"Reg Edwards" wrote
The number one reason for attenuation being higher is because the
conductor diameter is smaller and, as a consequence, its resistance is
higher. The exact simple mathematical relationship is -
Line attenuation = 8.69*R/2/Ro dB.
Where R is the resistance of the wire and Ro is the real component of
line impedance, all in ohms
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Is skin effect accounted for in your equation? For example at 10MHz, skin
effect confines most of the current to the outer ~21 µm of the conductor.

RF