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Old 11-07-17, 05:34 PM   #6
palmic
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Default Real navigation - Perfectly controlled approach

Quote:
Originally Posted by Xall View Post
Palmic,
Thank you for your post, I been reading this and trying to figure it out.

Can you healp me with my situation?

So my case is the following:

I got a contact report, that a convoy is heading SW, so true course 225.
I made visual contact while on course 210,
and using the Raofb I estimated the distance to the target to be around 11,000 meters (11Km.)
The targets bearing is 274.

Using your formula:

Step 1.
225 (Targets Course) - 210 (Our Course) = 15 [Angle A]

And here is where I am getting issues:

Step 2.
----------- Our course 210 - target bearing 274 = -64 [Angle B] -----------

If i continue your method with this negative -64 I get:

Step 3.
180 - 15 [Angle A] - 64 [Angle B] = 101 [Angle C]

But this is incorrect since it does not check out.
Using the Attack Disk to determine the True course of the Convoy,
I entered the AOB we plotted, and the course does Not check out to 225

I get Course 203 instead.

Setting the Attack disk to the true course of the target (225) I get an AOB of 79.

So in Step 3 I should get an AOB of 79, but I believe I am doing something wrong in Step 2.


If I do 210 (Our Course) PLUS 360 MINUS 274 (Target Bearing)
I get = 296

If I do Step 3 then:
180 - 15 - 296 = -131

Something is off and I need help with it.

Hopefully you know where I am going wrong



Edit:
Is it possible that in step 2. I need to have my course (210) then add 360, and then subtract the targets bearing (274) = 296 and convert this into degrees by doing 360 - 296 ?= 64
Because then it makes sense, 180 - 15 - 64= 101.

I just don't understand why the attack disc does not add up to course 255 if i use 101 as the aob... Maybe I am doing something wrong on the attack disc, but can anyone confirm that I am doing step 2 correcty?
Basically I am getting lost when my course is a smaller value when subtracking a larger bearing value, what do I do then? Add 360 before subtraction like in step 1 (Angle A)?

Please reply...


Hi Xall, sorry for my delay, i noticed today....

You are very close, but you are making mistakes, because you have no experiece with this, thats ok, this way of conversation is not very good for this teaching, but as i wrote, you are very close....

Just to clarify - what you are doing in step 2 is getting angle in triangle whose angles are crosses of your tracks (angle A - or angle of tracks - that you have right from step 1) and the second angle - at your current position, which you are calculating in step 2 from your course - 210 and target bearing 274 - thats just half right.
The third angle is obviously at position of your target.
That means no negative angles are possible here (no triangle has any negative angle, they are just differences - use always absolute value from subtraction.
That being said, we can continue:

Your mistake is you are using your relative bearing to target (274), thats wrong, its absolute bearing to target (your course + relative bearing), because when you look on that equation, your course is absolute value at map and your target course also. You cannot mix into equation relative bearing...
You just have to imagine that triangle at map, maybe draw it to sh5 map, youll get what you are doing wrong...

===
So Your input data for step 2 is your course 210 and absolute bearing to target.
Absolute bearing => 210+274=484 (-360=124)
210-124=86 => thats your second angle. (Honestly i do this in few seconds in my head - just look at your bearing scale - its your 360 - bearing 274 = 86.
(270 is 90 to the left from 360 yes its so simple! :-) )
Always keep this 4 bearing in head and train it with experience - 90,180,270,360, this way youll get much better situation awarnes just like that in your mind and youll start to calculate this not like 360-274, but like 4 degrees from 270, which is your relative "west" - 90 degrees on left..

Third angle (AoB) => 180-86-15=79.

Youre done sir


Bottom line:
I know it looks dificult at the begining, but its realy most simple and quick way how to get this. 99% of that post is just to clarify the facts which you have to absorb by 3-10. successful solutions.
This everything is just repeating of few little rules. Just keep it simple and dont try to be digitaly exact. It is realy not needed.
Most important things:
- if you and your target dont change your course, and you are quick enough to move them (or simply just their smoke 1 degree behind at your horizon, you've just managed to change their AoB by 1 degree - 1:1)
All inner angles of triangle are always 180 - ever. So just keep in mind triangle is always angle a+b+c and its always 180.
So if you can get difference of your courses (like you did) and you know where they are at your bearing, you have even the 3rd. - 180 - both of them.
- Their distance from you at 90AoB if they dont change their course will always be their current distance * sin(current Aob).

And my golden rule: Most of the times i always have my target about 10km away. Ideal firing distance is about 1/10 - 1km away.
So I would like to get my target into situation, where i submerge into periscope depth and wait for them to be about 1 km away before me.
1/10 is always sin(6), so i know i ALWAYS want to get before them into 6 degrees AoB.
And that ALWAYS means bearing where i want to see their smoke/hull is always 180-6-angleOfTrack (angle between our courses).

So from your example it would be 180-6-15=159 on your port.
159 on your port is your 360-159=201 (simply 159 degrees to the left from your 360).

So if youll manage to get them into your 201, submerge and turn your sub into course 225-90=135 (because its their course - right angle) youll have absolutelly perfect firing position.
Theyll be their current distance / 10 and they will be right before you just perpendicular to you - perfect 90 AoB.

This is something which youll master very quickly by your own experience, just go step by step, example by example, just finish your first and second example next time, i swear youll get into this by your own experience very quickly. Again - This seams to be complex, but its not, its just repeating of very simple triangle equations from elementary school which we forgot...

Last edited by palmic; 11-08-17 at 05:56 AM.
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