Either you didn't read the remainder of what I wrote, or I failed to
explain it clearly.

I was speaking of antennas of a constant length in terms of wavelength
as frequency is changed, for example half wavelength dipoles.

If you cut the frequency in half, the skin depth increases by a factor
of the square root of two, so (assuming a conductor at least several
skin depths in radius) the resistivity decreases by the square root of
two. But to maintain a constant antenna length in terms of wavelength,
the wire length doubles. So the total wire resistance at the lower
frequency is greater by a factor of the square root of two. In other
words, if you make two half wavelength dipoles out of the same diameter
and kind of wire, and cut one for 1 MHz and the other for 2 MHz, the 1
MHz dipole will have about 1.4 times the resistance of the 2 MHz one.
That's why you're more likely to see the loss of steel wire in lower
frequency antennas.

Roy Lewallen, W7EL

Richard Harrison wrote:
Roy, W7EL wrote:
"---the loss with a given wire size gets greater as you go lower in
frequency,---.

Effective resistance to r.f, is approximately proportional to the square
root of the frequency due to "skin effect" as Roy mentioned in
describing how current penetrates the conductor less completelty due to
inductance deeper in the wire. So, loss is greater at higher frequency
due to reduced effective cross-section in the wire. Conversely, the loss
with a given wire size gets lower as you go down in frequency.

Best regards, Richard Harrison, KB5WZI