Quote from: TinselKoala on July 27, 2012, 03:21:28 PM
For clarity? You are kidding me now, I know. I am talking about the same cases that you are talking about. Since you started at Level 1 and compressed, in your first setup, then your level 2 must be lower than level 1. Let's stick with that convention please, since we are trying to be clear. Now... is the piston FREE, or is it LOCKED? In case A the piston must be free and frictionless. Right? The work you put in is returned instantly when you stop compressing the piston.  In case C..... you first have the piston FREE, to compress at level 1 (call this C1). Then you must LOCK the piston otherwise it will expand back out to neutral buoyancy. Then you sink to level 2, with the piston LOCKED (C2). This sinking comes "for free" since it is a result of reduced buoyancy; in fact you will have to do work to make it stop sinking at level 2. At level 2, you unlock the piston (C3). What happens? The piston expands until the pressures on both sides are equal.... that is, to lesser volume than it was at level 1 because the external pressure is greater, and you haven't changed the mass of air, just its volume. Now you lock the piston again, but your apparatus is still just a little negatively buoyant, since it is not expanded back to the same volume as it had at level 1 but still masses the same (C4). Hence it will require work to get it back up to level 1. Now, at level 1 again, you can unlock the piston, let the chamber expand, reestablish the correct volume for neutral buoyancy (C1) and  NOW get that work back.
But where did the work come from, that raised the negatively buoyant apparatus from C4 to C1?   Did you not just reverse the order of level 1 and level 2? In your first setup and in my discussion, level 1 is above level 2, is it not? Let's please remain clear about this point. Let's point out that the Cartesian diver is a case where the "piston" is free to move, that is, unlocked, so that it is the water pressure only that changes the diver's volume. There is no "Mr. Hand" doing the initial compression of the internal volume, then locking it against expansion regardless of the external water pressure, as there must be in your a-b-c scenario. In your "b" case, the piston could be locked, as you postulate, or free to move, as in the Cartesian diver. But not both, one on the way up and the other on the way down.
Are we agreed? Where did the work come from that raises your negatively buoyant apparatus from C4, piston locked, back up to C1? In order for things to equal out, for you to get this work back at the top, you first have to put it in from the bottom.

Are you going to tell me that this work comes from the other Zed, at the top of its cycle?

I see where a mixup might be coming from.
Level2 is higher than level 1.
I also stated:
“””Now we can also take the Cartesian diver at level 1. Let’s say neutrally buoyant. Nudge it up and let it move to level 2.””
I ‘d think that the readers follow the movement from level1 to a higher level2, but you assumed Level 2 was lower. I apologize for the confusion if it wasn’t fully clear. (also to remotely refer  diagram post 373 level1 is lower level and move up to higher and then back down etc)

Ok,
For case a it is irrelevant if level1 is lower than 2.
For case b where the piston/cylinder is locked still irrelevant.  You can move it anywhere there is not net energy gain/loss. Are you stating differently?
“Translation of a buoyancy neutral body within the medium requires no net energy.  Agreed?”
So for case a and b there is no net energy for one full cycle correct?
Case a, no movement between levels, cylinder is held fixed,  buoyancy neutral,l piston compressed and released, expanded, everything we put in we get out.
Case b piston locked with cylinder, with enough air inside to make the whole thing buoyancy neutral.
Move it anywhere from anywhere, but for this case from level1 up to level2 and back to 1.
No energy net gain/loss right? I’m only looking at translation of a neutrally buoyant body inside water fully submerged.

Ok back to case c.
Your statement:
“In case C..... you first have the piston FREE, to compress at level 1 (call this C1). Then you must LOCK the piston otherwise it will expand back out to neutral buoyancy. Then you sink to level 2, with the piston LOCKED (C2). This sinking comes "for free" since it is a result of reduced buoyancy; in fact you will have to do work to make it stop sinking at level 2..”
“….But where did the work come from, that raised the negatively buoyant apparatus from C4 to C1? 

Again, the model presented is based from level1 (c1) to higher level2 (your c4?),
But no problem with moving lower, actually we can extract work while it is sinking, and that work can be used to bring it back up again. No net gain/loss. It is sinking so there is a net force on it, moving for some distance, then it can produce work. Integral of df*dx.  As I said earlier that work equals to the amount needed to bring it back up. No free lunch there either.

For case c, all I’m saying is, there is no net energy gain/loss moving from level1 to level 2 and back to level 1. I cannot see why we do not agree on this. Conservation of energy holds.
Nowhere so far I’ve stated or shown any net gain or net loss for one full cycle.
I’m only attempting to establish the basics, and simplify as much as possible.

From level1 to higher level2:
At level1 we have a piston/cylinder with enough air to make it buoyancy neutral. Net forces on the system are zero. This is where we start.
1..We can lock the cylinder/piston relative to each other, move it up to level 2 (zero energy in or out) unlock the piston at level2 and it will expand to a volume greater than the one at level1
Of course, after the unlock and piston is free to move, it will no longer be buoyancy neutral.
The translation gave nothing, but the expansion gave something more than the original at level1.
That’s not free energy, we moved up to a lesser outside pressure.
Now to move it back down to level1 it will require some work. That work equals to what we got from the extra expansion up at level 2. Right?
Or 2.. we can just nudge it up (piston unlocked) and it will start and move up to level2 where we stop the movement.
Now as per previous post - last paragraph Cartesian diver-
“As it moves up it can increasingly lift more weight until we stop it at lets say level2.
To move it back down to 1 is the reverse process (weight substation to stabilize it back to level1) and from 1 to 2 to 1 the net gain/loss is zero,(integral dw*dx for each path and so on) right?”

Your statement:
“”Are you going to tell me that this work comes from the other Zed, at the top of its cycle?””

I haven’t involved any ZED or a two systems interconnected anywhere so far


I think we are making this unnecessarily complex…

Bottom line is for cases a,b,c and for each full cycle there is no net gain/loss yes or no?

In your first post of a-b-c, you said this:

Quotec.   Now, lets do both. Compress at level 1 move (translate) to level 2 decompress at level2 and move back to level 1. Net energy is zero, the translation contributes/subtracts nothing, the compression/decompression cycle is also net zero.

Does this not imply that level 1 is higher than level 2? You are compressing the volume, reducing the buoyancy, then going to a lower level, are you not? So I still think you have flipped your level definitions on me. But no matter, let's use your redefinitions, where you are now using Level 1 to be lower than Level 2.

You start at Level 2, neutrally buoyant. You perform work C1 pushing in the piston to reduce the volume. If your piston is FREE it pops right back out. If your piston is LOCKED, it is not a Cartesian diver. So now you have a locked piston, that is, a FIXED volume. Now your apparatus sinks to Level 1 and you must do work--- or rather oppose the effective unbuoyed weight --  to prevent it from sinking further, since it has been negatively buoyant since you compressed the air and it will continue to try to sink as long as this is true. Now you release the piston and we have a Cartesian diver again: the air expands until the pressure on both sides of the piston is equal. But this now happens at a smaller volume than before. You still have negative buoyancy, and you are not yet back to the Cartesian diver neutral equilibrium, in fact you are in the position where the diver wants to sink, not rise. So where does the work come from to get you back up from Level 1 to the higher Level 2? You are not yet at the end of the cycle so there hasn't been any work returned yet.
One of us is missing something. I am perfectly willing for it to be me, but so far you haven't convinced me.

In the Cartesian diver, where the piston is always free and the compression and diver volume change happens naturally from without, as the outer water pressure is changed, then yes, I agree with your conservative scenario. I do not think this is equivalent to the situation you have described, though, where you are compressing the apparatus without compressing the external water, and where you are locking and unlocking pistons.

(Sorry, edited some confusion about level one and two.)

Instead of starting with a neutrally buoyant syringe of air... let's just use vacuum. After all it is the volume that controls buoyancy, and the mass of a bit of air is negligible, it might as well be vacuum from a weight/mass standpoint in these small volumes.

So we have our "diver" with neutral buoyancy and a locked piston, at the higher Level 2. We now unlock the piston, and the vacuum "sucks" the piston in, reducing the volume. We lock the piston, fixing the volume. Now obviously the diver sinks, and we catch it at the lower Level 1. Now we can lock or unlock the piston..... but how do we now return up to Level 2?

Quote from: TinselKoala on July 27, 2012, 06:09:23 PM
In your first post of a-b-c, you said this:
Does this not imply that level 1 is higher than level 2? You are compressing the volume, reducing the buoyancy, then going to a lower level, are you not? So I still think you have flipped your level definitions on me. But no matter, let's use your redefinitions, where you are now using Level 1 to be lower than Level 2.

You start at Level 2, neutrally buoyant.

No.  You start at Level 1.

You start at Level 1 and then you RISE (not sink) to Level 2.  The rise is not due to a change in buoyancy.  It is caused by an outside force being applied.

M.

Sorry, weird forum thingy...