I'm tempted to conclude from the above that you're not really "getting it." You do need to stop talking about the thermodynamics of dissolving objects. There's no reason that this problem can't be CONSIDERED and DISCUSSED as two distinct processes: The mechanical process; and the chemical process. And other than your repeated statements that (paraphrasing) "It can't work, because the laws of thermodynamics SAYS it can't work," I don't feel you are DIRECTLY ADDRESSING the issues.

I feel like I'M being explicit, and YOU'RE being vague. Let me now be as explicit as I can POSSIBLY BE. I wonder if you will respond in kind. I think the apparatus I describe can be DISCUSSED as being composed of FOUR processes--each of which can be discussed independently. They are:

(First a reminder about the physical components of the apparatus itself: a) A tube of indeterminant height standing in a gravitational field, and filled with water plus the molecular/atomic constituents that can chemically be combined into an object less dense than water, b) A special "factory" at the bottom of the tube, able to collect the atomic/molecular components from the water, and synthesize the object. c) A special factory at the TOP of the tube, able to destroy the object and return it's atomic/molecular components into the water--in such a manner that they are freely diffusible. d) Some method of collecting the work-energy generated when the buoyant object travels up the tube.

Process #1: The bottom factory does it's work. The cost of that work WILL vary with the water pressure of it's environment--but I think for the discussion now, that variation can be ignored (I think it will be a small variation, I think it may find increased pressure an energy BENEFIT, and I think I have NO IDEA how it would actually function). I think (although this is a point that can be argued) that it's roughly accurate to think of it requiring about the SAME amount of energy to collect the diffuse materials and synthesize--no matter the height of the tube, i.e., no matter the pressure--within...a realistic range. Certainly, if you agree that the energy required for the synthesis might very well DECREASE with greater pressure, then we don't NEED to worry about this quantity.

Process #2: The buoyant object floats to the top of the tube. Clearly, the work generated is linearly proportional to the height of the tube.

Process #3: The factory at the top of the tube destroys the object and returns it's atoms into the water. The energy requirements for this process won't change from the height of the tube.

Process #4: Each atom diffused into the water at the top will EVENTUALLY make a visit to the bottom of the tube, in the natural process of diffusion/dispersal of atoms in a liquid. No energy is required for that kind of diffusion (we will assume that the temperature of the water remains constant).

I claim that logically, the sum of the energy requirements can (logically, NOT thermodynamically, NOT practically) add up to a net positive source of energy, based on this: Processes 1, 3, and 4 will require a (more or less) CONSTANT input of energy REGARDLESS of the height of the tube; while process 2 can output a VARIABLE amount of energy--which increases with the height of the tube. If input stays the same, but output can be increased arbitrarily, then output can become more than input.

I can't accept insistence that "output can't be more than input." I can only accept arguments intended to show qualitative errors in the energy requirements I describe for the separate processes (or, perhaps, that there are more processes of significance that I've overlooked).



The potential energy of an object in a system of forces isn't a property of an object, it's the property of the object's POSITION in the system. And yes, the ENTIRE CRUX of this discussion involves the question of whether it is possible to MOVE an object from a position of one potential to a position of another potential--and BYPASS the work (the difference in potentials) that would "normally" be required. My answer is RANDOM DIFFUSION of the ATOMS of the object in a liquid. Really, it is THERE, and nowhere else, where this idea must be undone, if it can be.

Now, of course, I'm not saying that diffusion and collection will not involve energy. But I am saying that the SAME ENERGY will be required no matter WHERE the diffusion starts and the collection begins, within the apparatus I described. Show me how I'm wrong about that, and I think we can proceed.



Not at all. I'm simply saying that I THINK the diffusion of atoms in a liquid occurs without energy cost (of travel--of course there are entropy changes). I accept that moving a SOLID object to positions of different potential energy requires the work well understood by thermodynamics, but moving the ATOMS of the object, via diffusion, to areas of different potential energy would NOT require ANY work. Maybe I'm wrong; and surely entropy is involved in diffusion and recollection of those atoms. But I don't see that entropy quantity varying by the DISTANCE between the point of diffusion and the point of collection.






Well. If I'm talking about HARVESTING energy, then it's not a closed system--but it's open in the REVERSE manner from usual--it doesn't take energy IN, and puts energy OUT.

You can't keep saying that it doesn't work because it violates the laws of thermodynamics. I'm saying that it DOES violate them. You have to show WHY it doesn't, not that it can't because it can't.

The only way forward in this discussion is to show at WHAT POINT my analysis is wrong--not to state that it's wrong "overall" because it violates the laws of thermodynamics.

Again, to simplify and hopefully move this discussion forward, I propose we temporarily make a tangent to address these DEFINITIVE QUESTIONS:

1) Is there an "energy cost," i.e., is "work done" by atoms diffusing throughout a liquid (I say absolutely not--but there are entropy changes, of course, by allowing those atoms to become disordered, and then collecting them to an ordered state?

2) Can atoms diffuse into areas of DIFFERENT PRESSURE? I think they can. If they couldn't, well, you couldn't even have a simple solution in a glass--every atom would go to either the top or the bottom and stay there. Convection also plays a role, and that inserts a slight bit of confusion. There's some O2 even at the bottom of the ocean.

If an OBJECT cannot spontaneously move from a position of LOW potential energy to a position of HIGH potential energy, but the ATOMS that comprise that object CAN do that very thing, then that explains how the idea could, in principle WORK.

Why not discuss this issue, and see if we can progress with it?


EDIT: BTW, this could work JUST AS WELL with the bowling ball example. Drop your bowling ball, vaporize it, allow the gas to diffuse, collect it up high, reconstitute your bowling ball, and drop it again. And in that case I would make the same argument (but it's even easier to understand here): The cost to vaporize the ball, and to reconstitute it would be the same, no matter at WHAT HEIGHT one decides to re-collect the vaporized bowling ball. But the potential energy DOES change based on the height one collects the atoms. Therefore, one could expect that, if one collects them at high enough altitude, the cost of vaporization and reconstitution could be overcome.

I would agree. The diffusion results in a very significant increase in entropy, so it's safe to assume that diffusion is energetically favorable.



Again, I would agree. Pressure should not play a significant role in diffusion of dissolved molecules.



Here's where I have to disagree. When you reconstitute the dissolved object, that's when you're going to have to pay the cost for the energetic difference. Whether you displace the water by pushing the object back down to where you want it or displace the water by reconstituting the object in the place that you want it, the energy cost will be the same, as I understand it. (Incidentally, reconstituting the object cannot be expected to be energetically favorable because the entropy of the system will decrease significantly, but as far as I can see, that fact isn't the crucial barrier that would prevent the idea from working.

And I know you're not interested in thermodynamic objections, but I can't help pointing out a relevant parallel, take it or leave it: internal energy is a path-independent variable. It doesn't matter how we get from point A to point B, the integral of infinitesimal changes along any path between those points will be equal to the difference in energy of those two points.



Okay. Let's forget thermodynamics for now and stick with classical mechanics. I'll leave aside the first law of thermodynamics, though I am still concerned with the law of conservation of energy. It seems to me that you're saying the law of conservation of energy does not apply with respect to the buoyant force, that the energetic cost of reconstituting the object will be based solely on the properties of the object, and not on the environment in which it is reconstituted.

I would disagree, based on the objection I outlined above. When you reconstitute the object, you are displacing water. The depth at which the water is displaced will have a direct influence on the energy required to reconstitute the object at that depth. I also have a second objection: the size of the reservoir (the minimum volume of which is dictated by the depth of the tube and the dimensions of the object) will also have an influence on the energetic cost; the larger the size, the greater the decrease in entropy will be (and thus the greater the energetic cost will be) to reconstitute the object.

To be honest, I don't have the time or inclination to refresh my memory of fluid mechanics enough to do a back-of-the-envelope calculation to demonstrate this. But I think that's the next logical step if you believe you can show that conservation of energy doesn't apply in the field of hydrostatics.


With regards to the bowling ball example, I'm not sure that's quite the same (miscible phases versus dissolved particles), but my objection is essentially the same: the potential energy of the air where you reconstitute the object will affect the cost to displace that air.

I completely agree with this, and I see it as an un-doing of the idea. THANKS!! I offer a bow

Now...what about the bowling ball scenario? In that case, one doesn't have to displace much of anything.

To remind: One drops an object from a height, harvests the work, vaporizes the object, and re-constitutes it at a great height. It's the same problem, but now there's no "displacement" issue. In fact, we can use ice as the object. Drop it, vaporize it, let it diffuse as a gas to a great height, and then freeze it and drop it again. BTW, if this were done in a mostly empty chamber (no air resistance) one could really achieve WICKED accelerations when dropped, and harvest a lot of energy from the fall (instead of losing it to the friction of air resistance).

Ah, now that's a little more of a brain-twister. You're right, the potential energy of a given volume of air at a given height will be less than the potential energy of an equal volume of a denser-than-air solid at the same height, so that won't pose the same issue.

I'll have to take some time to think about this. I'm sure there must be some mechanisms which would tie the energy of reconstituting the object to the difference in potential energy at its initial and final states...

I think the answer can be found in Einstein's derivation of the Boltzmann factor by considering the energetics of Brownian motion, but I'd have to think on it (and find a copy of this derivation) to be sure.

Diffusion of a gas is purely a randomization. The movement of gas molecules to fill any container is simply the direct expression of each molecule's kinetic energy based on its temperature--and it's fair to stipulate a constant temperature of the entire apparatus (sort of).

The gas WILL (just like the water) be at higher concentrations lower in the gravitational field and lower concentrations higher in the field. But there will still be constant MOVEMENT (on average) THROUGHOUT the container--as long as the molecules are above absolute zero.

Furthermore, if the gas molecules ARE bouncing around and losing heat to the container, that's only a BENEFIT, when the next step in the process is condensing the gas to freeze (although I don't claim to actually SOLVE the problem of how one would collect and freeze the gas efficiently--just that the molecules will APPEAR at the top of the apparatus EVENTUALLY, where they COULD be collected).

Forget buoyancy. This seems to be a more direct expression of the concept of "moving" an object to an arbitrarily higher potential energy without doing work proportional to the potential energy difference.

Ya know what? I think I'm going to concede.

I realize now that a gas disbursing in a container really IS only an idealized picture, and that, obviously, atoms can't break any rules anymore than objects can.

So, atoms can only disperse against a gravitational field as much as their KINETIC ENERGY will allow them to. In fact, if a "dispersing atom" of a certain kinetic energy were headed AGAINST a gravitational field, it would in fact "run out of steam" (no pun meant!), slow, and stop. At that point it would begin to fall in the field, and heat up to as high a temperature (kinetic energy) as it could in the space it had to fall.

That's no different than would be the case with an object. So, the atoms would only (on average) rise as high as the HEAT one put into them when they were vaporized at the bottom. The apparatus couldn't be infinitely tall.




HOWEVER...hmmm. This DOES lead me back to the ORIGINAL APPARATUS: The tube of water and the buoyant object. IF one COULD get that buoyant object out from the top of the tube at low energy cost, one DEFINITELY COULD (I'm gonna stick to this) harvest as much energy as one pleases--free energy. There's just no doubt (and here's I'm talking about the air pressure derived tube--ten meters tall, bottom of the tube open in a pool of water, and top closed).

So...why couldn't one get the buoyant object out? After all, the pressure on the object, at the top of such a tube isn't "zero." The object is being "pushed" away from the water, still. What's needed is some sort of...anus (hehe), that sort of "craps" the object out at the top, without allowing water to get out. You might need a little push to get the object out (as, in fact, you need in the actual human process of elimination--even though the baseline pressures are equal, inside and outside the body).

So, I'm saying: Original apparatus. And the only energy requirements to take the free ride up to the top of the tube are: 1) The small amount of energy needed to push the buoyant obejct under water. 2) The small (ideally NONE) energy to "crap" the buoyant object out the top, through some sort flexible valve-type apparatus.

And there would be the small added bonus of now being able to harvest work as the buoyant object then FALLS the ten meters, as well.

Sorry, but I really do take this seriously. I'd love it if someone could show me where I'm wrong, but I'm stuck on seeing this as TRULY PLAUSIBLE.

I'm very fascinated with the watertube concept too. I've been making sketches.


So you have a water-container with a tube in it, a valve-system ontop and a somewhat heavy, yet buoyant object.
If the object falls, next to the tube, a sort of guidance rail in the water-container below could guide the object back under the pipe-entrance.
If the object is heavy enough, the power off the fall will push it under water and the rail/track will guide it
back under the pipe-entrance before it starts floating up again. That would require no energy.

If that works then the only energy you'd need to put into it is that needed to get the object out of the top of the tube.
I don't imagine that needs alot of energy.

I imagine the valve ontop of the tube to be some kind of airlock.
I shall sketch a few valve designs and see if anything sensible comes up.

It's almost at a pure calculation stage. You could imagine a ball of styrofoam (and find the density of styrofoam).

1) Energy required to push it underwater by its diameter.
2) Energy created by it travelling up the tube.
3) Pressure difference between the water at the top of the tube and the air outside
4) Pressure exerted by the water on the styrofoam when it's at the top of the tube
5) For the sake of "theory," calculate the energy needed to push the ball entirely into the air from the water (assuming a "perfect" "anus-type" valve on top).
6) Energy created when the ball is dropped.

Practically, of course, the valve is going to require more energy than just the difference in pressures (which might actually be negative: i.e., the pressure of the water on the styrofoam might be GREATER than the outside air pressure--so the ball would "poop out" the top without any energy added. And, of course, if THAT'S the case, then there's NO QUESTION that this would be a free energy generating apparatus--and one that wouldn't lose water out of the top when removing the ball, which would eventually make it unusable without "repairs."

EDIT: In fact, it seems to me that the force that the ball would exert on the anus valve would be identical to the force required to push it underwater initially. No kidding, this is starting to seem interesting. It might even be possible to DESIGN such a valve that would, sorta kinda, almost work just as I describe--and the ball would ALMOST push itself right through, with no air coming in!

Another Note: The "anus valve" could actually be a flat IRIS-TYPE mechanism! Top of tube is flat, iris plate is flat, ball is a sphere--and it will be pressing on the iris from the water pressure. As iris opens ball will come through it, and then begin to close at the half-way point. In principle the water will have EXACTLY ZERO tendency to rise UP out of the tube, because it will be at EXACTLY the maximum height that can be held up with air pressure, and the ball WILL have pressure to come up away from the water vs. air pressure--just as it did at the bottom of the tube.

I really igmire your enthusiasm, SWImfriend. I'm very passionate and positive about this too.

The Iris could be forced shut by springs that will give way under the upward, buoyant pressure of the floating ball, but close immediately again after crapping out the ball.
The water in the tube won't have any tendency to come out of the tube, but what you'd need to assure is that no air whatsoever can enter the top of the tube as the ball is crapped out.
If the Iris opens and, besides letting the ball out, sucks in some outside air, the pressure would be lost and the water in the tube would sink back into the container in which
the tube is suspended.
So if this Anus is to work it better be completely airtight.

I was thinking of an airlock myself. An extended piece of tube,on top, shut off by 2 valves and filled with air.
Valve 1 opens; floating ball enters airlock. Valve 1 closes. Valve 2 opens; Allowing floating ball out of the airlock and out of the pipe to
fall down and drive an electric generator on the way, before falling into the watercontainer and being guided back under the entrance of the pipe.
I guess you could operate this valve system with no energy-input at all if the floating ball were to be filled with a Metal core and magnets were used in the valve system.

Well...

Just be aware that others have claimed similar capabilities with the use of "magic valves." Of most interest to me would be actual calculations showing a net energy gain--and those calculations depend on the working of the valve, which is difficult to quantify.

In IMAGINATION, it does seem fair when I picture the buoyancy of the object serving to push the ball through an imaginary valve--and if that really happened in such a way, there's no question whatsoever that energy would truly be produced by this apparatus. To really convince people in a visceral way, it would be great if one could actually MAKE such a valve, that maybe only required minimal extra energy to function.

LOTS of technology is being modeled on biology--and this would be a useful venue for that, IMO. So such a valve could be very flexible, slippery silicon/rubber type material, with one or two rings of "muscle" embedded in it (in an iris fashion) that could be relaxed and squeezed shut as the ball passes.

Theoretically though, it would be nice if the basic physics/calculations could show that pressure would favor the ball actually spontaneously exiting such a valve, based on buoyancy pressure. I THINK that's true--but, honestly, I can't claim to be certain about it.

calculations, speculations, and theories aside...have you experimented and taken measurements?
what was the control?

No. Of course not. If you hadn't read the entire thread, I got started on this last week when viewing a youtube video of an upright tube of water, held up by air pressure. The fellow in the video placed a buoyant object into the bottom of the tube, and, of course, it floats to the top--which involves doing work.

The Traveler earlier pointed out that the undoing of this machine is the difficulty of getting the buoyant object OUT. The guy in the video had the idea of having a valve BELOW the top, and below the buoyant object after it had risen. You close the valve and take the object out--then do it all again. The problem is that the object will be DISPLACING A QUANTITY OF WATER, so there's NO POSSIBLE WAY to avoid allowing some small amount of air in WITH EACH TRIAL (even though the valve BELOW the ball prevents air from whoshing in and sinking the whole show).

So it would seem that if the buoyant object could somehow...DISAPPEAR...that you could in fact gain work by such an apparatus.

Most of the thread involved a tangent in which I considered that one could in fact CHEMICALLY disintegrate and reintegrate the ball at the bottom--after the atoms had dispersed there. Of course that would hardly work for a 30 meter tube (the max that air pressure can hold up), but it MIGHT work for an "arbitrarily tall" tube, that was sealed at the top AND bottom--whereas the chemical work would remain the same. But that idea sort of fizzled for a couple of reasons (one being that it would be HARDER to physically displace water at the bottom of a taller tube rather than a shorter--even if the chemical energy expenses stayed the same.

But then I realized that indeed the buoyant object would put some PRESSURE on the top of the tube--because water will continue to push on it (because it's simply less dense than water), and it might be reasonable to assume that it could sort of "poop out" of the right kind of valve (almost an "anus type" apparatus, or some sort of iris type of apparatus)--without significant amounts of work being done to run the valve (since I think the actual PRESSURE differential would FAVOR the object being "pushed" out of the water and into the atmosphere.

Mostly it's just a thought experiment at the moment (but the more I think about it, the more interesting it seems it might be to actually begin to LOOK INTO what would be involved in trying to design such a valve).

In any case, the idea of a CHANGEABLE object within a field of static forces has made me think of ANOTHER scenario (that I haven't discussed) where, I get the idea, it might VERY POSSIBLY WORK! Again, the main idea that intrigues me is that the "unbeatable" situation of getting no more work from the movement of an object than you have to use to RESET the apparatus could be "circumvented" in some ways, if the OBJECT ITSELF can bend some rules regarding its own "existence."

I guess a transparent pipe with an inner diameter of minimally 10 cm and a length of at least 1,5 meters would be ideal to experiment with.
A large bucket or an aquarium could serve as the watercontainer.

A metal ball with a styrofoam ball around it, made watertight with latex paint could be the floating object.
It will be buoyant, yet have mass enough to fall with enough speed to yield significant dynamic energy.
The metal core should also provide enough weight so that when the ball falls down it plunges underwater.
Then it just needs to be guided slightly sidewards under the pipe-entrance before it floats up again.
A rail going underwater could guide it like that.

I'm not quite sure about the top-valve part yet. That's why I'm going to keep theorising for a while till I come up
with a proper working valve-system.
Then I will know exactly what materials I need and how to build it.

It's not necessary to have a metal ball in the styrofoam. The object produces work on the way UP and falling down outside the pipe. The LIGHTER it is, the more work it could produce on the way UP, and the heavier, the more work on the way down--they sum to the same figure (not counting inefficiencies in collecting the work, of course).

The heavier weight will be easier to push underwater initially, but the light weight will be easier to lift out the top (or will put more pressure on a valve). These factor, too, will even out (I'm pretty sure).

Swimfriend,

If you approach this problem as a physicist, scientist, or arguably just as a rational person, creating energy from nothing is simply not possible. The second law of thermodynamics is not important to show this. In the very first line of the first response to this thread Trav pointed to the fundamental reasons why this cannot work. More explicitly, the potential energy in the system with the tube full of water and the ball at the bottom is Mball *g *Hbottom + the sum Pwater * Vwater *g *Hwater up the columb where M is mass, H is height, P is density, and V is volume. after the ball has floated to the top the potential of the ball is Mball *g *Htop and the potential energy of the water is the same as before except subtract the mass of the water displaced by the ball from the top of the columb and add it to the bottom. ie, the total potential energy has decreased by the difference of the mass of the ball and the water it displaced. This is the only energy available to open valves, pull pulleys, magnets or whatever you can imagine. From a physics perspective you simply cannot get more energy from this process than this, it is the law of conservation of energy.

Now when the ball drops back down, the potential energy of the system is lowered again still. In this case the maximum energy you could use is Mball *g *(Htop - Hbottom). If you repeat this process the water will drain out of the tube completely. The total energy available before the water is totally depleted is simply the potential energy stored in the system at the very start.

Now, ignoring the second law of thermodynamics allows a perpetual motion machine, as in the ball could potentially just keep circulating if the kinetic energy of the ball and water moving were used to keep refilling the water in the columb. You have confused a perpetual motion machine with a spontaneous energy generating machine, which is even more absurd. The second law of thermodynamics implies that the ball cannot circulate in this manner perpetually even if you so called "harvest" zero energy from the system because some of the energy will inevitably be lost as heat.

I never learned the physics of objects disappearing and reappearing whenever it's convenient so I won't comment on that scenario.

ouro, I'm not sure you're understanding the setup.

And yes, I use the phrase "perpetual motion" as a shorthand for an apparatus that "creates" energy. And yes, I'm fully aware that such a position contravenes the laws of thermodynamics.

I have an updated version in mind, that nicely solves the water loss problem. A bit too late to post it tonight. When I get around to posting it I invite you to consider it--if you can resist the boring response that it "can't work because it violates the laws of thermodynamics." Everybody already knows that answer. If you can show SPECIFICALLY why it can't work, then do--it would be much appreciated. As it stands, I don't think you're really understanding it. You say that the trip up the column is to a lower potential energy (that's right, going from higher to lower potentially performs work, i.e., "creates" energy), and falling outside the apparatus also goes to a lower potential energy, and that's right, and more energy is created. Each of those steps "create" energy--which could be harvested. The only energy that would be used (with the valve on top I propose) is the very small energy to initially push the object underwater and into the tube (MUCH less than the amount released over the LONG travel up the tube), and the energy needed to "lift" the object up a small bit as it exits the valve (MUCH less than the amount released as it falls the length of the long tube).

I'm sorry that "disappearing" is a concept that confounds you. I use the term only to hint at what I see is the "special" part of the apparatus. As it is, the "special valve" (which I will update tomorrow) allows the object to "escape" the position of low potential energy (in relation to the water), into the position of high potential energy (in relation to the gravitational field, in air).

SWIMfriend, you're not just claiming to contravene the laws of thermodynamics, you're claiming to contravene perhaps the most fundamental law in physics: conservation of energy. Consequently, I'm quite confident that if you were to actually calculate the net energy that could be derived from the system you describe, you'll find that you are not getting free energy at all, but (at best) merely extracting a finite quantity of energy from the system.

I think I've spotted the logical hole in the butt valve notion. The buoyant force will only push the object so far, leaving a portion of the object submerged; as I recall, the mass of the object divided by the submerged volume is equal to the density of the fluid. Now, recall that the pressure maintains the water at a particular height in the tube. If the butt valve is constructed to maintain the pressure (and thus the height of the water), then to lift the object out of the water will require the input of enough force to pull the volume of water that the object has displaced from the tube... and the energy to pull that water all the way from the surface of the reservoir to the top of the tube will be greater than or equal to the potential energy of dropping the buoyant object that same distance.

No matter how many clever modifications are made to the idea, you're always going to run into this barrier; you can't just pull energy out of nothing, it always comes from somewhere. I very strongly doubt that you can outwit conservation of energy, and certainly not on the macroscopic scale.

Sure. I do too. But one only LEARNS by coming to understand. When a question arises for me, I'm compelled to ANSWER IT--in a manner more convincing than just saying "it can't work."

More later on today when I have some more time to post. I think you'll like my new valve idea! I hope The Traveler will comment on it (but I think he's been very busy with his personal/business stuff for the last week or so), because it relates directly to his initial analysis.

OK, I think I see where my error has been (because I thought of a REALLY COOL sort of "anus" valve setup that seemed would HAVE to work...)

Then I realized: in "removing" the object from the top of the tube, one would have to OVERCOME air pressure as well. That no matter how you arranged it, pulling the object INTO the air, FROM the tube, would be like pulling the object through a tube against a VACUUM on the other side.

When you pull an object out of water "normally" you are ASSISTED by air pressure flowing in from all sides. If you're in a situation where that can't happen, then you must lift AGAINST air pressure--and that's a LARGE FORCE. Large enough, I imagine, to be the thing that "balances the work equation" that I just couldn't envision previously.

You can get around it, of course, with the original setup of valves beneath the object, but then you can't avoid air entering the system and eventually "using up" its capacity to do work--in fact you would (I'm sure) "use up" the work stored by the weight of water in the column, with each bit of air that gets in. When you try to AVOID THAT (by a special valve) then you must move your object out AGAINST AIR PRESSURE, which is really quite a large force--and I can easily imagine that undoing the scenario.

Of course....nobody stepped up and POINTED THAT OUT. It's not enough to simply say to these things "it can't work because of conservation of energy." The undoing is showing WHERE the balancing forces lie.

So, entropymancer's instinct at one point was correct: When you initially lift the column of water, you are "storing" work, and running the apparatus would "use up" that work--in this case by allowing air in. In you DON'T allow the air in, then you CAN'T get work from it (I'm assuming with some confidence), because MORE FORCE THAN EXPECTED is necessary to remove the object from the top of the column without allowing air in.

This is how I am. I simply can't allow things like this to rest until I FIGURE THEM OUT!

EDIT: Just to help those who might not easily see it...Imagine a flexible membrane lying atop water; no big deal. But imagine a flexible membrane lying atop the opened column! In that case the membrane would STRONGLY deform downward. The only reason the water of column can exist is because air is pressing on it FROM ONLY ONE DIRECTION. If the opportunity arises for air to press at the top of the column, the pressure would be VERY large.