SUBSIM Radio Room Forums - View Single Post - Checking if your interception course is right

palmic · 02-20-16, 10:47 AM

You can do it even in old fashion way (like real sub captains did) with good ol' paper and pen like this:

As your approach to the ship is triangle (as basically everything what you will be solving in u-bote

), you can describe it like triangle ABC where A is your position, B is collision point and C is current position of your target.
In this triangle opposite side to target position (C) is side c - your path to collision point (B).

Your task here is to solve angle at point A (your position) when you know target course and speed and you pretend you want to choose your speed.

In this example lets say target course is 100 and speed is 8kts

Your REAL bearing at which you hear it is 20. (real bearing means bearing at hydrophone + your current course)
The speed we want to maintain is 12kts (it means we want to flow at medium speed at surface).

So what we have here is triangle ABC with a=8, c=12. A=?, B=?, C=100-20=80 degrees

The law of sines says that we now have all mandatory input to find our desired A angle which solves our desired course to collision with the ship.
Its: "If you know either (1) or (2), you can use the law of sines to solve the triangle." We actually knows (1).

So: a/sin(a) = b/sin(b) = c/sin(c).
This means:
1) a/sin(A) = c/sin(C)
2) 8/sin(A) = 12/sin(80)
3) 8/sin(A) = 12/0.98
4) 8/sin(A) = 12.24
5) sin(A) = 8/12.24
5) sin(A) = 0.65
6) A angle is 40.6 degrees
7) our desired course has to be real target bearing + A angle => 20+40.6 = 60.6!

So if you have good target speed and course and you want to crash that ship directly to bow, order your navigator course 61 and wait few hours until you see the vessel!

If you want to be there in advance, just calculate with position of the target the same time in advance (when you'll get into this, you'll start to use trigonometry to find out at which bearing you'll here it next hour pretty simply)

HINT: If you want to re-orient the equation at 1) to solve the angle at once, its like that:
sin(A) = target_speed / your_desired_speed * sin(angle_on_bow (C) ) (which is basically equation to solve it all!!!)

Your task now her Kaleun is to solve the distance you will have to reach to collision point by the law of sines or law of cosines.
This means you'll have available even time to reach the target. And then you could start to grow your beard for sure

(and you can even stop to use navigator fixes, cuz you'll have better precision

)
For this, we need to have another input data and thats current distance to target at that bearing we hear it. Let's just say its 20.000 meters for now.
What will be distance to collision point?

BTW: As every inside angles of triangle summarized gives always 180 degrees, you can solve angle B pretty easily as 180-(60.6+80)=39.4

If you tried to attack ship in non-right angle by torpedo, you should already know, that its basically angle of track which could solve your torpedo solution if you are already in attack phase!
So this simple equation could literally gives you TDC inputs solution at max realism!

02-20-16, 10:47 AM	#2
palmic Grey Wolf Join Date: Jan 2014 Location: 50.1° N, 14.4° E Posts: 834 Downloads: 82 Uploads: 5	You can do it even in old fashion way (like real sub captains did) with good ol' paper and pen like this: As your approach to the ship is triangle (as basically everything what you will be solving in u-bote ), you can describe it like triangle ABC where A is your position, B is collision point and C is current position of your target. In this triangle opposite side to target position (C) is side c - your path to collision point (B). Your task here is to solve angle at point A (your position) when you know target course and speed and you pretend you want to choose your speed. In this example lets say target course is 100 and speed is 8kts Your REAL bearing at which you hear it is 20. (real bearing means bearing at hydrophone + your current course) The speed we want to maintain is 12kts (it means we want to flow at medium speed at surface). So what we have here is triangle ABC with a=8, c=12. A=?, B=?, C=100-20=80 degrees The law of sines says that we now have all mandatory input to find our desired A angle which solves our desired course to collision with the ship. Its: "If you know either (1) or (2), you can use the law of sines to solve the triangle." We actually knows (1). So: a/sin(a) = b/sin(b) = c/sin(c). This means: 1) a/sin(A) = c/sin(C) 2) 8/sin(A) = 12/sin(80) 3) 8/sin(A) = 12/0.98 4) 8/sin(A) = 12.24 5) sin(A) = 8/12.24 5) sin(A) = 0.65 6) A angle is 40.6 degrees 7) our desired course has to be real target bearing + A angle => 20+40.6 = 60.6! So if you have good target speed and course and you want to crash that ship directly to bow, order your navigator course 61 and wait few hours until you see the vessel! If you want to be there in advance, just calculate with position of the target the same time in advance (when you'll get into this, you'll start to use trigonometry to find out at which bearing you'll here it next hour pretty simply) HINT: If you want to re-orient the equation at 1) to solve the angle at once, its like that: sin(A) = target_speed / your_desired_speed * sin(angle_on_bow (C) ) (which is basically equation to solve it all!!!) Your task now her Kaleun is to solve the distance you will have to reach to collision point by the law of sines or law of cosines. This means you'll have available even time to reach the target. And then you could start to grow your beard for sure (and you can even stop to use navigator fixes, cuz you'll have better precision ) For this, we need to have another input data and thats current distance to target at that bearing we hear it. Let's just say its 20.000 meters for now. What will be distance to collision point? BTW: As every inside angles of triangle summarized gives always 180 degrees, you can solve angle B pretty easily as 180-(60.6+80)=39.4 If you tried to attack ship in non-right angle by torpedo, you should already know, that its basically angle of track which could solve your torpedo solution if you are already in attack phase! So this simple equation could literally gives you TDC inputs solution at max realism! Last edited by palmic; 02-20-16 at 11:50 AM.