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Jeff Schroeder wrote:
>
> Can someone point me to spherical trig (great circle sailing) formulas
> for determining destination Lat & Lon, if given departure Lat & Lon,
> true course, and distance travelled?

That's the same basic problem as solving a celestial sight for Hc and
Zn: you know two sides and the included angle. In sight reduction, the
sides are latitude and declination. Their included angle is LHA.

If you think of your starting position as the "pole" and the pole as
the "observer", the endpoint of the great circle can be found as if
you're doing a sight reduction. The two known sides extend from the
"pole". They are latitude of the starting point, and distance. Convert
distance to angle measure, then subtract it from 90 degrees. That's
necessary so it acts like declination. In other words, if distance is
1 degree, "declination" is north 89. If distance is 91 degrees,
"declination" is -1, or south 1 degree.

Wherever a trig function of declination appears, you can substitute
the co-function (use cos in place of sin, etc.) to eliminate that
subtraction from 90.

Reducing the "sight" yields "altitude", in reality, latitude of the
destination. If "altitude" is negative, latitude is south. The other
output is "azimuth". That's actually the difference in longitude. It
increases westward from the starting longitude.

   

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