From: rclark@lpl.arizona.edu (Richard Clark x4971)
Newsgroups: sci.astro
Subject: Re: Distance of horizon
Date: 20 Nov 92 08:52:05 GMT
Organization: Lunar & Planetary Laboratory, Tucson AZ.

In article <ETHANB.92Nov19004023@ptolemy.astro.washington.edu> ethanb@ptolemy.astro.washington.edu (Ethan Bradford) writes:
>In article <1992Nov19.021430.13833@sfu.ca> palmer@sfu.ca (Leigh Palmer) writes:
>
>   In article <lglhj3INNb0c@appserv.Eng.Sun.COM> fiddler@concertina.Eng.Sun.COM
>   (steve hix) writes:
>   >Anyone have handy a function for figuring the distance of the
>   >horizon from a viewer based on the viewer's height from the
>   >surface?
>			       -1
>   Try d = R arccos ( 1 + h/R )
>
>       d = horizon distance
>       h = height above MSL (assuming horizon is at sea level)
>       R = radius of Earth
>
>This is a complicated approximation to a simple function.  The exact
>answer is
>  d = sqrt(h^2 + 2 R h) \approx sqrt(2 R h)
>
>This can easily be derived by drawing a right triangle with the base
>being a line from the observer to the center of the earth and the two
>sides being (1) the line from the center of the earth to the horizon
>and (2) the line from the observer to the horizon (lines 1 and 2 meet
>at a right angle because line 2 is tangent to the surface of the
>earth).
>
>Since the cosine of the angle between the base and line 1 is R/(R+h)
>and since sin(\theta) \approx \theta for small \theta, the other
>formula will not be off by any significant amount.
>
>Leigh Palmer also writes:
>   I tried to find a series expansion for arccos ( 1 + z ) in my
>   tables, but there ain't one there. I think I see how to figure one
>   out, but I've got to go home now.
>
>You won't be able to figure one out; there isn't one.  The derivative
>of arccos(1+z) is infinite at z=0.  You will note that the exact form
>doesn't have a Taylor expansion either.
>
>-- Ethan


If you are only interested in small h (from the tops of towers or hills...)
The approximation d = sqrt(2 R h) is perfectly good for kids and friends.
But this is the geometric horizon without the affects of atmospheric
refraction.
Under typical conditions and fairly close to sealevel this is:
	d = sqrt(a R h)
where a:
	2		geometric
	2.31		visible
	2.58-2.67	radar/microwave
This is with d, R, h all expressed in the same units.
If you express h and d in units of convienence such as m/km or ft/miles
and absorb R into the wavelength factor you get
d = sqrt(b h)
with b:
			ft/miles	m/km
geometric		1.50		12.7
visible			1.74		14.7
radar			1.97		16.8

These values have been collected from such sources as The_Nautical_Almanac
(the one for sailors, not the old version of the astronomers' almanac with
the similar title), The_American_Practical_Navigator, and some notes from a
radar meteorology class many years ago.
I believe these refraction values are fairly constant throughout the lowest
couple thousand meters of the atmosphere but are highly subject to unusual
temperature profiles.

An interesting approximation used in radar meteorology is that of the '4/3
ficticious Earth'. Elevation calculations (given a range and elevation angle)
can be performed by replacing R with R' = 4 R / 3. This 'straightens out' the
refracted radar signal. This is useful for any elevation angle, not just
for looking out at the horizon. I've never done the numbers for the equivelent
factor at visible wavelengths.

Richard Clark
rclark@lpl.arizona.edu