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I am trying to find the general equation for space curves which have constant curvatures throughout their length. In general I am interested for curves of more than 3 dimensions.

Assuming that all curvatures are constant for the entire length of the space curve, can I use the frenet serret formulae to derive the most general representation of such a curve?

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are there Frenet Serret formulas for more then three dimensions? The formulas I know are very much three-dimensional in nature. I'm interested to see what the community has to offer here. –  James S. Cook Dec 24 '12 at 19:28
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@James: Wikipedia credits Jordan with the higher-dimensional generalization of the Frenet-Serret formulas. –  Rahul Feb 28 '13 at 21:38
    
@RahulNarain neat, I had not seen those before. Should make nice homework for some course. –  James S. Cook Mar 1 '13 at 13:58

1 Answer 1

I assume the curvature $\kappa$ is not $0$. For one of your curves, the tangent vector $T$ moves on the unit sphere in an arbitrary way, constrained only by $$\left| \dfrac{dT}{ds} \right| = \kappa$$ Thus its path can be any $C^1$ curve on the sphere, which you traverse at constant speed $\kappa$ (with respect to the parameter $s$). To get the actual curve in space, you then integrate: $$ X(s) = \int_0^s T(t)\ dt$$

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Thanks for the answer. I was wondering though, is any curve on the sphere a valid curve? Arent only circles the curves with constant curvature? –  George Dec 26 '12 at 8:59
    
The curve that the tangent vector traces out on the sphere does not have constant curvature. It is just a curve that can be traversed at constant velocity. The space curve with constant curvature is obtained by integration. –  Robert Israel Dec 26 '12 at 9:48

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