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    Re: Photo sextant sights
    From: Andr�s Ruiz
    Date: 2008 Jul 30, 15:53 +0200

    And this:
    
    Photographic sextant sights + moon horns
    http://www.starpath.com/cgi-bin/ubb/ultimatebb.cgi?ubb=get_topic;f=31;t=000016#000000
    
    
    -----Mensaje original-----
    De: NavList@fer3.com [mailto:NavList@fer3.com] En nombre de Andres Ruiz
    Enviado el: mi�rcoles, 30 de julio de 2008 15:37
    Para: NavList@fer3.com
    Asunto: [NavList 5951] RE: Photo sextant sights
    
    
    See also: 
    http://www.starpath.com/cgi-bin/ubb/ultimatebb.cgi?ubb=get_topic;f=31;t=000040
    
    Andr�s
    
    ________________________________________
    De: NavList@fer3.com [mailto:NavList@fer3.com] En nombre de Wolfgang K�berer
    Enviado el: mi�rcoles, 30 de julio de 2008 15:16
    Para: NavList@fer3.com
    Asunto: [NavList 5950] Photo sextant sights
    
    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 K�berer
    Wolfsgangstr. 92
    D-60322 Frankfurt am Main
    Tel: + 49 69 95520851
    Fax: + 49 69 558400
    e-mail: koeberer@navigationsgeschichte.de
    
    
    
    
    
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