Decoupling Per-Cycle Momemtum Yields From RPM

A Bessler, gravity, free-energy free-for-all. Registered users can upload files, conduct polls, and more...

Moderator: scott

Post Reply
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

Post by MrVibrating »

Google Translate seems to make a semi-coherent effort here, if only for this single passage:
"This nota bene add:
He will be called a great artist
Who can throw a heavy thing up easily,
And when a pound falls a quarter
It jumps four pounds to four quarters. x
Who can speculate on this,
Will soon perpetuate the course;
But if you don't know this yet,
All industry is in vain"
If it needed any reminder, as Google Translate improves heuristically over time, it can't yet begin to aspire to match the skills and knowledge of a human expert such as Mike Senior; he's interpreted 'quarters' as a vernacular for "ounces", assuming continuity of the context of 'weight'. Moreover, the formula as he phrases it seems to state a more definite proposition - consistently between the two sentences "1 lb vs 4 oz's / 4 lb vs 16 oz's".

Yet if 'quarters' aren't 'weight' but some other field property by which weight is multiplied, such as height or velocity, then the actual proportions being described may be a factor of 16.

If they're units of momentum, then perhaps it's referring to an amount of output momentum from G*t that can be ideally secured with 16x less input momentum? Or is it energy? Maybe it's mixing units and dimensions, who knows..

Bessler obviously thinks the riddle's solvable, so i'm sticking with it for now..
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

Post by MrVibrating »

OK, so we know you can't have an I/O GPE asymmetry - so, you can't lift on the cheap, for example; G, M and H are conservative at every instant in a GPE interaction.

IOW, you cannot buck the input GPE workload - if a mass is rising, work's being done against G to the M & H, period.

Unless it's counter-balanced by equal descending GPE, in which case no net work's done.

If it's counter-balanced by some other force or field, then again, work must be done there..

..however it's at least now possible that work / energy may scale differently in this other, counter-balancing field..

If this other, counter-balancing field isn't also GPE, then it can only be some kind of inertial interaction..

Yet, are 'quarters' units of height, or even inclusive of height? Could 'velocity' or some other field better fit a potential solution?
User avatar
agor95
Addict
Addict
Posts: 7704
Joined: Wed Sep 24, 2008 8:09 pm
Location: Earth Orbit
Contact:

Post by agor95 »

MrVibrating wrote:... making energy is as simple as applying a reactionless rise in momentum; eg. two 1 kg masses both at 1 m/s = 1 J of KE, now apply a 100% asymmetric 1 m/s inertial interaction between them, thus only accelerating one, so now one mass is at 2 m/s, the other still at 1 m/s.. hence 2 J + 0.5 J = 2.5 J, a rise of 1.5 J.. that internal acceleration only cost ½ J,
Can you consider two 1 kg rods pivoted on the same axis?

The Rod 1 is rotating by one end and Rod 2 by it's middle.

What you are doing by calculating the energy is a good thing to do.

Know have Rod 2 not moving at 55 degrees. Just calculate the MOI for that.

Rod 1 is horizontal to the left. This is pivoted on the left hand end. With a different MOI.

The effect of gravity rotates and speeds up Rod 1 as it falls. PE to KE as per normal.

Now decelerate Rod 1 as it nears the bottom and accelerate Rod 2. A spring might help.

The questions are what is the rate of rotation of Rod 1 before deceleration?
Also what is the rotation rate after Rod 2 is accelerated?

This is where the energy calculations become useful.

Cheers
Last edited by agor95 on Wed Nov 11, 2020 4:07 pm, edited 1 time in total.
[MP] Mobiles that perpetuate - external energy allowed
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

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

Post by MrVibrating »

Robinhood46 wrote:
Yes, it's a massive over-simplification, yet really does convey the raw mechanics of what's going on in an OU interaction:
Some bloke once said "If you can't explain it simply, you don't understand it well enough".
Mech. OU = A recoil-less water pistol on a moving roller skate; the KE of the water jet exceeding the input energy to the trigger / pump per ½mV², where 'V' is the skate velocity you should've lost, but didn't.


All other mech. OU solutions are just variations on accumulating reactionless momentum rises.

Bashically, scrub N3 and you can't move without making energy.. In practice tho, there's always an energy cost involved in accomplishing this... so our only hope is the prospect of input energy / PE to that field evolving differently with respect to some other conserved field property such as time or velocity.. and the solution to the riddle must presumably somehow speak to that issue..
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

Post by MrVibrating »

agor95 wrote:
MrVibrating wrote:... making energy is as simple as applying a reactionless rise in momentum; eg. two 1 kg masses both at 1 m/s = 1 J of KE, now apply a 100% asymmetric 1 m/s inertial interaction between them, thus only accelerating one, so now one mass is at 2 m/s, the other still at 1 m/s.. hence 2 J + 0.5 J = 2.5 J, a rise of 1.5 J.. that internal acceleration only cost ½ J,
Can you consider two 1 kg rods pivoted on the same axis?

The Rod 1 is rotating by one end and Rod 2 by it's middle.

What you are doing by calculating the energy is a good thing to do.

Know have Rod 2 not moving at 55 degrees. Just calculate the MOI for that.

Rod 1 is horizontal to the left. This is pivoted on the left hand end. With a different MOI.

The effect of gravity rotates and speeds up Rod 1 as it falls. PE to KE as per normal.

Now decelerate Rod 1 as it nears the bottom and accelerate Rod 2. A spring might help.

The questions are what is the rate of rotation of Rod 1 before deceleration?
Also what is the rotation rate after Rod 2 is accelerated?

This is where the energy calculations become useful

Cheers
Gotta get ready for work now but MoI for a rod about its center is 1/12 * mass * length², and about its end, 1/3ml².

You're absolutely playing in the right toy box, and would obviously enjoy simming.. much easier that solving manually.. if you could knock up a rough sketch of the interaction i could try simming it later, unless anyone else were to step in in the meantime..

Just for playing with the numbers tho, would it not suffice to simply assign whatever values you want, and then think about embodying any interesting propositions with actual moving masses? IOW, work backwards - find a mathematical solution first (i've shown how simple it can be), then try to 'mechanise the maths'..
Robinhood46
Devotee
Devotee
Posts: 1637
Joined: Tue Oct 31, 2017 9:22 am
Location: Lot, France

re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

There are many forces sourrounding a moving mass, time will allow some of these forces to have an effect on the moving mass, but time itself will not.
Does this not imply that my simple explaination is wrong, because, the KE of one closed circuit needs to be brought forward in time with regard the time frame of the other closed circuit?
Therefore;
Two closed circuits are rotating around the same axis.
They both respect the laws of physics.
One is trying to accelerate away from the other and in doing so it causes the other to accelerate.
Switching the energy from one circuits time frame to the other, sounds good to me.
User avatar
agor95
Addict
Addict
Posts: 7704
Joined: Wed Sep 24, 2008 8:09 pm
Location: Earth Orbit
Contact:

re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

I find keeping ideas as simple as possible. It helps in creating a firm foundation.

So I try not to think of forces and time frame/s.

Of cause time is a measurement of one process against another.
Nothing more or less.

Likewise forces are a measurement of acceleration.
Nothing more or less.

Both are created by man and are not real.

An inertial mass can change it's velocity by interacting with another inertial mass.

We need the rod length so lets use 1 metre.

So use the formula Rod 1 end rotation 1/3, 1 kg and length 1 m.
= 1/3 MOI

Rod 2 formula 1/12, 1 kg and length 1 m = 1/12 MOI

That would imply when Rod 1 decelerates Rod 2 will be accelerated to 4 time as fast.

Think about the 1 pound to 16 ounces quote.
Look at these as a force [Newtons] down when they were weighed.

Rod 1 exerts a force, if you like, to accelerate Rod 2.

So 16 ounce of force lifts 1 pound were the MOI is played out as about.

Cheers
[MP] Mobiles that perpetuate - external energy allowed
Robinhood46
Devotee
Devotee
Posts: 1637
Joined: Tue Oct 31, 2017 9:22 am
Location: Lot, France

re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

I agree entirely with keeping things simple. I try very hard to keep it as simple as possible.
The problem is, explaining things simply allows us to convey what we are thinking, but it doesn't allow us to convey the complexity of what is going on.
If you were to explain very simply the principal of a nuclear reaction, i would not have any difficulty believing your explaination to be realistic, for the simple reason i accept that nuclear reactions are possible.
Had nuclear reactions not yet been invented (discovered) and for the last 300 years the whole world had decided they were totally impossible, and anyone spending any time trying to do so was an idiot, i would not accept your simple explaination. I would think you are talking complete nonsense.
Take two really small things and wack them together, this will cause a reaction that can power a city.
Take two closed circuits and keep wacking them together, they can turn a wheel.
User avatar
agor95
Addict
Addict
Posts: 7704
Joined: Wed Sep 24, 2008 8:09 pm
Location: Earth Orbit
Contact:

re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Hi Robinhood46

LOL

How so true.

P.S. Thorium Reactors just saying. https://www.youtube.com/watch?v=H6mhw-CNxaE
[MP] Mobiles that perpetuate - external energy allowed
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

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

Post by MrVibrating »

Robinhood46 wrote:There are many forces sourrounding a moving mass, time will allow some of these forces to have an effect on the moving mass, but time itself will not.
Does this not imply that my simple explaination is wrong, because, the KE of one closed circuit needs to be brought forward in time with regard the time frame of the other closed circuit?
Therefore;
Two closed circuits are rotating around the same axis.
They both respect the laws of physics.
One is trying to accelerate away from the other and in doing so it causes the other to accelerate.
Switching the energy from one circuits time frame to the other, sounds good to me.
Well it's 'one circuit' - EMGAT; everything's rotating together, in the same rotating frame of reference, but it's an open system because the input and output workloads are in different FoR's - and increasingly so as RPM's build..

This can easily be understood by first considering any conventional kind of motor, which, due to N3, requires torque to be applied against some fixed external body - a 'stator' - and as the the 'rotor' gains speed relative to the stator / earth, the input torque required to maintain a given acceleration has to increase by the square of the elapsed angle... thus causing input PE / work done to perfectly track output KE.

In short, PE:KE symmetry is being enforced by the fact that both reside in the same - ground - FoR; the rotor is being accelerated - ie. work is being done - relative to the ground, and likewise, the resulting rotational KE is also, almost by definition, relative to the ground (it'd be of dubious value otherwise).

So both the input / PE workload, and the resulting output KE, are in the same, terrestrial, frame of reference - and so we're measuring 'elapsed angle' in the PE term (torque * angle) along with 'velocity' in the KE term, as relative to our static ground FoR, and this is the principle cause of energy unity.

The system is 'closed' due to the ubiquity and immutability of N3 - it's one big inertial system of interacting inertias, in which the net system momentum never wavers.

And as for motors, so for all other types of machinery..

For just about any system you can envisage, whatever forms the 'input' and 'output' workloads, there's a direct causal sequence of collisions / inertial interactions between their moving masses that's conserving the net system momentum.

Thus this foundational condition that the net system momentum always remains constant (if not a flat 'zero') - ie. N1 - emerges from every discrete interaction in the causal sequence of input and output workloads individually respecting N3.


And so if we break that causal chain of collisions interconnecting all actions within our closed system, it'll effectively become opened - its net momentum is going to change over time..

If its net mass is constant but net system momentum is changing, then net velocity is changing (because 'momentum' = mV).

However relative accelerations within that accelerating system are now decoupled from their KE value in the ground (ie. static) FoR - so for instance accelerating 1 kg from relative 'stationary' to relative 1 m/s still only costs ½ J, per the standard KE equation, yet the actual velocity change - and thus energy / KE value - in the static FoR is now a function of the accelerating system's velocity, either adding or subtracting from it.

This is why a recoil-less water pistol coasting on a rollerskate is OU - 'recoil' ensures that the internal velocities resulting from any inertial interaction remain energy-symmetrical from any other FoR - so a static observer would usually see the rollerskate decelerate in response to squirting the water forwards, reducing the jet's absolute velocity accordingly; and this is what's causing the work done by the trigger / pump to be equal to the KE of the water jet (minus losses of course).. and likewise, the cause of OU when that recoil is omitted.

So an OU wheel must gain statorless momentum, to have any chance of keeping its cost of production relative, rather than absolute - the EMGAT principle.

Inertial isolation.

It's own inner momentum must come into being, as from within; it cannot be torqued up conventionally, applying momentum externally, as from without, to paraphrase B. - that would once again be hard-coupling the I/O FoR's, closing the system.

See how it works? The game we wanna be playing it "motion's relative, therefore velocity's relative, therefore PE and KE are relative to whatever respective inertial FoR" - usually, that's the same frame, because N3 / N1, but gravitating inertial interactions effectively circumvent N3 & N1 (actually playing 'em against one another), per kiiking or classic over-balance, where gaining statorless angular momentum from gravity and time is seen to be trivial; thus the only outstanding hurdle is to fully capitalise on the statorless condition / EMGAT and fix the bleedin' input energy cost of accumulating momentum to its internal, relative speed metric, rather than the external 'ground' reference of 'zero' velocity, in which we'll be harnessing the resulting KE.

I hesitated from impulsively dubbing it "KE gain" there, as it isn't, really - true OU can only be an input work / PE discount - its value simply being inflated by the 'velocity' component of the inertially-isolated system's net momentum rise.

Because KE is a function of velocity relative to the ground FoR, any system such as a wheel can only ever have precisely the right amount of KE for its given inertia and speed. Any notion of 'excess KE' a misnomer..

Quite simply, accelerating 1 kg to 1 m/s is half a Joule of work. We only ever pay the 'relative' cost - the 'force' and 'displacement' components oblivious to the rising net system velocity, and thus the absolute KE value in the lab FoR.. so if the system's already at a steady 1 m/s and holds it, that internal 1 m/s asymmetric inertial interaction has accelerated 1 kg from half a Joule up to 2 J, yet by only performing half a Joule of work.

If the system were at, say, 10 m/s then the absolute acceleration to 11 m/s would be a KE rise of 10.5 J, yet again, from only 0.5 J of work done. The bigger the N1 break / velocity divergence, the greater the OU efficiency, because input energy is summing linearly with elapsed cycles / rising system RPM, whilst KE is squaring per ½mV². Thus net input energy plots as a straight-line diagonal, whilst net output energy follows the V² exponent, inevitably intersecting the former and pulling up ballistically. All energy under the front of the curve intersection is loss, and everything behind it, gain.

No riddles from me: accelerate a 1 kg system to 10 m/s in ten discrete 1 m/s accelerations with a relative internal cost of ½ J each, and you've spent a net input of 5 J, yet for 50 J of resulting KE.

We don't make energy, FWIW - inertia does; it's what constitutes the stopping power of a KE, its work potential. The energy gain is paid for in its currency of withdrawal - the energy 'source' being whatever constitutes 'inertia' (0.2% the Higgs, the rest being relativistic mass of the proton's component quarks), and the momentum gain's obviously from gravity and time.

The task is to get the universe to 'create energy', by exactly the same terms it usually does. Just shifting the goalposts, fixing the clocks, a little bait'n'switch.. a logic trap for nature..
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

Post by MrVibrating »

"Who can throw a heavy thing up easily,
And when a pound falls a quarter
It jumps four pounds to four quarters. x"



I like this Google Translation, since it applies no contextual interpretations..

IOW, as Rainer pointed out to Stewart in the post Fletcher referenced on the previous page, Bessler isn't using any units..

..but he's not assigning 'quarters' any dimensions either!

'Four quarters up' = a human presumption on intended continuity of the already-mentioned dimensions; no argument that he's talking about propelling a mass upwards / a GPE input, yet it's a riddle, so the failure to explicitly restate the dimensions of 'height' in relation to 'quarters' likely has a calculated purpose - it's a kind of 'logic switch' - possible a bait'n'switch, a fork in the path, potential for misdirection.. two doors and a ghost that never lies.. ie. it's a riddle..

'Quarters' might pertain to height..


..quadrants of a circle..

..inertia ratios, momentum yields, radius, CF force, some really oblique concept that only Bessler would think up, anything, at this stage... but he doesn't expressly imply they're necessarily units of height, or even displacement per se..

Also: "..it jumps / shoots four pounds to four quarters" seems to imply that there's an inertial interaction occurring here; ie. that a force is being applied between the 1 lb falling ¼ and 4 lbs (or 4 * 1 lb) 'going to four quarters' (not necessarily 'rising').


And again, "lightly / easily" accomplishing a GPE input only seems consistent with effective counter-balancing.. in which case, the solution to the riddle must pertain to some means by which a rising GPE can be counter-balanced by less work done in rendering the counter-balancing force.. presumably inertial torque or summink (tho sure i must've tried almost every permutation by now)..


This would feed into the perennial question of whatever the under-balancing weights were doing in the tied-off and stationary one-way Gera wheel; if the system's under static OB torque, then either the returning GPE has already been lifted, or else it's waiting to be lifted, presumably under stored elastic PE. Assuming EMGAT, perpetual OB only seems possible with some kind of radial lift system.

Point is, the Gera wheel thus couldn't've been actively cancelling OB torque whilst tied off stationary, right? So actively counter-balancing a lift doesn't seem like a consistent class of exploits in regards of the Gera wheel's actual properties..

The seeming necessity of that 'classic OB' system of radial lifts and angular drops in the one-way wheels precludes active weight cancellation as a whole class of potential exploits, i think..

Further, i suspect Occam would argue that, rather than an inner, distinct 'prime mover' gaining energy that then powers the peripheral zero-sum GPE game, that the process of causing the OB itself is also the same actions causing the inertial asymmetry / net momentum kicks.. ie. that the momentum of the 'under-balancing' weights, being innately conserved, must form at least a portion of the momentum heard being imparted by those falling on the descending side, per Wolff's impressions..

IOW the 'perpetual motion structures' are likely somehow part and parcel of the GPE cycling system / OB cycle, rather than a kind of mechanical mitochondria, or individual 'PSU' type-deal that merely powers it..

One question that really keeps bugging me tho is; can a gain scenario be formulated in which the gain is manifested directly as GPE, rather than KE? (Or even, just EPE for that matter?)

Just because i ain't twigged such a solution yet don't mean it ain't there.. one of the drawbacks in gaining KE is the inherent loss phase below the break-even velocity; easily overcome by pre-accelerating up to the threshold speed by any conventional means prior to initiating an asymmetric inertial interaction, and hence instantly getting 'OU in one shot' as it were; and indeed, in that instant the output could certainly be collected in the form of a GPE input..

..but is it possible to formulate a more direct GPE-from-inertial asymmetry? Pushing right up against what constitutes an effective 'GPE asymmetry' here, but honestly not looking to defraud G, M or H.. however i can only see PE:KE asymmetries accumulating from diverging I/O momentum cost / value metrics.. whereas, the riddle does excplicitly set the context of the gain, in the first verse, as GPE, no?

Baffling innit...


Still, sun's up, should prolly think about calling it a night..
Robinhood46
Devotee
Devotee
Posts: 1637
Joined: Tue Oct 31, 2017 9:22 am
Location: Lot, France

re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by Robinhood46 »

It will take me hours and hours (days) to get this sim anywhere near presentable. I don't think it is possible to do everything needed with Algadoo.
So to show clearly what i explained a few posts back, here are a few images of the desired progression of the weights.
The wheel has 8 sections with 8 big weights. one big weight is represented by that orange blob.
There are only 7 small weights, because they step progressively forward during rotation, they are shown as small red blobs. Only 4 are present in the images.
The first image shows the accumulation of the 3 small weights applying force to lift the orange blob. Each weight is free to swing forward with relation to the wheel, so it applies it's force to the weight in front which all adds together to force the big weight to lift.
The second image shows the ridiculously over exagerated orange weight, that just shot up miraculously at about 16 times the speed of light, in its new position closer to the center. The following big orange blob (not shown) has stopped the two falling weights to continue accelerating faster than the wheel, they are applying force to the wheel. The leading weight is free to pull away and continue its path forward. It will be raised to 12 by the one way spring loaded levers that restrict any backward movement (black).
Image 3 shows the leading weight pulling away and the top weight coming over the top to add its efforts to lifting the next big orange blob.
I think a green blob that miraculously lifts at 24 times the speed of light is just as likely to not bloody work.
I can sort of see the peacocks tail opening and closing, although it isn't very convincing.
Is it possible to have blocks of 4 weights, where the leading weight of each block steps forward to become the last weight of the preceeding block? Or more likely, the last weight falls back to become the leading weight of the following block?
Attachments
N°3
N°3
N°2
N°2
N°1
N°1
User avatar
agor95
Addict
Addict
Posts: 7704
Joined: Wed Sep 24, 2008 8:09 pm
Location: Earth Orbit
Contact:

Post by agor95 »

MrVibrating wrote:"Who can throw a heavy thing up easily,
And when a pound falls a quarter
It jumps four pounds to four quarters. x"
If we rotate at the same rate Rod 1 [end rotation] and Rod 2 [middle rotation] then Rod 2 would need less work [heavy thing easily]. There is the 4:1 MOI ratio remember.

This is caused by the ends of the Rods accelerating at differed speeds.

The ends of Rod 2 is 0.5 metres from the axis and the end of Rod 1 is 1 metre.

Therefore distance traveled along the arc has a 1:2 ratio. [circumference of an arc].

Lets treat quarter as a time period of 4:1.

So Rod 2 final speed should be 4 times the initial speed of Rod 1; after the interaction.

We decelerate Rod 1 which is 1 kg and traveling at twice the speed of Rod 2 in 1 seconds. [a pound falls a quarter]

Just make the rotation rate a quarter of an arc per second.
[change the duration rate of deceleration if you wish]

The 1 kg will feel heavier and a strain gauge will record this point.
So we need the measured weight to be 4 times for Rod 2. [MOI agrees]

Now we transfer the Kinetic Energy from Rod 1 to Rod 2 causing Rod 1 to stop rotating.

Rod 2 accelerates and for this riddle it needs to be moving at 4 times the speed of Rod 1 before the deceleration phase.

The acceleration of Rod 2 should make the end mass feel like 4 time as heavy.

P.S. This is my best guess what this riddle means.

Cheers
[MP] Mobiles that perpetuate - external energy allowed
MrVibrating
Addict
Addict
Posts: 2877
Joined: Sat Jul 31, 2010 12:19 am
Location: W3

Post by MrVibrating »

..just thought - a seemingly-obvious candidate solution might be a weighted vMoI:

• 4 radially-sliding masses in 'cross' (or 'X') config, interconnected to counter-balance one another (so moving one likewise moves all)

• a 5th equal mass overbalances the system, landing at horizontal and thus applying 90° of OB torque down to BDC..

• ..the resulting rise in CF force dragging the four masses out to 'four corners' of an 'X'

Note that the original print ends that sentence with one of those mysterious, seemingly-neurotic X's.. does the riddle frame one quarter turn of an 'X'-shaped vMoI, perhaps?

So 'quarters' being 'quadrants', bashically ..

This does boost the momentum yield over the ambient drop, however the increase is in the 'MoI' component, rather than velocity where da KE iz.. thus necessitating some means of trading it up at profit; every effort i've made at thus far, unsuccessful (just making unity minus losses)..

Must be other, better potential interpretations tho..

Back later..
User avatar
agor95
Addict
Addict
Posts: 7704
Joined: Wed Sep 24, 2008 8:09 pm
Location: Earth Orbit
Contact:

re: Decoupling Per-Cycle Momemtum Yields From RPM

Post by agor95 »

Must be other, better potential interpretations tho..


I am sure there are many. I look forward to a better version.

As Inertial mass [momentum] and the effect of gravity are believed to be conservative.

The riddle gets you looking at MOI and naturally varying it from one to the other.
So 'quarters' being 'quadrants', basically ..
Naturally the normal meaning of quarters is valid for other reasons.

Cheers
[MP] Mobiles that perpetuate - external energy allowed
Post Reply