On Wed, 7 Feb 2007 00:08:19 -0500, "Stefan Wolfe"
wrote:


"Richard Clark" wrote in message
.. .
On Tue, 6 Feb 2007 22:35:04 -0500, "Stefan Wolfe"
wrote:

2. The effect of the 80m hill 10Km away is negligible. The arc tan is only
.008 degrees, thus the transmitter hardly "sees" it.

Actually, Stefan, the transmitter cannot see through it at all.

Exactly . And taking into account the 15 degree bend of the radio horizon
(even at 10 GHz) vs the arctan of .008 degrees for the the hill, reduced
somewhat by the 20 foot transmitter tower to .006 degrees, the hill itself
is invisible at a 30km far field. The bend of the propagating waves which
extends the radio horizon clearly mitigates any possible effects of the 80m
hill.


Hi Stefan,

You seem to be simultaneously agreeing and disagreeing. The hill is
either in the way, or it is not. It is in the way. To count on an
intermediary, such as suggested by Jimmie, knife-edge propagation or
bend of the waves, is probably not in our student's syllabus. Besides,
I have seen neither you nor Jimmie offer the attenuation presented by
such refractions (and the attenuation is not marginal). Without
quantifiables, the path budget cannot be calculated.

The problem, as stated, has a clear answer in looking over the hill by
raising one antenna, the question informs us that is the answer and
that is simply resolved with trig (albeit, including the radius of
earth and accounting for its curvature).

73's
Richard Clark, KB7QHC