Which solids are characterized by their orthographic projections?

Question

If I know the orthographic projections of a given solid in Euclidean 3-space onto the $xy$, $xz$ and $yz$ planes, under which circumstances can I reconstruct the solid based on that information alone?

Clearly, it is necessary that the solid be convex, otherwise it might have inside surfaces that are not visible from the outside at all and thus will not leave a mark in any of the projections. But is that also sufficient?

Answer 1

This mathoverflow question is related: Is the sphere the only surface all of whose projections are circles?

In particular, they give an example of two distinct objects which have the exact same set of orthogonal projections (i.e. every projection of one is a projection of the other in some, possibly different direction).

This can be adapted for your situation; in fact, it's even simpler. Cutting off cross caps in the first octant from a sphere does not change any of the three projections.

Edit: The paper "How Many 2D Silhouettes Does It Take to Reconstruct a 3D Object?" by Aldo Laurentini examines this question in great deal, including describing families of surfaces that are exactly reconstructible from their sillhouettes.

Answer 2

Given a compact convex set $K\subset{\mathbb R}^3$ with $0$ in its interior the support function $H:\>S^2\to{\mathbb R}_{>0}$ of $K$ is defined as follows: $$H(u):=\max\{u\cdot x\>|\>x\in K\}\qquad(u\in S^2)\ .$$ $H(u)$ is the distance from the origin of the supporting plane of $K$ having outer normal $u$. The function $H$ can be pretty arbitrary apart of a convexity condition. Exhibiting the projections $\pi_i(K)$ onto the coordinate planes $x_i=0$ means giving $H(u)$ on the three great circles in these planes. Maybe there are some special examples where $K$ is uniquely determined by these data.

Answer 3

Let us define the three orthographic projections as follows:

$$\pi_x(K) = \{ (y, z) \in \mathbb{R}^2 \mid \exists x.(x, y, z) \in K \}$$ $$\pi_y(K) = \{ (x, z) \in \mathbb{R}^2 \mid \exists y.(x, y, z) \in K \}$$ $$\pi_z(K) = \{ (x, y) \in \mathbb{R}^2 \mid \exists z.(x, y, z) \in K \}$$

Given the projections, the problem is to reconstruct $K$. As pointed out in the comments, there exist $K \neq K'$ such that $\pi_x(K) = \pi_x(K')$ and $\pi_y(K) = \pi_y(K')$ and $\pi_z(K) = \pi_z(K')$. In other words, given the projections, there are multiple possible reconstructions.

One canonical choice of reconstruction is to form the largest solid that is consistent with the projections. If any reconstructions are possible, this choice will be the union of all of them. In that case, the canonical largest solid can be defined by

$$R(K_x, K_y, K_z) = \{ (x, y, z) \in \mathbb{R}^3 \mid (y, z) \in K_x \land (x, z) \in K_y \land (z, y) \in K_z \}$$

This choice of reconstruction is arbitrary, but sensible. If we decide to use it, then the condition for $K$ to be correctly reconstructed from its projections is simply $K = R(\pi_x(K), \pi_y(K), \pi_z(K))$. This statement is almost vacuous!

It is notable that the solid does not need to be convex. Here are some interesting examples of reconstructable solids:

• Any set of the form $K = A \times B \times C$, where $A, B, C \subseteq \mathbb{R}$.

• The union of a unit cube centred on $(-1, -1, -1)$ and another centred on $(1, 1, 1)$.

• The union of unit cubes centred on $(0, 0, 0)$, $(\pm2, 0, 0)$, $(0, \pm2, 0)$, $(0, 0, \pm2)$.