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    Re: Snellius Construction questions
    From: John Karl
    Date: 2009 Jan 5, 19:44 -0800

    One way this Snellius construction can be understood is by first
    understanding the case a single LOP that results from the horizontal
    angle between two markers B & C (buoys, etc.).  If you measure a
    horizontal angle A between the two markers, your ship is on a circular
    LOP with radius R = a/(2sin A), where a is the distance between the
    markers.  The three locations, B, C, and the ship, all lie on this
    circular LOP.
    You can can find the center of the LOP circle by striking arcs of
    radius R from both markers; then draw the LOP with radius R centered
    on this point.
    Alternatively, as the website shows, the center of the LOP circle can
    be found by the construction line they laid out at angle A from the B-
    C connecting line. By drawing the special case where the line from one
    marker (say B) to the ship passes through the center of the circle,
    you can see that a right triangle is formed, B-ship-C, with the right
    angle at C.  You'll also see that the perpendicular bisector of the B-
    C line must passes through the center of the LOP circle.  By studing
    the angles in this figure, you can understand their construction.
    It's tough without graphics and hand waving,
    John Karl
    On Jan 5, 2:43�pm,  wrote:
    > I have found an intriguing example of triangulation from a 16th century 
    mathematician, Willebrord 
    Snellius(http://www.sailingissues.com/navcourse4.html). I have searched other 
    web sources on the web and so far, have had no success in finding a straight 
    forward explanation of how this method of triangulation works. Have any of 
    you actually used the "Snellius Construction" technique, and could you help 
    me understand it as a practical navigation application?
    > Thank you,
    > Todd Frye
    > -----------------------------------------------
    > [Sent from archive by: fryefamily-AT-sisna.com]
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