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    Re: A Lunars Game
    From: Trevor Kenchington
    Date: 2003 Dec 17, 22:13 +0000

    I have a nasty feeling that you and the other lunarians are going to
    tell me that I am way off (as in "stuck on an island in a different
    ocean"). However, to the limit of my abilities, I would place your
    contestant as follows:
    His zenith is roughly 45% of the difference in declination from Canopus
    to Sirius. I take it that means 45%, not 40% or 50%, so we could say
    somewhere between 42.5% and 47.5%. The difference in declinations
    between those stars is 36 degrees, give or take a couple of minutes,
    while Canopus has a declination of 52?42'S (to the nearest minute). So
    his latitude is between 35.5 and 37.5 degrees South, rounded outwards to
    the nearest 30 miles.
    With Venus and the Moon low in the sky, the parallels of declination
    passing through them will make an angle with the observer's vertical
    that is equal to his latitude. Thus, looking at the sky (or your
    photograph), your contestant can approximately draw in the grid of
    declinations and GHAs in the vicinity of those two bodies. From that,
    the apparent GHA of the Moon (without allowing for parallax) is less
    than that of Venus by about 40% of the angular distance between them.
    Your contestant has estimated the separation of those two bodies at
    about 2.5 degrees, so the difference in GHA (still ignoring parallax) is
    about one degree. By inspection of the almanac, that cannot have
    occurred as early as 1500 GMT on the 24th, since the Moon was then still
    west of Venus, nor as late as 2100, when the difference in GHA would be
    over 2 degrees. So that would place sunset on the observer's island at
    around 1800 GMT, give or take an hour or so.
    Also from the Almanac, sunset at his latitude on January 24 should
    occur at 1920 local mean time, putting him (without regard to parallax)
    at 20 degrees East longitude, give or take 15 degrees (i.e. 5 degrees to
    35 degrees East). There are, of course, no islands in the block of ocean
    between 35?30'S 5?E and 37?30'S 35?E. From first principles, however,
    parallax must make the Moon appear lower in the sky than its true
    celestial position. Thus, relative to Venus, it must have a greater
    difference in GHA, making the time of sunset later and the contestant's
    longitude more westerly.
    Your photograph of the sky does not show the horizon but, assuming that
    it is only a little way below the clouds (lit as they are by the setting
    Sun), and taking the estimate of 2.5 degrees of arc between the Moon and
    Venus to provide a scale, then the altitude of the Moon, as perceived by
    your contestant, is very roughly 10 degrees. HP for the Moon at about
    the time of the observation is 58', to the nearest minute, and to that
    degree of precision is steady throughout the hours that the observation
    could have been made. The parallax in altitude is, therefore one degree,
    to the sort of precision we are dealing with. (And it would not be any
    different if the Moon's apparent altitude was anywhere from zero to well
    above 20 degrees.)
    Correcting for parallax would therefore push the difference in GHA
    between Moon and Venus out from around one degree to very roughly two
    degrees. That puts the time of the observation at about 2100 GMT, rather
    than 1800 or 0000. In turn, the estimate of the contestant's longitude
    would move west to something between 10 and 40 degrees west.
    The only islands in the block of ocean between 35?30'S 10?W and 37?30'S
    40?W are the Tristan da Cunha group at about 37?30'S 13?W. Since you
    described a helicopter ride following your contestant's flight
    but did not mention any urban development on the island, I'll guess that
      you did not mean Tristan itself but one of its small sister islands
    (Inaccessible and Nightingale?).
    Maybe I am way off in my estimations. Even if I am not, I dare say that
    the lunarians on this list can get a neater and more precise longitude.
    However, given your choice of latitude and the placement of your
    contestant on an island, even these very crude eyeball estimates seem
    sufficient to answer your puzzle.
    Trevor Kenchington
    Trevor J. Kenchington PhD                         Gadus{at}iStar.ca
    Gadus Associates,                                 Office(902) 889-9250
    R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
    Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555
                           Science Serving the Fisheries

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