Thanks for your explanations! smile

> A big stumbling block is not knowing how much extra energy is
> needed to produce HHO at higher pressures.

Thats the point: you need at least as much energy as needed to overcome to surrounding pressure and to enhance the volume.

Thats the whole point of my mechanical box: to make it easier to calculate this needed energy.

The surrounding pressure at the bottom, the buoyance and the falling distance are directly related to each other.

> d) If you had an atmosphere that was >978 miles high, but had
> a pressure at ground level the same as our atmosphere, then it
> looks like you come out better than even.
> This atmosphere would have to be made of a gas less dense than
> air, and a tiny bit denser than HH

The pressure at groundlevel is directly depending on the density of the atmosphere and the gravitional pull.

pressure = height * density_ath * g
energy_gain = energy_gain_up + energy_gain_down

energy_gain_down = height * g * volume_sinking * density_ath

energy_gain_up = height * g * volume_floating * density_ath

energy_loss = pressure * (volume_floating - volume_sinking)

total_energy = energy_gain - energy_loss

As you can see the total amount of energy can be calculated by this simple formula which is only referring to the basic parameters of your system:

= (height * g * volume_sinking * density_ath) +
(height * g * volume_floating * density_ath) -
(height * density_ath * g * [volume_floating - volume_sinking])

since "height * g * density_ath" is always the same constant factor k, we can write:

= ( k * volume_sinking ) + ( k * volume_floating ) - ( k * [volume_floating - volume_sinking])

I hope it is clear this formula will always result in 0 !
It doesn't matter how big or small this factor of k is!

You can cange the density of the atmosphere, the gravitational pull and the height of your tower in ANY COMBINATION, it will only result in some different constant factor "k".