Has An Important Property Of Fluids Been Overlooked ?

[Fletcher]

My proposal with the symmetrical structures in rotary format is that forces can be redirected - this would bring into question one of Newton's Laws, specifically the one of an equal & opposite reaction - equal forces in magnitude but not opposite, if I am right.

The gist being that if forces can be redirected to one side of a pivot then that symmetrical format will receive a repeating impulse of force giving it momentum & KE - because it is symmetrical [effectively] it will act like a flywheel & coast for large portions when not receiving an impulse of force.

EDITED TO ADD:

It is a symmetrical system therefore it is a flywheel able to store momentum & KE.

There is no appreciable mass movement in a F x D [work done joules] sense - it stays fairly well symmetrical with its CoM [as we describe it] located at the Center of Rotation [CoR].

Non compressible fluids [liquids] under these sorts of pressure maintain uniform distribution of mass & their volume does not change [appreciably], aka density is constant - that means there is no lowering of system CoM & loss of PE, aka symmetrical.

When a force is applied to a contained fluid it builds pressure in that fluid - in hydrostatics that means an increase in the energy density of the liquid i.e. energy per unit volume - the reason that liquid density [& volume] remains constant [at non extreme pressures] is varied but I think of it like this - liquid molecules are tightly packed into a volume with no spaces between them, but they are able to move in all directions [fluids are isotropic] - so when they are squeezed by applying a force they can't easily move closer together - pressure creates heat & that energy must be radiated into the environment - so the radiation pressure exactly equals the forces trying to push molecules closer together [like the sun thermonuclear system in equilibrium] & so we have equilibrium of forces.

If the application of a mass weight force increases pressure in the liquid & this increases the liquids energy density then some things become apparent - first off, the liquid sealed piston with mass on top must be in equilibrium of forces with the pressure on the piston face, so the buoyant force equals the mass weight force - this is in synchronicity because the greater the mass the higher the pressure etc [self adjusting to find equilibrium of forces] - the second thing is that if the increase in liquid pressure has increased the liquids energy density per unit volume [volume didn't change] then effectively the mass's PE is absorbed into the liquids increased energy density - I don't have a better term for it at the minute but it conveys an idea.

Next, this concept is one about redirection of forces on one side of a pivot - Newton's Laws of equal & opposite reactions might have an exception i.e. equal force but not always opposite.

I am attempting to redirect the ascending side [left side] mass's weight force into the liquid medium at the axle due a special type of buoyancy which is just an extension of the hydrostatic paradox of which Archimedes paradox is an example.

So, if forces can be redirected into the fluid medium [MUST BE UNDER PRESSURE FROM THAT MASS] then the rotary format will be given an impulse force so that it gains momentum & KE - when not receiving that impulse force because it is effectively symmetrical it coasts like a flywheel - so you have a flywheel that self accelerates thru periodic & repeating force impulse.

The crux is to find a verifiable experiment to prove that a mass/piston analogy pressurizing an enclosed fluid acts & behaves just like a buoyancy force weight force equilibrium as per Archimedes volume displacement method.


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I added the bell housing compression chamber & gas bags etc into the concept presented because some people have a hard time grasping that pressure can be built in a contained liquid without the piston actually moving [ f x d ] - it was the same reason for pictorial purposes that I show a small air gap by each end mass - it's more familiar.

Additionally, whilst liquids maintain their volume under temperature change, metal casings/containers etc do not - they have different thermal expansion rates etc - I could add alcohol [methanol & ethanol etc], to say water, to stop it freezing & expanding & blowing the system I could also just be prudent & allow for naturally occurring flexing & small changes that might lock things up otherwise, hence the compression/expansion chamber - btw this could be located anywhere almost providing they are symmetrically balanced around the CoR.

[Tarsier79]

Gday Fletcher. I have done that exeriment in the past. The problem is you are providing that force. Your finger will be buoyed up by the same force applied to the scales, just as in my experiment, it is the pressure in the bag pushing down on the scales.

[Fletcher]

Yes, Newton's Laws - you will note that it doesn't matter where in the beaker you put your finger in, the scales still read the same weight [higher than before you put your finger in] - your immersed finger produces a higher pressure on the bottom of the beaker & since pressure is the same all over the bottom of the beaker that can be represented as a single force vector [p = f / a => f = p x a] right in the center.

The scales is accumulating forces & interpreting them as weight force i.e. weight force of beaker & water + buoyancy force uplifting against your finger - the upshot is that buoyancy force is an artefact of gravity & the greater the pressure differential top & bottom surfaces of an object the greater the buoyancy force on that object - normally this occurs in a fluid where pressure increases linearly with depth - it doesn't matter how the pressure underneath the object got there, just that it is high enough in relation to the top pressure to cause a buoyancy force equal to that body's weight force.

Next I ask your cooperation in another experiment.

[daanopperman]

Hi Fletcher ,
If you stand on a scale and submerge your finger in your beaker , you will find the scale you stand on has lost weight equal to the amount of weight gained by the beaker scale , therefor you have put weight in the beaker and it has nothing to do with pressure inside the beaker , the only pressure increase will be the extra head the water has because of the volume of water displaced by your finger .

[Tarsier79]

So, what experiment do you think might help?

What about the following?

ADD: two separate containers (tubing), with equal horizontal distribution, but increased pressure on one side only.

[Fletcher]

Tarsier ..

I'll get back to you as soon as I can.

What do you think will happen in your drawing ?

I have changed the fulcrum from underneath to a hanging solution which is easier to build & far more forgiving - you just need to see a tilt.

[rlortie]

Either I have missed an inning or two of this game or I am confused with what the issue is!

Question:

If I install a regulator in the left tube just to the left of the vertical column and a counter weight on the right side to make up for the weight of the regulator, the device is back in balance.

With same amount of liquid (incompressible mass) in both right and left tubes, applying pressure to the right, how does this create an imbalance other than the density of the compressed air or gasses making up the compression factor?

Ralph

[Fletcher]

It doesn't create an imbalance Ralph & that was the point.

Below I've simplified the pic & assumed we have pressurized the right hand side section.

[Tarsier79]

So how is this different to your proposal? The verticql column of water increases the pressure, not unlike a weight increasing the waters pressure. I don't think there will be a virtual displacement of the COM.

[Fletcher]

See the following diagrams - do they make sense to you ?

First there is Archimedes volume displacement buoyancy - you can put the floated mass anywhere & maintain balance.

N.B.1. note that the the mass was floated up from the bottom of the tank by addition of water - if the mass in the first few pics were on the bottom before water was added it would be unbalanced.

Then I narrow the container until we have the Archimedes Paradox [the ship floated in a bathtub] i.e. where the fill volume is less than the displaced volume.

N.B.2. the Archimedes Paradox is just the Hydrostatic Paradox.

Then I fill a container where there is no gap wide enough for water molecules to rise up the sides of the float/piston - water is filled thru a snorkel - it still floats just the same - I can add more water & the piston rises - [side pressures are nullified anyway].

Can we agree this far that buoyancy is pressure differential related, at least as presented in these examples ? If not, why not ?

EDITED: to fix the pic - I'm flat out these last few days & for the next few so mistakes & response delays can happen ;7)

[rlortie]

So how is this different to your proposal? The verticql column of water increases the pressure, not unlike a weight increasing the waters pressure. I don't think there will be a virtual displacement of the COM.

I think I am starting to understand this epic debate and have two questions. I may be sticking my neck into something that may backfire, but I have broad shoulders and not much between the ears.

1. An increase in a vertical column will raise the PSI on the bottom, but how does the displacement of a buoyant weight increase water pressure other than by elevation?

2. I believe Fletcher has done a good job with his six diagrams, I do however ask about #6, it appears to me that the fluid poured into the snorkel will simply back up to a point of overflow?

A buoyant weight displaces its weight with water, there is no change in pressure but if it is setting on the bottom then it is displacing only its physical mass and water pressure still does not change.

Ralph

[daxwc]

I agree all 6 are right. It is your next assumption that COM will move without the mass displacement. My opinion is it will compress till the mass is held back by the fluid, at that very instant the weight will be felt where the mass is.

It is an easy enough test ....this whole debate could have been over by now; if I was home I would do it.


I think Kaine's test should have gave some indication if it were true. His other test with the fluid head would have proved it one way or the other.

[Tarsier79]

That is exactly my point Dax. I was wondering if Fletcher could see the correlation between my proposed experiment and his "virtual mass displacement" mechanism.

Yes Fletcher, I agree all 6 diagrams are correct, and in number 6, depending on the mass, the fluid height will eventually cause it to shift. Personally, I always examine these designs as a mass displacement problem, not a pressure problem.

I agree with Ralphs take on buoyancy mass increase. The three diagrams at the top are balanced, because the mass distribution is also perfectly balanced. The floating block weighs as much as the water it displaces.

[Fletcher]

1. An increase in a vertical column will raise the PSI on the bottom, but how does the displacement of a buoyant weight increase water pressure other than by elevation?

2. I do however ask about #6, it appears to me that the fluid poured into the snorkel will simply back up to a point of overflow?

Ralph

1. water pressure increase via displacement is linear in an open system Ralph - the how I've already mentioned but will go over again.

2. You start with the conditions being near frictionless then add back frictions later, like viscosity Ralph - if the piston is near frictionless, say a ceramic sleeve & piston head [very good tolerances] then the water won't back up much if at all - if the piston is sticky then yes extra fluid head will be required to get it moving then it will settle lower i.e. stiction, static & dynamic forces.

I agree all 6 are right. It is your next assumption that COM will move without the mass displacement. My opinion is it will compress till the mass is held back by the fluid, at that very instant the weight will be felt where the mass is.

Yes Fletcher, I agree all 6 diagrams are correct, and in number 6, depending on the mass, the fluid height will eventually cause it to shift. Personally, I always examine these designs as a mass displacement problem, not a pressure problem.

Dax & Tarsier .. those propositions are what I'm testing - you'll have to give me a little more time.

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Comments or disagreements on the drawing below please ?

The piston is floated - its density increases to that above water density at say 1000 kg/m^3 & the volume decreases - still the pressure differential exists to keep it buoyed - we know this because the snorkel doesn't overflow & forces are in equilibrium.

[Fletcher]

Restated: The hypothesis is that buoyancy force is solely dependent on pressure differentials & not on volume displacement & uniform density doctrine.

The proposition briefly is that in a symmetrical system as described that mass 1 induces pressure increase in the structure fluid [increases energy density] - when that pressure is vectored into thrust [up-force against the piston face] then mass 1 is effectively buoyed & its weight force is in equilibrium with the up-thrust force from the fluid pressure beneath it [the vertical component of gravity] - this is akin to buoyancy - because of Pascal's Principle mass 1 creates the internal pressure increase by applying its weight force in the form of pressure increase which is spread to all parts of the fluid equally - therefore the proposition is that the structure will have one centralized position where vector forces act & that will be at or near the Center of Rotation [as opposed to the downward side where mass 2 has no influence on the fluid pressure relationship so is independent of it] - furthermore, since the entire structure is symmetrical & designed to rotate all parts have a degree of freedom of movement - at any stage whilst in motion the internal fluid pressure against the piston head will be in flux, waxing & waning at the small level, so some lag occurs & equilibrium of forces can not be established unless it is stopped.