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In the biography "King of infinite space: Donald Coxeter, the man who saved geometry" by Siobhan Roberts, the following passage describes an aspect of the subject's relationship with Buckminster Fuller:

Coxeter was not the first to be frustrated by Fuller. Fuller was neither an architect nor engineer nor mathematician by training, and he became a controversial figure among experts in all those fields. He made awkward geometric mistakes, such as how many spokes are needed on a wheel to hold it rigid (Fuller said twelve instead of seven).

I was intrigued by the claim that seven spokes are needed on a wheel to hold it rigid. Searching Google does reveal that Fuller seems to have thought the answer was twelve, but I cannot find any other claim that it should be seven.

I cannot figure out a (mathematical) model in which seven spokes is the answer (nor twelve, for what it's worth). It certainly seems that the spokes are assumed to be "tension" only, as of course there are three-spoke bicycle wheels made of solid (rigid) spokes. In other words, I assume the subject under discussion is wire wheels, where the spokes "function mechanically the same as tensioned flexible wires".

I have constructed a few models, definitely unrealistic, where the minimal number of spokes is three. Can anyone give a mathematical model of tension-spoked wheels where the minimum number of spokes is seven (or twelve, or perhaps any finite number greater than three)?

[The "inverse problem" in the title refers to the fact that in this situation, we know the answer but not the question!]

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    $\begingroup$ I am confused by this question. There certainly are rigid wheels with only $3$ spokes. What am I missing. Needs Details/Clarification $\endgroup$
    – Mike
    Jan 30 at 19:55
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    $\begingroup$ @Mike - it seems that details and clarification are precisely what OP is asking for $\endgroup$
    – Dustin G. Mixon
    Jan 30 at 20:02
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    $\begingroup$ Apparently the correct model for a spoke here is just a thread of large but finite strength that can bend easily but cannot elongate and the wheel consists of a cylindrical hub of noticeable length and diameter (say, 1/4 of that of the rim) and the coaxial rim that is just a 1D circle, with the spoke allowed to join any point on the hub with any point on the rim. You should exclude any possible motion of the rim with respect to the hub in 3D. But it is just my guess... $\endgroup$
    – fedja
    Jan 30 at 22:36
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    $\begingroup$ @fedja - Is 7 optimal for that model? If so, can you present the argument? $\endgroup$
    – Dustin G. Mixon
    Jan 30 at 23:08
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    $\begingroup$ The argument for twelve spokes appears in A Fuller Explanation, published 1987; the book's text can be found on archive.org, for example. $\endgroup$
    – Théophile
    Feb 1 at 20:13

1 Answer 1

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The space of possible forces between the rim and the hub is $3$-dimensional, and the space of possible torques between the rim and the hub is also $3$-dimensional. So the spokes need to, in some sense, span a $6$-dimensional space. You said the spokes can only pull, not push, so we're considering conical combinations rather than linear combinations. An $n$-dimensional space is not the conical hull of any $n$ vectors, but is the conical hull of $n+1$ vectors. Hence... $7$.

And if the vectors are required to be paired oppositely, then the $n$-dimensional space needs at least $n$ pairs and thus $2n$ vectors to span it. But I don't know if this was Fuller's reasoning to get $12$.

For an example, you can put $3$ spokes in triangular symmetry on one side of the wheel, and $4$ spokes in rectangular symmetry on the other side of the wheel. (The triangular symmetry has $3$ mirror planes, and the rectangular symmetry has $2$ mirror planes, all containing the wheel's axis.)

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  • $\begingroup$ It all seems to make perfect sense, but I am still not seeing what rigid motion will be possible in the first order approximation if we do 3 and 3 on each side with 60 degrees rotation between them. Do you see it? $\endgroup$
    – fedja
    Mar 25 at 22:34
  • $\begingroup$ If all spokes pass through the wheel's axis, you have no way to produce (or counteract) torque around that axis. $\endgroup$
    – mr_e_man
    Mar 26 at 22:54
  • $\begingroup$ Yep, I understood that much. I meant a bit more complicated configuration, but I finally figured out what the admissible (in the first order) movement is there too. $\endgroup$
    – fedja
    Mar 31 at 1:19

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