Pete,
The exact formula for the angle you described is
atan((h_t - h_a)/d_f), where h_t is the height of the terrain in feet, h_a is
the height of the antenna in feet and d_f is the distance in feet.
I like a neat ballpark estimate (accurate within about 10%) of
(h_t-h_a)/(d_m*100), where h_t and h_a are the same as above, d_m is distance
in miles and the result is in degrees.
For instance if in your example the height of the antenna is 50 ft, the exact
formula would give us atan((500-50)/15*5280) = 0.325 deg, the estimate gives
((500-50)/15*100) = 0.3 deg.
73, Denis - K7GK
> Date: Sun, 23 May 2010 09:44:19 -0400
> From: n4zr@contesting.com
> To: xdavid@cis-broadband.com
> CC: cq-contest@contesting.com
> Subject: Re: [CQ-Contest] Contest QTH
>
> I hope that Dean will chime in here, because I recall an e-mail
> conversation a few years ago in which he said he felt that About 3 miles
> was a practical limit. I once built a terrain file that had too many
> points for HFTA to process, which suggests to me that you need a
> compromise between the range covered and the amount of detail in the
> profile.
>
> My trig is way too rusty to attempt the math, but if you start with an
> antenna at a given altitude, go out 15 miles and identify a point that
> is maybe 500 vertical feet above that plane, and draw a line between
> them, as well as a horizontal line (we're talking flat earth here
> hi)from the antenna to the obstruction, what angle above the horizontal
> will just clear the ridgeline? I know the number is pretty small, just
> not *how* small.
>
> 73, Pete N4ZR
>
> The World Contest Station Database, updated daily at www.conteststations.com
> The Reverse Beacon Network at http://reversebeacon.net, blog at
> reversebeacon.blogspot.com
>
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