Quote from: LarryC on June 19, 2012, 08:35:45 AMThat would be double dipping. The pod is Archimedes and is calculated using volume or pressure differential formulas, please see my reply in 563.

Let's reiterate over facts:
Only the outermost riser is affected directly by the weights. All the other risers are affected by air pressure of the riser above them. Essentially the risers are "sitting on each other" on air cushions - which is described by the pressure.
The pod is lifted by displacement of water (not air) - if and only if it were submerged fully in air, would it be "double dipping" (if I understand that term correctly) - which is what you did in your former spreadsheet (adding forces with different directions, which is just wrong)
Buoyancy is the upwards force exerted on a submerged body and is equal to the weight of displaced water. In this case, you can simplify it just by using pressure over area. Since the pod is not fully submerged in water, you need to account for the pressure exerted on the pod from above (from the weight of other risers).
Don't believe me? Grab 4 cups, manometer and a force meter and see for yourself (or you can just look at the drawings and remember simple physics).
The drawing I have attached to post #574 shows the net forces acting on those risers and the pod.

Edit. Let me simplify:
The buoyancy volume calculations can only be used as the net force when the pod is fully submerged.
When the pod's top is above the water's surface, you need to subtract the pressure exerted on the pod's top to get the net force exerted on the pod.

Also, regardless of the pod's height, only it's submerged volume counts and since it is supposed to do work and lift weights, the submerged volume will gradually decrease. As approximation, you can use the average submerged volume.

And as a last note, if you want to be precise, you can calculate riser pressures as

QuoteP2 = P1 + h·ϱ·g

where h is the height difference in the U-column of water in two neighbouring risers;
ϱ is water density (999.7026 kg/m^3 (at 10 °C));
g is gravity (9.80665 m/s^2);
(again, see the attached image in my post #574 )

Quote from: telecom on June 18, 2012, 09:14:27 PM
Thank you,
but were is the proverbial air bubble from the post 108?
If its in air gaps, is the air pressurized?
Also, to collapse quickly, is the air gap supposed to disappear?

The air bubble from the simple cup experiments is now the air gaps in the charged ZED construction.  And yes it is under pressure to equal the pressure of the water head created throughout the system.

Think about starting with the ZED on the right in the previous illustration but having only water, and no air in or around each riser.  Then you pump air into the bottom center chamber.  That air would have to be pressurized to overcome the water pressure.  And as it is pumped in it would eventually fill all the air gaps just as shown in that illustration until bubbles would be rising in the outermost chamber.  So the air gaps will naturally balance to the correct pressures to enable the configuration of air and water in the illustration.  Of course, the risers need to be weighted or held from rising further up while this air pre-charge is being implemented.

The air gaps do not disappear.  As water is allowed to leave the bottom of the right ZED, the water head pressure drops.  So each column of water begins to drop and pushes the air through the serpentine channels between risers and the separating rings from the outside most towards the center.

M.

@KanShi, as the air pressure increases in each chamber, so does the WATER pressure.  The air pressure at the surface of each water column is equal to the water pressure at that surface.

If that is not clear, please consider this:  Actual atmospheric pressure is ~14.7 PSI.  So any buoyant device place in water has that 14.7 PSI pushing down on it, right?  Why doesn't everything sink?  Because it has exactly 14.7 PSI pushing back up at the surface of the water.

You do not need to consider the pressure of the air on top of a body of water when determining the buoyancy of an object in that water.  You need only consider the pressure differential from any vertical displacement of that water.  This is known as the head pressure.  It can be calculated as (Inches of Vertical Water Displacement)/27.71.  At least at fairly minor depths like we are discussing here.

M.

Quote from: mondrasek on June 19, 2012, 10:27:02 AM
@KanShi, as the air pressure increases in each chamber, so does the WATER pressure.  The air pressure at the surface of each water column is equal to the water pressure at that surface.

That is correct. I said the same thing - maybe in more complex terms, I am not used to express myself in simple terms as most of my colleagues and students usually understand.

Quote from: mondrasek on June 19, 2012, 10:27:02 AMYou do not need to consider the pressure of the air on top of a body of water when determining the buoyancy of an object in that water.  You need only consider the pressure differential from any vertical displacement of that water.

Exactly. Please read my posts carefully, every word counts. The pod is NOT FULLY SUBMERGED - that is the whole point, if it were, you could simply calculate the net force as the volume of displaced fluid (which is what I've actually stated in my post). I have stated explicitly I am talking about the pressure exerted on the pod's top not on water surface.

@KanShi,

All that needs to be considered is the height of the water displaced by the portion of the Pod submerged in the water.  That "height" / 27.71 = the water head.  That times (x) the surface area of the Pod is your Buoyant Force.  The air pressure above the Pod plays NO part in the calculation.

M.