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This post was posing the question "The Calippo™ popsicle has a specific shape ... Does this shape have an official name?", and the accepted answer was that it is a Right Conoid which in fact is much resembling.

However, a conoid is not developable and so cannot be economically made out of a cardboard sheet.

Now, a cardboard cylinder, pinched and sealed at one end, will look like a toothpaste tube, and the pinched end will be larger than the mouth diameter (by $\pi /2 $).
That does not fit to the calippo's look, whose tube is in fact fabricated from a cone (made from a plane circular sector) pinched at one end (and reinforced at the opposite circular rim).

And a related subject is that, a cylinder with both ends pinched at 90° relative to each other make up a tetrahedral "milk carton" (berlingot in French).
In this Mathcurve page after stating that
"le berlingot est la surface réglée non développable .. ",
referring to a picture of a real tetrahedral milk carton is noted:
"Attention, un berlingot comme ci-dessus fabriqué avec du papier est une surface développable, fabriquée avec un patron de tétraèdre en arrondissant les arêtes... "
Also refer to the Coin Conic page.

Suppose that the material is inextensible, and that a grid of directrices and orthogonal circles is marked on the cone/cylinder, then it is clear that upon pinching the directrices will not appear straight. Since a developable surface implies that it be ruled, then either the ruling lines have moved, or the surface has undergone some deformation.
Therefore the question is:

Can the calippo (and bottom of toothpaste, and milk) package be a ruled developable surface , in case, admitting that the shape of the pinched edge be not exactly linear ? or is some deformation of the material demanded ?

---- P.S: -----

Thanks to the answer, with expressive sketch, received by Rahul, complemented with that provided by C.Blatter to the previous post, we can conclude that such a surface is a plausible model.
So, scaling all the dimensions to have the base circle of radius $1$, height $=h$ and length of the pitch $=2c$

calippo3

we can parametrise the surface as $$ \eqalign{ & {\bf q} = \left( {sc,\;0,h} \right) - \left( {\cos \alpha ,\;\sin \alpha ,0} \right) = \left( {sc - \cos \alpha ,\; - \sin \alpha ,h} \right) \cr & {\bf p} = \left( {\cos \alpha ,\;\sin \alpha ,0} \right) + t\,{\bf q} \cr} $$ where $$ \left\{ \matrix{ \alpha = - \pi /2\quad - 1 < s < 1\quad 0 \le t \le 1 \hfill \cr - \pi /2 < \alpha < \pi /2\quad s = 1\quad 0 \le t \le 1 \hfill \cr \alpha = \pi /2\quad - 1 < - s < 1\quad 0 \le t \le 1 \hfill \cr \pi /2 < \alpha < 3\pi /2\quad s = - 1\quad 0 \le t \le 1 \hfill \cr} \right. $$

Then we can develop half of the surface as indicated, by putting for the flattened conic part that $$ q(\alpha )d\beta (\alpha ) = d\alpha \quad \left| \matrix{ \,q = \left| {\bf q} \right| \hfill \cr \,s = 1 \hfill \cr} \right. $$

G Cab
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  • The other suggestion in that question, the convex hull of a line segment and a circle, is a developable surface. Does that answer your question? –  Apr 27 '17 at 06:25
  • On the other hand the tetrahedral milk carton will actually form exactly a tetrahedron with four flat triangular faces. –  Apr 27 '17 at 06:27
  • @Rahul what do you mean by convex hull ? made by straight lines ? if so, how do you relate (map) the end points $P(s)$ on the segment with $Q(r \alpha)$ on the circle? under a linear map I am not sure it will be developpable. That's the core of the question. – G Cab Apr 27 '17 at 10:09
  • @Rahul One matter is to make up a tetrahedron by taking a piece of paper, draw the four triangles connected on a hedge, fold it and solder on three edges. Another is to start from a cylinder, pinch and solder it at the ends, making them more or less straight, in case trimming straight the joints. Just think and cut open a milk carton .. – G Cab Apr 27 '17 at 10:26
  • Careful: the curved sections do not unfold to sectors of a circle, because they are not portions of right circular cones. After all the three angles at the apex of each isosceles triangle should add up to $180^\circ$. –  Apr 27 '17 at 22:46
  • oohps, you are .."right" ! the half cones are not right : their median generatrix is (if the pinch equals the diameter). But I do not get your remark about the triangles. – G Cab Apr 27 '17 at 23:30
  • I mean there should not be two "corners" where the arcs meet; they should form a single differentiable contour. After all if you look at the two sliver triangles on both sides of the large one in my discretized figure, all three are almost coplanar, and their angles at the shared vertex add up to almost $180^\circ$. In the limit this tells you something about the angle between the contour's tangents on either side. –  Apr 27 '17 at 23:46
  • A better approximation might be a single circular arc centered at roughly $(0,-2)$ in your figure. –  Apr 27 '17 at 23:48
  • @Rahul: so I re-elaborated the situation and edited the P.S., and in fact the sector is almost circular (for $h>>c$ ) – G Cab Apr 28 '17 at 13:35

1 Answers1

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The convex hull of a circle and a line segment is developable. I don't know if this is the unique developable surface with the given boundaries.

enter image description here

(Ignore the extra chords drawn in the circle.)

I'm fairly sure that if you start with a cylinder and pinch it flat at both ends, a perfectly flat tetrahedron is the only developable surface you can get. The curved nature of the real-world milk cartons is because the material is slightly extensible (it has a coating of plastic to make it liquid-proof). Better intuition can be gained by trying the experiment with a sheet of plain paper instead.

  • this solid, with the right proportions chosen, can cast a shadow of a circle, a square, or an equilateral triangle. From a book I read as a child. – Will Jagy Apr 27 '17 at 17:16
  • @WillJagy: Usually people have a slightly different shape in mind for that problem, though this one works too. However I think it is impossible to have a circle, a square, and an equilateral triangle as shadows, because the altitude of the triangle is too short. –  Apr 27 '17 at 17:20
  • I guess you are right, if you tilt it just a little you get an elliptical arc as part of the shadow. – Will Jagy Apr 27 '17 at 17:22