Best Answer:
If he's moving away from you or toward you it's easy.

The trick is when he's going in the cross-wise direction.

Let's say you're a Coast Guard cutter and spot a suspicious

vessel that is distance D away and moving at V knots on a

heading perpendicular to the line between your two boats.

Your boat can do W knots (W>V) at full throttle.

I would use a fixed coordinate system to keep things simple.

Let's call the origin your position when you first made

contact. Assuming he stays on course and doesn't see you

coming, then his coordinates are <Vt,D> where t is time

since contact. You would go in a straight line on

heading h to intercept so your coordinates would be

<Wtsin(h),Wtcos(h)>, where h is the angle relative to the

original line connecting the two boats.

At intercept time T both of the coordinates would match, so

WTsin(h) = VT so sin(h) = V/W

WTcos(h) = D so cos(h) = D/WT

We solve for T by squaring both sides and adding equations

sin(h)^2 + cos(h)^2 = 1 = (V/W)^2 + (D/WT)^2

so T = D/Wsqrt(1-(V/W)^2)

Therefore to catch this guy in the minimum time possible,

you would go W knots at a heading of arcsin(V/W), and expect

to board his ship at time T=D/Wsqrt(1-(V/W)^2). Total distance

covered by your pursuit will be WT=D/sqrt(1-(V/W)^2).

It gets more complicated if he changes course, so you'll have to

make adjustments when they happen. But assuming that you

always can know his position, heading, and speed, you can

repeat the above calculation on the fly to adjust your own

heading and estimated time to board.

Source(s):