

In four replies, I've seen no hard numbers, so here goes:  For traces
on the outside (not inner layers), expect, in 1 oz copper, at 15 amps,
the following temperature rises versus width.  This is from Sams'
"Reference Data for Engineers," seventh edition, Pg 5-30, which claims
in turn to be from MIL-STD-275C...

    width   Temp rise
    inches  degrees C
    ------  ---------
     .125	100
     .15	 75
     .17	 60
     .20	 45
     .24	 30
     .33	 20

At 10 amps, the rise for the .125 width is only about 30 degrees.
Power goes as square of the current, plus the copper resistance goes
up as temperature goes up...  Certainly .20" (~5mm) traces should be 
ample for what you want to do.  And 2 ounce copper almost cuts the 
required width in half.  (I'd do 2 oz at about 0.08" width myself, 
I think...given that the _rated_ current is 10 amps and the 15 is
a transient or fault condition.)