Quote from: webby1 on March 16, 2014, 01:11:16 AM
ST1=>ST2=>ST3=>ST1=>ST2=>ST3,,,,,,,,,,

Yes MarkE,, ALWAYS,, as long as there is a positive pressure within AR1 the fluid will vent.

Lets take state 0 to end state 1, but stop the risers 4mm short and then close the vents.  Now finish sinking the risers, what happens.

There is less displaced water than in the old State 1, but the coefficients of the force gain equations do not change.  So the neutral position of State 1X is now closer to the new State 1 than before.

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The air\water volumes can not move any where except one direction, that is, they will shift from AR1 out through AR7,, so when the water is up the outside of AR7 you have sunk the risers.

If you think the risers sink if you stop short 4mm you are simply wrong.   You have no mass to load the thing down, and you still have displaced water.  The three stipulations: incompressible and massless "air", and massless risers guarantee that the neutral position after forced submersion is always above the forced submersion level.    With my soda bottle experiment due to the mass of the water bottle there is a minimum submersion depth.  But not so with these ideal stipulations.

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Lets look at end of state 2, where are the air\water columns,, they are shifted from AR1 out and through AR7, so in essence putting in water in AR1 is sinking the risers, you have pushed the risers down by filling up AR1.

If state 3 were a lift condition by the addition of more fluid into AR1 and the risers were allowed to move up with that volume increase,, then the lift force value is a constant.

In English please?  If you add more water in State 2, then you change the ending force in State 2.  You do not change the force versus distance gains.  More force at the end of State 2 requires more lift distance at the same force vs distance rate in the transition to State 3.

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3 times more input for twice the 325% output,, you push down on the risers to make them move up.

The system emulates a linear compression spring.  The dimensions that you pick set the spring rate.  The prefill levels set the force at the end of the respective prefill states:  State 1, and State 2.  You really need to wrap your arms around this slide I posted earlier.

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Going up, r2 is pushing on r3,, going down r3 is pushing on r2,, that negative head thing.

Quote from: webby1 on March 16, 2014, 01:11:16 AM
ST1=>ST2=>ST3=>ST1=>ST2=>ST3,,,,,,,,,,

Yes MarkE,, ALWAYS,, as long as there is a positive pressure within AR1 the fluid will vent.
...

Correction: AR1 fluid vents if the net pressure of water at the bottom of AR1 against the pod chamber floor is positive.  The net pressure at the bottom of in AR1 at the end of State 3 is the sum of the two values:  ST3_R1_PRESSURE and pWater*G₀*ST3_AR1_Height.  Take a look at what they are in your example. 

It may be that you confuse the buoyant pressure that acts on the pod for this value.  They are not the same.  The pod is subjected to ST3_R1_PRESSURE on its top surface as is the water in the AR1 column.  ST3_R1_PRESSURE cancels out with respect to lifting force on the pod.  But it does not cancel out in terms of net pressure on the water in the pod chamber.  It adds to that pressure.  When ST3_R1_PRESSURE is more negative than the pressure that the AR1 column would form in isolation, as it is in your example, then: Opening the valve at the bottom of the pod chamber draws in "air".  It does not dump water.

ST3_POD_Pressure   15179.9613985   Pa  This is the net pressure pushing up on the pod.  It is NOT the pressure acting against the bottom of the pod chamber floor.
ST3_R1_Pressure   -25971.3758487   Pa
ST3_AR1_Height   1555.175926   mm
Pod Chamber Floor Pressure = ST3_R1_Pressure+ST3_AR1_Height*mm_to_pressure = -10747.76177 Pa

The pod chamber floor pressure is negative and opening the floor valve draws in "air".

In contrast the Mondrasek "ideal ZED" dimensions create a different set of circumstances:
ST3_POD_Pressure   216.0815190   Pa This is the net pressure pushing up on the pod.  It is NOT the pressure acting against the bottom of the pod chamber floor.
ST3_R1_Pressure   24.6763852   Pa
ST3_AR1_Height   24.664393   mm
Pod Chamber Floor Pressure = ST3_R1_Pressure+ST3_AR1_Height*mm_to_pressure =  266.1160825 Pa

The pod chamber floor pressure is positive and opening the floor valve causes the pod chamber to drain.

Quote from: webby1 on March 16, 2014, 08:32:50 AM
MarkE,

I find it a little awkward when you go back, after there has been an exchange, and edit your posts.  I would prefer that you make a new post with the changed information and reference the old post.

Thank You.

Where have I done that?  I added posts that show that you have created a condition in State 3 where the pressure on the floor of the pod chamber is negative.  Opening the valve will not let water out.

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Back to this

Back to what?  You wanted to open AR1 in State 3.  Now you want to open it in State 1.

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Leave the AR1 open, push down on the risers and close the vents for the risers only, release the risers.

What stops the risers from moving back up to the state0 position,, this is with AR1 open,, they will not make it up to the start position of state 0.

Opening AR1 at State 1 means that air can enter the system from both AR1 and AR7 instead of just AR7. This reduces but does not eliminate the feedback gain.  It relaxes somewhere between the State 0 and State 1X positions.

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Maybe it is the concept of pressure that is getting in the way, maybe it would be better to see that it is the weight of two water columns, like weights on a string,, that are at work.

AR7 pushes down,, AR6 pushes down, the riser wall and the ring wall focus all that pushing and leaves the only place that something can move is the bottom and top of the riser, R3.

I believe that AR7,5, and 2 go down, 6, 4, and 3 go up.

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If I pull up on AR6, how much of all that is left to push up on the bottom of the R3,, the same pressure but not the same volume,, then how do I pull up on AR6?, that would be by pulling down on AR5 and then the whole inside TOP of R3 acts like a fixed pulley,, so with pressure I would have the lift pressure and the sink pressure,, or if it is just weight then AR6 becomes the sink weight and AR7 becomes the lift weight, and these two functions are relative to there height, so the only time that R3 can raise up is if AR7 is above AR6.

How do you pull up on AR6 in isolation?  The fluids are incompressible.  You have allowed the fluid volume under Riser 1 to change freely by opening AR1.  The three water volumes, and the air volumes under R3, and R2 are still fixed.  What you have done is to change the feedback coefficients.  If you want to use the block and tackle analogy, you have changed from one free and one fixed end to two free ends.

If you want to use a block and tackle analogy, then with AR1 closed, the line is anchored at that end.  If you open AR1 then you have let go of that side, and nothing is anchored.  Equilibrium results from balance of the weights versus gain of shrinking pulley sizes going from AR2 towards AR7.

So what does any of this do but keep reconfirming that the scheme lifts and drops water, and in doing so emulates a 1000 times or more smaller linear compression spring?

Quote from: webby1 on March 16, 2014, 02:36:30 PM
Not so,,

The string is anchored to the bottom of AR4-AR5 because of the internal volume thing,, this anchor makes R3 go heavy and R1 go light.

The volume of R3 is much larger than that of R1,, which one sucks in more per unit distance of travel,, which way will the string shift the most???

The water masses exert force due to their weight.  I do not see an anchor possible except at the ends.  The fluids are free to move through the serpentine.  We know that the amount of vertical movement shrinks as we move from center outward, just as conversely the relative force gain increases.