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03-14-16, 11:17 AM | #7 | |
Officer
Join Date: Aug 2010
Location: USA
Posts: 237
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Quote:
Example: You are traveling true heading 020°. You have a contact that is at relative bearing 300° with an AoB of 70° to STB. The equation would be 180-(70+60) = 50. 70 from the AoB, and 60 from the bearing (300 would be the angle outside of the triangle, 60 would be the angle inside (360-300). Your course + 50° or 20° + 50° = 70° target course. So for bearing you need the degrees from your bow using the shortest path (60° to port points to the target). Does that make sense? Hell I think I confused myself while re-reading it. It's a lot easier to illustrate.
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