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    Re: Stellarium for sight reduction math practice?
    From: Paul Hirose
    Date: 2014 Sep 05, 21:27 -0700

    > I've been playing with Stellarium and it altitude values to practice the 
    math of a sight reduction. I'm assuming I don't have to do any sextant 
    corrections if I use the apparent versus the geometric values for 
    Stellarium's "Alt" generate my H_o? Is the only difference between the two 
    that the apparent incorporates refraction (A2 Tables of the NA)?
    
    In astronomy, "geometric" refers to where a body actually is at a given
    moment. "Apparent" refers to where it *appears* to be. The difference is
    due to:
    
    1) The time for light to travel to the observer. This is ignored for
    stars since light time is implicitly included in star catalog data.
    
    2) Light deflection by the gravitation of other bodies. These bodies are
    often (usually?) ignored except the Sun, and that too is ignored unless
    great accuracy is necessary.
    
    3) Aberration of light due to the observer's velocity through space.
    This causes an cyclic annual variation in the apparent coordinates of
    every body by a maximum of 20″ from the "un-aberrated" coordinates. The
    effect can be ignored when modest accuracy is adequate, as in a long
    term almanac.
    
    In addition to geometric and apparent, there are also astrometric
    coordinates, which are adjusted for light time only. (In the case of a
    star, the catalog coordinates are adjusted for parallax and proper
    motion only.)
    
    None of these includes refraction. In fact, The Astronomical Almanac
    definition of "topocentric place" specifically includes aberration and
    excludes refraction.
    
    I don't know if Stellarium uses these terms in the way I described.
    

       
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