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Take a flat, flexible ring, such as a rubber washer. It has an inner radius for its inner circle, and an outer radius for its outer circle. Fold it so that the inner circle of the washer joins up with itself in a half-circle, and the outer circle remains loose. Then pinch the inner circle shut so that it becomes a line segment, still leaving the outer circle loose.

enter image description here

Is there a name for the resulting saddle-like surface?

A version of this surface can be plotted in polar coordinates as $f(r, \theta) = sin(2 \theta)$.

A 3d plot on sage cell (click and drag to move the figure): here

enter image description here

causative
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  • Connected: https://math.stackexchange.com/q/1561008 – Jean Marie May 01 '21 at 18:06
  • There's no name I'm aware of. Intrinsically, it's flat away from the semicircle arc, the integral of the curvature is $\pi$ at each cone point (i.e., the angular defect at each is $\pi$), and the integral of the curvature over the semicircular arc is $-2\pi$. If it matters, this is not isometric to the graph in the edit, and is only homeomorphic if we pinch off two pairs of "adjacent" sheets along the vertical axis leaving a V (the vertical interval traced twice), but not if we include the vertical segment over the origin, gluing all four sheets. – Andrew D. Hwang May 01 '21 at 18:37
  • @AndrewD.Hwang the semicircle arc was only there in an intermediate step, it gets glued to itself to form a line segment gluing all four "sheets" if I'm understanding you correctly – causative May 01 '21 at 18:55
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    The point about the semicircle is, even though it's been "sewn up", the surface has negative "Gaussian curvature" along the seam, with the integral of the curvature along the arc being $-2\pi$. (The Gaussian curvature is infinite pointwise, but its integral along an arc is finite.) The resulting surface is topologically a disk. <> If it matters, the graph $z = \sin\theta$ perhaps looks more like the annulus with inner boundary identified, assuming I've understood your description, though for that surface as well, the portion of surface on the vertical axis has curvature $0$. – Andrew D. Hwang May 02 '21 at 03:26
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    @AndrewD.Hwang Oh, I got it, thanks. It makes me curious if there's some sense "between" the geometric sense and the topological sense, where the "pinches" matter and this is different from a disk. – causative May 02 '21 at 04:50
  • It's a good question whether there's a "useful" structure between a metric structure on one hand and a conformal structure on the other. Locally, a cone point is conformally equivalent to a disk by a branch of a complex power function, though metrically a cone point and a smooth/flat disk are distinct. – Andrew D. Hwang May 02 '21 at 13:51

2 Answers2

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This is Plucker's conoid (with $n = 2$).

Mathematica graphics

See also https://mathcurve.com/surfaces/plucker/plucker.shtml

Eric Towers
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  • Congratulations !!! Now a question: is there a family of parametrized surfaces evolving from say a half-torus to this surface ? – Jean Marie May 02 '21 at 11:52
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Radially pinched Developable helicoid?

Some other names are in:

Developable helicoid

e = 1; q = 1;
p1 = ParametricPlot3D[{  Cos[u] - q v Sin[u], Sin[u] + q v Cos[u], 
   e (u + v)}, {u, 0 , 2 Pi}, {v, 0, 3}, PlotStyle -> Green, 
  Mesh -> {12, 8}]
p2 = ParametricPlot3D[{  Cos[u] - q v Sin[u], Sin[u] + q v Cos[u], 
   e (u + v)}, {u, 0 , 2 Pi}, {v, 0, -3}, PlotStyle -> Pink, 
  Mesh -> {12, 8}]
Show[{p1, p2}, PlotRange -> All]

The helical line of stiction traces a cuspidal edge when sweeping out a single nappe of zero Gauss curvature K surface shown below.

If we further pinch/glue a developable helicoid along two radial lines I suppose it would make a surface homomorphically equivalent to the one you sketched.

enter image description here

Narasimham
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