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Originally Posted by klh
Quote:
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Originally Posted by jimmie
Now, what's it's like for the diverting course triangle? I actually had problem finding out target's speed when I wasn't at collision course..
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It helps me to draw a picture. Consider a target that is heading away from you on a constant relative bearing.
You are right that your course and the target's course no longer form a triangle. But if the target is remaining on a constant bearing, then you both would have occupied a single point behind you (think of a collision in the past). The lines from the stern do form a triangle. The problem is still workable, only we must now calculate the angles A' and B'. A'=180-A, and B'=180-B. Therefore, the target's speed is... Target speed = Own speed [ sin (180° - B) / sin (180° - A)
In the above example, A=150, B=80, and the target is traveling at 3.9 knots.
Does that help? |
YES!! Thank you VERY much!
I will try this in the game to digest well!
I think you should include this somewhat "advanced" (?) example in the already well-written tutorial to make it yet better!