In message , Michael A.
Terrell writes

Ian Jackson wrote:

OK.

At 450MHz and temperature T1 centigrade, your 5000 foot of coax will
have an attenuation of A1 dB.

[The value of A1 obviously depends on the characteristics of that
particular type of coax, and on its length. For a different frequency /
temperature / length / type of coax, the value of A1 will be different.]

At temperature T2, the change of attenuation will be approximately
(T2-T1) x 0.02 x A1 dB.

At 11GHz and temperature T1, your 5 foot of coax will have an
attenuation of A2 dB.

[Again, for a different frequency / temperature / length / type of coax,
different value of A1.]

At temperature T2, the change of attenuation will be approximately
(T2-T1) x 0.02 x A2 dB.

Essentially, you don't have to know anything about the frequency /
length / type of coax. Although these do determine the amount of initial
attenuation, all you need to know is the actual value of the initial
attenuation, and the change of temperature. The higher the initial
attenuation (for whatever reason), the greater will be the change of
attenuation.


You still need to know the length. I used to have to do the
calculations for CATV system design on a four banger calculator, before
PCs were common. Levels at the highest & lowest temperature was one
limiting factor of the amplifier spacings. Some cables that were almost
identical at first glance in the catalogs could be 100 feet difference
in maximum length. At that time, pressurized, fused disk was the best
.750 cable available, but it was so fragile we wouldn't use it. Then you
had to wade through all the various foam types, the DC resistance per
100 feet, maximum certified frequency and a dozen other numbers.

A 5000 foot cable will have twice the change of a 2500 foot cable
forthe same temperature change.


You still haven't stated if your .02 dB
change/degree is per fook, per 100 feet, or per mile.


I have! It's 'per dB'.

I'm sure you must have used a similar rule-of-thumb formula yourself
(but maybe in Fahrenheit). Maybe you use one which requires you to know
the length and attenuation per unit length (which will give you the
actual attenuation). It all amounts to the same thing in the end.
Somewhere, you need to input numbers for (or which give you) the cable
attenuation.

Please read carefully:
The formula states that the cable attenuation changes by appx .02dB per
dB per degree C. Note the 'PER dB'. That 'dB' the actual cable
attenuation. A 5000 foot cable will have twice the attenuation of a 2500
foot cable so, as you rightly say, a 5000 foot cable will have twice the
change of a 2500 foot cable for the same temperature change.

To use that formula, the important thing is to know the ATTENUATION of
the cable. If you don't know this, you have to find out. You may know
this from actual measurement, or by calculation from the spec figures
for the loss per unit length at the frequency of interest, and the
actual length.

All I can say is "Thank heavens for optical fibre!"
--
Ian