Lecture 10: Surfaces

A solid is bounded by surfaces. We now have everything needed to define one, with almost no new mathematics. The trick is called the tensor product: a surface is a curve whose control points move along curves.

From a moving curve to a net

Take a curve $C(v)=\sum_jN_{j,q}(v)Q_j$. Now let each control point move along its own curve, with a common parameter $u$ and a common basis:

$$Q_j(u)=\sum_iN_{i,p}(u)P_{ij}.$$

Substituting gives

$$S(u,v)=\sum_{i}\sum_{j}N_{i,p}(u)\,N_{j,q}(v)\,P_{ij}.$$

That is a tensor-product B-spline surface: a rectangular control net $P_{ij}$, degree $p$ and knots $U$ in $u$, degree $q$ and knots $V$ in $v$, over the rectangle $[u_p,u_{n+1}]\times[v_q,v_{m+1}]$. With weights it is a NURBS surface, the homogeneous points $(w_{ij}P_{ij},w_{ij})$ obeying exactly this polynomial formula upstairs:

$$S(u,v)=\frac{\sum_{i,j}N_{i,p}(u)N_{j,q}(v)w_{ij}P_{ij}}{\sum_{i,j}N_{i,p}(u)N_{j,q}(v)w_{ij}}.$$

What makes this powerful: fix one parameter, and the rest is a curve algorithm. Fix $u=u_0$, and $S(u_0,v)$ is a B-spline in $v$ with control points $Q_j(u_0)$. This is an iso-curve (or iso-parametric curve). So:

• Evaluate: de Boor along $u$ on each column of the net gives one row of points; de Boor along $v$ on that row gives the point. The order is irrelevant.
• Split: insert the same knot in every row. This is what “works on rows” meant in Lecture 9.
• Convex hull: with non-negative weights, the products $N_{i,p}N_{j,q}$ are non-negative and sum to one. So $S(u,v)$ lies in the hull of the active $(p+1)(q+1)$ net points.
• Boundary: for clamped knots, the boundary of the patch is four curves whose control points are the outer rows of the net, and the corners are the corner control points.

The demo shows the two-stage evaluation for a bicubic Bézier patch: sixteen control points, cubic in each direction.

Things to check in the demo: move $v$ only, with $u$ still: the orange curve does not change. Choose the corner $P_{00}$ and raise it: the surface’s corner follows exactly. Raise an inner point $P_{11}$: the surface follows only part of the way.

Partial derivatives

Differentiate the double sum with respect to $u$, holding $v$. By Lecture 7’s derivative formula, applied to every column, $S_u$ is again a tensor-product surface, whose net is differences along $u$:

$$S_u(u,v)=\sum_{i,j}N_{i+1,p-1}(u)\,N_{j,q}(v)\,\frac{p\,(P_{i+1,j}-P_{ij})}{u_{i+p+1}-u_{i+1}}.$$

For a Bézier patch on $[0,1]^2$, this is $p\sum B_i^{p-1}(u)B_j^q(v)(P_{i+1,j}-P_{ij})$. The same goes for $S_v$. Second derivatives come from differencing twice: $S_{uu}$ along $u$ twice, $S_{vv}$ along $v$ twice, and $S_{uv}$ once each way. A key fact is that all these nets are finite data, so their hulls bound the derivative over a whole box of $(u,v)$. Tangent directions and normals of a piece of surface can be bounded by proof, not by sampling.

For a rational surface one differentiates upstairs and applies Leibniz in two variables, as for curves. With $A=wS$:

$$S_u=\frac{A_u-w_uS}{w},\qquad S_{uv}=\frac{A_{uv}-w_{uv}S-w_uS_v-w_vS_u}{w}.$$

The normal

$S_u$ and $S_v$ span the tangent plane at a point, if they are independent. The unit normal is

$$n=\frac{S_u\times S_v}{\lvert S_u\times S_v\rvert}.$$

Two things to keep in mind.

Its direction has meaning, and the parametrization decides it. Swapping $u$ and $v$, or reversing $u$, flips the normal. In a solid, the normal must point out of the material. The kernel has no orientation flag on a face: “Orientation has no flag. It lives in the surface’s parametrization” (from the topology crate page). reverse_face mirrors the surface in $u$ and its pcurves with it.

It can fail. Where $S_u\times S_v=0$ the formula is $0/0$. It may be that the surface itself has a crease or a tip. It may equally well be that a perfectly smooth surface has a bad coordinate system there. We will now look at the latter, the most important kind.

Fundamental forms and curvature

Everything a kernel needs to know about the local shape of a surface comes from two quadratic forms. They are worth deriving rather than memorizing.

First fundamental form. Move in the domain at a velocity $(\dot u,\dot v)$. The velocity in space is $S_u\dot u+S_v\dot v$, of squared length

$$\lvert\dot S\rvert^2=E\dot u^2+2F\dot u\dot v+G\dot v^2,\quad E=S_u\cdot S_u,\ F=S_u\cdot S_v,\ G=S_v\cdot S_v.$$

So $E,F,G$ measure lengths and angles on the surface, using only the parameters. A patch of surface of domain size $\mathrm{d}u\,\mathrm{d}v$ has area $\sqrt{EG-F^2}\,\mathrm{d}u\,\mathrm{d}v=\lvert S_u\times S_v\rvert\,\mathrm{d}u\,\mathrm{d}v$.

Second fundamental form. Now take a curve on the surface, $\gamma(s)=S(u(s),v(s))$, with arc length $s$. Its acceleration $\ddot\gamma$ has a component in the tangent plane, which says how the curve turns within the surface, and one along $n$, which says how much the surface forces it to bend. Since $\dot\gamma\cdot n=0$ everywhere, differentiating gives $\ddot\gamma\cdot n=-\dot\gamma\cdot\dot n$, and expanding $\ddot\gamma$:

$$\ddot\gamma\cdot n=L\dot u^2+2M\dot u\dot v+N\dot v^2,\quad L=S_{uu}\cdot n,\ M=S_{uv}\cdot n,\ N=S_{vv}\cdot n.$$

This $\ddot\gamma\cdot n$ is the normal curvature $\kappa_n$ in the direction $(\dot u,\dot v)$. It depends only on the direction, not on the curve. For a general (non-unit) velocity:

$$\kappa_n=\frac{L\dot u^2+2M\dot u\dot v+N\dot v^2}{E\dot u^2+2F\dot u\dot v+G\dot v^2}.$$

As a direction turns, $\kappa_n$ varies between a maximum and a minimum, the principal curvatures $\kappa_1,\kappa_2$. They are the eigenvalues of $\mathrm{I}^{-1}\mathrm{II}$, where $\mathrm{I}=\begin{pmatrix}E&F\\F&G\end{pmatrix}$ and $\mathrm{II}=\begin{pmatrix}L&M\\M&N\end{pmatrix}$. Their determinant and half trace are the Gaussian and mean curvature:

$$K=\kappa_1\kappa_2=\frac{LN-M^2}{EG-F^2},\qquad H=\frac{\kappa_1+\kappa_2}{2}=\frac{LG-2MF+NE}{2(EG-F^2)}.$$

$K>0$ means the surface locally stays on one side of the tangent plane, like a dome. $K<0$ means a saddle. A surface with $K=0$ at every point, like a plane or a cylinder, is called developable: it can be unrolled to a flat sheet without stretching.

A worked example

Take the bicubic form of $z=xy$: $S(u,v)=(u,v,uv)$ on $[0,1]^2$ (with the control heights $\frac i3\frac j3$; try the “saddle” preset, which has this shape up to scale). Then

$$S_u=(1,0,v),\quad S_v=(0,1,u),\quad S_{uu}=S_{vv}=0,\quad S_{uv}=(0,0,1),$$

$$S_u\times S_v=(-v,-u,1),\quad n=\frac{(-v,-u,1)}{d},\quad d=\sqrt{1+u^2+v^2}.$$

$$E=1+v^2,\ F=uv,\ G=1+u^2;\quad EG-F^2=d^2;\quad L=N=0,\ M=\frac1d.$$

So $K=\frac{-M^2}{d^2}=-\frac{1}{d^4}$ and $H=\frac{-2MF}{2d^2}=-\frac{uv}{d^3}$. At $(0,0)$: $K=-1$, a true saddle. These two formulas are what book/scripts/demo_tests/surface-patch.test.mjs checks the demo against.

The surfaces of a CAD system, as NURBS

Everything a mechanical part is made of is exactly representable.

A plane or planar rectangle is a bilinear patch: degree (1,1), four control points, weights 1 (NurbSurface3D::bilinear). It is the simplest tensor product, and its normal is $(P_{10}-P_{00})\times(P_{01}-P_{00})$.

A ruled surface joins two curves by straight lines: $S(u,v)=(1-v)A(u)+vB(u)$. If $A$ and $B$ have the same degree and knots, the net is simply their two control polygons side by side: degree $(p,1)$. Making two arbitrary curves compatible (the same degree and knots) is by degree elevation and knot insertion, NurbCurve::compatible in nurb_curve/compatible.rs, and then NurbSurface3D::ruled. An extrusion is a ruled surface between a curve and its translate, NurbCurve::sweep in nurb_curve/sweep.rs. So a cylinder is a circle extruded: degree (2,1).

A surface of revolution turns a profile $P(v)=\sum_jN_j(v)w_jP_j$ about an axis. Each profile control point describes a circle, which Lecture 8 tells us how to write exactly:

$$\text{row } j\text{ of the net: the arc through the turn of } P_j,\text{ with weights } w_j\cdot(1,\ \cos\tfrac\theta2,\ 1,\ \cos\tfrac\theta2,\ \dots).$$

So the net is a product, and the weights multiply. NurbSurface::revolve in nurb_surface/revolve.rs does this: $u$ is the turn over $[0,1]$, with rational quadratic pieces of at most a quarter turn, and $v$ is the profile’s own parameter, so the degree is $(2,q)$. A cone revolves a straight line that meets the axis. A sphere revolves a half circle. A torus revolves a circle clear of the axis. They are not kinds in the kernel, only control points.

Poles: where the coordinates fail

Take a revolved surface whose profile touches the axis: a sphere’s pole or a cone’s tip. The profile’s control point on the axis does not move as it turns. So a whole row of the net collapses to one point. The surface at that boundary is the same point no matter what the turn $u$ is, so $S_u=0$ there, and $S_u\times S_v=0$: the formula for the normal fails.

The surface may be perfectly fine. For a sphere, the pole is a point like any other, with a good tangent plane; a different coordinate system would not notice it. This is the same situation as the Earth’s north pole in latitude and longitude: every longitude names one point, and no step in longitude moves you, but the ground is flat there as anywhere else.

AGENTS.md’s “Degenerate parametrizations: the surface is fine, the coordinates are not” gives three real bugs, and each comes from confusing the coordinates with the geometry:

• A seed on the pole never leaves it. NurbSurface::project skipped any Newton step whose Jacobian could be singular. From a seed on the pole every step is singular, so it returned the pole for every target. The fix is to step along the parameter that still moves the point, down a meridian: which meridian is a free choice, so take the one that heads most towards the target (off_pole in nurb_surface/project.rs).
• Decide tangency where the geometry is, not at moving foot points. Two quarters of a sphere meet at its pole, as one smooth surface. Comparing normals at foot points clamped to each patch’s domain made them two nearly equal conditions, and a singular system.
• Straight in $(u,v)$ is not preserved by offsetting. A sphere’s meridian cut by a plane comes out as a small circle on the inner copy.

So how does the kernel get a normal at a pole? From the net, directly. NurbSurface::normal first tries $S_u\times S_v$. If that cannot be normalized, it looks for a pole on a boundary row the point touches (rows_touching, pole, pole_normal). Leaving a pole, the derivative across the row is a positive combination of the spokes, the vectors from the pole to the next row’s control points. So the pole has a tangent plane exactly when those spokes are coplanar, and its normal is that plane’s. If the spokes are not coplanar, as for a cone’s apex, there is no single normal, and it is an error: “not an arbitrary pick among them”.

Check that against a sphere and a cone. A sphere’s next row near the pole lies in the tangent plane at the pole, because a circle’s control polygon leaves along the tangent: the spokes are coplanar. A cone’s next row is a circle further down, and the spokes form a cone. So the sphere’s pole has a normal, and the cone’s apex has none, by the same test, with no branch on the kind of surface.

What does a loop on such a face look like? Lecture 18 answers that. A collapsed row becomes a coedge that lies on a vertex, and the pcurve runs along the row.

What the hull does for surfaces

We close with the property that makes all this searchable. Each net cell’s hull encloses a piece of surface. Split in $u$ and $v$ and the net refines, and the distance from the net to the surface falls with the square of the cell’s size, as for curves. So a question like “could this surface contain this point?” comes down to a question about a finite array of numbers. That is surface_could_contain in core/geop-core-geometry/src/contains/surface.rs, with fat-line clipping in place of plain bisection, as Lecture 13 will explain. For the cheapest case there is NurbSurface::aabb.

Recap

Exercises

1. Write the net, weights and knots for a quarter of the unit cylinder of height 2, as a NURBS surface of degree (2,1).

2. For the quarter cylinder of exercise 1 at $u=0$, compute $S_u$, $S_v$, $n$, $E,F,G$ and $L,M,N$, and hence $K$ and $H$. Use $S_u(0)=2w_1(P_1-P_0)$, and find $S_{uu}$ from the circle: at parameter speed $\sigma$, a unit circle’s second derivative has normal component $\sigma^2$ towards the axis.

3. Show that for $F=0$, $\kappa_n$ along $u$ is $L/E$. Compare with curvature_radius in nurb_surface/curvature.rs, and say exactly when its result is a true principal curvature.

4. Why must a surface’s two knot vectors be separate, with a single net? A designer asks for a patch with an extra row of control points only in one part of it: what happens? (This is a real limit of tensor products. It is why trimmed faces and multiple patches exist.)

5. A revolved profile has one control point on the axis, and its other control points clear of it. Which row of the net collapses? Which partial vanishes at that row, and which does not? Compute $S_v$ at the pole for a sphere of radius 1, with $v$ the parameter along a quarter-arc profile from the pole to the equator. Does its direction depend on $u$?

6. Show that a tensor-product Bézier patch with an axis-aligned box has its normal bounded by the cross products of the difference nets. How would you bound the normal over a sub-box? Give a condition under which the patch cannot fold over onto itself.