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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

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  • $\begingroup$ Connected: math.stackexchange.com/q/1561008 $\endgroup$
    – Jean Marie
    May 1 at 18:06
  • $\begingroup$ 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. $\endgroup$ May 1 at 18:37
  • $\begingroup$ @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 $\endgroup$
    – causative
    May 1 at 18:55
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    $\begingroup$ 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 $0$. $\endgroup$ May 2 at 3:26
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    $\begingroup$ @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. $\endgroup$
    – causative
    May 2 at 4:50
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This is Plucker's conoid (with $n = 2$).

Mathematica graphics

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

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  • $\begingroup$ Congratulations !!! Now a question: is there a family of parametrized surfaces evolving from say a half-torus to this surface ? $\endgroup$
    – Jean Marie
    May 2 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

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  • $\begingroup$ It looks rather far from the targeted surface. $\endgroup$
    – Jean Marie
    May 2 at 11:56
  • $\begingroup$ Resemblance can come only after cutting and gluing together the two radial cut edges perpedicular to the cuspidal helix. $\endgroup$
    – Narasimham
    May 2 at 13:05

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