Trolling Motor Wire Size: 12V, 24V, 36V, 48V Trolling Motors
Trolling motors are powered by onboard batteries, and they draw relatively large currents, up to 5060 Amps. To keep the energy losses to a minimum and to prevent burns and even fires, trolling motor wires must be of sufficient thickness.
When choosing extension wires for your trolling motor, be sure to know the trolling motor voltage, maximum current, allowed/recommended voltage drop, wire/cable length, and similar.
Published: March 27, 2023.
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3% Voltage Drop and 140°F (60°C) Wire Surface Temperature vs. 5% Voltage Drop and 221°F (105°C) Wire Surface Temperature
When looking for electric trolling extension wire/cable, one of the most important parameters is energy efficiency, voltage drop percentage, the maximum allowed wire surface temperature, and wire storage  wires can be enclosed or suspended in the air.
Recommended allowed voltage drop is usually 3%, although some electric trolling motor manufacturers recommend 5%  personally, go for 3% since 5% can lead to overheated wires/cables.
Allowed maximum wire surface temperature is usually 60°C/140°F, @75°C/167°F, @90°C/194°F, or 105°C/221°F. Personally, always go for 60°C/140°F maximum allowed temperature since wires heated to 60°C/140°F one can hold for a few seconds for whatever reason, while wires heated to, for example, 105°C/221°F can cause skin burns.
Enclosed wires are not efficiently cooled as wires suspended in air, which is the reason why wires suspended in the air have higher Ampacity values. However, people often forget that thick wires exposed to the sun can overheat easily, even when they are not in use.
Personally, always use values calculated for enclosed values  many people buy wires using ratings for wires suspended in air but very soon place them in protected channels or similar enclosed areas.
So, when looking for electric trolling motor extension wires, it is recommended to use the following parameters:
 Voltage Drop: 3%
 Maximum Wire Surface Temperature: 60°C/140°F
 Enclosed or Suspended in the Air Wires: Enclosed Wires.
Also, when looking for trolling motor wire extensions, one has to check the Ampacities of the wires and go for the actual Ampacity of the wire, which is for safety reasons, 80% of the theoretical Ampacity.
Copper Relative Resistivity: Pure Copper vs. Copper Wire vs. Aluminum Wire
Some of the most commonly used materials for wires are copper and aluminum, while wires can be solid wires or stranded wires  solid wires are cheaper, but stranded wires are easier to work with, despite their somewhat larger diameter for the same Ampacity due to some empty space (air) between wire strands.
The electrical resistivity of these materials is:
 pure copper: 16.78 nΩm at 20°C,
 pure aluminum: 26.5 nΩm at 20°C.
However, due to the impurities, the "real life" electrical resistivity is usually higher  for example, copper's actual resistivity is usually around ~17.24 nΩm (instead of 16.78 nΩm).
Thus, many manufacturers offer "Oxygen Free Copper Wires," which have less impurities than standard copper wire, but also such wires have a higher price.
Personally, go for good solid or stranded wire, and if unsure, choose somewhat thicker wire.
Since copper feature density of 8.96 g/cm^{3} and aluminum feature density of 2.70 g/cm^{3}, wires:
 of the same dimensions: copper wires have better conductivity,
 of the same weight: aluminum wires have better conductivity.
So, choose according to your own needs and preferences.
12V Trolling Motor Wire Gauge
The following chart lists 12V electric trolling motor wire gauges according to the wire length and maximum current through the wire  the maximum allowed voltage drop is 3.0% (0.36V), and the maximum allowed wire temperature is 140°F (60°C) for the wires suspended in the air.
Also, the battery cable consists of two wires, thus 10ft wire equals the 5ft battery cable!
Solid Copper Wires Suspended In Air:
American Wire Gauge (#AWG) Copper Wires Suspended In Air 

Length  Maximum Current (Amps) @12V (Max. 0.36V Voltage Drop) 

30  40  50  60  70  
10ft; 3.05m  10  8  6  6  4 
15ft; 4.57m  8  6  6  4  4 
20ft; 6.1m  6  6  4  4  4 
Solid Enclosed Copper Wires:
American Wire Gauge (#AWG) Enclosed Copper Wires 

Length  Maximum Current (Amps) @12V (Max. 0.36V Voltage Drop) 

30  40  50  60  70  
10ft; 3.05m  8  6  4  2  2 
15ft; 4.57m  8  6  4  2  2 
20ft; 6.1m  6  6  4  2  2 
Note: for the values that were very close, the thicker wire was chosen. Also, the actual Ampacity (80% of theoretical Ampacity) of wires suspended in the air was taken into account. If the wires are going to be enclosed, calculate the actual Ampacity (80% of theoretical Ampacity) for enclosed wires and NOT for wires suspended in the air.
Little Bit Of Math
Since there is great confusion about electric trolling motor wires and cables, we will show You how one can calculate actual wire gauges for various voltages and currents.
Wire: 10ft (3.05m), Curent 30 Amps:
First, we want to calculate the maximum allowed wire resistance for a maximum current of 30 Amps and a maximum voltage drop of 0.36V:
R_{max} = U_{max} / I_{max} = 0.36V / 30A = 0.012Ω = 12mΩ
Now, we want to calculate the maximum relative resistance of the 10ft (3.05m) wire for the maximum current of 30 Amps and a maximum voltage drop of 0.36V:
R_{rel} = R_{max} / L_{wire} = 12mΩ / 3.05m = 3.9344 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
So, to keep the maximum voltage drop below 0.36V, we must choose a wire gauge that has relative resistance of 3.9344 mΩ/m or less, which is 10 gauge wire.
Note: For more info on this topic and wire gauge charts, please check our Wire Gauge Charts: AWG to mm2 Conversion Charts and Wire Sizes article.
Now, let's check the maximum voltage drop if we use 10 gauge 10ft wire for 30 Amps current  it must be less or equal to 360mV:
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 3.05m * 30A = 299.8455 mV = ~300 mV ≤ 360mV
Last check  actual Ampacity, which is, for safety reasons, 80% of theoretical Ampacity (again, 10 gauge solid copper wire, suspended in air, max. temperature 140°F (60°C)):
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A > 30A
For enclosed wires:
Ampacity_{80%} = Ampacity * 0.80 = 30A * 0.8 = 24A < 30A
So, 10 gauge copper enclosed wire features an actual Ampacity below the required 30A. Thus, we need a wire that supports 30A:
Ampacity = 30A / 0.8 = 37.5A → 8 gauge wire with Ampacity of 40 Amps for enclosed wire!
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 30A, we can take 10 gauge copper wire. For the enclosed copper wire, go for 8 gauge wire.
Wire: 15ft (4.57m), Curent 30 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 30A = 12 mΩ
R_{rel} = R_{max} / L_{wire} = 12mΩ / 4.57m = 2.6258 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 4.57m * 30A = 282.5631 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48A
For enclosed wires:
Ampacity_{80%} = Ampacity * 0.80 = 40 * 0.8 = 32A (close, but good enough)
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 30A, we can take 8 gauge copper wire. For the enclosed copper wire, go for 8 gauge wire, too.
Wire: 20ft (6.1m), Curent 30 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 30A = 12 mΩ
R_{rel} = R_{max} / L_{wire} = 12mΩ / 6.1m = 1.9672 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 6.1m * 30A = 237.168 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A
For enclosed wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 30A, we can take 6 gauge copper wire. For the enclosed copper wire, go for 6 gauge wire, too.
Wire: 10ft (3.05m), Curent 40 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 40A = 0.009Ω = 9mΩ
R_{rel} = R_{max} / L_{wire} = 9mΩ / 3.05m = 2.9508 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 3.05m * 40A = 251.442 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48A
For enclosed wires:
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A < 40A
The actual Ampacity for 6 gauge enclosed wire is:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A > 40A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 40A, we can take 8 gauge copper wire. For the enclosed copper wire, go for 6 gauge wire, too.
Wire: 15ft (4.57m), Curent 40 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 40A = 0.009Ω = 9mΩ
R_{rel} = R_{max} / L_{wire} = 9mΩ / 4.57m = 1.9693 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 4.57m * 40A = 236.9088 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 40A, we can take 6 gauge copper wire. For the enclosed copper wire, go for 6 gauge wire, too.
Wire: 20ft (6.1m), Curent 40 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 40A = 0.009Ω = 9mΩ
R_{rel} = R_{max} / L_{wire} = 9mΩ / 6.1m = 1.4754 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 6.1m * 40A = 316.224 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 40A, we can take 6 gauge copper wire. For the enclosed copper wire, go for 6 gauge wire, too.
Wire: 10ft (3.05m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 50A = 0.0072Ω = 7.2mΩ
R_{rel} = R_{max} / L_{wire} = 7.2mΩ / 3.05m = 2.3606 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 3.05m * 50A = 314.3025 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48A
So, although 8 gauge wire ensures a voltage drop of ~315mV, the actual Ampacity of such wire is 48A, which is less than the maximum current of 50A  we have to go for a thicker wire, which is a 6 gauge wire. If we check the actual Ampacity of 6 gauge wire, we get the following:
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A > 50A (max. current)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For the enclosed copper wire, go for 4 gauge wire.
Wire: 15ft (4.57m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 50A = 0.0072Ω = 7.2mΩ
R_{rel} = R_{max} / L_{wire} = 7.2mΩ / 4.57m = 1.5754 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 4.57m * 50A = 296.136 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For the enclosed copper wire, go for 4 gauge wire.
Wire: 20ft (6.1m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 50A = 0.0072Ω = 7.2mΩ
R_{rel} = R_{max} / L_{wire} = 7.2mΩ / 6.1m = 1.1803 mΩ/m → R_{rel} (4 gauge wire) = 0.8152 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 0.8152 mΩ/m * 6.1m * 50A = 248.636 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 105A * 0.8 = 84A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 50A, we can take 4 gauge copper wire. For the enclosed copper wire, go for 4 gauge wire, too.
Wire: 10ft (3.05m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 60A = 0.006Ω = 6mΩ
R_{rel} = R_{max} / L_{wire} = 6mΩ / 3.05m = 1.9672 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 3.05m * 60A = 237.168 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A (very close to max. current of 60 Amps, but it is OK)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire, although 4 gauge copper wire is recommended, just in case. For enclosed copper wire, go for 2 gauge wire.
Wire: 15ft (4.57m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 60A = 0.006Ω = 6mΩ
R_{rel} = R_{max} / L_{wire} = 6mΩ / 4.57m = 1.3129 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 4.57m * 60A = 355.3632 mV ≤ 360mV (very, very close)
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A (very close to max. current of 60 Amps, but it is OK)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire; however, 6 gauge copper wire would have energy losses very close to the maximum allowed when maximum current is run through the wire.
Thus, although 6 gauge copper wire can be used, we recommend using 4 gauge wire in this case. For the enclosed copper wire, go for 2 gauge wire.
Wire: 20ft (6.1m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 60A = 0.006Ω = 6mΩ
R_{rel} = R_{max} / L_{wire} = 6mΩ / 6.1m = 0.9836 mΩ/m → R_{rel} (4 gauge wire) = 0.8152 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 0.8152 mΩ/m * 6.1m * 60A = 298.3632 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 105A * 0.8 = 84A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 60A, we can take 4 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 10ft (3.05m), Curent 70 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 70A = 0.0051428Ω = 5.1428mΩ
R_{rel} = R_{max} / L_{wire} = 5.1428mΩ / 3.05m = 1.6861 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 3.05m * 70A = 276.696 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 64A
So, although 6 gauge wire ensures a voltage drop of ~276mV, the actual Ampacity of such wire is 64A, which is less than the maximum current of 70A  we have to go for a thicker wire, which is a 4 gauge wire. If we check the actual Ampacity of 4 gauge wire, we get the following:
Ampacity_{80%} = Ampacity * 0.80 = 105A * 0.8 = 84A > 70A (max. current)
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 70A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 70A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 70A, we can take 4 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 15ft (4.57m), Curent 70 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 70A = 0.0051428Ω = 5.1428mΩ
R_{rel} = R_{max} / L_{wire} = 5.1428mΩ / 4.57m = 1.1253 mΩ/m → R_{rel} (4 gauge wire) = 0.8152 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 0.8152 mΩ/m * 4.57m * 70A = 260.782 mV ≤ 360mV
Ampacity_{80%} = Ampacity * 0.80 = 105A * 0.8 = 84A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 70A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 70A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 70A, we can take 4 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 20ft (6.1m), Curent 70 Amps:
R_{max} = U_{max} / I_{max} = 0.36V / 70A = 0.0051428Ω = 5.1428mΩ
R_{rel} = R_{max} / L_{wire} = 5.1428mΩ / 6.1m = 0.8430 mΩ/m → R_{rel} (4 gauge wire) = 0.8152 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 0.8152 mΩ/m * 6.1m * 70A = 348.090 mV ≤ 360mV (very, very close)
Ampacity_{80%} = Ampacity * 0.80 = 105A * 0.8 = 84A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 70A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 70A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 12V trolling motor with a maximum current of 70A, we can take 4 gauge copper wire. However, if You generally push your 70 Amps trolling motor at 100% throttle most of the time, consider using 2 gauge wire, or go for a 24V, 36V, or 48V electric trolling motor. For enclosed copper wire, go for 2 gauge wire.
24V Trolling Motor Wire Gauge
The following chart lists 24V electric trolling motor wire gauges according to the wire length and maximum current through the wire  the maximum allowed voltage drop is 3.0% (0.72V), and the maximum allowed wire temperature is 140°F (60°C) for the wires suspended in the air.
Also, the battery cable consists of two wires, thus 10ft wire equals the 5ft battery cable!
Since 24V trolling motors allow larger voltage drop in the wires for the same efficiency (usually 3%) if You are unsure about the gauge or You already have "old" cables, feel free to go for 12V cables of the same length and current, assuming the insulation is sufficiently good (it is sufficient in at least 99.999% cases).
Solid Copper Wires Suspended In Air:
American Wire Gauge (#AWG) Copper Wires Suspended In Air 

Length  Maximum Current (Amps) @24V (Max. 0.72V Voltage Drop) 

50  60  
10ft; 3.05m  6  6 
15ft; 4.57m  6  6 
20ft; 6.1m  6  6 
Solid Enclosed Copper Wires:
American Wire Gauge (#AWG) Enclosed Copper Wires 

Length  Maximum Current (Amps) @24V (Max. 0.72V Voltage Drop) 

50  60  
10ft; 3.05m  4  2 
15ft; 4.57m  4  2 
20ft; 6.1m  4  2 
Note: for the values that were very close, the thicker wire was chosen. Also, the actual Ampacity (80% of theoretical Ampacity) of wires suspended in the air was taken into account. If the wires are going to be enclosed, calculate the actual Ampacity (80% of theoretical Ampacity) for enclosed wires and NOT for wires suspended in the air.
Little Bit Of Math:
Wire: 10ft (3.05m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 0.72V / 50A = 0.0144Ω = 14.4mΩ
R_{rel} = R_{max} / L_{wire} = 14.4mΩ / 3.05m = 4.7213 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 3.05m * 50A = 499.74 mV ≤ 720mV
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A
Although 10 gauge wire ensures a voltage drop below the required 720mV, it would be running very hot.
So, if the maximum current through the wire is 50 Amps, using the 80% Rule, the theoretical Ampacity of the wire should be at least:
Ampacity = 50 / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 24V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 15ft (4.57m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 0.72V / 50A = 0.0144Ω = 14.4mΩ
R_{rel} = R_{max} / L_{wire} = 14.4mΩ / 4.57m = 3.1509 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 4.57m * 50A = 470.9385 mV ≤ 720mV
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48A
Again, although 8 gauge wire ensures a voltage drop below the required 720mV, its actual Ampacity (80% of theoretical Ampacity for copper wires suspended in air) is just below the required maximum current of 50A.
Thus, we should take the next larger wire and check the Ampacity:
Ampacity = 50 / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 24V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 20ft (6.1m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 0.72V / 50A = 0.0144Ω = 14.4mΩ
R_{rel} = R_{max} / L_{wire} = 14.4mΩ / 6.1m = 2.3606 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 6.1m * 50A = 628.605 mV ≤ 720mV
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48V
Again, although 8 gauge wire ensures a voltage drop below the required 720mV, its actual Ampacity (80% of theoretical Ampacity for copper wires suspended in air) is just below the required maximum current of 50A.
Thus, we should take the next larger wire and check the Ampacity:
Ampacity = 50 / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 24V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 10ft (3.05m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 0.72V / 60A = 0.012Ω = 12mΩ
R_{rel} = R_{max} / L_{wire} = 12mΩ / 3.05m = 3.9344 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 3.05m * 60A = 599.691 mV ≤ 720mV
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A
Although 10 gauge wire ensures a voltage drop below the required 720mV, it would be running very hot.
So, if the maximum current through the wire is 60 Amps, using the 80% Rule, the theoretical Ampacity of the wire should be at least:
Ampacity = 60 / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 24V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 15ft (4.57m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 0.72V / 60A = 0.012Ω = 12mΩ
R_{rel} = R_{max} / L_{wire} = 12mΩ / 4.57m = 2.6258 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 4.57m * 60A = 565.126 mV ≤ 720mV
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48A
Although 8 gauge wire ensures a voltage drop below the required 720mV, it would be running too hot when the trolling motor is pushed to its max.
So, if the maximum current through the wire is 60 Amps, using the 80% Rule, the theoretical Ampacity of the wire should be at least:
Ampacity = 60 / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 24V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 20ft (6.1m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 0.72V / 60A = 0.012Ω = 12mΩ
R_{rel} = R_{max} / L_{wire} = 12mΩ / 6.1m = 1.9672 mΩ/m → R_{rel} (6 gauge wire) = 1.296 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 1.296 mΩ/m * 6.1m * 60A = 474.336 mV ≤ 720mV
Ampacity_{80%} = Ampacity * 0.80 = 80A * 0.8 = 72A
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 24V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
36V Trolling Motor Wire Gauge
The following chart lists 36V electric trolling motor wire gauges according to the wire length and maximum current through the wire  the maximum allowed voltage drop is 3.0% (1.08V), and the maximum allowed wire temperature is 140°F (60°C) for the wires suspended in the air.
Also, the battery cable consists of two wires, thus 10ft wire equals the 5ft battery cable!
Similarly to 24V wires, if You are unsure about gauges for 36V trolling motors or You already have "old" cables, feel free to use wires for the same current intended for 12V and 24V trolling motors.
Solid Copper Wires Suspended In Air:
American Wire Gauge (#AWG) Copper Wires Suspended In Air 

Length  Maximum Current (Amps) @36V (Max. 1.08V Voltage Drop) 

50  60  
10ft; 3.05m  6  6 
15ft; 4.57m  6  6 
20ft; 6.1m  6  6 
Solid Enclosed Copper Wires:
American Wire Gauge (#AWG) Enclosed Copper Wires 

Length  Maximum Current (Amps) @36V (Max. 1.08V Voltage Drop) 

50  60  
10ft; 3.05m  4  2 
15ft; 4.57m  4  2 
20ft; 6.1m  4  2 
Note: for the values that were very close, the thicker wire was chosen. Also, the actual Ampacity (80% of theoretical Ampacity) of wires suspended in the air was taken into account. If the wires are going to be enclosed, calculate the actual Ampacity (80% of theoretical Ampacity) for enclosed wires and NOT for wires suspended in the air.
Little Bit Of Math:
Wire: 10ft (3.05m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 1.08V / 50A = 0.0216Ω = 21.6mΩ
R_{rel} = R_{max} / L_{wire} = 21.6mΩ / 3.05m = 7.0819 mΩ/m → R_{rel} (12 gauge wire) = 5.211 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 5.211 mΩ/m * 3.05m * 50A = 794.67 mV ≤ 1.08V
Ampacity_{80%} = Ampacity * 0.80 = 30A * 0.8 = 24A
Although 12 gauge wire ensures a voltage drop below the required 1.08V, it would be running very hot.
So, if the maximum current through the wire suspended in air is 50 Amps, using the 80% Rule, the theoretical Ampacity of the wire should be at least:
Ampacity = 50 / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 36V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 15ft (4.57m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 1.08V / 50A = 0.0216Ω = 21.6mΩ
R_{rel} = R_{max} / L_{wire} = 21.6mΩ / 4.57m = 4.7264 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 4.57m * 50A = 748.79 mV ≤ 1.08V
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A
Ampacity = 50A / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 36V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 20ft (6.1m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 1.08V / 50A = 0.0216Ω = 21.6mΩ
R_{rel} = R_{max} / L_{wire} = 21.6mΩ / 6.1m = 3.5409 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 6.1m * 50A = 999.48 mV ≤ 1.08V
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A
Ampacity = 50A / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 36V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 10ft (3.05m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 1.08V / 60A = 0.018Ω = 18mΩ
R_{rel} = R_{max} / L_{wire} = 18mΩ / 3.05m = 5.9016 mΩ/m → R_{rel} (12 gauge wire) = 5.211 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 5.211 mΩ/m * 3.05m * 60A = 953.613 mV ≤ 1.08V
Ampacity_{80%} = Ampacity * 0.80 = 30A * 0.8 = 24A
Ampacity = 60A / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps (close, but good enough)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 36V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 15ft (4.57m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 1.08V / 60A = 0.018Ω = 18mΩ
R_{rel} = R_{max} / L_{wire} = 18mΩ / 4.57m = 3.9387 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 4.57m * 60A = 898.55 mV ≤ 1.08V
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A
Ampacity = 60A / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps (close, but good enough)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 36V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 20ft (6.1m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 1.08V / 60A = 0.018Ω = 18mΩ
R_{rel} = R_{max} / L_{wire} = 18mΩ / 6.1m = 2.9508 mΩ/m → R_{rel} (8 gauge wire) = 2.061 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 2.061 mΩ/m * 6.1m * 60A = 754.32 mV ≤ 1.08V
Ampacity_{80%} = Ampacity * 0.80 = 60A * 0.8 = 48A
Ampacity = 60A / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps (close, but good enough)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 36V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
48V Trolling Motor Wire Gauge
The following chart lists 48V electric trolling motor wire gauges according to the wire length and maximum current through the wire  the maximum allowed voltage drop is 3.0% (1.44V), and the maximum allowed wire temperature is 140°F (60°C) for the wires suspended in the air.
Also, the battery cable consists of two wires, thus 10ft wire equals the 5ft battery cable!
Similarly to 24V and 36V wires, if You are unsure about gauges for 48V trolling motors or You already have "old" cables, feel free to use wires for the same current intended for 12V, 24V, and 36V trolling motors.
Solid Copper Wires Suspended In Air:
American Wire Gauge (#AWG) Copper Wires Suspended In Air 

Length  Maximum Current (Amps) @48V (Max. 1.44V Voltage Drop) 

50  60  
10ft; 3.05m  6  6 
15ft; 4.57m  6  6 
20ft; 6.1m  6  6 
Solid Enclosed Copper Wires:
American Wire Gauge (#AWG) Enclosed Copper Wires 

Length  Maximum Current (Amps) @48V (Max. 1.44V Voltage Drop) 

50  60  
10ft; 3.05m  4  2 
15ft; 4.57m  4  2 
20ft; 6.1m  4  2 
Note: for the values that were very close, the thicker wire was chosen. Also, the actual Ampacity (80% of theoretical Ampacity) of wires suspended in the air was taken into account. If the wires are going to be enclosed, calculate the actual Ampacity (80% of theoretical Ampacity) for enclosed wires and NOT for wires suspended in the air.
Little Bit Of Math:
Wire: 10ft (3.05m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 1.44V / 50A = 0.0288Ω = 28.8mΩ
R_{rel} = R_{max} / L_{wire} = 28.8mΩ / 3.05m = 9.442 mΩ/m → R_{rel} (14 gauge wire) = 8.286 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 8.286 mΩ/m * 3.05m * 50A = 1263.615 mV ≤ 1.44V
Ampacity_{80%} = Ampacity * 0.80 = 25A * 0.8 = 20A
Although 14 gauge wire ensures a voltage drop below the required 1.44V, it would be running very hot.
So, if the maximum current through the wire suspended in air is 50 Amps, using the 80% Rule, the theoretical Ampacity of the wire should be at least:
Ampacity = 50A / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 48V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 15ft (4.57m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 1.44V / 50A = 0.0288Ω = 28.8mΩ
R_{rel} = R_{max} / L_{wire} = 28.8mΩ / 4.57m = 6.3019 mΩ/m → R_{rel} (12 gauge wire) = 5.211 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 5.211 mΩ/m * 4.57m * 50A = 1190.713 mV ≤ 1.44V
Ampacity_{80%} = Ampacity * 0.80 = 30A * 0.8 = 24A < 50A
Ampacity = 50A / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 15ft wire (7.5ft cable) suspended in the air with max. temperature of 140°F (60°C), 48V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 20ft (6.1m), Curent 50 Amps:
R_{max} = U_{max} / I_{max} = 1.44V / 50A = 0.0288Ω = 28.8mΩ
R_{rel} = R_{max} / L_{wire} = 28.8mΩ / 6.1m = 4.7213 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 6.1m * 50A = 999.485 mV ≤ 1.44V
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A < 50A
Ampacity = 50A / 0.8 = 62.5A → 6 gauge wire Ampacity = 80 Amps
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 50A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A > 50A
Long Story Short: for 20ft wire (10ft cable) suspended in the air with max. temperature of 140°F (60°C), 48V trolling motor with a maximum current of 50A, we can take 6 gauge copper wire. For enclosed copper wire, go for 4 gauge wire.
Wire: 10ft (3.05m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 1.44V / 60A = 0.024Ω = 24mΩ
R_{rel} = R_{max} / L_{wire} = 24mΩ / 3.05m = 7.8688 mΩ/m → R_{rel} (12 gauge wire) = 5.211 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 5.211 mΩ/m * 3.05m * 60A = 953.613 mV ≤ 1.44V
Ampacity_{80%} = Ampacity * 0.80 = 25A * 0.8 = 20A
Ampacity = 60A / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps (close, but good enough)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 48V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 15ft (4.57m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 1.44V / 60A = 0.024Ω = 24mΩ
R_{rel} = R_{max} / L_{wire} = 24mΩ / 4.57m = 5.2516 mΩ/m → R_{rel} (12 gauge wire) = 5.211 mΩ/m (very close, but...)
U_{drop} = R_{rel} * L * I_{max} = 5.211 mΩ/m * 4.57m * 60A = 1428.85 mV ≤ 1.44V (very close, but...)
Ampacity_{80%} = Ampacity * 0.80 = 30A * 0.8 = 24A
Ampacity = 60A / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps (close, but good enough)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 48V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
Wire: 20ft (6.1m), Curent 60 Amps:
R_{max} = U_{max} / I_{max} = 1.44V / 60A = 0.024Ω = 24mΩ
R_{rel} = R_{max} / L_{wire} = 24mΩ / 6.1m = 3.9344 mΩ/m → R_{rel} (10 gauge wire) = 3.277 mΩ/m
U_{drop} = R_{rel} * L * I_{max} = 3.277 mΩ/m * 6.1m * 60A = 1199.382 mV ≤ 1.44V
Ampacity_{80%} = Ampacity * 0.80 = 40A * 0.8 = 32A
Ampacity = 60A / 0.8 = 75A → 6 gauge wire Ampacity = 80 Amps (close, but good enough)
For enclosed 6 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 55A * 0.8 = 44A < 60A
For enclosed 4 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 70 * 0.8 = 56A < 60A
For enclosed 2 gauge wires:
Ampacity_{80%} = Ampacity * 0.80 = 95 * 0.8 = 76A > 60A
Long Story Short: for 10ft wire (5ft cable) suspended in the air with max. temperature of 140°F (60°C), 48V trolling motor with a maximum current of 60A, we can take 6 gauge copper wire. For enclosed copper wire, go for 2 gauge wire.
How To Wire a Trolling Motor
Electric trolling motors are powered by DC (Direct Current) electricity, usually 12V, 24V, 36V, and 48V.
With a standard battery voltage of 12V, in order to achieve higher voltages, one has to connect several batteries in series:
 24V: two 12V batteries in series,
 36V: three 12V batteries in series,
 48V: four 12V batteries in series.
Thanks to the advancements in technology, many brands offer Lithium Iron Phosphate (LiFePO_{4}) deep cycle batteries featuring not only 12V nominal voltages but also 24V, 36V, or even 48V, simplifying electric trolling motor wiring.
Also, in order to achieve higher capacities, one has to connect batteries in parallel.
Note: when connecting batteries in series and parallel, especially lithium batteries, always connect them as recommended by the manufacturer.
Electric trolling motors can be wired directly to the battery, which is not recommended  if You have a single deepcycle battery, place it in the battery box, also known as the "Trolling Motor Power Center," that features enough room to store one Group 27 or Group 31 battery, but also features an electric breaker of suitable amperage and an electric master switch.
Larger batteries or battery banks should be stored on the boat in permanent or semipermanent positions and have wiring that includes required switches and electric breakers  safety first!
For example, if You have a 48V electric trolling motor and You want to connect eight (8) 12V batteries, the batteries must be connected in 4S2P configuration:
So, if You have eight (8) 12V 100Ah batteries connected in 4S2P configuration, You will get a 48V 200Ah battery pack.
Note: 48V wire must be connected to the battery "H" and NOT to the battery "D" in order to avoid imbalances in the battery pack.
For more information about this topic, feel free to check our How To Connect Batteries in Series and Parallel: Series vs. Parallel article.
Further Reading
If You are interested in this topic, feel free to check these articles:
 Wire Gauge Charts: AWG to mm2 Conversion Charts and Wire Sizes
 How to Choose the Best Trolling Motor Battery
 Best Kayak Trolling Motor Battery
 100Ah Lithium Battery  12V, 24V, 36V, and 48V Batteries
 200Ah Lithium Battery  12V, 24V, 36V and 48V Models