Mathematical Analysis of an Ideal ZED

[MarkE]

MarkE,

Don't have to,, already done that a long time ago,, Jr. High school science.

I got sidetracked and forgot to answer your other question,, sorry about that.

What you have shown is a single piston assembly,, not a nested system, as well as the volume usage thing,,  get the same results with only using 1\10 the volume while having all the forces the same,, that would be distance pushed down, weight of water lifted blah blah blah,,

The physics of the single piston don't change when one goes to multiple pistons.  Nesting folds a larger effective assembly into a smaller volume, at a cost of lost efficiency.  It does not create energy gains for reasons that should be amply visible from the demonstration that you call middle school science.  Ergo, none of this should be news to you at all.

I am not saying you screwed with anything,, I explained most likely what was happening and you came along and conferred with my guess as to causality.

All my pictures are taken by hand,, you can imagine that if I tried to do a collage what it would look like

Anyway,, your simple little testbed does exactly what it should, that is you push a float down add some weight and it comes back up to the point where the buoyant force balances out with the weight.  Your input work pushes the water up,, and then some of that fills the little bottle and the rest falls back down pushing the little bottle back up.

Great!  Then we are on the same page.  Do that with more floats and more weights and what do you get?  Just a more complicated implementation of the same thing.  There is no change in where the energy comes from.  There is no change in the behavior of gravity.

I understand that you think this is the  same as a nested system with pod,, that is what the discussion is all about.

It demonstrates all of the fundamental behaviors one needs to understand to see the ZED claims as the false nonsense that they are.

[MarkE]

Minnie, if I felt that MarkE's intentions were to learn anything, I might be inclined to do differently.  But he clearly was able to present a correct State 2, but refused, at first, to present a State 3.  Instead he simply "waved his hands" and said that for reasons previously explained, it cannot work.  I found that to be disingenuous at first.  But he did agree to continue.  It is at that point he deviated to a Volume in = Volume out approach, instead of the clearly stated method of checking if Energy in = Energy out.  And he has had many "errors" ever since.  Including the ridiculous claim that the buoyancy Force is calculated by the ID and not the OD.

So I find that MarkE is not working in good faith to help with anything.  He is only disrupting now.  In fact, he is trying, once again, to make us go back to the beginning.  Why do you think that is?  I know why, do you?

If anyone who displays a desire to learn what I presented and cannot on their own (I hope many can) I would be happy to help.  But MarkE's demands will not stop.  And nothing that I present will ever be satisfactory.  And he will never let the Analysis be completed if he is involved.

So, why bother?

Again, this is what you asked for:

All, please check the math.  I would appreciate if you can point out any mistakes in the math, assumptions, logic, and conclusions. 

That was your request.  You deliberately have refused to publish the work that you said you want checked.

[MarkE]

MarkE,

Attached first is my example that you said had the input work wrong.


Second is a picture of the spreadsheet integration that shows the same results as the example.


Third is the actual integration spreadsheet. Where it is using P * V at each increment.


Correction: the results are the same as yours, after I correct your .65 to .43 for the psi/ft.

In the spreadsheet that you had sent you average pressure across multiple columns and then multiplied that pressure by volume to find energy.  That's wrong.  I have explained why it is wrong.  It is wrong because:  N*(X/N)² = X/N. Given two identical columns:  One column filled to 2m holds 4X the energy of one column filled to 1m, or twice the energy of two columns each filled to one meter.  Under the special circumstances of a single column, one can prove (I've shown the math several times) that:  0.5*P_MAX*V = 0.5*density*G₀*Area*H².   The latter is the actual solution for the stored energy for each H.  When you have multiple heights and you average pressure you can easily get the wrong answer as in the two column example I just offered.

[MarkE]

MarkE,

I was wondering, is a buoyant lift an ID thing or an OD thing?

So if someone were using the ID and pressure then would it be a buoyant lift, or would it be something else?

The upward force is equal to the weight of the displaced fluid.

[MarkE]

MarkE,

Attached first is my example that you said had the input work wrong.


Second is a picture of the spreadsheet integration that shows the same results as the example.


Third is the actual integration spreadsheet. Where it is using P * V at each increment.


Correction: the results are the same as yours, after I correct your .65 to .43 for the psi/ft.

Larry, if you correctly calculate the reflected force back to column 1, which is in this case (1+(4-2))ft*pWater*G₀*Area/ft = 3*pWater*G₀*Area/ft  and apply that to the net force:  F = 0 + K_FORCE*H_COL1  then when you evaluate the integral you get:  0*H + 0.5*K_FORCE*H_COL1²  then you get the exact answer without iterative calculations.   We can observe that for these circumstances of starting from zero, that P_AVE*V_{COL1_ADDED} yields the same result.  The math works because the force and pressure were related back to one column.  In this case you stacked columns side by side.

When you set up with concentric rings of fixed diameter steps, then when you calculate the force, such as back to the pod chamber for the State 1 to State 2 transition in Mondrasek's "ideal ZED", then the opposing force per unit height inside AR1 is:  sum(1+AR1area/AR2area + AR1area/AR3area ... AR1area/AR7area)*pWater*G₀*AR1Area/AR1_unit_height.  You can of course relate the force and pressure by the area.