Yes, I can but I do not see any reason to name drawings that represent what Webby is describing after myself.  I think that would be confusing.

yes , it will be confusing,
or  "Webby is describing to markE" .  no bargain

I am not picky either ,, but it is not my idea.

MarkE, the videos did not show a "new" effect and IIRC Tom called the displacement replacement effect the "Travis" effect,, like other kind of things are not treated in a similar fashion. <= sarcasm

Scientists stated, that heavier than air craft can not fly, scientist stated that bumble bees should not fly,, this is all part in expanding our understanding of things, what we thought was this way ends up being that way instead.

This is about how I got myself to my understanding of Wayne Travis's device, that is what I am trying to walk MarkE through, however, his comprehension of my writing and my ability to write in a comprehensible way are at odds.


The payload is the max that A can lift and, as I stated, I find that this makes for an 83 percent efficient lift from the potential that is stored in B.  After A has lifted then A becomes the store for B,, back and forth it goes BUT the transfer pump needs to be changed to make this happen much better.

Since MarkE has a very hard time interpreting what I write this may get even more out there,, but I will try.

If the transfer pump were a pressure balancing system, that is if it could take in a high pressure at a low volume and convert that into a lower pressure higher volume then the process of making that exchange becomes much more efficient.

A simple approach is to think of 3 discs that have a good thickness to them and they are stacked on top of each other making like a pyramid shape, and they move in a housing as such that at first the smallest disc sees the incoming pressure and moves the whole assembly and after a distance of motion that first disc moves away from its seal and exposes the next larger disc, it moves and opens the chamber up to the last disc, this allows for a pressure drop and volume increase and then a volume increase with the pressure drop to match closer to the potential coming from the stored potential in the cylinder.

Webby it would be best to stay on topic and finish defining the arrangement.  At the moment I am waiting for you to define a payload weight and SG.  Surely if you have already been through this you have values that you like.  For the apparatus as it is currently described I can recommend a 2.934g dry weight payload with an SG of 10.0.  That payload will have an underwater weight of 2.641g, 0.01g less than the weight of water that can occupy the annular ring between the piston and the cylinder with the cylinder all the way down as shown for the "A" side at the starting condition you have stated while the  "B" side completely filled with "air" resting up against the stop.  The difference of 0.01g is so that with the "air" displacing all the water in the "A" annular ring we will have a small net buoyant force to lift the payload initially above the piston.  Once the cylinder clears the piston so that there is "air" between the piston and the cylinder underside, then the force will of course initially jump and then decay somewhat as the cylinder moves up and more "air" fills the expanding piston to cylinder underside cavity.  As long as we remain net buoyant, and under these conditions we do, then the cylinder successfully lifts the payload by 15mm, where we can remove the payload and count the work done. 

Our remaining task then is to return the apparatus to the State 2 condition you have stipulated as the cycle starting point where the "A" cylinder is down on the piston once more, there is no "air" on the "A" side and the "B" side is back to holding all the "air" that it can.  Please clarify how you would like to return to that condition:  For instance by pumping "air" from the "A" side back to the "B" side.  Running the transfer pump in reverse works for me.  If you want to vent and use the surface pumps that's fine too.  I just need to know what you have in mind.

Unless there is something particularly important to you about the internal design of the transfer pump, why don't we simply assume that the transfer pump is 100% efficient?

If you share Marsing's objection to the filenames, let me know and I will change the names of new postings to something you like.

forget about it markE.  you are right,  and thank you     

MarkE,

If you are OK with calling the transfer pump 100% that is fine, it is just that in the real world I have not been able to achieve that and as a general rule I round up my input and round down my output,,

Is not the volume of the cylinder 26.5072 cubic cm? and so the payload should be very close to that value.

There is of course the filler that is displacing most of that volume, but it is the volume of displaced water in total and not just what has been added, volume wise, by the air.  So once the air is put into the cylinder so that there is a very thin layer of air on top the filler and down the annular gap, the equivalent volume of displaced water is the entire volume of the cylinder, as if it were an open chamber filed with air.

After the air has been transferred from B into A then B can be moved into the down position that A was at, hence, A becomes B and B becomes A.

I am not overly concerned about the buoyant value to the weight, the weight could be located out of the water as long as it was in communication with the cylinder.  Using weight is a more convenient method that using resistance, but in the end it is the resistance of the weight that allows for work to be done.

I really do not care about file names, whatever is easiest for you is fine,, I am trying to keep things easy and simple,, as you are aware, my communication skills are not the best

Webby, unless you start with a bubble between the top of the "A" piston and the underside of the cylinder, you do not get hydraulic force gain. Under those circumstances, the upward force that transmits to the underside of the cylinder is only that of the water displaced from the annular ring.  That restricts you to a submerged payload weight less than 2.651g, or else the cylinder never separates from the piston.  Once it does separate, then the force jumps because then you get the hydraulic force gain of the ratio of the entire area on the underside of the cylinder, versus the area of the annular ring.  For this reason I recommended a submerged weight of 2.641g as 2.934g dry weight with an SG of 10.0.  The submerged weight difference of 0.01g is incidental to the energy in the problem: < 0.5%, and meets the condition that the cylinder rises above the piston.  (We are ignoring things like surface tension.)

You need to define a whole cycle with a starting condition and an ending condition that are the same.  I am fine with making the cycle more complicated, but we will not be done until we execute a full cycle.  So, if after lifting the payload on the "A" side you want to pump the remaining "air" out of the "B" side so that we have the mirror image of State 2, that is fine by me.  We can then load a payload on the "B" side, pump air back from the "A" side, let that weight rise, remove the weight, pump the remaining "air" from the "A" side and we will have a full cycle starting and ending with the State 2 condition.  By symmetry, we should expect that the total work applied and total work done is exactly twice what it takes to get from State 2 to the mirror condition.

Mondrasek, if the goal is to get a bubble in there, then what would do that is to either lengthen the cylinder slightly, or shorten the piston slightly.  I propose to shorten the pistons to 14.99mm.  That will leave a 0.01mm high volume that is initially water.  Once we equalize the two sides the water will be pushed out, and we will get the 10:1 hydraulic piston force effect.  That will let us lift the heavier payload.  It will also change what happens as the piston rises.  Whereas with a smaller payload and no initial bubble the upward force jumps 10X and then falls off with the rise, under these circumstances, the force will just be enough to become positively buoyant.  As the cylinder rises, the bubble above the piston gets larger shortening the length of the bubble in the annular ring, bringing the system back to neutral buoyancy.  In order to remain positively buoyant we will have to keep pumping "air" from the "B" side all the way to the stop.  The amount of work that we will have to perform doing so will be the difference in P*V of the cylinder right before it starts rising and the cylinder up against the stop.  The good news is that when we are done there won't be any air left on the "B" side.  We can then give the "B" side a little nudge and the "B" cylinder will fall back to the bottom of the vessel.

Please don't forget about the Cartesian Diver. The nested Zed risers have some "CD" effect happening _due to the compressiblity of air_. The allegation that a Zed system would "work" with 2 incompressible fluids has never been shown, in practice, to do the same thing as the same system with air and water.

The change in pressure on the open-bottom risers causes change in air pocket volume, hence displaced water volume, and this affects buoyancy, enough to make a "CD" float, hover, or sink.

http://www.youtube.com/watch?v=ljvp-iR18Ko

   Hi,
       I get the feeling this particular gravity wheel is shuddering to a halt.
                                    John.

"... A simple three layer system that is clearly overunity by itself."
--Wayne Travis


So, according to Travis himself, you do not need "2 lift systems" to demonstrate his overunity scheme. Why are we not shown this simple, three layer system that is clearly overunity by itself? Why aren't Webby and MarkE analyzing this simple, clearly overunity system that has only three layers _by itself_?

I know why.