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there's my question:

Given 2 regular plane curves (let's say $\mathcal{C}^1$) in the 3D space, is there always a developable surface which contains both curves ?

Thanks, anders

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Sure. If you have parametric equations for both your curves, and the parameter ranges for both curves are the same, you can then consider the surface drawn out by a moving straight line whose endpoints are at the two curves. – J. M. Jul 26 '12 at 16:44
Do you really want the surface to contain the curves, or to have them as parts of the boundary? Are the curves disjoint? Is the surface required to be embedded (without self-intersections)? – user31373 Jul 26 '12 at 21:16
@J.M.: It seems to me you are proving that you can build a ruled surface from two curves, but nothing proves such built surface is developable. – anderstood Oct 13 '13 at 16:39

I think the answer is no.

Take two skew lines, A and B as the two curves. Take some point P on line A. Using the "ruled surface" approach mentioned by J.M. this point P would be connected by a linear "ruling" to some point Q on line B. In order for the surface to be developable, the surface normal has to be constant along each such ruling. But clearly it won't be if the lines A and B are skew (not coplanar).

This isn't a rigorous proof. It only deals with attempts to construct a developable surface using the linear ruling approach. I have shown that the linear ruling approach does not always work. Conceivably there could be some other way to construct a developable surface. I don't think there is another way, but I haven't proved this.

On the positive side, here is a construction that often works. At a given point P on curve A, construct the tangent line, L. Take a plane containing this tangent line, and rotate it around L until it touches curve B (tangentially), say at a point Q. This may not be possible, but it often is. Take the line between P and Q as a ruling. The ruled surface constructed this way is developable.

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I am not sure your proposition yields a "globally"-developable surface, even if the point Q is always defined ; I think it only ensures "local"-developability. Do you agree? – anderstood Oct 13 '13 at 16:43

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