Rosch taking orders on OU Bouyancy device.

[d3x0r]

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.

what is mgh? 
I'm stating that 100% of the energy from the input is entirely lost disregarding any bouyancy effect.  So given 0 input energy used to move the floats it's obvious that input has nothing to do with output.

[d3x0r]

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.

but whether there is a chamber there or not the water level is raised as the air causes displacement at the bottom.
And it is displaced further as the bubble rises and is not under such great pressure, causing further displacement.
Yes that work is where the energy on the input goes.... 100% to displacing water. 
0% to moving the chamber.

[tinman]

. 

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?

Lol,the guy that designed it and own's it IS the big power company owner lol.
Oh,and it's no where near the ocean.

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

I found no full report on building 7,and in the !what they call a report!,they actually admit to not testing for explosives . In fact,everything was shipped of that fast to china to be melted down,no real investigation was done. All the reports you put forth are government backed lol. You havnt shown one report from an independent body of any type. The reason for that is-all the real engineers,and true experts in the field all know the buildings were brought down by explosives-not fire. You cant even understand as to why building 7 came down slightly faster than free fall speed lol. The reason you dont understand is because the only way building 7 could fall slightly faster than free fall speed go's against your blind faith.

They aren't hard to find.  They are not difficult to read and understand.  No magic nano thermite or dustifying beams required.

Im sorry Mark,but i am the one that has so far provided credible evidence-along with scientific fact,that the buildings did not fall due to fire. And dont forget-they didnt test for explosive residue at all in your reports.

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

No Mark, a stock 757 and 767 just cannot do 500 and 570MPH at sea level without falling apart. The only thing you presented is a Mach.86 speed-->which is at cruising altitude lol-not sea level.

You call your self a man of science and fact's,and yet here you are backing rubbish lol. You shun people for not being able to read a scope,you say there is no excuse for some one of that stature to get it wrong,and yet a simple circuit analysis seems to be beyond your limit's.
You know what they say-people that live in glass houses

[tinman]

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.

Only you dont have to put that much air in the bucket when the bucket is at the bottom of the column of water. If your bucket has say a 10ltr capacity,then you only need to displace 1ltr of water with your compressed air. As the bucket rises,the pressure around it decreases,and more water is displaced from the bucket making gain bouyancy. This also raises the head level of the water in the column,and thus the bucket can provide more work. The energy used to put the air into the bucket is regained as the bucket makes it's way to the top-along with extra energy,as the head of water has been raised as the bucket displaced more water on it's way up. Then the buckets fill with water and make there way back down,and although slight,there is an energy gain there as well,but what energy was used to fill the buckets with water?-gravitational ?.

[Farmhand]

That is a bit of a confusing post to read with no back reading. But if you displace one liter of water with air while the bucket is at the bottom then the air remains compressed to some degree and as the bucket rises in the water column the air will become decompressed and expand to more than one liter due to a lowering of external pressure which would in turn displace more water as it rises up.

It would seem some input energy remains in the compression of the air until the air reaches the atmosphere or equivalent pressure to it.

The decompression of the air is what would displace more water even if no more air is added. The water at the bottom being under more pressure means the air at the bottom is also under more pressure, and a relation would exist between the water pressure and the air pressure right from the get go.

All needs to be considered, the energy used to compress the air, the energy needed to generate the electricity to compress the air as well.

A stand alone system does not connect to the grid or other power source. We pay money because of the cost associated to the generation of electricity and it's distribution the cost is due to the energy involved in the pre generation work such as providing the fuel and plant. And that is all part of the energy considerations.

We cannot begin our calculations at the compressed air cylinder or even at the wall plug, true cost also considers fuel and generation losses.
For a transformation system we can evaluate efficiency at the input to the device and the output of the device. But for a complete system all energy must be accounted for that is involved in producing the end result.

Compress the air by hand and see how free the energy is. Or if using a battery the energy required to put the battery back to it's original SOC would be the input. Not what is actually measured coming from the battery, that is just the usable input.

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