[makman94]

Originally Posted by TorpX

@ makman94,

Unless I miss my guess, we are using different terms to arrive at the same result. I was puzzled by your "lb" variable, but looking at it, it seems like this is the same as the track angle. This is what the U.S. Navy calls the angle measured from the target's track to the sub's track, starting from ahead of the target. This means:

sin Aob = sin (Atr - Abg) [where Atr is the track angle]

I will provide an example to illustrate:

ship length.............320 ft.
time......................29 sec.
sub speed.............3.78 kts.
bearing angle.........100.5 deg.
angle on bow...........25 deg.

*target speed...............15.30 kts.

With the above example the track angle is 125.5 deg.
Tell me if you get a different answer.
(The numbers are odd because I drew this test case out on graph paper to check this.)

To be honest, I've barely touched the TDC lately. I've been using manual fire control without the TDC. (S-boat style)

Bye

hi TorpX

yes , TorpX ...with these data the calculated speed is ,indeed, 15,3kts.

and yes ...angle lb is , at your example, 125.5 degrees(i named it that way...meaning the angle owncourselinetargetbowdirection....look at pic 1 to see what angle i name as lb....maybe this angle is called track angle ...don't know)

now, what i told you at the previous message is that the 25 degrees angle (at your example) is NOT trully the AoB but an angle that its value is very close to the AoB (the AoB at the time you start your stopwatch).look at pic 1 , when you say that bearing angle is 100,5 degrees you are talking for angle φ (this angle is not changing during the procedure) .when scope is at this angle(φ).... the 25 degrees angle is ,in fact, the angle θ and NOT the AoB (meaning the AoB at the momment of starting the stopwatch...look at pic 2...at AoB1).
the angle θ (also is not changing during the procedure) is the angle that you are using in your calculations(and leads to the above formula) and this angle can be found only through the angle lb (which ,also is not changing but this means that we have to know the target's course relative to our course).from pic 1 : θ=lb-φ ( θ = 125,5 - 100,5 = 25 degrees)

as you can see from pic 2 ,the AoB is changing during the procedure.The only situation that AoB is not changing is when you are on a collision course...but ,in this case, the target would never 'pass' your scope's vertical line

ps1: the back side of attack disc is(one of its uses) for helping you to get target's with the above data given.no tdc....etc
ps2: i have upload at my FF page (look at my sig) the proof for the spoken formula for you and whoever else is interest in.

bye

PIC1:

PIC2: