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nubse · 11-02-09, 02:12 AM

I've figured out a way to calculate the AoB (at least I think so) but you need to know the targets course. Here's an example:

Your course is 21 degrees
Your target is in your scope bearing 25 (that's at 46 degrees)
Your targets course is 245 degrees

Roughly you're heading NNE (between 0 and 90 degrees) while your target is heading WSW (between 180 and 270 degrees)

The Angle-on-Bow is a simple mathematical solution. You already have one angle, 25 degrees, which is the angularoffset of your periscope in relation to your course.

Since the AoB can not be more than 180 degrees and the angular sum of a triangle is 180 degrees, you can subtract 25 from 180, leaving you with 155 degrees left for the two other angles.

Because that we already have both the sub-course and the target-course, we can calculate the 2nd angle by adding to 90 degrees, the target-course offset from 270 degrees and the sub-course offset from 0 degrees, which is 90 + 25 + 21 degrees, so our 2nd angle is 136 degrees.

This leaves us with these results
Angle 1: 25 degrees (our periscope offset relative to our course)
Angle 2: 136
Angle 3 (AoB) = 180 - 25 -136 = 19 degrees. Jeez :p

Maybe a little more work can reduce the complexity of these reductions... other, maybe larger, triangles and compass-quadrants: A for Sub & B for Target: in my example, A is I & B is III and vice versa.

11-02-09, 02:12 AM	#2
nubse Nub Join Date: Nov 2009 Location: Denmark Posts: 4 Downloads: 50 Uploads: 0	AoB calculation I've figured out a way to calculate the AoB (at least I think so) but you need to know the targets course. Here's an example: Your course is 21 degrees Your target is in your scope bearing 25 (that's at 46 degrees) Your targets course is 245 degrees Roughly you're heading NNE (between 0 and 90 degrees) while your target is heading WSW (between 180 and 270 degrees) The Angle-on-Bow is a simple mathematical solution. You already have one angle, 25 degrees, which is the angularoffset of your periscope in relation to your course. Since the AoB can not be more than 180 degrees and the angular sum of a triangle is 180 degrees, you can subtract 25 from 180, leaving you with 155 degrees left for the two other angles. Because that we already have both the sub-course and the target-course, we can calculate the 2nd angle by adding to 90 degrees, the target-course offset from 270 degrees and the sub-course offset from 0 degrees, which is 90 + 25 + 21 degrees, so our 2nd angle is 136 degrees. This leaves us with these results Angle 1: 25 degrees (our periscope offset relative to our course) Angle 2: 136 Angle 3 (AoB) = 180 - 25 -136 = 19 degrees. Jeez :p Maybe a little more work can reduce the complexity of these reductions... other, maybe larger, triangles and compass-quadrants: A for Sub & B for Target: in my example, A is I & B is III and vice versa. Last edited by nubse; 11-02-09 at 04:50 AM.