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I have a piecewise linear path in two dimensions (with finitely many pieces). I would like to approximate it by a curve whose radius of curvature is bounded away from 0 (i.e. I specify the bound a priori). Any suggestions for an algorithm?

I don't have any specific metric in mind -- this is in order to produce a nice visualization, not a theorem -- but, for the sake of argument, let's say I want an approximation that minimizes the Skorohod J1 distance from the piecewise linear path.

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You may need conditions on the path in order to succeed. I can imagine paths which approximate certain spirals or sin(1/x) which can prevent such approximations by smooth curves with curvature bounded away from infinity. For example, can you guarantee an alpha > 0 such that any two consecutive segments of your path form an angle greater than alpha? (As well as something similar for every collection of n consecutive segments whose lengths are bounded by, say, 1/n.) Gerhard "Ask Me About System Design" Paseman, 2011.06.08 – Gerhard Paseman Jun 8 '11 at 8:43
Sorry, I took the wrong inequality. It may be that you might take some conformal transformation to turn your problem into the one I mention above. Gerhard "Used Binoculars To Shrink Images" Paseman, 2011.06.08 – Gerhard Paseman Jun 8 '11 at 8:48
I've edited my question to clarify that my path has finitely many pieces, which rules out problem cases like sin(1/x). I am not looking for an arbitrarily good approximation, just the best approximation given a bound on curvature, so I do not believe I need to impose any guarantees on the angles between segments. – Damon Jun 8 '11 at 9:27
If you can guarantee such an $\alpha$, then splines should work. If you can't, then how are you going to approximate a very flat spiral? – Douglas Zare Jun 8 '11 at 10:03
I do not know what Skorohod J1 distance is, but by far the easiest way to approximate a piecewise linear path is to replace a neighborhood of each vertex by a circular arc of the approximate angle. This will bound the radius of curvature away from zero. – Deane Yang Jun 8 '11 at 12:01

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