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    Clearing a lunar
    From: Bruce Stark
    Date: 2003 Apr 28, 17:56 EDT

    Several weeks ago I started a posting that was intended to show how my method
    of clearing the lunar distance works. Other things intervened, and the effort
    was abandoned. Coming back to it, I've decided to divide the explanation into
    two postings. This one will deal with the general idea behind the method of
    clearing. A later one will focus on the Tables.
    
    Like the methods devised by Dunthorne, Borda, and Kraft, the method used in
    the Tables is rigorous. The distance is cleared with an oblique spherical
    trigonometry equation. The equation was derived by combining two other
    oblique spherical equations. Those two can be found, in one form or another,
    in navigation manuals back to, and including, Maskelyne's Requisite Tables.
    
    But I'll put individual methods aside for now, and use this posting to give a
    quick review of the lunar distance, and the general idea behind the rigorous
    methods of clearing. How you convert a cleared distance into GMT will be
    explained in the second posting.
    
    Pick an object more or less in line with the moon's monthly circuit through
    the celestial sphere. The angle between the moon and that object will be
    changing about one minute of arc every two minutes of time. And every
    measurement of the angle will correspond to a particular instant of GMT.
    
    Let's say the object is a star. With your sextant, bring the moon's
    enlightened limb to the star, letting the two brush past each other in the
    horizon glass to make sure there's a contact, but no overlap. Note the watch
    time of the contact. Since this is a bare-bones explanation, we'll use this
    single measurement rather than the average of a number of contacts.
    
    First, adjust the measurement for sextant error. That is, apply the index
    correction and any correction given on the certificate in the sextant box.
    Next, apply the moon's augmented semidiameter. Augmentation has to do with
    the fact you are generally a bit closer to the moon than the center of the
    earth is, so she appears slightly larger to you than she does from the
    Nautical Almanac's perspective. When the moon is directly overhead you are
    closer by the full radius of the earth. At such times, augmentation can
    amount to as much as three-tenths of a minute of arc. Add or subtract the
    augmented semidiameter, according to whether the moon's enlightened limb
    faces toward, or away from, the star.
    
    You now have the apparent distance. The question is: How do you adjust it for
    refraction and parallax?
    
    Refraction and parallax operate in the two vertical circles, one of which
    passes through the moon and your zenith, the other through the star and your
    zenith. If it just happens there is only one vertical circle, that is, the
    moon, star, and zenith are all in same great circle, refraction and parallax
    can be applied directly. They will have 100% effect on the distance. At the
    other extreme, suppose the plane of the distance is perpendicular to one of
    the vertical circles. Let's say the angle at the moon, where the plane of her
    altitude and the plane of the distance meet, is 90�. The moon's refraction
    and parallax will have little, if any, effect on the distance.
    
    Clearly, the angle formed at a body, where the vertical circle and distance
    circle meet, is the main factor determining what percent of its refraction
    and parallax show up in the distance. The approximate methods of clearing
    focus on these angles, calculating corrections, and corrections to
    corrections, to apply to the distance. Rigorous methods, on the other hand,
    calculate the true distance itself. For either method you need altitudes but,
    fortunately, the accuracy of the altitudes is not critical. Within five arc
    minutes will do fine, and larger errors seldom cause trouble.
    
    Suppose you want to clear your distance the rigorous way, but without using
    one of the procedures designed for the job. You have the three sides of the
    apparent triangle formed by your zenith, the moon, and star. The two apparent
    altitudes, or rather, their complements, form two sides. The apparent
    distance forms the third side. With these, calculate the angle at your
    zenith. Obviously that angle isn't changed when the altitudes of the moon and
    star are adjusted for refraction and parallax, so the zenith angle is the
    same for both apparent and true triangles.
    
    With the zenith angle and the true altitudes you now have the three parts you
    need to solve the true triangle. Solve it for the third side. That will be
    the true, or cleared, distance.
    
    Solving two oblique spherical triangles one at a time is not the easiest way
    to clear a distance. But it's a transparent operation, and shows the logic of
    the rigorous approach.
    
    How the cleared distance is converted to Greenwich time can be explained in
    relation to the Tables for Clearing, which I hope to write about in another
    posting.
    
    Bruce
    
    
    

       
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