My math is a little rusty but I'm thinking I have about 4.80 amps with this 65 watt light. Was wondering what size wire to use from the cockpit to the the tip of the RV9A wing? I thought 18 guage sheilded 2 conductor wire and a 7 amp breaker. I'm thinking 150% 0f the amps used is the breaker size.
There are several things to consider.
1) The current carrying capacity of the wire must be sufficient for the load current. See AC 43.13-1B table 11-9. On this criteria, 18 AWG is ok. But...
2) The resistance of the wire should not produce an excessive voltage drop. About 0.5V drop is generally considered reasonable in a 12V system. Resistance is a function of wire gauge and length. Again see AC 43.13-1B table 11-9. From instrument panel to wing tip is let's say about a 15-20 ft wire run, so 30-40 ft of wire round trip (assuming you're also running a ground wire of the same gauge). 40 ft of 18 AWG wire will have a total resistance of approx 0.25 ohm, so you'll get a voltage drop of 4.8 A * 0.25 ohm = 1.2 V. That's a bit high, and will mean your landing light would shine a bit dimmer. If you want to keep the voltage drop to about 0.5V, then use 14 AWG wire, which will have about half the resistance and therefore half the voltage drop of 18 AWG wire. (Judgment call here. I'm explaining the general concept so you can decide.)
3) The circuit breaker exists to protect the wire, and its current rating should be less than the current carrying capacity of the wire (that's the idea anyway... some fudge factor in reality). Or just see AC 43.13-1B table 11-3 for generally acceptable breaker values per wire gauge. The breaker's current rating also must obviously be high enough to support the load current with some margin, but it will be anyway if you've sized your wire correctly as described above.
4) There's no reason to use shielded wire for an incandescent lamp.
Is there a site I could go to figure this all out.
The AeroElectric Connection explains these concepts pretty well.
AC 43.13-1B chapter 11 is also a useful reference.