Hello Jason,
I agree with Flux that the loss calculators do not tell the whole story. What is important is not that you have a system "without too many losses", but that the system works for what you want it to do. The following assumes you are at 12V nominal; if not, scale up (for 24V, multiple voltages by 2; divide currents by 2).
A typical PV panel puts out a very slowly decreasing current up to about 14-15V, at which point it starts decreasing faster and hits its maximum power point (where the current is decreasing at the same rate at which the voltage is increasing) at about 17V, and then the current falls off precipitously above that. A lead-acid battery needs up to 14.4V to top it off, but at that point the current is being limited by the charge controller anyway.
If you don't have a Maximum Power Point Tracking (MPPT) charge controller, as long as you are not losing so much (calculated) voltage that the resulting (calculated) voltage is lower than the battery voltage, you're not really losing much power.
For example, let's say you have a 12 ga wire (~0.0017 ohms/ft), so your total resistance is 0.34 ohms (2 x 100' x 0.0017 ohms/ft). If your 120W panels are typical, they put out about 7A at 17V (7A x 17V ~= 120W). So your voltage loss is 7A x 0.34 ohms = 2.4V, leaving you 14.6V, enough to charge your batteries, so your loss is minimal (less than the I^2R loss of 7^2 x 0.34 ~= 16W would have you believe; you've already lost more than 16W by not using MPPT).
However, the voltage of the panels will go down when they are really hot, so if it's 100F (~40C) outside, the panels might be 140F (60C), and they might lose about 1-1.5 volts, leaving you about 13.1-13.6V, which is probably enough to keep your current from dropping too much if the batteries are below 80% full. On the other hand, if your battery's at 14.4 volts, your charge controller is probably limiting the current anyway, so it's rather moot. Note that this is an approximation, and you would have to use the actual I-V curves at the cell temperature to determine the current, but only do this if you like math and analysis and don't really have much else to do.
If you are charging batteries, the heat loss is probably moot, because you have extra hours of sun in the summer anyway, and your batteries will stay charged (unless you are using your solar power for irrigation or another normally summer only activity). 12 ga wire is pretty cheap and available, so that's what I'd use. This will accommodate around 600W if you eventually go to MPPT at 48V nominal, so you've got a bit of breathing space. If you want the highest efficiency in the summertime, use 10 ga.
Sorry this was so long, but I haven't yet figured out how to write concisely.
Best of luck with your system,
Dave
