instantaneous 1:1 exchange - to all intents and purposes, the momentum gained by the fixed mass is the same unitary quantity of momentum being lost by the sliding mass; it thus fulfills the appearance of a transfer of the conserved quantity from one body to another. To argue otherwise would seem like meaningless semantics.

Yet consider what happens in the reverse scenario, where the inbound sliding mass is much greater than the fixed-radius mass: if we pull in 10 kg at 1 mm/s over 10 m radius from 1 RPM, while 1 kg remains at 10 m, then the sliding 10 kg mass begins with around 10.47 kg-m/s, to the fixed mass's 1.047 kg-m/s, then rising to a peak ratio of around 17.5 : 7.5, so net system momentum has risen from 11 kg-m/s up to around 25 kg-m/s - a 2.27-fold increase... before plummeting back down to 11 kg-m/s as the sliding mass enters the center, motionless, with all of that system momentum embodied on the 1 kg mass at 10 m fixed radius.

So think about the implications of that... the only remaining mass in motion in the system has 11 kg-m/s, yet the system momentum was previously more than twice that, and 17.5 kg-m/s of it was manifest on the inbound sliding mass...

So this result unequivocally crushes any misconceptions about "momentum transfer".... there is no such thing! Momentum is never "transferred" between interacting bodies. It's not just impossible, but oxymoronic - momentum is an instantaneously unique property of the particular mass in motion that it is constituted from. "Transfer" would be an implicit de-facto result, and thus a valid concept, only in the context of an assumed constant system momentum - ie. it's founded in a blanket assumption and robotic application of conservation of momentum always holding, in a temporally-invariant manner.

Yet our simple system here can perfectly confirm that illusion, or just as easily leave it in tatters. If momentum cannot be meaningfully "transferred" in the latter interaction, then it was never being "transferred" in the former..

Rather, momentum is induced...

And the conditions mediating that induction, are most usually encountered under equitable exchanges... esp. in linear interactions. But in angular interactions we can control these conditions, and so modify the relative induced quantities, in magnitudes and comparative rates of change. We can break that momentum symmetry by which momentum is presumed to be "transferred", and show it to be false. Here, the momentum is clearly not conserved, constant or thus "transferred". Ergo, it never was in the first place. Rather, the conditions determining instantaneous and equitable momentum exchange have finite limits that we can overcome - simply and easily - to induce more or less momentum than either mass or the net system began with.

So this very basic interaction, of elementary simplicity - one mass changes radius while another doesn't - seems to delve effortlessly and inexorably into an issue of profound significance, with implications from quantum to cosmic.. from 'ambient' momentum, to galactic..

Which of course is a little beyond our current remit... but all i wanna do for now is map out the contours of this asymmetry. What is the optimal mass ratio for maximising the net momentum delta? For instance a 1:1 ratio of translating-to-fixed mass seems to cause a net momentum delta of around 50%, while at 10:1 it's 227%, but it would seem worthwhile plotting momentum deltas over a range of mass ratios for a given initial RPM and radius. There is no essential limit, besides practical considerations, but is there a constant multiplier or is it a non-linear function? Obviously, "optimum" for our purposes is also a function of the KE over momentum deltas - the cost / benefit ratio and thus trading rates for buying and selling momentum.

Still, in the meantime, i think we've busted the myth of "momentum transfer" fairly conclusively here.. in other words, in a straightforward linear elastic collision between two masses, no momentum is transferred directly 'between' them - rather, there is a menage-a-trois between the masses and the vacuum activity endowing them with that mass and inertia; some momentum gets transferred from 'mass A' to the vacuum, and in turn some momentum is transferred from the vacuum back to the 'interacting' masses 'A' & 'B'.. proportionately to their unique spatiotemporal components of motion.. which are the usually-symmetrical parameters, but which evidently, do not strictly need to be.

The net system momentum we gain, on the way in or on the way out, has not come from some other mechanical momentum. It can only have come from the vacuum - ie. the same place all momentum comes from and goes to.

Ergo, all momentum interactions are inherently open thermodynamic systems. The division of momentum between interacting masses is nonetheless usually equitable and constant, and thus outwardly indistinguishable from a closed system, yet it clearly doesn't have to be, and we can source and sink momentum directly to and from the vacuum, in flagrant defiance of Newton's 3rd law, using only internal PE.

With any luck we're on the home straight, and all that remains is to tease out an energy asymmetry between that internal PE vs external GPE values. For now though, it seems kinda cool that such a trivial interaction can have such challenging implications - the "ice-skater effect" is usually given as a pithy demonstration of CoM in action - acceleration and deceleration as a function of varying MoI... but look a little closer and you realise that much of her mass isn't changing radius - only three limbs are - and thus the rest of her body is accelerating and decelerating at fixed radius, and thus may or may not be conserving net system momentum, depending on her BMI index! Giver her a dumbbell in each hand and she'd undoubtedly start channeling raw ambient momentum to and from the vacuum, totally laying waste to the attempted de-mystification she's supposed to illustrate.. instead proving a transient violation of N3!

This simple interaction proves that creation and destruction of momentum is actually the rule, rather than the exception! Usually the amounts created and destroyed are equal, but they don't need to be - at least, not with regards to angular interactions. Of course, it's not really being "created" so much as induced, nor "destroyed" so much as dissipated, albeit not in the regular sense of thermodynamic entropy.. And when i say "usually" - in truth, we must be surrounded by every-day motions that include radial displacements under angular translation, and so may or may not be holding constant system momenta..

That an interaction so simple and ordinary can have such seemingly extraordinary implications is pause alone for thought..
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re: Into the Vanishing Point..

Post by Fletcher »

hmm .. so correct me if I'm off the reservation here.

The system total momentum (thru your very specific range of translation conditions) forms a quadratic parabola shape loosely like an n, waxing and then waning to a datum.

But the currency for doing Work is Energy and not momentum (as currently defined in physics) - so does the energy curve show the same peculiarities albeit a taller thinner curve ?

If there is a specific rise in system energy by specific translation conditions then would not the cost of obtaining that translation against cf (the reset) be the expense side of the ledger ? And if the system energy spike is greater than the energy expenditure would not this be the gross revenue side of the P & L ?

If there is a Net Surplus of energy after costs then (in theory) energy symmetry from momentum induction has been theoretically broken ?

All that would remain would be to engineer efficient mechanics to aid the translation conditions that give a theoretical surplus of energy and momentum ?

Am I visualizing this correctly ? Thanks.
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Post by MrVibrating »

If we're to take as read Bessler's assertion that his wheels obtained further advantage from applied loads, then this clue deserves some consideration with regards to the current findings:

- an additional load applied at the axle adds its inertia to that of the wheel's own non-radially translating mass, increasing its overall ratio to the radially-translating mass. We see in the above discussed trends that raising the amount of fixed inertia relative to the variable inertia, up to a certain point, increases the amount of additional momentum induced, but beyond that point, decreases the amount of extra momentum induced, therefore if the advantage to which Bessler refers is the quantity of momentum generated per cycle, then we can surmise that the applied loads needed to be geared to tune the total non-translating inertia to translating inertia ratio so that the momentum profile was poised towards the bottom of the main acceleration curve, such that further increases in inertia (as from 'sticky spots' innate to the applied loads) cause an increase in induced momentum, rather than the decrease that would otherwise occur if the total non-translating inertia was already at the upper limits of the ratio conducive to further momentum increase.

For example, consider the square sprocket around the Archimedes screw in the Kassel engraving - as the square rotates around to feed the rope from its corners and then in turn from its sides, this cyclic acceleration modulates the amount of water being pumped per angle of rotation, and hence a variation in the ratio of total non-translating inertia to internally translating inertia. If however that ratio was already at its upper limit when the rope was feeding on and off the sides of the square, then the acceleration associated with its riding around the corners would momentarily raise the amount of non-radially translating inertia too far, braking the angular velocity and thus momentum of the internally translating mass, and restricting or even reducing the momentum boost.

However this interpretation is predicated on the assumption that these transient momentum gains are the key - and as discussed previously, it might alternatively be the losses associated with translating all the way into the center that are the key to success - paying to dispose of reaction momentum or some other permutation...

Either way though, this clue might be worth bearing in mind here..
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