WOLFRAM|DEMONSTRATIONS PROJECT

Some Unistable Polyhedra

​
number of faces is 2m+1
m
9
angle correction ϵ
0.
0.05
0.1
0.15
0.2
0.25
0.3
consider the mass of the polyhedron to bethe subset consisting only of its
vertices
edges
faces
interior
a
12
b
4
show segment of the line z = h y - a
and perpendicular from mass center
to one prism's base
opacity
0.7
show
polyhedron
cross section
2D net
unistable polyhedral…
… solid with 19 faces
… surface with 15 faces
… skeleton with 13 faces
zoom
y coordinate of the mass center = 0.306938
number of faces = 19
A polyhedral solid with uniform density is unistable (or monostable) if it is stable on exactly one face. Besides these polyhedra, this Demonstration also considers polyhedra as surfaces, skeletons, and as a set of vertices.
In 1968, Guy constructed a unistable polyhedron with 19 faces. It is a kind of truncated prism. Half of its cross section consists of
m=9
right-angled triangles (see cross section); each of the angles at the coordinate center equals
π/m
. The construction here allows the first and the last angle to be a little larger, namely
π/m+ϵ
(
ϵ
is called the angle correction), and the others to be a bit smaller. If truncation makes the center of mass fall below the coordinate origin, only the bottom face (see cross section) and possibly two truncated bases are stable. But the last two can be made unstable by making the length
2b
of the shortest lateral edge be sufficiently large. In this case, the perpendicular from the center of mass (green line) to the corresponding prism's new basis must not intersect it. So the blue line on the basis and the green line must meet outside of the solid.
For example, to find a unistable polyhedral surface with 15 faces, choose
m=7
, consider the polyhedron as a set of faces, and set
a=29
,
b=1.83
. To get a unistable polyhedron, its center of mass (red point) must be below the origin (black point). The meaning of the parameters
a
is:
z=±a
, which are the
z
coordinates of intersections of the polyhedron with the
z
axis. So the slope of the blue line is
h=(a-b)/r
, where
r
is the distance of the origin to the shortest lateral edge of the polyhedron.