Checking if your interception course is right

[kstanb]

Originally Posted by Nathaniel B.

TBH, I've never played any version of SH, so I can't answer that.

No worries, still any chance you can post a clean version of the board?

Also, if you have time, would you solve the following scenario in the maneuver board:
same idea as before, the contact is at 20 degrees heading, 20K meters away, course is 100 and the speed is 8
your maximum speed is 15 knots, your cruise speed is 12 knots

how would you intercept it, optimally, if you want to be at least 8km away until you reach your desired periscope position (this so that it won't spot you during your cruise?

thanks again!!

[palmic]

Originally Posted by kstanb

No worries, still any chance you can post a clean version of the board?

Also, if you have time, would you solve the following scenario in the maneuver board:
same idea as before, the contact is at 20 degrees heading, 20K meters away, course is 100 and the speed is 8
your maximum speed is 15 knots, your cruise speed is 12 knots

how would you intercept it, optimally, if you want to be at least 8km away until you reach your desired periscope position (this so that it won't spot you during your cruise?

thanks again!!

When you already see your target and you know their course, you dont need to calculating anything.
Just set the same course with higher speed and put him behind enough to have time to prepare and then set course to get closer and closer to his track.
When you think its close enough, turn the sub to have perpendicular course to its course to prepare sub to best firing position and tune distance from periscope depth as he is closing..
(for instance if their course is 100, you'll have to prepare sub to course 10 and wait until he will be before you)



Until you see your target
So here you have this triangle:
AoB (angle C) = 100 degrees (angle between his course and the bearing of 20)
his speed (side b) = 8
your speed (side c) = 12


What you want here is the angle between the bearing which you see the target and your desired course (angle B)
The law of sines for angle B here is that:
b/sin(B) = c/sin(C)
8/sin(B) = 12/sin(100)
8/sin(B) = 12/1
8/sin(B) = 12
sin(B) = 8/12
sin(B) = 0.7
B = 45 degrees

So your desired course to meet your target by the speed of 12 kts is the current absolute bearing + 45 => 20 + 45 = 65.

You can see here that since you have your target with AoB about 80-100 degrees, you can calculate the sine of angle between the bearing and desired course simply as his speed / your speed.
Its about 10 seconds to get your desired course..
You can round sines to whole numbers, theres no need to be very precise since your input data are also rough. Try it and youll see it works.

[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.

[palmic]

Nathaniel B.: I belive you did something wrong.
If the target is at 20° (NNE) and its course is 100 (EES). approach course could never be 314° (WWN), its just opposite..

[Sean C]

Originally Posted by palmic

Nathaniel B.: I belive you did something wrong.
If the target is at 20° (NNE) and its course is 100 (EES). approach course could never be 314° (WWN), its just opposite..

314° is not an approach course in my example. It is what the bearing of the target will be when we reach the closest point of approach while keeping 8km distant.

[Delgard]

Like in doing a back azimuth from a known point. So many known points out in the middle of the Atlantic.

[palmic]

Ok, thanks for info

[Sean C]

Originally Posted by Delgard

Like in doing a back azimuth from a known point. So many known points out in the middle of the Atlantic.

I understand you were just trying to be funny. But, just to ensure there is no confusion on the part of anyone reading this, I feel I should explain:

The CPA or closest point of approach is not a specific point relative to the ocean. It is the point at which two ships will be closest to each other relative to their respective positions, speeds and courses. The CPA is described by a distance and bearing from each ship.

It is the job of the navigator (and ultimately of the captain) to know exactly where other ships and objects are in relation to their own ship in order to avoid a collision. Calculating a CPA is the first step in determining if a danger of collision exists. It is one of the easiest tasks to perform on a maneuvering board. For example: if the line drawn through successive plots leads directly to the center, you are on a collision course with the object being plotted.

In fact, you don't even need the maneuvering board to realize this. If the bearing of an object does not change over time, but the range continually decreases, then the CPA is 0 yds. In other words: you are going to collide. It's really just common sense.

And actually, there are many known points in the middle of the Atlantic. Technically there are an infinite number of them. For example: 27°31.7' N, 43°12.1' W. The main job of the navigator is to determine at which of these points the ship currently is. Modern tools such as GPS and ECS make this job a snap. But more traditional methods such as celestial navigation at sea and bearings via pelorus or compass and soundings during coastal navigation were good enough to get the job done for hundreds of years.

However, it is best practice to use all available sources of information to determine one's position. Only a fool would go to sea without a GPS today, relying only on celestial navigation. And, more than one accident has been caused by complacency and over-reliance on a GPS or ECS which may be inaccurate for various reasons. Information from one source should be continually cross-checked with information from another. Sometimes it's even as simple as lifting one's head from the glowing display and looking out the window.

[LTJGBeam]

Originally Posted by Nathaniel B.

I understand you were just trying to be funny. But, just to ensure there is no confusion on the part of anyone reading this, I feel I should explain:

The CPA or closest point of approach is not a specific point relative to the ocean. It is the point at which two ships will be closest to each other relative to their respective positions, speeds and courses. The CPA is described by a distance and bearing from each ship.

It is the job of the navigator (and ultimately of the captain) to know exactly where other ships and objects are in relation to their own ship in order to avoid a collision. Calculating a CPA is the first step in determining if a danger of collision exists. It is one of the easiest tasks to perform on a maneuvering board. For example: if the line drawn through successive plots leads directly to the center, you are on a collision course with the object being plotted.

In fact, you don't even need the maneuvering board to realize this. If the bearing of an object does not change over time, but the range continually decreases, then the CPA is 0 yds. In other words: you are going to collide. It's really just common sense.

And actually, there are many known points in the middle of the Atlantic. Technically there are an infinite number of them. For example: 27°31.7' N, 43°12.1' W. The main job of the navigator is to determine at which of these points the ship currently is. Modern tools such as GPS and ECS make this job a snap. But more traditional methods such as celestial navigation at sea and bearings via pelorus or compass and soundings during coastal navigation were good enough to get the job done for hundreds of years.

However, it is best practice to use all available sources of information to determine one's position. Only a fool would go to sea without a GPS today, relying only on celestial navigation. And, more than one accident has been caused by complacency and over-reliance on a GPS or ECS which may be inaccurate for various reasons. Information from one source should be continually cross-checked with information from another. Sometimes it's even as simple as lifting one's head from the glowing display and looking out the window.

Originally Posted by Nathaniel B.

In fact, you don't even need the maneuvering board to realize this. If the bearing of an object does not change over time, but the range continually decreases, then the CPA is 0 yds. In other words: you are going to collide. It's really just common sense.

CBDR. Constant Bearing, Decreasing Range. Naval Shorthand...

[Tonci87]

You can also have a look at this.



Plotting an intercept course is not difficult and does not take much time

[Sean C]

Originally Posted by LTJGBeam

CBDR. Constant Bearing, Decreasing Range. Naval Shorthand...