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    Re: Great Circle Computation
    From: William Trayfors
    Date: 2000 May 23, 7:16 AM

    The following, taken from one of my old navigation courses, may be helpful:
    A great circle course is the shortest distance between any two points on
    earth.  It is of interest to the navigator because he can often cut the
    total distance sailed from point to point considerably by following a great
    circle course.
    Great circle courses always traverse higher latitudes than do rhumb line
    courses, i.e., further north in the Northern Hemisphere and further south
    in the Southern Hemisphere.
    Where obstructions or navigational hazards lie in the path of the great
    circle course, a combination of great circle sailing and rhumb line or
    traverse sailing can often be used to advantage.  This is called composite
    It is possible to solve great circle course and distance problems in
    several ways.  By using a gnomic projection chart, great circle routes can
    be plotted as straight lines.  A series of points on this chart can then be
    transferred to a regular mercator chart -- say at intervals of 5 degrees of
    longitude -- and then the rhumb lines between each point can be taken
    directly off the mercator chart and used for navigation.  Using rhumb line
    sailing between each 5 degree point on the great circle track introduces
    only a small error, but requires less frequent course changes than would be
    necessary to follow a great circle track precisely.
    Another practical way to solve great circle course and distance problems is
    by the use of mathematical formulae.  This method is fine if you're handy
    with a scientific calculator or have a computer handy, particularly when a
    gnomic projection chart is not available.  In the following examples, two
    simple equations and a table of natural trigonometric functions (such as
    Table 31 in Bowditch) are used.  It is only required that the longitude and
    latitude of the points of departure and destination be accurately known.
    These can be taken from charts, Sailing Directions, or from Appendix S in
    The equations shown were taken from the A.R.R.L. Antenna Book, pp 284-285.
    They can also be used to compute radio bearing and correction factors.
    Equation:   cosD = sinL1 sinL2 + cosL1 cosL2 cosDLo    where...
      D = great circle distance in degrees of arc
     L1 = latitude of departure
     L2 = latitude of destination
     DLo = difference in longitude between departure and destination
    Example:  It is desired to compute the great circle distance between the
    entrance to Chesapeake Bay (Cape Henry Light) and Bermuda (North Rock Light).
    Solution:  From Bowditch, Appendix S, obtain the maritime positions of the
    departure and destination points.
            Cape Henry Light = 36 deg 56 min North  76 deg 00 min West
         North Rock Light = 32 deg 28 min North  64 deg 46 min West
            L1 = 36 56 N     sinL1 = .60089     cosL1 = .79934
         L2 = 32 28 N     sinL2 = .53681     cosL2 = .84370
         DLo = 11 deg 14 min                cosDLo = .98084
         Substituting in formula:
            cosD = .60089 x .53681 + .79934 x .84370 x .98084
         cosD = .32256 + .66148  or  cos D = .98404
         By inspection of trig. tables, .98404 corresponds with the
         cosine of 10 deg 15.0 min.  Therefore, D = 10 deg 15 min or
         10 x 60 + 15 = 615 nautical miles.
    Note:  Equation is used as shown above only when L1 and L2 are of the same
    name, i.e., when both lie in the northern or southern hemisphere.  When
    they are of opposite names, the value of sinL2 is minus.  For example, if
    in the above illustration L2 were 32 deg 28 min South instead of North, the
    substitution would be made as follows...
        cosD = .60089 x (-.53681) + .77934 x .84370 x .98084
        cosD = -.32256 + .66148  or cosD = .33892
        D = 70 deg 11.5 min or 70 x 60 + 11.5 = 4211.5 nautical miles
    When extracting the value for D, be careful of the ambiguity, particularly
    when the value is fairly high.  In this case, .33892 could be either the
    cosine of 70 deg 12 min or of 109 deg 49 min.
            Finding Initial Course
    Equation:   sinC = cosL2 cscD sinDLo     where...
               C = initial great circle course (true)
              L2 = latitude of destination
            D = great circle distance between departure and destination
             DLo = difference in longitude between departure & destination
    Example:  A navigator wishes to determine the initial true course
              along the great circle track from Cape Henry Light (36                deg
    North, 56 min West) to North Rock Light, Bermuda
              (32 deg 28 min North, 64 deg 46 min West).
    Solution: (Assumes D is known or has been previously computed.)
            L2 = 32 28 N     cosL2 = .84370
            D = 10 deg 15 min (615 miles)    cscD = 5.61976
            DLo = 11 deg 14 min       sinDLo = .19481
    Substituting in equation...
            sinC = .84370 x 5.61976 x .19481
         sinC = .92367
               C = 112.32 deg True
    Caution:  .92367 is SIN of both 112.32 degrees and of 67.28 degrees; care
    must be exercised to be sure the correct one is selected.
    Hope this helps,
    At 04:04 PM 5/23/00 +0200, you wrote:
    >Frank Dinkelaar wrote:
    >> In regards to the attachment (navigation formulas) send by Ed Kitchin
    >> the formula in #1 of the great circle course
    >> it seems to me it looks like this
    >> cos D =  abs(sinL1 - sinL2) - abs(cosL1 x cosL2 x cos Dlo)
    >> it would save me many hours of experimenting if somebody
    >> could confirm or send me the formula in a clearer form.
    >> thanks
    Bill Trayfors 
    The Washington Decision Support Group, Inc.
    Specialists in Advanced Information & Communications Technologies
    2401 South Lynn Street, Arlington, VA 22202
    Office (703) 838-8784   Tech Support (703) 573-WDSG   FAX (703) 838-0019

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