# Math Help - [SOLVED] Finding an intercept bearing

1. ## [SOLVED] Finding an intercept bearing

Ok this one is a bit complicated to describe so i hope this works.

Code:
                           _
|\
\
_                        \ Vb
Va  /|                        \
/                           \
/ a                        b  \
A*----------------D--------------*B

I am sitting at point (A) while another is at point (B) wich is distance (D) away. The person at point (B) will travel in some direction (b) at velocity (Vb). What direction (a) must I travel in at Velocity (Va) to intercept this person?

With this in mind we will always know a few of the variables.

1) the x,y positions of both (A) and (B).
2) the distance (D)
3) the bearing of the target (b) and his velocity (Vb)
4) my Velocity (Va)
5) the bearing from (A) to (B). even tho this picture just shows them on an even plane, this bearing may not be due east weast as in this picture.

what i need to know is a formula that i can plug the current values of these variables into that will return me a value of (a). keep in mind this drawing is set in a nice flat isosceles triangle. however it does not have to be this way. the angle (b) may be acute or obtuse, and the target does not necessarily start on an even plane to me. what i need is a formula that is dynamic eneugh to handle any situation. i can see i have more than eneugh information to solve this but i cant seem to get it to work.

if it helps you to know the real world application it goes like this. i have a submarine who knows his GPS grid position. he has a sonar, and can therefore find the bearing and distance to his target. with this information he can calculate the target's GPS location. after he records a series of these locations, he can then calculate the target's bearing and speed. now what he needs to do is calculate a fire solution based on the fact that he will know the speed of the various torpedoes he can deploy. assuming his target will not alter his course or speed what angle must he fire his torpedo at to hit the target? of course he will know the speed of his own torpedo based on wich ever one he selects.

i can calculate the target's GPS locations ok, and then calculate his bearing and speed, however the next step never seems to work for me.

if anybody can help me with this it would be great. also if you can render down the final solution to it's simplest form, because the "computer" used to do this is not an electronic computer i can just type a complex formula into and let it solve it. it is a mechanical computer, and therefore the simpler the final formula is the bigger the chance of success. think WWII pre computer days. and you get the idea. if you do not understand mechanical computers do not worry, it can do most anything i need it to do. for instance it can do basic math, it can solve sin, cos, tan, arctan, as well as a few other higher math functions. powers, square roots etc.

***** EDIT *****
hmm the BBS ruined the formatting on my drawing. any idea how i can get my ASCII drawing to retain its shape?

ok fixed the formatting issue. so if you can take a crack at a solution it would be great

2. displacements in the y-direction will be equal ...

$v_a\sin(a) \cdot t = v_b\sin(b) \cdot t$

$\sin(a) = \frac{v_b \sin(b)}{v_a}$

$a = \arcsin\left(\frac{v_b \sin(b)}{v_a}\right)$

if you need to know the time ...

$v_a\cos(a) \cdot t + v_b\cos(b) \cdot t = D$

$t[v_a\cos(a) + v_b\cos(b)] = D$

$t = \frac{D}{v_a\cos(a) + v_b\cos(b)}$

3. Actually this is much simpler than i was making it. if i always take the bearing to the target and then figure his course as a delta from that as you say the "Y" value will always remain the same. this makes the equations so much simpler. glad i posted here to get this fresh point of view. i was stuck in my original frame of mind and could not seem to break free.