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    Re: Question re. A and B factors (corner cosines) lunars
    From: Magnus Sjoquist
    Date: 2012 Jan 24, 10:03 +0100


    Fog lifting, Clear views 360 degrees. Thank You!

    Now creeping out through the back door, wet tail close to floor.


    When, or rather if, I regain some self confidence I will ask You what the ?Q? factor is,

     but until that happens I will stick to house cleaning, snowshuffling and writing my memoirs. The latter quite an easy job, since I clearly do not remember very much.

    Above all, I will shut up.



    /Magnus Sjoquist

    ----Ursprungligt meddelande----
    Från: FrankReed@HistoricalAtlas.com
    Datum: 2012-01-24 05:26
    Till: <NavList@fer3.com>
    Ärende: [NavList] Re: Question re. A and B factors (corner cosines) lunars

    Sure, I would be happy to help you out with that.

    In the case of a lunar, we have two objects in the sky, the Moon and the Sun (or a star, but usually the Sun). If we correct for semi-diameters in the usual way, we can treat the Sun and Moon as mathematical points on the celestial sphere, call them S and M. We also need the zenith, Z, since changes in the position of the Sun and Moon due to refraction and parallax are almost exactly in the vertical direction. We directly measure the angle between the Sun and the Moon and we also measure the zenith distances of both objects (of course z.d. is just 90-measured altitude). Since we've got all three sides, we can calculate the cosines of the angles at any of the corners. The "A" factor is the cosine of the angle ZMS between the arc MZ (from the Moon to the zenith) and the arc MS (from the Moon to the Sun --the "lunar arc").

    When considering the geometry of any case, it can be convenient to imagine the Sun pinned to a specific spot on the celestial sphere at some azimuth (might as well be zero) and some convenient altitude, like 45 degrees high. Now slide the Moon around to different positions and consider how the shape of the triangle changes and how the angle ZMS changes. In particular, consider the cases where the distance has some fixed value, like 30 degrees. If the Moon is more or less at the same altitude as the Sun (it actually has to be a bit lower in altitude), either east or west of the Sun, the angle ZMS will be near 90 degrees and its cosine would be zero. If the Moon is directly above the Sun (its altitude would then be 75 degrees), the angle ZMS is zero and its cosine is +1. If the Moon is directly below the Sun (necessarily at an altitude of 15 degrees), the angle ZMS is 180 degrees and its cosine is -1.

    Now forget about all that, and let's consider a case on the surface of the globe that would be mathematically identical. Suppose I have looked up somewhere the distance to the nearest tenth of a mile between the center of Chicago and the center of Paris. This is a distance between two points on a sphere (the nearly spherical Earth) and it is analogous to the measured distance between the Sun and the Moon on the celestial sphere. But now suppose I am doing a presentation for a group in Waukegan (sixty nautical miles due north of Chicago) and I want to tell them the distance to Paris from their location. I could start from scratch and do a complete calculation (and that's easy if you have time and computing available) or I could at least get a good estimate by knowing the relative bearing of Paris from Chicago. If Paris was due north of Chicago, then travelling north by sixty miles would directly reduce the distance to Paris by those sixty miles. If Paris was southeast of Chicago (such that the great circle route began on a due east course leaving Chicago), then travelling north sixty miles would hardly change the distance at all since that motion is perpendicular to the great circle route to Paris. In fact Paris bears halfway between northeast, close to 50 degrees true azimuth, on a great circle route so travelling north sixty miles would reduce the distance by just about 60*cos(50) or roughly 39 nautical miles. If we had the cosine of the bearing accurately to three digits (better yet four), the change in distance would be accurate to a tenth of a nautical mile.

    Does that help?


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