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    Re: Aldebaran occultation
    From: Lars Bergman
    Date: 2017 Mar 7, 11:55 -0800

    Brad, you asked
    "Would you kindly comment more on the procedure and methodolgy?  Is it lunars based?"

    It has similarities with lunars, when the star disappears or reappears its true local distance to the Moon center equals the augmented semi-diameter of the Moon.

    With an approximate GMT find Moon's SHA, declination, semi-diameter and HP. Calculate the local hour angle (t in my paper). Then you calculate the true "altitude", or rather the complement to the true geocentric zenith distance (H), using the geocentric latitude (psi); and the parallactic angle (s). With an approximate value of parallax in altitude (P'), calculate the true local altitude, to find a better value of P'. Then calculate parallax in declination P'delta=P'·cos(s) and parallax in siderial hour angle P'SHA=P'·sin(s)·sec(delta).

    Now calculate the angle (A) at the pole between the declination circles through the star and the local true center of the Moon by solving the spherical triangle with sides

    Moon's augmented semi-diameter (sd')
    Moon's true local polar distance (c)
    Star's polar distance (b)

    SHA star +/- A then is the Moon's local SHA. To this add, or subtract, P'SHA to find SHA Moon. Find GMT by interpolation between known SHA values.

    100 years ago SHA was not tabulated in NA, but right ascension was used.

    Found an error in my paper, lower right, sin(A/2)=...=5.24' shall of course be A/2=...=5.24'. I didn't trust my hav table in Burton's with such small angles so I calculated this angle separately.

    Which leads me into your question of sines of small angles. The sine of a small angle is equal to the angle, if the angle is expressed in radians. To convert an angle in minutes of arcs to radians, multiply by pi/(60·180). The log of this multiplier is 6.46373 (-10) which I calculated separately (lower half right). To find log sin(1.74') look up the log of 1.74 and add 6.46373, i.e. 0.24055+6.46373=6.70428. This has been well known for centuries. 

    Lars

       
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