Gravity Mill - any comments to this idea?

[prajna]

How do you reason that, Stefan?

Think of it this way:  You can store 1 litre of water in a container that has an area of 1m² and a height of just 1mm.  Or you can store that same litre of water in a container that is 10cm by 10cm by 10cm, or, indeed, in one that is 5cm by 5cm by 20cm, or even one that is 1cm by 1cm by 1m.

Aah, your previous post has just arrived.  Yes, we have to account for the weight of water in the tube as a weight that our shuttle has to lift.

In my example we want to lift 1 litre of water a height of, say, 1m.  In our exit tube - let's say it has an area of 1cm² - we have 1 litre (i.e. 1 kilo) of water.  Now, if our shuttle has a positive boyancy of just 1.1 kilos then it will push that 1 gram (1cc) of water out the top of the 1m pipe and it will continue to do that, with an increasing rate of flow, as the shuttle rises (because the boyancy is increasing as it rises up the pipe).  The total amount of water pumped is the same whether we pump it (slowly) out of a narrow tube 1m tall or quickly out of a wider tube 10cm tall.

If you are working with your figures you can still discount the weight that is causing you so much trouble by using a thin tube from water level up to where you want the water.  You come adrift when you use the same diameter exit tube as the pumping tube; then, sure, you will have a lot of extra weight to lift.

[hartiberlin]

Again, neither tbird nor I understand what you are referring to when you say hydrostatic paradoxon do you mean 'paradox' as in http://scubageek.com/articles/wwwparad.html?

Yes, that?s it,
also have a look at this:
http://de.wikipedia.org/wiki/Hydrostatisches_Paradoxon

There you can also see the forces in the lower diagramm.

The main problem is, as the main pipe is connected to the smaller
diameter exit pipe, the hydrostatic pressure ontop the water column
is the same if you have only 10 Liters of water or 785,4 Liters of water
inside the exit pipe , the weight the water column see is always
785,4 Liters = 785,4 kg !
So the shuttle volume must be very big at 10 Meter deepth to overcome
this additional 785,4 kg and this makes the whole concept unvalid !

So only a nozzle at seawaterlevel will work in my opinion,
which sprinkles the water to an upper reservoir and avoiding
the negative impact of the hydrostatic pressure paradoxon !

http://de.wikipedia.org/wiki/Hydrostatisches_Paradoxon

[hartiberlin]

Have a look at this and also watch the Quicktime movie !

Understanding the hydrostatic pressure paradoxon is a MUST before
you understand the gravity mill !

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Pressure/HydroStatic.html

[hartiberlin]

Have a look at this:


Case C is the case for the connection of our pipe into the smaller exit pipe.

The water column would see as the weight the whole big diameter of water, although there
is just this small diameter water column ontop of it !
So instead seeing the real 10 Liters= 10 Kg  of water the main
pipe sees 785,4 Kg of weight, which must be additionaly lifted and which
puts a hard burden onto the volume size of our shuttle and energy
requirement !

[tbird]

hi stefan,

looks like you are back online again, so i'll try again.

as per your "hydrostatic pressure paradoxon" pressure at any given point is the same in all directions. so, the water has as much pressure on the bottom of the shuttle as is on top, but only to the water level.  the only extra pressure is the amount of water above water level.  now if your shuttle has a positive pressure created by displacing more water weight than it weighs itself, it will rise until the weigh of the water (contained in a tube) above weighs as much as the extra displacement. then it will be netural (static).  like the drawings show, it can be short and fat or tall and skinny, as long as the weight is not more than the shuttle pressure can handle.

tbird