Cartesian diver, overunity concept ?

[hartiberlin]

Mh, perhaps you are right. I still cannot imagine that you are able to get the diver at 20m depth (perhaps also hold it there) and if its getting at 21m the diver is lost...

Yes, at 21 Meter it would be lost, as you can not overcome anymore the hydrostatic pressure of all the water column above it via the pressure release of the syringe like
upper piston.
You have to calculate it this way:

P1 x V1 = P2 x V2 = P3 x V3= P4 x V4

Inside water we have due to the air pressure and water column weight pressure:
P1= 1 bar at 0 Meters
P2= 2 Bar at 10 Meters
P3= 3 bar at 20 Meters
P4= 4 bar at 30 Meters

Now if we go with 4 Liter starting volume of the shuttle we have
for 0 meter, 10 Meters and 20 Meters

1bar x 4Liter = 2bar  x 2 Liter= 3 bar x 4/3 Liter= 4 bar x 1 Liter

Now if we compress at the top already with additional 3 bar we get
for 0 Meters, 10 Meters, 20 Meters and 30 meters

4 bar x 1 Liter= 5 bar x 4/5 Liter= 6 bar x 4/6 Liter= 7 bar x 4/7 Liter

So at 20 Meters we have the 2 different pressures and
volume of:
3 bar x 4/3 Liter versus 6 bar x 4/6 Liter.

So if we attach a 1.2 Kg weight to the
3 bar @ 4/3 Liter= 1,33 Liter  shuttle it could
still rise.
But at the top with the 4:1 compression:

4 bar x 1 Liter
the shuttle
would still sink with the 1,2 Kg weight attached to it.

But if you would go down to 30 Meters, where we have only 1 Liter
volume in the uncompressed state, so we would
not be able to rise the shuttle with its attached 1,2 Kg weight !
Due to the big water volume above it the air stays too much compressed,
so the volume is only 1 Liter= 1 Kg upwards force, so the 1,2 Kg weight is heavier
and the shuttle is lost at 30 Meters deepth.

[hartiberlin]

Hi Stefan

Here is a little drawing.
Take a syringe, cut of the front part so that you can see its own cross-section and glue this on a bottle neck.
With this experimental setup you can simulate how much force you have to apply on the syringe and how far you have to push for make the diver move up or down.

From my point of view the force (pressure) transmision between the syringe and the "bottle neck" of the diver is 1:1 one (in the case of my drawing A1 = A2). Now look at the diver self. It is bottle-shaped and on the widest cross-section it is 3 times (A3) wider than the bottleneck, so there is a ratio from 1:3.

And here is the paradoxon again  . If you push now the syringe 1 cm down it will transmit the force/pressure 1:1 to the bottleneck of the diver, so the water will rise in the bottleneck 1 cm too.
When the water now rises up to the upper cross-section in the diver, remember 1:3 between the cross-sections, the force/ pressure transmission will be 3:1, because of the paradoxon. That means that you have to push the syringe 3 cm down in order to let the water rise 1 cm in the diver.

Nevertheless this setup looks very interesting, because the dimension of the weight and the distance it must been moved depends only of the dimension of the diver and not of the whole watercolum.

CU
2Tiger

@2Tiger
I think you are very right with your calculation.
So if we have inside the shuttle the surface Area 3 times
bigger, then we must press the syringe piston also 3 times
the longer distance.
This is exactly also the case inside a hydrostatic press.

This is also true, cause it obeys the law of energy conversion,
so if you can multiply the force with it, it moves only a
shorter way with the multiplied force as:

W= F1 x s1 = F2 x s2

So for instance if F1= 1 Newton and F2 = 10 Newton
we get for S1= 10 Meters thus follows s2 = 1 Meter only.
so we have=
Energywork= 1 Newton x 10 Meter= 10 Newton x 1 Meter

It is like a see-saw.

Now the question is,
if we make the areas both the same inside the shuttle and
at the syringe piston, will
the shuttle then be able to do  more work:

W= (Force-buoyancy) x (movement-distance-inside-water)

than is needed to put a weight onto the syringe piston
and remove the weight  again from the syringe piston ?

Am pretty busy right now and wil think about this later.
Thanks for clarifying this.

Regards, Stefan.

[hartiberlin]

Okay, let?s stay with the above example I had given:
at the top at sealevel 4 Liter shuttle volume
of the Cartesian diver  versus 1 Liter volume
when the
syringe piston compression is done via placing a weight
onto the syringe piston.

That means when we use a quadratic and not a round
surface,
we must compress 10cm x 10 cm x 40 cm= 4000 cm^3 =4 Liter
to
10cm x 10 cm x 10 cm= 1000 cm^3= 1 Liter

4 Liters shuttle volume will have a buoyancy of 4 Kg, so we could probably
just place a bit more than 4 Kg onto the upper syringe piston
to compress the water and the volume inside the Cartesian diver shuttle
would go down from 4 Liters to 1 Liter. Hmm, is it really this easy ?
Or must the weight be different ?
I am not sure, but let?s continue to calculate this way:

Then the 4 Kg weight would have moved 30 cm, so we would
later have to move it up the same distance and thus
do the work:
W= mass x g x height
W= 4 Kg x 9,81 x 0,3 Meter= 11,77 Wattseconds or Joules.

Now the question is, what force multiplied via distance we get,
when we sink and rise again the shuttle.
Both sinking and rising must be added for the energy output.
As the force is not constant over the 20 meters deepth
example, it must be integrated over the distance.
Also it has to noted, that we need to attach a 1,2 Kg weight
onto the shuttle to be able to sink the Cartesian diver at the
1 Liter buoyancy at the top, so at the top the downwards force
is just 0,2 Kg x 9,81= about 2 Newton.

Maybe someone else can calculate this,
as I am pretty busy right now.
Many thanks.

Regards, Stefan.