RL Bot — Aerial Kinematics
Solving for the orientation and boost a Rocket League car needs to reach a target point in the air at a given time.
For the bot to hit an aerial it has to juggle a lot of factors at once: ball height, ball speed, car speed, distance, angle, remaining boost. The idea is simple — point the car at the right place and boost — but getting it robust is fiddly. These notes set out the underlying kinematics.
Define the quantities of interest:
- \(c(t)\) — the car’s position
- \(f(t)\) — the unit vector in which the car is facing
- \(B(t)\) — acceleration due to boost
- \(P\) — the target location
All positions and directions are world vectors (X, Y, Z).
While airborne, the only accelerations available are boost (along the facing direction) and gravity, so the car’s motion is governed by the differential equation:
\[\ddot{c}(t) = B(t)\, f(t) + g\,\hat{z}, \qquad \dot{c}(0) = v_0,\quad c(0) = x_0,\quad f(0) = f_0\]\(\hat{z}\) is the unit vector in the z-direction; in Rocket League \(g\) is negative (downward) with magnitude \(650\frac{uu}{s^2}\).
Suppose we want to arrive at \(P\) at time \(T\), and we start solving at elapsed time \(\Delta t\). Integrating twice with constant boost acceleration \(B_0\) along a fixed direction \(f^*\), and rearranging for that boost term, gives:
\[B_0 f^* \;=\; \frac{2}{(T-\Delta t)^2}\left(P - x_0 - v_0\,\Delta t - \tfrac{1}{2} g\, \Delta t^2\, \hat{z}\right) \;=\; \bar{A}\]\(T\) is the total time at which we want to arrive at the target; \(T - \Delta t\) is the time remaining.
Both sides of this equation have units of acceleration — so the right-hand side \(\bar{A}\) can be read as the acceleration required to be at target \(P\) in the time remaining, after accounting for where the car already is, where its current velocity is carrying it, and what gravity will do along the way.
From there the solution separates neatly into a magnitude and a direction:
\[B_0 = \lVert \bar{A} \rVert, \qquad f^* = \frac{\bar{A}}{\lVert \bar{A} \rVert}\]The Simple Answer
Align the car with the required direction \((f = f^*)\) and feather the boost so the average acceleration matches \(B_0\). Because the ball — and therefore the target — keeps moving, this isn’t solved once: the bot re-evaluates \(\bar{A}\) every tick and keeps re-aligning, so the aerial is really a continuous correction loop rather than a single launch calculation.
Two practical checks fall straight out of the maths before committing to an aerial at all:
- If \(B_0\) exceeds the boost acceleration the car can actually produce, the target is unreachable in the time available — don’t jump.
- If \(f^*\) points somewhere the car can’t rotate to face quickly enough, the time budget needs to include the reorientation, or the aerial will arrive late.
Related: aligning shots on target — because arriving at the ball is only half the job.