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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 749654 times)

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1035 on: March 16, 2014, 02:21:20 AM »
What am I ignoring,, you said that it will fall apart like a house of cards and then you said

There is pressure still within the pod chamber so the fluid will exit,, as the fluid exits what happens, the system returns,, this is what I interpret you to be saying.

Besides,,  what if state3 were a proper lift,, that is what if you put in 2 more times the input energy and took the lift distance out as a constant full force value,, then what happens if you do not let the risers "POP" and instead take the fluid back out of the pod chamber?
Lots.  You have chosen to take things out of context:  The house of cards (actually I recall saying "collapsing balloon" )referred to the specific geometries of Mondrasek's model a week or two ago.  You quote but ignore what you quoted:

Quote
Heads in odd AR#'s add to the reflected force, heads in the even AR#'s subtract and there are additional terms due to the nested risers.  The thing will drain so long as the reflected sums of the other annular ring forces and the weight in the pod chamber is positive. 
 

Do you fail to understand what that quote means in plain English? 

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1036 on: March 16, 2014, 02:47:29 AM »
I tried that and it did not work,, then there were all those _1 ends,, so I just did it the way I did.
Are you using Excel?  I highlighted the various lines, cut them, and then inserted the lines where you had them.  The only thing that is different is that I created a single cell IF(...) statement to detect underflow.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1037 on: March 16, 2014, 03:15:45 AM »
Yes I do understand what a quote is,,

Like I said there is still pressure within the pod chamber,, so the fluid will come out.

I thought that the additional input is from going from end of state 1 to end of state 2,, which is B117,, is not state 1 a "build" condition and only gets put in once?
You quote my answer to a question posed on March 6, regarding the specific Mondrasek dimensions and then apply it as though it is my answer to your different general case question of today, when I carefully stated the dependence on the reflected forces.  Kindly tell me how you have done anything other than attempt to create a straw man.

Why you choose to compare energy output from State 2 to State 3 with energy input going from State 1 to State 2 as some sort of cyclic energy balance beats me.  Go from S2 to S3, and back.  In order to get back you have to at least input the ST2_ST3_ENERGY_LOSS which is greater than the ST2_ST3_External_Work_Performed.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1038 on: March 16, 2014, 03:17:37 AM »
No.

I do not have Excel so I am using Open Office.

I will look at the file later,, thanks.
I recommend you use Libre Office instead of Open Office.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1039 on: March 16, 2014, 04:55:16 AM »
What within the system changed, other than the physical dimensions that would change what the outcome should be.  If there is no such change then if the outcome is what I have shown then there is an "engineering" component that can be used to optimize the setup which Mondrasek"s dimensions do not meet.

I do NOT need to go back to end of state 2 in one step,, I can go back to end of state 1 and from there go to end of state 2 which leads us once again into state 3.

Since there is a positive difference in water height between ar2 and ar3 and since ar6 is pulling up on ar5 when ar1 goes down and removes that positive pressure add ar4 will not be pushed down helping to lift ar5,  it is ar5 pulling on ar6 that is raising the water level in ar6, that will make ar5 fall down and ar4 move up and that will increase the negative buoyant value under R3, this will continue to move until all water levels are back to being even, back to end of state 1.

This is not a straw man anything, this is simple from this point for reset, as you said it may not be intuitive, not right off of the bat.  All the water weight and or height is balanced in only one condition after the risers are sunk.  The force to sink is a setup cost and is only paid once, after that the volumes *that can not change* keep the system with the lowest state of energy when freely allowed to move being end of state 1.

It may be hard to see a negative value as a source of positive motion, but any change from end of state 1 is a positive input, push down on the risers and they will come back up, pull up on the risers and they WILL go back down,, they were pushed up by the additional input going from end of state 2 to end of state 3 so the opposite reaction will be that they will pull themselves back down.

If you stretch your spring will it not pull back,, if you compress your spring will it not push back.
Oh, please, how convoluted do you want to get?  You emphasized "ALWAYS" in all caps in your current question and I stated that the result is determined by the coefficients.  You quoted but ignored my perfectly clear answer and wish to refer to my answer to a different more limited question. 

State 1 is not a rest state.  It requires external force to maintain.  State 1X is the rest state after sinking the risers and venting.  If you don't believe me, then go repeat the soda bottle experiment.

As you have found out by playing with the spreadsheet you can create a State 3 that is at a higher or lower energy state than State 1.  A passive system requires external energy input to traverse from a lower energy state to a higher energy state.  If you do not believe me about this, at least as far as it applies to the "ideal ZED", then use the spreadsheet to explore.  If you want to pretend to yourself that:  ST1 => ST2, and ST2 => ST3 constitutes a cycle, then who am I to disabuse you of your self-delusion?  If you want to go from ST1 to ST3 and back use the spreadsheet to determine the actual work traversing the loop.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1040 on: March 16, 2014, 07:15:59 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.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1041 on: March 16, 2014, 11:02:38 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.



MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1042 on: March 16, 2014, 06:23:47 PM »
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.
Quote

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.
Quote

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.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1043 on: March 16, 2014, 07:24:48 PM »
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?

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1044 on: March 16, 2014, 08:32:19 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.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1045 on: March 16, 2014, 08:35:55 PM »
Dumb question time,,

If I have a value of ZN and it is being moved over a distance of Ymm would I take the Ymm and multiply that by 0.001 and then divide N by that?
1 Joule = 1 Newton * 1 meter.  E in Joules = Z N * Y mm * 1m/1000mm.

minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1046 on: March 16, 2014, 08:44:59 PM »



  Webby,
             what are you trying to achieve? Whatever you do you can't breathe
  life into a hydraulic cylinder, it's just a dead thing!
    As  for ZED to ZED transfer, I'm afraid we learned the answer on page one.
                    John.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1047 on: March 16, 2014, 09:20:18 PM »
Thanks,,

I keep blowing brain farts with which is what when I am looking at what is doing which,,

N/m has nothing to do with distance, but I see the m and I am thinking distance,, I am trying to think of it as torque right now,, just to keep it straight.

−10791.415  I am working through this in AR1 and 2520cc and twisting it further by asking myself what happens if the risers are held still,, you know all the normal questions one might ask oneself on a Sunday :)
N/m is the force gradient per unit distance.  We see it in a couple of different conditions:  The force per unit height of a given area column, or the restoring force per unit movement.

You can always construct a model.

minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1048 on: March 17, 2014, 11:52:52 AM »



    Sadly this particular ZED seems to be dead.
       My wife and  daughters are overjoyed because I keep looking at
     this forum and it was a waste of time as far as they were concerned.
         Mark has shown us how to present  work in a proper scientific
     manner. Koala has introduced some lovely words like "kludge" and
     "Buffoon" to name but two.
         As for me, I think I've got withdrawal symptoms.
                            John.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1049 on: March 17, 2014, 12:34:34 PM »
I think that this thread has pretty much run its course.  I don't believe that any further attempt to defend a ZED as over unity using claims of a physical analysis will be forthcoming from the pro-HER/Zydro camp. 

If certain events are in motion they have bigger things to worry about.