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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 749616 times)

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #570 on: March 05, 2014, 11:22:45 PM »
I object to this improper claim that buoyancy forces are "turned on".  Force builds from zero linearly and incrementally as water pumped in is forced around the obstacles in its path.  There is no "on" state or contrary "off" state.

MarkE, you can object to this description of "turned on" as it is not an instantaneous change between a state of 0 or 1.  But it is the vernacular used by the majority of the population.  I was only trying to point this out.

Ever "turn on" a CRT device and wait for the picture to appear as the tube "warms up?"  Same thing.  "Turn on" does not have to mean instantaneously.  There is a delay in the on and off state of every device, no matter how high the switching frequency.

Forces don't resolve.  Acceleration stops when net force reaches zero.

Yes, acceleration stops in this single ZED system when all the buoyancy Forces sum to zero.  If they are left at any other value then those Forces are: a) unresolved (my term), or b) (please tell me how to properly express this condition here).

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #571 on: March 06, 2014, 12:19:02 AM »
MarkE, you can object to this description of "turned on" as it is not an instantaneous change between a state of 0 or 1.  But it is the vernacular used by the majority of the population.  I was only trying to point this out.
I challenge you to show where this idea of "turned on" is accepted in industry or academia.  It is a bull shit suggestion by our own HER/Zydro.  It is part of their misdirection.
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Ever "turn on" a CRT device and wait for the picture to appear as the tube "warms up?"  Same thing.  "Turn on" does not have to mean instantaneously.  There is a delay in the on and off state of every device, no matter how high the switching frequency.
No it is not.   The buoyant force builds linearly from zero as water is pumped in.  It has no time dependency.  It has no state dependency.  More displacement = more force.
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Yes, acceleration stops in this single ZED system when all the buoyancy Forces sum to zero.  If they are left at any other value then those Forces are: a) unresolved (my term), or b) (please tell me how to properly express this condition here).
Unbalanced forces mean net force.  Newton's Second Law still applies:  F = mA.  This example like your two riser and three riser before it is fundamentally lossy.  You start by supplying work to create State 1.  Then you add work, ideally without loss adding potential energy to get to State 2.  Then without extracting any useful work, you lose more than 2/3 of the potential energy you added to get to State 2 by going to State 3.   So, this scheme is less efficient than a brick.  And yet it is the "ideal ZED".  That means that real ZEDs with real friction can only underperform this machine that is already less efficient than a brick.  The best ZED is therefore no ZED at all.  HER/Zydro's claims to extra energy by using pods and risers are by your example of the ideal device: completely refuted. 

Now, what I want to know is that having last night looked back at the hydro differential thread and having seen that among others Kan Shi explained all of this almost two years ago while you were engaged, why do you still resist what you obviously have the capacity to understand?



MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #572 on: March 06, 2014, 12:30:11 AM »
I tried not to misrepresent anything.

Actually MarkE,, you need to take that up with TK, he says he is the one to first use that term, and he demonstrated that the virtual water provided the same change in scale reading as the real water.

Still, it only took a small volume of real fluid to make that big change.
I highly doubt that TK being the learned person he is ever claimed to be the discoverer of Archimedes' paradox.  I am quite confident that he explained the paradox, which really isn't a paradox at all when one thinks about it carefully.  We covered this before when we discussed Grimer's cement volume derivations.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #573 on: March 06, 2014, 12:40:00 AM »
Newton's Second Law still applies:  F = mA.

Yes it does.  And lets look at that relationship.

F = mA

m is mass.  Mass is a property of a physical material that does not change for the accepted IDEAL conditions of a constant temperature and obvious absence of a state of matter change.  Therefore m is a CONSTANT.

A is acceleration.  In this case it is the acceleration due to gravity.  It is also a CONSTANT.

So the Force (F) in F = mA is a mathematical fact which the calculation of cannot be disputed.  It is the product of two CONSTANTS (and yes, TK, one is a vector so the result is a vector).

So, what is Energy?  It is a resultant of the prior mathematical fact that is Force.  It is F*ds (where ds is distance).

F is Force which is the product of two constants.

ds is distance which is another calculable (or measurable) physical fact and therefore a CONSTANT.

Ergo, you must solve for Force before you calculate the Energy.  And regardless of the outcome of that Energy value, it must be correct.

In your first attempt at this Analysis you solved first for Energy Balance.  This was erroneous and resulted in a physical State 3 that could not actually exist due to "unresolved" Forces in the system that did not sum to zero.


MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #574 on: March 06, 2014, 12:43:45 AM »
MarkE,

Why would you NOT extract the useful work that can be extracted, sure if I run my car and don't go anywhere I will get ZERO MPG.

Common sense would say that if you can extract with no other effects then maybe you should, what part of this is it that you do not understand.
Webby, you are stuck on the same potential transfer problem that seems will vex you forever. 

For each um of upward movement we can calculate the net force on the riser.  It goes from its maximum starting value to zero.  It does so as a linear function.  Therefore we can express the force as:  F = F0 - K2*Z.  (K2 because I do not wish to confuse this constant with the pi/4*pwater*G0 K1 I often use.)  In other words the system operates like a cocked spring.  And now comes the problem with the ZED:  It is the way that the spring is utilized that is fundamental to the energy loss.  The potential energy stored in this "spring" is transferred into other potential energy in a direct process.  As a result, the process always suffers from the tyranny that:  N*(X/N)2 = X/N.  The only way to break even is to restrict N to 1.0, which means that you cannot redistribute potential energy from one store to another by any direct transfer mechanism without loss.  Gears, levers, pulleys, or in this case the riser, IE anything that translates static force and distance to another static force and distance constitutes making N greater than zero, and you lose: period.  In the case of the ZED, the less upward movement of the riser, the better.  The optimum is zero.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #575 on: March 06, 2014, 12:49:33 AM »
MarkE, regarding your note inserted in the graphic that states:

3) No work extracted going from State 2 to State 3

We have already gone over this?  There is an ASSUMED non-physical device that MUST restrain the ZED from rising unimpeded (and wasting all that Energy rather than collecting it) that would account for the Work/Energy you keep throwing away.

This is an IDEAL Analysis and so a physical device should not need to be presented.  But if you insist that one does, I will oblige.  Please let me know if you need to see a physical manifestation of a "Worked on Device" or if you can agree that the Energy "lost from the system" due to the lift could have been collected.  Obviously we have all the correct ingredients:  A Force (from buoyancy) and a ds (distance that the riser lifts)?

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #576 on: March 06, 2014, 12:53:35 AM »
Yes it does.  And lets look at that relationship.

F = mA

m is mass.  Mass is a property of a physical material that does not change for the accepted IDEAL conditions of a constant temperature and obvious absence of a state of matter change.  Therefore m is a CONSTANT.

A is acceleration.  In this case it is the acceleration due to gravity.  It is also a CONSTANT.
Yes, in this case with your stipulations for the "ideal ZED" m of the riser and "air" is constantly zero.  Therefore the kinetic energy is constantly ... wait for it: zero.
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So the Force (F) in F = mA is a mathematical fact which the calculation of cannot be disputed.  It is the product of two CONSTANTS (and yes, TK, one is a vector so the result is a vector).

So, what is Energy?  It is a resultant of the prior mathematical fact that is Force.  It is F*ds (where ds is distance).

F is Force which is the product of two constants.
No, F is whatever function defines it over the traversed distance S that it will be evaluated.  In the other thread I thought I read you saying that you work with CFD.  How could you work with CFD and misrepresent these fundamental concepts? Are you trolling?
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ds is distance which another calculable (or measurable) physical fact and therefore a CONSTANT.
I find it hard to believe that you flunked calculus.  But, if you want to represent that you did, who am I to argue?
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Ergo, you must solve for Force before you calculate the Energy.  And regardless of the outcome of that Energy value, it must be correct.
You must solve for the force at each incremental point over a path in order to solve for the energy applied.  Drag a heavy object with a real coefficient of friction for a distance and you apply real work.  That work all converts to heat.  You end up with zero kinetic or potential energy in the thing you dragged.
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In your first attempt at this Analysis you solved first for Energy Balance.  This was erroneous and resulted in a physical State 3 that could not actually exist due to "unresolved" Forces in the system that did not sum to zero.
No, I calculated the energy loss for the change in internal energy.  You may rightly contest that I did not correctly solve the equilibrium height, because I did not. But the energy loss calculated was correct for the calculated movement, and only gets worse going to the higher true equilibrium height.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #577 on: March 06, 2014, 01:00:18 AM »
MarkE, regarding your note inserted in the graphic that states:

3) No work extracted going from State 2 to State 3

We have already gone over this?  There is an ASSUMED non-physical device that MUST restrain the ZED from rising unimpeded (and wasting all that Energy rather than collecting it) that would account for the Work/Energy you keep throwing away.
Your assumption that some device can magically collect the lost energy is a fallacy.  Place any mechanism that you like in communication with the risers and show that you don't lose energy.  You cannot.  But go ahead and prove me wrong.  Every um of movement by the riser results in permanently lost energy.  I have shown the physical basis for this and the associated math.  If you want to argue differently, do more than exclaiming "No it isn't."
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This is an IDEAL Analysis and so a physical device should not need to be presented.  But if you insist that one does, I will oblige.  Please let me know if you need to see a physical manifestation of a "Worked on Device" or if you can agree that the Energy "lost from the system" due to the lift could have been collected.  Obviously we have all the correct ingredients:  A Force (from buoyancy) and a ds (distance that the riser lifts)?
I do insist, because it is fundamental.  You cannot collect what you lose lifting, because the very act of lifting changes the N in N*X/N2 to a value greater than 1.0. 

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #578 on: March 06, 2014, 01:13:33 AM »
In the other thread I thought I read you saying that you work with CFD.

Sorry, but you must be remembering someone else.  I do no work with CFD, and have not since the days when grad students waited for their time slots on a NASA supercomputer in the early morning hours to run their sims.  But I was not directly involved with their work.  I was an undergrad at the time.

mrwayne

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Re: Mathematical Analysis of an Ideal ZED
« Reply #579 on: March 06, 2014, 01:13:48 AM »
Sometimes simple things get missed.

It doesn't help if the missed things are counter intuitive.

If you assume a system can only be 100%, you must conclude that any loss means no Net Energy.

If the assumption holds true - no Net Energy.

................

Yet what happens when the system is 105% then 160% or 340%

Can you as a designer choose to use components that have some losses and still have NET.

and more over - if a standard car engine is 33% efficient - and powering a 330% efficient ZED -

Pretty unlikley to have enough losses that result in no Net gain.

...................

Just saying - you need to open the Box a bit.

















orbut 3000

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Re: Mathematical Analysis of an Ideal ZED
« Reply #580 on: March 06, 2014, 01:17:46 AM »
Now - Orbo - "an obvious lie and a false claim"

What do you base that off?

Wayne


Wayne, good question, you're almost there. My observation is based on the following statement:
"Our Technology produces clean energy Mechanically, by altering the the once believed conservative field of gravity - allowing us to supply endless and abdundant clean Energy."


That statement is obviously wrong.


You should re-read the thread more carefully. User @MarkE has explained very well why your claim is at odds with reality.
Don't give up yet, you're almost there.


orbut



mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #581 on: March 06, 2014, 01:24:31 AM »
Your assumption that some device can magically collect the lost energy is a fallacy.  Place any mechanism that you like in communication with the risers and show that you don't lose energy.  You cannot.  But go ahead and prove me wrong.  Every um of movement by the riser results in permanently lost energy.  I have shown the physical basis for this and the associated math.  If you want to argue differently, do more than exclaiming "No it isn't."I do insist, because it is fundamental.  You cannot collect what you lose lifting, because the very act of lifting changes the N in N*X/N2 to a value greater than 1.0.

Disregarding any unexpected events, I will draw up the simplest physical device I can conceive of that shows the lift is performing Work and provide it to you tomorrow.

Meanwhile, I would still like to learn your method to find the final end of lift state that resolves the positive buoyant Forces in the current State 3.  Like I said before, I could only imagine to do that iteratively.  But I believe you could find a way to reduce the calculus to an equation that would give the final (net zero buoyant Forces) lift height.  You do have mad math skills.  Please understand that this is not a demand or requirement for our current analysis.  I am just eager to learn if there is a simple way to do what I currently find horrendously difficult.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #582 on: March 06, 2014, 01:41:34 AM »
Sometimes simple things get missed.

It doesn't help if the missed things are counter intuitive.

If you assume a system can only be 100%, you must conclude that any loss means no Net Energy.
You have offered no evidence that the First Law of Energy is wrong.
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If the assumption holds true - no Net Energy.
It is not an assumption, it is a law developed from countless careful observations.  You have offered zero counter evidence.
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................

Yet what happens when the system is 105% then 160% or 340%
What happens if Peter Pan and Godzilla get into a smack down?  What happens if Benjamin Netanyahu declares he is a Sunni muslim?  We can hypothecate fantasies all day long.  Perhaps someday you will understand that when it comes to energy efficiency, the only value greater than 100% is undefined.
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Can you as a designer choose to use components that have some losses and still have NET.
No, energy is conserved.  Again, see the First Law of Energy.  You may want to practice that because paragraph 0008 of your patent application will trigger a rejection for lack of utility because of the claim to a First Law violation.
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and more over - if a standard car engine is 33% efficient - and powering a 330% efficient ZED -
Don't forget to include Captain Hook pushing from behind ...
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Pretty unlikley to have enough losses that result in no Net gain.
No, it is completely impossible that you have a First Law violation.
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...................

Just saying - you need to open the Box a bit.
The box that we can all hope opens once long enough for you to enter is 6' x 9'.  Then you can have plenty of time to explain your ideas to Bubba.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #583 on: March 06, 2014, 01:56:00 AM »
Disregarding any unexpected events, I will draw up the simplest physical device I can conceive of that shows the lift is performing Work and provide it to you tomorrow.

Meanwhile, I would still like to learn your method to find the final end of lift state that resolves the positive buoyant Forces in the current State 3.  Like I said before, I could only imagine to do that iteratively.  But I believe you could find a way to reduce the calculus to an equation that would give the final (net zero buoyant Forces) lift height.  You do have mad math skills.  Please understand that this is not a demand or requirement for our current analysis.  I am just eager to learn if there is a simple way to do what I currently find horrendously difficult.
The method should be easy to understand:  Determine the initial net up force.  Then determine the up force as a function of lifted distance.  Solve for a change in up force equal and opposite to the initial up force. 

Where you and I diverge is this:  I am concerned only with your original problem statement that concerns your stated belief that a two riser system loses energy, while a three riser system gains energy.  I have shown that any upward movement by the riser results in net total energy loss, as necessitated by the:  E = k*N*X/N2 relationship.  As demonstrated, there are enough details that must be accounted for in determining the equilibrium height that it is easy to make a mistake getting the value.  But whatever the value is, no energy is imparted to the massless, incompressible fluid, nor the massless riser(s) in changing their heights.  If you load the riser with a weight, you do three things:  You reduce the height of the new equilibrium state, you reduce the amount of lost internal energy, and you convert a fraction of that smaller lost energy into change in GPE of the raised mass.  The larger that you make the mass the smaller the net loss until you reach a limit at zero movement, zero energy imparted to the mass that then doesn't lift at all, and zero internal energy lost.  IOW, the best you can do is to never leave State 2 which means that the best that you can do is to emulate a brick.

The moment you allow the risers to lift by any amount, you lose net energy.  Since the ZED/Zydro scheme relies upon this lift to operate, it loses energy.  Adding a second, a third or M such assemblies only multiplies the loss.  The claims by HER/Zydro that they have realized any improvement in hydraulic or hydraulic / pneumatic or hydraulic / pneumatic / buoyancy schemes is false.  The claims that they have discovered any new physics is false.  The claim that they can switch buoyancy on and off is false.  The claim that they can generate energy ex nihilo is false.

mondrasek

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Re: Mathematical Analysis of an Ideal ZED
« Reply #584 on: March 06, 2014, 02:10:21 AM »

BTW, I did say:
ds is distance which is another calculable (or measurable) physical fact and therefore a CONSTANT.

And you replied with:
I find it hard to believe that you flunked calculus.  But, if you want to represent that you did, who am I to argue?


Distance is an indisputable fact.  Calculus does not apply.  Distance is simply distance.   It is a measurement that has units of length.  In SI the unit of length is the meter.

Once a distance is calculated or measured it is a CONSTANT that can in no way be in dispute.