My argument concerns just the cross-section of the region around one particular tetrahedral edge of the surface, and cut off by the extended neighbouring faces of the tetrahedron. The Minkowski average combines a Reuleaux "cusp" with a Meissner cyclide: its cross-section is the mean of a pair of adjacent arcs with a single arc, yielding some kind of non-circular curve with a rounded apex of greater curvature, and only its endpoints in common with either summand. The Roberts comprises an lower portion shared with the Releaux arcs, capped higher up by a circular arc. The two curves are evidently incongruent; therefore so are the completed surfaces. Finally, while I'm prepared to believe that there may be some obvious or well-known reason why this global Minkowski surface should retain the constant curvature of its summands, the matter remains currently beyond my comprehension --- can anybody cast any light here? Fred Lunnon On 1/11/13, Dan Asimov <dasimov@earthlink.net> wrote:
But the Minkowski average in question does not involve the Reuleaux tetrahedron as summand.
It is the sum of two types of Meissner tetrahedra: One (M_V) where 3 Reuleaux arcs that share a vertex are modified, and one (M_F) where 3 Reuleaux arcs that share a face are modified.
--Dan
On 2013-01-10, at 7:57 PM, Fred lunnon wrote:
Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all.
So the Minkowski yields a distinct surface.
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