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

[Tarsier_79]

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.

[Novus]

@ Tarsier, thanks for taking an interest in this topic and providing valuable feedback.

It 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.

Archimedes' 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.

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.

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?

That 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?

[Tarsier_79]

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.

[panyuming]

 
Rough calculations.

[Tarsier_79]

Is it? The pontoons still displace the same amount of water.

How do you seal the pontoons against the shield without creating massive friction?