Mathematical Analysis of an Ideal ZED

[TinselKoala]

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]

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]

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. 
K_FORCE = 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]

    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]

     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.