Decoupling Per-Cycle Momemtum Yields From RPM

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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by MrVibrating »

Tarsier79 wrote:And do you believe a sim program enough to rule out your theory?
Well the way i hope it'll fall into place is in spitting out this 25% accumulator / 5 cycs to OU efficiency envelope, just as an idealised mathematical model.

Then it's a case of reducing that 'spherical cow' down into a series of sims or experiments that can mechanically reproduce the key steps involved..

Specifically, decoupling the p/c momentum yield from the ever-diminishing p/c G-time.. this singular objective seeming eminently testable, yes.

If you follow the physics, Bessler succeeded before us, and this is the only way forwards, so don't look at me - i'm as disillusioned with simming as anyone.. but we've come this far..

On the one hand experiments don't lie, but they don't spit out 32,000 data points for any arbitrary action or period either, and if all we want for now is answers (and inspirations), high-fidelity data is the key.

Once we understand the energy-gain cycle, then we can design a simple engine..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

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Art wrote:Bessler Quote "And when a pound falls a quarter,
It jumps four pounds to four quarters. x"

MrVib Quote "Yet if 'quarters' aren't 'weight' but some other field property by which weight is multiplied, such as height or velocity,.."

Got me to wondering about the meaning of "quarters" in the expressions - "give no quarter" and "close quarters" .

Both of these expressions seem to have originated in 17th century about and before Besslers time .

https://www.phrases.org.uk/meanings/close-quarters.html

https://www.phrases.org.uk/meanings/giv ... quarter%27

Quarter in these two contexts seem to carry the meaning of :-

- Quarter (or quarters) = lodging , - Space , - Position.


A possible ( "far out " meaning ? ) of Bessler's poem :-

"And when a pound falls from its position ,
it jumps 4 pounds to four positions."
Good one (possibly brilliant)! Certainly hadn't occurred to me.

All i know is, the solution to the riddle has to pertain to asymmetric inertial interactions; their causative principles, and/or their results, if only by default.. nowt else delivers the goods, but besides, it's all he bangs on about, what with 'EMGAT' and 'momentum from within, not without' etc.

Obviously tho, if the solution involved moving four masses outwards radially - thus increasing MoI, perhaps a functional action in a gain sequence - this is not an elementary four-factor! It'd be arbitrary - ie. why not just move two counterposed masses outwards, and make 'em twice the mass to begin with? Same MoI change, same everything, functionally..

..whereas, the art of the riddle is a poser with many possible resolutions, but only one stand-out, check-all-boxes perfect match for its carefully-crafted sophistry.. hence i rather think we're looking for some inherently quadratic relationship or function..
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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

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Robinhood46 wrote:That gives me another idea Agor.
The one weight falling a quarter and the leepfrogging of the weights (or whatever it was leepfrogging).
John believes the number 5 to be very important too.
How about 5 weights or multiples thereof, with one weight leepfrogging the other 4 and causing them to lift on the way? Each taking it turns to do the leepfrogging.
I also thought "jumping" = "leap frogging" - perhaps hinting at 'snakes and ladders' of sorts.

I am most certain however than the significance of the number 5 is that 'five quarters' equals 125% of unity.

Three quarters, OTOH - per the 'AP wheel' - represent 75%; on the one hand, a loss, but also the efficiency of each individual 'quarter' (they're momentum yields, basically).

One such momentum yield is 25% efficient, wasting 75% of input energy.

Two in succession (again, each losing 75%) gives 50% net efficiency, dissipating the other half.

Three consecutive helpings of these individually-75% lossy momentum gains brings us to a net efficiency of 75%.. so, a nice symmetry exists here..

A fourth serving brings us to unity. That is, four successive 75%-inefficient accelerate-and-collide cycles produces mechanical unity, where system KE = input PE. This however ignores the dissipated losses, by which we're now comfortably OU..

Still, a fifth, identical, 75% inefficient asymmetric inertial interaction brings us up to 125% of mechanical unity.

So where the AP wheel represents an interesting (if perhaps slightly ominous) nexus of key, functional, loss ratios, the five cycles of asymmetric CW / CCW torques displayed on the Toys page directly mathematically accumulate to an I/O efficiency of 125%.

So these 'quarters' fall out of a mathematically-idealised system; effectively, 'elements' of a minimally-complex system: a reductive symbolism, as employed by Carl Sagan in designing the Pioneer plaque - the key issue being how to communicate in a trans-cultural, universal syntax? Bessler resigned his IP, along with any chance of eventual exoneration, truly, to posterity.. evidently having little faith in the prospects for those in his own timeline, or even their cultural descendants.. looking ahead to anyone, of any tongue, of discerning mind only.. Sagan's 'elements' were things like hydrogen's wavelength, C, and binary, from which he constructed basic 3D maps and scales showing Sol's position and human forms etc., while Bessler's were, inevitably, these fixed-price momentum yields, that effectively decoupled the PE:KE cost / benefit ratio.

Again, because a single reactionless acceleration at any ambient speed could instantly swing up a nice GPE gain, yet Bessler is instead plainly framing an interaction that is optimally and elementarily-efficient only when involving either 1:1 or 3:1 inertia ratios and inevitably, collisions between mutual accelerations (ie. 'closed-loop inertial interactions', asymmetric or otherwise), we can safely assume he wasn't able to crack asymmetric accelerations either, and instead, is painting this interaction involving discrete accelerations, collisions, and a 25% per-cycle efficiency accumulator on the KE value of the rising angular momentum relative to its cost of production, a la 'quarters'.

It's one big jigsaw, but the pieces are beginning to fit..


Here's another swerve-ball for yuz:

• what if the initial "one pound" is also one of the "four" that jump four quarters?

IOW it performs an action that acts upon itself, in concert with another three masses..

Note he only explicitly tells us about four masses - never expressly intimating that the initial 'one' is actually a fifth.. merely an assumption we project..?

Because, then, you know, we'd be looking at an inertial interaction in a 3:1 ratio..

..and you all know how the maths pan out by now - you just add the inertia ratio, so, 3 + 1 = 4, to get the 'unity threshold' - the no. of elapsed cycles at which the KE = PE, provided no one's counting losses, in which case we technically have mechanical unity but with free heat from nowhere, tho we can only guess if Bessler was aware of this.. Point is tho that a 3:1 inertia ratio would indeed furnish the 25% p-c efficiency accumulator, matching the Toys page interaction cycle..


If a 'cross piece' is one complete minimal mechanism, comprising four masses, and one of these undergoes a normal, reactive, acceleration, which is then arrested with an asymmetric deceleration against the combined inertia of the other three in an inelastic collision, then we waste 75% of input PE in the collision, and retain the other 25% as KE. That's the first rung on the Toys page ladder.. do the maths, you get 50% on the 2nd cycle, then 75% on the 3rd, unity at the 4th, 125% at the fifth, 150% at the 6th, 175% @ the 7th, 200% @ the 8th etc. etc.

So, maybe there's only four pounds / masses in total - the 'four' including the 'one', rather than the ostensible 'four plus one'.

It's a physics problem, as much as a semantics game..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

A one pound weight falling from 10 to 7 has fallen a quarter, so too has one falling from 2 to 5.
If this falling weight can raise a four pound weight from the rim at 6 to 2 inches above the rim at 6, lightly, and in a flash, and the weight stays there for a whole turn it would have been raised for four quarters.
If the falling weight can raise a 2 pound mass at 6 and a 2 pound mass at 12, 2 inches each., and they both rotate a half circle, 2 pounds x 2 and 2 quarters x 2 = four pounds four quarters.
Two pounds of sugar and two pounds of flour is a four pound mass.
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Hi Robinhood46

Just unpacking your post - being the more concise post on this thread.

You have a falling 1 pound mass and it has fell 1 quarter.

What distance did it fall or what radius did it have from the axle?

You have used the KE to lift up 2 pound at 6 & 12 Of The Clock [i.e. bottom and top]
up 2 inches.

These paired 2 pound masses rotate 2 quarters each.
That is your 4 pound, 4 quarters.

Could you clarify who you get back to the start position?

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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Hi MrVibrating

It's good too see you are back and well rested.

The post with the example of 'reactionless rise in momentum'.

You appreciate two masses moving in the same direct with the same mass do not interact.

If they do then both have a change in direction or speed.

When a mass interacts with another mass and it's speed and direction [velocity] has not changed. Then it has not interacted.

The term 'asymmetric inertial interactions' is conservation of momentum were work is done to pull in their masses.

'Instead of asymmetric accelerations, asymmetric decelerations, or collisions.. '
What's was that alternative called/definition?

'instant of an inelastic collision' any instantaneous event should be change to one that takes a finite period length of time. It will improve accuracy and agree with the real world.

P.S. I stopped reading when I hit my buffer overload.

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Re: re: Decoupling Per-Cycle Momemtum Yields From RPM

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Robinhood46 wrote:A one pound weight falling from 10 to 7 has fallen a quarter, so too has one falling from 2 to 5.
If this falling weight can raise a four pound weight from the rim at 6 to 2 inches above the rim at 6, lightly, and in a flash, and the weight stays there for a whole turn it would have been raised for four quarters.
If the falling weight can raise a 2 pound mass at 6 and a 2 pound mass at 12, 2 inches each., and they both rotate a half circle, 2 pounds x 2 and 2 quarters x 2 = four pounds four quarters.
Two pounds of sugar and two pounds of flour is a four pound mass.
EMGAT, unfortunately - anything even momentarily lagging the rotation will require re-accelerating, incurring N3; a weight parked in a 'hanging' position - even if only for part of a turn - is effectively an internal gravitating stator.. and as RPM's rise, inevitably so will these required internal accelerations, their PE costs likewise squaring with rising velocity and so tracking unity.. besides, isn't it "..to four quarters", rather than "..for.."?

Nice try tho.. keep at it, this is meant to be solvable..
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Post by MrVibrating »

OK here's the beginnings of a plan:

• we need to shift the inelastic collisions into the free-fall FoR..

• ..thus taking the wheel's 'zero momentum frame' with them:

IE. the hypothesis being that a momentum distribution that is symmetrical in the ground FoR is asymmetric from the free-fall FoR, but equally, vice versa..


.."they come two and two" = two pairs, not one! A minimum mech. comprising four masses..

So how about this:

• one pair are 'classic OB' weights, undergoing a closed cycle of radial lifts and extensions on the under-balancing side, and resting still on the over-balancing side

• the second pair are for colliding with the first..

• ..taking turns to be accelerated (presumably reactively, respecting N3), before colliding with the current OB weight whilst it's descending / over-balancing the system

That's basically all i've got thus far - tho it seems to put MT 133 & 134 in better perspective, and with them, the Toys page interaction..

It's conceptually simple - basically playing whack-a-mole with over-balancing weights..

..but also, may resolve the riddle:

• a 'pound weight' falling 'one quadrant' onto an over-balancing weight, thus causing a momentum asymmetry in the ground FoR, providing the energy gain to..

• jump / shoot / shift three other pound weights plus itself between the 'four quadrants' of a full 360° system rotation

..you get the picture.

So i need to do some simple experiments testing that central hypothesis - that inelastic collisions in the free-fall / gravitating reference frame cause effective N3 breaks in the ground / KE frame..

Work now, back later..


ETA: lol - basically comes off as a desperate, last-ditch attempt at classic OB - having devolved into trying to physically 'hammer' the process into perpetuating.. "for the love of (whack!) ..turn! (whack!) bleedin' rotate you bastard!"

..and yet might just be crazy enough..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

I can assure you that shouting and swearing at the thing will not make it turn.
I've tried it loads of times. When the bloody thing doesn't want to rotate there is nothing we can do to change it's mind. Best bet is to rip it to pieces, that'll teach it to not do what it should..
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Well distorting the inertial frame of space and time. Who in there right mind would do that.

Distorting a Bessler quote/clue is an erzate thing we can all do; but we don't.

Then whacking those cute fury rodents. That just going to far.

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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Hi MrVibrating

Putting the jollity to one side.

Yourself and Fletcher's post do make a difference.

They inform and exercise the minds of other members and guests.

In a way you are a good intentioned mentor and enthusiastic text typing tor-mentor.

I would like to thank you; for you did drive me to look at what I thought I know and check.

You did create a WHY in me about the relationship of 1/2 of values and squaring values.
They appear to be linked.

This is were I went to get an understanding of a effective force we all know, miss-spell and get confused about :)

https://en.wikipedia.org/wiki/Centripetal_force

Notice the Local Coordinate section and this text
The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force.
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Post by MrVibrating »

Yes, these key geometric relationships controlling CoE and CoM - and thus, their prospects for manipulation - are simply and easily demystified by just considering their respective causes:

• output KE - and inversely, usually also input PE - squares with velocity; why? What is actually squaring is the elapsed displacement over which a given force must be maintained in order to continue a given acceleration - be that meters or radians - so to accelerate something to double the speed, a given force must be applied across four times the displacement..

..and since forces can't be unilateral, they can usually only be applied between inertias - and so the displacement between the thing you're pushing and whatever it's being pushed against is squaring.

What's actually happening 'in the physics' to cause the squaring? It's a kind of 'hysteresis' effect - if you're already at 1 m/s then a further 1 m/s acceleration will see you at a new speed of 2 m/s, which is really just '1 m/s relative to the 1 m/s you already had previously'; hence it's a simple, compounding scaling factor.

Thus it's easy to see how PE:KE symmetry - CoE - is bound to CoM / N3 / N1 / relative stasis of the zero momentum frame... and thus, implicitly, what's involved in breaking that symmetry..!

This same simple dynamic also underpins rotating systems:

• Tangential displacement squares with radius!

Say it a thousand times, but file it alongside the squaring relationship of displacement to velocity, because it's the same simple geometry!

Doubling radius quadruples the displacement of a point mass for a given RPM.

Rotation at constant speed is constant acceleration (constant rate of change of planar direction).

Hence doubling the radius of a given point mass quadruples the amount of space it's being accelerated thru per unit angle..

..the same mass being accelerated thru four times the space in equal time thus presenting four times the angular inertia..

..hence, MoI = mr²..

..CF force = mass * angular vel. squared * radius..

..multiply that by the radial velocity of a radially-moving mass, and you get a term with the rather nifty sounding dimensions of 'centrifugal power'..

..the time integral of that plot (area under curve) gives the radial work done / input PE / change in output rotKE..

..and rotKE = ½MoI*RPM².

At the root of it all lies 'inertia'..

..the concise definition (covering both angular and linear terms) being a function of how much mass has been accelerated thru how much space in how much time..

..and it is that, singular, metric, in relation to the inherent spacetime geometry of the 3+1 manifold, that enumerates and substantiates all mechanical interactions, their conservation of momentum, and energy..

..our 'wriggle room' being the distinction between relative vs absolute speeds, as regards CoM and relative stasis of the zero-momentum frame.. thus any 'effective' N3 break allows you to pay the same energy for a 1 m/s acceleration when already moving as you would from a standing start..

..hence, in an idealised (and almost certainly unattainable) example, the min. poss. cost of accelerating 1 kg or 1 kg-m² by 1 m/s or 1 rad/s is half a Joule..

..maintain that efficiency over a succession of say, ten cycles, and you've spent 5 J of PE, yet have 50 J of KE.


There's no mystery to any of it.. all follows succinctly from first principles, no gaps or uncertainties. Finite complexity, easily grasped in whole. The explicit instructions for breaking CoE & CoM are implicit within their respective terms of conservation..
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Post by MrVibrating »

OK, spent a lil' while just passively dropping a radial lever-weight onto an acting OB weight.

Obvs, MoI ratio between the falling weight and the wheel determines relative speed = needs a high ratio to work reliably under passive gravitation.. tho less of an issue if active torque were applied..

You do see a good kick when it inelastically lands, tho nothing's metered up yet..

Need a day off or two to really get stuck in, still, the mission remains to try and effect some kind of relative vs absolute N3 / N1 break... at RPM-invariant PE cost.
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re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Hi MrVibrating

An interesting read and the notion of 'hysteresis' effect & 'centrifugal power'.

I have this link to assist those reading that need a catch-up tutorial.

https://www.youtube.com/watch?v=RIGMaw8gsic&t=2529s

Note.

I choose to attribute K.E to a local FoR. This appears to address some of your terms
and addresses the CoE & CoM breaks.

P.S. You are looking in the right area. Thanks for keeping the text down.
You are helping your readers in absorbing your message.

EasterEgg [ https://www.skeptical-science.com/enter ... ster-eggs/ ]

https://www.youtube.com/watch?v=xb0FU8rSisU

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Post by MrVibrating »

Bugger.. slipped at work earlier, bust me right arm above the wrist - the sticky-outy kind of fracture unfortunately, so had to go get it reset (tried doing it meself but couldn't pull hard enough with only the other arm)..

6 weeks off (yay!) w/o income (eek!) w/ only me left arm for everything.. and i'm a rightie, so.. prolly not up to much..

Pain killer's starting to wear off so here comes the sting.. s'cuse me while i go writhe sleeplessly in bed a while..
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