How to measure the radius of a sphere?

Question

It is easy enough to determine the radius of a circle, using a minimal number of "tools". For instance, one could take any two points on the circle, build a right angle triangle and, by Thales's theorem, the hypotenuse of said triangle would also be the diameter of the circle.

However, I would like to know if there are any simple methods of accurately measuring the diameter/radius of a sphere, without cutting it or altering it in any other way, and using a minimum number of tools and intermediate measurements (to reduce the final error).

One method that comes immediately to mind is to pin down the sphere between two flat surfaces (blocks) and measure the separation between these blocks; but that method introduces many possible sources of measurement errors, and requires the use of additional equipment (the blocks themselves).

tl;dr : Is there a simple way to determine the radius of a sphere without using anything else than a (bendy) ruler?

Answer 1

Are you aware of Eratosthenes' experiment?

A variation of this would involve taking a wedge formed by two planes and putting the sphere between the planes and measuring the distance from the corner of the wedge. Could you work the math for this example?

Answer 2

If you can measure the "straight-line" distances accurately but not allowed to move the end-points of your ruler to maximize the distance and determine the diameter, you can pick any $4$ points $\vec{p}_1, \vec{p}_2, \vec{p}_3, \vec{p}_4$ on the sphere which approximately forms a regular tetrahedron, the circumradius of the tetrahedron will be the radius $r$ of the sphere you want.

To determine the circumradius $r$, you can measure the 6 distances between the points

$$d_{ij} = | \vec{p}_i - \vec{p}_j | \quad\text{ for }\quad 1 \le i < j \le 4. $$

The radius $r$ will be the unique positive value which make following determinant vanishes:

$$\det\begin{bmatrix} 0 & 1 & 1 & 1 & 1 & 1\\ 1 & 0 & r^2 & r^2 & r^2 & r^2\\ 1 & r^2 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2\\ 1 & r^2 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2\\ 1 & r^2 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2\\ 1 & r^2 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0 \end{bmatrix} = 0\tag{*1}$$ If one expand this mess out, you will find this is an linear equation in $r^2$.

Geometrically, the determinant above is the Cayley-Menger determiant of a $4$-simplex formed by the center of the sphere and the 4 vertices $\vec{p}_i$.

It is known that the Cayley-Menger determinant for a $n$-simplex is proportional to the square of the $n$-volume of the $n$-simplex. So the vanishing of determinant is $(*1)$ is nothing but the obvious fact that the $4$-simplex formed by above $5$ points is degenerate and has zero hyper-volume in $4$-space.

Answer 3

It is easy enough to determine the radius of a circle. There is another experiment rather

Using a compass, draw four identical circles, each having a flat radius $r$ (known value), on the surface of the sphere with unknown radius say $R$ such that these circles exactly touches one another at six different points on the spherical surface. (see in the diagram)

In this case of tangency of four circles, the radius of sphere $R$ is simply given as $$R=r\sqrt{\frac{3}{2}}$$