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Old 07-30-2011
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Originally Posted by kwindancer View Post
Hello All,

Hope you can answer and explain the answer to this question:

Let's say my point of departure is at point A and destination is to point B.

The track (Course to Make Good) from A to B is 145 degrees magnetic.

The winds are west at 22 knots when I depart point A at 1530. Let's say the leeway is 4 degrees and the boat speed is 6 knots. What course, corrected for leeway, should I steer from point A to point B?

Thanks so much.
The 'solution' lies within your question.
A—B = 145°M
leeway = 4°
therefore steer 145° + 4° = 149° to arrive at B

rigorous solution:
SIN 4° = 0.06975
1 mile = 6080 ft.
1 mile X SIN 4° = 6080 X 0.06975 = 424 FEET
therefore aim 424 FEET (per nautical mile per total distance travelled .... or add 4°) to the 'right' (windward side) of your course and you will be 'ded-on' at your arrival: 145 + 4 = 149°M
If the wind would be from the east and causing the same value of leeway, then subtract 4°; 145° - 4° = 141°M
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