Hi all,
http://www.lis.pitt.edu/~sochats/TELE2058/awg.html has a table that lists
AWG sizes 10 to 29 (that's a strange size, by the way, you'll have a hard
time finding ANY odd gauge wires), and the resistance per 1000 feet of the
stuff.
The reason they don't list 'current rating' is that it's really up to you
how much power loss you want to incur through the use of a particular gauge.
Ohm's Law says V = I * R, where V is the voltage drop, I is the current (in
amps) and R is the resistance in ohms, of any part of an electrical circuit.
So let's say we're going to wire some headlights. For convenience I'll
assume a 60W filament, and a 12V operating voltage. This implies a 5A
current draw for the (one) light. So if we used 16 gauge wire, and there was
10 feet of wire between the fuse block, to the switch, then to the
headlight, and then to the ground point (gotta consider everything involved
in making the light go), then we would lose 5 amps * (10 feet * 4.10 / 1000
ohm/foot) = 0.2 Volts. Not too bad, if your 16 gauge wire was in good shape
through its entire length. Even using 20 gauge wire would only cost you
about half a volt, and would probably save you about five pounds in copper
if you applied this thinking through the entire car...
Another thing to consider is the heating effect of the resistance of the
wiring. The power dissipated in an electrical thing (for DC, resistance-type
loads - AC stuff is a little more complicated) is P = V * I, where V is the
voltage DROP across the load, and I is the current through it. So our 16
gauge wire dissipates 0.2 volts * 5 amps = 1 watt. Since this occurs over 10
feet of length, the heating effect is quite negligible, but for higher
current things (heaters and wiper motors, for example) it bears
consideration, and to be really safe you should actually use the fuse rating
for the circuit in your loss calculations instead of the nominal current.
Since Lucas didn't think it necessary to fuse the headlights we can't use
that as an example, but let's say we had a short in that 16 gauge, unfused,
headlight circuit. The resistance is 0.041 ohms, and we can re-arrange the
two equations to show that P = V^2 / R. So shorting our wire gives us P = 12
* 12 / 0.041 = 3512 watts. This is enough power to cause several non-linear
effects in the wire, one of which is its melting and making a big mess
before it actually breaks the circuit. The worst part is that having this
happen can wreck a lot of other wiring in your car as well, even if it
doesn't actually cause a fire..
Regards,
Theo
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