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Frank:

The following, taken from one of my old navigation courses, may be helpful:

GREAT CIRCLE SAILING

INTRODUCTION

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
sailing.

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
Bowditch.

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.



        SOLUTION OF GREAT CIRCLE DISTANCE PROBLEMS

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.

GREAT CIRCLE SAILING

        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,

Bill


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
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