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### Author Topic: Mathematical Analysis of an Ideal ZED  (Read 643746 times)

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #45 on: February 23, 2014, 05:34:48 PM »
Of course.

*sigh*  Would you be so kind as to describe the correct method in the form of a mathematical relationship?

Now you have lost me. Neutral buoyancy means that the mass of the displaced water is equal to the mass of the displacing object. Adding additional force pressing down on the "neutrally buoyant" object makes it sink, so is this extra force to be included in the figuring?  Since your risers are "massless" I think you are once again up against a place where your assumptions are non-physical and may be leading you off the correct track.

Please refer to the start condition and the charged condition shown in the first two diagrams here:

http://www.overunity.com/14299/mathematical-analysis-of-an-ideal-zed/msg387854/#msg387854

A specific volume of water is being added to the pod chamber (AR1).  If the ZED is acting strictly like a simple ideal hydraulic cylinder, then the outer riser should need to rise by an amount that would be equal to having received that same volume of water.  That volume is calculated by measuring how much the outer riser lifts, I think.  It is the same way we correctly measure your U tube diagram:  Draw a box around the ZED.  Then see what enters and exits that box.  Since we are using incompressible fluids the volume of water entering "the box" must be equal to the volume of the outer riser that exits "the box."  Is that correct?

When the charged ZED is released to rise, it will rise due to buoyant force until it is neutrally buoyant.  If the above paragraph is correct, that rise amount should be the measured volume of the outer riser that lifts up (and out of "the box").  So the actual rise * cross sectional area of the outer riser = volume that exited "the box."  That volume must be equal to the volume of water added during the charge.

I propose to draw the ZED as if it has risen to satisfy having stroked by the exact same volume that has been added by the charge.  The position of the fluids will be redistributed correctly due to the constraints of being incompressible.  If the ZED is acting exactly as a simple ideal hydraulic cylinder, then it should come to rest at this condition, and so equalize the buoyant forces caused by the charge and be neutrally buoyant.  If it is found to be NOT neutrally buoyant in this condition, it would have to stroke less or further and definitely not be acting like a simple ideal hydraulic cylinder.
« Last Edit: February 23, 2014, 09:46:16 PM by mondrasek »

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #46 on: February 23, 2014, 06:27:14 PM »
MarkE, the volume of water admitted is only enough to fill the pod chamber to 37mm height.  That is shown in the second diagram here:

http://www.overunity.com/14299/mathematical-analysis-of-an-ideal-zed/msg387854/#msg387854
Thanks.  That helps.  37mm will not cause an underflow / overflow problem.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #47 on: February 23, 2014, 06:31:01 PM »
Do you (or anyone else) know how to calculate the amount of energy that crosses into a system when a specific volume of water is introduced, starting at a pressure of zero and building linearly to a final pressure of Pin?

Here is what I would propose to try next:  Using the incompressible fluids in the model.  Is it correct to say that the Volume input (volume of water admitted into the bottom of AR1) should be equal to the Volume output as measured by the height change of the outer riser * the cross sectional area of the outer riser?

If so, a ZED that is drawn with that exact amount of rise and with the fluids re-distributed correctly should be neutrally buoyant, right?  If it is acting exactly as a simple ideal hydraulic cylinder?
The work calculation is still performed as the integral of F*ds.  You can normalize by using the area of the feed pipe ID.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #48 on: February 23, 2014, 11:16:19 PM »
*sigh*  Would you be so kind as to describe the correct method in the form of a mathematical relationship?
Say you have a syringe full of water, connected by a tube to the input port of your device. You use a spring scale to press the syringe plunger into the syringe. You monitor the distance the plunger has travelled (s), you record the instantaneous force readings from the scale (F), and you calculate the integral of the spring scale's instantaneous force readings over the distance the plunger has travelled. (Integral from 0 to s of F ds.) This results in an input energy (or work) value. You can press the empty syringe the same distance to get a value for the syringe frictional work that isn't injected, subtract the latter from the former, and you then have the actual work injected into the system.
Quote

Please refer to the start condition and the charged condition shown in the first two diagrams here:

http://www.overunity.com/14299/mathematical-analysis-of-an-ideal-zed/msg387854/#msg387854

A specific volume of water is being added to the pod chamber (AR1).  If the ZED is acting strictly like a simple ideal hydraulic cylinder, then the outer riser should need to rise by an amount that would be equal to having received that same volume of water.  That volume is calculated by measuring how much the outer riser lifts, I think.  It is the same way we correctly measure your U tube diagram:  Draw a box around the ZED.  Then see what enters and exits that box.  Since we are using incompressible fluids the volume of water entering "the box" must be equal to the volume of the outer riser that exits "the box."  Is that correct?
Maybe. Without knowing exactly what you mean I'm not sure. Recall that the surface area of a cylinder isn't cut in half when you cut the cylinder itself in half across its height, since the ends are still the same area as before... only the "wall" area has been cut in half.
Quote

When the charged ZED is released to rise, it will rise due to buoyant force until it is neutrally buoyant.
Nope. The buoyancy of an item does not change as it rises; if it was buoyant at the start, it will rise until the mass of the displaced water is equal to the _entire mass_ of the object.  Neutral buoyancy means that the entire volume of the item displaces the same mass of water as the item itself masses.  If you now only want to count the part of the item that still remains under the water line in your "neutral buoyancy"... I don't think this is legitimate.
Quote
If the above paragraph is correct, that rise amount should be the measured volume of the outer riser that lifts up (and out of "the box").  So the actual rise * cross sectional area of the outer riser = volume that exited "the box."  That volume must be equal to the volume of water added during the charge.
Maybe. How _far_ does the riser need to rise in order for that to be true? More, or less, than the distance you depressed the plunger of the syringe? Much less, I'll wager, since the syringe is of smaller cross sectional area. How much work is performed by that lift, though?

Quote
I propose to draw the ZED as if it has risen to satisfy having stroked by the exact same volume that has been added by the charge.  The position of the fluids will be redistributed correctly due to the constraints of being incompressible.  If the ZED is acting exactly as a simple ideal hydraulic cylinder, then it should come to rest at this condition, and so equalize the buoyant forces caused by the charge and be neutrally buoyant.  If it is found to be NOT neutrally buoyant in this condition, it would have to stroke less or further and definitely not be acting like a simple ideal hydraulic cylinder.

Automatic bollard. Is the retracted 300 pound bollard nearly "neutrally buoyant" because it only takes a few ounces lift to raise it all the way up? Compressible fluids store energy like springs, in the compression. Incompressible fluids can only store energy by "head height". But when you have raised up the head height of an incompressible fluid you have stored energy there, which can be returned, just as the spring in an automatic bollard helps you lift it easily and lower it gently. Once the bollard is "precharged" by assembling it with its compressed spring, you can lift and lower over and over again as many times as you like, making complete cycles, with only a few ounce-feet of work put in each time. And this is a single-layer system! So you've input a small amount of work (say one pound x lift distance) by reducing the input required to perform the task of lifting the bollard, and you've recovered a huge amount of work (the 300 pounds x lift distance) and so your "net" is just the difference between the two, say 299 pounds x lift distance. Right? So now all you have to do is take some of this "net" and use it to power another automatic bollard. And _THAT_ is where the difficulty lies.

#### LarryC

• Hero Member
• Posts: 911
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #49 on: February 24, 2014, 01:42:45 AM »
In the JPG below, I copied some of the important results from the 2 Flow Analysis and did comparisons to highlight some important performance differences between the Zed and Archimedes single and dual versions.

In the original 2 spreadsheets I added a separate column for the important 'PSI differentials' values (peach color), they are key to the efficiency of the Zed. It was previously part of the Input Ft Lbs formula.

Any ideas, questions or suggestions to improve the comprehension of the Zed process with these spreadsheet would be appreciated.

[size=78%]Thanks, Larry [/size]

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #50 on: February 24, 2014, 03:09:14 PM »
Say you have a syringe full of water, connected by a tube to the input port of your device. You use a spring scale to press the syringe plunger into the syringe. You monitor the distance the plunger has travelled (s), you record the instantaneous force readings from the scale (F), and you calculate the integral of the spring scale's instantaneous force readings over the distance the plunger has travelled. (Integral from 0 to s of F ds.) This results in an input energy (or work) value. You can press the empty syringe the same distance to get a value for the syringe frictional work that isn't injected, subtract the latter from the former, and you then have the actual work injected into the system.

Sorry to put you through that explanation.  My question was supposed to be tongue in cheek.  In this case (assumed linear pressure rise starting at zero Pa) that Integral resolves to Pin average * Vin.  When solving for the supposed Vout using an Energy Balance the equation is of the form similar to PinVin = PoutVout.  Which is also a representation of Boyle's law.  But just because it has the same form as Boyle's law does not mean it is correct to site that law for this case.  I was incorrect to do so.

Nope. The buoyancy of an item does not change as it rises; if it was buoyant at the start, it will rise until the mass of the displaced water is equal to the _entire mass_ of the object.  Neutral buoyancy means that the entire volume of the item displaces the same mass of water as the item itself masses.  If you now only want to count the part of the item that still remains under the water line in your "neutral buoyancy"... I don't think this is legitimate.

When the ZED is released from it's position in the "charged" state, the pod and risers will rise.  When that is happening, the water levels in the various annuli begin to change.  This affects the head level on each of those moving members and their buoyancy changes as they move.  It does not remain constant.  And so the pod and risers will rise until they are in a position where the sum of the buoyancy forces on each individual member resolves to zero.  This is what I am trying to represent when I describe the system rising until it is neutrally buoyant.  Neither the pod or any of the risers actually resolve to an individual neutrally buoyant state in the analysis being performed so far.  In the final state shown in the 2-layer analysis the outer riser is actually negatively buoyant, by a lot!  But the pod and inner riser are still pushing up.  The sum of these forces is a net negative buoyancy and would require that the system sink back down in order to achieve the neutrally buoyant system condition required for no further motion.  And since that appears to require less energy than predicted by the input, I conclude that the energy balance for the 2-layer ZED is showing underunity.

Maybe. How _far_ does the riser need to rise in order for that to be true? More, or less, than the distance you depressed the plunger of the syringe? Much less, I'll wager, since the syringe is of smaller cross sectional area. How much work is performed by that lift, though?

Of course the riser moves a much shorter linear distance than the syringe plunger, assuming a syringe with diameter smaller than the ZED's outer riser.  That is why I proposed to compare volumes.

The work performed by the lift is again the Integral of Pout * Vout.  The Integral of Pout is the average of the starting P due to the buoyant lift force (applied to the outer riser cross sectional area) and the ending P which must be zero.  Vout must equal Vin if the ZED is acting identical to a simple ideal hydraulic cylinder.

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #51 on: February 24, 2014, 04:27:55 PM »
Mondrasek, yes the levels are going to rearrange.  Ignoring the base there are four things that matter:

1) The surrounding atmosphere, and also trapped fluid.  We have defined this to be massless, incompressible material.
2) The riser cylinders.  We have defined these to be massless SG=0 materials.
3) The water.  SG=1.  I don't know what you are using for acceleration due to gravity, or density of water.  I use 9.80665N/kg for G0, and 0.9982g/cc for the density of water at 20C.

The cylinders do not exert any downward force of their own.
The trapped fluid does not exert any downward force due to weight.
Only the water exerts a downward force due to its weight, and a counter upward force on whatever displaces some volume of same.  That is the buoyant force.  Unless you change the problem by adding weights to the cylinders, only the water can be neutrally buoyant.

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #52 on: February 24, 2014, 04:52:27 PM »
3) The water.  SG=1.  I don't know what you are using for acceleration due to gravity, or density of water.  I use 9.80665N/kg for G0, and 0.9982g/cc for the density of water at 20C.

MarkE, I had used values of 9.81N/kg and 1.0g/cc for simplicity.  I was doing some cross checking with a conversion program that used those values and so also used their roundings for this simple study.  Feel free to use your values or change them as suits your preference.

#### mrwayne

• Hero Member
• Posts: 975
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #53 on: February 25, 2014, 02:45:40 PM »

I can remember when you were quite opposed to the ZED system, you were almost hostile - but you helped another engineer "Do the Math" and you asked me very hard questions.

I impressed with your intelligence and character, you did the math.

Our systems do not defy the math - and you are doing a great job presenting that.

Logically, that is obvious - a person should be able to prove or deny with the "math".

The right questions have to be asked - and the wrong prejudices have to be put on hold.

I hope you are able to teach others - You have certainly earned my respect.

Wayne

#### MarkE

• Hero Member
• Posts: 6830
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #54 on: February 25, 2014, 03:08:41 PM »

I can remember when you were quite opposed to the ZED system, you were almost hostile - but you helped another engineer "Do the Math" and you asked me very hard questions.

I impressed with your intelligence and character, you did the math.

Our systems do not defy the math - and you are doing a great job presenting that.

Logically, that is obvious - a person should be able to prove or deny with the "math".

The right questions have to be asked - and the wrong prejudices have to be put on hold.

I hope you are able to teach others - You have certainly earned my respect.

Wayne
Isn't it funny then that you have shown neither math that works, nor a unit that works in six years of selling this snake oil?

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #55 on: February 25, 2014, 03:52:52 PM »
Come on, Travis, show your electric bills for the last six months.

And then explain why they are non-zero, since you claim to be able to build a 20 kW free energy self running generator plant that will fit in the footprint of a garden shed. After all, you have all that empty space in your building. Where are the ZEDs pumping out free electricity? You have plenty of room.

I know where... and so do you.

How's that for math?

(sound of crickets chirping....)

Don't forget, Wayne old boy, that we have the PowerPoint demonstration where you claimed to your investor prospects that you would put a 50 kW free energy generating plant at your CHURCH in three months after receiving funding from the investors. And this PPT was made in November of 2010, according to data in the file.

I imagine the Church elders are pretty disappointed in you, since they have had to spend many thousands of dollars on electricity since they first read your promise back in 2010.

#### LarryC

• Hero Member
• Posts: 911
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #56 on: February 25, 2014, 04:14:58 PM »
Isn't it funny then that you have shown neither math that works, nor a unit that works in six years of selling this snake oil?

Interesting, then could you please show why the math showing an efficiency of 153.94% in the attached spreadsheet is incorrect? The efficiency is in field B16. The drawing shows the process cycle.

Thanks, Larry

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #57 on: February 25, 2014, 04:25:50 PM »

Interesting, then could you please show why the math showing an efficiency of 153.94% in the attached spreadsheet is incorrect? The efficiency is in field B16. The drawing shows the process cycle.

Thanks, Larry

Interesting, then could you please show why Travis, and everybody else, has been unable to produce a device that actually works and demonstrates a gain of energy? The failure to produce a real powerplant is evident in Travis's (and your) electric bills. The image in my last post above shows that he promised something years ago and has never accomplished it yet.

Thanks, TK

(Of course I know that Larry and Travis both have me on their "ignore" lists.... so I don't expect any kind of real answer to my questions. By the way, the word for a person who IGNORES things is.... ignoramus.)

#### mondrasek

• Hero Member
• Posts: 1301
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #58 on: February 25, 2014, 05:39:11 PM »
so I don't expect any kind of real answer to my questions.

Oh come on!  And you also gave _no real answer_ to LarryC's question.  He specifically asked how his MATH is incorrect.  To do so does not require any more words.  It requires math.

LarryC has presented evidence by the accepted method of science:  A mathematical solution/analysis.  If it is correct, it is correct.  If it is incorrect, it is incorrect.  There is no gray area.  There is only one correct solution to the math.  Checking his math and process is the only correct way to move forward.

I have also presented evidence by the method of a mathematical analysis.  I have also asked for my math and process to be checked.  I thank MarkE for his assistance so far and offer to double check.  TK, your assistance with the math is also appreciated.

If anyone would like to work on LarryC's math question instead of my own, that is fine, as his work appears to show the same anomaly that I am trying to find the reason for.

Thanks,

M.

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Mathematical Analysis of an Ideal ZED
« Reply #59 on: February 25, 2014, 05:44:22 PM »
Here's an incontrovertible fact: If your calculations show "overunity" performance, then you are making an error somewhere. It may be in the math, but it is more likely to be in the underlying assumptions that go into setting up the model. If you or LarryC want to pay me my usual daily consulting fee I would be more than happy to waste, er, spend the necessary time to go over things with a fine toothed comb. Otherwise, I think that spreadsheet analyses without real experimental data going into them, are worth approximately what they cost: nothing.

Meanwhile, real experimentation has never revealed any "gains" in work or energy from anything Travis, you, webby, LarryC, dale, RedSunset, etc have ever presented. That has got to be telling you something. After all, if LarryC gets 153.94 percent OU (lol) from two Zeds, think of the OU he would get if he chained six of them together. Easy and simple to get hundreds or thousands percent OU, right? But where, then, are the functioning models? FFS, it's just a bunch of stacked tubes. How complicated can that be, to build and demonstrate, if it's true? But we are in the position where _ALL_ valid experimentation has failed to show any OU at all, and in fact the experiments show massive losses and inefficiencies. Therefore... the theoretical models are failing, because they must be wrong somewhere. Digging out the place where the models are wrong is actually the responsibility of the person putting forth the model, not mine.

It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.
--Richard P. Feynman

ps: Just why should I respond reasonably to a strawman question from someone who is deliberately ignoring me? I respect people who respect me, like you. LarryC and Wayne Travis.... I don't respect them, because they have both insulted me and placed me on their ignore lists.

And if that weren't enough... I still believe the automatic bollard is showing the same "anomaly" that you think you have found. Nobody has "done the math" to prove me wrong about that. Why not? Do you deny that the 300 pound bollard can be raised to its full height, and lowered back down again, with just a few pound-feet of work? Reducing the input, getting the big output, subtracting the input from the output to get the "net" gain in work .... it's all there in the automatic bollard. So?