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    Re: lunars with and without altitudes
    From: Frank Reed CT
    Date: 2006 Nov 12, 17:50 EST
    George H, you wrote:
    "That didn't fit my expectations, so I cooked up some trial values. I
    presumed an observed star altitude, above the true horizontal, of 70
    degrees, an observed Moon altitude (of its centre) of 80 degrees, and
    an observed lunar distance (between centres) of 10 degrees."
    Sorry, George, I should have reminded you that there is also an altitude factor of cos(h). If you do lunar observations with either object above 60 degrees altitude, the required accuracy in altitude is reduced by a factor of two, and above 80 degrees, by a factor of six. The portion of the sky above 60 degrees altitude is, of course, relatively small compared to the entire sky --about 13%.
    But rather than write it all up again, I'll point you to my post on the topic back in 2004:
    where you will find these two equations for the required accuracy in the observed altitudes for lunars:
    AccuracyBody = 6' * sin(Distance) / cos(BodyAltitude)
    AccuracyMoon = 6' * tan(Distance) / cos(MoonAltitude).
    I wrote this up again and described how to see it graphically using the old lunars graphs by Margetts from c.1793 in NavList 134 in May:
    This last one, by the way, I think is pretty interesting, if I do say so myself. :-)
    For a specific case, suppose the altitudes of both objects are 20 degrees and the lunar distance is 88 degrees. Then the required accuracy in the star's altitude is about 6 minutes of arc. Any greater error will yield an error of 0.1 minutes of arc or more in the clearing process. For the Moon, the required accuracy is about 183 minutes of arc or three degrees. So the measured altitude of the Moon hardly matters at all in a case like this (I called it a "90 degree miracle" because there's no particularly good reason why this should happen --it is an "accidental" cancellation of the dependence of the Moon's parallax on altitude and the other geometric factors. I believe it helps to explain, in part, why historical lunar distances were often preferred when the Moon was around first and last quarter).
    For another specific case, suppose the Moon and star are both 30 degrees high and the lunar distance is 3 degrees. Very tempting to measure this one and late 19th century French astronomers recommended such short lunar distances (since they're easier to hold steady in the sextant's field of view). If you calculate the altitudes, there shouldn't be any problem. If you measure them, then by the above formulae, the required accuracy in the star's altitude is 0.36 minutes of arc and since the tangent and the sine are nearly identical for small angles, you need the same accuracy in the Moon's altitude. This is not impossible, but it's clearly demanding. Also, note that if both altitudes have the same error, maybe from a bad estimate of dip, it is nowhere near as important.
    Finally, for the specific case that you tried, where both objects are very high, the required accuracy in the star's altitude is 3' and the required Moon altitude accuracy is 6'. So by picking a case very close to the zenith, you've managed to cancel out the distance factors almost perfectly. This does indeed happen, but only in that small window near 90 degrees altitude.
    42.0N 87.7W, or 41.4N 72.1W.

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