ANGLE ON THE BOW: A DEFINITION

[Daniel Prates]

SUBSIM's Angle on Bow guide (https://www.subsim.com/tips/wolfpack/aob.htm)


I've been reading some conflicting things about the 'angle on the bow' concept.

In my perception, target solution can be acheived if you have (besides some more data which are irrelevant to my questioning) at least two angles inside a triangle. Actually the equation demands all three angles inside the a triangle, but since all angles inside a triangle adds up to 180 degrees, if you have two, the third can be easily deducted. Check the drawing below: it lists two angles, being them the AOB and "deflection". Cool, everybody knows that. The TDC calculates the firing solution by the means of such equation, where all its data depicited below is supplied either by you (range etc), or by the machinery itself (such as bearing).

Thus:

http://upload.wikimedia.org/wikipedia/commons/1/1d/FiringGeometry.png (http://upload.wikimedia.org/wikipedia/commons/1/1d/FiringGeometry.png)


Anyway, everybody knows this. This is not the point. The thing is: Is the AOB the angle depicted above (that is, made between the ships course and your line of sight), OR RATHER, IT IS THE ANGLE FORMED BETWEEN YOUR COURSE AND THE TARGET'S COURSE?

In my point of view, AOB is a constant angle (presuming nobody changes course). This works if we define AOB as being the angle between your course, and the targets course notwitstandingly the passage of time. But if the AOB is the angle between your line of sight (or line of bearing) and the ship's course, well... then you have a constantly-changing angle.

How do you possibly input an AOB that is changing all the time? don't you have to input a constant AOB?

So this is my question. For you, which angle is the AOB? The angle depicited on the above drawing, or the angle to the top-left, that is, the interception angle between your course and the target's course?

SUBSIM's Angle on Bow guide (https://www.subsim.com/tips/wolfpack/aob.htm)

[razark]

How do you possibly input an AOB that is changing all the time? don't you have to input a constant AOB?
You input what it is at the time you observe it. The TDC keeps track of it over time.

So this is my question. For you, which angle is the AOB? The angle depicited on the above drawing, or the angle to the top-left, that is, the interception angle between your course and the target's course?
The AoB is the angle between the target's course and the target's line of sight to your position. On the drawing, it's listed as theta[bow] at the top-left corner. The top-right is the intercept of your course and the target course. If you use that angle, you will be using the wrong value.

[Rockin Robbins]

Angle on the bow is nothing more than your bearing from the target's perspective, expressed as an angle from 0º at the bow and 180º off the stern and a designation starboard or port to separate the two 180º ranges.

So if a watchman on the target sees you 90º off his starboard bow, your AoB is 90º starboard.

From your point of view, if you're looking at the target's starboard side, the AoB is also starboard.

[Daniel Prates]

Great, everybody. Thanks a lot. I'be been doing it differently, but with little impact on precision since I always tend to shoot as close as possible - but clearly i've been doing it wrong! Thanks a lot.

[kajonas]

The AoB is the angle between the target's course and the target's line of sight to your position. On the drawing, it's listed as theta[bow] at the top-left corner.

This last paragraph... PERFECTO! :Kaleun_Wink:

(keep in mind YOU are Submarine in diagram, and target is Target)

[yubba]

From a video I watched ,,angle on the bow is where the other guy is looking at you from his bridge if he could see you.

[ETR3(SS)]

From a video I watched ,,angle on the bow is where the other guy is looking at you from his bridge if he could see you.:sign_yeah: That's what I was taught as a lookout in the bridge. A much simpler way of explaining it.