Decoupling Per-Cycle Momemtum Yields From RPM

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Georg Künstler
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Georg Künstler »

MrVibrating wrote:
it's about having two independent frames of reference
I know, but how do you couple it ?

There are many ways, the two independant frames must be able to interact.
From physic point the two frames are to close together, one is arranging the other.

A wheel in wheel construction will not work, even if you have two independent frames of reference.
So the two frames must allow also an sidewards movement.
Best regards

Georg
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Post by MrVibrating »

Here's a pair being turned at a constant 2 rad/s by a motor:

Image


So now i wanna add a pair of hefty vMoI masses, vertically to these levers, and start playing around with the timing..

But just consider for a moment what could happen here - as the wheel's MoI drops it'll undergo a reactionless acceleration that will further load the springs due to the inertia of the bobs, that will slow down their flight in the direction of rotation, increasing their 'soak time' under gravity and so gaining more momentum... that's the idea anyway.


Might also increase the 'taper' on the springs, they seem a tad wobbly there..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Georg Künstler wrote:MrVibrating wrote:
it's about having two independent frames of reference
I know, but how do you couple it ?

There are many ways, the two independant frames must be able to interact.
From physic point the two frames are to close together, one is arranging the other.

A wheel in wheel construction will not work, even if you have two independent frames of reference.
So the two frames must allow also an sidewards movement.
What defines the reference frames is that motion - and work done - within them is measured in relation to them, as if they were stationary.

FoR's need to be masses. More to the point, inertias.

As said, the output FoR is the KE FoR is the ground FoR. In any symmetrical (conservative) interaction, it would also be the FoR of the input energy / work / PE... but we don't want a symmetrical interaction..

So we need the FoR of our input energy to somehow be accelerating around with the system, such that the input workload always begins each cycle in relative stasis, and thus starting out from the bottom of the w² multiplier in the ½Iw² KE equation. This is why, in any true PMM, everything must, of necessity, go around together; there can be nothing involved in it that remains stationary upon the axle. Emgat.

That Bessler was able to accurately friggin' generalise the principles of mechanical OU decades before any of his contemporaries fully grasped the relationship between momentum and KE, CoM and CoE, gives the Lord's seal to his claims, and we should pay attention..

Statorless operation was neither a prima facie nor pro forma detail; it's the key to mechanical OU, because your input PE to the rotor is no longer bound to the ever-rising relative velocity of a stator, hence no longer necessarily squares with rotor velocity 'w'.


Imagine you're riding a skateboard or roller-skates; whilst rolling forwards, you throw a heavy mass forwards - the counter-momentum decelerates you, and momentum and energy are conserved (PE=KE).

Now do that again, but without the recoil; now PE<KE. A superhuman throw!

Try another example - suppose you have a bat and ball, connected by string; if you could somehow hit the ball without incurring recoil, when the string goes taut the ball's momentum will be shared back with you and you'll start gaining forwards momentum, by the same increments each cycle, for doing the same amount of work each cycle - striking the same ball the same way with the same bat.. so your input work / energy is scaling linearly; simply summing. However the KE value of each of these little net accelerations of the system is not accruing linearly; rather, it's squaring with velocity... so even though you may only be gaining 1 mm/s in speed each cycle, the KE value of those 1 mm/s accelerations keeps climbing, for ever, even though all you're spending is the constant per-cycle input energy times the number of elapsed cycles.


In a true PMM, the FoR of the input energy workload is accelerating with the system.

As such, it might be regarded as a 'divergent' inertial frame, since it's undergoing anomalous (unilateral) acceleration; any autonomously-accelerating FoR breaks energy symmetry with all other FoR's in the universe, ie. internally, if you've spent 10 J then you've achieved 10 J of work, but to any external observer you may have actually done much more or less (not 'appear to have done', since it's not an illusion).

What maintains CoE consistency between all 'normal' FoR's is CoM, courtesy of N3. N3 can't be beaten, 'because mass constancy'. But momentum gained from gravity and time circumvents N3 - we can gain angular momentum without torquing against a stator.

So the implicit objective - and the only kind of mech OU that can exist - is to fix the unit energy cost of momentum, so that its cost simply sums over successive cycles, while the system KE naturally squares up..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

MrVibrating,
I think i understand everythng you are saying.
There are two points where i see a slight difference. Are they in the way we are looking at what we think should be happening or are they in the way we express the same thing we are thinking?
So we need the FoR of our input energy to somehow be accelerating around with the system, such that the input workload always begins each cycle in relative stasis, and thus starting out from the bottom of the w² multiplier in the ½Iw² KE equation. This is why, in any true PMM, everything must, of necessity, go around together; there can be nothing involved in it that remains stationary upon the axle. Emgat.
I see it more like, nothing involved in it that remains stationary with "it's reference to" the axle.
In a true PMM, the FoR of the input energy workload is accelerating with the system.
And this, In a true PMM, the FoR of the input energy workload is advancing faster than the system.
RH46
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Georg Künstler »

MrVibrating wrote:
In a true PMM, the FoR of the input energy workload is accelerating with the system.
agreed with your findings.
It is exactly what I am doing with my developments,
you explain it with other words.
I am shifting the FoR with the help of gravity so that we get torque.
Best regards

Georg
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Post by MrVibrating »

Consider the most rudimentary 'vMoI' rig:

• a plain disc for a rotor, balanced with an opposing pair of radially-sliding masses


I want to draw your attention to the way in which angular momentum is conserved as the MoI changes while rotating:

• only some of the rotating mass is radially mobile - the MoI of the wheels' disc is constant

• as the masses are drawn inwards, they're actually losing angular momentum, to the disc, which is gaining it; the point being that net momentum is only conserved on the condition that the momentum of one MoI is increasing in direct and instant proportion to the decrease of the other..


If we instead made the disc from aerogel (ie. with vanishingly-low MoI), it'll exert little resistance (MoI) to the inertial torque caused by the inbound masses, and RPM's will go ballistic as the masses approach the center of rotation..

..and at the other extreme, if the wheel disc was much heavier than the MoI of the radially-sliding masses, the net acceleration will be that much lower.

In either case however the net momentum is a constant sum of the two respective MoI * RPM products.

And that's the basics of CoAM.


But now consider the following fly in that ointment:

• what if the disc's MoI were spring-buffered?


In other words, you can see that these leaf springs soak up a bit of delay in the angular accelerations applied to them and the wheel, due to the inertia of the bobs..

..ba-ding! See it yet?


Is the net system AM still constant, or else time-buffered - ie. time variant - due to the delayed acceleration caused by the loading and unloading of sprung PE?

Could this be the key to a transient CoAM exception / manipulation?

I've just firmed up the spring gradients, with a factor of four multiplier between the thick and thin ends.. all i need to do now is add a pair of vMoI masses to the wheel..



If i meter everything up - sprung PE, KE, and individual momentum distributions for the vMoI and wheel, will the net angular momentum be changing in time..?

It'll have to, surely?

Grasping such an exploit by the horns could be truly momentous, potentially turning this thing around on a sixpence..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Robinhood46 wrote:MrVibrating,
I think i understand everythng you are saying.
There are two points where i see a slight difference. Are they in the way we are looking at what we think should be happening or are they in the way we express the same thing we are thinking?
So we need the FoR of our input energy to somehow be accelerating around with the system, such that the input workload always begins each cycle in relative stasis, and thus starting out from the bottom of the w² multiplier in the ½Iw² KE equation. This is why, in any true PMM, everything must, of necessity, go around together; there can be nothing involved in it that remains stationary upon the axle. Emgat.
I see it more like, nothing involved in it that remains stationary with "it's reference to" the axle.
In a true PMM, the FoR of the input energy workload is accelerating with the system.
And this, In a true PMM, the FoR of the input energy workload is advancing faster than the system.
RH46
The FoR is the mass / inertia that is considered as 'stationary' for the purposes of measuring accelerations / states of motion of other bodies.

In any conventional motor, we find a 'rotor' and 'stator' portion; such as an engine block and crank shaft or DC motor or whatever - torque (and force generally) can only be applied between inertias (no unilateral forces), so the part that spins up does so against the relative stasis of the 'stator'; usually, ultimately, the Earth.

As speed / RPM's increase, it does so in relation to the 'stator' portion of the interaction - ie. the ground FoR.

Since we're relying on the stator to continue being able to further accelerate the rotor, the relative speed between them necessarily increases - and is, manifestly, the same speed as the 'system speed' - so if it's spinning at say 10 RPM, then that's also the relative velocity between the 'rotor' and 'stator' halves of the ongoing interaction.

As RPM's rise, the angular distance over which a given constant torque must be applied in order to continue constant acceleration is squaring with speed - so for instance if i have to apply torque over 90° of angle to achieve a 1 rad/s acceleration, a second 1 rad/s will require applying that same torque over four times the angle - so fully 360° - and since input work is torque * angle (or force * displacement more generally), doubling the RPM requires four times the input energy..

This is also the 'KE' FoR - if we want to harness our wheel's energy, then the instantaneous energy available is equal to half its MoI times its RPM squared, in relation to the ground. There's obviously nothing we can do, or should want to, about that - so any manipulations / exploits we'll want to apply upstream..

Now suppose that we could switch off - or else, just temporarily lower - the MoI of our reaction mass / stator - so we could torque against it, applying equal opposing angular momentum to it per the current MoI distributions, but then change those distributions before braking the rotor against it - thus a closed-loop 'spin and brake' cycle would gain angular momentum..

It would effectively be a system bootstrapping itself into rotation by accelerating a little at a time, and jerking its 'stator' around with it; because rotor and stator are now co-rotating, the relative speed between them never increases.. so for instance an internal 1 kg-m²-rad/s costing ½ J would always cost ½ J no matter the net system RPM relative to the ground, because the internal FoR between rotor and 'stator' always begins each cycle in relative stasis, right at the bottom rung of the 'velocity' multiplier that otherwise causes PE costs to square symmetrically with KE values.


If it helps clarify the matter further, Bessler's wheels were like these attempts you see at 'inertial motors' - usually piston / solenoid type devices that try to generate net thrust without ejecting any reaction mass (example); whereas they always turn out to be simple friction motors (at best), Bessler's gravity exploit actually worked..
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Post by MrVibrating »

Imagine you're riding a skateboard or roller-skates; whilst rolling forwards, you throw a heavy mass forwards - the counter-momentum decelerates you, and momentum and energy are conserved (PE=KE).

Now do that again, but without the recoil; now PE<KE. A superhuman throw!

Try another example - suppose you have a bat and ball, connected by string; if you could somehow hit the ball without incurring recoil, when the string goes taut the ball's momentum will be shared back with you and you'll start gaining forwards momentum, by the same increments each cycle, for doing the same amount of work each cycle - striking the same ball the same way with the same bat.. so your input work / energy is scaling linearly; simply summing. However the KE value of each of these little net accelerations of the system is not accruing linearly; rather, it's squaring with velocity... so even though you may only be gaining 1 mm/s in speed each cycle, the KE value of those 1 mm/s accelerations keeps climbing, for ever, even though all you're spending is the constant per-cycle input energy times the number of elapsed cycles.


In a true PMM, the FoR of the input energy workload is accelerating with the system.

Anyone genuinely interested in researching Bessler's wheel should consider actually doing the above exercises.

Begin with linear examples:

• momentum (p) = mass * velocity

• KE = ½ mass * velocity²

• PE or work done = force * distance, or displacement

If 'force' in the PE calc is constant then it's a simple multiplication (ie. avoid considering variable forces for now as that requires integration).

Just keep accelerating 1 kg by 1 m/s increments, ie. buying 1 kg-m/s of momentum at a time, while plotting input and output energies.

Straight away you'll start to see numbers that will become as familiar as the notes of a scale, and then, from that solid foundation, you can finally begin to peer around and actually see, visualise in an empirical way, the possibilities for mechanical magic that simply cannot be appreciated by visual imagination alone - 'mechanical intuition' is so limited and limiting that it's taken us a geological epoch just to understand the relationships between momentum and energy - ie. the three formulas above. We're ignorant idiots with blinkered and naive grasps of physics, usually barely sufficient to keep us alive..

All i can give you is this: the energy cost of momentum usually squares with velocity, matching its KE value, ultimately due to N3 and the usual necessity of a stator. But momentum sourced from a +/- G*t asymmetry doesn't rely on a stator. Bessler's exploit also depended upon being statorless. Ergo, he could only have been fixing the input energy cost of momentum-from-gravity..

Ecce homo, ergo elk.
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Georg Künstler »

note that we have 3 formulas for energy.

PE and KE as you mentioned and the third one from Einstein
E = m* c * c
when we substitute the constant for light c by v then you see that we are missing the other half.

This energy is always transferred to the earth.
Half the energy to the object, the other half to the earth.

So if we do not direct the energy directly to the earth, we have won.
This will happen with a moveable carrier.
Therefore Bessler has used the wheel, a moveable carrier.
Best regards

Georg
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Johndoe2 »

Awesome work !
You are still missing step one though. &#128540;
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Georg Künstler wrote:note that we have 3 formulas for energy.

PE and KE as you mentioned and the third one from Einstein
E = m* c * c
when we substitute the constant for light c by v then you see that we are missing the other half.

This energy is always transferred to the earth.
Half the energy to the object, the other half to the earth.

So if we do not direct the energy directly to the earth, we have won.
This will happen with a moveable carrier.
Therefore Bessler has used the wheel, a moveable carrier.
The energy distribution is a function of the mass / inertia distribution..

So for example if i expend ½ J accelerating a 1 kg rock, then i've imbued it with 1 kg-m/s of momentum, and also applied 1 kg-m/s of momentum to earth. The resulting counter-acceleration of the earth is found by dividing that metric '1' by earth's mass in kg - which'll give you something like 14 zeros before you see a number, an infinitesimal (but real and non-trivial) change in the planet's velocity. Applying the KE formula to that, would multiply half earth's mass by that 1e-14 infinitesimal, squared, which would be a KE below femtojoule range; this fleeting energy is subtracted from the KE of the rock, such that its real KE is only 0.5 J to say twelve zeros or so.. but you only perform exactly 0.5 J of work.

So no, energy distributions aren't automatically equal and opposite, precisely because momentum distributions are, and whereas momentum scales linearly with speed, KE squares.. so a non-linear relationship.
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Post by MrVibrating »

So here's a baseline example of the simple momentum transfer interaction i want to try to thwart using springs..

Image


The green MoI is a 2 kg uniform disc with a 1 m radius.

The red vMoI is 2 * 0.5 kg masses at 1 m radius, then cyclically going to zero and back.

The MoI calc for the green disc is ½mr², hence an MoI of '1'.

For the red vMoI masses it's just mr², hence also a base MoI of '1'.

The system begins coasting at 1 rad/s.

The angular momentum of both green and red inertias is varying by 1 kg-m²-rad/s, the former simply absorbing that of the latter as its MoI and thus orbital angular momentum goes to zero as the masses enter the center.


The net system momentum never so much as flutters, respecting N1.




The question that has been prompted by playing with leaf-spring weight-levers is, what happens when that momentum-transfer process is spring-buffered?

To test this we don't even need leaf springs - we could just colour the green 'base' rotor there red to match the masses, call the whole thing a vMoI, and then place another uniform disc atop it, coaxially, and connected by a rotary spring..

..so the system would again begin in uniform motion, but the inertial acceleration caused by the reducing vMoI will first have to load the spring before it can transfer any momentum across to the plain disc rotor.. seemingly implying that the net system momentum may vary in time..!?

Presumably i'm making some crass error, as usual, expecting something fantastic that just isn't going to happen, for some overlooked reason.. but this is why we test eh..

If my hunch proves right however then we'll effectively have a transient N1 violation.. yet gravity's not even active here!

Interesting proposition eh?
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Post by MrVibrating »

Well, negative result there:

Image


Little surprise tho.. once again, then, the only working N1 violation so far is the +/- G*t delta. AKA kiiking.

So if leaf springs have any use at all, it's going to be with regards to accomplishing that singular objective..

Nowt else to try now but adding a vMoI to the leaf-spring wheel.. and looking for some kind of advantage wrt isolating +/-G*t yields from RPM.
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Johndoe2 »

So everything seems to be positive. Im not sure what you are looking for.
It appears to me what you are saying is (long story short) the math supports pm. Which i completely agree with. My last comment basically was in regards to the prime mover or the initial gravitational motive force seems to be missing. So having the math and sim supportive of the conclusion ( pm is possible ) is nice but still brings us back to the starting point of a working model. If you are trying to use the sim to point you in the direction of a where to find the correct configuration i think it will be very hard although possible because it is like looking for a needle in a haystack. For my next build after my last unsuccessful build (although promising) i went back to basics ie square 1 and simplified it ad much as possible. Basically i asked myself what i criteria did i need to meet and what was the simplest way to achieve it. The window for pm is very small and requires very precise design and implementation so the simpler the better considering i do not currently have access to a complete machine shop and am working out of my garage.
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Johndoe2 »

From looking at your sims if you are adding any external power source you might be able to find areas of increased efficiency or unity but as far as gravitationally powered pm i think your shooting yourself in the foot because it opens an already large window and increases the search area making an already difficult problem only more difficult.
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