Cylinder between two circles in $\mathbb{R}^3$

Question

Let $\mathbf{p}, \mathbf{q} \in \mathbb{R}^3$, $r_p, r_q \in \mathbb{R}^+$ and $\mathbf{n}_p,\mathbf{n}_q \in S^2$. Further let $\mathbf{n} = \frac{\mathbf{p}-\mathbf{q}}{||\mathbf{p}-\mathbf{q}||}$.

Suppose I have two circles $(\mathbf{p}, r_p,\mathbf{n}_p)$ and $(\mathbf{q}, r_q,\mathbf{n}_q)$ sufficiently far from each other. I can join them up to create a cylindrical shape. The volume enclosed by this tube is the convex hull of the two circles.

1. Is there a natural way to parametrize this surface (or the solid)? A "natural" map $S^1 \times (0,1) \to \mathbb{R}^3$?

2. Is there a name for this shape? Or perhaps a linear map to "straighten" it into a more regular shape?

3. What happens if the surface is extended? Does it self-intersect in a single point (cone) or does something more complicated happen?

4. How can I calculate its sections with planes $\{ \mathbf{r} : \mathbf{n} \cdot \mathbf{r} = a\}$? Are they ellipses?

Answer 1

I will attempt to answer these in order 2,1,4,3. Disclaimer: I don't have that deep knowledge in algebraic geometry but since noone has answered yet I will take a stab at it.

The convex hull is a nontrivial developable surface + the two end caps of the circles. In the below, I will ignore the end caps and focus on the developable surface.

2: This surface, let's call it teres surface (I like the suggested name), is a ruled surface: https://en.wikipedia.org/wiki/Ruled_surface. Some people would call it a scroll.

A ruled surface $S$ is such that through every point of $S$ there is a straight line that lies on $S$.

I think there are nice maps in higher dimensions to "straighten out" such a surface but not with a linear map directly in 3D.

1: If the circles are parameterized with, say angle $u$, then a ruled surface (not the actual teres) connecting the two circles can be expressed as: $\mathbf{x}(u,v)=(1-v)\mathbf{c}(u)+v\mathbf{d}(u)$ where $\mathbf{c}(u)$ and $\mathbf{d}(u)$ are the parametric equations for the circles (directrices) and $v$ is the parameter running along the generator lines.

To get the convex hull, or rather developable surface (the interesting non-flat part, and not the end cap surfaces), you can then rotate and scale (nonlinearly) the parametrization of one of the circles to align it in such a way that the resulting surface is a developable surface (and hence part of the convex hull). The actually parameterization you would need to use might be highly nontrivial to derive, see an example further below. For most parameterizations you don't get a convex hull but a surface with negative Gaussian curvature. In addition, there is no guarantee that there exists a rational parameterization of this surface, again see below for a non-rational parameterization.

4: The only surfaces of algebraic degree 2, or quadric surfaces, that are ruled surfaces are circular or elliptical cones and cylinders, hyperbolic paraboloids, or hyperboloids of one sheet. https://en.wikipedia.org/wiki/Quadric

I am ignoring all degenerate cases.

The hyperbolic paraboloid and hyperboloid of one sheet actually have the property that for every point of $S$ there are two straight lines that lies on $S$. The teres is definitely not one of these two.

Could the teres be an elliptical cone? Indeed, elliptical cones contain circles: https://en.wikipedia.org/wiki/Circular_section and we could truncate an elliptical cone in two circles to get an oblique circular cone. However, circular sections of an elliptical cone only exist for certain configurations of the intersection planes so the teres cannot be an elliptical cone in the general case. However, it can be a cone in special cases.

This means the surface is, in general, of algebraic degree 3 or higher, see below.

$S$ is said to be torsal if its line generators have a unique tangent plane (the same tangent plane to the surface all along the line). Some ruled surfaces have only a finite number of torsal lines. I have attached a visualization of the Gaussian curvature of a particular attempt of me to produce an approximate teres surface (by various numerically methods). Here, blue represents low values (in this case negative values of approximately -0.001, so everything is close to zero here) and red represents high. The Gaussian curvature of a ruled surface is nonpositive and in the figure deep red is zero. In the figure one gets a hint that there might be two or three torsal lines among the generating lines. Note that the edge curves are actually circles lying in two different planes that are not parallel. However, due to the perspective they will inevitably look like ellipses. However, this is just a near-teres surface. A real teres surface, which by the question is part of the convex hull surface, needs to have zero Gaussian curvature, for an exact teres surface see further below. See also: https://mathoverflow.net/questions/103982/when-is-the-hull-of-a-space-curve-composed-of-developable-patches In other words, the true teres surface is a developable surface.

However, the above near-teres comes very close.

The generic planar sections are not ellipses for a general teres surface since its algebraic degree is higher than 2. I have attached a picture of an orthographic projection of a generic section of the above near-teres suface and one can kind of see that it is not an ellipse. Note that since the section curve might have cubic degree it may not be rationally parameterizable (this is a theorem for elliptic curves). The viewing angle doesn't matter since an ellipse will always look like an ellipse under projection. To find a nice equation for a planar section intersection curve is probably hopeless and would need to be handled numerically.

3. This reference as well as others: https://www.researchgate.net/publication/257314770_Developable_Surfaces_Their_History_and_Application states that any developable surface of this kind might be a generalized cone or a tangent developable. If it is a generalized cone, then this means that although the section curves can be complicated they will all intersect at a point. Think of a line with that singular point as a fulcrum tracing out some interesting shape. The teres surface is characterized by, among other things, having (exactly?) the desired two circles and the surface cannot be a generalized cone for two circles in general position in space. This means that the teres surface is a tangent developable surface for some curve. The curve is then to be found nontrivially somewhere along the surface, outside of the region between the circles, beyond the smaller circle if the circles have different radii. For an example, see below.

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Updated Information:

I worked a bit more on this fascinating problem (applied algebraic geometry). For a counter example showing that in general the intersection is not a point, but rather a curve belonging to a tangent developable surface, consider the following example: The developable surface beweteen a circle of radius $1$ centered at $x=y=z=0$ and rotated by $\phi=\pi/6$ (30 deg) around the y-axis and a circle of radius $2$ centered at $x=y=0$ and $z=5$ and parallel with the xy-plane.

To find a parameterization of this is a bit tricky and I used the computer. Start by parameterizing the cone correponding to the first circle parallel with the xy-plane ($\phi=0$).

We then have:

$\mathbf{x}(u,v)=(1-v)\mathbf{c}(u)+v\mathbf{d}(u)$

with:

$\mathbf{c}(u)=(c_x(u),c_y(u),c_z(u))$

$c_x(u)=\frac{1-u^2}{1+u^2}$

$c_y(u)=\frac{2u}{1+u^2}$

$c_z(u)=0$

and

$\mathbf{d}(u)=(d_x(u),d_y(u),d_z(u))$

$d_x(u)=2\frac{1-u^2}{1+u^2}$

$d_y(u)=2\frac{2u}{1+u^2}$

$d_z(u)=5$

Now, to get a more general ruled surface (not developable yet), rotate the first circle around the y-axis using the matrix (with $\phi=\pi/6$):

R=$\begin{bmatrix} cos(\phi) & 0 & sin(\phi)\\ 0 & 1 & 0\\ -sin(\phi) & 0 & cos(\phi)\\ \end{bmatrix}$

This gives

$c_x(u)=cos(\phi) \frac{1-u^2}{1+u^2}$

$c_y(u)=\frac{2u}{1+u^2}$

$c_z(u)=-sin(\phi) \frac{1-u^2}{1+u^2}$

Combine into the ruled surface:

$x(u,v)=(1-v) cos(\phi) \frac{1-u^2}{1+u^2} + 2 v \frac{1-u^2}{1+u^2}$

$y(u,v)=(1-v) \frac{2u}{1+u^2} + 2 v \frac{2u}{1+u^2}$

$z(u,v)=(1-v) (-sin(\phi) \frac{1-u^2}{1+u^2}) + 5 v$

Now this will only give us a ruled surface with negative Gaussian curvature everywhere. This surface can be implicitized, which gives a surface of degree 4 as follows:

$99 x^4 + (196 \sqrt{3} - 388) x^3 z + (1000 \sqrt{3} - 2020) x^3 + (-200 \sqrt{3} + 574) x^2 y^2 + (-364 \sqrt{3} + 630) x^2 z^2 + (-4000 \sqrt{3} + 6960) x^2 z + (-10500 \sqrt{3} + 18354) x^2 + (1748 \sqrt{3} - 3092) x y^2 z + (-6760 \sqrt{3} + 11500) x y^2 + (-116 \sqrt{3} + 204) x z^3 + (-360 \sqrt{3} + 660) x z^2 + (2108 \sqrt{3} - 3516) x z + (5040 \sqrt{3} - 8580) x + (-200 \sqrt{3} + 475) y^4 + (-3292 \sqrt{3} + 5706) y^2 z^2 + (10080 \sqrt{3} - 17440) y^2 z + (-6900 \sqrt{3} + 11950) y^2 + (132 \sqrt{3} - 229) z^4 + (920 \sqrt{3} - 1600) z^3 + (-416 \sqrt{3} + 678) z^2 + (-7160 \sqrt{3} + 12280) z + 7100 \sqrt{3} - 12425=0$

Now, to "twist" this into a developable surface, consider $w$ as a function of $u$ and set $w=w(u)$ and set:

$\mathbf{d}(w)=\mathbf{d}(w(u))$

We have that for a surface on the form:

$\mathbf{x}(u,v)=\mathbf{\gamma}(u)+v\mathbf{\zeta}(u)$

it is a developable surface if and only if

$det(\dot{\gamma}(u),\zeta(u),\dot{\zeta}(u))=0$

For our surface we have:

$\gamma(u)=\mathbf{c}(u)$

$\zeta(u)=\mathbf{d}(w)-\mathbf{c}(u)$

with $w=w(u)$ ($w$ is a function of $u$)

The determinant condition where we use the chain rule for $w=w(u)$

gives

$-8 \frac{d w(u)}{du} \frac{5 w(u)^2 \sqrt{3} u - 2 w(u)^2 u - 9 w(u) u^2 - 5 \sqrt{3} u + 11 w(u) - 2 u}{(u^2 + 1)^2 (w(u)^2 + 1)^2}$

from which we can see that for the determinant to vanish we need:

$5 w(u)^2 \sqrt{3} u - 2 w(u)^2 u - 9 w(u) u^2 - 5 \sqrt{3} u + 11 w(u) - 2 u$

to vanish.

Solving for $u$ now gives, as one of two roots:

$w=\frac{45 \sqrt{3} u^2 + 18 u^2 - 55 \sqrt{3} - 22 + \sqrt{6399 u^4 + 1620 \sqrt{3} u^4 + 6794 u^2 + 1720 \sqrt{3} u^2 + 9559 + 2420 \sqrt{3}}}{142 u}$

And substituting back into:

$\mathbf{d}(w)=\mathbf{d}(w(u))$

gives us the developable surface.

This is not a rational parameterization but a rational parameterization in radicals. There may still be some way of rationally parameterizing this surface but my guess is that this is not possible in general. See also: http://www.imus.us.es/2cji/archivos/pdf/2cji-MatComp-Sevilla.pdf

Although it is not rational, the radical parameterization means there is an algebraic relationship between $x,y,z,u$, and $v$ and that the surface is algebraic. To see this, just manipulate $w=w(u)$ to eliminate the radicals and get an algebraic relationship between $w$ and $u$:

$5 w^2 \sqrt{3} u- 2 w^2 u- 9 w u^2- 5 \sqrt{3} u+11 w-2 u=0$

This hints that the degree of the corresponding developable surface is probably higher than the previously implicitized ruled surface's $4$. Implicitation of this surface (to get the surface on the form $g(x,y,z)=0$) is difficult. It would be interesting to know its degree. Perhaps someone out there knows a good way to determine the degree. Trying to implicitized the section curve at $z=5$ hints that the degree is $8$. The fact that the oloid is an $8$th degree surface, https://mathcurve.com/surfaces.gb/orthobicycle/orthobicycle.shtml, means the teres is almost certainly an $8$th degree surface. Note that the above argument should hold for any types of conic section curve pairs (any end-curves) due to "projective equivalence".

Some visualizations:

The tangent developable surface between the two circles (magenta). Isoparametric lines are shown (blue) as well as the cuspidal curve (green). This particular parameterization is quite nonuniform and doesn't cover the entire algebraic surface (the Zariski closure of the surface), hence the "gap". The parameterization becomes singular along a line and the "gap" is surrounding this line for visualization-numerical reasons:

The cuspidal curve. The wanted surface is swept out by the tangents of this curve:

A zoomed in view of the isoparametric net of original ruled surface and the net of the developable surface. The ruled net is red and the developable is blue. The "rotation" of the isoparametric curves is visible. The original ruled surface and developable surface are almost indistinguishable. For a more general positioning of the circles the difference would probably be greater:

A visualization of the ruled surface (green) and the developable surface (gray)]. They are almost overlapping. However, the Gaussian curvature of the ruled surface is going as low as $-0.005$ whereas numerical evaluation of the developable surface shows the magnitude of the Gaussian curvature to smaller than $10^{-10}$. However, the surface was symbolically derived so we know it to be exactly zero:

Transparent close up view of the interesting area: