Poss. Symmetry Break?

[MrVibrating]

A two-body system in which the inertia of one body is passively time-variant is able to violate both CoE and CoM.

It circumvents the usual limitations of mass constancy and Newton's 3rd law.

It's right there in the illustration, provided we're looking at intrinsic "heaviness", as directed by Bessler.

Not "hidden" or encoded in any way whatsoever. On the contrary, meticulously detailed in a plain diagram.

A black & white map of the direct route to the treasure - the asymmetric pendulums are the X marking the spot.

Modulating "intrinsic heaviness" IS modulating net energy. Covariant parameters - lowering I raises E.

Due to the increase in V. Which is enforced by none other than CoM itself - IOW applying the conservation laws against themselves - and so fully consistent with, and dependent upon the classical laws of conservation.

IOW evidence of Wolff's noted open thermodynamic system and consequent vacuum energy, rather than disproof of CoE / CoM (which are intrinsically, logically impossible to disprove).

So far, this bears all the hallmarks of a genuine symmetrey break. MoI, it seems, can be freely or cheaply modulated. This is Bessler's exploit.

It's the only thing it could be, and he's repeatedly carved diagrams and offered carefully-worded explanations painstakingly labouring this precise phenomenon.

Something tells me we're warm...

[MrVibrating]

A quick recap for any stragglers:

- the cyclical GPE / GKE load of the pendulums is always energy-symetrical - the masses and gravity don't change, so energy-wise, they're just pendulums. So, since their net energy is zero... just subtract it all from consideration - they may as well be oriented horizontally. All that concerns us here is their angular inertias - the fact that the unvarying masses move varying distances either side of their BDC's, which are offset from vertical by their offset axes, and that the subsequent varying inertias are synched by the crank (ie. hard-coupled) to alternate 180° arcs of the wheel's rotation.. hence the net angular inertia (MoI) of the moving system is varying passively.

This opens the door to a symmetry break.

[MrVibrating]

Moving swinftly on, and staying with this image:

https://www.lhup.edu/~dsimanek/museum/p ... essler.htm

Please turn you attention to the relationship between the box of bricks on the right, and the stampers on the left.

If the box is being raised, then the stampers are bieng dropped, and vice versa.

So in the former case - if this image really depicted a real demonstration, then in order for the rotation to raise the box, the stampers would have to "fall" upwards...

Winding in the rope raising the box, pushes the stampers down, not up. The rotation can't raise them back up. They'd have to fall up.

If OTOH they're being raised to fall against gravity, then the system is rotating in the opposite direction, and the box of bricks is being lowered.

Hence it is a constant rate of GPE input.

A consistent unit of drop causes a consistent angle of rotation.

So the conversion of GPE to RPM is linear.

And for a constant MoI, so is the conversion of RPM to RKE.

But not so for a varying MoI.

IOW, the effective output value of the input energy, as manifested in the standard terms of RKE (1/2 MoI * RPM^2), is cyclically non-unity. At a constant rate of RPM, energy that was input at higher MoI pays compound interest when MoI goes down. If displacement doesn't increase (constant RPM = constant angle * time) then torque has (because energy is force * displacement).

So the box is the input load, and the stampers the output load.

However here we have a conundrum - each stamper is raised twice per cycle, there's four of 'em, so two are raised by high inertia / low energy, and the other two are raised by low inertia / high energy.

Yet all four are just dropped again - so the system is inherently OU, but doesn't seem to be closed-looping for some reason... anyone have any thoughts on this?

[MrVibrating]

Maybe the stampers are synched to the swing phases (upswing, downswing, asymmetric CW vs CCW displacements hence four distinct phases?)

Time for a long soak in the bath with a favourite tipple.. need a eureka moment..

[MrVibrating]

OK, suppose that weird bracket & pivot at the far left indicates the system's center of rotation (as i keep banging on).

Linear inertia of the stampers exerts torques on the net system.

So the rising and falling of the stampers is trading radial for axial inertias, with the latter (MoI) rising and falling each 180° in turn..

If such a system was balanced about its center of rotation, with one of these radial-for-axial inertial exchanges every 45° of angle between the inner and outer mechanisms, the system is effectively unbalanced by the varying MoI (as opposed to the radial distribution of the stampers and conventional OB systems.)

At any angle, there is an unbalanced inertia on one half of the wheel - and it's moving around the wheel as a function of the external to internal displacements. This inertia is equivalent to gravitating mass (per relativity), so even if it begins at the lower half of the wheel, as it climbs up, the wheel becomes effectively overbalanced and rotates the net system.. forever chasing a shifting equilibrium.

Do i win the £5 yet? (i'm gonna buy SO many Kinder eggs).

[MrVibrating]

So the excess impetus is an effective mass, not a real one, so never actually needs raising.

How friggin' smug is that?

The system's trying to balance against an effective mass acceleration as if it was a gravitating body.

If we simplified the situation and presumed the linear inertia of the stampers is equal to the peak MoI of the rotor and pendulums (flails?), then each 45° of external angle implies a 90° net rotation between inner and outer mechanisms.

IOW if the wheel turns 45° clockwise, the internal mechanism has been shifted 45° anti-clockwise, for a net 90° displacement, only half of which is externally apparent.

So if the net system rotates 45° in the attempt to balance, the thing it is trying to drop has instead climbed 45°

Thus, whatver the actual balance of linear to angular inertias, the net system must accelerate until the external velocity matches the net velocity..!

At this point, the internal velocity can effectively be at rest with respect to the external frame..!

So let's imagine what that image would look like:

We'd see the outer wheel spinning steadily, while the "stampers" inside formed a stationary square - the vertical stampers on the left and right would be pumping up and down in alternate step, and the horizontal ones at top and bottom would be oscillating left and right in alternate turn. All physical GPE's are irrelevant and sum to zero - only the cyclically varying balance of radial or linear vs axial inertias matters.

To all intents and purposes, it is the unabalanced inertia that both gravitates, as it climbs back up, remaining perpetually outside the center of gravity.

There's only four stampers... but there's five effective weights, one of which is simply the "virtual weight" of an inertial acceleration caused by a cyclically attenuating MoI.

In essence, at constant operating RPM this 5th virtual weight is an emergent torque, equivalent to an extra mass always perched at 90° to the steady-state rotation.


Am i talking complete kack or is any of this hitting the mark? Dunno, but it all seems to jive in 2D for now so should be simmable..

[daanopperman]

Mr V ,

If you calculate the period of the pendulum in J.B. drawing , what do you get if you use the basic formula for the calculation of T when you use the basic formula for simple pendulum .
When I look at the bottom of the 2 drawings that a link was given for , I would think that the pendulum was +- 10ft in dia . Since the weight of the bob does not contribute to the period we need not at this stage have a need for them . ( 3X no. 33 )
If we use this formula T only changes by 1% if theta is about 20deg . That is very small and I think that the swing in J.B.'s wheel is less than 25 deg from rest .
The reason for this is that if the period of the pendulum does not match the rpm of the wheel then all hell will break loose .

[MrVibrating]

All hell you say? My my, one can only shudder at the possibilities..

Much like such a wheel would..

Because its MoI would be all over the place...!

Not constant!

And energy is a function of inertia.

Bessler's one-way wheel was seen to exert a cyclical 'kick' on its supports, which would be consistent with such an unbalanced system.

To test this i've taken forward a principle used in recent experiments, whereby a pendulum is arranged that has equal angular inertia to a wheel - an impulse applied between them induces equal opposite accelerations.

Here however a conrod / crank has been added, so the wheel and pendulum are hard-coupled, as Bessler kept dementedly illustrating.

Finally, a constant torque is applied to the wheel's axle, and its acceleration plotted, resulting in a trace of the varying inertia as a function of angular accel. / time.

Pay close attention to this plot at the bottom of the sim - expand it to full screen if poss - and note the balance of integrals above and below the zero line..

Note also that this pendulum is not hung asymmetrically - yet a stable asymmetric distribution of inertia / angle is established.

We're leading inexorably to a stunning, profound, yet kick-yourself-obvious revelation here - that it is not the effective weights of the weights themselves that is being modulated, but that of the wheel's own inertia!

!

The wheels own inertia is modulated, in relation to the constant GPE of the input and output loads. Work input at high MoI is worth more at low MoI.

Again, this is akin to Fletcher's concept of trading a given momentum's mass and velocity to create KE - mass constancy precludes this in linear interactions but its rotational equivalent, MoI, is evidently variable. And where MoI goes down, velocity and thus KE must rise to conserve momentum.

So although in a linear system we can't drop a mass when it's heavy and pick it up when it's light, we can input GPE when the wheel is heavy, and pick it up when it's light...



TL;DR
_____


The weights don't get heavier and lighter, the actual wheel does instead.

(who saw THAT coming eh?)



ETA: this image shows the balance of instantaneous net MoI / angle for a single full cycle at constant input torque (no point running the sim up to silly velocities).

[eccentrically1]

The weights don't get heavier and lighter, the actual wheel does instead.

(who saw THAT coming eh?)

Things that make you go hmm.

[MrVibrating]

Here the rotation's reversed to CCW (because it makes the plots easier to read), but it's the same system.

Gravity's disabled, and the pendulum is hung symmetrically. The amplitude's ever-rising because we're applying constant torque, in order to reveal the corresponding resistance to that acceleration, ie. the effective instantaneous MoI. However the interesting detail is that what this reveals is a directional asymmetry in the MoI...

Note the alternating positive peaks in the sequence - these represent the MoI when the pendulum is swinging right-to-left. The smaller positive peaks in between indicate the MoI when swinging back left-to-right.

Aside from the required vertical offset between the axes, the system is horizontally symmetrical. It's just a pendulum and a flywheel, being accelerated against their combined rotational inertia, by a constant torque applied at the wheel's axle. Yet apparently that MoI is directional - a left-to-right downswing has a lower peak acceleration than the same phase in the return direction.

If F=MA then A=M/F and if F is constant (as here) then a varying M (ie. inertia) is the only remaining culprit for the variation in A.

In short, the MoI of the system varies with the changing direction of the pendulum - each half of a full swing, and thus each 180° half of a full rotation.

For half the rotation the wheel's heavier, and then for the other half it's effectively lighter. Not its linear weight, obvioushly.. but how "heavy" it feels to turn the axle..

This is almost a bullseye prediction of the hypothesis - the only detail i got wrong was thinking asymmetric pendulums were required... but on that count, i couldn't understand why the 3rd pendulum on the far right wasn't also hung off-center. Perhaps the resolution of this is that the image is hinting at this inherent inertial asymmetry in an otherwise symmetrically-hung pendulum... as apparently evinced here. **

Not there yet - bit all over the place really, still more predictions to test, but looks like MoI could be our wildcard..

Could really use some means of plotting torque / angle, WM doesn't seem to support this and accel / time skews the picture slightly.. (clarified by clicking thru the sim frame-by-frame with one eye on the trace and the other on pendulum angle, but torque / angle's more intutitve to read)..


Sorry to text bomb like this, i'm honestly enthralled at these new possibilities. Soon as you start playing with MoI, it's hard not to find anomalies. This is really exciting progress, with a growing sense of inevitability...

FWIW everything in this current system is depicted in the image below:



...constant torque + variable MoI = variable KE. A prime mover, not hidden or encoded in any way, but offered in complete form with all pertinent details in plain black and white images. Repeatedly.

Again, the pendulums needn't be interpreted literally - the key point is simply that there's other ways of manipulating MoI besides symmetrical radial translations. pendulums seem to be one route, but the image showing an Archimedes screw appears to be using the square wheel to similar effect..

Next, i want to try flinging masses inwards - weight pole pivoted to rim, its angular acceleration flings the weight outwards from the rim, inwards towards the wheel's axle, exerting a counter torque on the wheel while lowering its MoI. Probably be symmetrical, but worth checking..

Basically, anything that could affect MoI is worth revisiting. Bessler's success likely means that it's possible to effect a passive modulation of MoI, capitalising on a free variation in torque - for a constant velocity, where MoI goes down, torque goes up. By sheer process of elimination, we must be nearing our goal.. and this looks like a serious contender from left field..





ETA: **on reflection, another possible intepretation of the two asymmetrically-hung pendulums is that they're not actually hung, due to the deliberate occlusion errors on their hangers, instead hinting at an emergent asymmetry from an otherwise symmetrical pendulum as shown on the right, properly affixed? IOW the image tells us to look for a horizontal asymmetry in a simple pendulum & flywheel system under constant torque.. they're kinda like "ghost" pendulums. Same goes for the disconnected ones in the background of the water screw image - completely redundant, but for the fact they're depicted with unequal horizontal sections...

So this is starting to tick an eerie number of boxes (seriously, hairs standing on end here). Lotta potential resolutions on the table.. like directionality of the basic concept, per the Gera wheel.. drop a weight against low MoI and pick it up with high MoI and we'd destroy energy.

But it seems certain his wheels didn't use pendulums (i still believe everything rotated together, or even depended on this), so it's still early days. I'm sure there's something useful here though - forget gravity and read these illustrations in the context of inertia only, and varying output KE is their direct and inescapable conclusion. A special, key principle is being repeatedly laid out in these diagrams - and not disguised, just thinly veiled as depitions of actual demonstrations. First and foremost though, they appear to be technical instructions for how to circumvent mass constancy..

[John doe]

Excellent post! Very enlightening!

[Fletcher]

Grav is OFF

- Aside from the required vertical offset between the axes, the system is horizontally symmetrical. It's just a pendulum and a flywheel, being accelerated against their combined rotational inertia, by a constant torque applied at the wheel's axle. Yet apparently that MoI is directional - a left-to-right downswing has a lower peak acceleration than the same phase in the return direction.

If F=MA then A=M/F and if F is constant (as here) then a varying M (i.e. inertia) is the only remaining culprit for the variation in A.

In short, the MoI of the system varies with the changing direction of the pendulum - each half of a full swing, and thus each 180° half of a full rotation.

For half the rotation the wheel's heavier, and then for the other half it's effectively lighter. Not its linear weight, obviously.. but how "heavy" it feels to turn the axle..


Could really use some means of plotting torque / angle, WM doesn't seem to support this and accel / time skews the picture slightly.. (clarified by clicking thru the sim frame-by-frame with one eye on the trace and the other on pendulum angle, but torque / angle's more intuitive to read)..

Just some random thoughts as I read thru your post Mr V ..

Like I found in the Jack experiments on my thread where I was looking for the formulas to explain a lack of acceleration - I found that to arrive at the correct acceleration I had to factor linear inertia (m) by the square of the gearing relationship. In your rotary example here the crank will be an effective gear wouldn't it, depending on angle to pendulum etc ? So as MOI is effected by mass distribution to the CoR in a squared relationship so was linear acceleration using gears. So would this example perhaps ??

On that note down load the file - it is yours but more generic - adjust the tables to find the Torque / Angle plot you want. Not sure I used the angle you were after ?

[MrVibrating]

For thoroughnesss, just tried asymmetrically-hung pendulums..

The previous, symmetrical plot looked like this:



...we get this emergent drop in the net inertia in one direction, but concentrated around a single change of direction (from CW to ACW about its pivot).


However when the pendulum is hung off-center, we get this pattern:



...and now both direction changes of the lower-MoI half-cycle are attenuated - enhancing the inertial asymmetry.

Extrapolating that result, adding a second opposing pendulum would presumably further increase this asymmetry...


Also need to take on board Daan's point about tuning the pendulum length.. almost completely arbitrary at mo. (equal to wheel radius cos i was balancing their respective MoI's as a function of bob mass w/o gravity).

[MrVibrating]

Just some random thoughts as I read thru your post Mr V ..

Like I found in the Jack experiments on my thread where I was looking for the formulas to explain a lack of acceleration - I found that to arrive at the correct acceleration I had to factor linear inertia (m) by the square of the gearing relationship. In your rotary example here the crank will be an effective gear wouldn't it, depending on angle to pendulum etc ? So as MOI is effected by mass distribution to the CoR in a squared relationship so was linear acceleration using gears. So would this example perhaps ??

On that note down load the file - it is yours but more generic - adjust the tables to find the Torque / Angle plot you want. Not sure I used the angle you were after ?

Yes you're right, the crank radius determines the power ratio, and this asymmetric inertia between directions arises do to the conrod / crank connection crossing the axis - the increased horizontal distance at 180° causing a decrease in the vertical travel. IOW the usual issue of linear-to-angular conversion efficiency as a function of crank angle. This causes an ordinary symmetrically-hung pendulum to travel a slightly shorter distance when the conrod is effectively shortened by crossing the axis diagonally. An asymmetrically-hung pendulum can increase or offset this variation.

If indeed the pendulums are just metaphors for the general principle of MoI variation then we shouldn't expect too much from this system - the analogy will have its limits.

One has already occurred to me earlier, which is that without a common axis about the wheel (ie. turning with it), the pendulums are a kind of "outboard" mass, whereas to convert an effective mass variation to a velocity change, you'd think everything would need to be able to accelerate together.. not sure tho, all the mass in this system can still accelerate, so i still need to test for energy symmetry. It's possible that a MoI variation by any means would suffice, but Bessler seemed very clear that full rotation about a common axis was a prerequisite rather than a design prefernce.

And cheers for polishing the sim - all the meters besides rot. accel were leftovers from the previous round of N3 tests (using gravity to reverse the direction of the reaction mass between equal opposite inertias). The torque / angle plot i wanted was for the main rotor - the inertia / angle as inferred from the accel / time plot is rotationally squished by the constant time rate, whereas the raw inertia / angle plot is time invariant and shows a consistent angle increment along the x axis... i began doing an integration then realised this was invalidated by the speed variation / time.

Not quite sure torque / angle is ideal either - really want inertia / angle since that's what this is all about - an integration of MoI / angle for 360° will either confirm or refute the hypothesis. If the MoI really does have a 180° variation then so will our RKE integral for a full cycle at constant velocity.

WM can't measure inertia directly, but it can do torque, momentum, KE, time and angle etc... so it should be possible to get a fairly direct indication of inertia / angle, ideally in as unambiguous a format as possible.

Maybe momentum / time for a constant velocity would do...

[daanopperman]

Hi Mr V

The reason for the diifference in symmetry break is that the center of movement of the 2 pins is not squere to each other . That is the only reason .If you draw 2 lines parralel appart and place the start and end positions of the pins on this line so u have 4 points , 2 on each line , then to have symmetry the center of the 2 top points must be squere to the center of the 2 bottom pins .

Daan.