Links to related topics | Glossary | References

formed is not illuminable from every point. The polytopal room in figure 1, for example, does not admit a pool shot from

Ao to A1.

**Figure 1 - Cylinder made by sweeping a non-illuminable
region**

Given any point x in C, let us examine the orthogonal projection of
the rays out of x onto the base S. Consider a ray r out of point x that
hit one of the walls at a point p in such a manner that the reflecting
plane of r at p is perpendicular to the base S. Then, he projection of
r onto S preserves the the angles. This implies that reflection is preserved
by the projection.

Now consider rays out of x that create reflecting planes not parallel
to S when they hit the wall. Because the wall is perpendicular to the base
S, the orthogonal projection will preserve reflection even if it does not
preserve the angles (see figure 2).

**Figure 2 - The projection of the ray r on
P preserves reflection**

Note also, that a ray leaving x and reflecting on base S (or the opposite
top surface) will be projected as a simple ray. The projection conserves
only its motion in the plane parallel to S. When it will hit a wall, after
having bounced off S

(maybe several times) this would look like the first reflection in
the projection plane.

Therefore, any ray out of x reflecting on the wall can be reduce to
a projection onto S.

We can therefore conclude that the path taken by rays in three dimensional
cylinder C, reverts to an orthogonal projection on base S. Since the shape
of S is not illuminable, there exists a part of C, projecting orthogonal
onto S that will never be reached from a point p. For example, in
figure 1, any ray leaving A0 will never reach A1.

Figure (3b) shows a polytopal room which is not illuminable. This room is built by following the following rule. Any edge not attached to vertex A must remain an edge.

**Exercise** - prove claim 2 using the
projection argument

**Figure 3 - (a) In a cube, no ray will leave
point A to be reflected back to point A.**
**
(b) A non-illuminable region built by intersecting many cubes.**
**
There is not ray leaving a point labeled A in one cube that will reach**
**
another point labeled with an A.**

We obtain the object shown in figure (4b). Then there are no rays leaving A or D that will be reflected back to A or D. The latter shape is particularly interesting since it suggests some possible applications in acoustics or thermodynamics.

**Exercise** - prove that the object
of figure 4b is not illuminable from either its vertex A or D.
**Exercise** - build other non-illuminable
solid using smooth surfaces. Is there a three dimensional counter part
to the Penrose mushroom which is non cylindrical?

**Figure 4 (a) No ray leaving point A
will ever reach point D**
**
(b) Non illuminable polytopal object build by rotating the shape in (a)
around the axis made by AD.**

Links to related topics | Glossary | References