Look at it like this. The power delivered to the load doesn't matter. Only the power dissipated in the wire.
Here's an extreme example.
I have wire rated for 1 ampere with insulation rated for 1000VDC.
The first circuit uses 1000V into a load of 1000 ohms, for a current of 1 ampere. It's delivering 1000 watts to the load. (More or less.)
simulate this circuit – Schematic created using CircuitLab
The second circuit uses only 10 volts into a 1 ohm load. That's 10 amperes of current for a load power of 100 watts.
The load power is lower, but that poor 1 ampere wire is going to melt its insulation off if not straight up catch on fire - despite delivering less power to the load.
The load power doesn't matter.
It is the power dissipated in the wire that matters, and that depends only on the current and the wire properties (resistance of the wire itself.)
Calculating the actual power dissipated in the wire requires knowing the resistance per foot for the wire and the length of the wire. You'd need a table for various lengths as well as for the various wire sizes - and it wouldn't tell you anything more useful than the table with only amperes and wire size.
Yes, I ignored the losses in the wires when calculating the load power while depending on the losses in the wire for the heating. It's a simplification.