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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: A Lunars Game
From: Trevor Kenchington
Date: 2003 Dec 17, 22:13 +0000
From: Trevor Kenchington
Date: 2003 Dec 17, 22:13 +0000
Frank, 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@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 http://home.istar.ca/~gadus