[sswee]

According to my notes:

Plugging your numbers into an equation for the 9.4% electrolyte -
For a current density of 12 ASF, and a thickness of 0.0005" (0.5 mils); 720 / 12 x 0.5 = 30 minutes.
The actual current you would apply is 1.75 sq,ft.(252" sq.) x 12 ASF = 21 Amps.
or 12A / 144 = .0833 per in^2 * 252 = 21A setting
The required peak voltage is 0.95 Ohms per sq.ft / 1.75 sq.ft. = .54 Ohms. .54 Ohms x 21 Amps = 11.34 Volts. This assumes no voltage drops due to bad connections, and you haven't increased the actual surface area with racking.

Excerpt from other post:
For 20 ASF, 9.4% electrolyte concentration by volume, and 70 deg. F., this resistance is about 0.95 Ohms per square foot. Its value is inversely proportional to surface area (2 sq.ft. = 0.475 Ohms). It varies a little with alloy as well.

The calculations in the 720 calculator are based on the 5% concentration. The resistance in the circuit is an average or approximate but there are no places to change this in relation to CD. Therefore, I would think that if it has been determined that .95 OSF is the resistance of 20 ASF with a 9.4% acid concentration, it should be close enough for calculations at other CD's with the same acid concentration. With all the different parameters effecting the circuit resistance, it would take several pages to figure it all out.

I'll keep reading through my note and looking around for more. Maybe Neilfj will stop in and help clarify it all.
SS