OK, first time ever to write a HOW TO. I have no pictures but have installed 3 amps. I helped Archer and Bbob, I believe members here, to troubleshoot their amps and people have PM'd me for lots of help. I will help anybody BUT this is alot to type every time. I should have saved it before this.
So lets try this:
Things you need.
12 feet of power cable (10 gauge--I THINK)
round eyelets and butt connectors big enough for cables
A fuse holder with as big a wire as the power wire and a 25 amp fuse for in it
Approx. 4 or 5 feet of simple speaker wire with butt connectors
I'm going on the idea that everyone can access the battery and take off the fairing.
Take 12 feet of power cable and cut it in half. 6 feet each, one will be for the power + from the battery. The other will be for ground - from the battery. Used some black electrical tape on each end of the one you will use for ground so you can tell them apart later. Connect fuse holder to the power wire side. Connect eyelets to the very ends to attach to battery later. DON'T HOOK UP YET!!!!! Now run wires from battery up the left side to behind tank then toward front into fairing. Screw power and ground into the Rockford plug. (Tighten and loosen a few times each, every time the connection will get better.) It's the style of plug, you'll understand when you see it. You are kind of smashing the wire flat inside the plug. This is good.
“NEC has established that a 2% maximum voltage drop is acceptable.
Using stranded copper wire for a 12V system with 6 feet of wire for 42A would be calculated as below:
The formula is CM = (25 x 42A x 6')/0.24v = 26,250 circular mils.
CM = Circular Mils
25 is for 25°C…the Fusing Current (melting wire) is estimated based on 25°C, (77°F) ambient temperature.
42A is the Amp draw in a 12 volt system using a 6 foot stranded copper wire.
.24v is the NEC standard 2% voltage drop in a 12V system… 2% or .02/12 = .24v
Hence the formula:
CM = (25 x 42A x 6')/0.24v = 26,250 circular mils
This would require a 6 AWG wire according to NEC standards. Using the "voltage drop" method, if the user wanted less than 1% voltage drop on this wire (assuming a nominal 12 v potential) in this setup then a 6 AWG wire is required. If the user can tolerate a 1.5% drop then an 8 AWG wire would suffice. A 10 AWG wire would have a 2.2% voltage drop, just over the NEC standard.”
CM = (25 x 25A x 6')/0.24v = 15,625 circular mils
…. 8 AWG = 16,510 CM the voltage drop for 6 feet at 25A is .095 so .095/12*100 = .79% voltage drop.
10 AWG = 10,383 CM the voltage drop for 6 feet at 25A is .1525 so .1525/12*100 = 1.27% voltage drop.
NEC standard is a 2% voltage drop:
6’ of 8 AWG for 25 amps would be less than a 1% voltage drop, (.79%).
6’ of 10 AWG for 25 amps would be a 1.27% voltage drop still within the NEC standard, but I’d probably go with 8 AWG of stranded copper.
However, 10 AWG is also ok to use, good write up.