This optional companion to Chapter 6 derives the interpolation used by the pit route. Work through the equations, check the numerical example, then compare them with spline.cpp in the final checkpoint.
At each knot we know a longitudinal distance s, an offset d, and its slope \(m=dd/ds\). Between two consecutive knots, we want a function that matches both values and both slopes.
Distances and offsets are in metres. Slopes are metres per metre, hence dimensionless. Here s is distance along the track, not world X, and d is a lateral offset, not world Y.
Choose a cubic on one interval:
\[P(s)=a s^3+b s^2+c s+e.\]
Four coefficients let us impose four conditions:
\[\begin{aligned} P(s_i)&=d_i,&P(s_{i+1})&=d_{i+1},\\ P'(s_i)&=m_i,&P'(s_{i+1})&=m_{i+1}. \end{aligned}\]
We could solve that system directly, but a local normalized coordinate makes the algebra simpler and avoids carrying large absolute track distances through the polynomial.
Define:
\[h=s_{i+1}-s_i>0,\qquad t=\frac{s-s_i}{h},\qquad s=s_i+ht.\]
Now t runs from 0 to 1. Set \(Q(t)=P(s_i+ht)\) and write:
\[Q(t)=At^3+Bt^2+Ct+D.\]
The value conditions become \(Q(0)=d_i\) and \(Q(1)=d_{i+1}\). The derivative conditions need the chain rule:
\[\frac{dQ}{dt}=\frac{dP}{ds}\frac{ds}{dt}=h\frac{dP}{ds}.\]
Therefore \(Q'(0)=hm_i\) and \(Q'(1)=hm_{i+1}\). The supplied slopes describe change per metre; derivatives in t describe change over the normalized interval. Their units and numerical values differ.
Differentiate the normalized cubic:
\[Q'(t)=3At^2+2Bt+C.\]
At t = 0, two coefficients follow immediately:
\[D=d_i,\qquad C=hm_i.\]
At t = 1:
\[\begin{aligned} A+B+C+D&=d_{i+1},\\ 3A+2B+C&=hm_{i+1}. \end{aligned}\]
Let \(\Delta=d_{i+1}-d_i\). Substitute C and D:
\[\begin{aligned} A+B&=\Delta-hm_i,\\ 3A+2B&=h(m_{i+1}-m_i). \end{aligned}\]
Subtract twice the first equation from the second to obtain A, then substitute back to obtain B:
\[\begin{aligned} A&=-2\Delta+h(m_i+m_{i+1}),\\ B&=3\Delta-h(2m_i+m_{i+1}),\\ C&=hm_i,\\ D&=d_i. \end{aligned}\]
All four normalized coefficients have offset units, metres. The parameter t is dimensionless.
Check the solution by substituting t = 0 and t = 1 into Q and Q′. All four endpoint conditions should reappear.
Collect terms multiplying the two values and the two normalized slopes:
\[\begin{aligned} Q(t)={}&H_{00}(t)d_i+H_{10}(t)h m_i\\ &+H_{01}(t)d_{i+1}+H_{11}(t)h m_{i+1}, \end{aligned}\]
where:
\[\begin{aligned} H_{00}(t)&=2t^3-3t^2+1,\\ H_{10}(t)&=t^3-2t^2+t,\\ H_{01}(t)&=-2t^3+3t^2,\\ H_{11}(t)&=t^3-t^2. \end{aligned}\]
Each basis function supplies one endpoint condition while contributing zero to the other three. For example, H₀₀ has value 1 at t = 0, value 0 at t = 1, and zero derivative at both endpoints. H₁₀ has zero endpoint values, derivative 1 at t = 0, and derivative 0 at t = 1. Verify the other two in the same way.
The basis and coefficient forms are the same polynomial. The former makes the endpoint roles clear; the latter is convenient for evaluation.
Take \(s_i=0\), \(s_{i+1}=20\,\mathrm{m}\), offsets 0 and 5 m, and zero slopes at both ends. Then h = 20 m and:
\[A=-10\,\mathrm{m},\quad B=15\,\mathrm{m},\quad C=D=0.\]
Thus \(Q(t)=5(3t^2-2t^3)\,\mathrm{m}\). At s = 5 m, t = 1/4:
\[Q(1/4)=5\left(3\cdot\frac1{16}-2\cdot\frac1{64}\right) =0.78125\,\mathrm{m}.\]
At the midpoint, s = 10 m, the offset is 2.5 m. To recover the physical slope, divide Q′ by h:
\[\frac{dd}{ds}=\frac{Q'(t)}{h} =\frac{30t(1-t)}{20}.\]
The midpoint slope is 0.375 m/m, while both endpoint slopes are zero. Linear interpolation through the same endpoints would have a constant slope of 0.25 m/m.
The positive offsets in this example move toward the road's left. Negating the offsets and slopes gives the mirrored right-side construction.
The knots are stored in increasing distance order. interval uses binary search to find the adjacent pair surrounding the query. At an interior knot, either adjoining interval supplies the same value and first derivative when they share the same knot data; the implementation selects the interval to its right. At the final knot it uses the last interval.
Horner's scheme evaluates Q with nested multiplication:
\[Q(t)=((At+B)t+C)t+D.\]
The first derivative with respect to track distance is:
\[\frac{dd}{ds}=\frac{(3At+2B)t+C}{h}.\]
These expressions correspond directly to HermiteSpline::evaluate. The code computes the local interval coefficients, then selects the value or derivative. It does not solve a global linear system on every driving update.
With nonzero endpoint slopes, a constant extension changes derivative at the domain boundary. Our pit route uses zero endpoint slopes, so that boundary policy is consistent with its centerline/lane joins.
Sharing the value and slope at each knot makes the piecewise offset function \(C^1\): its value and first derivative are continuous. It does not require the second derivatives of adjoining intervals to agree.
For the normalized interval:
\[\frac{d^2d}{ds^2}=\frac{6At+2B}{h^2}.\]
Follow the worked interval with a second one from s = 20 to 50 m, bringing the offset from 5 m back to zero, again with zero endpoint slopes. At the join, both offsets are 5 m and both first derivatives are zero. The second-derivative limits are −0.075 1/m from the left and −1/30 1/m from the right.
This is piecewise cubic Hermite interpolation, not a natural cubic spline whose slopes are found by enforcing second-derivative continuity and boundary conditions. Both are useful constructions, but they impose different conditions.
Arbitrary supplied slopes can produce values outside the endpoint range. With zero slopes, each individual interval is a scaled smoothstep and is monotone between its endpoint values. That prevents offset overshoot on that interval, but it does not prove that a vehicle following the world-space route fits through the road and pit geometry.
Check the path between knots, not only at the knots. Account for the footprint, barriers, steering limits, and the speed at which the car attempts the turn.
The route uses the track centerline \(\mathbf c(s)\) and its left unit normal:
\[\mathbf r(s)=\mathbf c(s)+d(s)\mathbf n(s).\]
For a smooth planar centerline parameterized by arc length, let κ be its signed curvature and t its unit tangent. Differentiating gives:
\[\mathbf r'(s)=(1-\kappa d)\mathbf t+d'(s)\mathbf n.\]
This explains why a zero offset slope gives a locally parallel path when \(1-\kappa d>0\). It also shows why the offset graph alone does not describe the world-space heading or curvature. Track curvature changes, large offsets, and the vehicle's dynamic response matter. Smooth d(s) does not by itself guarantee smooth world-space curvature—or safe passage.
The driver evaluates a point ahead on this route and steers toward it. The car does not jump onto the mathematical curve or stop at p3 merely because a knot exists there. Speed control, capture conditions, and traffic decisions complete the behaviour.
Checks: (2) 4.21875 m; (3) A = −8 m, B = 9 m, C = 4 m, D = 0; (4) the endpoint constraints may conflict, and changing h changes the problem. Identical semantic milestones with the same value and slope can instead be represented by one knot before constructing the spline.
Return to Chapter 6, or ask a question.
1.14.0 