Editing Astronomical coordinate systems (section)

=== Equatorial ↔ horizontal ===
Azimuth ({{mvar|A}}) is measured from the south point, turning positive to the west.<ref>
{{cite book
 | last1 = Montenbruck
 | first1 = Oliver
 | last2 = Pfleger
 | first2 = Thomas
 | title = Astronomy on the Personal Computer
 | publisher = Springer-Verlag Berlin Heidelberg
 | year = 2000
 | isbn = 978-3-540-67221-0}}, pp 35-37</ref>
Zenith distance, the angular distance along the [[great circle]] from the [[zenith]] to a celestial object, is simply the [[Complementary angles|complementary angle]] of the altitude: {{math|90° − ''a''}}.<ref>
{{cite book 
 | last1 = U.S. Naval Observatory
 | first1=Nautical Almanac Office
 | first2 = H.M. Nautical Almanac Office
 | last2 = U.K. Hydrographic Office
 | title = The Astronomical Almanac for the Year 2010
 | publisher = U.S. Govt. Printing Office
 | year = 2008
 |isbn = 978-0160820083
 |page=M18}}
</ref>

:<math>\begin{align}
  \tan\left(A\right) &= {\sin\left(h\right) \over \cos\left(h\right) \sin\left(\phi_\text{o}\right) - \tan\left(\delta\right) \cos\left(\phi_\text{o}\right)}; \qquad \begin{cases}
    \cos\left(a\right) \sin\left(A\right) = \cos\left(\delta\right) \sin\left(h\right) ;\\
    \cos\left(a\right) \cos\left(A\right) = \cos\left(\delta\right) \cos\left(h\right) \sin\left(\phi_\text{o}\right) - \sin\left(\delta\right) \cos\left(\phi_\text{o}\right)
  \end{cases} \\[3pt]
  \sin\left(a\right) &= \sin\left(\phi_\text{o}\right) \sin\left(\delta\right) + \cos\left(\phi_\text{o}\right) \cos\left(\delta\right) \cos\left(h\right);
\end{align}</math>

In solving the {{math|tan(''A'')}} equation for {{math|''A''}}, in order to avoid the ambiguity of the [[arctangent]], use of the [[atan2|two-argument arctangent]], denoted {{math|atan2(''x'',''y'')}}, is recommended. The two-argument arctangent computes the arctangent of {{math|{{sfrac|''y''|''x''}}}}, and accounts for the quadrant in which it is being computed. Thus, consistent with the convention of azimuth being measured from the south and opening positive to the west,
:<math>A = -\operatorname{atan2}(y,x)</math>,

where
:<math>\begin{align}
  x &= -\sin\left(\phi_\text{o}\right) \cos\left(\delta\right) \cos\left(h\right) + \cos\left(\phi_\text{o}\right) \sin\left(\delta\right) \\
  y &=  \cos\left(\delta\right) \sin\left(h\right)
\end{align}</math>.

If the above formula produces a negative value for {{math|''A''}}, it can be rendered positive by simply adding 360°.

:<math>\begin{align}
  \begin{bmatrix}
    \cos\left(a\right) \cos\left(A\right) \\
    \cos\left(a\right) \sin\left(A\right) \\
    \sin\left(a\right)
            \end{bmatrix} &= \begin{bmatrix}
    \sin\left(\phi_\text{o}\right) & 0 & -\cos\left(\phi_\text{o}\right) \\
    0                              & 1 &  0                              \\
    \cos\left(\phi_\text{o}\right) & 0 &  \sin\left(\phi_\text{o}\right)
  \end{bmatrix}\begin{bmatrix}
    \cos\left(\delta\right)\cos\left(h\right) \\
    \cos\left(\delta\right)\sin\left(h\right) \\
    \sin\left(\delta\right)
  \end{bmatrix} \\
                          &= \begin{bmatrix}
    \sin\left(\phi_\text{o}\right) & 0 & -\cos\left(\phi_\text{o}\right) \\
    0                              & 1 &  0                              \\
    \cos\left(\phi_\text{o}\right) & 0 &  \sin\left(\phi_\text{o}\right)
  \end{bmatrix}\begin{bmatrix}
    \cos\left(\theta_L\right) &  \sin\left(\theta_L\right) & 0 \\
    \sin\left(\theta_L\right) & -\cos\left(\theta_L\right) & 0 \\
    0            &  0            & 1
  \end{bmatrix}\begin{bmatrix}
    \cos\left(\delta\right)\cos\left(\alpha\right) \\
    \cos\left(\delta\right)\sin\left(\alpha\right) \\
    \sin\left(\delta\right)
  \end{bmatrix}; \\[6pt]
       \tan\left(h\right) &= {\sin\left(A\right) \over \cos\left(A\right) \sin\left(\phi_\text{o}\right) + \tan\left(a\right) \cos\left(\phi_\text{o}\right)}; \qquad \begin{cases}
    \cos\left(\delta\right) \sin\left(h\right) = \cos\left(a\right) \sin\left(A\right); \\
    \cos\left(\delta\right) \cos\left(h\right) = \sin\left(a\right) \cos\left(\phi_\text{o}\right) + \cos\left(a\right) \cos\left(A\right) \sin\left(\phi_\text{o}\right)
  \end{cases} \\[3pt]
  \sin\left(\delta\right) &= \sin\left(\phi_\text{o}\right) \sin\left(a\right) - \cos\left(\phi_\text{o}\right) \cos\left(a\right) \cos\left(A\right);
\end{align}</math>{{efn|Depending on the azimuth convention in use, the signs of {{math|cos ''A''}} and {{math|sin ''A''}} appear in all four different combinations. Karttunen et al.,<ref name=Karttunen/> Taff,<ref name=Taff/> and Roth<ref name=Roth/> define {{math|''A''}} clockwise from the south. Lang<ref name=Lang/> defines it north through east, Smart<ref name=Smart/> north through west. Meeus (1991),<ref name=Meeus/> p.&nbsp;89: {{math|sin ''δ'' {{=}} sin ''φ'' sin ''a'' − cos ''φ'' cos ''a'' cos ''A''}}; ''Explanatory Supplement'' (1961),<ref name=ExplSupp/> p.&nbsp;26: {{math|sin ''δ'' {{=}} sin ''a'' sin ''φ'' + cos ''a'' cos ''A'' cos ''φ''}}.}}

Again, in solving the {{math|tan(''h'')}} equation for {{math|''h''}}, use of the two-argument arctangent that accounts for the quadrant is recommended. Thus, again consistent with the convention of azimuth being measured from the south and opening positive to the west,
: <math>h = \operatorname{atan2}(y, x)</math>,

where
:<math>\begin{align}
              x &= \sin\left(\phi_\text{o}\right)\cos\left(a\right) \cos\left(A\right) + \cos\left(\phi_\text{o}\right)\sin\left(a\right) \\
              y &= \cos\left(a\right)\sin\left(A\right) \\[3pt]
  \begin{bmatrix}
    \cos\left(\delta\right)\cos\left(h\right) \\
    \cos\left(\delta\right)\sin\left(h\right) \\
    \sin\left(\delta\right)
  \end{bmatrix} &= \begin{bmatrix}
     \sin\left(\phi_\text{o}\right) & 0 & \cos\left(\phi_\text{o}\right) \\
     0                              & 1 & 0                              \\
    -\cos\left(\phi_\text{o}\right) & 0 & \sin\left(\phi_\text{o}\right)
  \end{bmatrix}\begin{bmatrix}
    \cos\left(a\right) \cos\left(A\right) \\
    \cos\left(a\right) \sin\left(A\right) \\
    \sin\left(a\right)
  \end{bmatrix} \\
  \begin{bmatrix}
    \cos\left(\delta\right) \cos\left(\alpha\right) \\
    \cos\left(\delta\right) \sin\left(\alpha\right) \\
    \sin\left(\delta\right)
  \end{bmatrix} &= \begin{bmatrix}
    \cos\left(\theta_L\right) &  \sin\left(\theta_L\right) & 0 \\
    \sin\left(\theta_L\right) & -\cos\left(\theta_L\right) & 0 \\
    0                         &  0                         & 1
  \end{bmatrix}\begin{bmatrix}
     \sin\left(\phi_\text{o}\right) & 0 & \cos\left(\phi_\text{o}\right) \\
     0                              & 1 & 0                              \\
    -\cos\left(\phi_\text{o}\right) & 0 & \sin\left(\phi_\text{o}\right)
  \end{bmatrix}\begin{bmatrix}
    \cos\left(a\right) \cos\left(A\right) \\
    \cos\left(a\right) \sin\left(A\right) \\
    \sin\left(a\right)
  \end{bmatrix}.
\end{align}</math>