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    Photo sextant sights
    From: Wolfgang Köberer
    Date: 2008 Jul 30, 15:16 +0200

    Yesterday I received "The Navigator's Newsletter" issue 97 - 99. The last issue contains an article by David Burch (from his book "Emergency navigation") about "Photo sextant sights". As this had been discussed on the list not too long ago I read the article right away. It relates the example of a "Photo sextant sight" taken in Florida at 27deg 12,2 min N, 80 deg 13,4 min W.

     

    Now being in the legal profession I lack the knowledge or the brains (or both) to invent new methods of celestial navigation, but like to try their practical usefulness. The article sums up: "had we not known time or longitude, we would have found our longitude this way to within 53 min ". This would mean a possible error of about 47 miles - which is probably acceptable.

     

    Then I did a quick "Sumner": I varied the input to see what happens to the result. On the assumption that there is no error in latitude (having bent my sextant only after taking the height of Polaris or a star on the meridian) and that my time is correct, I varied my assumed longitude and got a perplexing result: Using Frank's program and assuming I am about 150 miles out in the Atlantic (= assumed longitude 77deg 10 min W) I got the following result:

    "Error in Lunar: 0 min

    Approximate Error in Longitude: 0 deg 00.4 min"

    which - following David Burch – I interpret as saying I am almost at my assumed position. As the "Photo Sextant Sight" supposedly was taken 3 degrees further west something doesn't fit. What did I get wrong? It's not the input, I checked it several times.

     

    Dr. Wolfgang Köberer
    Wolfsgangstr. 92
    D-60322 Frankfurt am Main

    Tel: + 49 69 95520851
    Fax: + 49 69 558400
    e-mail: koeberer@navigationsgeschichte.de

     


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