This is actually two questions about the Wolfram MathWorld article on torsion here.

1: What is a geometric interpretation of torsion? For curvature I understand it as the reciprocal of the radius of the circle around which a point is moving at a given instant. But the description in the link says torsion "is the rate of change of the curve's osculating plane". I know what the osculating plane is, but what does it mean by its "rate of change"?

2: This is a small point, but how do they get from the equation $$\tau = \frac{\left|\dot{x} \,\ddot{x} \, \dddot{x}\right|}{\left|\dot{x} \times \ddot{x}\right|^2}$$ to this? $$\tau = \rho^2 \,\left|\dot{x} \,\ddot{x} \, \dddot{x}\right|$$ (where $|x\,y\,z|$ is the triple product and $\rho$ is the radius of curvature)

Is the following correct? $$\rho = \frac{\left| \dot{x} \right|^3}{\left|\dot{x} \times \ddot{x}\right|}$$

It seems like there is a factor of $\left| \dot{x} \right|^6$ unaccounted for.


1 Answer 1


$\newcommand{\Vec}[1]{\mathbf{#1}}\newcommand{\Tgt}{\Vec{T}}\newcommand{\Nml}{\Vec{N}}\newcommand{\Bnm}{\Vec{B}}$Here's one way to make "rate of change of the osculating plane" precise. A path $\gamma$ parametrized by arc length may be expanded as a Taylor polynomial in the arc length $s$. Using subscripts to denote evaluation at $s = 0$, assuming $\kappa_{0} > 0$, and letting $\{\Tgt_{0}, \Nml_{0}, \Bnm_{0}\}$ denote the Frenet frame at $\gamma_{0}$, $$ \gamma(s) = \gamma_{0} + s\, \Tgt_{0} + \kappa_{0} \tfrac{1}{2}s^{2}\, \Nml_{0} + \kappa_{0} \tau_{0} \tfrac{1}{6}s^{3}\, \Bnm_{0} + \text{higher order terms.} $$

In answer to your second question, probably the Wolfram page assumes the path is parametrized by arc length.

  • $\begingroup$ That's an interesting identity - any idea on where I could learn more about it? I was looking for something more geometric. Would this be correct? "Torsion is the rate of change of angle of the instantaneous axis of rotation per unit length" $\endgroup$
    – user142299
    Commented Dec 14, 2015 at 13:37
  • $\begingroup$ It's a good (and not difficult) exercise to derive it; start with the ordinary Taylor expansion $$\gamma(s) = \gamma(0) + s\gamma'(0) + \tfrac{1}{2!}s^{2} \gamma''(0) + \tfrac{1}{3!}s^{3} \gamma'''(0)$$and use the Frenet equations. Your verbal description sounds correct, as per the third Frenet equation. :) $\endgroup$ Commented Dec 14, 2015 at 21:57
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    $\begingroup$ Ok thanks. By the way I like your website! $\endgroup$
    – user142299
    Commented Dec 15, 2015 at 1:38
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    $\begingroup$ When I attempt to derive it I'm getting extra terms of the $s^3$ order. Specifically, I have $\frac{d\kappa}{ds}N + \kappa(-\kappa T + \tau B)$ multiplied by $s^3/3!$, whereas you just have the $\kappa \tau B$ part. Am I doing something wrong? $\endgroup$
    – user142299
    Commented Dec 15, 2015 at 3:06
  • $\begingroup$ Your calculation is correct; my glib use of "higher-order terms" requires further explanation. :) The cubic approximation can either be decomposed into Frenet components, or "graded" by powers of $s$. The stated expansion is decomposed into Frenet components, and only the lowest-order (i.e., largest) term in each component is retained. The extra terms you note are third-order in $\Tgt$ or $\Nml$, so they're omitted. $\endgroup$ Commented Dec 15, 2015 at 11:48

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