Wikipedia: There is an exception to Archimedes' principle known as the bottom (or side) case. This occurs when a side of the object is touching the bottom (or side) of the vessel it is submerged in, and no liquid seeps in along that side.
Picture 1
A ('brown') container contains a fluid.
Submerged in de container is a trapezium shaped mass 'A'
The trapezium consists of two separate parts which fit air tight and can slide laterally. As a result mass 'A' can change in volume.
The sides of 'A' are kept airtight to the wall of the 'brown' container by some kind of guiding rail system without liquid seeping in between (= exception to Archimedes' principle)
Tube 'B' connects 'A' with the air outside of the container.
In picture 1 the combined density of the walls of mass 'A' and the volume of air trapped inside is buoyancy neutral.
The width of 'A' is 1. The width of the 'brown' container is more than 1.
The downward buoyance force on top of 'A' = Fb1 = h(height)*p(density)*g(gravitation constant)*a(area) = 6 * 1 * 9.8 * 8 = 470.4
The upward buoyance force on the bottom of 'A' = Fb2 = h(height)*p(density)*g(gravitation constant)*a(area) = 8 * 1 * 9.8 * 6 = 470.4
Fb1 = Fb2
Picture 2
'A' has increased in volume. The combined density of the walls of mass 'A' and the volume of the air trapped inside is now positive buoyant.
The downward buoyance force on top of 'A' = Fb1 = h(height)*p(density)*g(gravitation constant)*a(area) = 4 * 1 * 9.8 * 10 = 392
The upward buoyance force on the bottom of 'A' = Fb2 = h(height)*p(density)*g(gravitation constant)*a(area) = 6 * 1 * 9.8 * 8 = 470.4
Fb2 > Fb1 + positive buoyancy results in 'A' moving upwards.
Each interval between picture 1 en 2 will results in an increase in positive upwards buoyancy force.
Picture 3
The 'guiding rail system' is disconnected and the sides of 'A' are no longer kept airtight to the wall of the brown container.
The two separate parts of trapezium 'A' are locked.
Fb2 > Fb1 + positive buoyancy results in 'A' moving upwards.
Picture 4
The two separate parts of trapezium 'A' are unlocked.
The force on the sides of 'A' reduces the volume of the trapezium to the same size as in picture 1.
The combined density of the walls of mass 'A' and the volume of air trapped inside is buoyancy neutral.
'A' will 'float' down to the same position as in picture 1.
How are you changing the volume of 'A'?
Without some mechanism (and energy input) the transition between steps 2 & 3 will not occur in the manner as stated.
"unlocking" the width of A, will simply compress 'A' to its' minimum width at any point in the diagram. Water pressure at every depth = > 1 ATM
This forces air out through the tube and 'A' will remain at its minimum volume.
The key point, as per the exception to the Archimes principle, is that there are no lateral forces in step 1 and 2 and therefore object A will not compress to its minimum volume even when the 2 parts of the trapezium A are not locked.
In your opinion will object A move up, down or remain stationary based on calculating Fb1, Fb2 and any other factors which you believe are relevant?
Thanks to Novus for reminding Archimedes buoyancy calculation.
Imagine that this graph can rotate itself?
Thank you!
https://www.youtube.com/watch?v=PZ4Ri2l4W8w
This guy tried to use the same principle... No pressure on the sides. It didn't work.
The important thing about buoyancy, it is the result of water moving downward with gravity. If Mass moves downwards, it will move. If it can't, it won't. Mass can't move downwards throughout the entire cycle unless you are filling it back up at the top and releasing some from the bottom.
Identify when in your system mass moves downwards and when water mass moves upwards. Then you will know when it isn't going to move.
https://www.youtube.com/watch?v=PZ4Ri2l4W8w
@ Tarsier, nice find. Impressive build. The principle is indeed the same, however I guess it will not work because any 'gain' on one site will be cancelled by an equal 'loss' on the opposite site.
See below for the principle on how I understand buoyancy calculations works.
The question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.
some observations...
panyuming design would function given that..
1. Seeping of the fluid into in between the fixed wall and the outer
surface of the pontoons can be prevented without creating to much friction.
2. The exception to Archimedes's is a valid exception.
A rubber like surface on the outside surfaces of the pontoons and upon the
inner face of the fixed wall might prevent the fluid's seepage.
The design referenced by Tarsier-79 in the video
@ https://www.youtube.com/watch?v=PZ4Ri2l4W8w
is only, in part, the same principle as is the Novus's design.
The design referenced by Tarsier-79 in that video does not
incorporate the bottom or side case exception to Archemedes's principle,
in that the sides are missing from the floats, at all times during its operation.
i.e. during both the submersion and the rising.
Quote.... The principle is indeed the same, however I guess it will not work because any 'gain' on one site will be cancelled by an equal 'loss' on the opposite site.
See below for the principle on how I understand buoyancy calculations works.
The question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.
In the video, his motor runs in the wrong direction. The larger surface area on the tubes should be rising, not falling. There used to be a video of this device test running in water.
Explain to me the "equal loss"? as it relates to your principle.
Your Buoyancy calculations match this I think:
(https://image.slidesharecdn.com/lecture03-archimedes-140114101600-phpapp01/95/lecture-03-archimedes-fluid-dynamics-3-638.jpg?cb=1389694713)
Also, if we use your buoyancy calculations on the device in the video, won't it work?
QuoteThe design referenced by Tarsier-79 in that video does not
incorporate the bottom or side case exception to Archemedes's principle,
in that the sides are missing from the floats, at all times during its operation.
i.e. during both the submersion and the rising.
Does that actually make any difference?
Quote from: Tarsier_79 on April 30, 2023, 08:38:47 PM
Does that actually make any difference?
Do you mean...
1. is the Archimedes principle exception valid ?
or
2. that, assuming the Archimedes principle exception is valid, "does that actually
make any difference ?"
Answer to # 1. I don't know if the Archimedes principle exception is valid or not.
It is an idea I myself pondered in the past, but didn't have enough balls to consider
that I might actually be correct / had not heard of it from any source / had not the means
to test the idea.
It makes sense to me.
Answer to # 2. Yes, it makes a difference,
In the Panyuming design @
https://overunity.com/19459/buoyancy-calculations-making-use-of-an-exception-to-archimedes-principle/msg577132/#msg577132
the aluminum pontoons touching the fixed wall, will be less buoyant than the
pontoons that have all sides exposed to the liquid. The wheel will have an
imbalance in the forces causing pontoons to rise and therefore, the wheel will
have a tendency to rotate clockwise, if the Archimedes principle exception is valid.
https://www.youtube.com/watch?v=PZ4Ri2l4W8w
QuoteThe principle is indeed the same, however I guess it will not work because any 'gain' on one site will be cancelled by an equal 'loss' on the opposite site.
QuoteExplain to me the "equal loss"? as it relates to your principle
The design in the video is essentially a symmetrical circular 'wheel' type design whereby my 'guess', without making any calculations, based on experience, is that this will never work since any 'gain' on one site is offset by an identical 'loss' on the opposite site.
The Panyuming design is clearly different in the sense that the volume of the pontoons do not change. It might have merit but should probably be presented as a seperate topic.
QuoteThe question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.
QuoteThe question remains if the object in the example given in my earlier post will move up, down or remain stationary, preferably supported by calculations rather then general concepts on buoyancy, water displacement etc.
Anyone knows the answer to this question?
Let me know if the question is not clear and/or if relevant information is missing.
Thanks.
Willy, I don't think panyuming's design will work. The pressure removed from the equation is on the circumference, and missing force is 90 degrees to the rotation, so won't make any difference to the rotation.
Novus, I don't know the answer. I suspect if there was no seal on the sides it would lift. Creating and removing the seal might be a problem. I suspect if you don't interrupt the seal, it will stay put, but if you let water in it will float.
Your other problem, to get it down to start with you need to make the assembly denser than water. It sinks to the bottom and wedges itself against the sides and somehow we seal it. Then to move back upwards, it needs to be less dense to float up and expand..... I think that is a killer of your design.
QuoteNovus, I don't know the answer. I suspect if there was no seal on the sides it would lift.
Given that the density of the object is less than the density of the fluid the standard Archimedes principle is applicable and the object would rise.
QuoteI suspect if you don't interrupt the seal, it will stay put,
I find this hard to believe since it would need to apply to any given depts, surface areas and density of the object in relation to the density of the fluid.
The following factors relate to the question raised in relation to the example scenario;
1. There are no lateral forces on the trapezium shaped object when the exception to Archimedes' principle applies to both sites?
2. The force on the bottom of 'A' = h2*p*g*A which in the example given equates to Fb2 = 6*p*g*8 = 48pg
The force on the top of 'A' = h1*p*g*A which in the example given equates to Fb1 = 4*p*g*10 = 40pg
Since Fb2>Fb1 the result would be a net upwards buoyancy force? Since 'the density of object A is slightly less than the fluid' the object would start to rise
and increase in volume? As per below picture 1 we would exchange a loss in Fb for an increase in volume?
3. Any other forces/factors which are applicable to the scenario?
QuoteYour other problem, to get it down to start with you need to make the assembly denser than water. It sinks to the bottom and wedges itself against the sides and somehow we seal it. Then to move back upwards, it needs to be less dense to float up and expand..... I think that is a killer of your design.
See picture 2 below based on the initial post which would result in a net gain when the answer to the example would be that the object rises upwards
and increases in both volume and buoyancy.
As for the practical implications and the challenges with sealing the sites water tight and lock and unlock the seals I propose to leave this aside for now and focus on why this should fail even in theory.
Information on the internet on the bottom (and side) case exception to Archimedes' principle which I have been able to find is limited with no example found when the exception applies to both sides of an object:
https://arxiv.org/pdf/1110.5264.pdf
"The existence of exceptions to Archimedes' law has been observed in some simple experiments in which the force predicted by AP is qualitatively incorrect for a body immersed in a fluid and in contact to the container walls. For instance, when a symmetric solid (e.g., a cylinder) is fully submerged in a liquid with a face touching the bottom of a container, a downward BF is observed, as long as no liquid seeps under the block [2, 13, 14, 15, 17]. Indeed, the experimental evidence that this force increases with depth (see, e.g., Refs. [2, 15, 16, 17]) clearly contrasts to the constant force predicted by AP. These disagreements led some authors to reconsider the completeness or correctness of the AP statement, as well as the definition of BF itself [2, 14, 15, 16, 18], which seems to make the things more confusing yet."
@ Tarsier (and anyone else who wants to join the discussion) – can you maybe have an other look at this since we both agree that a gravity/buoyancy based design will never work.
Looks like this is a valid exception to Archimedes principle...
In that
some particular mathematical approaches to solving for buoyancy do not
work in specific circumstances, i.e. the bottom and side exceptions.
In addition to that...
Real world experimentation demonstrates that there is
NO VARIATION IN BUOYANCY
in the circumstances prescribed for a bottom and / or sides exception.
I, (living in the good old U.S. of A and all) easily acquired the near perfect apparatus for
testing the "exception".
Real world... THERE IS NO EXCEPTION THERE AFTER ALL.
PS
Did you get some local help with the math ?
I don't completely agree Willy. If you exclude water from the bottom of an upturned cup and place it on the bottom, it will stay there. It is no exception to the bouyancy rule though. Without pressure pushing up from the bottom, the pressure pushing down is going to win.
Novus. A big issue is the displacement of water. If your "exception" mechanism sits at the bottom position sealed at both sides, lets pretend that it is positively buoyant looking at your fig1 and fig2. As it rises, we have about 14 units of water that drop 2 spaces and 4 units that are displaced all the way to the top of the water surface, about 5 or 6 spaces. So around 22 units of Potential energy working against you vs 14 units moving down 2 spaces...28 units of PE. That is if the buoyant container volume is equal to air. If it weighs less than 6 units of water, it will float, any more (ie if it is neutrally buoyant), it will not rise. I don't think this "exception" will overcome this, but you would need to test.
@ Tarsier, thanks for taking an interest in this topic and providing valuable feedback.
QuoteIt is no exception to the bouyancy rule though
I agree that the bottom case scenario seems pretty obvious (with an increasing downward Fb depending on dept). Below links provide an interesting read on why apparently it not so obvious?
https://www.scirp.org/journal/paperinformation.aspx?paperid=75679
https://www.scielo.br/j/rbef/a/w7VfCBmYgN46Wm77ttMmQ7d/?lang=en
The reason why it is an exception to the AP is on how the principle is worded. For the bottom and site case the wording would need to be changed to; a body
fully immersed in a fluid, however this would exclude the scenario in which a body is partially submerged where AP is applicable.
QuoteArchimedes' principle (also spelled Archimedes's principle) states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces
In our case ^Fb will decrease with dept untill Fb is zero (as per picture 1 in the initial post) and becomes increasingly negative (vFb) at more dept.
QuoteAs it rises, we have about 14 units of water that drop 2 spaces and 4 units that are displaced all the way to the top of the water surface, about 5 or 6 spaces. So around 22 units of Potential energy working against you vs 14 units moving down 2 spaces...28 units of PE.
I believe your calculations are correct, however not sure why this would be 'against you'. A downwards displacement of 28 units would be sufficient for a lift of 22 units?
QuoteThat is if the buoyant container volume is equal to air.
If this is correct it obviously will not work. However I don't understand why only a slightly buoyant object A wouldn't be able to rise with a downwards water displacement of 28 units exceeding the upwards lift of 22 units?
Think about it. if it were neutrally buoyant and it moves upwards, there is no downwards displacement of mass, as the assembly weighs the same as the water, so we just displace water to the top, lifting PE. Those calcs (which might not be perfect) were based on the assembly weighing the same as air.
:)
Rough calculations.
Is it? The pontoons still displace the same amount of water.
How do you seal the pontoons against the shield without creating massive friction?
Quote from: Tarsier_79 on May 04, 2023, 05:42:14 AM
How do you seal the pontoons against the shield without creating massive friction?
The principle is the foundation, everyone determines that the principle is correct,
and then how to make the second layer.
It may be possible to use the nano fluff on the surface of the lotus leaf.
There are some situations that are correct but cannot be implemented, such as the space ladder.
@Panyuming, interesting design. Thanks for sharing. Maybe it fails based on below site case scenario which whould be applicable to H1 in your design (aand probably partially to H2 and H3). It looks like Fb not only changes direction but also increases?
.....
Quote from: Tarsier_79 on May 03, 2023, 09:26:26 PM
I don't completely agree Willy. If you exclude water from the bottom of an upturned cup and place it on the bottom, it will stay there.
So ? A playing card will keep the water in a over turned cup as well.
Try it with something other than a cup.
Question. is a buoyant in water, 5 sided pyramid, changed in buoyancy when its point faces down ? Of course not.
Note...
Precision mated, ground glass surfaces are what is needed in the ferris wheel design.
The viscosity of water is too great for water to leak through the plunger and barrel fit
in my syringe. Even so, the sliding action is very low friction.
Quote from: Novus on May 04, 2023, 06:27:17 AM
@Panyuming, interesting design. Thanks for sharing. Maybe it fails based on below site case scenario which whould be applicable to H1 in your design (aand probably partially to H2 and H3). It looks like Fb not only changes direction but also increases?
I also don't think it's possible that simple to succeed. ;) :P ::)
But...
Seriously and sincerely, thanks for bringing the information and idea to
the forum.
Very cool !
P.S.
Overunity devices do exist.
I know, I have one that sits on my bench at my home.
Quote from: Novus on April 29, 2023, 01:07:27 PM
The key point, as per the exception to the Archimes principle, is that there are no lateral forces in step 1 and 2 and therefore object A will not compress to its minimum volume even when the 2 parts of the trapezium A are not locked.
This assumption only holds true while the inner vessel is in contact with the outer vessel
It is not a true exception to Archimedes principle.
But rather that multiple vessels must be taken into consideration
QuoteThis assumption only holds true while the inner vessel is in contact with the outer vessel
It is not a true exception to Archimedes principle.
But rather that multiple vessels must be taken into consideration
I won't argue you're point. One thing is for sure that, at least to my calculations, Fb behaves in a strange fashion in the sense that it decreases with dept, than at some point is equal to zero and becomes negative (similar to the bottom case scenario) at greater depts.
https://arxiv.org/pdf/1110.5264.pdf
QuoteThis downward force then increases linearly with depth, which clearly contrasts to the force predicted by AP, but agrees to experimental results [2, 15, 17]. In fact, the increase of this force with depth has been subject of deeper discussions in recent works [2, 15, 16, 33], in which it is suggested that the meaning of the word 'immersed' should be 'fully surrounded by a liquid' instead of 'in contact to a liquid', which would make the 'bottom' case, as well as all other 'contact cases', out of scope of the Archimedes original propositions, as well as AP modern statement [2, 15, 33]. Note, however, that this redefinition is deficient because it excludes some common cases of buoyancy such as, for instance, that of a solid (e.g., a piece of cork) floating in a denser liquid (e.g., water). In this simple example, the body is not fully surrounded by a liquid and yet AP works! More recently, other authors have argued that the definition of BF itself should be changed to "an upward force with a magnitude equal to the weight of the displaced fluid" [16]. However, I have noted that this would make AP a definition for the BF and then, logically, AP would not admit any exceptions at all. In face of the downward BF experiments already mentioned, it is clear that this is not a good choice of definition. Therefore, I would like to propose the abandon of such redefinitions, as they are unnecessary once we admit some exceptions to the AP, which is the natural way to treat the exceptional cases not realized by Archimedes in his original work.
QuoteI know, I have one that sits on my bench at my home.
Could you elaborate a little?
In all my years of searching, I have never seen a successful OU device, mechanical, magnetic or electrical.
QuoteCould you elaborate a little?
In all my years of searching, I have never seen a successful OU device, mechanical, magnetic or electrical.
It is discussed on the Free Energy RANT CAFFE tread.
Tarsier79
Quote
Could you elaborate a little?
In all my years of searching, I have never seen a successful OU device, mechanical, magnetic or electrical.
End quote
Quote from: Novus on May 04, 2023, 04:22:57 PM
It is discussed on the Free Energy RANT CAFFE tread.
End quote
—-///—-///—-///—-///——-
Bumped Willy/ Floors topic
Here https://overunity.com/19272/more-clairifaction-of-floors-twist-drive/msg577334/#new (https://overunity.com/19272/more-clairifaction-of-floors-twist-drive/msg577334/#new)
Sorry for intrusion
Respectfully
Chet K
Thanks for the link.
Quote« Tarsier Reply #17 on: May 04, 2023, 10:59:12 AM »
Quote
Think about it. if it were neutrally buoyant and it moves upwards, there is no downwards displacement of mass, as the assembly weighs the same as the water, so we just displace water to the top, lifting PE. Those calcs (which might not be perfect) were based on the assembly weighing the same as air.
Thanks for your feedback explaining why this can not work. However, I 'm not yet ready to give up on this one. See below revised schematics.
I enlisted on a science forum and posted below question;
Will the object in below scenario (see picture 1) move upwards, downwards or will it remain stationary?'
An object is submersed in a container 'A' which contains a fluid 'B' (e.g. water)
The trapezium shaped object consists of two separate parts which enclose each other in such a way that no water can seep in and, as a result, can expand in a horizontal plane.
The submersed object consists of solid area's 'C' with the same density as water and a volume of air 'D' which can expand or compress in a horizontal plane. The volume of air 'D' is connected to the air outside of the container via air tubes.
It is given that, without going in further detail that "The sides of the object are 'embedded' water tight on a 'sliding system' inside the container wall" The object can move up or down whereby, as a result, the volume of air will either increase or decrease.
Each square in the picture is 1 cm2. The object has a width of 1 cm. The width of the container is > 1 cm.
The top of the object is at a depth of 4 cm and the bottom at a depth of 6 cm. The area of the top of the object is 10 cm2 and the area of the bottom of the object is 8 cm2.
The answers as a result of the ensuing discussion can be summarized as follows;
* 'I agree that there are no lateral forces. And no vertical forces from the sealed and slanting sides.'
* 'Because the object is not surrounded by fluid on all sides, Archimedes law does not apply. The net force from fluid pressure must be calculated from first principles rather than from the weight of the displaced fluid.'
* 'As the trapezium moves up and expands, water is also flowing down, from above to under the trapezium...' ' Just because the water level rises, doesn't necessarily mean that the water's center of mass (and thus PE) also rises...' 'When air bubbles rise they also expand and so does the water level, but the total PE of the water decreases'. (I've calculated the water's center of mass for the scenario and for the object having moved upwards 2 cm – including a subsequent rise in the water level - and found that the center of mass for the water (and thus the P.E.) decreased)
From scientific articles on the bottom-case scenario we find that;
...However, this law, also known as Archimedes' principle (AP), does not yield the force observed when the body is in contact to the container walls, as is more evident in the case of a block immersed in a liquid and in contact to the bottom, in which a downward force that increases with depth is observed.
https://www.researchgate.net/publication/51958652_Using_surface_integrals_for_checking_Archimedes'_law_of_buoyancy
...A question that normally comes up during discussions of Archimedes' principle is that when an object in the form of a rectangular block rests on the bottom of a container with no fluid under it, where does the upward buoyant force come from? In fact, in this case because of the fluid pressure on top of the block, the net hydrostatic force on it would be downward, resulting in the apparent weight of the block to be greater than its true weight. But this conclusion is in complete contradiction with all observations since even in this case the apparent weight of the block is less than its true weight by the weight of the fluid displaced.
https://file.scirp.org/Html/1-1720827_75679.htm
The question as stated in the forum can now be answered as follows (see also picture 2);
Solving Fb for the total volume of the object (note that in this case calculating Fb as the difference between the forces on the top and bottom area's does not work); Fb = V * p * g = 18 cm3 * 1 kg/m3 * 9.81 m/s2 = 0.1766 N. For the downwards hydrostatic force on top of the slanted area's; Hf = p* g * h * A = 1 kg/m3 * 9.81 m/s2 * 4 cm * 2 cm2 = 0.0785 N. The resulting Fb(res) = Fb - Hf = 0.0981 N. W = m * g = 0.014 kg* 9.81 m/s2 = 0.1373 N. The object will therefore move in a downwards direction.
Note that this is my calculation and conclusion as posted on the forum which was neither confirmed nor rejected possibly because at the same time questions were being raised if this was not connected to an attempt 'to build a PMM which is against the forum's rules'.
If above statements and calculations are correct...
Below revised scenario (squares are 1x1 cm)
All forces are upwards buoyancy forces. The total force of 1+2 is identical to 3+4. With the appropriate counterforces the total design would be in equilibrium (which to some extend 'proofs' that the underlying force calculations could be correct).
Note that by closing the 'seal' from 4 to 1 the upwards force drops from 0.07848N to zero.
Not to overcomplicate the scenario it should be assumed that the excess volume of water from the expanding objects' volume, with the movement from 1 to 2, flow into a sufficiently large horizontal reservoir on top of the container not to materially impact the calculations.
So everything is as expected; ...equilibrium...
However...
We have an excess energy gain from the compression of the volume between 2 and 3...which can be used to move the cycle from 1,2,3,4,1...
It is probably easier to understand that the scenario is in equilibrium by increasing the mass of the solid area's of the object (currently 1000 kg/m3) with a factor of 1.6666 after which we get;
1. F = -/- 0.07848N
2. F = + 0.07848N
3. F = 0N (neutrally buoyant)
4. F = 0N (neutrally buoyant)
As for the change in force from 4 (F = -/- 0.07848N) to 1 (F = 0N) by closing the seal - see below simplified bottom-case example.
The compression of the object between 2 and 3 generates 'free energy' (depending on the question if the forces are calculated correctly)
As part of the discussion on the science forum I received multiple questions on how the object could be sliding up and down while expanding/decreasing in volume. Below a possible solution as posted on this forum.
The airtube should probably be in a vertical direction rather then horizontally through the container wall.
As for achieving a water-thight 'seal' this becomes less of a challenge when we realize that rather than trying to prevent any water seeping through the waterpressure needs to be removed.
The pulley/counterweight construction is probably not necessary by properly calibrating the combined density of the sliding panel and the object.
Quote« Tarsier Reply #15 on: May 04, 2023, 03:26:26 AM »
QuoteNovus. A big issue is the displacement of water. If your "exception" mechanism sits at the bottom position sealed at both sides, lets pretend that it is positively buoyant looking at your fig1 and fig2. As it rises, we have about 14 units of water that drop 2 spaces and 4 units that are displaced all the way to the top of the water surface, about 5 or 6 spaces. So around 22 units of Potential energy working against you vs 14 units moving down 2 spaces...28 units of PE. That is if the buoyant container volume is equal to air. If it weighs less than 6 units of water, it will float, any more (ie if it is neutrally buoyant), it will not rise. I don't think this "exception" will overcome this, but you would need to test.
In the revised scenario (post 34) 8 cm3 of water moves to the top of the container as a result of the expanding/upwards movement of the object between 1 and 2, however the water's center of mass (and thus PE) drops by 2.29 cm.