Electronic – Working out thickness of a copper wire

Convection does not scale linearly with surface area, it's a fluid dynamics problem rife with Nusselt numbers and Rayleigh coefficients. Since it's not my field, and since I'm more likely to use CFD than first principles in such a situation anyway, I'll point to a calculator that might allow some feel for the problem here (assuming it's actually implemented correctly, of course).

If I stick in reasonable numbers like 100°C for the wire 1, 2, 10mm for the wire diameter then I get total heat loss scaling with roughly the diameter (and thus the surface area) to the power of 2/3.

Preece was looking at fusing current, which would be at a much higher temperature than typical operation of an electrical wire.

The different resistances of those incandescent bulbs are made with different length and different diameters of tungsten wires. A very long and thin wire is wound to a filament, sometimes even a double filament. But the values calculated from power and voltage are the resistances in hot condition. The resistance at cold condition is much lower. I have here a bulb with 120 W at 230 V and a resistance at room temperature of 29 Ohm. The resistance in hot condition at 2950 K is 440 Ohm. If we assume a diameter of 0.1 mm for that tungsten filament wire, we can calculate a length of 4.3 m for a resistance of 29 Ohm. I used the specific resistance of tungsten 0.0528 Ω ⋅ mm²/m for that calculation.