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    Re: Destination from course and distance
    From: Bill B
    Date: 2005 Mar 11, 15:33 -0500

    > I actually succeeded in using Paul Hirose's suggestion to "rotate" the
    > triangle, substitute values, and use complements of values (90 - A).
    > But I failed miserably in trying to simplify the formulae by
    > substituting the complementary trig functions.  If I knew what I was
    > doing (or received several more hints) I'd bet there was a much simpler
    > formula hiding in there somewhere.
    
    If understood your initial question, it was:
    
    If I start out at a known position, and sail a constant course for a given
    number of miles (nautical), where will I wind up?  You desire your new
    location to be given as latitude and longitude.
    
    Susan P Howell's book, "Practical Celestial Navigation" gives three methods
    under the chapter heading, "Mercator Sailing."
    1.  Formula
    2.  Simpler trig formula and Bowditch meridional-parts table
    3.  Totally tabular with Bowditch traverse table
    
    You asked for a formula, so I will try my best to relate it via text.
    
    L1 = starting latitude
    L2 = finishing latitude
    l1 = starting longitude
    l2 = finishing longitude
    C = Course
    D = Distance
    ln = natural log
    
    NOTE:  East and South are negative
    
    L2 = [(D cos C)/ 60] + L1
    
    l2 = l1 - tan C {180 [ln tan (45 + .5 L2) - ln tan (45 + .5 L1)] / pi}
    
    In the reverse of the this formula (given L1, l1, L2 and l2, find course and
    distance) Howell notes that the you must inspect the direction between the
    two points to be sure course is measured eastward from north.  C is labeled
    north or south depending on direction between L1 and L2. C is labeled east
    or west depending on direction between l1 and l2.
    
    Then the C derived is corrected to actual course (Cn) via the following:
    Cn = NE
    Cn = 360 - NW
    Cn = 180 - SE
    Cn = 180 + SW
    
    She does not state if course in the above formulas need to be adjusted from
    Cn to C.  I will leave that to you to discover.
    
    
    I am bad at converting formulas to ASCII, and worse at typing, so will scan
    the page and email to you at your request.
    
    Since I am already using a calculator to do the above, for a distance of
    less than 500 nm in my neck of the woods, I use a derivation of mid-latitude
    sailings and the polar-to-rectangular feature on my TI-30Xa (which is very
    intuitive).  I can enter the course in cardinal degrees and tenths and
    obtain my corrections to L1 and l1 in nautical miles.  Latitude corrections
    can be taken as minutes with no conversion.  The main tricks here are to
    remember x and y distance answers are swapped as we are not in using trig
    coordinates (y will become longitude, and x latitude); and longitude
    distance must be converted from nautical miles to degrees and minutes
    longitude.  This is basically as simple of adding L1 to L2, dividing that by
    2, and taking its cosine.  Invert the cosine (1/X key), then multiply that
    by the nm distance for minutes longitude (convert to degrees/minutes if
    necessary) to add or subtract from l1.
    
    If interested, I have the above polar-to-rectangular (as well as
    rectangular-to-polar) written up with illustrations in Microsoft word, and
    would be glad to send them to you at your request.  It sounds painful at
    first, but after you do it a few times, it is almost automatic.
    
    Hope the above is of some use.
    
    Bill
    
    billyrem42@earthlink. net
    
    
    

       
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