How this instrument works
A circular mil is the wire industry's unit of cross-sectional area: the area of a circle exactly one mil, one thousandth of an inch, across. American Wire Gauge tables are built on it because doubling a wire's circular-mil area — not its diameter — is what actually halves its resistance per foot, and gauge steps are sized to move by clean multiples of area rather than clean multiples of diameter. Sizing a conductor in circular mils, then, is sizing it by the one property that actually governs how much of the supply the wire itself eats before current reaches the load.
The first line just turns a percentage into a number of volts, and that step is where 12 V systems get punished. Three percent of a 230 V mains circuit is 6.9 V of headroom; three percent of 12 V is 0.36 V — twenty times less room to work with for removing the same resistance. The second line is Ohm's law rearranged around wire geometry: resistance runs as K·L ⁄ CM, current out and current back both travel the conductor, so the length doubles, and solving for CM against an allowed voltage drop gives the minimum area that keeps the loss under budget for a given current and distance.
K, the constant in that line, is not copper's bare laboratory figure. Pure copper's resistivity works out to close to 10.4 ohm-circular-mils per foot at a cool 20 °C bench; the 12.9 used here sits above that on purpose, to cover a real installation — a stranded conductor rather than a solid lab sample, run somewhere hotter than a bench, since copper's resistance climbs with temperature. What the formula still leaves out is everything that isn't the two lengths of wire: fuse holders, crimped lugs, switch contacts, and the battery's own internal resistance all add loss the calculator cannot see, so a wire sized exactly on the line has less real margin than the number suggests.
- Enter Circuit current — the actual draw of the load in amps, not a breaker or fuse rating.
- Enter One-way wire run length, ft — the distance from source to load measured once; the formula doubles it for the return path internally.
- Set Allowed voltage drop, % — the fraction of the 12 V supply you're willing to lose in the wire; 3% is the usual conservative target for accessory circuits.
- Read Allowed voltage drop, V — the absolute figure your percentage becomes once multiplied against the 12 V supply.
- Read Minimum wire size, circular mils, then round up to the nearest standard AWG size whose rating meets or exceeds it — never down.
Worked example — a 20 A circuit over a 15 ft run
A 20 A accessory load — a compressor fridge or a winch control circuit is typical — sits 15 ft one-way from the battery bus, and the installer wants to hold the loss to 3% of the 12 V supply. First line: VD = 12 × 3 ⁄ 100 = 0.36 V, the entire budget in volts. Second line: CM = 2 × 12.9 × 20 × 15 ⁄ 0.36. Working it through, 2 × 12.9 is 25.8, times 20 A is 516, times 15 ft is 7,740, and 7,740 divided by 0.36 V lands on exactly 21,500 circular mils.
Standard AWG 8 wire carries only about 16,510 circular mils — short of the requirement — so the correct choice rounds up to AWG 6, rated near 26,240 circular mils, comfortably past the 21,500 minimum. Note what would happen on a mains circuit instead: the same 3% budget at 230 V is 6.9 V, nearly twenty times more headroom, so an equivalent 20 A, 15 ft run would need only a fraction of this cross-section. The fat cable a 12 V accessory panel demands is not overcaution; it is what the arithmetic actually requires at that supply voltage.
Questions
Why does the run length field ask for one-way distance, not the round trip?
Because the formula already doubles it for you. The CM ⁄ VD line multiplies length by 2 internally to account for the return conductor, so entering the full there-and-back distance would double-count that leg and make the calculator recommend wire roughly twice as thick as the circuit actually needs.
Why is a 3% drop budget so much tighter at 12 V than at mains voltage?
Because 3% is relative, but volts are absolute. Three percent of 12 V is 0.36 V; three percent of 120 V is 3.6 V; three percent of 230 V is 6.9 V. The percentage rule is identical, but the low-voltage system has to remove far more resistance to stay inside its much smaller absolute allowance for the same current and distance — which is exactly why 12 V wiring runs look oversized next to household cable.
Where does the constant K = 12.9 come from?
Copper's own resistivity gives roughly 10.4 ohm-circular-mils per foot at a cool 20 °C bench reading. This instrument's 12.9 sits above that deliberately, building in margin for a stranded rather than solid conductor and for real operating temperatures — an engine bay or a sun-heated console — that run hotter than a lab sample, and copper's resistance rises as it warms.
The result is 21,500 circular mils. Which wire gauge do I actually buy?
Round up to the next standard AWG size whose circular-mil rating meets or exceeds your result, never down. AWG 8 offers about 16,510 circular mils, which falls short of 21,500, so the correct purchase is AWG 6 at roughly 26,240 circular mils — the smallest standard size that still clears the requirement.
Does this account for fuse holders, connectors, or the battery's own resistance?
No. The formula models only the two straight lengths of copper conductor between source and load. Every crimped lug, switch, fuse holder, and battery terminal in a real circuit adds resistance the calculator cannot see, so a wire sized exactly on the calculated minimum carries less real-world margin than the figure implies — leave room, especially on long or high-current runs.
What happens if the allowed voltage drop percent is set to zero?
The calculation breaks down on purpose. Zero percent asks for zero volts of drop while still carrying current, which needs infinite cross-sectional area — no finite wire qualifies. That is why the instrument requires a positive percentage; the closer that figure sits to zero, the larger, and more expensive, the resulting wire becomes.