Silhouette curve for isometric surface of revolution

Question

I am trying to figure out how to convert a flat representation of a curve into the silhouette of a surface of revolution in a isometric projection. In essence in want the planar cut of the surface edge to be converted to the silhouette.

Image 1: Starting point.

The geometry in this case is a Bézier curve although if somebody has solutions for any functions or higher order curves would be fine. So in essence im open to conics and NURBS also.

Image 2: End result. How to transform the magenta curves?

The way I've done this now is by sampling several forms and in this case draw the new curve over the points. But id like to do this automatically and more precise. Please note: I am perfectly aware on how to do this with discrete data, I am looking for a more analytical solution if possible.

Image 3: The shape was derived using an old draftsman's trick.

Please note: Most applications polygonize NURBS or Bézier surface for rendering. Also trivial root finding does not really work well as result should not be pixels but a curve. Let us to avoid this, as I am perfectly capable of discretisizing the solution and project the edges on the back face culling edge, and doing a secondary fitting, or even fitting on the NURBS underworld and then to 2d*. That is exactly the same solution as I am using now. Seem to me there should be a somewhat analytical transformation possible.

Image 4: I am perfectly capable of doing this with subdivision/discretisation of the domain.

* Though if you know a robust way to do the projection to a curve on surface I'm interested in that too.

Answer 1

The silhouette curve of a quadric surface is a conic (ellipse, parabola, hyperbola, etc.). Suppose you have the quadric $\mathbf{X}^T \mathbf{M} \mathbf{X} = 0$ and your eye is at the point $\mathbf{Y}$. Then the silhouette curve is the intersection of the quadric and the so-called polar plane of $\mathbf{Y}$, which is $\mathbf{Y}^T \mathbf{M} \mathbf{X} = 0$.

For anything more complex than quadrics, things get nasty in a hurry, and you won't find any analytic solutions; you'll need numerical methods to trace the curve. Even the silhouette of a torus is pretty complex. So, in essence, the best you can do is compute discrete points along the curve, and use these to construct either a polyline or a spline of some sort. But, of course, you can always get a polyline just by tesselating the surface and finding the silhouette of the tesselation.

One approach would be to approximate your surface of revolution by quadrics, but that may not be any better than tesselating. You'd get a piecewise conic curve, instead of a polyline, which might be an improvement.

Note that if your surface is not G2, then it's silhouette will not be G1 (i.e. it may have sharp corners). Tracing a curve with corners is tricky.

There are silhouette generation functions in geometry kernels like Parasolid and ACIS. Maybe in OpenCascade, too.

To illustrate the difficulty, suppose we have a surface with parametric equations $(u,v) \mapsto \mathbf{S}(u,v)$, and we are viewing it from an eye- point $\mathbf{E}$. Let $\mathbf{N}(u,v)$ be the surface normal at parameter values $(u,v)$, which we can calculate as the cross product of partial derivatives: $\mathbf{N}(u,v) = \mathbf{S}^u(u,v) \times \mathbf{S}^v(u,v)$. If a point $\mathbf{P}$ is on the silhouette curve, then the surface normal at this point is perpendicular to the vector $\mathbf{P} - \mathbf{E}$. So, the silhouette curve is the locus of points where $$ (\mathbf{S}(u,v) - \mathbf{E}) \cdot \mathbf{N}(u,v) = 0 $$ Substituting for $\mathbf{N}(u,v)$, this becomes $$ (\mathbf{S}(u,v) - \mathbf{E}) \cdot (\mathbf{S}^u(u,v) \times \mathbf{S}^v(u,v) ) = 0 $$ If $\mathbf{S}(u,v)$ is even a very simple surface, like a torus or a biquadratic or bicubic patch, the silhouette equation will not have any analytic solution.