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I have to find the curvature and torsion of a curve (parametrised by arc length), given only the Binormal vector.

Whilst I understand how to find these if I have the curve, I cannot for the life of me work out how to go in this direction.

Any help would be appreciated

Stephen Palmer
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  • You cannot. Fix any (constant) binormal vector $v$, this defines a plane $\Pi$. Take any curve on such plane, this will have binormal $v$. The curve will have zero torsion, but arbitrary curvature. – Pedro Sep 04 '16 at 03:35

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Assume that your curve is differentiable enough. If you know the binormal unit vector $B(s)$, then the Frenet-Serret formulas tell you that $B'(s) = -\tau(s) N(s)$ and so you can recover the absolute value of the torsion as $|\tau(s)| = ||B'(s)||$.

If the torsion is zero, then all you can deduce is that the curve lies on a plane whose normal is $B(s)$ (which is constant). If the torsion is everywhere non-zero, you can reconstruct you the normal vector (up to a sign) as $N(s) = \frac{B'(s)}{||B'(s)||}$. Using Frenet-Serret again, we have

$$ k(s) = ||N'(s) - \left<N'(s), B(s) \right> B(s)||. $$

levap
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