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Author Topic: Mathematical Analysis of an Ideal ZED  (Read 749315 times)

TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #990 on: March 13, 2014, 06:44:24 PM »
Let us not also forget that Honest Wayne Travis spoke of his "patent" many times in the early days. He had to be shown that it might actually be illegal to try to sell something, claiming it is "patented", when all there really is, is a patent _application_ that has been filed, but not yet granted.

This distinction between an actual granted patent, and a mere patent application (especially a WIPO app) seems to be lost on a lot of "free energy inventors". Travis's podmate Rosemary Ainslie did the same thing, claiming a patent that she never actually held.

I think it was actually  minnie, who first pointed out the implausibility of using water heads to make the kind of power outputs that Honest Wayne has claimed. Once we got to looking at the issue, I then was more interested at the hydraulic fluid volume flow rate and pressure needed to turn an ordinary hydraulic motor driving an ordinary PM alternator to make the multi-kW output power he touted. Sure, make your magic with Zeds, whatever. How do you get the high pressure, high volume flow of _hydraulic fluid_ to turn the ordinary parts downstream of the Zeds? There is nothing in any of what we have seen that can produce the flow of hydraulic fluid necessary to turn a generator at 65 HP or even 5 HP.

Pirate88179

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Re: Mathematical Analysis of an Ideal ZED
« Reply #991 on: March 14, 2014, 02:40:45 AM »
"Zed is dead baby".

(Bruce Willis's character at the end of Pulp Fiction)

Bill

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #992 on: March 14, 2014, 02:50:31 AM »
MarkE,

Is F147 correct?
Yes.

F147 is the air volume under Riser 1 in state 3.  The equation used is: 

=(Wall1Height-ST3AR1HEIGHT)*AR1CirArea+  //This is the air that is in the annular ring between the pod and innermost ring wall up to the top of the ring wall
AR2CirArea*(Wall1Height-ST3AR2HEIGHT)+ //This is the air from in between the ring wall and the riser wall
(ST3HeightIncrease*(AR1CirArea+AR2CirArea+AR1_2CirArea))+ //This is the air above the ring wall but below the original 1mm high "attic" under the ID
VerGap*AR2OD^2 //This is the original 1mm high attic under the ID




MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #993 on: March 14, 2014, 05:00:27 AM »
What is inescapable in all of this is that pressing down on the assembly lifts up a quantity of water and letting go allows that quantity of water to fall back down.  there is no magic.  There is no cheat on gravity.

orbut 3000

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Re: Mathematical Analysis of an Ideal ZED
« Reply #994 on: March 14, 2014, 05:22:02 AM »
I think the idea is to outsmart gravity by overly complicated plumbing. The problem is that gravity is too stubborn and unimaginative to be outsmarted by such efforts. Investors and cultists are not, unfortunately.

TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #995 on: March 14, 2014, 06:11:38 AM »
I think the idea is to outsmart gravity by overly complicated plumbing. The problem is that gravity is too stubborn and unimaginative to be outsmarted by such efforts. Investors and cultists are not, unfortunately.
Sounds exactly like my bathroom, actually.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #996 on: March 14, 2014, 06:53:36 AM »
Basically, if you take a deep dive into it what you have is a set of force gains set up by the nested risers.  There is a term for each:

1. Pod 0 before there is water in the pod chamber
2. Riser 1 underside ID area from pressure conveyed through the "air" underneath
3. Riser 1 top side OD area from pressure conveyed through the "air" above
4. Riser 1 side wall area from pressure in the water under the wall
5. Riser 2 underside ID area from pressure conveyed through the "air" underneath
6. Riser 2 top side OD area from pressure conveyed through the "air" above
7. Riser 2 side wall area from pressure in the water under the wall
8. Riser 3 underside ID area from pressure conveyed through the "air" underneath
9. Riser 3 side wall area from pressure in the water under the wall

Each of these terms is itself the result of several other coefficients.

For any given "charged" state, each of these components add together to form a final, single height dependent restoring force versus distance coefficient.  For any given "charged" state we can replace the whole affair with a compression spring with an appropriate rate constant and preload distance.

If we were to create a State 1.X where we do not restrain the system after venting in State 1, then just as in my soda and water bottle demonstration:  even though the water levels are even we have to apply downward force to maintain those even displaced values.  As we saw in the photographs, sealing the vents and letting go of the down force causes the system to find a new equilibrium where the reflected weight of lifted water matches the weight of water displaced by the pontoons, or in the case of the "ideal ZED": the riser walls.  As long as the outer annular ring:  AR7 can freely communicate with the outside atmosphere as stipulated in the problem, the assembly would rise up a short distance until the buoyant up lift on the riser walls is offset by the reflected weight of water.  We have again a device that emulates a spring between the State 1, and State 1.X states, just as it does albeit with different rate constant and preload distance between State 2 and State 3.

Whether he realized it or not, Mondrasek scuttled his own claim of an overunity result when he came up with his clever variable rate payload.  His payload has the same characteristics as a linear spring as an exactly matched load to the transfer function of the "idea ZED".  Therefore Mondrasek acknowledges that his three layer "ideal ZED" has the transfer function of a linear spring.  Linear springs are not OU, and therefore his three layer "ideal ZED" cannot be OU.  In reality it must be under unity even in the ideal case, because the payload force must be less than the "ideal ZED" lifting force over at least the initial part of the travel, or the payload will not move.



MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #997 on: March 14, 2014, 07:38:09 AM »
Pop math quiz:  How many hp*hours does it take to fill, or can be extracted by emptying a volume of water equal in size to an Olympic sized pool?

An Olympic sized swimming pool is 50m x 25m x 2m. 
The pressure at the bottom of the pool is:  2*9789 Pa/m = 19,578Pa. 
KFORCE = 9789N/m2/m*1250m2 = 12.236E6N/m
E=0.5*12.236E6N/m*4m2 = 24.5MJ

Alternately, we can use 0.5*19,578Pa*2500m3= 24.5MJ.

1hp = 745.7W
1h = 3600s
1hp*h = 2.68MJ

24.5MJ = 9.12hp*h.

Extra credit question:  How many Olympic sized pools of water would we need to empty or fill to exchange 5hp for 48 hours?

A:  5hp*48h = 240hp*h/9.12hp*h/pool = 26.3 Olympic sized pools.

Special bonus question:  How much water are we talking about?

A: 26.3*2500m3 = 65,800m3, or about 53 acre feet of water.


We now know why the Dansie demonstration has been delayed:  HER/Zydro must still be looking for land to acquire.

minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #998 on: March 14, 2014, 08:24:01 AM »



    I can imagine the scene in the Travis household. Sandy says "Wayne you're getting
   yourself into deep water with this stupid forum-you're  banned"!
        Which is pretty much the same as what goes on at my house, the women  keep
    things in order.
         I've been struggling with my sewage disposal machine. I have a fairly big family
    so it's a big 'un. 50 arse power. I put on a new motor three years ago and used a
    thermal overload that was too big so the motor ended up as toast. What amazed
     me was the unloaded current-it was very little less than what the trip had to be set
    for. Heaven knows how these mo-gen things are supposed to work
                     John.

minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #999 on: March 14, 2014, 08:50:41 AM »



     Webby,
               Imagine a u tube with one side say 10m tall and 1m diameter and the other side
   say the same height but 1cm diameter. There is a valve at the bottom which is closed.
    Fill any one side and open the valve. Can you think of any way of restoring the original
   potential to its original value without a bit of pumping?
                John.

TinselKoala

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1000 on: March 14, 2014, 09:48:25 AM »
Just like that,, nope,, but then things are not just like that.

It is kind of like TK carrying on that the rams must stroke for feet and feet and feet,, hogwash,, never heard of a lever??
Hogwash... so that's the secret of the ZED! It's full of hogwash!
Yes, Webby, I've heard of levers. Did you know that the product of force times distance is the same for both ends of the lever? Move one end a long way with little force and the other end moves a short way with great force. Or vice versa. You can have small travel and high pressure at one end, but the other end is going to have large travel... and low pressure.
http://www.youtube.com/watch?v=_HuxmF_1Z90
Quote

How much REAL water does it take to lift 65,000lbs of virtual water?

How long does it take to lift that virtual water and dump it?
Meaningless questions because you aren't specifying initial conditions.
http://www.youtube.com/watch?v=rEeNmSOEgQY
Quote
Some of the arguments against the ZED reminds me of the comic Jimmy Jones,, the one about getting pulled over for speeding on the A1,, 60mph?? but I have only been out for 10 minutes,, If you remember back that far you know why I did not write the actual joke :)
Isn't that Wayne's line? He only needs ten minutes input to get a net of an hour's output, it sounds like. Or  maybe it's the other way around.... he puts in an hour, recycles that hour over and over and gets a net ten minutes out every cycle. Sounds like he's using mental health hours, which are only 50 minutes long to begin with.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1001 on: March 14, 2014, 09:50:10 AM »
This is a fun little piece of trivia about this,, you most likely already know it,, rough numbers of course :)

The volume of the whole system at end of state 3 is

116cc

The volume of the whole system in end of state 2 is

112cc

The equivalent volume of displaced water in the end of state 2 is

147cc

The volume of water in the system at the start of state 2 is

22cc

The added volume to make all this happen is

2.5cc

So playing with 24.5cc we get 147cc worth of fun :)
Another way to look at it is this:
A complicated device with a cylindrical volume compressed of ~131cc compressed, and  ~136cc extended can be replaced by an ~$1. compression spring that occupies a volume of:  0.08cc compressed and 0.1 cc extended.  The latter device works in any orientation, does not require complicated procedures to prepare, does not change its properties due to evaporation, leakage, or contamination.

minnie

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1002 on: March 14, 2014, 10:12:27 AM »



  Webby,
           I think that your load is 65,000lbs of displaced water so the answer must be that
   times the distance you raise it. It's just the paradox.
                      John.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1003 on: March 14, 2014, 10:28:08 AM »
MarkE,

Are you going to post your fixed spreadsheet?

I might also suggest that you make the riser OD self setting instead of hard coded.
Oh sure, here it is along with the five state graphics.  I added an additional state:  State 1X which is a state that would occur if after reaching State 1, the risers are not restrained.

MarkE

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Re: Mathematical Analysis of an Ideal ZED
« Reply #1004 on: March 14, 2014, 11:10:19 AM »
Thanks MarkE.

If you have automagic riser OD's then use a HorGap of 3 and a pod of 400 and drop the fill down to 33
You can change the geometries and the spreadsheet will still calculate correctly.  Do you really want to increase the pod from a diameter of 20mm to 400mm???