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Halley1986 (talk | contribs) m →Methodology: +Zn comment |
Halley1986 (talk | contribs) →Methodology: +clarify the diff. between Zn and Z (which was in initial draft and current figure, but re-written). |
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: <math>\sin(Hc) = \sin(lat) \cdot \sin(dec) + \cos(lat) \cdot \cos(dec) \cdot \cos(LHA) </math>
: <math>\begin{align}
\tan(Zn, Zn \pm 180) &= \frac{\sin(LHA)}{\sin(lat) \cdot \cos(LHA) - \cos(lat) \cdot \tan(dec)} Z &= arctan(tanZ) \\
Zn &= \begin{cases}
Z & \text{if }LHA > 180, \text{ North Latitude} \\
360-Z & \text{if }LHA < 180, \text{ North Latitude} \\
180-Z & \text{if }LHA > 180, \text{ South Latitude} \\
180+Z & \text{if }LHA < 180, \text{ South Latitude} \\
\end{cases} \\
\end{align}</math>
:The adjustment from Z to Zn (which is in <math>[0,360]</math>, and measured from North) has two reasons:
::(1)The angles in [0,360] with the same <math>\tan</math> is not unique (since <math>\tan (X) = tan (X \pm 180)</math>), but <math>\arctan</math> is defined only in <math>[-90,90]</math>.
::(2) The LHA, which is used for disambiguation, is measured from South, unlike the North Azimuth, Zn.
or, alternatively,
: <math>\cos(\pm Zn) = \frac{\sin(dec) - \sin(lat) \cdot \sin(Hc)}{\cos(lat) \cdot \cos(Hc)}</math>
Where
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:''Hc'' = Computed altitude
:''Zn'' = Computed azimuth (Zn=0 at North)
:''Z'' = preliminary result for Zn (in nautical almanac)
:''lat'' = Latitude
:''dec'' = Declination
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