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This question was inspired by this thread I just saw on Space Exploration Stack Exchange:

https://space.stackexchange.com/questions/39163/did-feynman-cite-a-fallacy-about-only-circles-having-the-same-width-in-all-direc

where an anecdote is mentioned regarding how that the famous physicist Richard Feynman realized that part of the fault for the 1986 Space Shuttle Challenger disaster lie in not realizing that "a circle is not the only figure which has the same width in all directions".

Basically, it seems, the inspectors had tried to verify that the fuel tank sections, which are tubes and need to have cross-sections which are almost exactly perfect circles at the points at which they join together, in order to ensure a good fitting-together thereof with no leaks, by naively measuring the diameter over and over in different places and seeing that all the diameter results seemed to agree, and hence concluded the tanks were "circular".

And the problem with this is that there are non-circular shapes which have the same "diameter" at all points - a famous example being the so-called Reuleaux triangle which, if used to form the cross-section of a drill bit and then such suitably mounted, can be used to quickly and usefully make squarish holes with rounded corners.

And this, to me, raises a question: is there any way to, using only the ability to measure between points on the object's circumference, and no other points, determine if that object is or is not circular?

In more formal mathematical terms, if we have a closed plane curve $C$, then what are the conditions on the distances

$$d(a, b)$$

between points $a, b \in C$, that are necessary and sufficient for circularity of the curve?

The_Sympathizer
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  • Constant diameter (width) is how we typically define roundness, btw. Circularity requires constant radius. Why do you bring up the Reuleaux triangle; why is it even a thought? I ask because it seem as though you allow yourself to use the ruler to make measurements of the diameter. You recognize that constant diameter doesnt equate to circle, and thats why you bring up those special shapes. If you are allowed to use the ruler to measure out diameters, why then can you not measure out radii, thus answering your own question? – SquishyRhode Oct 04 '19 at 03:58
  • @SquishyRhode : As said, because to do so, you need the center point. If all you have on hand is a ruler and pen (which is the rules I'm imagining), how are you going to find and then track that accurately? – The_Sympathizer Oct 04 '19 at 03:59
  • You recognize that you need a center identified to measure out radii. This is true. But you need that anyway if you measure out diameters, do you not? I seem to be confused by what you are and are not allowed to use/know because you dont seem to be consistent on that. – SquishyRhode Oct 04 '19 at 04:00
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    @SquishyRhode : A diameter can be defined more generally as a segment between two points on the periphery whose length is maximal. Thus, you can identify it by fixing one end of the tape measure and then varying the other end until you find the maximum. There is no need to know the center. – The_Sympathizer Oct 04 '19 at 04:05
  • Ah. yes, I understand what youre saying. This allows for objects with a certain "asymmetry" (loosely used). If you sample more than one of these "diameters" and fail to find a single point of convergence between all such diameters, then you know your shape is not "symmetric". Thats what you do. Measure out a large sample of these "largest diameters". Firstly, if any one of them is different from the others then its not a circle. Secondly, if any pair of diameters intersect at a different point from the rest, it isnt a circle. Thirdly, the rest is a statistical confidence game. – SquishyRhode Oct 04 '19 at 04:13
  • Since a circle is the only figure in a plane with constant (nonzero) curvature, we should be able to measure the curvature at each point along the tube and check that it's (within tolerance of) constant. This comes down to approximating second derivatives in the x and y directions, which you can do with a tape measure (and too much free time) – HallaSurvivor Oct 04 '19 at 05:40
  • Since you have a tape measure rather than a ruler it might be long enough to locate two diameters simultaneously. Then you could read off the distance from their intersection - i.e. the centre of the circle - and verify constant radius (you might however need four hands for this). – Tom Collinge Oct 04 '19 at 07:36
  • @Tom Collinge : Finding the center isn't necessarily the problem. The problem is how you mark that, to high accuracy, with only a tape measure and a big gaping void around where that center is located. – The_Sympathizer Oct 04 '19 at 07:44
  • If you can find two diameters simultaneously you can measure the distance from their intersection to the edge without needing to mark the point. – Tom Collinge Oct 04 '19 at 07:48
  • @Tom Collinge : You may need to measure more than two diameters, though. I should specify that only a single tape measure is allowed. – The_Sympathizer Oct 04 '19 at 07:52
  • A long enough tape measure can bend and find two diameters at once giving the "radius" to four points: then repeat. – Tom Collinge Oct 04 '19 at 07:53
  • @Tom Collinge : Perhaps, if the "tape measure" is also suitably flexible, i.e. like the cloth ones and not the metal ones. But I've also edited the question to hopefully clarify what I am after. – The_Sympathizer Oct 04 '19 at 07:55
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    With two tape measures, we can exploit the Power of a Point theorem. Fix $A$, $B$, $C$ on the curve, and let $D$ be a variable test point, say, on the $\stackrel{\frown}{AB}$ not containing $C$. Stretch tape measures from $A$ to $B$ and from $C$ to $D$; let them meet at $P$. Provided $P$ lies in the curve's interior (if it doesn't, then you have don't even have convexity), then $D$ lies on the circumcircle of $\triangle ABC$ if and only if $|AP||BP|=|CP||DP|$. Test other arcs similarly. (Note: This doesn't require finding a center or diameter.) – Blue Oct 04 '19 at 09:02

1 Answers1

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A closed curve is a circle with points $a,b$ as diameter if and only if all points $c$ on the curve satisfy $d(a,c)^2+d(c,b)^2=d(a,b)^2$.