Rosch taking orders on OU Bouyancy device.

[d3x0r]

Your premise fails right when you start conflating force and energy.

well... they are both expressed as answers to
1) how much work is done by bouyancy
and
2) how much work is required to compress air


I realize that the result is in different units; but which equation is wrong? 

[MarkE]

You seem to have forgotten that your compressed parcel of air needs to be moved down in the water or, if you compressed it on the bottom the water above it is displaced upwards.

The end result is that the energy expended in sinking or creating a buoyant object is exactly the same amount that can be recovered by allowing the object to rise.

And let's not forget, you can only reclaim all of the energy if:  There are no viscous losses and you take infinite time to reclaim the energy.

[MarkE]

so if I compress to 1.82 atmospheres 4x the air it's 13.9 ft-lb and the output is still nearly 3x that... (4kw in 12kw out is in the ballpark of product performance)


taking out 1/4 of the volume of it results in 1.36 atm (75% of 1.82atm), and only 1.05 atm is required with current setup of 0.5 inch 1 foot tubes.


so... it's still probable.

It is no more probable then compressing a spring returns free energy, or hoisting a weight attached to a counter weight via a rope over a pulley.

[MarkE]

well... they are both expressed as answers to
1) how much work is done by bouyancy
and
2) how much work is required to compress air


I realize that the result is in different units; but which equation is wrong?

Force and energy are expressed in different units because they are unique from each other.  They do not equate with one another.

1) Zero, nada, de minimus, squat. 
2) A lot more than anyone has ever been able to recover letting it expand again.

So from 1) you get:  The machine can't perform any net useful work. And, from 2) you get:  The machine cannot transfer work between a source and a load efficiently.  It is a very expensive room ornament / heater.

[LibreEnergia]

well see that's why I'm asking how does the pressure relate to that?  I mean... if the pressure is more, it's going to displace... the whole point in needing the excessive pressure over 1atm is the weight of the water... but the weight of the water turns out to be a very small percentage of 1atm additional... so at 1.053atm it can displace 21 inches of water easily.


(increasing the pipe size at this time does improve additional gain... to go 0.5inch which is 2.6x to 3inch is from 4.86x and an 8 inch pipe is 7.5x... increasing the length doesn't help (the height of water is computed based on the pipe diameter) but then requires much more pressure.. and almost 2HP to compress...)


so at 8 inches the height is 10.5 feet, and the pressure required is 1.32atm minimum... (10.5feet is 3.2m and 10m is +1 atm)

The amount of energy you will need to impart to the air will be (at minimum) the amount of energy required to lift a mass of water represented by the column of water with the horizontal component of the surface area of the float times the depth that it is submerged to. Fairly easy to calculate.

Using air will require more than that due to the losses to heat when compressing the air, and depends on the efficiency of the compressor.