"Bob Bob" wrote in message
...
Hi all.

No doubt I could get this reading lots and lots of text books. It might
however be interesting to air a discussion on it..

Have been reading the free space pathloss formula from the ARRL antenna
handbook;

Loss in dB = K + 20logf + 20logD

Where f is the freq in GHz and D is the distance in miles

K is a constant of around 96.6dB

I have known about a "fixed K" loss in an antenna system for ages. It even
made mention I think in this NG recently when talking about a passive
repeater system. From my own exposure to path modelling (EDX/Pathloss etc)
I noted a very high dB loss per distance rate in the first (say) 100
wavelengths when looking at graphs of same. (I wasnt doing the actual job,
just providing data to the engineer to compare measured with predicted.
Fascinating stuff!)

What I wonder is where roughly does the 96dB odd "come from". More
interestingly can it be reduced by any appreciable amount. Is it maybe a
antenna to "air" coupling loss, maybe even that a RX antenna cant possibly
extract all of the energy from the wave as it goes past. I would
appreciate any input on this.

My initial forays (as a young ham) into LOS paths went through the
isotropic/point source radiators and looking at the surface area of the
covered "sphere" containing all of the radiated power idea. Then the RX
antenna "aperture" area was used to calculate the actual received power.
Needless to say it never met the actual measured values!

Cheers Bob VK2YQA (in W5)

Take the ratio of the destination antenna aperture (0.13*lambda^2 for a half
wave dipole), to the surface area of a sphere, of a radius equal to the
distance between the source and destination antennas (i.e. Pr/Pt). The
assumption is that the source is isotropic. Massaging the above expression
will produce a constant and the sum of two logs.

Regards,

Frank