Talk:Developable surface

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"what" "is" "the" "deal" "with" "all" "the" "quotation" "marks" —Preceding unsigned comment added by Monguin61 (talk • contribs) 05:11, 17 March 2009 (UTC)[reply]

In mathematics, a developable surface is a surface with zero Gaussian curvature.

Wondering if having everywhere zero curvature implies that the surface is developable? If so is their a proof of the result? --Salix alba (talk) 19:55, 27 February 2007 (UTC)[reply]

By definition, surface is developable if and only if it has zero Gaussian curvature (sum of angles of any triangle on that surface always equals to 180°). Admiral Norton ^(talk) 20:57, 19 February 2008 (UTC)[reply]

Complete classification?

I've just removed the following reference from the article: Stoker, J.J. (1961), Developable surfaces in the large. Comm. Pure Appl. Math. 14(3), 627--635, doi:10.1002/cpa.3160140333. A previous editor cited this reference in support of the statement "The developable surfaces which can be realized in three-dimensional space are...", implying that this is a complete list of developable surfaces. In fact the paper quotes this as an erroneous statement, then goes on to explain that there exist other developable surfaces. I don't know whether it's possible to give a complete list of 'all developable surfaces that exist in three dimensions.

For reference, here is the first paragraph of Stoker's paper:

In various textbooks on differential geometry one finds the statement
that the developable surfaces in three-dimensional space can be classified as the plane, the cylinder, the cone, or the tangent surface of a space curve. As a consequence, it would be clear that the only developables without singularities-if they are continued far enough along generators-would be the plane and the cylinder. If one examines the usual proofs of the statement, however, one finds that a quite restrictive assumption is made, i.e. that all points of the surface are either planar points (that is points such that all coefficients of the second fundamental form are zero) or that all points are parabolic points. Obviously, developables exist which contain points of both kinds. For example, a plane triangle to which circular cylinders have been attached to the sides is a case in point; and furthermore, such a surface
could be so defined as to have derivatives of arbitrarily high order.

Jowa fan (talk) 02:27, 5 December 2012 (UTC)[reply]

I don't quite get the triangle example it seems that it would need to be a surface with a boundary or self intersection. In effect his construction is union of generalised cylinders and planes. If you allow such piecewise construction then you could end up with a large set of posible shapes. --Salix (talk): 09:06, 5 December 2012 (UTC)[reply]