long 12 volt outlet runs?

Don't worry about feet & meters (meters are a bit longer than yards, and 3m = ~10'), but I'll use that Merkan AWG table anyhow... Do you have any suggested or target max voltage drops - ie, <0.5V from MPPT to battery; <1V from panels to MPPT; less than 1V from battery to outlets at 5A etc? I won't do it now, but then I can redo those calcs - maybe in very verbose form so that you can later DIY by substituting new values (distance, current, etc).    It is "simply" (ha!) an exercise on Ohm's Law, ie: V = IR. aka "the voltage drop V along something equals the Current (I) going through it TIMES the total Resistance of it". (Units are Volts, Amps & Ohms. Though often V, mA & kΩ are used - the milli- and kilo- cancel each other out.) [ BTW: Ω = Ohms, but many use the "R" instead because we & I couldn't be bothered looking up and pasting the Ω symbol, or writing it, etc. And we know when the R means Ω as opposed to as in V=IR. BTW, in V=IR, the IR is short for I*R or IxR (I times R). And I is the symbol for Current (Amps). And if V=IR, then R = V/I where / means "divided by". That's obtained by dividing both sides by "I", ie: V = I * R hence V/I = I/I * R     hence V/I = R (because any number divided by itself is one - aka "the I's cancel".) ] The modification is that you are using lengths of cable. But a cable of certain size has a resistance per length. EG - in that linked table, see the "Ohms per 1000 ft." column. For simplicity, 13G is 2.003 Ohms per 1000 ft. Lets call that 2 Ohms per 1,000', and I'll use "R" instead of "Ohms". So 100' = 0.2R. 10' = 0.02R. 1' = 0.002R = 2mR (milli-Ohms). That's just dividing each side (2R = 1000') by 10 each time. EG - if you had 30', then that's 2R x (30/1000) = 2 x 0.03 = 0.06 Ohms. (check: if 1' = 2mR, then 30' - 30 x 2mR = 60mR = 0.06R. Phew!) So at 10 Amps, V = IR = 10A x 0.06 Ohms = 0.6. So 30' of 13G at 10A will have a 0.6V drop. That means 1.2V (2 x .6) taking the ground path into consideration (ie, it is then 2x30' = 60' at 0.002R per ' = 0.12 Ohms. 0.12 Ohms x 10A = 1.2V. (Great - checked ok again!) Rats! I started my "not now" rant already. But can you follow that?   Or maybe the notation is confusing eg, 2 Ohm / 1000' = 2 Ohms divided by 1,000' which is like saying "2 Ohms per 1,000'. More later - including other answers to your last....

oldspark wrote: But can you follow that?   Or maybe the notation is confusing eg, 2 Ohm / 1000' = 2 Ohms divided by 1,000' which is like saying "2 Ohms per 1,000'.

No, I cannot to be honest. All I understood, as I understood beforehand is that wires have resistance which in turn impede the flow of electrons which in turn heats up your wires, maybe even turn them into candle sticks. That's all I know. I am math stupid. When you start reciting all that gobbledegook my eyes glaze over and my brain goes into "huh?" mode. I just need to know the most efficient/safe wire gauge to use in my 12-volt outlet run, about 35' to 40' from the battery bank.

For the most efficient wire gauge to use for you purpose, I need the values/parameters I asked for in my last. But otherwise, I can figure something out and provide the parameters. Do you want to understand the gobbledegook, or just want an answer? The math itself uses basic primary school rules (things called the commutative law etc - eg, (AB)C = A(BC); A*1=A; A/A=1).