Limiting voltage drop is particularly important in lower voltage power distribution, and especially in 12-Volt DC circuits, which are typically about the lowest voltage of electrical power that is distributed. The reason for greater concern with 12-Volt DC power distribution is those circuits often use high-current. The voltage drop in a conductor is directly proportional to current. Since good design calls for a limit of wiring voltage drop to 3-percent or less, it is clear that with 12-Volt power a limit of 3-percent means only 0.36-Volts of drop in the wiring can be tolerated.
In order to know the voltage drop in a conductor, only the current (in Amperes) and the resistance (in Ohms) are needed. The load current is generally known by the design of the system, or the current can be measured with an Ammeter. The resistance of the wire is usually so small that direct measurement of the wire resistance is impossible without very sensitive and carefully calibrated instruments; you cannot just use a $50 DVM to measure resistance values that are much, much less than 1-Ohm.
The resistance of copper wire can be assumed from the size of the conductor. In the USA, wire conductors are usually characterized by the American wire Gauge (AWG). The resistance value is usually given for a 1,000-feet length of wire, as this avoids having numbers with many decimal places that begin with a string of zeros. For example, a copper wire of 10-AWG is specified to have a resistance of just about exactly 1-Ohm/1,000-feet.
Values for copper wire are usually given in tables from reliable references. Here is an excerpt of one table of values:
AWG Ohms/1000-feet
0000 0.048
000 0.0618
00 0.078
0 0.0983
1 0.124
2 0.1563
3 0.197
4 0.2485
5 0.3133
6 0.3951
7 0.4982
8 0.682
9 0.7921
10 0.9989
12 1.588
14 2.525
16 4.016
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.htmlThe table is based on the AWG cross section area and uses a resistivity value of copper of 1.724 ×10-8 Ohm-meters at 20-degrees-C. Actual wire will vary somewhat based on the purity of copper. The table was originally published in "Electric Circuit Fundamentals," 2nd Edition, Thomas L. Floyd, (a 661-page reference book that retails for $170), and I assume it is a reliable source.
I don't bother with wire sizes smaller than 16-AWG because according to USCG federal regulations and ABYC standards, the smallest wire size that can be used in boat wiring for power distribution is 16-AWG.
A very common instance for a battery to be relocated to be further from its load in boating is for an outboard engine starting battery to be moved out of the stern of boat to reduce weight there, often based on increasing weight of the outboard engine itself. Because the current flow in an engine cranking motor is extremely high, typically hundreds of Amperes, there is very great concern about voltage drop in the wiring. For this reason, the conductors must be increased in size whenever their length is increased.
The precise value of current that will flow or the absolute maximum voltage drop that can be tolerated are generally not known for outboard engine starting circuits, but the engine manufacturer usually publishes guidelines about what wire size must be used for certain wire lengths. Here is a table from Evinrude in which they specify wire size for the battery cables for various lengths:
Recommended Conductor Size for Engine Starting
Feet Minimum AWG
1 to 10 4-AWG
11 to 15 2-AWG
16 to 20 1-AWG
Source: "Evinrude E-TEC Installation and Predelivery" guide, page 16.
We can make an inference about the maximum value of resistance in the battery cables just based on these recommendations. We can compute the resistance in the cables (for one conductor) based on the length and resistance-per-1000-feet for the wire gauge recommended.
For a maximum of 10-feet, 4-AWG can be used. The resistance of that conductor will be
(0.2485-Ohms/1000-feet) × 10-feet = 0.002485-Ohms
For a maximum of 15-feet, 2-AWG can be used The resistance of that conductor will be
(0.1563-Ohms/1000-feet) × 15-feet = 0.0023445-Ohms
For a maximum of 20-feet, 1-AWG can be used, The resistance of that conductor will be
(0.125-Ohms/1000-feet) × 20-feet = 0.0025-Ohms
Looking at the three values of resistance, we see that the highest value occurred at the longest length, 0.0025-Ohms at 20-feet. This indicates that the resistance maximum must be 0.0025-Ohms. Any cables used must have equal or less resistance.
If the battery cables are to be made longer than 20-feet, the wire gauge must be further increased. The method to calculate the wire gauge based on a maximum permitted resistance is also easy. We know the resistance at some shorter distance, so we just scale up the distance to 1,000-feet to find the resistance at 1,000-feet as used in most tables.
For example, if the wire length must be increased to 25-feet, what conductor size will produce a resistance of 0.0025-Ohms or less?
For the given values, 0.0025-Ohms and 25-feet, we need a conductor whose resistance per foot will be:
0.0025-Ohms/25-feet = (0.001-Ohm/1-feet) × (1000/1000)
= 0.1-Ohm/1000-feet
We enter the table and look for a wire gauge with a resistance-per-1000-feet of 0.1-Ohm or less. The suitable wire is 0-AWG with a resistance per 1,000-feet of 0.0983-Ohms. Now we can deduce that for the particular engine (an Evinrude E-TEC) to be properly wired with 20-feet-long cables, the wire size must be increased to 0-AWG.