# Self-tightening knot

Is there a way, for some finite L>1, to tie two 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 have length L, but I am open to other models.

If rope 1 has ends A and B, rope 2 have ends C and D, we tie B and C together, then 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?

-
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. – Kevin Walker Jan 8 '12 at 14:34
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 . – Joseph O'Rourke Jan 8 '12 at 15:03
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. – Matt Brin Jan 10 '12 at 5:00

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

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.

-
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 ? – Mircea Sep 19 '12 at 15:05
@Mircea, yes, this one might work. – Anton Petrunin Sep 20 '12 at 4:49