    Order with Radio work delayed 3-7 days
 The AWG - American Wire Gauge - is used as a standard method of denoting wire diameter, measuring the diameter of the conductor (the bare wire) with the insulation removed. The higher the number - the thinner the wire. This AWG table is for a single, solid, round wire. Small gaps between the strands in a stranded wire make it so it has to be slightly larger to keep the same current-carrying capacity and electrical resistance as a solid wire. Because a thick wire has less electrical resistance it will carry more current with less voltage drop than a thin wire. AWG Diameter (mm) Diameter (in) Circular Mils Square (mm2) Resistance (ohm/1000m) 14 1.63 0.064 4110 2.0 8.54 12 2.05 0.081 6530 3.3 5.4 10 2.59 0.10 10380 5.26 3.4 8 3.25 0.13 16510 8.30 2.2 6 4.115 0.17 26240 13.30 1.5 4 5.189 0.20 41740 21.15 0.8 2 6.543 0.26 66360 33.62 0.5 1 7.348 0.29 83690 42.41 0.4 0 8.252 0.33 105600 53.49 0.31 00 (2/0) 9.266 0.37 133100 67.43 0.25 000 (3/0) 10.40 0.41 168800 85.01 0.2 0000 (4/0) 11.684 0.46 211600 107.22 0.16

What good is all this data? I put it here so you can calculate the proper wire size for your
mobile installations. (yes I only included the wire sizes I thought were important)
The formula for voltage drop is:

VD = 2 x K x I x D/CM.

• K = Direct-Current Constant. K represents the dc resistance. (for a 1,000-circular mils conductor that is 1,000 ft long, at an operating temperature of 75�C).
K is 12.9 ohms for copper and 21.2 ohms for aluminum.
• I = Amperes: The load in amperes at 100%
• D = Distance:  Where we specify distances here, we are referring to the conductor length.
• CM = Circular-Mils: The circular mils of the circuit conductor as listed in NEC Chapter 9, Table 8.

Let�s do an example. An amplifier rated 13.8VDC and 75A is wired
to the Batteries with 10 ft lengths of 4 AWG stranded wire.

 Copper Wire K = 12.9 ohms, copper I = 75A D = 10 ft CM = 41740 VD = 2 x 12.9 x 75 x 10 / 41740 =  .46 vdc dropped (So the voltage at the amplifier will be 13.34 vdc) Aluminum Wire K = 21.2 ohms, Aluminum I = 75A D = 10 ft CM = 41740 VD = 2 x 12.9 x 75 x 10 / 41740 =  .76 vdc dropped (So the voltage at the amplifier will be 13.04 vdc)

Just one more to those 16 pill-ers out there using 15 feet of 1/0 wire....this same formula for
350 amps would leave a voltage drop of 1.32 vdc and would leave 12.48 vdc at the amplifier.
1.32vdc x 350A x 65% = 300 watts lost

Reference : National Electrical Code (NEC)