Quote:
Originally Posted by palmic
i never got how this board works  |
Here is the solution to your example problem on a maneuvering board.
The red line (em) is the target's true course and speed: 100°, 8 kts. (The speed scale used is 2:1, so the line is drawn to the 4 circle.) The blue circle is our desired speed to intercept: 12 kts. The dashed purple line is the target's bearing: 20°.
We want to move in the direction of the target, so we draw a line parallel to the dashed purple line from the end of the targets course and speed vector: this is the solid purple line (rm), which is the desired
relative motion vector.
We then draw a line from the center (e) to where the rm line intersects our desired speed: this is the green line (er), which is our required course at 12 kts. As you can see, it is indeed 61°.
Now, if we measure the length of the relative motion line (rm), we can determine our
relative speed. If you look at the bottom of the 2:1 scale on the far left, you will see that the relative speed is about 7.6 kts.
Supposing the target is 20,000 meters away, that equals about 21,800 yards. Using the nomogram at the bottom, we draw a line from the relative speed (7.6 kts.) through the distance to close (21,800 yds.) and arrive at a time of about 85 min., or 1 hour and 25 minutes.
The actual distance traveled can also be solved on the maneuvering board, but I didn't want to clutter it up too much.