Gravity Mill - any comments to this idea?

[hartiberlin]

P.S. As 0,0981 Liter that is about 0,1 Liter of water
has about 100 cm^3 of volume, in a 1cm^2 area exit
header tube we would have a height then of:
100 cm^3 / 1 cm^2 = 100 cm = 1 Meter.

Hmm, this is quite a lot as I just see,
now it depends on how big the shuttle must be
at 100 Newton positive buoyancy force.

Regards, Stefan.

[prajna]

Stefan,

Head height = ((max displacement - weight) + (weight - min displacement))  / 2.  Let's put some figures in and see if this works: max displacement = 1.5 litres, min displacement = 1 litre, weight = 1.25kg so at the top the pressure on the header tube = (1.25 - 1) = 0.25kg/cm² and at the bottom the pressure on the header tube = (1.5 - 1.25) = 0.25kg/cm².  To get the average pressure we calculate (0.25 + 0.25) / 2 = 0.25kg/cm².

We can ignore the pressure at the sides of the tube because both the tube and the shuttle are rigid (if not then we have a problem because the shuttle will stick or water will leak around it).

The hydrostatic paradox has bitten you again:  In calculating the static head we need account for only the pressure on top of the shuttle and underneath the shuttle.  Indeed this is precisely what the hydrostatic paradox is about.  We need only take account of a column of water.

The height of the shuttle has absolutely no effect on the head because the shuttle is always submerged.  If we are talking about floating bodies then the aspect ratio might come into it (haven't given it a great deal of thought, probably it doesn't even effect it then.)  The shuttle can be 30cm x 10cm x 10cm or 10cm x 10cm x 30cm but so long as it is submerged then it will always displace exactly displacement - weight of water.  If you think you know better then think better.  I believe you are getting confused by this paradox thing because once a body comes to the surface it must support the weight of any part that is above the surface.  That doesn't come into play in our system.  That is why I said in a much earlier message "The hydrostatic paradox doesn't apply in this system", or at least it doesn't as far as the floating bodies part is concerned and that is the part which is causing you so much confusion.

[tbird]

hi all,

i'm not sure how to start, so i'll just get right to the answer you've all been waiting for.

it doesn't work.  i was wrong and i owe stefan and 2tiger, in particular, a huge apology. I'M SORRY!!  i feel like the kind of guy i was fussing about from time to time.  making definitive statements without having the definitive science to back me up.  in my heart, i believed it.  i wasn't trying to just be disruptive.

the definitive answer comes from the formulas pertaining to pumps, head pressure in particular.

http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html

this site gives all the needed formulas to prove the point.  not to mention my test that failed.  as a result i came back to the computer to find the truth.  armed with these formulas and a bit of time, i saw the light.  to me, the small exit pipe (when filled with the amout of pressure [water] it is exposed to) serves as a cork to what is otherwise already closed.

true to my word, this will be my last post, unless popular opinion thinks i should continue.

good luck to all!!

tbird

[FreeEnergy]

hi all,

i'm not sure how to start, so i'll just get right to the answer you've all been waiting for.

it doesn't work.  i was wrong and i owe stefan and 2tiger, in particular, a huge apology. I'M SORRY!!  i feel like the kind of guy i was fussing about from time to time.  making definitive statements without having the definitive science to back me up.  in my heart, i believed it.  i wasn't trying to just be disruptive.

the definitive answer comes from the formulas pertaining to pumps, head pressure in particular.

http://www.engineeringtoolbox.com/pump-head-pressure-d_663.html

this site gives all the needed formulas to prove the point.  not to mention my test that failed.  as a result i came back to the computer to find the truth.  armed with these formulas and a bit of time, i saw the light.  to me, the small exit pipe (when filled with the amout of pressure [water] it is exposed to) serves as a cork to what is otherwise already closed.

true to my word, this will be my last post, unless popular opinion thinks i should continue.

good luck to all!!

tbird

[prajna]

I concur with tbird.  I have finally discovered the error in my calculations that made it appear to work.  If you would like the difinitive answer then you can see it at http://declarepeace.org.uk/elsa/.  Make sure that you read all the way down to the end of the main text (past the output power calculations).  I declare ELSA a non-runner, unless anybody would like to find fault in my detailed analysis.