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    Re: Figuring Course given Lat/Long of destination
    From: Tony S
    Date: 2000 Feb 27, 7:00 PM

    When you say that "there is the error of the Macerator thing", can you be
    more specific?  Did you use Bowditch Mercator sailing by tables?  This
    should work out OK.
    Actually, just using Plane sailing with mid-latitude should be quite close
    because the distance is relatively short; only earth eccentricity is ignored.
    Why the problem suggests also GC (great circle) does not make much sense.
    There would be less than a mile difference.  I did check the results by
    computer and they are OK.  [ Sometimes they are not. ;) ]
    Tony    in San Francisco
    > Ed Kitchin wrote:
    > An interesting problem appears in the latest issue of "Ocean Navigator" Which asks that you figure
    > the course to a destination given origination and destination. It would seem easy to determine the
    > difference in lat. (The destination was over several degrees of lat.), but deg. of long. differ in
    > length as you change lat. One could simply take the mean of the two given long. and use that, but
    > that bothers me as not being all that accurate. There is the error of the Macerator thing. You
    > could use universal plotting sheets and construct using a vertical representing diff./lat., then
    > draw a horizontal from the top of the lat. fig., representing the long. at the destination, and
    > draw a hypotenuse as the course line. (???) Are there any mathematicians out there to
    >  give me a good formula to learn for this task? Thank you.
    > Ed Kitchin

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