Past exam of the computer science course of the University of Oxford / 2015 / Geometric modelling / 2

For a fixed $t\in [0,1]$, De Casteljau's algorithm first takes the same fractional position along both adjacent sides of the control polygon:It then applies another linear interpolation with that same $t$:This is the degree-two Bernstein basis representation of a quadratic Bézier curve. Its coordinate polynomials areThe degree is exactly two, since its second derivative $(-4,-4)$ is nonzero. As a check on the construction, at $t=1/2$ the two intermediate points are $(-1,1/2)$ and $(0,1/2)$ and their midpoint is $F(1/2)=(-1/2,1/2)$.

Three useful geometric properties, with their concrete consequences here, are:

These give the compact answerThe same sum-to-one calculation also explains affine equivariance of a geometric basis: applying an affine map to the handles applies that map to the curve.

Use a row of descending monomials $T(t)=[t^{2}t1]$ and a matrix with one control point per row. Expanding $(1-t{)}^2,2t(1-t),t^2$ gives their coefficients in the three columns of the basis matrix:The matrix multiplication isIn particular, $T(t)M_B=[(1-t{)}^{2}2t(1-t)t^2]$, so multiplying the geometric matrix takes exactly the weighted control point sum in part (a). The two output entries are $F_x(t),F_y(t)$. The basis matrix depends only on the degree and ordering convention; $G_B$ holds the particular geometry.

Choose a quadratic uniform B-spline segment on a unit spline knot interval. This choice is useful because adjacent pieces can later share de Boor points at simple, equally spaced spline knots. Its three active basis pieces on local $t\in [0,1]$ areThey follow from the Cox-de Boor recurrence for degree two and consecutive integer spline knots. If the three de Boor points are $D_0,D_1,D_2$, then the equivalent Bézier curve handles areOne can verify this conversion simply by substituting these expressions into the Bernstein basis formula and collecting each $D_i$. Solving givesThe desired matrix representation is thereforeThe left row is again the monomial row, the middle matrix contains the monomial coefficients of $b_0,b_1,b_2$, and the right matrix lists the de Boor points. Unlike the Bézier curve basis matrix, the constant row averages two handles. A clamped B-spline with repeated end spline knots could instead reuse $A,B,C$ directly; the simple uniform convention above specifies which representation is being used and supplies the shared handles required for the extension.

Rotating the geometry through $180^{\circ }$ sends every point $p$ to $-p$. To start the extension at the origin, also reverse the parameter:Then $E(0)=(0,0)=F(1)$ and $E(1)=(2,0)$. Merely using $-F(t)$ would trace the same rotated image toward the origin, instead of away from the join.

The reversed and rotated de Boor points are $-D_2,-D_1,-D_0$, namely $(0,1),(0,-1),(4,1)$. The first two are already $D_1,D_2$, so there is only one new de Boor point, $\boxed{D_3=(4,1)}$. Using the same uniform basis matrix,Equivalently, the Bézier curve handles are $C=(0,0)$, $-B=(0,-1)$ and $-A=(2,0)$, which gives the same polynomials immediately.

Assign one unit parameter interval to each piece. ThenThere is no ambiguity at $u=1$, since both pieces give the origin. For a single B-spline representation take degree two, the four de Boor pointsand the equally spaced spline knot vector $\tau =(-2,-1,0,1,2,3,4)$. With $N_{i,2}$ denoting the normalized degree-two B-spline on this vector, the complete valid parameter interval is $[{\tau }_2,{\tau }_4]=[0,2]$ andOn $[0,1]$ the active handles are $D_0,D_1,D_2$, and on $[1,2]$ they are $D_1,D_2,D_3$. Shared handles and a simple interior spline knot are what merge the two pieces into one uniform B-spline. The exterior spline knots support the basis; they do not extend the interval being drawn.

In particular, the requested horizontal pieces and their vertical counterparts are

The join has matching positions and first derivatives:Because both pieces have unit parameter intervals, these are also the one-sided derivatives of $S$ at $u=1$. Thus $S$ has first-order parametric continuity, and hence first-order geometric continuity, with no corner or cusp. Its nonzero tangent vector points vertically downward.

The second derivatives do not agree:So this is $C^1$ but not $C^2$ parametric continuity. The signed planar curvature $\kappa =(x^{\prime }{y}^{\prime \prime }-y^{\prime }{x}^{\prime \prime })/(x^{\prime 2}+y^{\prime 2}{)}^{3/2}$ has one-sided valuesThe unsigned curvatures agree, but the bending switches sides; there is no second-order geometric continuity through a regular orientation-preserving reparametrization. The join is a reversal of bending with a curvature jump, rather than a smooth zero-curvature inflection.

For the “gradient” compute $dy/dx=y^{\prime }/x^{\prime }$ away from the join:HenceThe zero horizontal derivative is not a stationary point of the parametric curve, since its vertical derivative is $-2$. It means that the locally regular curve has a vertical tangent line and cannot have a finite graph derivative as $y(x)$ there.