Is there a way, for some finite $L>1$, to tie two length $L$ pieces of rope together, such that any finite force is not enough to pull them apart?

The type of rope I have in mind is something like cylindrical with radius 1, unbreakable, unstretchable, perfectly flexible, non-self-intersecting and has length $L$, but I am open to other models.

Suppose that rope 1 has ends $A$ and $B$, rope 2 has ends $C$ and $D$. Tie $B$ and $C$ together. Pull $A$ and $D$. Is there a knot that holds for every coefficient of friction $e>0$ and every force $F>0$ applied to the two ends?

  • 8
    $\begingroup$ This is an interesting question, but I think you would need to provide more details of the mathematical model of rope you have in mind before it becomes answerable. For example, how do you reconcile "radius 1" with "perfectly flexible"? If the center line of the rope can make arbitrarily tight turns (perfectly flexible), then the rope would have to intersect itself (because of of radius 1 >> 0). Perhaps someone will be able to point you to papers where similar questions have been studied. $\endgroup$ Jan 8, 2012 at 14:34
  • $\begingroup$ A model of a rope with a curvature constraint that addresses Kevin's point was used in an earlier MO question, "Coiling rope in a box": mathoverflow.net/questions/26525 . $\endgroup$ Jan 8, 2012 at 15:03
  • 6
    $\begingroup$ Sounds like you want Ashley's big book of knots rather than a math reference. It's worth owning no matter what your mathematical interests. Along with drawings of around 1000 knots, there is a description of when he was advising weavers on how to best make unobtrusive yet stable knots between threads in woven cloth. The experiments he describes and the descriptions of the resulting knots create a worthwhile read. Most applicable are the parts where he talks about finding good knots for very slippery nylon. In the meanwhile, your question needs to be better posed. $\endgroup$
    – Matt Brin
    Jan 10, 2012 at 5:00
  • $\begingroup$ As a starting point, you might want to look at this expository article that I wrote: lightandmatter.com/article/knots.html . Clearly $L>1$ isn't a sufficient condition, since you can't tie a knot unless the length is considerably greater than the radius. For greater lengths, the answer is certainly yes, but the state of the art is not such that anyone can prove this sort of thing rigorously. A big roadblock to progress is characterizing the geometry of an idealized knot. Such a characterization is lacking for even simpe cases like two U shapes linked together. $\endgroup$
    – user21349
    Sep 4, 2017 at 16:50

1 Answer 1


The answer to your question is obviousely "NO", since you want $L$ to be fixed. So let me consider the following question instead:

Given the coefficient of friction $e>0$, is there a knot that holds any force?

I would bet that the answer to this question is "YES". Obviously we should have $L\to\infty$ as $e\to 0$.

A right way to proceed would be to take the Ashley's big book of knots suggested by Matt and look for a knot which admits a sequence of iterations of some kind. Even if you made right guess for iterated knot, actual proof that it holds any force might be difficult.

Now let me explain why I would bet for "YES". Assume in addition you are allowed to make any metal ring with zero coefficient of friction with the rope. Then this

alt text
(source: psu.edu)

would solve your problem. The metal ring is marked by black, the green and red ropes alternate; i.e., they go under and over the black ring in turn. (The width of the ring should be about 1.

  • 2
    $\begingroup$ what is the difference between the case with the metal ring and the ring-less uni-knot here: mouches.free.fr/pagesus/articles/knots.html ? $\endgroup$
    – Mircea
    Sep 19, 2012 at 15:05
  • $\begingroup$ @Mircea, yes, this one might work. $\endgroup$ Sep 20, 2012 at 4:49
  • $\begingroup$ The link in this answer has broken. $\endgroup$
    – Lee Mosher
    Sep 4, 2017 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.