Converts a RA/DEC (J2000) coordinate to approximate Alt/Az (J2000) coordinate. This uses Greenwich **Mean** Sidereal Time to compute the approximate horizon coordiates, if you need
the coordiates also corrected for nutation, you will need to modify this to use Greenwhich **Apparent** Sidereal Time. It is assumed that all corrections the user has
interest in have already been applied to the RA/Dec coordinates.

You will likely come across different sets of equations for the RA/Dec to Alt/Az conversion. I have included three below from three different sources. The first is derived from spherical trigonometry and is from the Explanatory Supplement to the Astronomical Almanac 3rd ed. The second is from Astronomical Algorithms, using this in combination with the atan2 function included in most programming languages, is usually the easiest to implement. And the last is a matrix form, which is useful if you already have your coordinates in matrix form, or want to generalize the transformations.

The matrix form is likely the most useful for most applications since you will likely need to perform the conversion to rectagular coordinates for other purposes. But, if you need to provide both the RA/Dec and Alt/Az coordinates, Meuus' method will save you a few steps.

This implementation uses the Astronomical Algorithms set, but an implementation of the other is available in the set of test data which validates both sets of equations, and can be used as an example of how to implement the Explanatory Supplement style. I have altered the algorithm from the method it is described in the book. I have converted it so that Longitudes to the West are negative (like GPS coordinates), and so that 0 Azimuth is North (like a compass). I believe these changes make it behave more like a user would expect.\( \textbf eq1\ \ \ \cos \mathit{a} \sin A_{z} = -\cos \delta \sin \mathit{h} \\ \textbf eq2\ \ \ \cos \mathit{a} \cos A_{z} = \sin \delta \cos \phi - \cos \delta \cos \mathit{h} \sin \phi \\ \textbf eq3\ \ \ \sin \mathit{a} = \sin \delta \sin \phi + \cos \delta \cos \mathit{h} \cos \phi \)

Where \(h\) is the hour angle, \(\delta\) is the declination, \(A_{z}\) is azimuth, \(a\) is altitude, \(\phi\) is the lattitude. You will need to check the signs of \(\cos A_z\) and \(\sin A_z\) to determine the propper quadrant:

if \(\cos A_z\) > 0 and \(\sin A_z\) > 0: use either eq1 or eq2

if \(\cos A_z\) > 0 and \(\sin A_z\) < 0: use eq1

if \(\cos A_z\) < 0 and \(\sin A_z\) < 0: use \(A_z\) = 360 - eq2

if \(\cos A_z\) < 0 and \(\sin A_z\) > 0: use eq2

\( \tan A = \dfrac{\sin H}{\cos H \sin \varphi - \tan \delta \cos \varphi} \)

\( \sin h = \sin \varphi \sin \delta + \cos \varphi \cos \delta \cos H \)

Where \(H\) is the hour angle, \(\delta\) is the declination, \(A\) is azimuth, \(h\) is altitude, \(\varphi\) is the lattitude.

\( \\ \mathbf{l}(h,\delta)= \begin{bmatrix} \cos \delta\cos h \\ \cos \delta\sin h \\ \sin \delta \\ \end{bmatrix}\\ \\~\\ \mathbf{l}(A,a)=R_2(90^{\circ}-\Phi )\ \mathbf{l}(h,\delta) \)

Where \(h\) is the hour angle, \(\delta\) is the declination, \(A\) is azimuth, \(a\) is altitude, \(\Phi\) is the lattitude.

\(\mathbf{l}(A,a)\) will be in cartesian coordinates (\(x, y, z\)), to convert to RA/Dec:\( r = \sqrt{ x^2+y^2+z^2} \\ \cos A = z / r \\ \tan a = y / x \\ \)