Of course, a sphere is not Rupertable (in any dimension). But in 3D, any generic ellipsoid (axes A,B,C with 0<A<B<C) is Rupertable. Indeed ellipsoids in any dimension are generically Rupertable. The expansion factor is (at least) MIN(j) Axis[j+1]/Axis[j] if the axis lengths are assumed pre-sorted into increasing order. For the same reason hyperrectangles are generically Rupertable. Any 2D convex object with diameter>width is Rupertable. Therefore every 2D convex polygon is Rupertable. But the "Reuleaux triangle" in 2D, made of 3 circular arcs, https://en.wikipedia.org/wiki/Reuleaux_triangle is NOT Rupertable since like the circle it has diameter=width. There exist an infinite number of 2D "constant width convex objects" which all are non-Rupertable. An upper bound on the Rupert expansion factor of any convex object in any dimension, is diameter/width. This bound is tight in 2 dimensions. It is usually weak, but always valid, in higher dimensions. Meanwhile a lower bound is, of course, 1. (Definitions: diameter is furthest distance between two points. Width is closest distance between two parallel hyperplanes, which object lies between.) A usually-better upper bound on the Rupert expansion factor of any convex object Q in any dimension, is as follows. Define OP(Q,d) to be the orthogonal projection of Q in direction d. The upper bound is diam(Q) / MIN[over all directions d](diam(OP(Q,d))). (Definition: a "direction" is a unit-length vector.) But this upper bound also often seems very weak. Regard an n-simplex as determined by its (n+1)*n/2 edge lengths. In 2D any triangle is Rupertable. Does there exist any non-Rupertable simplex? In any dimension? Hell: does there exist any non-Rupertable convex polytope? Not knowing that, is pathetic. Here is an attempt to make one. Make a convex polyhedron in 3D with lots of vertices (say at least V=10^12) all lying on the sphere x^2+y^2+z^2=1, fairly uniformly randomly distributed. It is their convex hull. Any orthogonal projection, is going to be a polygon with many (say, at least about 10^6) vertices, that closely approximates unit circle but generically lies inside it. The number of combinatorially-distinct types of orthogonal projections is going to be at least about 10^24. No matter which one you choose, if seems unlikely that there will exist another which can be moved to lie strictly inside it. Because there are only going to be about 10^12 "different" ways to move it. Point is, it feels like you do not have enough control to repair about 10^6 problems with only 3 degrees of freedom worth of control. Too many constraints, too few variables. So I suspect polyhedra like that in 3D, are usually going to be non-Rupertable when V is large enough. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)