Hi,

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

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.

Don't know that particular expression but plug in very low
values for 'f' and 'D' and you come up with something like
Loss(dB) = K(dB) -(~100dB) since the log of numbers less than 1 is
negative.

The formula then is simply of the form -

y = 20log(f) + 20log(D) + C

So when 'f' and 'D' are both equal to 1, the path loss = 96dB
and, when much less than that, it approaches zero.


Cheers - Joe