"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.

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 ?

--
Michael Melland, W9WIS
Winneconne, WI USA
http://webpages.charter.net/w9wis

Bravo and kudos to Mr. Robbins for his comments below.

To elaborate, the equation to calculate the loss in a specific length of
coax when its loss per 100' is known for a given frequency is:

Loss in decibels = (published loss per 100 feet) (decimal number of
hundreds of feet in the system)

When the total loss of the coax in decibels is known...

Coax system efficiency = 1/10^(loss, total/10)

These equations are based on 1:1 SWR within the coax, and its terminations.
While this is not likely in the real world, nevertheless it is the basis
upon which the FCC sets and licenses the operating powers of AM, FM & TV
broadcast stations in the US.

RF

Visit http://rfry.org for FM broadcast RF system papers.

____________________________

"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.

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.

All depends on the starting point data which may not be easy to obtain.


For EXACT classical calculation of loss along coaxial, solid polyethylene
insulated lines for given conductor dimensions, for any terminating
impedance, any SWR, from power frequencies to UHF, download program COAXPAIR
from website below. This program may be used to check accuracy of
simplified formalae available in handbooks.


For power and voltage ratings and temperature rise above ambient, download
program COAXRATE. Information is much better than is generally provided by
manufacturers.


Approximately, at HF,

Attenuation = 4.34*R/Zo decibels per metre.

Where R is the HF resistance per metre of the inner plus outer conductors
and Zo is line impedance.


R = Sqrt( F.MHz ) * ( 1/d + 1/D ) / 12 ohms per metre.


Where d = inner conductor diameter, mm

and D = outer conductor inside diameter, mm

--
.................................................. ..........
Regards from Reg, G4FGQ
For Free Radio Design Software go to
http://www.btinternet.com/~g4fgq.regp
.................................................. ..........

Now THERE'S an interesting rule of thumb I hadn't picked up on in the last 40
years. Thanks, Tom.

Jim


(Tom Bruhns)
shared these priceless pearls of wisdom:

-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.)
Jim Weir, VP Eng. RST Eng. WX6RST
A&P, CFI, and other good alphabet soup