What's wrong with this

[Floor]

@ T.K.

For sure, I'll let you know if I run across any negative friction
bearings.

So the "Magic Valve" Is described on the last page of the previous PDF  file (OUfloater 3). and also
reproduced in the  (MagicValve.pdf) attached below.

As you have probably guessed, it's not realy magic ! But it can let an object in or out of a
fluid or gas filled chamber with out letting fluid or gas, in or  out.

Quote "Don't forget that as you push up the bottom floater through the bottom magic valve, you are not only lifting it against gravity but you are also pushing out the top floater through its magic valve, lifting it up against gravity as well "End Quote

Between the bouyancy of the two floats, only the weight of one of the floats will be lifted, (before losses).

        floor

[MarkE]

If Ou can be done any where, then it can be done any where ?

Please find the attached file OUfloater 3.pdf

Cheers
       floor

It doesn't work for reasons that have been explained concerning buoyancy schemes before.

At the bottom of the tank the float entering has zero impetus to enter the fluid section, magic valves or not.  The only way to get that float into the fluid section is to displace a volume of fluid equal to the float volume.  The fact that one is doing work lifting another float out the top does not help.  It just complicates the bookkeeping.  The problem can be simplified to a representation of the linear sum of two processes:  A first process of inserting a buoyant float into the bottom of a fluid column and allowing that float to rise up through the fluid, and a second process of withdrawing the float from the top of the fluid column and recovering the added energy by dropping the float.  Even in the ideal case the first process is always lossy as shown in the attached diagram.  In the ideal case the second process can at best be break even.  The sum of the two processes is therefore always lossy even before any practical problems of magic valves or other special pleadings are addressed.

[Floor]

@Mark E

I see some similarity,  (a fluid and a float)

But

1. There is fluid dispalcement in what I presented, but there is no fluid rise above the
starting fluid level in what I have presented.

2. There is no drop in the fluid level, or falling of the fluid in what  I presented.

3. There is no air lock in what I presented.

Quote "Even in the ideal case the first process is always lossy as shown in the attached diagram. "End Quote

4. The diagram "first process" you presented, does not apply to the example (except to illustrate the the volume
exchanges are equal ?)

5. "Lossy" or not, the insertion of the float into the fluid vessel and the rise of the float (before losses) are
equal.

6. You made no representation of the energy in the free fall of the float (in air or vacuum),
in comparison  to the equal exchange (before losses), of the first two process I presented
(the insertion of the floats against pressure and floatation).

I respect your experience and knowledge, and I wish your input had been more relevent
to my query.  If you will please review my presentation with closer scutiny, I would
welcome the more relevant input.

                       Sincerely
                            floor

[MarkE]

@Mark E

I see some similarity,  (a fluid and a float)

But

1. There is fluid dispalcement in what I presented, but there is no fluid rise above the
starting fluid level in what I have presented.

2. There is no drop in the fluid level, or falling of the fluid in what  I presented.

3. There is no air lock in what I presented.

Quote "Even in the ideal case the first process is always lossy as shown in the attached diagram. "End Quote

4. The diagram "first process" you presented, does not apply to the example (except to illustrate the the volume
exchanges are equal ?)

5. "Lossy" or not, the insertion of the float into the fluid vessel and the rise of the float (before losses) are
equal.

6. You made no representation of the energy in the free fall of the float (in air or vacuum),
in comparison  to the equal exchange (before losses), of the first two process I presented
(the insertion of the floats against pressure and floatation).

I respect your experience and knowledge, and I wish your input had been more relevent
to my query.  If you will please review my presentation with closer scutiny, I would
welcome the more relevant input.

                       Sincerely
                            floor

The principles at work are the same.  Go ahead and make a state diagram for one cycle and track the energy in out and stored at each state.  If you can show a gain in energy over the course of one cycle.  For the reasons I have already explained you will not be able to do that.  Let's review again:

1) No float can rise in a containing fluid in any finite time without a greater mass of fluid falling.  Ergo the system always loses stored energy between the state with the float at the bottom and the state with the float at the top.

2) No energy can be gained by removing a float from the top of a tank and dropping it.  In the ideal case the GPE lost by the float is completely transferred outside the system as useful work and the there is zero sum gain.

Ergo buoyancy machines consume rather than generate net energy each cycle.

[Floor]

@MarkE

I'll do some diagrams as you suggest.  Thanks, untill then.


                   floor