[Sean C]

Sorry for the late response, I've been busy. Anyway, here is a solution on a maneuvering board:

Example 2

A few things to note first:

• There are two different plots here: one consisting of the "speed triangles" (emr and emr') and another consisting of relative positions (R ["reference" ship, or our own ship] and M ["maneuvering" ship, or the ship we are tracking])
• This assumes we want to arrive ahead of ship M and will be traveling at max speed (15 kts.)
• This assumes ship M does not change course or speed
• Speed scale is 2:1 and distance scale is 3:1
• All distances are in yards (20km ≈ 22,000yds.; 8km ≈ 8,750yds.)

Now, here are the steps for plotting this solution:

1. Plot the position of M at 22,000yds, 20°.
2. Draw a circle around own ship (R) with radius 8,750yds., the distance we want to keep from M
3. Draw a line from M, tangent to the 8,750yd circle (note that if we drew the tangent to the other side, we would end up behind ship M). This is the desired relative movement line. By inspection, we see that the direction of relative movement (DRM) is 223.6°
4. Draw a line (em) representing M's course and speed (100°, 8 kts.)
5. From the end of em, draw a line (rm) parallel to the DRM which stops at our desired speed of 15 kts. (the 7.5 circle at 2:1 scale)
6. Draw line er to find the required course at 15 kts. of 70°. The length of rm is the speed of relative movement (SRM): ≈9.2 kts.
7. Draw a perpendicular to the DRM which intersects R at the center. The point where this line intersects the 8750yd circle is the closest point of approach (CPA): 8750yds. @ bearing 314°
8. Using the 3:1 scale, measure the distance from M to the CPA: 20,000yds.
9. On the nomogram, draw a line from 9.2 kts. through 20,000yds. to find the time to CPA: ≈66 minutes

Now, if (for example) we then want to turn to intercept M at a speed of 12 kts., we can do the following:

1. Draw a line (r'm) parallel to M's bearing at the CPA (314°) which stops at the 12 kt. circle (6 at 2:1 scale)
2. Draw line er' to determine the required course at 12 kts.: 335.5°
3. The length of r'm is the SRM: ≈18.0 kts.
4. On the nomogram, draw a line from 18.0 kts. through 8,750yds to find the time to intercept: ≈15 minutes

But, as you asked earlier, how did we determine that M's true course and speed was 100° @ 8 kts. in the first place? Take a look at this example:


Example 3


Suppose we first spotted M (by radar or other method) at 1600hrs., bearing 05.8°, range 26,100 yds. Six minutes later, we spot M at 12.2°, 23,900 yds. Another six minutes go by and we spot M at 20°, 22,000 yds.


We plot these positions on the maneuvering board, draw a line through them and find that M has a DRM of 136.5°. In six minutes time (one tenth of an hour), M has moved about 3,500 yds., so we draw a line on the nomogram from 6 minutes through 3,500 yds. to find that M has an SRM of 17.6kts.


Now, our own ship is moving on a course of 340° at 12kts., so we draw line er to represent that. We then draw line rm parallel to the DRM with a length matching 17.6kts. And finally, we draw line em and find M's true couse and speed of 100°, 8 kts.


Pretty much any maneuvering problem can be solved quickly and intuitively on a maneuvering board, once you become familiar with how it works. The manual I linked to earlier even shows how to solve complex problems, such as maneuvers involving multiple ships at one time with different speeds and distance constraints.