of a 1×1 square and a stick in the shape of a b×(1/b) rectangle (with b>1>(1/b)). As b increases, the area of the enclosing cube (≈b2) increases, but the area of the enclosed cube remains constant (=1) and the total area of the shape also remains constant (=2). Thus the enclosing-cube-slimness can grow arbitrarily while the enclosed-cube-slimness remains constant (=√2). See this GeoGebra page for a demonstration. On the other hand, consider a rectilinear 'snake' with width 1/b and length b, that is entirely folded within a square of side length 1. As b increases, the area of the enclosed cube(≈1/b2) decreases, but the total areas of the snake and of the enclosing cube remain constant (=1). Thus the enclosed-cube-slimness can grow arbitrarily while the enclosing-cube-slimness remains constant (=1). With both the lollipop and the snake, the two-cubes-slimness grows arbitrarily, since in general:enclosing-ball-slimness ⋅ enclosed-ball-slimness = two-balls-slimnessenclosing-cube-slimness ⋅ enclosed-cube-slimness = two-cubes-slimnessSince all slimness factor are at least 1, it follows that if an object o is R-fat according to the two-balls/cubes definition, it is also R-fat according to the enclosing-ball/cube and enclosed-ball/cube definitions (but the opposite is not true, as exemplified above). Balls vs. cubes The volume of a d-dimensional ball of radius r is: , where Vd is a dimension-dependent constant: A d-dimensional cube with side-length 2a has volume (2a)d. It is enclosed in a d-dimensional ball with radius a√d whose volume is Vd(a√d)d. Hence for every d-dimensional object: enclosing-ball-slimness ≤ enclosing-cube-slimness ⋅ . For even dimensions (d=2k), the factor simplifies to: . In particular, for two-dimensional shapes V2=π and the factor is: √(0.5 π)≈1.25, so: enclosing-disk-slimness ≤ enclosing-square-slimness ⋅ 1.25 From similar considerations: enclosed-cube-slimness ≤ enclosed-ball-slimness ⋅ enclosed-square-slimness ≤ enclosed-disk-slimness ⋅ 1.25 A d-dimensional ball with radius a is enclosed in a d-dimensional cube with side-length 2a. Hence for every d-dimensional object: enclosing-cube-slimness ≤ enclosing-ball-slimness ⋅ For even dimensions (d=2k), the factor simplifies to: . In particular, for two-dimensional shapes the factor is: 2/√π≈1.13, so: enclosing-square-slimness ≤ enclosing-disk-slimness ⋅ 1.13 From similar considerations: enclosed-ball-slimness ≤ enclosed-cube-slimness ⋅ enclosed-disk-slimness ≤ enclosed-square-slimness ⋅ 1.13 Multiplying the above relations gives the following simple relations:two-balls-slimness ≤ two-cubes-slimness ⋅ √d two-cubes-slimness ≤ two-balls-slimness ⋅ √d Thus, an R-fat object according to the either the two-balls or the two-cubes definition is at most R√d-fat according to the alternative definition. Local fatness The above definitions are all global in the sense that they don't care