Quote from: TinselKoala on July 27, 2012, 08:14:31 PM
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?

I know, I was looking similar questions among the many, when I got into analyzing this.

Actually we don’t have to exceed the -more than neutrally buoyant level of expansion- at level 2. At level2 even at that amount of expansion, it has expanded more giving some extra energy. So at levl2 we got out what we put in as compression at level 1 plus some extra. The extra can be used for raising it… So now at level 2 it is neutral buoyancy, it only needs an nudge to fall back down, and it can also produce some work as it is doing so. ( the more it lowers the more it weighs  and so on), up until it arrives  back down let’s say to a support or stop at level 1, end of cycle.
We  can compress it just enough at level1 so when it reaches level2 is at exactly buoyancy neutral state..

Bottom line is: what we put in at level 1 as added compression we get out at level2 plus some, and the -plus some- helps/pays for the lifting cost of the less than buoyancy neutral body (after compression at level1)

Next cycle same as the first,  we recompress  ( using the “what we got out is what we put in” at level2) and repeat. Again no net gain or loss out of the full cycle. Conservation of energy holds. Keeping it simple that’s all.

Side note on scenario c :
There are many ways but all result in the same bottom line.
For example,
We can compress, lock, lift, expand, push if needed back down(depending on expansion amount at level2, and repeat.
We can start buoyancy neutral at leve1, no lock, nudge it, it lifts by it self to level2. Producing some work as it ascends, we use that work to compress it (at level2 back to the volume of level1) and lock it. Then it can sink with no added energy, (locked in buoyancy neutral now), back to level1 release the lock, it goes nowhere, nudge it up and so on. Ok maybe have we talked enough on this.

Bottom line is still the same, no net gain or loss, what ever is put in it is taken out during a full cycle.

Quote from: Seamus101 on July 28, 2012, 02:53:50 AM
I agree that all three a,b,c scenarios are energy neutral. No gain or loss in a fictionless cylinder during a cycle. 
The' bottom line' comment is bordering on not being true though. There is no 'plus some' that could ever  be extracted as usable work. It can only ever be exactly the the same effort input or extracted. I'm not sure if you meant that or not.

So far we have described a system that could move about with no energy input but without the ability to extract energy from it. What happens next?

Correct, that’s what I meant.
That -plus some- stays inside the system, is not extracted, it is not over unity, it is part of the cycle. Again as I noted earlier, that was the reason it was left out in the beginning so we can keep it simple. The initial pressurization at level1 stays in the system and just gets recycled. All we did for the cycle is, compress, move up, decompress, move down.
So far no net energy extraction has taken place, no external weights lifted, no external mechanisms are driven. It is obvious, no debating should have been be required.

Great, we are in agreement on a, b, c

Ok let’s move next the one pod one riser model.


Now to one pod one riser.  Use as ref the patent with one riser only, also refer to video 5 but there use straight walls, the inverted cup is the riser,  the cement is the pod, its top area Ap is smaller than the cup area (riser) Ar,     Ap<Ar.
(the riser is the cylinder and the pod is similar to the piston/rod but with Ap<Ar)

At Level1
Everything is fully submerged, there is some air under the riser (air pocket) to counter balance the riser’s weight. The riser is buoyancy neutral. Let’s immobilize the riser at this level.
The pod is designed to also be buoyancy neutral. Next insert the pod under the riser so the pod’s top surface is at the same level as the water level under the riser (cup). The pod has not broken that surface, no added pressure there, the riser is still buoyancy neutral and has not moved. The pod is also buoyancy neutral inside  the water. No energy need moving it inside the water.


The riser is immobilized so it cannot move vertically. It stays at level1 for now.
Next push up the pod, lets say half way up for simplicity into that air pocket.
What is going to happen?
We compress the air (precharge) and also create a head differential that equals to air compression at pressure P greater than what was there before.
The head differential is at the gap between the pod wall and inner riser wall. (remember Ap<Ar)The riser, (cylinder) has not moved yet. We are holding it in place, immobilized with e.g. a pair of large pliers from its outside sidewalls etc.

At this state, let’s do a simple free body diagram on the pod and then on the riser:
The pod, we are pushing up with a force Fp and that force is P*Ap where Ap is the surface area on top of the pod. Fp=P*Ap plus some exposed pod weight above that water surface.
The riser sees a force Fr=P*Ar where Ar is riser area and Ar is greater than Ap and Fp<Fr 

The riser is held in place by pliers so the reaction at the plier contacts is also Fr, the riser has not moved yet.

Now lets drop/place a weight W equal to Fr , W=Fr on top of the riser, then release the pliers. Nothing moves.

The riser as it stands now is again buoyancy neutral inside the water. It is still weightless inside the water.
So if we push up the pod with just an infinitesimal additional force the whole thing will move up lets say to level2 .

Now what just happened? With Fp<Fr we moved W (Fr) from level1 to level2 (distance x)
For the move we have EnergyIn = Fp*x and EnergyOut = Fr*x=W*x
But Fp<Fr then EnergyIn<Energy out.

At level 2 we clamp the weight W, and we clamp the riser. The riser is now at level 2 and cannot move anywhere.
Then we release the Fp so it expands to the same position relative to the riser similar to before precharge at level1. (top of pod moves back down to water surface inside the pocket, head moves up to its original position and so on)
Now we are back to the same state as level1, unclamp the riser, next we move the riser/pod back to level1 ( they are buoyancy neutral similar to scenario c) (yes the added expansion due to x comes in place but adds or subtracts nothing, as we discussed before, the exposed pod weight is also returned no net there so we leave all these out)
With every cycle we are gaining (Fr-Fp)*x 
(If we run actual numbers the net gain is small with one riser, but can be increased with more risers, Ar total becomes greater for the same Ap, and so on) The assistance comes from the head differential which moves along throughout the x movement without any additional outside help. In this model, the head is created by precharge at level1 and given back at level2. That whole move along is cost free energywise but the travel is limited. The head differential is max at level1 and smaller at level2. Look also at diagrams post373 and patent. The change is not linear, we could let it go higher than levl2 but at earlier portion of the travel we get the higher advantage. 
The precharge  itself is intact energy wise, whatever we put in it was given back and so on.
The pre-charge is not used as an energy source during the cycle. Whatever is compressed at the lower level is given back when it expands at the higher level. Same as the piston/cylinder example earlier posts.

If a simple theoretical model/math does not work, then physical models built should not work either.
On the other hand, and to keep it simple, if a simple theoretical model/math does work so should the physical models built.

Ok over and out for now.

Quote from: webby1 on July 28, 2012, 12:39:04 PM
small sidetrack from the current discussion

I am still working on my test build testbed, I am mounting it onto a base and all that stuff and will be attaching the head extenders after that is done and checked for leaks.

One thing I have done to make it easier to distinguish between risers is to color the bottoms of the risers, hopefully that will also allow me to see the difference between air and water.

Would be nice to some pics or even video when its done.

I am sorry, but I keep choking on some points. Please help me to understand before rushing on.

Quote from: phwest on July 27, 2012, 11:10:22 PM
I know, I was looking similar questions among the many, when I got into analyzing this.

Actually we don’t have to exceed the -more than neutrally buoyant level of expansion- at level 2.

Wait a minute. So you are stopping the piston before it reaches its full travel, then, leaving the pressure on the inside greater than the outside.If the device was neutrally buoyant at a certain volume at the lower level, you cannot let it expand to a greater volume at the higher level without also making it positively buoyant. Therefore you must stop the piston before the pressures inside and out are equal. You no longer have a Cartesian diver if the piston is locked, so you don't get your dynamically unstable "hover". The CD's behaviour depends on its "piston" being free and all volume change a result of applied external pressure from the surrounding water.

QuoteAt level2 even at that amount of expansion, it has expanded more giving some extra energy. So at levl2 we got out what we put in as compression at level 1 plus some extra. The extra can be used for raising it… So now at level 2 it is neutral buoyancy, it only needs an nudge to fall back down, and it can also produce some work as it is doing so. ( the more it lowers the more it weighs  and so on), up until it arrives  back down let’s say to a support or stop at level 1, end of cycle.

Not if the piston is locked, which it must be for your first condition to hold.

Quote
We  can compress it just enough at level1 so when it reaches level2 is at exactly buoyancy neutral state.. 

When, in the limit of zero difference between levels? If it's neutrally buoyant at volume V0... it will not be neutrally buoyant at any other volume. The amount of water displaced, hence the buoyancy, depends on the volume doing the displacing. Change the volume by moving the piston in any direction.... which is required to compress or decompress the gas... and you are no longer neutrally buoyant. And if you lock the piston so the outer water pressure does not equalize the inner air pressure by moving the piston and changing the buoyancy, you don't have the dynamically  unstable CD and your height changes don't come for free.

Quote
Bottom line is: what we put in at level 1 as added compression we get out at level2 plus some, and the -plus some- helps/pays for the lifting cost of the less than buoyancy neutral body (after compression at level1)

Next cycle same as the first,  we recompress  ( using the “what we got out is what we put in” at level2) and repeat. Again no net gain or loss out of the full cycle. Conservation of energy holds. Keeping it simple that’s all.

Side note on scenario c :
There are many ways but all result in the same bottom line.
For example,
We can compress, lock, lift, expand, push if needed back down(depending on expansion amount at level2, and repeat.
We can start buoyancy neutral at leve1, no lock, nudge it, it lifts by it self to level2. Producing some work as it ascends, we use that work to compress it (at level2 back to the volume of level1) and lock it. Then it can sink with no added energy, (locked in buoyancy neutral now), back to level1 release the lock, it goes nowhere, nudge it up and so on. Ok maybe have we talked enough on this.

Bottom line is still the same, no net gain or loss, what ever is put in it is taken out during a full cycle.

Bottom line for me is that there are still some equivocal points in your analysis that I'd like cleared up before charging ahead.  I've tried to keep it simple by sticking to your original description (once I got it straight) but now you are playing around with stopping pistons before they've fully expanded, and other things that seem non-physical to me.
How do you change the pressure inside the syringe without changing its volume hence its buoyancy? It seems that your above explanation requires this, in addition to requiring CD behaviour even though the piston must be locked.

If you want to specify clearly that the initial compression is part of your full cycle, and that ALL components, water levels, volumes, diver levels and internal and external pressures arrive back at the starting point of your full cycle, then I'll happily stipulate that you are correct even if your details are a bit muddled ... so far. But I'd really like to see you put that statement out there, explicitly, listing all conditions that have to repeat exactly to count as a full cycle.

In other words... how do I tell if a cycle, or tens of cycles, have occurred, between observation T1 and the later observation T2, both taken at the same point in a cycle?

Quote from: webby1 on July 28, 2012, 03:16:58 PM
But of course!

One thing I think I should mention about this device I am playing with,, it is very sensitive to the way it is setup.

I have found,,by accident so far,, that there are sweet spots in the setup.  When you find one of these the lift value is much higher and the water pressure on return seems to be more for a longer return value.

Also, there is a difference in the lift compared to just letting the system lift right away and holding the system still until full lift pressure is met.

Yes, good for you. May I suggest that you also make yourself a simple Cartesian Diver and meditate upon its workings, if you haven't? It's a simple situation where the buoyancy depends on the pressure from the water, changing the volume of the air that's inside of the "lifting" element. Sound familiar?