### C3 - THE NAVIGATION TRIANGLE USING NAPIER'S ANALOGIES - 10/03/2004

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When sailing from point 1 (L1, LO1) to point 2 (L2, LO2)**
X and Y are intermediate values to simplify the calculation.
t = DLO = difference in longitude (degrees)
C1 = the course angle at point 1
C2 = the course angle at point 2 (reversed))
D = distance from point 1 to point 2 (degrees)
tan X = ctn (t/2) * cos ((L1-L2)/2) / sin ((L1+L2)/2)
tan Y = ctn (t/2) * sin ((L1-L2)/2) / cos ((L1+L2)/2)
C1 = X + Y
C2 = X - Y (usually not required)
tan (D/2) = tan ((L1-L2)/2) * sin X / sin Y
or
tan (D/2) = ctn ((L1+L2)/2) * cos X / cos Y
Note that if L1 = L2, then Y = 0 and C1 = C2 = X.

To convert D from degrees to nautical miles

multiply D by 60.
The angle (C2) that is calculated above is the internal angle of
the spherical triangle at the destination point,but the actual course angle at that point
would be opposite by 180 degrees.

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