Big try at gravity wheel

[MarkE]

     Webby,
              all we need from you is a diagram of a single ZED that gives back more than
   the pre-charge. If you can't get more out than is put in that's the end of the story.
   period.
            John.

Minnie, webby has long stated that the precharge energy is just an initial condition.  That is actually OK.  As long as cycle to cycle the precharge energy returns to the same initial value, then what that value is does not change whether the machine is over or under unity. 

As the problem has been constructed with one cylinder being precharged, (see the State 2 drawing) at the start of a half cycle, and then the other cylinder being identically precharged at the end of a half cycle, the precharge is maintained.  What webby completely misses is the amount of work that has to be performed to get through a half cycle.  It is almost four times as much work as is extracted by raising the payload weight, leaving the resulting efficiency for a single cycle substantially less than 1/3.  Webby insists that he gets 83%.  He has declined so far to show his work getting to that vastly inflated, yet still under unity number.  Yet, this is what Wayne Travis trumpets as an example of someone who supposedly appreciates the style, quality, and craftsmanship of the emperor's new ZED.

[MarkE]

Webby, the total volume of the "air" is constant because the "air" is incompressible as we stated.

The volume of air in each cylinder changes as "air" is transferred from the "B" cylinder to the "A" cylinder as required to operate the machine as you have specified.

After transfer of the product of the annular ring area:  pi/4*(15mm^2 - 14.23mm^2), times 67.5mm height, the meniscus in the "A" cylinder has moved down 67.5mm and the meniscus in the "B" cylinder has come up 67.5mm at which point the two meniscus are at that same point.  At this point the volume of "air" is not equal in the two cylinders.  The meniscus of each cylinder is at the same depth.  The volumes do not become equal until some time later in the transfer process when the meniscus on the "B" side has risen to 0.75mm above the top of the piston.

Prior to this point, the pressure at the meniscus in the "B" cylinder is greater than the pressure at the meniscus of the "A" cylinder, and no external work is required for the transfer pump to operate.  Beyond this point, work which will incrementally become greater each and every stroke until the "A" cylinder begins to rise, will have to be applied to the transfer pump in order for it to operate.  Additional work is applied throughout the lift phase of the "A" cylinder, but on an incremental basis decreases slightly from the bottom to the top of the "A" lift phase.  It is all of this additional externally applied work that ultimately restores the potential energy stored, and lifts the payload.

As both cylinders are fixed in their positions up to this point, the only two potential energy stores that can change up to this point are the volumes of "air" underneath the cylinders.  Before we started pumping "air" from "B" to "A" the total stored energy was:  7.408mJ.  At this point it is: 3.899mJ.  If you still don't see what has happened, consider that the 67.5mm high column of "air" in the annular ring of the "A" cylinder is in the upper half of that cylinder, while the 'lost' matching column section of the "B" cylinder was in the bottom half.  If you remain blind to the rest of the physics, it should still be intuitively obvious to you that the energy stored in that newly formed higher bubble under the "A" cylinder has less energy than the like 'lost bubble' of the same volume removed from the bottom of the "B" cylinder.  It should also be obvious that now that the meniscus has come up in the "B" cylinder that the pressure acting on the remaining "air" in the "B" cylinder has dropped from 1471Pa to 809.5Pa.

You are again:  welcome to try and get to and through this transient state in a way that does not lose the ~3.5mJ of stored potential energy.  You are welcome to diagram your version of states in the scheme that you propose and account for the energy.  Surely, you performed some set of calculations to reach your claimed 83% efficiency.  Surely, you based those calculations on some set of conditions in the system at a set of states that you evaluated.  Surely, you can show your work as you promised to do Feb. 3.  But, if you don't, you don't.  You'll just be stuck arguing with empty hands like Red and Wayne Travis.

[minnie]

   MarkE,
             is there any way we could do an analogy with an ordinary see-saw?
    ie, what would correspond to pre-charge? After all we're just dealing mainly
    with weights-(displaced water) aren't we?
                       John.

[MarkE]

   MarkE,
             is there any way we could do an analogy with an ordinary see-saw?
    ie, what would correspond to pre-charge? After

Minnie, we can by constucting a see-saw of appropriate dimensions and placing appropriately sized container columns on either side of the see-saw.  However,  I am afraid that it may not be very intuitive.  Then again it may more clearly illustrate the absolute futility of HER/Zydro's buoyancy claims which all revolve around lifting and dropping weights.

In order to approximate the buoyancy machine: the payload weight and a tall column that we would precharge with mass (assume water) by filling it would be on the same side of the see-saw.  The column height would be 10X the fulcrum height.  The free vertical travel of the see-saw needs to be 1/10th the column height.  The see-saw arms are ideally many times longer than the column height.   The payload weight that we will lift will be just a smidge less than the weight of the water that goes into a filled column, and have a high SG such as 10.

We would start by placing our payload weight on one side of the see-saw and then "charging" the system by filling up the column of water on the same side.  If that sounds dumb, it is.  But that is really what goes on when in webby's example we "charge" the system by pumping "air" into one cylinder to make it buoyant when we are going to lift a weight that is on another cylinder.  Then we connect a pump between the first cylinder and the second to move our "charge" water from the first cylinder to the second one.  After we have pumped half the water over, we can tally our stored energy and found out that just as in webby's example, we have lost just about half of what we had.  Now we keep pumping water over until the other side has all the water, and exerts a small net force in excess of the first side raising the payload weight.  We can remove the payload weight, load one on the other side of the see-saw that is now in the down position and repeat the half cycle. 

This inane and inefficient machine is no more silly or inane than the HER/Zydro devices and webby's proposal.

[minnie]

   MarkE,
            thankyou  for that explanation. Down on the farm we use lots of hydraulic
   machines. An actual hydraulic system isn't that efficient but is very convenient.
               John.