OK, I hope you people are satisfied! I had to dig out my engineering books for this one!
To compute the relative resistance of two foot strand of 18 AWG copper wire;
R= 10.37 x 2/1624 [specific resistance of copper at 20*C x 2 feet / circular mil]
R= 0.0128 Ohms
Relative conductance of two foot strand of 18 AWG copper wire
G=1624/(10.37 x 2)
G= 78.30 Mhos
Now to compute the maximum current carrying capability of
the copper;
I=E/R
I=5V/0.0128 Ohms
I=390 Amps (max amperage before copper core meltdown!)
Now the copper core will heat up and it's this heat build up that will meltdown the insulation long before the copper melts. Below is a cut/paste of a chart in my electrical engineering book (
but is also referenced here)
Allowable Carrying Capacities of Copper Wire
(Regulations of the National Board of Fire Underwriters)
Size,...Diameter, mils........Cross section,....Amperes..........R
B&B....(1 mil = 10^-3in.).....circular mil....Thermoplastic.....ohms/1000 ft
Gauge.....................................................insulation
10........101.9........................10380...............30..................1.0
12........80.81........................6530.................25..................1.6
14........64.08........................4107.................20..................2.5
16........50.82........................2583.................10..................4.0
18........40.30........................1624.................5....................6.4
Sorry if the table looks like crap.
Basically, 18AWG wire has a cross section of 1624 circular mils and 6.4 Ohms per 1000 feet.
And....
It looks like for 18 Gauge wire, 5 Amps is the max;
and for 16 Gauge wire, 10 Amps is the max.
The max power draw for a hard drive occurs at initial spinup. So if the max current draw rating of these drives are 2.8A then using 16AWG wire you should be safe connecting three drives (2.8A x 3 = 8.4A)
Maybe four or five drives if they're rated less.
Hope this answers everyones questions.
My brain hurts now. I hope you're happy!
