Language: 
To browser these website, it's necessary to store cookies on your computer.
The cookies contain no personal information, they are required for program control.
  the storage of cookies while browsing this website, on Login and Register.

GDPR and DSGVO law

Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding.
Amazon Warehouse Deals ! Now even more Deep Discounts ! Check out these great prices on slightly used or just opened once only items.I always buy my gadgets via these great Warehouse deals ! Highly recommended ! Many thanks for supporting OverUnity.com this way.

User Menu

A-Ads

Powerbox

Smartbox

3D Solar

3D Solar Panels

DC2DC converter

Micro JouleThief

FireMatch

FireMatch

CCKnife

CCKnife

CCTool

CCTool

Magpi Magazine

Magpi Magazine Free Rasberry Pi Magazine

Battery Recondition

Battery Recondition

Arduino

Ultracaps

YT Subscribe

Gravity Machines

Tesla-Ebook

Magnet Secrets

Lindemann Video

Navigation

Products

Products

WaterMotor kit

Statistics

  • *Total Members: 83536
  • *Latest: sockoman

  • *Total Posts: 514436
  • *Total Topics: 15310
  • *Online Today: 44
  • *Most Online: 103
(December 19, 2006, 11:27:19 PM)
  • *Users: 0
  • *Guests: 15
  • *Total: 15

Author Topic: Mathematical Analysis of an Ideal ZED  (Read 432086 times)

Offline mondrasek

  • Hero Member
  • *****
  • Posts: 1300
Re: Mathematical Analysis of an Ideal ZED
« Reply #660 on: March 06, 2014, 06:57:40 PM »
Have you downloaded the spreadsheet?  Everything you want should be in it.  The ZED is a piece of useless junk.  I don't know where you obtained OU numbers, but it certainly is not reflected in the ZED.  The "ideal ZED" using incompressible fluids, once you've paid all the energy to prepare it and charge it up, just acts like a linear spring.  If you make one with  compressible "air" it acts like a variable rate spring with additional loss.

MarkE, AFAIK you have ignored (thrown away) the Energy released by the ZED system when it is allowed to rise in every one of your attempts so far.  You have finally correct that in the simple no-pod, single riser example after being shown that there is Energy leaving the system due to the rise.  That Energy is F*ds which resolves to Fave*S for this specific case.  That same Energy value has NOT been calculated for any of your 3-layer attempts.  I am asking you to do so.  Without adding that correction to your previous work the way you have for the no-pod, single riser example, those proofs are wrong.  You ignore Energy that clearly leaves the system when the ZED rises.  Energy that can be used to do real Work.

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #660 on: March 06, 2014, 06:57:40 PM »

Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #661 on: March 06, 2014, 07:15:46 PM »
MarkE, AFAIK you have ignored (thrown away) the Energy released by the ZED system when it is allowed to rise in every one of your attempts so far.  You have finally correct that in the simple no-pod, single riser example after being shown that there is Energy leaving the system due to the rise.  That Energy is F*ds which resolves to 0.5*Fave*S for this specific case.  That same Energy value has NOT been calculated for any of your 3-layer attempts.  I am asking you to do so.  Without adding that correction to your previous work the way you have for the no-pod, single riser example, those proofs are wrong.  You ignore Energy that clearly leaves the system when the ZED rises.  Energy that can be used to do real Work.
Mondrasek, I have shown that these contraptions once they are set up reduce to linear springs.  I have explained the N*(X/N)2 tyranny issue of potential energy stores, and that includes linear springs and things that act like linear springs.  For non-linear devices it is worse.

Kindly read the spreadsheet.  Everything you need is in there. If you will not be bothered to look at the work I see no reason to pay attention to your demands for additional spoon feeding.  The single layer and three layer only differ in their coefficients.   Their completely under unity, linear spring emulating behavior is the same.  The machines are worthless, overly complex, lossy junk. 

The spreadsheet is very complete.  It audits correctly.  I have been very patient with you.  You have yet to show the calculations that you relied upon to reach the erroneous conclusion that a three riser system is OU.  Springs are not OU.  The three riser like the single riser behaves as a linear spring.  The ZED claims of over unity cannot be met by their emulation of ordinary springs.  QED.

Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #662 on: March 06, 2014, 07:24:32 PM »
MarkE,

This is kind of,, maybe off the discussion,, but I was looking at the new drawing for the end of state 3 and I find myself asking what would happen if I opened the valve for the pod chamber after the risers and pod lifted?

Would the water just kind of dribble out, would it be able to come out at all or would it come out with some force behind it?
There is at the end of State 3 still more energy in the system than at the end of State 1.  Why would you think that the system would not return to State 1?

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #662 on: March 06, 2014, 07:24:32 PM »
Sponsored links:




Offline mondrasek

  • Hero Member
  • *****
  • Posts: 1300
Re: Mathematical Analysis of an Ideal ZED
« Reply #663 on: March 06, 2014, 07:38:04 PM »
Mondrasek, I have shown that these contraptions once they are set up reduce to linear springs.  I have explained the N*(X/N)2 tyranny issue of potential energy stores, and that includes linear springs and things that act like linear springs.  For non-linear devices it is worse.

Kindly read the spreadsheet.  Everything you need is in there. If you will not be bothered to look at the work I see no reason to pay attention to your demands for additional spoon feeding.  The single layer and three layer only differ in their coefficients.   Their completely under unity, linear spring emulating behavior is the same.  The machines are worthless, overly complex, lossy junk. 

The spreadsheet is very complete.  It audits correctly.  I have been very patient with you.  You have yet to show the calculations that you relied upon to reach the erroneous conclusion that a three riser system is OU.  Springs are not OU.  The three riser like the single riser behaves as a linear spring.  The ZED claims of over unity cannot be met by their emulation of ordinary springs.  QED.

No, MarkE.  And after checking, I think we need to return to your "correction" to post #667.  There you introduced a "rate" into that Energy equation.  You will need to explain that.

Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #664 on: March 06, 2014, 08:13:16 PM »
I thought that it would, but was not sure since riser 2 is at a neutral buoyant condition and riser 3 is at a negative buoyant condition leaving riser 1 and the pod positive.  I was not sure if the atmosphere pushing down on riser 3 and the pressure left within the system would do anything.
There would in the hypothetical set-up be no atmosphere.  In practice there of course would be.  However, the water still has weight.  All of the columns would be happy to equalize if they could.  They can't because of the restriction that air can only enter at AR7.  Open up the drain plug under the pod chamber and the restriction no longer exists. 

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #664 on: March 06, 2014, 08:13:16 PM »
Sponsored links:




Offline mondrasek

  • Hero Member
  • *****
  • Posts: 1300
Re: Mathematical Analysis of an Ideal ZED
« Reply #665 on: March 06, 2014, 08:38:07 PM »
MarkE, you introduced a rate into an Energy calculation.  Rate does not apply when calculating the Energy leaving the system as the piston rises in this case. 

The correct Energy = F*ds reduction for the case of the riser lifting the water on top of the piston is simply Fave*S.

The Force starts at the maximum buoyancy Force that can be calculated from State 2.  It then declines linearly to exactly zero which is the case in State 3 since the water is flowing off of the piston as it is rising.  This is a straight forward calculation and involves no rates.

Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #666 on: March 06, 2014, 08:38:21 PM »
No, MarkE.  And after checking, I think we need to return to your "correction" to post #667.  There you introduced a "rate" into that Energy equation.  You will need to explain that.
Do you or do you not understand how to integrate in order to obtain energy?

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #666 on: March 06, 2014, 08:38:21 PM »
Sponsored links:




Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #667 on: March 06, 2014, 08:40:31 PM »
Not to sound to silly,, but would that then suck the water up AR6 and over into AR5?
No, opening the pod chamber would allow water to flow back from AR6 into AR7.  It's just a collapsing balloon. 

Offline TinselKoala

  • Hero Member
  • *****
  • Posts: 13971
Re: Mathematical Analysis of an Ideal ZED
« Reply #668 on: March 06, 2014, 08:41:20 PM »
Don't forget that the atmosphere outside the system is pushing with a constant pressure. Nothing the apparatus does can affect this pressure. The outer surface area of course changes, the volume that displaces air outside changes, and there IS measurable "air buoyancy" from the displaced outer air.... but the outer air is not an "ideal gas" in the sense that the apparatus cannot store energy in the outer air, nor extract energy from it. This is because changes in the apparatus volume do not cause changes in the pressure of the outside air. And if the inner fluids of the apparatus are incompressible, then also changes in the outside air pressure do not cause changes in the apparatus volume. By having two incompressible fluids in your apparatus you have killed all chances of taking advantage of the Cartesian Diver effect, and also the submarine's manner of altering  buoyancy by pumping compressed air into and out of tanks to empty and fill them with water.

So.... how are you going to "turn buoyancy on and off"?

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #668 on: March 06, 2014, 08:41:20 PM »
3D Solar Panels

Offline mondrasek

  • Hero Member
  • *****
  • Posts: 1300
Re: Mathematical Analysis of an Ideal ZED
« Reply #669 on: March 06, 2014, 08:41:45 PM »
Do you or do you not understand how to integrate in order to obtain energy?

MarkE, we appear to have cross posted.  Please see my post previous to yours.

Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #670 on: March 06, 2014, 08:46:02 PM »
MarkE, you introduced a rate into an Energy calculation.  Rate does not apply when calculating the Energy leaving the system as the piston rises in this case. 

The correct Energy = F*ds reduction for the case of the riser lifting the water on top of the piston is simply Fave*S.

The Force starts at the maximum buoyancy Force that can be calculated from State 2.  It then declines linearly to exactly zero which is the case in State 3 since the water is flowing off of the piston as it is rising.  This is a straight forward calculation and involves no rates.
What are you putting into your coffee?  F*ds is not energy.  The integral of F*ds is energy.  The equation for up lift force as the riser goes up is just as I wrote it, and conforms to your description.  Similarly, the solution of the integral that I wrote is correct for these circumstances.  The second term in the integral is identically half the first term.

Free Energy | searching for free energy and discussing free energy

Re: Mathematical Analysis of an Ideal ZED
« Reply #670 on: March 06, 2014, 08:46:02 PM »
3D Solar Panels

Offline MarkE

  • Hero Member
  • *****
  • Posts: 6830
Re: Mathematical Analysis of an Ideal ZED
« Reply #671 on: March 06, 2014, 09:09:05 PM »
Alrighty,, it is kind of funny looking because riser 2 is neutrally buoyant and the water is "hanging" on riser 3 and pulling the water in AR5 up, so I was wondering about what happens when the push from the water from the other side goes away, it looks like riser 2 would also have water hanging on aka a negative buoyant condition,, all of that of course since the "head" around the pod will need to drop down before the pod can lower allowing the system itself to lower,, well it is a case of the balancing act,, naturally the pod will drop down, or maybe be pushed down,,
It may seem non-intuitive to you, but the fluid in the pod chamber is drove the system to the State 3 condition.  That fluid remains under pressure.  The pressure is countered via the coupled force on the various sides of the risers.  If you open the pod chamber, the whole thing will collapse without hesitation. The pod and Riser 1 have net upward pressure on them, meaning that they are net pushing back down on everything in the pod chamber and AR2.  Open the pod chamber at the bottom and that positive pressure will drive fluid out.  That will unbalance the rest of the assembly and the risers will collapse down and the water volumes will equalize.  The reachable lower energy state is State 1:  32.5mm in AR[2-7], and no fluid in the pod chamber.

Offline mondrasek

  • Hero Member
  • *****
  • Posts: 1300
Re: Mathematical Analysis of an Ideal ZED
« Reply #672 on: March 06, 2014, 09:16:06 PM »
What are you putting into your coffee?  F*ds is not energy.  The integral of F*ds is energy.  The equation for up lift force as the riser goes up is just as I wrote it, and conforms to your description.  Similarly, the solution of the integral that I wrote is correct for these circumstances.  The second term in the integral is identically half the first term.

MarkE.  I apologize for leaving out the word "integral" again.  It is a bad habit.

I think I saw this integral reduction shared by you.  It is the one we are using for the Energy in the water introduced during the change from State 1 to State 2.  That integral of F*ds is 0.5(Pstart-Pend)*V for the special case where the start or end Pressure is zero.  Is that correct?

Offline minnie

  • Hero Member
  • *****
  • Posts: 1244
Re: Mathematical Analysis of an Ideal ZED
« Reply #673 on: March 06, 2014, 09:31:44 PM »



   Hi,
      if you imagine a see-saw, equally weighted, try taking say one eighth of the weight from
 one side. The now lighter side will stubbornly rise, that's what will happen if you try
 this with your ZED.
     The pod can be discounted because it's just Archimedes, and the rest of the thing will
 just behave like any ordinary telescopic ram, if you have massless, incompressible air.
    Anyone who seems to be getting more out than is put in has seriously got to hunt for
 glitches in their work.
     Anyone proving that you really can get free energy out of this sort of device is going to
 become a very famous person!
   I'm not saying that it can't be done, but it sure is one hell of a task
                            John.

Offline mondrasek

  • Hero Member
  • *****
  • Posts: 1300
Re: Mathematical Analysis of an Ideal ZED
« Reply #674 on: March 06, 2014, 10:19:48 PM »
Sure:

Take a volume where we are going to eject water replacing it with an incompressible fluid, where:

H is the height of the volume.
A is the cross-sectional area of the volume.
Ge is the acceleration due to gravity on earth.
pW is the density of water.
pX is the density of the incompressible fluid.

The pressure difference from bottom to top of the volume varies from 0 to H*Ge*(pW-pX).
The force required varies from 0 to H is A*Ge*(pW-Px)*H.
The work done is the integral of F*ds: = Integral( A*Ge*(pW-Px)*H dh)
The solution of the integral from 0 to H is of the form:  Kh*(H2^2 - H1^2) + F0*( 0.5*A*Ge*(pW-pX)*H^2

For pX = 0:  = 0.5*A*Ge*pW*H^2
And since the weight of water in the volume would be: Wdisplaced = A*Ge*pW*H, we get:  E = H/2*Wdisplaced.


From page 1 of this thread (Valentines day no less).  So the Energy integral resolves to H/2*Wdisplaced.  Or 0.5*H*Wdisplaced.  We have changed the variable that represent each value along the way, but I believe that Wdisplaced is the Fstart value we calculate from the buoyancy Force.  And H is our lift height, or S as we have recently called it.  And so the Energy integral does resolve properly to 0.5*Fstart*S.

With one problem.  "For pX = 0:  = 0.5*A*Ge*pW*H^2."  We are not pushing on the water in the piston directly with air (pX = 0).  We are pushing on it with the physical piston.  But is it okay to say that the piston is being moved by the air that is re-distributing in the ZED during the rise from State 2 to State 3 and therefore this equation applies?

 

OneLink