10 AWG wire good for 30 Amps of continous load, thats more than enough

 

Up to about 7 feet wire length 10 gauge will work for up to 35 amps current.

 

According to NEC(National Electric Code) Section 240-3c (Small conductors "#10AWG Copper W/THHN insulation") "Overcurrent protection shall not exceed 30 amperes for ambient temperature"

You are right that you can get 35 AMPS from #10AWG but only if it is the lone conductor in a approved conduit(Ridgid Metal Electrical Conduit, Or EMT Electrical Metallic Tubing) which nobody ever runs in a car. If your wondering why I know this its because I'm and electrical enginner and deal with these questions all day long.

If you have anyother question please feel free to ask.

 

So was this post just to let everyone know the amperage rating of 10awg wire??

 

I honestly thought it was a question. So rather than doing any math I took the answer directly from a table. I am not sure if the calculations required rely soley on amperage, voltage, or if it is the component of power. For 0.5 voltage drop or less ten gauge is rated up to 7 feet with 35 amps of current. Whatever this was, a statement of fact, a question, whatever. I would expect an engineer to include all the neccessary units.

 

As you can see that was my first post. It was actually supposed to be a reply to someone else’s post. But you kindly responded to it and caught my attention and so I had to reply. Well here are the calculations you request to back up my statement. If you need anymore please feel free to ask, we're all here to help one another.

The reason why all those calculations weren't included is because unless you plan on running the power feed around the car 10 times voltage drop is negligible. You said you want units so here ya go.

All calculations show below were done using Direct-Current Resistance Conductor Properties.

Assuming your using 10AWG insulated copper stranded wire rated for 75 degrees C(standard) the resistance per 1000ft is .321ohms.
Ok so therefore .321ohms/1000ft=.000321ohms/ft
Lets say we use 20 ft of wire (well be generous)
.000321ohms/ft X 20ft= .00642ohms for 20 feet of wire
Now since V=IR (Voltage=Current Multiplied By Resistance)
V=30A X .00642ohms
Oh wow that gives us a 0.1926 VDC drop
And that is considering that you are pulling a full 30 amps thru that wire.

Just for kicks now well do it for 7Ft of 10AWG wire at 35Amps
.000321ohms/ft X 7 ft= .002247ohms
V=IR
V=35A X .002247
Your gonna like this one 0.078645 VDC drop for 7 feet at a full draw of 35 amps

So now we can see why I did not include all the calculations. But you are right that I should have to make my point valid. But if you are really concerned with voltage drop then I suggest using 250 MCM wire, that will give you an almost 0VDC voltage drop yet cost you a pretty penny.