Ok, I have tried to talk about this before, and as I said my math was lacking. So this time I when to the UTA's (University of Texas at Arlington) math department and got some help. So I am going to start over from the top! First I will post two images of the situation. In both images, you can see that the slant range is the line R, and H is the height of the antenna above the ground. Therefore the line going down from the plane to H is the altitude above the antenna. But wait a minute the antenna is not on the ground (sea level) it's arbitrary. So now you have to figure out the rest. If you draw a line down from the antenna to the center of the earth, and then to the plane you now have an angle α with a length of *r _{e}*.

So this gets us:

**R**^{2}=r_{e}^{2}+(r_{e}+H)^{2}-2r_{e}(r_{e}+H) cos α*So now*

*you ask what the heck is r*? Well,

_{e}*r*is equivalent to the radius of the earth. "Under the assumption that the earth is a sphere, from the angle α, the part of the circumference can be calculated with a simple ratio calculation from the

_{e }total earth circumference.

**360**^{°}⋅ R_{topogr }= α⋅2πr_{e }And with a little math magic, this can now be written as:

*r*_{e }= 360^{°}⋅ R_{topogr}/α⋅2πThis is the trigonometrical connection without consideration of the earth's bend.

In part 2, I will deal with the software.

UPDATE:

- April 23, 2023: I forgot to show you how to come up with
**R**_{topogr}**R**_{topogr}= R⋅ COS ε - April 24, 2023: I decided to add information about WGS84
**, "**meridian of zero longitude is the IERS Reference Meridian, 5.3 arc seconds or 102 meters (335 ft) east of the Greenwich meridian at the latitude of the Royal Observatory. (This is related to the fact that the local gravity field at Greenwich doesn't point exactly through the Earth's center of mass, but rather "misses west" of the center of mass by about that 102 meters.) The longitude positions on WGS 84 agree with those on the older North American Datum 1927 at roughly 85° longitude west, in the east-central United States.

The WGS 84 datum surface is an oblate spheroid with equatorial radius a = 6378137 m at the equator and flattening f = 1/298.257223563. The refined value of the WGS 84 gravitational constant (mass of Earth's atmosphere included) is GM = 3986004.418×10^{8}m^{3}/s^{2}. The angular velocity of the Earth is defined to be ω = 72.92115×10^{−6}rad/s.

This leads to several computed parameters such as the polar semi-minor axis b which equals*a*× (1 −*f*) = 6356752.3142 m, and the first eccentricity squared,*e*^{2}= 6.69437999014×10^{−3}." This is from the Wikipedia