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### Author Topic: Mathematical Analysis of an Ideal ZED  (Read 704757 times)

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #735 on: March 07, 2014, 07:13:07 PM »
Well MarkE,

There may be some factual basis to your statement, but the ZED is not a spring, it may have some attributes of a spring, but it is not "only" a spring.

Work does not only happen in one direction,, work can happen in any direction,, work really is not fixed to a direction,, a force over a distance, whether a straight line or a curve, is work.

I take it that you agree with my assessment in that you need to include the lever for the proper analogy with a spring to a ZED.
Webby1 we have been told to disregard all the set-up steps as one time affairs.  That's fine.  That leaves us with the behavior between State 2 and State 3, and then back from State 3 to State 2.  That behavior from a black box standpoint of the "ideal ZED" is indistinguishable from the behavior of a linear compression spring: a very complicated, and large, low energy compression spring, but a linear compression spring just the same.

Unless you specify operation between states other than just State 2 and State 3, the "ideal ZED" can for purposes of external behavior, IE black box be completely replaced by a 0.48N/mm 2.492mm relaxed to maximally compressed linear compression spring.

Where did I say that I agree?  You can add levers externally and define a new composite machine.  It will not fix the N*(X/N)2 efficiency problem that plagues this compression spring emulating device.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #736 on: March 07, 2014, 07:19:52 PM »
Of course I am using the OD instead of the ID.  I am calculating the lift force due to BUOYANCY.  The correct water displacement value is achieved when you consider the water displaced by the air inside the riser AND the wall thickness.  That means you use the OD.

If you want to calculate the buoyancy Force on a fully submerged 1m3 cube that has a wall thickness of 100mm throughout, is it not displacing 1m3?  Or do you want to calculate the displacement using only the remaining internal volume?
Mondrasek, State 1 is in equilibrium is it not?  And why is that?  It is because the meniscus under the ID is at the same level as the water outside the riser.  The water displaced by the riser wall plays no role in the force balance.  What is true for State 1 is true for State 2, and State 3.  If you close off the bottom of the riser so that the meniscus cannot rise up underneath it, then you change the problem to where the OD determines the displacement.  Are you really going to argue that the riser wall that contributes no up force in State 1, suddenly changed its behavior because we introduce water into the pod chamber in State 2?

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #737 on: March 07, 2014, 07:25:55 PM »
4.653 is the single riser.  I was answering your question with respect to the three riser.

Untrue, MarkE.  Here is what you wrote in post 708:

So you get a movement S = FSTART/KRATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ.

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

In the first paragraph you are discussing the no-pod, single layer.  See where you twice use the value for that lift of 0.004653m?

In the second paragraph you now reference the lift as only 2.492mm.  And include internal energy values from where exactly?

You then calculate an erroneous efficiency using these mismatched and false values.

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #738 on: March 07, 2014, 07:43:04 PM »
Mondrasek, State 1 is in equilibrium is it not?  And why is that?  It is because the meniscus under the ID is at the same level as the water outside the riser.

The equal level of the two mensci results in a HEAD difference of zero.  Buoyancy Force = Weight of the Displace Water = (Density of Water)* HEAD * (Cross Section Area).  If HEAD is zero, buoyancy Force is zero.

The water displaced by the riser wall plays no role in the force balance.  What is true for State 1 is true for State 2, and State 3.  If you close off the bottom of the riser so that the meniscus cannot rise up underneath it, then you change the problem to where the OD determines the displacement.  Are you really going to argue that the riser wall that contributes no up force in State 1, suddenly changed its behavior because we introduce water into the pod chamber in State 2?

Complete hogwash.  And now I will clearly state that it is MarkE's intention to misdirect and provide false information in this thread.  He has done this several times in the past few days.

Buoyancy Force = Weight of the Displaced Water.  For the riser what is displacing the water is the Air inside, and the Material of the riser itself.

So yes, I will argue that the riser wall has no relevance at all on up Force in State 1 dues to no water HEAD.  But once a HEAD is introduced, yes, the wall thickness is included when calculating the Weight of the Displaced Water.  The riser wall did not change its behavior.  The water HEAD condition changed.

Why are you purposely trying to stall and now misdirect this Analysis?

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #739 on: March 07, 2014, 07:55:17 PM »
Quote
Quote
Quote from: MarkE on Today at 01:54:01 AM

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift...

Okay, I am lost here.  The lift was 4.68536mm.  Did I miss something?

Or 4.653mm by your math.  Probably due to the use of different constants for water density and gravity?

Mondrasek, as you see in #745 you quoted where I talked about 2.492mm movement.  2.492mm movement applies to the 3 riser system just as I responded.  It looks like in #708 I wrote the 3 riser movement value of 2.492mm where I should have used the 4.653mm value.  However, the 4.653mm and only the 4.653mm value was used in the calculations.

#708
Quote
Re: Mathematical Analysis of an Ideal ZED
« Reply #706 on: Today at 01:54:01 AM »

Quote

Quote
Quote from: mondrasek on March 06, 2014, 10:46:18 PM

Another way to look at this is to calculate the Energy in the column of water that is on the piston.  That Energy is dissipated as the piston rises and the water spills out.  So that Energy must be equal to the Energy that is released when the ZED is allowed to rise.  That equation also resolves to 0.5*Fstart*S.

For the circumstance that you set-up that is correct:

So you get a movement S = FSTART/KRATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ.

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

This is the tyranny of N*(X/N)2.  Whenever you directly transfer energy between two potential energy stores you run into the N*(X/N)2problem.  In order to avoid the problem you have to retain N=1.0 and still transfer all of the energy.  You can only do that by converting the form of the energy from potential to kinetic.  That is how a pendulum works.  The insipidly stupid ZED design cannot do that.  It translates potential from one place to another.  There is even the "equalization phase" that HER/Zydro say is necessary.  Every time you take energy from one potential store and redistribute it to more stores than the original without first converting the form of the energy, you suffer losses.

Now let's talk about the quote above where you claim I am misrepresenting the physics.
Quote

Untrue, MarkE.  Here is what you wrote in post 708:

Quote
So you get a movement S = FSTART/KRATE 0.30078N / 64.649N/m =  0.004653m and an output energy of
0.5*0.30078N*0.004653m   0.700mJ.

The energy expended, is the loss of potential energy in the annular columns as they move from the State 2 condition to the State 3 condition with the riser at 2.492mm lift:  The internal energy goes from 2.560mJ to 1.536mJ which equals 1.024mJ internal energy loss.

The efficiency is:  0.700mJ/1.024mJ = 68.3%.

In the first paragraph you are discussing the no-pod, single layer.  See where you twice use the value for that lift of 0.004653m?
Of course I did.  Apparently, you are still having difficulty understanding how to calculate energy even when I laid it out there for you to see.  The energy depends on the height squared.  You can get that by going through the detail work of writing out the integral, or given that form of integral has been solved many times, you can skip to the end and use the maximum pressure and volume values, that ... wait for it ... result in squaring the height.
Quote

In the second paragraph you now reference the lift as only 2.492mm.  And include internal energy values from where exactly?

You then calculate an erroneous efficiency using these mismatched and false values.
I calculated the values correctly, and showed you the work so that anyone could follow along.  Apparently, you were not able to follow and latched onto the typographical error without checking any of the work.  Are you able to follow along that the distance S is a function of the starting force and the rate at which that force changes per unit distance?  Do you suffer difficulty verifying that S = FSTART/KRATE 0.30078N / 64.649N/m =  0.004653m does in fact result in the correct displacement value?  pWater used is 998.2kg/m3, G0 used is 9.80665m/s2.

Here again is the graphic for the single riser with the calculations as originally posted.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #740 on: March 07, 2014, 08:04:07 PM »
The equal level of the two mensci results in a HEAD difference of zero.  Buoyancy Force = Weight of the Displace Water = (Density of Water)* HEAD * (Cross Section Area).  If HEAD is zero, buoyancy Force is zero.

Complete hogwash.  And now I will clearly state that it is MarkE's intention to misdirect and provide false information in this thread.  He has done this several times in the past few days.
Then you make a false accusation and make a fool of yourself in the process.
Quote

Buoyancy Force = Weight of the Displaced Water.  For the riser what is displacing the water is the Air inside, and the Material of the riser itself.

So yes, I will argue that the riser wall has no relevance at all on up Force in State 1 dues to no water HEAD.  But once a HEAD is introduced, yes, the wall thickness is included when calculating the Weight of the Displaced Water.  The riser wall did not change its behavior.  The water HEAD condition changed.
Then you need to go back to your text books, or perform an experiment for yourself, or both.
Quote

Why are you purposely trying to stall and now misdirect this Analysis?
Why are you unable to follow along?  Why are you throwing hissy fits?

ETA: we are going to need to calculate using the OD.  I concede that point.  Will post later.  Off to lunch.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #741 on: March 07, 2014, 08:08:16 PM »
Nitpicking time.

When state 2 is reached in Wayne's system, more input is used to make the lift,, there is NO state 3 change for production, state 3 change is part of the recovery process, but the risers are NOT allowed to "pop"
That's a nice nonsequitor.  I referred to Wayne's earlier comments that he made today with respect to Mondrasek's "ideal ZED".

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #742 on: March 07, 2014, 08:14:00 PM »
Actually MarkE, there is a huge difference.

Can you make your spring extend further by compressing it with more force?  Or is it that the spring can only move the same distance it was compressed.
Since everything up to and including State 2 was declared part of the set-up, only the resulting behavior after State 2 has been reached matters.  The behavior is that of a linear compression spring.  The parameters used getting to State 2 define the coefficients for the equivalent linear compression spring.

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #743 on: March 07, 2014, 08:45:14 PM »
Mondrasek, as you see in #745 you quoted where I talked about 2.492mm movement.  2.492mm movement applies to the 3 riser system just as I responded.  It looks like in #708 I wrote the 3 riser movement value of 2.492mm where I should have used the 4.653mm value.  However, the 4.653mm and only the 4.653mm value was used in the calculations.

4.653 is the single riser.  I was answering your question with respect to the three riser.

This misdirected me to look at the 3-layer.  You did not say it was a transcription error or typo.  You said you were now talking about the 3-layer.

As for the rest of the Energies due to columns of water that you then refer to, I did not see them on your graphics or my 3-layer calcs that I thought you had directed me to move over to.  But I also did check my calcs for the no-pod, single layer and also did not find them.  I see that you are able to point out these values on your graphic, but again, we have a problem.  The Energies due to the columns of water do not match my own calcs for the no-pod, single riser Analysis.  My calcs and yours do match exactly for the 3-layer, so I am confident we are both calculating those the correct way.  But I do not get your values at all for the no-pod, single layer.  I have checked them several times now.  Could you please double check your calcs for the Energy in the system due to just the water columns for State 1?  The value I get is ~.879mJ.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #744 on: March 07, 2014, 10:07:57 PM »
Fair enough.

Still, your spring can only undo as far as you do it up.

I wonder what would happen if the risers only had 1\2 the maximum lift value holding them down while going from state 1 to end of state 2,, I know it is out of the area of discussion.
Then during the State 2 pumping exercise the whole assembly would rise part of the distance that it rises in State 3.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #745 on: March 07, 2014, 10:35:43 PM »
The Zed that does not produce anything is not Travis's Zed. Where is the production, what form does it take, when in the cycle is it taken off so that part can be re-used as "Flow Assist" and the other part used off-line? Where, exactly, is any of this explained?

Something, somehow, is "produced", taken away from the single Zed somewhere in the cycle, for use. Part of this production is used to "Flow Assist" either this same Zed or the other Zed to reset it. The other part.... the MAJOR part.... of what is taken away from the True and Holy Zed is used _outside the system of the two Zeds_ to perform usable work.

Of what does the production consist, exactly? Is it pressure of air, water, hydraulic fluid? Where exactly is this production removed from the Zed under analysis? Where and how does it become "flow assist" and what is done with the rest of the production when it done performing its work outside the Zeds? If you use hydraulic pressure off-line, that pressure is gone, out of the system. If you use water.... gone. Hydraulic fluid itself..... gone, except for the bit of "flow assist" that you put back in, a far smaller bit according to honest Wayne Travis, than you can use off-line. A weight, removed at the top and replaced at the bottom? Where is your weight transfer? How does it become "flow assist"?

So? Why are you analyzing a Zed that has _no production output_ ? It is not the "real" zed and when you are done, this is what Travis will tell you. You are proving that a rock will fall in gravity.... Travis is telling you he has circumvented gravity. You are describing a spring that re-sets with no production, Travis is claiming a system that resets itself by using part of its own production. Ergo, your analysis is moot.

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #746 on: March 07, 2014, 10:51:42 PM »
So? Why are you analyzing a Zed that has _no production output_ ? It is not the "real" zed and when you are done, this is what Travis will tell you. You are proving that a rock will fall in gravity.... Travis is telling you he has circumvented gravity. You are describing a spring that re-sets with no production, Travis is claiming a system that resets itself by using part of its own production. Ergo, your analysis is moot.

TK, my Analysis of the Ideal 3-layer ZED still shows excess Energy.  I have been trying to come to terms with MarkE on the way I performed that Analysis.  We switched gears to the no-pod, single riser, as it was easier for me to make clear how a single riser acts, and that Work can be performed as it rises.  Once MarkE can confirm his values that do not conform to mine for the Energy inside this system at each State due only to the columns of water, we should be on the same page.  Then we can go back to the 3-layer ZED and evaluate the potential Work that system could have done while lifting.  That has never been done, AFAIK, and is the reason that MarkE's Analysis does not show the excess Energy that mine does.

Once excess Energy is confirmed to be present or not, then we can see if we have anything more to talk about.  The math does not lie.  And there is only one correct solution.  If the Analysis does not show excess Energy, then I see no reason to continue.  But if it does show excess Energy, then I will discuss the way that Mr. Wayne has described that he can harvest and utilizes that net gain.  I say "can" because he described only one way in the Patent App. and in his diagrams and pictures/videos of a physical test bed.  That is the only way I have considered.

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #747 on: March 07, 2014, 11:04:33 PM »
I understand what you are doing, and why you are doing it, and I approve. I am not trying to discourage you.

What I am doing is pointing out that you are leaving honest Wayne Travis a huge hole to wriggle out of.

Just like you tell me that the Cartesian Diver is irrelevant to the real Zed (it isn't) and that the spring-loaded automatic bollard doesn't do what you have been showing the single Zed does (it does)..... honest Wayne Travis will look at your completed analysis, and will respond in one of two ways. Or maybe both.

If your "OU" numbers hold up in the correctly-modelled spreadsheet (they won't) he will say "I told you so all along" and if your numbers do NOT hold up once you've modelled correctly and re-inserted reality in the form of masses, compressibility, viscosity and so on.... he will simply say, "Thanks fellas... but the real Zed, which is now horizontal and which we call the Rotary Taz.... doesn't work that way at all, you have left out some important steps and it's really too bad you aren't worth your salt. I'll pray for you anyway."

#### TinselKoala

• Hero Member
• Posts: 13958
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #748 on: March 07, 2014, 11:14:41 PM »
In the real ZED the production is the lift from the risers, in the version in the videos that lift is resisted by a hydraulic ram that charges up the external accumulator.

So the internal forces raise the risers, the risers moving up move the production ram and compress the fluid in the ram and that fluid passes into the accumulator and gets stored, from there it is distributed to the flow assist rams and they apply there force against the lever connecting the two bags together.

In the "ideal ZED" this part is not being considered.

You left something out, Webby. Can you find it in the text below?

paejnrcnan oe ohropa rThatPortionOfProductionThatIsUsedOutsideTheZedsnanrf;o'ian ca hfn'

Whatever pressure or leverage or force that is used OUTSIDE, the part that isn't returned in the Flow Assist, the 600 percent OU part that runs external generators.... there will be a pressure drop once the hydraulic fluid has done its outside work. Right? So when this now lower pressure fluid is.... well, just what is done with it? Where is it in your model?

#### minnie

• Hero Member
• Posts: 1244
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #749 on: March 07, 2014, 11:18:22 PM »

"The math does not lie" As far as we know this is a passive device.
If we have no losses,what would the expected answer be?
If there is excess energy, where does it come from?
John.