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.

Make a note of it in your notebooks.

And, hopefully, that should be the end of the matter. But, knowing
you lot, it probably won't be. ;o)
----
Reg, G4FGQ

================================

To you all.

As predicted, I appear to have stirred up a hornet's nest.

First of all, give credit to where credit is due. The simple equation
is not due to me but to Oliver Heaviside, 1850 - 1925. May God rest
his soul. And mine!

It applies from DC to VHF where the predominent loss is due to
conductor resistance including skin effect. At higher frequencies, say
above 0.5 GHz, loss in the dielectric material begins to play an
important part.

The complete equation is -

Attenuation = R/2/Ro + G*Ro/2 Nepers

where G is the conductance of the dielectric, which is small for
materials such as polyethylene and Teflon. And 1 Neper = 20/Ln(10) =
8.686 dB.

The Neper is the fundamental unit of transmission loss per unit length
of line, familiar to transmission line engineers. It is named after
Napier, a canny Scotsman who had something to do with the invention of
Logarithms around the 18th Century.

Attenuation is simply the basic matched loss of a particular line,
unaffected by SWR and all the other encumbrances which amateurs such
as W5DXP ;o) worry about. KISS.

Incidentally, the additional-loss versus SWR curves, published in the
ARRL books and copied by the RSGB, for many years, are based on an
incorrect mathematical analysis. But they are near enough for
practical purposes.

Not that SWR matters very much. SWR meters don't measure SWR on any
line anyway. You are all being fooled. ;o) ;o) ;o)
----
Reg, G4FGQ