User:ThatDonGuy

To aim a satellite dish, you need to know the latitude of the dish, and the difference between the satellite's and dish's longitude. Since longitudes are based on an arbitrary zero meridian, any meridian can be defined as "zero longitude," so to make calculation easier, assume the dish is at zero longitude.

If LAT is the dish's latitude and LNG is the satellite's longitude relative to the dish's:

Azimuth change (the left-right direction) = $$\arctan \left ( \frac{\tan LNG}{\sin LAT} \right )$$

In the northern hemisphere, this is in degrees left (positive) or right (negative) of due south. In the southern hemisphere, this is in degrees right (positive) or left (negative) of due north.

Altitude change (the up-down direction) = $$\arctan \left ( \frac {\cos LNG\ \cos LAT - 0.1513}{\sqrt{cos^2 LNG\ sin^2 LAT + sin^2 LNG}} \right )$$

This is in degrees up from the horizon. (If the number is negative, the line of sight goes through the surface of the Earth.)

The number 0.1513 is the distance from the center of Earth to geostationary satellites divided by the radius of Earth.

The Math
The following terms are defined: Positive latitude is north; positive longitude is east Xs, Ys, Zs - the X, Y, and Z coordinates of the satellite Xd, Yd, Zd - the X, Y, and Z coordinates of the dish In both cases, point (0, 0, 0) is the center of Earth; +X is toward latitude 0, longitude 0; +Y is toward latitude 0, longitude +90; +Z is toward latitude +90 R - the radius of Earth A - the distance from the center of Earth to the satellite

First, convert latitude and longitude to rectangular coordinates:

$$X_s\ = A\ \cos LNG$$

$$Y_s\ = A\ \sin LNG$$

$$Z_s\ = 0$$

$$X_d\ = R\ \cos LAT$$

$$Y_d\ = 0$$

$$Z_d\ = R\ \sin LAT$$

Next, "rotate the universe" through the Y-axis until the dish is at latitude +90 (the satellite is rotated within the plane of its Y value):

$$X_s\ = A\ \cos LNG\ cos (90 - LAT) = A\ \cos LNG\ \sin LAT$$

$$Y_s\ = A\ \sin LNG$$

$$Z_s\ = A\ \cos LNG\ cos (90 - LAT) = A\ \cos LNG\ \sin LAT$$

$$X_d\ = 0$$

$$Y_d\ = 0$$

$$Z_d\ = R$$

By putting the dish on the North Pole, the angle of altitude from the horizon can be calculated much easier.

Next, move the universe distance R in the negative-Z direction:

$$X_s\ = A\ \cos LNG\ \sin LAT$$

$$Y_s\ = A\ \sin LNG$$

$$Z_s\ = A\ \cos LNG\ \sin LAT\ - R$$

$$X_d\ = 0$$

$$Y_d\ = 0$$

$$Z_d\ = 0$$

The azimuth and altitude angles are:

$$\tan Azimuth = \frac{Y_s}{X_s} = \frac{A\ \sin LNG}{A\ \cos LNG\ \sin LAT} = \frac{\sin LNG}{\cos LNG\ \sin LAT} = \frac{\tan LNG}{\sin LAT}$$

$$\tan Altitude = \frac{Z_s}{\sqrt{X_s^2 + Y_s^2}} = \frac{A\ \cos LNG\ \sin LAT\ - R}{\sqrt{(A\ \cos LNG\ \sin LAT)^2 + (A\ \sin LNG)^2}}$$

$$= \frac{A\ \cos LNG\ \sin LAT\ - R}{\sqrt{A^2 \cos^2 LNG\ \sin^2 LAT + A^2 \sin^2 LNG}} = \frac{A \left (\cos LNG\ \sin LAT\ - \frac{R}{A} \right )}{A \left ( \sqrt{\cos^2 LNG\ \sin^2 LAT + \sin^2 LNG} \right )}$$

$$= \frac{\cos LNG\ \sin LAT\ - \frac{R}{A}}{\sqrt{\cos^2 LNG\ \sin^2 LAT + \sin^2 LNG}}$$