Constant Deceleration

To simplify matters for the moment, we will assume constant deceleration and ignore the contribution of wind resistance. We will also, for the moment, pretend that braking traction is independent of cornering traction. In reality, if the car is at , all of its traction is used up keeping it from sliding sideways; no traction would be available for braking.

The equations for position and velocity as a function of time under constant acceleration are

$$x(t) = x_0 + v_0t + \frac{1}{2}at^2$$ (21)

$$v(t) = v_0 + at$$ (22)

Since we're describing braking, the number that we plug in for $a$ will be a negative number. To get $v(x)$ we note that $v^2 = v_0^2 + 2v_0at + a^2t^2$ and that $2a(x - x_0) = 2v_0at + a^2t$. Substituting and solving gives

$$v(x) = \sqrt{v_0^2 + 2a(x - x_0)}$$ (23)

Here, $x_0$ is the position of the car when braking starts. For the remainder of the discussion we will set $x_0 = 0$ and interpret $x$ as the distance traveled since braking started. The initial speed $v_0$ is the car's speed when braking started. Our final drag-free braking equation is

$$v(x) = \sqrt{v_0^2 + 2ax}$$ (24)

Equation 24 defines the boundary between reachable and unreachable points in the $x$-$v$ graph for track positions ahead. This boundary curve must not exceed or the car will slide off the road. We can ensure that it does not by checking the curve during each timestep in the simulation. If the curve touches it's time to brake. Figure 8 illustrates the process.

Figure 8: Optimal speed through turns 4, 5, and 6 for a car that can accelerate and brake at 0.6g. At the exit of turn 4 (point A) the car undergoes maximum acceleration until it reaches in turn 5. At point C, the braking curve stays below , so the car maintains . Braking begins at point D to avoid exceeding in turn 6. Braking ends at point E. From here the car maintains until it can accelerate at the exit of turn 6.