Hydro Differential pressure exchange over unity system.

[Red_Sunset]

My answer. The air in the gap between the lid of the outer riser and the lid of the next riser. gave extra lift. But this extra lift was cancelled out by that same air pocket pressing downwards on the lid of the next riser.
Anyone agree/disagree?

Neptune,
The same lift ?? yes and no,  it all depends on the layer ratio,
Before the outer layer was the biggest lift area with the LOWEST pressure, now the outer area has a way smaller area with the same low pressure, the center pod area will become the largest area with the HIGHEST pressure. This means that the pod will produce the most lift and control the lift. I would imagine that the control will become more sensitive and possibly more difficult since the buffering up the water column will be way less than before
Michel

[see3d]

Data (as it is):
Mass lifted:  ~2.5 lbs
ZED stoke: ~ 11.5 mm.
Water removed and introduced again to obtain repeatable(ish) stroke as measured by a "rain gage" from my yard: ~87.5mm.
Overall change in the head in the elevated fill tube during the cycle is ~ 190 mm.

I think the volume of water in your rain gage per mm height would be needed -- to get the mass.  There may be more dimensions needed to calculate the work input.

[mondrasek]

I think the volume of water in your rain gage per mm height would be needed -- to get the mass.  There may be more dimensions needed to calculate the work input.

Yeah, I was considering that before posting.  But I "assumed" that a rain gage must be some kind of standard.  Like, 1" in the rain gage means 1 in^3 of water.  And so 1 cm in the rain gage means 1 cm^3.  I guess I can double check the scale on the side with a tape measure pretty easy.  And if it turns out to just be a "relative" scale, I can get the ID for volumetric calcs.

M.

[TinselKoala]

LarryC's drawing above shows water flowing from one unit to the other as a result of the hydraulic assist.

But earlier he said,

Regards, Larry

PS: The actual system doesn't transfer the water, since each retains it's own water supply. It just transfer the force to move each Zed's water supply.

How is that drawing compatible with that statement? Just what is being transferred between the Zeds, and how is it being transferred, if no water is transferred? Just what is the physical nature of this "D" that is being recycled in the diagram that shows A=B and C=B,  but A not equal to C?

And now it appears that MrWayne is telling us that a single Zed system, no matter the layers, can't be all that efficient.... 105 percent? Easily in the measurement noise floor, just like when I ran the set of good numbers from Webby.


@mond:
Looks great, I see that illumination from underneath works quite well.  You could try different color LEDs, positioned to illuminate the different internal layers from underneath. You might get lucky and be able to have different colors in the different layers due to diffusion and internal reflections. Or it could just mix into a mess... but it will still look nice sitting next to the LavaLamp, anyway.

I like your data set, but can you please describe in painful detail just how a "trial" is conducted and just how repeatable the start and finish positions are, etc. And yes, I believe we need the cross sectional area of your input cylinder rain gauge... or if it's one of those silly tapered ones..... some other means of determining the mass (or volume really) of the total water you inject and recover during a trial. (The mass is important to find the pressure but it's the volume, I believe, that does the work in the Zeds.) This, combined with the pressure computed from the elevation of the center of mass of the raingauge water, will allow an input work figure to be calculated.

[see3d]

On another note,would I be the only one assuming that Red_Sunset and Mr Wayne are one and the same person?

Your powers of observation need some serious sharpening if you are confused about that... LOL