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. 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.)

I was going by the before your edit, I just refreshed and saw the edit note the original is lost but here is the reply:

True, hat statement in itself does not explicitly state level1 higher or lower than 2 but others as I note in the previous post do and/or refer so.

It always started at Level1 for the full cycle. I cannot see where I posted starting at level two for cases a,b,c.
The model presented, always uses level2  higher than level 1. (it works the other way too but let’s stick with level2 higher than level1 for consistency)

Level1, we start at buoyancy neutral and then we compress the piston then lock it and then move up. The move up requires energy since some buoyancy is lost due to compression. (loss of neutrality as we both agree)

At level 2 since the outside pressure is less, if we unlock it, It will give all the compression energy put in at level1 plus some due to the lower outside pressure, more expansion etc.
So we had to provide energy to move it up, but we got out some extra due to the higher volume expansion, the extra equals the energy we used to move it up. That extra expansion energy we can use to move it up. Nothing gained or lost

Similar in reverse if we move it back down from 2 to 1 so the full cycle is completed. Net energy is zero from level1 to level2 to level1.

All I’m saying is either locking it first and moving it from level1 to level2, or starting uncompressed (some air but buoyancy neutral) and then nudge up, after the full cycle nothing is gain or lost (friction and rest excluded)

Ok so we agree as you stated on the energy conservation scenario, but the issue is with technical description of what is happening. As I said I left that out earlier so it will be simple.

Then we are in agreement for a,b,c scenarios in one full cycle net energy gain or lost is zero.

I’ll wait to hear from seamus1 also before going to the single riser single pod next.

Maybe we can pick up the Cartesian diver analysis if we have time in another thread.

All right, I misunderstood you at first, then, since I could not imagine why you might want to compress, thus doing work to reduce buoyancy, then RAISE the volume, doing extra work. I still can't, off the cuff, but I don't have time to rethink it now, so please excuse my earlier misunderstanding and carry on.

(We have all this time  considered adiabatic compression and expansion, with no heat loss to the surrounding cylinder walls or water, right? These losses will of course mean that, even if I do agree with your conservative ideal a-b-c scenario so far (reserving judgement here) in the real world this heat loss will require outside energy to replace.)

OK, so you start at lower level, with free piston, neutral buoyancy B0, and internal pressure P0 = external pressure E0. You compress the piston and lock it, performing work on the system by reducing the volume, raising the internal pressure to P1 > P0, and decreasing the buoyancy to B1 < B0. Then you perform further work to raise the syringe to higher level. The external pressure is now lower than E0, so when you unlock the piston it will go out further than it was at the lower height. Then you lock the piston. You now are displacing more volume than you were at the lower level, so you are more buoyant than neutral, that is, B > B0. How do you get back down to the lower level?

Quote from: TinselKoala on July 27, 2012, 08:00:51 PM
All right, I misunderstood you at first, then, since I could not imagine why you might want to compress, thus doing work to reduce buoyancy, then RAISE the volume, doing extra work. I still can't, off the cuff, but I don't have time to rethink it now, so please excuse my earlier misunderstanding and carry on.

(We have all this time  considered adiabatic compression and expansion, with no heat loss to the surrounding cylinder walls or water, right? These losses will of course mean that, even if I do agree with your conservative ideal a-b-c scenario so far (reserving judgement here) in the real world this heat loss will require outside energy to replace.)

No problem. Misunderstandings are usually part of these exchanges. It’s all in good fun.
Yes, on frictionless, adiabatic vs isothermal, ideal gas approximation etc. I think I noted these conditions in the very early posts. Agreed, for the purposes of the model constructed, heat losses, transfers in out, would be set aside, but in the real world they are part of the package even if they can be very small.

Quote from: Seamus101 on July 27, 2012, 08:00:56 PM
I think  see where you are going with this and initially I'd considered it a candiate for a 'wow' this is the working principle that explains this. Alas, I think you have missed something that does not allow it but however...

I'd agree that translation of a neutrally buoyant object  in a frictionless environment is energy neutral. That in itself is no use to us, but I'd predict you are about to introduce the concept of somehow 'capturing' buoyancy at a one level, using this energy neutral translation to transfer this 'capture' to another level and thus this 'captured' buoyancy could then be used to exert a force between level one and level 2. If that could be made to happen then that would be the working principle and this device would work.

Correct?

No wow moment shown so far. I understand where you are coming from , I was there also on this, but that is for another discussion alltogether.
All I’ve been attempting to do is to establish base lines using simple physics, understood by all who would follow, so we can build on them for the next steps. Your prediction, yes in the ball park I think. But you can be a better judge of what you exactly describe when we continue later on. So far no laws have been “broken” as we know them, physics hold well

So you do agree with a,b,c scenarios in one full cycle, net energy gain or lost is zero?