Posts: 2752
Location: Mauston, Wisconsin | Hi guys- The Alanwire website is talking about AC voltage drop for your house or office- very different than a DC circuit with a 12V battery. So just for grins- I checked the DC resistance of #10AWG coated copper wire. It's 1,29 Ohms / 1000ft. That doesn't seem like to much. Howver, I also took a look at a typical boat like mine- the batteries are in the rear and the TM is on the bow. So the total distance for my 17.5 ' boat is maybe 20' once you consider all the turns & bends to get to the plug for the TM. Lets say the TM draws 25A.
Now the total circuit length is going to be ~ 40ft (pos 20ft added to neg 20')
Total circuit resistance is the total distance divided by 1000, then times the ohms/1000ft:
R = (.04/1000) * 1.29 = .0516 Ohms.
The voltage drop is current times resistance = 25A * .0516 = 1.29V
The power loss in the circuit is the voltage drop times the current = 1.29V * 25A = 32.25 watts
We can also calculate it via current squared times the resistance. Either way the number is the same. What this means that while I'm running the trolling motor at 25A my cables are giving off ~32 watts of heat. That's wasted power that's not getting to my trolling motor to do useful work, i.e., get me on the spot while fighting the waves.
Keep in mind that a 12V battey is considered fully discharged at ~10.5 volts. A fully charge 12V battery is about 12.7V - this means we normally have a working voltage window of about 12.7 - 10.5 = 2.2V . Now if I subtract the voltage drop from the 12.7V, I get ~ 11.41V. My trolling motor is going to be seeing a working voltage of 11.41V even when my battery is at 100% charge. The voltage drop severely reduces the available voltage/power at the trolling motor.
Now if I upgrade the cables two sizes to #6AWG . It has a DC resistance of .510 Ohms/1000ft. So once again (40/1000) * .510 Ohms= .0204 Ohms
The voltage drop is going to be 25A * .0204 Ohms = .51V We've reduced the voltage drop by .78V or .78/1.29 = 60% The cable losses are now .51V * 25A = 12.75W or 32.25 -12.75/32.25 = 60%
The lesson here is: use the biggest cables that you can if your batteries are in the back of your boat. Usually that's based on the largest gauge wire you can fit into the trolling motor connectors.
You can also use even larger cables by attaching a short pig tail to the cable ends, f.ex if the connector will only accept a #8 AWG - attach a 6" piece of #8 to the #6AWG and insulate it well - you will get the voltage drop of the # 6AWG - some guys just cheat and cut a few strands of the #6AWG until it fits into the #8AWG termination. This works also.
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