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.