Has An Important Property Of Fluids Been Overlooked ?

[daanopperman]

Hi all ,
Substitude the mass with equal amounts of water (same mass ) and look again . If the water in the containers cannot increase it's com because it is constrained by the container the weight of the mass is where it is put on the arm . In a open container the com of the water will rise equal all over .

[daxwc]

Fletcher, I understand you want to use it as a COM switch, but does Com really move? Isn’t this a type of Roberval balance?

Does no Work on the system per se but creates a real system CoM change almost instantaneously [the switch], while there is a virtual displacement of mass occurring internally.

We have just redistributed mass.


I guess strangely it is like saying I drove my car to town and back but it did no work because it is back in the parking stall.

A see-saw that moves for free? But is it free, you still have to unlatch the mass.

[Fletcher]

Hi guys ..

Roberval balance ? - essentially that's what I been thinking about - a system that is balanced in one orientation which can be unbalanced in another.

dax & mickegg .. does the CoM move ? - I don't know but that's what I'm trying to find out.

If the mass in contact with the membrane [plunger, piston etc] creates pressure which is increased all thru the fluid then the mass's force [weight force] is now acting over a large internal area [force as pressure] & not directly down beneath it [as in the descending side mass not in contact with anything except the cage] - so you could say its weight force has been redistributed else where & evenly - i.e its nullified by having its net influence at the pivot, I think.

This would be like shifting the mass to the center of the structure [if it can happen this way] - since the descending mass is not in contact with the fluid its weight force acts straight down on an arm - so this arrangement would be a sort of switch which displaces weight force on one side turning it into pressure thru the fluid.

Dan .. we know a floating object displaces fluid which rises in an open container & the system CoM rises - in this case I am trying to substitute open systems with buoyant objects [where their mass is redistributed evenly] with a no volume change environment but where pressure changes as a result of contact with the mass & can its influence be felt elsewhere just like the floating example ? - the upshot being not CoM raising but shifting sideways ? That's the virtual displacement I talk about.

.....

It's a hard concept to visualize & a hard one to test.

[daxwc]

Although you could get it to tilt and fall by latching, it might not rise back up in all cases (single ball)

This is because of where you are applying your pressure (unlatch tilted and weight on the left), it will not be at surface. You will have to fight the hydrostatic pressure head of the true vertical depth from left to right. Imagine that true vertical depth as just a long vertical pipe filled with water. If you wish to inject a pressure at the bottom to equalize it you need to overcome your hydrostatic head first. This is because it is not a surface applied pressure now.


One ball and you are fighting hydrostatic pressure and might need a reservoir at the end at the opposite end. So your better system would be the two balls but problem you actually have to lift the ball to get the weight to latch in as the caged ball will be down
.

[jim_mich]

Fletcher, as I understand your concept... you start with a balanced fluid system. Then you add the weight to one side. This causes an unbalanced system before any fluid flows. But as the weight sinks into the fluid, by way of the membrane/piston/plunger, it causes fluid to flow over to the other side and the system becomes once again balanced.

What you seem to be proposing is to let the weight push on the fluid, but at the same time prevent the fluid from flowing over to the other side. If you do that then the system will remain unbalanced. It would act like the water was frozen. Which would act like any solid lever-type structure. Which is easy to mentally visualize.

Stated differently, if you prevent the weight from sinking into the water up to its natural buoyant level, then the back pressure that is keeping the weight from sinking, is the excess force that keeps the system unbalanced.

Hydraulics is simply a fancy leverage system where force and distance is converted into volume and pressure and then the volume and pressure get converted back into force and distance. You seem to be trying to eliminate the distance part while keeping the force.

I don't know if I'm making any sense here?

[Trevor Lyn Whatford]

Hi Jim, Fletcher,

It makes perfect sense, but still will balance though!

Regards Trevor

Edit, More correctly it will move to a balance position, but far from make the reset!

Edit again, I have seen this in action on my lever operated reservoirs, wherein the fluid pressure lifts the lever with the worst angle to gravity’s pull, Like Bob said on another post, and you have said here, you have to switch or latch the system, that is why I went for hydraulic force transfer to use the stored leverage force when you wanted it, and the timing and latching would have been valve operated. I gave up on this out of balance stage, and just went for using the levers to drive hydraulic motors to drive the wheel Directly!

Edit again, Hi all, of topic a bit! I want to make it clear that I believe wheels like MT 114 can be closed loop out of balance wheel systems, if you do not get to greedy with them and try to move the fluid to high, to stand a chance you would have to move the fluid in and out on there relative sides with the correct timing and latching, with more of them, and the levers as short as possible! Just want you to Know that I still believe they can work!

[Tarsier79]

Fletcher, thinking about it, it should be very easy to test in a one shot experiment, using some simple plumbing. You dont even need the weight, increased static head of the water should be a good enough equivalent to test, provoding the hardware and horizontal tube are equivalent.

[Fletcher]

Thanks guys for all your thought out comments & taking the time to try & understand - I appreciate them all.

I'm a little short of time to answer individually today so I drew up a quick pic which I think explains things more clearly about what I'm trying to achieve & how I propose that it might be done - I made the relevant parts stand out much more & added some detail so its easier to see where I coming from & trying to head to - whether you agree or not with the proposition is another thing ;7)

In essence the the plunger is low mass on which rests the mass 1 [ascending side] - it acts like a hydraulic piston with very little volume movement of fluid but pressurizes the fluid considerably - the contention is that as pressure at the plunger face increases it makes the plunger mass 1 effectively buoyant because according to my reasoning buoyancy is due to Archimedes Laws of equal fluid volume displacement OR fluid pressure differential between the plunger face & the air gap [mine].

If it has a buoyancy effect it should be like ordinary buoyancy where an object floats in a fluid & its CoM is nullified because density at the water line is constant - in this case I'm proposing that the same thing might happen but due to pressure increase rather than fluid displacement.

[jim_mich]

This is way more complicated than needs be. You can eliminate the yellow mass, the gray piston, the cyan extended housing, and simply use the plunger as the mass, making it weigh whatever needed.

You will notice that as the plunger moves it causes fluid to move from one end to the other end. This is simple hydraulics. The right side will get heavier as the fluid flows to the right side. The problem is that this flow starts to happen just at the point when Mass_1 rises higher than Mass_2. The flow will start too late to be of any benefit. I've talked about this latency effect before. It happens whenever we attempt to use gravity to move weight.

BTW a condom makes for a most excellent membrane, plunger, piston if you size things correctly.

[daxwc]

Fletcher

in this case I'm proposing that the same thing might happen but due to pressure increase rather than fluid displacement.

At first I thought no, but in the latest diagram I see your added applied pressure at the surface and expecting the right side to rise. It might work you will have to test it, never heard anybody talk about it; your right it should equalize the applied weight of the mass and rise. It is a little counterintuitive, but so is the saucer in the pan.


It will be hard to get it to work both ways because of the upside down piston. A test without the right side piston and just compensated mass would be best for testing.

[daxwc]

Tarsier79

The Fletcher, thinking about it, it should be very easy to test in a one shot experiment, using some simple plumbing. You dont even need the weight, increased static head of the water should be a good enough equivalent to test, provoding the hardware and horizontal tube are equivalent.re going to be problems with friction and wall contact.

Yes, I think so; as long as you compensate the other side of the lever for the added weight of the hydrostatic head added and container with a solid mass. Then you don’t need any pistons to test the principle.

[Fletcher]

This is way more complicated than needs be. You can eliminate the yellow mass, the gray piston, the cyan extended housing, and simply use the plunger as the mass, making it weigh whatever needed.

Yes, I agree - the plunger/piston can be an all in one providing that no fluid gets up the sides.

You will notice that as the plunger moves it causes fluid to move from one end to the other end. This is simple hydraulics. The right side will get heavier as the fluid flows to the right side. The problem is that this flow starts to happen just at the point when Mass_1 rises higher than Mass_2. The flow will start too late to be of any benefit. I've talked about this latency effect before. It happens whenever we attempt to use gravity to move weight.

BTW a condom makes for a most excellent membrane, plunger, piston if you size things correctly.

I over exaggerated the conditions in the pic a tad - the idea is that very little fluid actually moves - just like a hydraulic press Input force pressurizes a fluid in a vessel before the Output piston moves & releases some of that pressure [the analogy being that static friction is greater than dynamic friction].

In this case there is no Output piston, just the Mass1 pressurizing the internally contained fluid until mass [weight force] is in equilibrium with underside pressure [force] - that pressure increase, for virtually no fluid movement, is transmitted all thru the fluid as per Pascal's Principle - pressure builds on the underside of the plunger/piston face - since one way to look at buoyancy is as a pressure differential then that means that the increased pressure at the face is far higher than the atmospheric pressure in the air gap - this should cause a buoyancy effect on the plunger/piston i.e. equilibrium of forces.

N.B. Archimedes buoyancy doesn't increase with depth in non compressible fluids at any depth because the pressure differential remains constant because of the constant density & linear relationship of pressure differential is maintain with depth.

In this case I am suggesting that buoyancy of piston isn't volume dependent but pressure differential dependent [the hydraulic press component] which is greater than Archimedes because of Pascal's Principle at work between pressurized fluid & atmospheric pressure - fluid density doesn't change however because we assume non compressibility.

EDIT: the idea came from studying force pumps.

[mickegg]

Hi Fletcher

As a stepping stone to possible acceptance (my brain's having a hard time with this one <grin>) ..I'm imagining your same setup but with a Tee piece extension doubling as the fulcrum/axle connected to a stand pipe via a rotating seal.

Now one could implement the principle by displacing balance fluid up the central pipe...I know you don't want to move the fluid, but this is helping me try to get to grip on the idea.

The weights mass is now tranferred to the center?

If so, can the balancing column of fluid be replaced by a piston and spring with the correct preload?.... and if so can this then be sited in the Tee extension doing away with the seal and stand pipe?

Regards

Mick

[Tarsier79]

Gday again Fletcher

I have reduced your thought experiment to (nearly) its simplest form. I think this should be an easy enough to build as well. You don't actually need the diaphragm, but could add weight to the RHS without the empty tower to balance the structure before adding water.

At the top.. obvious balance.

The bottom, we have increased the pressure by raising fluid head height. I think someone mentioned it before, you replace the weight with the equivalent water mass. If your theory is correct, it should still balance, correct?

[jim_mich]

If you seal the interface (i.e. use a leak-proof piston or such) then you have created a hydraulic lever. If both sides have equal piston face area then the leverage is 1:1. Any movement of one mass results in equal movement of the other mass. In the position which you have drawn, both weight mass will sink lower until stopped by some limiting factor. If the assembly were to tilt CCW until the CoG of one weight mass matched horizontally with the CoG of the other weight mass, then the hydraulic pressure of the two weights would balance. And the wheel would balance.

Now if very little fluid moves from side to side, then correspondingly there will be very little weight movement. The motion of a piston is always directly proportional to the flow of fluid.

As a side note, all fluids have greater pressures at lower points than at higher points. This is usually ignored. Water gains about 1/2 pound per foot of depth. The weight of about 30 feet of water will approximately equal atmospheric air pressure, thus you cannot suck water any higher than about 28 feet. If you have a 12 foot Bessler type wheel containing water, then the pressure at the bottom would be about 6 PSI greater than the pressure at the top. But I seem to be rambling...