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XLjedi · 06-13-08, 08:15 PM

The longer manual solution would look like this...

Now here you can see the parallel line drawn from the "r" position. In this case we're using the actual maneuvering board graphic to represent the speed vector. The "r" is on the 6th ring which in 2:1 would represent 12kts.

A parallel line to the DRM is drawn from the "r" position to the 9th ring which represents 18kts.

With that line you can plot point m1 which intersects the parallel line at the speed ring for 18kts. Notice now how the line plotted from the center of the board to m1 is pointing to the 262° bearing. So for an 18kt speed, "Cruiser A" must take course 262° to intercept the M2 station.

I don't recommend you actually use MoBo like this, just because it's faster to use the intercept tool; but you certainly can if you want to. Incidentally, this same plot example appears on page 11-16.

06-13-08, 08:15 PM	#7
XLjedi Ace of the Deep Join Date: Jun 2005 Location: Palm Beach, Florida Posts: 1,200 Downloads: 53 Uploads: 8	The longer manual solution would look like this... Now here you can see the parallel line drawn from the "r" position. In this case we're using the actual maneuvering board graphic to represent the speed vector. The "r" is on the 6th ring which in 2:1 would represent 12kts. A parallel line to the DRM is drawn from the "r" position to the 9th ring which represents 18kts. With that line you can plot point m1 which intersects the parallel line at the speed ring for 18kts. Notice now how the line plotted from the center of the board to m1 is pointing to the 262° bearing. So for an 18kt speed, "Cruiser A" must take course 262° to intercept the M2 station. I don't recommend you actually use MoBo like this, just because it's faster to use the intercept tool; but you certainly can if you want to. Incidentally, this same plot example appears on page 11-16. __________________ Last edited by XLjedi; 06-13-08 at 08:56 PM.