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?

  • 1
    $\begingroup$ Since you can compute the volume of a sphere ($V = \frac{4}{3} \pi r^3$), would submerging it in a graduated container of water be valid? It's probably prone to error and of course physical constraints of measuring spheres of "reasonable" sizes but it's a start. $\endgroup$
    – Xoque55
    Jan 3, 2015 at 19:49
  • $\begingroup$ Your application of Thales's theorem is incorrect. That would only work if you started with diametrically opposed points, in which case you already know the diameter. $\endgroup$
    – pre-kidney
    Jan 3, 2015 at 19:52
  • $\begingroup$ en.wikipedia.org/wiki/Micrometer $\endgroup$
    – Will Jagy
    Jan 3, 2015 at 19:53
  • 1
    $\begingroup$ @pre-kidney: You missed the point. Start with any two points $A$ and $B$ on the circumference; construct a line through $B$ perpendicular to $AB$; let $C$ be the intersection of this line with the circumference. Then $AC$ is a diameter. $\endgroup$
    – TonyK
    Jan 3, 2015 at 20:10
  • 1
    $\begingroup$ @TonyK: I think it is in the spirit of using a bendy ruler. Tha is why I made it a comment, as well $\endgroup$ Jan 3, 2015 at 21:04

5 Answers 5


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?

  • $\begingroup$ I like this method. Also, the maths is trivial if the planes are perpendicular to each other. $\endgroup$
    – TonyK
    Jan 3, 2015 at 20:17
  • $\begingroup$ But the op already mentioned this method and is looking for an alternative! $\endgroup$
    – sranthrop
    Jan 3, 2015 at 20:23
  • 1
    $\begingroup$ @sranthrop The method described by the op is different than mine. $\endgroup$
    – hjhjhj57
    Jan 3, 2015 at 21:47
  • $\begingroup$ @sranthrop: The OP's suggested method involved placing the sphere on a plane, and then placing a parallel plane on top of the sphere. This is fraught with practical and theoretical problems. In contrast, hjhjhj57's solution involves constructing (once and for all) a pair of perpendicular planes, rigidly attached to each other, and placing the sphere so that it touches both of them. Much easier in all respects! $\endgroup$
    – TonyK
    Jan 4, 2015 at 19:59

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.

  • $\begingroup$ @GlenWhitney your comment is detail enough and should be posted as a separate answer. $\endgroup$ Jul 24, 2018 at 6:43
  • $\begingroup$ @GlenWhitney In the complicated expression $r^2 = \frac14 \cdot \frac{a^2f^2 + \cdots}{a^2f + \cdots}$, $$\begin{align} a^2f^2 + \cdots &= -16\Delta^2(\sqrt{af},\sqrt{be},\sqrt{cd})\\ a^2f + \cdots &= -144 V^2 \end{align}$$ where $\Delta(u,v,w)$ is area of a triangle with sides $u,v,w$. and $V$ is the volume of the tetrahedron. The whole expression is equivalent to a result by A. L. Crelle. $$6Vr = \Delta(d_{12}d_{34},d_{13}d_{24},d_{14}d_{23})$$ $\endgroup$ Jul 24, 2018 at 7:41

This is really just a special case and practical refinement of @achille-hui's answer, but @achille-hui recommended I re-post it as a separate answer rather than comments to that other answer. So here goes.

Computing with Sage Math from the determinant formulas in that answer, if we let $a=d_{12}^2, b=d_{13}^2, c=d_{14}^2, d=d_{23}^2, e=d_{24}^2,$ and $f=d_{34}^2$, then $r^2$ is $$\frac14\cdot\frac{a^2f^2+b^2e^2+c^2d^2-2(bcde+acdf+abef)}{a^2f+af^2+b^2e+be^2+c^2d+d^2c+abd+ace+bcf+def-abe-abf-acd-acf-adf-aef-bcd-bce-bde-bef-cde-cdf}.$$ (Note that even though it may wrap to more than one line, all of the stuff under the fraction bar is a single sum of trinomials.)

Although I couldn't find any way to simplify the polynomials in the numerator and denominator (but see @achille-hui's comment for another perspective on that), we can geometrically recognize what terms appear. The edges corresponding to the squared lengths $a$ through $f$ come in three opposite pairs that are not incident at any vertex: $a\sim f, b\sim e$, and $c\sim d$. In the numerator we have the squares of the products for each opposite pair, minus twice the products of two different opposite pairs. In the denominator, we have for each opposite pair, their product times their sum, plus the products for each of the triangles in the tetrahedron, minus the products for any three edges that do not form a triangle and do not emanate from a single vertex.

To take practical advantage of this expression in the situation described by the OP, note that we can easily arrange to choose points so that $c=b=a$: Choose an arbitrary point $p_1$, set a compass to an arbitrary length $\sqrt{a}$ (chosen in the ballpark of 1/4 to 1/3 of the way around the sphere), and draw a circle with this setting centered at $p_1$. Choose three arbitrary points $p_2, p_3$, and $p_4$ on this circle (relatively equally spaced) and measure their pairwise distances and square them to get $d,e,$ and $f$. Then there is a great deal of cancellation in the formula and $r^2$ turns out to be $$\frac14\cdot\frac{a}{1+\frac{def}{a(d^2+e^2+f^2-2de-2ef-2df)}}.$$

However, we can simplify even further. It's also easy to arrange that $d=e=a$, as well. After drawing the first circle, choose a point $p_2$ on it arbitrarily, and then draw another circle of the same radius centered on $p_2$. The two points of intersection with the first circle are $p_3$ and $p_4$. Measure the distance between them and square it to get $f$. Then substituting $d=a$ and $e=a$ into the previous formula gives us that $r^2$ is now simply $$\frac14\cdot\frac{a(4a-f)}{3a-f}.$$

I have now tried both this variation of @achille-hui's solution, and @hjhjhj57's solution using a right angle, in practice. I found this method to be easier and more practical to execute. The primary difficulty is that it is hard to precisely observe and measure the points of contact of the sphere with the two planes in that solution, both because it's tricky to get a good view of the contact point and because near the point of tangency, the sphere looks very "flat" and lies quite close to the plane in a fairly large region, making the exact point of contact hard to identify. By contrast, in this method it's very quick to draw the two circles, and possible to measure the distance between their intersection points very accurately with calipers.


If you can measure the (ratios between the) six great-circle distances between four (not all co-linear) points of the sphere (say points $A$, $B$, $J$, and $K$) then calculate the largest real number $\lambda$ for which the corresponding Gram determinant vanishes:

$$\begin{vmatrix} 1 & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \right] & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{AJ}{AB}\right) \, \right] & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{AK}{AB}\right) \, \right] \cr \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \right] & 1 & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{BJ}{AB}\right) \, \right] & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{BK}{AB}\right) \, \right] \cr \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{AJ}{AB}\right) \, \right] & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{BJ}{AB}\right) \, \right] & 1 & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{JK}{AB}\right) \, \right] \cr \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{AK}{AB}\right) \, \right] & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{BK}{AB}\right) \, \right] & \text{Cos} \! \left[ \, \left(\frac{1}{\lambda}\right) \, \left(\frac{JK}{AB}\right) \, \right] & 1 \end{vmatrix} = 0.$$

The sought radius of the sphere is expressed as $\lambda \, AB$.

Cmp. S. L. Kokkendorff, Gram Matrix Analysis of Finite Distance Spaces in Constant Curvature.


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)

circles on the sphere

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

  • $\begingroup$ I don't see any practical method to find $r$ / draw the circles when $R$ is not already known. $\endgroup$ Jul 23, 2018 at 20:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .