how big my track should be to handle this current
Here's a VERY simplistic analysis:

An inch of a 20 mil wide trace of half-ounce copper will have a resistance of about 0.05 Ohm, and a surface area of 0.02 inch. If you ran an amp through it, you'd have 0.05 Volts, so 0.05 Watts, 50 mW. At a constant current, power would be proportional to length, and so would surface area.

For comparison, an old one Watt axial-leaded carbon composition resistor is about 0.6 inch long and 0.23 diameter. Call that 0.6 x 0.75 in surface area, or less than half a square inch. The circuit board will conduct heat away from your trace more effectively than air would from the resistor.

This really doesn't seem easier than an on-line calculator.
This one seems to work fairly simply: http://www.4pcb.com/trace-width-calculator.html
I put in your requirement of 0.5 Amp, assumed a standard 0.5 ounce copper, and allowed 20C rise (more than the default of 10). Assuming also you start at 25 C room temperature, you get a 15 mil wide trace, whether your trace length is one inch or ten.

You can order boards with heavier copper (1 and 2 ounce per sq ft are not unusual), and the resistance will go down appropriately.