82 ohms is probably OK, although for 13.8 Volts (a typical lead acid voltage) you're pushing 40 mA per string and developing a bit over 1/8 watt in the resistor. Best go to half watters. Fresh from the charger you may have a problem....
Otherwise, the math's the same for Lueons, 12 Volts minus 3.5 is 8.5 Volts 'across the resistor'. If we want to get one Amp (1000 mA), all we need to do is divide that voltage by that current; 8.5 divided by one is of course 8.5. That's how many ohms. Power is 'I squared R'. One times one times 8.5, 8.5 Watts *there*. Higher voltage will make it worse.
Conservative design calls for 1000 mA at 14.7 Volts 'hot off the charger' (and lower output at normal Voltages). So we're talking 14.7 minus 3.5 or 10.2 volts divided by one amp....10.2 ohms (ten probably close enough). One times one times one is ten, go with at least 15, better still 20 Watt. At 12 Volts then we'll 'only' be pushing 12 minus 3.5, 8.5 Volts (again) divided by ten ohms is .85 Amps (850 mA), still plenty bright. This circuit is much more stable with reasonable voltage changes than the 3 series string because of the larger voltage drop across the resistor. Small changes in battery voltage are a smaller percentage of the total.
More efficiency can be had from a second LED in series (like the Luxeon V suggestion), and recalculating for 7 Volts Vf.
Perhaps a better way to go is with a *regulator*? A (fairly) simple series regulator would hold the current steady and could drive the Luxeon at one Amp or the total of the 31 strings at a bit over 30 mA per string (reduce the series resistor in each string perhaps). A simple switch could drive one or the other. The bank or one (two or even three) LS.
Good luck, sounds like a fun project.
Doug Owen
