If the air volume in space one is pressurized from 4 PSI to 8 PSI then its volume has to be reduced by half, so half is not negligable even if the volume of air in space 1 is probably quite small, this is the law of ideal (perfect) gas (PV=NRT ) or boyle's law (P1V1=P2V2) and air is much much closer to an ideal gas that to an incompressible fluid.

Now let see what would happen if we replace the air by an incompressible fluid (liquid) of a lower density than water, then the space 1 pressure can be increased immediately, (i.e no need to inject a certain volume of water to compress a gas by half its volume to go from 4 to 8 PSI). The pressure will be transfered from that lower density space 1 to water column between layer 1 and layer 2 then to lower density space 2 and then to next space (no 3) and having identical pressures in all spaces meaning same force on both sides of surface 1, both sides of surface 2, surface 3 being at 8PSI (or whatever pressure) minus atmospheric pressure, the multiple layers effects seems to be lost and we are not anymore in the conditions described in Wayne ZED. (the above is assuming a 3 layers only)

Note that even with a compressible gas like air, with the understanding I have of the ZED, there is no added value from the vertical forces on the multiple horizontal surfaces (S1, S2, S3), the reason is that S1(P1-P2) + S2(P2-P3) + S3(P3-Atm) is identical to S1(P1-Atm) (all surfaces being practically same or to be very accurate if only 1 layer then S1 can be = S3) so the improvement of efficiency comes from something else than the vertical forces and from what I dont know. Anyone? I asked Wayne but still don't get it.

Sorry if I used only few numbers.

Quote from: see3d on September 03, 2012, 12:06:11 AM
In my simulation, the compressibility is so slight that it makes little difference to the operation.  First learn if you are talking about a first order issue or a second order issue in the particular application.  If you want to prove a point, do it with numbers to show what difference it will make.  If you are looking at 300% issues, then 3% issues can be ignored.  I do not know the answer for a multiple layer system yet, so I would not make such a statement.

Quote from: parisd on September 03, 2012, 11:00:57 AM
If the air volume in space one is pressurized from 4 PSI to 8 PSI then its volume has to be reduced by half, so half is not negligable even if the volume of air in space 1 is probably quite small, this is the law of ideal (perfect) gas (PV=NRT ) or boyle's law (P1V1=P2V2) and air is much much closer to an ideal gas that to an incompressible fluid.

The 4 and 8 PSI values are GAGE pressures.  The equations you are stating all require ABSOLUTE pressures.  So:

4 PSI (gage) = 4  + 14.7 = 18.7 PSI absolute

8 PSI (gage) = 8  + 14.7 = 22.7 PSI absolute

where 14.7 is an accepted value for atmospheric pressure.

The volume of air is not reduced by half as the pressure changes from 18.7 PSI to 22.7 PSI.

M.

You are correct I am not used to PSI and was thinking we used absolute pressures, but still air cannot be considered as incompressible. Does the rest of my post make sense ?

Quote from: mondrasek on September 03, 2012, 11:19:02 AM

The 4 and 8 PSI values are GAGE pressures.  The equations you are stating all require ABSOLUTE pressures.  So:

4 PSI (gage) = 4  + 14.7 = 18.7 PSI absolute

8 PSI (gage) = 8  + 14.7 = 22.7 PSI absolute

where 14.7 is an accepted value for atmospheric pressure.

The volume of air is not reduced by half as the pressure changes from 18.7 PSI to 22.7 PSI.

M.

Quote from: parisd on September 03, 2012, 12:37:02 PM
You are correct I am not used to PSI and was thinking we used absolute pressures, but still air cannot be considered as incompressible. Does the rest of my post make sense ?

I think the rest of your post makes sense.  Sorry that I can't give you an answer to your questions though.  I became stuck with my own analysis once the air compressibility and lift required an iterative or higher math solution.  So I was never able to find the OU myself.  Instead I look at the tools from LarryC and see3D and wait for more. 

And I wait for glue to dry...

M.

@Mondrasek. Re your post 1853. This is without doubt a very practical and informative post for model makers.
People in different parts of the world may have to experiment with different glues and   materials that are available locally. One thing I figured out is that the top caps of the risers do not need to be transparent. So alternative materials could be used here, including thin plywood if it is treated with a waterproof paint or varnish.
        In the past, I have used a cheap bench grinder as a substitute lathe when making plywood discs. I cut the disk approximately to size using a fretsaw or band saw. I then drill a suitable hole at the centre and mount the disc in place of the grindstone. I then take a long rasp, and placing one end on the bench, bring the middle of the rasp against the revolving disc. Naturally. i wear goggles. For safety reasons, you do this at your own risk, but it works for me.