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I drew two intersecting circles and then folded the 3D shape.

enter image description here enter image description here

Question.

Does this 3D shape have a mathematical name? is there a formula that describes this 3D shape?

Edit.

I ploted an astroid and circle.

enter image description here

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  • $\begingroup$ According to me it looks like the shape formed between 4 intersecting spheres. The name is the tricky part. But this is just my constrained imagination. It could be something else. $\endgroup$ – Mohammad Zuhair Khan Dec 26 '17 at 6:59
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    $\begingroup$ There's a bit of cheating involved: the side parts are cylindric, but you can't get a plane sheet of paper into a shape with Gaussian curvature $\neq0$ without distortions/creases, like the upper and lower parts of your shape.. $\endgroup$ – Professor Vector Dec 26 '17 at 7:42
  • $\begingroup$ The name will be the easy part. First let us obtain a mathematical description of this object, so that we can see whether it "exists" or not. @ProfessorVector: One can fold a piece of cardbord along a curved line. The structure of the cardboard then is only damaged along the fold, which should be acceptable. $\endgroup$ – Christian Blatter Dec 26 '17 at 14:39
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    $\begingroup$ It seems to me that this surface really exists, the top part consisting of a flat square and four cylinder segments (although one cannot make the whole shape from a single piece of paper without making distortions in process). It is mentioned at ics.uci.edu/~eppstein/junkyard/teabag.html (2 Dec 1998 message), and some origamists call it a pillow box ( mathcraft.wonderhowto.com/how-to/… ). $\endgroup$ – colt_browning Dec 26 '17 at 23:28
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It can be done without cheating; and, as colt_browning has remarked, such boxes can even be bought.

The box has a star shaped top and bottom, and four lens shaped sidewalls. The following figure shows the bottom star before the bending. The idea is to bend the four spikes cylindrically up, so that the grey lines become level lines, and the small central square stays flat. The bending is performed in such a way that the arc $\gamma$ in the figure becomes an arc $\gamma'$ rising according to the same law $s\mapsto z(s)$ as the lower rim of a lens put on edge.

enter image description here

The arc $\gamma$ begins at $\bigl(1-{\sqrt{2}\over2},1-{\sqrt{2}\over2}\bigr)$ and ends at $(1,0)$. It can be parametrized by arc length as $$\gamma:\quad s\mapsto\left(1-\sin\left({\textstyle{\pi\over4}}-s\right),1-\cos\left({\textstyle{\pi\over4}}-s\right),\ 0\right)\qquad\left(0\leq s\leq{\textstyle{\pi\over4}}\right)\ .$$ After the bending we have an arc $\gamma'$ given by $$\gamma':\quad s\mapsto\left(x(s),1-\cos\left({\textstyle{\pi\over4}}-s\right),\ 1-\cos s\right)\qquad\left(0\leq s\leq{\textstyle{\pi\over4}}\right)\ .$$ The function $s\mapsto z(s)=1-\cos s$ models the slope of the lower lens rim, and $s\mapsto x(s)$ has to be determined from the condition $\dot x^2+\dot y^2+\dot z^2=1$. One obtains $$\dot x^2(s)={1\over2}\bigl(\cos(2s)+\sin(2s)\bigr)\ ,$$ so that $s\mapsto x(s)$ will not be an elementary function.

In the end the lens will be glued to $\gamma'$. It is then standing on edge and will be cylindrically bent as well.

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