This is not an easy problem, to intercept a projectile with another one. You'd have to set there equations of motion equal to each other and figure out the missing info.

Or there's an iterative approach.

Lets say you have a function targetPath() which takes a parameter dt. targetPath() returns the position of the target after dt from it's intial position. so targetPath(0) is where the target is now, and targetPath(1) is where the target will be in 1 second. myFlightTime() takes a parameter target which then returns the time required to reach the target.

const float distThreshold = missileRadius; //interception accuracyfloat flightTime = 0;Vector target = targetPath(0);Vector error;for (;;){ //compute flight time and new target location flightTime = myFlightTime(target); Vector targetFuture = targetPath(flightTime); //determine error error = targetFuture - target; if (error.Length() < distThreshold) break; //keep searching target = targetFuture;}//shoot at "target", and you'll intercept with "error" accuracy// (assuming these functions will converge :P)

The trajectory of the target (modeled as a point) would be
rA(t) := pA + t * vA
where it is assumed to move on a straight line.

Similarly, the trajectory of the bullet would be
rB(t) := pB + ( t - T ) * vB, T>=0, |vB| == vB
where T denotes the time when shooting as mentioned in the OP.

A perfect hit would be given if
rA(t) == rB(t)

Now, in 3D you have 4 equations
rA,x(t) == rB,x(t)
rA,y(t) == rB,y(t)
rA,z(t) == rB,z(t)
|vB| == vB
but 5 variables
vB,x, vB,y, vB,z, t, T
and hence cannot solve uniquely.

Fortunately this seems me no problem since you can choose T for whatever you like. For simplicity we can choose it as "now" (T==0), so that with the abbreviation
pAB := pA - pB
the equation gives
pAB = t * ( vB - vA )

Splitting this into the 3 scalar equations, so that
vB,x = ...
vB,y = ...
vB,z = ...
are given, they can be inserted into the 4th equation
vB,x2 + vB,y2 + vB,z2 = vB2

Without self having done so, it seems me that this is a quadratic equation of t (after expanding with t2 on both sides), and hence can be solved with the p,q-formula to a t1 and t2, from which presumbly only 1 value is positive. The negative value, however, requires a negative vB, what is valid due to the square but nevertheless is nonsense for the given use case.

Now that a single, specific t is determinded, rA(t) can be computed by setting that t in, and since rA(t)==rB(t) also vB can be computed.


Hopefully I made no mistake ... ;)