Lecture 27: Sweeping along a path

Lecture 26 swept a profile along a straight line and round a circle, and both were exact. What about any other curve: a pipe’s route, a spring? We have the grid already. The new question is what the stations’ frames should be. That is, as the profile travels, which way should it face and how should it be rolled?

The first part is easy: for a sweep the profile usually stays square to the path. The second is the entire content of this lecture. A plane square to a curve at a point leaves one degree of freedom, a rotation about the tangent, and nothing about the curve alone seems to fix it.

What a frame is, and what can go wrong

A frame along a curve $C(t)$ is a choice, at every $t$, of an orthonormal triple $(T,U,V)$ with $T$ the unit tangent. It need not be Frenet’s. Sweeping a profile with coordinates $(x,y)$ gives the surface

$$S(t,x,y)=C(t)+x\,U(t)+y\,V(t).$$

The choice of $U(t)$ matters a great deal. If $U$ spins quickly about the tangent, a rectangular section winds round the path like a drill bit, even though the path did not ask for that. So we need a rule that chooses $U$ so that the sweep does not twist more than the path forces it to. Let us look at the obvious candidates.

The Frenet frame

Differential geometry offers a frame made from the curve alone. Writing $s$ for arc length:

$$T=\frac{dC}{ds},\qquad N=\frac{dT/ds}{\lvert dT/ds\rvert},\qquad B=T\times N,$$

with the Frenet equations

$$\frac{dT}{ds}=\kappa N,\qquad \frac{dN}{ds}=-\kappa T+\tau B,\qquad \frac{dB}{ds}=-\tau N,$$

where $\kappa\ge0$ is the curvature and $\tau$ the torsion. Equivalently the frame rotates with the Darboux vector $\omega_F=\tau T+\kappa B$: for each axis $a$, $da/ds=\omega_F\times a$.

The Frenet frame is canonical, which sounds ideal, and it fails in two ways.

At an inflection. Where $\kappa=0$, the direction $N$ is undefined. Take the planar curve $y=x^3-x$. On one side of the inflection $N$ points up and on the other it points down. Passing through, the frame flips by 180°, in one jump. A swept rectangle would be turned through half a turn in zero length.

On a straight piece. A line has no $N$ at all. A path of a line followed by an arc, the most common pipe in the world, has no Frenet frame along its first piece.

And it twists when it needn’t. Even where it is defined, Frenet follows the curve’s torsion into a roll that nothing physical requires.

Rotation-minimizing frames

Instead of asking what the curve says $U$ should be, specify how $U$ is allowed to change. Require that $U$ turn no more about the tangent than it must:

$$\frac{dU}{ds}\parallel T.$$

Such a frame’s angular velocity has no component along the tangent. It turns $T$ as required and nothing else, and the angular velocity with those properties is

$$\omega=T\times\frac{dT}{ds}=\kappa B.$$

So it is the Darboux vector with the $\tau T$ term removed. Starting from any $U_0$ square to $T(0)$, this gives a unique frame by solving an ODE. Bishop introduced this frame in 1975 and showed that it exists wherever the curve is $C^2$, including at inflections and on straight pieces, where $\kappa=0$ simply means no rotation. Mechanically it is a pipe of constant section being pushed along without being twisted.

The relation to Frenet comes straight from the two angular velocities. They differ only by $\tau T$, so the rotation-minimizing frame turns against Frenet’s about the tangent at the rate of the torsion:

$$\phi(s)=\pm\int_0^s\tau\,ds.$$

For a planar curve, $\tau=0$: so the plane’s normal, carried along, remains the plane’s normal, and the frame in the plane is simply “the other axis”. That is the exact statement in the code: “a rotation-minimizing frame, which for a planar path is the frame that keeps the path’s plane normal fixed”.

Double reflection

We need to integrate this frame along a curve. The textbook route would be a Runge–Kutta step on $\dot U=\omega\times U$, with renormalization. Wang, Jüttler, Zheng and Liu (2008) found something much simpler and better: compute each step as a product of two reflections.

Let $x_i,t_i$ and $x_{i+1},t_{i+1}$ be two consecutive points with their unit tangents, and $r_i$ the frame’s vector at $x_i$.

1. Reflect in the plane bisecting $x_i$ and $x_{i+1}$, whose normal is $v_1=x_{i+1}-x_i$: $$r^L=r_i-\frac{2\,v_1\cdot r_i}{v_1\cdot v_1}\,v_1,\qquad t^L=t_i-\frac{2\,v_1\cdot t_i}{v_1\cdot v_1}\,v_1.$$
2. Reflect in the plane that takes $t^L$ onto $t_{i+1}$, whose normal is $v_2=t_{i+1}-t^L$: $$r_{i+1}=r^L-\frac{2\,v_2\cdot r^L}{v_2\cdot v_2}\,v_2.$$

Why this is natural: the first reflection carries the point $x_i$ exactly onto $x_{i+1}$. Since the chord $v_1$ runs nearly along the tangent, that reflection almost reverses the tangent: $t^L$ is close to $-t_{i+1}$, so $v_2$ is close to $2t_{i+1}$. The second reflection, in a plane through $x_{i+1}$ nearly square to the curve, turns the tangent exactly onto $t_{i+1}$ and hardly touches anything square to it. On a circular arc, this product is the exact rotation about the arc’s axis (exercise 4). Two reflections compose to a rotation, so the frame stays orthonormal to rounding with no need to renormalize, and the method’s global error falls as the fourth power of the step. The demo below reports this: halve the steps and the error falls by about sixteen. The test converges_at_fourth_order in book/scripts/demo_tests/helix-frames.test.mjs measures a ratio in that range.

The helix: Frenet, RMF and screw motion

A helix of radius $r$ rising $p$ per turn is $$C(\theta)=(r\cos\theta,\ r\sin\theta,\ c\,\theta),\qquad c=\frac{p}{2\pi}.$$ Its speed is $\lvert C'\rvert=\sqrt{r^2+c^2}$, and the pitch angle $\alpha$ with $\tan\alpha=c/r$ is the slope of the tangent over the horizontal. A short calculation gives

$$\kappa=\frac{r}{r^2+c^2},\qquad \tau=\frac{c}{r^2+c^2}.$$

Every point of the curve looks the same: the screw motion “turn by $\phi$ about the axis and rise by $c\phi$” takes the helix to itself. That gives a third natural frame: carry a frame at the start by the screw motion. Then $U(\theta)$ can be taken as the radial direction $(\cos\theta,\sin\theta,0)$, and since the principal normal of a helix is minus the radial direction, this frame is the Frenet frame up to a fixed rotation.

Which frame is right for a coil spring? They differ. Combine the rate of Frenet against RMF with the closed form: per unit of the turning angle $\theta$, the twist between the screw frame and the rotation-minimizing frame is

$$\frac{d\phi}{d\theta}=\tau\,\lvert C'\rvert=\frac{c}{\sqrt{r^2+c^2}}=\sin\alpha.$$

So per full turn, a rotation-minimizing frame rolls $360°\sin\alpha$ against the screw frame. Take a spring of radius 10 rising $20/3$ per turn: $c=1.061$, $\sin\alpha=0.1055$, so $38.0°$ per turn. Over three turns the section would have rolled $114°$ against the coil’s own axis.

For a round wire that does not matter: a circle rolling about its centre is the same circle, though the surface parametrization does change. For a square-section coil it is very visible. With an RMF, the square’s sides gradually turn away from the axis, and with a screw frame they stay facing it. A designer drawing a square-section spring means the screw frame, and that is what the kernel gives a sweep along a helix, as we shall see: the profile keeps facing the axis the way it was drawn.

Closing a path

For a closed path a new problem appears. Carry a frame all the way round with the rotation-minimizing rule. It need not come back to where it started.

Here is the geometric meaning. The unit tangent $T(s)$ traces a closed curve on the unit sphere, the tangent indicatrix. Carrying $U$ with no rotation about $T$ is exactly parallel transport of a tangent vector on that sphere, because $U$ is perpendicular to the point $T$ and changes only along $T$. By Gauss–Bonnet the parallel transport round a closed spherical curve turns a vector by the enclosed area:

$$\text{mismatch}\equiv 2\pi-A\pmod{2\pi},$$

where $A$ is the area enclosed by the indicatrix. For a planar closed curve with no self-crossings the indicatrix is a great circle, covered once, enclosing $A=2\pi$: mismatch zero. That is why a sweep round a planar circle or a planar rounded rectangle closes without any care.

For a nonplanar closed path, a frame computed by the RMF rule generally arrives turned by some angle. A real sweep must then do something: distribute that angle evenly along the path as extra twist, or force an integer number of turns, or refuse. Which of these a kernel does is a design decision.

From frames to a surface

A frame at each of a finite number of stations does not yet give a surface. Consider which spans are exact.

A line moves the profile straight on. Span::Line, degree 1. Exact.

A circular arc turns the profile about the arc’s axis. Every profile point moves on a circle, so Span::arc gives a rational quadratic row, as in the revolve. It is the same kind of span, and the wall is a piece of torus. Exact.

So a path of lines and arcs yields an entirely exact sweep, and for such a path the rotation-minimizing frame is also exact: on a line nothing turns, and on an arc the frame turns about the arc’s axis.

A helix turns the profile about the helix’s axis as it rises: a screw motion. The kernel’s helix is a chain of rational quadratics of at most a quarter turn each, at double knots (helix_rows, below). cut_helices cuts such a curve into those pieces, its own control points three to a piece, not cut anew, which would round them, and each piece is then a span of its own, named after the curve: c8.q0, c8.q1, and so on, so that face names gain c8.q3. Each piece is carried as an arc that rises: Kind::Arc has a rise along the axis, zero for an arc in a plane, and Span::arc builds the wall. Every point of the profile stays on a cylinder of its own, so it keeps facing the axis as it was drawn. No rotation-minimizing frame and no skinned spline is involved. Only the height runs off the true helix between the span ends, by the same small amount as the path itself (below). As the module’s comment records, carried by a rotation-minimizing frame the profile “rolled over turn by turn, and its walls left their cylinders”.

One piece alone cannot be told from a helix or an arc: its legs fit a helix turning either way about one of two axes. Two can. screw_between looks at two neighbouring pieces, and uses that $P_0+P_2-2P_1$ lies square to the axis whatever the rise, and points where the piece is turning: two of them span the plane it turns in, which gives the axis. They must also have one middle weight and rise alike. kinds asks this of each piece and its neighbour. Where two pieces of a helix meet, no corner is treated: they go on into each other, to the last bit or not, since the rows they share were rounded each where it lies.

This holds for a plain sweep. A profile that twists, scales or follows rails is sampled on every span, as a spline’s is (Control), and one kept facing the way it was drawn (Orientation::FixedNormal) has every point travelling a copy of the path.

Any other curve has no swept surface that is a NURBS in general: the frame’s trigonometry is not rational in the curve’s parameter. So it is approximated: sections of the profile are placed at several parameters, and the frames between them are interpolated by cubic Hermite interpolation of the frames (hermite), giving a Span::Curve of degree 3 whose middle rows are frames a third of a step along the derivatives at the sections. Because a row is linear in the profile coordinates ($p=o+xe_1+ye_2$), interpolating the frames interpolates every profile point at once with the same degrees and knots, and so the grid of Lecture 26 is unchanged.

That leaves “is this curve a line, an arc or something else?”. kind in path_sweep.rs asks that on the control points: an arc is a rational quadratic Bézier with legs of one length and a middle weight the cosine of half the turn, with those checked by could_be_equal, “so a curve that could be one is taken as one”.

Note the tension with AGENTS.md’s Everything is a NURBS curve or surface: isn’t kind a type dispatch? It does not branch on a stored kind, and everything that leaves it is a NURBS, but it does choose a more exact representation when the control points admit one. This is where the kernel draws the line. The result of a line or an arc span has an exact swept surface available, which no amount of generality in the fitted path could recover, since a fit could only approximate it. The recognition chooses what to build, and no downstream algorithm looks at the outcome and takes a special path.

The screw sweep

There is a second way to sweep along a helix, and the difference is where the profile is drawn. A path sweep takes the profile as you drew it, and the path is a curve of a 3-D sketch. screw in lib/geop-lib-sweep/src/revolve.rs sweeps a profile drawn in $(r,z)$, a half-plane through the axis. screw_path gives stations with $e_1$ radial and $e_2$ the axis itself: the profile is not square to the helix but contains the axis, and it moves by the screw motion. That is what a thread is defined by, and thread standards give the tooth’s section in an axial plane.

The rows come from helix_rows in core/geop-core-geometry/src/nurb_curve/helix.rs: a helix is not a rational curve, since its height grows linearly with the angle, and no rational parametrization of a circle turns at a constant rate. So it is approximated, one rational quadratic per quarter turn at most, with the circle part exact:

The curve lies exactly on its cylinder: seen down the axis it is the kernel’s own circle. Only the height is approximate. It rises from span end to span end exactly as a helix does, the middle control row sits at the height halfway between […] and in between the height runs ahead of or behind the true helix by at most 0.53% of the pitch for quarter-turn spans.

The module quotes a measured $5.3\times10^{-3}$ times the pitch, tested there, and points out that for an ISO metric M6 thread (pitch 1 mm) that is 5 µm, well inside the thread’s own tolerance.

NurbCurve3D::helix builds the helix as a curve from the same rows, which is why its projection down the axis is exact. A helix of a 3-D sketch is such a curve (Lecture 25), and it is what cut_helices recognizes.

Corners

A path need not be smooth. Where two lines meet at an angle, a frame jump gives a problem: the two walls would not meet at the corner. The classical answer is a mitre: at the joint, the profile is projected along either line onto the plane bisecting the corner. Then the two straight walls meet there exactly, since each is a translation of that section, and a section projected along the line stays on the wall. Sweeping a square round a square gives a picture frame, with mitred corners. (The test sweep_round_a_square_is_a_frame builds exactly that.)

At any other corner, at an arc or a spline, the projected section lies on neither swept wall. So the walls either side are put on the surfaces the rigid profile sweeps along its own curve, and the body is rebuilt around them: a vertex moves to where the surfaces meet, and an edge to where they cross. That is the one rebuild of Lecture 30. Because geop-lib-sweep cannot depend on the shell crate (its tests sweep their solids), the rebuild is handed in as a Rebuild, and without one such a corner is refused by name.

Tight bends are a separate hazard. If the profile reaches across the axis of an arc, its walls would pass through each other, a self-intersection no boolean should be left to find. clear_of_bends refuses this at the outset, naming the curve of the profile and the bend. It decides on the control points of each curve’s quarters, which enclose the curve, so that the refusal is proven rather than sampled. This illustrates AGENTS.md’s Refuse what you do not support: early, by name, and say why.

What else travels

Control in path_sweep.rs lets the profile change on the way:

• Orientation::FixedNormal keeps the profile facing as drawn: every profile point then travels a translated copy of the path, so every span is the path’s own degrees, knots and weights, and is exact. It only requires the path to keep passing through the profile’s plane, which translated checks on control points.
• Twist and scale, growing evenly with the length travelled.
• Rails: where a guide crosses the section is where the profile’s point goes. A single rail turns and scales the section uniformly, and two give a linear map of the section’s plane.

The example horn in ops/geop-ops-extrude-revolve/src/examples.rs is a circle swept along a straight path whose size is set by a rail. Each such case is no NURBS in general, so every span is sampled, except a line along which the map changes linearly, and that remains exact. As you would now expect, “exact where exact is possible, enclosed by construction elsewhere” is the rule throughout.

Exercises

1. Compute the Frenet quantities of the helix $C(\theta)=(r\cos\theta,r\sin\theta,c\theta)$ and show $\kappa=r/(r^2+c^2)$ and $\tau=c/(r^2+c^2)$.

2. A planar curve in $z=0$ with $r_0=(0,0,1)$. Show that double reflection leaves $r$ unchanged at every step, exactly, whatever the step size.

3. Suppose a rectangular duct (not a circle) were swept along a helix of radius 40 and pitch 50 by an RMF over two turns, which the kernel does not do. By how much has the section rolled against the coil axis at the end?

4. Show that one step of double reflection on a circular arc, of any size, is exactly the rotation about the arc’s axis: compute $t^L$ and $v_2$, and follow the two vectors of the frame.

5. What is the mismatch, by the area formula, of a closed path whose tangent indicatrix is a small circle at colatitude 60° covered once? Is such a path possible with this exact indicatrix?

6. Find what Orientation::FixedNormal checks about the path in translated and why it is enough to check control points.