I'm building a little tower-defense game with some friends in Java. Now I'm assigned with the logic for towers and at the moment I'm trying to figure out how a tower has to turn to aim and hit a target monster. Because a monster moves on while the tower is turning and shooting, it needs to aim for a future position. I've implemented a function, which gives me the position of a monster at any time t and also a function, which gives me the smaller angle needed to turn to a monster, but now I'm confused because there are three unknown variables:
So I'm searching for a solution for:
min(t1+t2) = min(t3)
where the target monster is still in range of the tower. I already thought of calculating with the maximal needed turn and maximal possible range and then stepwise decrementing, but I'm curious if there is a "perfect" non-heuristic solution?
ADDED INFO:
I assume that a given monster has distance D to the tower, moves on the shortest path to the tower, and the tower starts turning towards the monster. This is the situation at t=0.
FIXED TYPOS:
If your tower turns with an angular speed omega, that is, the angle phi at time t is
phi = omega * t
So if you know that your tower has to turn an angle phi, the bullet will be shot at
t = phi/omega
From this on the distance which the bullet speed v has traveled is
s(t) = v * (t-phi/omega)
If your monster moves with a speed vm, the monster will be
at distance d
d(t) = D - vm * t
The bullet hits the monster if
s(t) = d(t)
This equation is easy to solve: just substitute d(t) and s(t) and rearange the terms for getting t:
t = (D + v * phi/omega) / (phi/omega + vm)
And the bullet will have traveled s(t) at this moment. If this value is negative, the monster was too fast and reached the tower before the bullet was fired