Intercept method: Difference between revisions

Revision as of 14:38, 2 July 2022

Revision as of 15:33, 2 July 2022

Line 33:

: <math>\begin{align}

\tan(Zn,) = \tan(Zn \pm 180) &= \frac{\sin(LHA)}{\sin(lat) \cdot \cos(LHA) - \cos(lat) \cdot \tan(dec)} \triangleq tanZ \\

Z &= arctan(tanZ) \text{ } \in [-90,+90] \\

Zn &= \begin{cases}

Z & \text{if }LHA >\in 180[0, \text{ North Latitude}90] \\

360-Z+180 & \text{if }LHA <\in 180[90, \text{ North Latitude} 270]\\

180-Z+360 & \text{if }LHA >\in 180[270,360] && \equiv \text{ Southmod Latitude360} \\

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 negative angle must be adjusted to positive angle.

::(2) The LHA, which is used for disambiguation, is measured from South, unlike the North Azimuth, Zn.

or, alternatively,

: <math>\begin{align}
\cos(\pm Zn) &= \frac{\sin(dec) - \sin(lat) \cdot \sin(Hc)}{\cos(lat) \cdot \cos(Hc)}</math> \triangleq cosZ \\

Z & = \arccos(cosZ) \text { } \in [0,180] \\

Zn &= \begin{cases}

180+Z & \text{if }LHA <\in [180, \text{ South Latitude}360] \\

-Z+360 & \text{if }LHA \in [0,180] \\

\end{cases} \\

\end{align}</math>

:The adjustment for disambiguating <math>\cos</math> values has similar reasons.

Where

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:''Hc'' = Computed altitude

:''Zn'' = Computed azimuth (Zn=0 at North)

:''Z'' = preliminary result for Zn (in some nautical almanacalmanacs)<ref name="US_Army_Training_Movie_CelNavTraining_Movie_CelNav">{{cite web

| title = Celestial Navigation

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