Got nailed,
I have been designing DC plants for a few decades now; and there is no easy answer to your question. NEC 310.16 is the bare-bones minimum you can get away with but for the most part entirely inadequate for most DC applications because of the low voltage nature of the beast.
Telephone companies have developed their own tables with minimum wire size vs breaker/fuse size. These charts are proprietary in nature, but far exceed any NEC requirement. For example a 20-amp breaker minimum cable size is 6 AWG. However the charts have distance limitation of around 50-feet
The only way to be precise is to use voltage drop calculation. To use the calculations you first have to have standards for maximum voltage drop allowed on the various nominal voltages of 12, 24, and 48 volt. To complicate it further is the distribution system has to broken down into components and voltages assigned to each segment like battery bables, battery buss, primary distribution, and secondary distribution.
You start by the industry standard of total allowable voltage drop from the battery terminals under full discharge current to the far end equipment. It looks like this:
12 = ? volt.
24 = 1 volt
48 = 2 volts
96 = 4 volts
As you may notice this is very tight tolerances. Now it gets worse because the part of the system you are working with is probable the secondary distribution, and you do not get the full voltage drop allotment, only a fraction of it because the upstream system get a fare share.
So if my assumption of secondary distribution is correct you max allowable voltage drop are:
12 = ? volt
24 = ? volt
48 = 1 volt
96 = 2 volts
So now all you need is a formual now that you know the max allowable voltage drop and the maximum discharge load current which is:
CM = (22.2 x I x D)/Vd
Where:
CM = Circular Mills of copper cable
I = Maximum load current at full discharge
D = One-way distance in feet
Vd = Maximum allowable voltage drop.
So lets say we have a 12-volt system on a 50-amp breaker that will draw 35-amps at full discharge voltage and the one-way distance is 60 feet. Plug in the values into the formual:
I = 35
D = 60
Vd = ?
And you get CM = 186,480 circular Mills. Go to NEC Chapter 9, Table 8 and look to see what size cable you need without being smaller comes out to 4/0. One final check go to NEC 310.16 and make sure the cable is equal to or larger than that required for the OCPD of 50-amps on the cable type of RHW which is # 8 AWG. 4/0 is larger than required so you are good to go.
Easy huh?