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    Lunars - Even Easier
    From: Frank Reed
    Date: 2008 Jul 01, 17:01 -0400

    I've discovered another "clearing miracle"...
    We all know that if the two bodies in a lunar observation are perfectly
    lined up vertically, then clearing the lunar distance is not substantially
    different from correcting ordinary sextant altitudes:
      |z1 - z2| = LD.
    The difference in the zenith distances is the lunar distance, so "clearing"
    a lunar distance is no different from correcting the altitudes (or zenith
    distances) of the two bodies. We have to be more careful with small details
    in the altitude corrections, as is always the case with lunars, but the math
    is essentially the same. No trig is required to correct altitudes.
    Likewise, though not quite as obvious, when the two bodies are in exactly
    opposite azimuths, the correction is nearly as simple:
      z1 + z2 = LD.
    The sum of the zenith distances is the lunar distance. This was the case
    with the lunar shot by Jeremy a few weeks ago, and as I've already
    described, it means we can clear it without trig.
    But sometimes a miracle happens. Even when the objects aren't perfectly
    aligned, the same almost trivial math applies. The math doesn't care whether
    the objects are really aligned so long as there is an equivalent case where
    they are in fact aligned. For example, the two objects could be separated in
    azimuth by 170 or even 165 degrees instead of 180 degrees, and under the
    right conditions we can "pretend" that they are separated in azimuth by 180
    degrees, and it all works out correctly.
    How can this be?! Some kind of crazy voodoo? No, just good old rigorous
    math. It works because the altitude of the Moon doesn't matter much,
    especially when the observed lunar distance is close to 90 degrees or the
    Moon's altitude is near the zenith. So we can ignore the Moon's real
    altitude (as observed) and substitute an altitude that turns the math into
    one of those special "aligned" cases.
    Assuming we have observed a lunar where the objects are nearly aligned, we
    use the mathematical condition of an aligned lunar to replace the Moon's
    altitude. Suppose we see the Sun and Moon roughly on opposite sides of the
    sky. Suppose we observe the lunar distance is 85 degrees (center-to-center)
    and the Sun's zenith distance is 45 degrees (after taking out dip and
    semi-diameter) and the Moon's observed zenith distance is 38 degrees (also
    after dip and SD). The sum of the zenith distances is 83 degrees which is
    less than the lunar distance so they're not aligned. But let's pretend
    they're aligned and drop the observed zenith distance of the Moon. We
    replace it by 40 degrees. Then we can work the clearing process by simple
    addition and subtraction of the altitude corrections --as if the two bodies
    were aligned perfectly. No trig at all!
    So how much of a change in the Moon's altitude can we tolerate? Well, we
    already know that. The allowable error in the Moon's altitude is given by
     dh = (6')*tan(LD)/cos(h_moon).
    This is the allowable error in the sense that if you make an error smaller
    than this in the Moon's altitude, the error in the clearing process will be
    smaller than 0.1 minutes of arc. If this doesn't look familiar to you, it's
    something I discovered a few years ago. It's not in the literature. You can
    prove it with a little calculus. Note that this formula applies under the
    assumption that refraction can be ignored compared to parallax. At very low
    altitudes, it doesn't work quantitatively, but the behavior is qualitatively
    similar (the zero error zone is skewed to larger distances).
    All of this implies that there are a rather large number of practical cases,
    at least in the tropics, where you can clear lunars without any tables at
    all, except the standard almanac altitude correction tables. So chew on
    that, kids!
    PS: the corresponding "allowable error" for the other body's error is
     dh = (6')*sin(LD)/cos(h_body).
    That's not relevant to this story and I'm including it here only for
    Navigation List archive: www.fer3.com/arc
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