While studying some physical problem we stumbled across a geometrical problem, to which we could not find any solution.

Consider a unit sphere in 3D and a collection of $N$ points on its surface. Now you choose a direction to project the sphere to a 2D-plane, which yields a unit circle and points inside it. We then calculate the maximal distance of the points to the center of the circle, which is what we call the "size of the projection" (i.e., the smallest radius $r$ such that all points lie inside the circle with radius $r$).

Now, for a fixed placement of $N$ points, one can find the direction of projection that that *minimizes* the *maximum* over the distances, see this figure:

We are now interested in the *worst* placement of the point, that maximizes this minimal distance.

More formally, if we parameterize the direction of the projection by a unit vector $\vec{n}$, the distance of a point $\vec{p}$ to the center of the projected circle is given by $$ d_{\vec{n}}(\vec{p}) = \sqrt{1-(\vec{p}\cdot\vec{n})^2} = |\vec{p} \times \vec{n}| $$ and we are looking for the quantity $$ B(n):=\max_{\{\vec{p}_i\}_{i=1}^N} \min_{\vec{n}} \max_i d_\vec{n}(\vec{p}_i). $$

Some edge cases (see figure 2):

For $N=1$, no matter how you place it, you can always project along its direction to reach distance zero, thus $B(1)=0$.

For $N=2$, the best placement is choosing the two points orthogonal, yielding $B=\frac{1}{\sqrt{2}}$.

If you have an infinite number of points, you can evenly distribute them around the sphere, and there are always points with distance one. Thus, $B(\infty)=1$.

In between, the function is monotonically increasing.

Interestingly, for $N=3$, the best placement is not given by a trihedron, with three pairwise orthogonal points ($0.816$), but by evenly distributing them on the equator ($B=\frac{\sqrt{3}}{2} \approx 0.866$). For $N=4$, however, a tetrahedronal placement ($B=\frac{4}{3\sqrt{2}} \approx 0.943$) is favored over a planar arrangement.

The question now is the following: Is there a closed expression for $B(N)$ for all $N$? If not, is there an upper bound?

Best, Herimon