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Consider a line L in R^3 in the shape of a trefoil knot. Consider the surface S that is the union of all unit circles that have centers on this line and whose tangent vectors are all perpendicular to the tangent vector of L at the cirle's center. S does not intersect itself.

What is the shortest possible length of L?

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To add a remark to Oliver's answer: Until the 2006 paper "Quadrisecants Give New Lower Bounds for the Ropelength of a Knot" (arXiv math.DG/0408026), it was unknown if a foot-long rope 1 inch in diameter could tie a trefoil. Their lower bound of 15.66 showed the answer is: No! –  Joseph O'Rourke Jan 31 '11 at 1:38
@Joseph I believe the answer to the foot-long root question was actually resolved a few years earlier in Y Diao, The lower bounds of the lengths of thick knots, J. Knot Theory Ramifications 12 (2003) math.uncc.edu/files/preprint/2002/2002_06.pdf –  Oliver Jan 31 '11 at 5:33
Oliver: I stand corrected! Thanks. –  Joseph O'Rourke Jan 31 '11 at 11:36
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1 Answer

up vote 19 down vote accepted

The invariant you are talking about is usually called the "ropelength" of the knot. You can find some basic stuff at the wikpedia page http://en.wikipedia.org/wiki/Ropelength which also gives some good references. (Note that some people use unit circles, while other people use circles of diameter 1, so the reported ropelength differs by a factor of 2.)

The exact value of the ropelength is not known for any nontrivial knot. However in the case of the trefoil, there are some pretty good bounds. It is between 15.66 and 16.372 if we define ropelength using circles of diameter 1. The upper bound is believed to be tighter.

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