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Gravity powered devices => Gravity powered devices => Topic started by: hartiberlin on September 01, 2006, 08:04:26 AM

Hi All,
I am opening up a new thread about this, so it will not get
confused with the gravity mill concept.
Okay, here we have the Cartesian diver concept again:
(http://upload.wikimedia.org/wikipedia/commons/thumb/2/2b/Cartesischer_Taucher_animiert.gif/180pxCartesischer_Taucher_animiert.gif)
(http://upload.wikimedia.org/wikipedia/en/thumb/c/cf/Cartdivr.jpg/180pxCartdivr.jpg)
Now here are some calculations:
Imagine having a 4 Liter shuttle case and attached to it a 1,2 Kg weight.
So the buoyancy force at the top is 2,8 Kg into direction top upwards.
Now we apply via a 1 cm^2 area at the top of the bottle a 3 Kg weight.
This will bring up the pressure at the top from 1 bar outdoor sealevel pressure to 4 bar.
Now due to the gas laws, the volume of the shuttle will be compressed to 1 Liter and
thus have not enough buoyance force to compensate the 1,2 Kg weight, so the shutle is
sinking.
The bottle is maximum 20 Meters deep and at that deepth there is without the 3 Kg weight at the top
a pressure of 3 bar and with the weight applied at the top we have down at 20 Meters a pressure of 6 bars,
due to the water column added above it.
Now when the shuttle has come down to 20 Meters the volume of the shuttle will now be:
P1 x V1 = P2 x V2
thus follows:
4 bar x 1 Liter = 6 bar x 4/6 Liter.
So at 20 Meters deepth the volume is only 4/6 Liter= 0,8 Liter of air inside the shuttle
at 6 bar,
due to the high pressure there.
Now, when we release the 3 Kg weight from the top opening, the pressure at 20 Meters deepth will
be again 3 bar , so that means the shuttle air volume will now be at 4/3 Liter, which is 1,33 Liter and
this gives more buoyance than the 1,2 Kg weight of the shuttle itsself and the shuttle again
rises up and the air in it expands, so that it will rise even faster and have some more upward force,
until it is at the surface, where the airvolume will be back at 4 Liters.
So by placing a 3 Kg weight onto the 1 cm^2 opening we can rise and sink the shuttle so we can
get a work energy out of Integral from 20 to 0 Meters of F(t) x ds
F(t)= Forces depended on time
ds= difference of distance
As the force changes over time with the height of the shuttle, it is probably
much more easy to calculate the output via the gas laws, which I will do next and see,
if there is coming more energy out, that needs to be put in when you just
lift the 3 Kg weight a few millimeters to apply the pressure.
BTW,do you think the 3 Kg weight must be lifted very high,
as the water has to compress the 4 Liters to 1 Liter inside the
shuttle ?
But when you have a big 20 meters deep bottle, I guess you have so much water in there, that it does not need
much pressure on the 3 Kg to move this further into the bottle, so the distance ds is pretty low.
As the energy for this is W= F x s = 3 kg x 9,81 x s, it really now depends on the distance s how big the input enery
must be... Can this calculated somehow ?
I there are e.g. 10000 Liter in the 20 Meter deep bottle, then how far will the 3 Kg weight go into the opening,
when it has to compress the 4 Liters of air down to 1 Liter ?
Many thanks.
Regards, Stefan.

Hi Stefan.
You know, first I was negative about your idea. Yesterday I had some time for gooing deeper into it. As water is nearly incompressable, you don't have to care how big the bottle is, as long as its a closed system. The generated pressure on it will always compress the air inside the diver. I also think, your system is not limited to 20m. Sure, the pressure down there is higher than at 5m, but that belongs to the whole diver and so with applying constant pressure to overcome the beginning buoyancy (make the diver havier than the water it displaces) it will sink down till the water absorbs most of the pressure generated (and as water is nearly incompressable that will be a long run). When releasing the pressure, the shuttle will become buoancy, also in 50m deep  as the extra pressure is always the same in the whole system.
Andi.

Hi Andi, there is the problem that the deeper you go with the shuttle the more adds up the compression of the air inside the shuttle via the deepth pressure of the water. In 50 Meters the deepth pressure is so big, you will not get the shuttle up again, cause the air volume is too much compressed ! Or you would need a compression of 10:1 already at the top, but that would need a very big weight also...

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...

BTW,do you think the 3 Kg weight must be lifted very high,
as the water has to compress the 4 Liters to 1 Liter inside the
shuttle ?
But when you have a big 20 meters deep bottle, I guess you have so much water in there, that it does not need
much pressure on the 3 Kg to move this further into the bottle, so the distance ds is pretty low.
As the energy for this is W= F x s = 3 kg x 9,81 x s, it really now depends on the distance s how big the input enery
must be... Can this calculated somehow ?
I there are e.g. 10000 Liter in the 20 Meter deep bottle, then how far will the 3 Kg weight go into the opening,
when it has to compress the 4 Liters of air down to 1 Liter ?
Hi Stefan
Here is a little drawing.
Take a syringe, cut of the front part so that you can see its own crosssection 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 bottleshaped and on the widest crosssection it is 3 times (A3) wider than the bottleneck, so there is a ratio from 1:3.
And here is the paradoxon again :o. 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 crosssection in the diver, remember 1:3 between the crosssections, 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

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.

Hi Stefan
Here is a little drawing.
Take a syringe, cut of the front part so that you can see its own crosssection 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 bottleshaped and on the widest crosssection it is 3 times (A3) wider than the bottleneck, so there is a ratio from 1:3.
And here is the paradoxon again :o. 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 crosssection in the diver, remember 1:3 between the crosssections, 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 seesaw.
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= (Forcebuoyancy) x (movementdistanceinsidewater)
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.

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.