For a 12V system your math is easy. Just divide your watt-hours by 12 to get that will be taken in amp-hours from the battery. In your case, 240/12 is 20AH.
One small caveat is that the efficiency of a lead-acid battery can be anywhere from 50 to 80 percent. So in order to recharge a lead-acid battery that you have taken 20AH from, you may have to put back the equivalent of anywhere from 25AH to 40AH from your charging source.
As for the PV panels, I would recommend (without having used them myself!) a couple of CIGS modules such as this one that will work well in low light conditions, and is attractively priced at $5 per watt.
http://store.altenergystore.com/Solar-Panels/51-to-99-Watt-Solar-Panels/Global-Solar-60W-12V-Framed-Solar-Panel/p5574/[/quote]
Wasn't even aware of battery inefficiencies, so thanks for that knowledge.
So if I base my calcs on cycling the battery only 20% deep AND want 2 days of autonomy at 20ah per day. Then I would need a 200ah battery based on my usage of 20ah a day.
To be on the very safe side, I would then need to make sure to put back 40ah a day, just to be safe.
AGAIN, my question is, what is really happening when I am the computer system is using 3.33ah of DC off the battery, when at the same time, the battery would be receiving 7.0ah of solar. Is ALL of the 7.0amps from the panels going into the battery? OR is my battery usage somewhat offset, by the solar panels, thus requiring a smaller battery?
thanks so much for your help...just about there..
James