Rosch taking orders on OU Bouyancy device.

[MarkE]

You have your eyes wide shut Mark,there is no doubt about that.
Oh,and i got some bad news for you in regards to bouyancy/gravity devices not being able to do work. How about a 10Kw unit thats been running since 2010,and was the show piece at the united Nations  COP15 conference in Copenhagen . Bet you would love to know all about that one.

Sure they did.  But being enslaved to big oil the magic bag of wet hammers machine hasn't got out.  It's been suppressed and only the valiant nutter community knows about it.  You do understand that a buoy moved up and down by tidal motion is powered by the tide and not buoyancy don't you?

Oh,could you direct me to the building 7 part in the NIST report-->i cant seem to find it?

Then you are saying that you did not read it?  WTC 1 and WTC 2 are covered in one report and WTC 7 in another.  They aren't hard to find.  They are not difficult to read and understand.  No magic nano thermite or dustifying beams required.

What planes can really do. But i doubt you will even give it the time.


https://www.youtube.com/watch?v=Rs5RQ_5nu4k

Your absurd claims as to what the planes could do have already been refuted.

[MarkE]

point 1 : good, then I get more pressure for the same volume; +1 goodness.  (edit: err alright I guess I have to increase pump character a little ... because I will end up with less volume of correct pressure at increased temp or the correct volume at slightly reduced pressure at increased temp... but the temp increase on nitrogen&oxygen ... which apparently is a constant that can be looked up is apparently neglegible in 90%+ of applications... because *spoiler* the temp is given in Kelvin, and is a small delta compared to either pressure of volume deltas.)
point 2 : after leaving the compressor I don't care if it sheds the heat to the environment, other than it will end up decompressing less.... it won't go below the initial ambient temperature... and slowly increasing heat in the water does make it less dense... but the overall molarity of the particles above will increase, slightly increasing the required pressure, but allowing more force from bouyancy.   (water expands above 4 degrees C... but the expansion is so slight it's got to be on the order of 0.00001x increase... )


http://en.wikipedia.org/wiki/Compressed_air_energy_storage (generalities...)


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Dug around a lot and read a lot of questions asking for 'how much tempuratue changes' mostly on physics forums... they almost always just go back to an adiabatic process and relate V1P1=V2P2 and disregard tempurature because ... what's gained in compression is returned during decompression without regard to heat transfer.... (if it needs heat after shedding some heat it will absorb it back from the system) ... so the bottom of the water vessel will be hotter than the top- other than the buckets end up mixnig it.


The other factor that affects tempurature increase in amount of volume change in time... only a very small pressure increase (relatively to tempurature which is in kelvin) is required in this system, and it's nowhere near 'fast' requirement... 1 foot/sec  (for a 5 inch diameter) is only 0.681818MPH... nowhere near the speed of sound (compressing with with the volume changing at the speed of sound induces excessive heat, since the air colliding with the compressing surface causes heat also).
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I also realized not 100% is lost, as the air in the bucket rises, it is expanding and causing increased displacement... and assuming there is no increased displacement only 0.75% is lost to that for my calculations.


So, less than 1% of the energy to compress the air is recovered.
The output is still grossly more than this.

What all this adds up to is that in the very best of overly optimistic cases you get zero sum gain and the machine simply transmits input power to the output without loss.  In real life it is much worse than that.

[d3x0r]

What all this adds up to is that in the very best of overly optimistic cases you get zero sum gain and the machine simply transmits input power to the output without loss.  In real life it is much worse than that.

No you mis-understand.  less than 1% of the input energy is used for any useful work.
the output energy has nothing to do with the input.
again saying output=(input-losses) is obviously incorrect... because output would be a negative amount, because output = (0 - losses) implies you would have to turn the output shaft yourself to get it to go.  the force available from the lift of bouyancy will definatly produce positive output.
output has nothing to do with input in this case.

[MarkE]

No you mis-understand.  less than 1% of the input energy is used for any useful work.
the output energy has nothing to do with the input.
again saying output=(input-losses) is obviously incorrect... because output would be a negative amount, because output = (0 - losses) implies you would have to turn the output shaft yourself to get it to go.  the force available from the lift of bouyancy will definatly produce positive output.
output has nothing to do with input in this case.

Either we are talking about two different things or you are very confused.  Buoyancy is basically borrowed gravitational energy.  Output work is only available from decreasing mgh of previously lifted atmosphere above the float.  Using water as the atmosphere:  work is performed lifting the water atmosphere, and a portion of that expended work is reclaimable as useful work.

[TinselKoala]

I'll vote for "very confused".

Let DXor consider the case where you have a vertical column of water in a tube... paint it white if you like. There is only one float chamber and it is at the bottom of its chain loop full of water, ready to be inflated.

Note the level of the _TOP_ of the water column by making a mark against the side of the tube.

Now, by any means you like, fill the float chamber with air, displacing the water that is in it. Don't let it go anywhere yet.

Climb back up to the top of the tube and measure the water level. What do you find? Is the mark you made, now submerged? Of course it is.  And, by clever calculations, you adeptly find that the height increase x the surface area equals _exactly_ the volume of the float that you have filled with air !! The float that is all the way down at the bottom of the tube!!

Do you see the consequences? You have _raised up_ a volume of water equal to the volume of the float, _all the way up_ to the top of the tube! You cannot do this without performing work, and you can calculate precisely how much work it costs to _raise up that volume of water_. This, then, is the INPUT that you must consider, and you must multiply this INPUT energy by the number of floats you fill! 

When the floats reach the top of their travel and flip over, filling once again with water, that volume of lifted water is "falling" and that is the only return of work that you will be able to get out of the system: What you put in to raise the water in the first place, minus losses.