Hydro Differential pressure exchange over unity system.

[KanShi]

All that needs to be considered is the height of the water displaced by the portion of the Pod submerged in the water.

For buoyant force, YES. For net force, NO. Again, please read the whole posts. I am talking about the net force (the total force acting on an object).

When you have a body that is partially submerged and it's total density is lower than water's, it would (by your thinking) float up and end up above surface. It does not happen in reality. The body ends up partially submerged because it's weight (Fg) pushing down is equal to the buoyant force (Fb) pushing up.

Now, consider this: the body is partially submerged (eg. on surface of a lake) and a hovercraft runs over it. The air pressure (which is higher than atmospheric pressure) of the hovercraft's air cushion will push the body down.

[mondrasek]

When you have a body that is partially submerged and it's total density is lower than water's, it would (by your thinking) float up and end up above surface. It does not happen in reality. The body ends up partially submerged because it's weight (Fg) pushing down is equal to the buoyant force (Fb) pushing up.

You are correct, the total force is the Buoyant Force - the Weight.  However, Weight also has NOTHING to do with the air pressure above the object.

Now, consider this: the body is partially submerged (eg. on surface of a lake) and a hovercraft runs over it. The air pressure (which is higher than atmospheric pressure) of the hovercraft's air cushion will push the body down.

This is incorrect.  The hovercraft is increasing the surface pressure of the water just as much as it is increasing the pressure of the air.  The partially submerged body will NOT be pushed down.

M.

[LarryC]

@KanShi,

Maybe this will help. The spreadsheet shows that 1 Travis 4 Riser system has the lifting force of  5 separate Archimedes system. Thanks again GreenHiker.
Your 'more efficient design' is only approximately equal to the lift force of the Riser 1 pod.


'P2 = P1 +  h·ϱ·g'
We could, but the goal is for all to understand.

I too, will be glad when you understand the system, as Wayne previously stated.

@M
Good Luck. It is to much like spitting into the wind for me.

Regards, Larry

[KanShi]

However, Weight also has NOTHING to do with the air pressure above the object.

Yes, it doesn't.

This is incorrect.  The hovercraft is increasing the surface pressure of the water just as much as it is increasing the pressure of the air.  The partially submerged body will NOT be pushed down.

Don't be hasty . This is actually true in the real world. And here is why:
We are dealing with forces and a small area. The lake is really big, the hovercraft is really small. The hovercraft exerts force on the water, BUT it is insignificant in the scope of the lake. Also, most of the force is directly pushing on the partially submerged body. Force = pressure over area.
Let's have P = pressure of the air cushion 200 kPa.
Area of the body is 5 m^2, area of the hovercraft is 6 m^2. Which means the exerted force on water is only 200 kN, while the exerted force on the body is 1 MN (do you see the difference?). The buoyant force is actually the same (for the stated reasons).

Now, if the hovercraft was as big as the lake, things would be as you say (this was just an amusing experiment to test your skills).

@LarryC: sure, if in your example the pod is submerged, then your calculations are correct (except you forgot to subtract forces on the risers - you just added them all up - that was the main point of my post). If you look at the drawings, it is not submerged - unless the height is the submerged height, then it is also OK. Also, my "more efficient design" has, for the same amount of required energy, the same lift as the Travis system (eg. 5x more than regular buoyancy with the same amount of water displacement).

[mondrasek]

Let's have P = pressure of the air cushion 200 kPa.
Area of the body is 5 m^2, area of the hovercraft is 6 m^2. Which means the exerted force on water is only 200 kN, while the exerted force on the body is 1 MN (do you see the difference?). The buoyant force is actually the same (for the stated reasons).

Now you are confusing apples and oranges, aren't you?  In this example the water under the hovercraft is supporting 200kN of the hovercraft mass.  The body is supporting 1 MN of the hovercraft mass.  And as you state, the buoyant force is still the same.  And the body will not be pushed down into the water.  It has the same buoyant force in any case, regardless of the air pressure at the surface of the water and over the top of the body.  No weight is applied to the body to make it sink.  This is because that weight you are imagining is actually a pressure that is applied equally to the body and the water around it and buoyancy is not affected.  Pressures are already accounted for in the spread sheet as written.

M.