Buoyancy calculations – making use of an exception to Archimedes' principle?

[Novus]

https://www.youtube.com/watch?v=PZ4Ri2l4W8w

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.

Explain 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.

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.

[Novus]

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.

Anyone knows the answer to this question?

Let me know if the question is not clear and/or if relevant information is missing.

Thanks.

[Tarsier_79]

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.

[Novus]

Novus, 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.

I 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?

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.

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.

[Willy]

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  ?