Hydro Differential pressure exchange over unity system.

[wildew]

I'll take a stab at "Ideal Lift" just to see if I understand the term.....
I think it could be re-written to be absolute or maximum lift; the highest point in the cycle dictated by the physical dimensions of the assembly and the setup. OR, the height at which the skirts blow....
Dale

[TinselKoala]

Exactly what I have been saying (or trying to say) for the past several posts.  No energy is gained or lost through one complete cycle, right?

Please do not try to equate this simple siphon setup to the ZED.  It has nothing to do with the complete ZED system.  It is only being used to show how the water input and removal at the fill tube can be considered an energy neutral cycle for simple ideal case analysis.  It in no way involves the ZED Pod/Risers where ZED system output is supposed to be measured.

M.

That's not what I thought you meant when you first brought it up.

TK, what you describe is not necessary, IINM.  The fact that I am removing water by allowing it to vent from the bottom of the Pod chamber is only due to the construction of the test system.  In this case, yes, the vented water must be raise to be reintroduced into the fill tube.  And that water is required to be vented in order for the ZED to sink and return to the initial pre-lift starting condition.  However, the venting of the water could also have taken place directly from the top of the water in the fill tube and at that same level if the system was designed for that more difficult venting option.  And if vented from the top of the fill tube it is clear that the vented water does not need to be raised or lowered (change of PE) to do so.

When you said this it seemed to me that you were not talking about a full cycle from start to finish, but only the step of actually raising the water (or somehow else replacing that pressure lost when you drain the water out).  You appeared to be saying in subsequent posts that you could pour water from one place into another without a gradient of some kind if you use infinitesimal steps and that work would vanish if the rate at which it was performed tended to zero.... both of which I find silly.

So please excuse me for misunderstanding... We agree that in these simple two-jug model systems, work in = work out for a full cycle, and I hope we also agree that removing the water from the top of the fill tube in your Zed, then replacing it at the same location, also requires just as much lifting of water as it does sinking.

So then... it appears that all model systems that produce unity actually have nothing to do with the complete Zed system, since Zeds are overunity. But only when combined, since each one is underunity itself. Except of course the simple three layer system that is clearly overunity by itself, the one that MrWayne has but refuses to discuss in any meaningful way, while we fool around with example systems that aren't Zeds and are clearly _not_ overunity by themselves but are somehow supposed to teach us how to make an overunity pair of Zeds.

I definitely need another cup of coffee.

[neptune]

OK, while I have my Village Idiot head on , how about this. Refer to the diagrams in post number 2137 on page 143 by MT . Look at the middle diagram, precharged state. On my screen, the pod is 10 units tall, and the space above it is one unit tall. Therefore, the space above the pod is About one tenth of the volume of the pod.
       Suppose we have a tank full of water the same shape and size as the space above the pod. A flexible hose joins this tank at the bottom and is connected to the main tank at a point level with the top of the pod in the precharge position. The top of the auxilliary tank is level with the top of the pod. We then lift the aux tank unti its bottom is level with the top of the main tank.  This auxilliary tank contains half a gallon , weighing 5 pounds. So we lifted 5 pounds a distance of 2 units.Let the volume of the pod be 5 gallons. So the upthrust on it will be 50 pounds. We release the pod , to allow it to rise under a load, and we allow the auxilliary tank to empty into the main tank.
So input =5 pounds lifted 2 distance units
output= 50 pounds over 1 distance units
To complete the cycle we lower aux tank to starting position, and it drains the main tank to the precharge state.


So output = 5 times input? Where did I go wrong?

[neptune]

@Wildew and Weby1. Many thanks for helpful replies. Any thoughts on my last post please?

[TinselKoala]

OK, while I have my Village Idiot head on , how about this. Refer to the diagrams in post number 2137 on page 143 by MT . Look at the middle diagram, precharged state. On my screen, the pod is 10 units tall, and the space above it is one unit tall. Therefore, the space above the pod is About one tenth of the volume of the pod.
       Suppose we have a tank full of water the same shape and size as the space above the pod. A flexible hose joins this tank at the bottom and is connected to the main tank at a point level with the top of the pod in the precharge position. The top of the auxilliary tank is level with the top of the pod. We then lift the aux tank unti its bottom is level with the top of the main tank.  This auxilliary tank contains half a gallon , weighing 5 pounds. So we lifted 5 pounds a distance of 2 units.Let the volume of the pod be 5 gallons. So the upthrust on it will be 50 pounds. We release the pod , to allow it to rise under a load, and we allow the auxilliary tank to empty into the main tank.
So input =5 pounds lifted 2 distance units
output= 50 pounds over 1 distance units
To complete the cycle we lower aux tank to starting position, and it drains the main tank to the precharge state.


So output = 5 times input? Where did I go wrong?

I have highlighed where you went wrong. Most of the "output" lift is coming from the precharge, I believe, which you do not account for, and simply "lowering" the aux tank won't reset the system, you will have to push the pod down somehow.

And... by the way.... "A pint's a pound, the world around". Actually 16 fluidounces per pint, and 8 pints per gallon, but a fluidounce of water is a little more than an ounce of weight; there are 473 mL in a pint, so a pint of water is actually just a bit over a pound (454 g)  in weight. But there are 8 pints to the gallon, not the ten implied by your numbers.