Three radar points, one parabola, one root and one launch time. Predict where each rocket lands and protect the cities with as few interceptors as possible.
Everything the game does on the right-hand panel comes from these seven ideas.
Ignoring air resistance, the horizontal speed stays constant and gravity pulls down with constant acceleration g:
Solve the first equation for t and substitute into the second: y becomes a second-degree function of x. The path is a parabola.
A parabola y = ax² + bx + c has three unknowns: a, b and c. Each radar echo (x, y) gives one equation. Three echoes give a 3×3 system with exactly one solution.
That is why the game waits for 3 echoes before drawing the dashed prediction.
The rocket hits the ground when y = 0. Solve ax² + bx + c = 0:
One root is where it was launched; the other is the impact point. Bonus fact: the vertex (highest point) is exactly halfway between the two roots.
To intercept at a point P, the interceptor must arrive at P at the same instant as the rocket. If P is d metres from the launcher and the interceptor flies at speed v:
If this time is already in the past, it is too late for that point. Choose a point further ahead on the path.
Interceptors are limited and expensive. Because the impact point can be predicted seconds in advance, there is no need to fire at rockets that will land in empty areas. Prediction turns a defence problem into an optimisation problem: protect the most with the least.
From wave 6, hypersonic rockets fly about twice as fast and maneuver in mid-flight. The 3 echoes still give an exact parabola, but it describes the path BEFORE the maneuver. After it, the prediction is wrong and the radar needs 3 new echoes.
With less time left, d/v may no longer fit: the launch time tP − d/v is already in the past.
Light travels at c = 300,000,000 m/s. To cross 600 m it needs 600 / 300,000,000 = 0.000002 s. In the formula, d/c is practically zero; what counts is the heating (dwell) time the beam must stay on the target.
