ANGLE ON THE BOW: A DEFINITION

Daniel Prates · 06-09-11, 03:00 PM

SUBSIM's Angle on Bow guide

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:

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

06-09-11, 03:00 PM	#1
Daniel Prates Grey Wolf Join Date: Apr 2011 Location: Curitiba, Brazil Posts: 938 Downloads: 65 Uploads: 0	ANGLE ON THE BOW: A DEFINITION SUBSIM's Angle on Bow guide 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: 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 Last edited by Onkel Neal; 02-14-20 at 11:38 PM.