Just as Dave wrote...

And it's easy to understand. If you have 100 feet, that's just two
50-foot sections in series. Each will act just like an attenuator,
and they will be the same. To get, say, 1.8dB total, it must be a
cascade of two 0.9dB sections. Remember that dB expresses a ratio (of
output to input). Or a cascade of 18 5.555 foot sections, of 0.1dB
each, etc.

Also worth remembering: for well-made coax operating at HF, and
generally well up into VHF/UHF, the attenuation in dB of a given
length goes up as the square root of frequency, so at 450MHz, it will
be (very close to) ten times the dB it is at 4.5MHz. That's because
the loss is almost all in the resistance of the conductors, which
(usually) varies as the square root of freq., because of the skin
effect, in this freq range. (At high enough freqs, dielectric loss
kicks in, and at low enough, the skin depth is greater than the
conductor thickness.)

Cheers,
Tom

"David Robbins" wrote in message ...
"Michael Melland" wrote in message
...
If a cable, any cable ..... for argument sake... say Andrew LDF2-50... has
a
specified loss at.... say 1000 MHz.... of 3.52 dB in 100 ft... is there a
way to calculate what the loss would be in different lengths ? Say for 50
feet ?

The cable above is an example.... but is there a rule for calculating loss
for cables of any kind under 100 feet ? Would loss at 50 feet be exactly
1/2 of that at 1000 feet ?


for a 1:1 swr the loss is a linear function of length. so the loss at 50'
would be 1/2 that at 100' (not 1000' as you typed above!). if the swr is
greater than 1:1 its not so straight forward.

oh, just fair warning... questions like this often attract nit-picky trolls
who will turn this thread into a never ending discussion of why swr isn't,
and how lossy cables change the world... ignore them for your own sanity.