Can anyone give me an intuitive explanation for the kind of twisting or bending that a curve in $\mathbb{R}^{3}$ undergoes when it's at a point of small curvature, but high torsion?

I understand that torsion measures the failure for a curve to be planar i.e. be contained in a plane. So a curve with a high torsion in some region must be twisting in various directions through various planes in 3-space. Curvature on the other hand measures how rapidly the tangent vectors are changing with respect to the arc length of the curve. So it seems like my intuition wants to think that high torsion implies high curvature, but that's certainly not the case. I was wondering if someone can explain this


We can create a curve with fixed constant curvature and torsion: this is congruent to a helix $$ r(s) = \left( \frac{\sin{\theta}}{\lambda}\cos{\lambda s},\frac{\sin{\theta}}{\lambda}\sin{\lambda s}, s \cos{\theta} \right). $$ It is easy to see that this is unit-speed parametrised, and a calculation shows it has curvature $$ \kappa = \lambda\lvert\sin{\theta}\rvert $$ and torsion $$ \tau = \lambda \cos{\theta}, $$ and we can then use this curve to see what a curve with any curvature and torsion looks like locally. In particular, for $\theta$ close to zero, the helix becomes very elongated.

Remark: if $\kappa>0, \tau \in \mathbb{R}$, another way of writing this helix is $$ r(s) = \left( \frac{\kappa}{\kappa^2+\tau^2}\cos{\sqrt{\kappa^2+\tau^2}s}, \frac{\kappa}{\kappa^2+\tau^2}\sin{\sqrt{\kappa^2+\tau^2}s}, \frac{\tau}{\sqrt{\kappa^2+\tau^2}}s \right), $$ which has curvature $\kappa$ and torsion $\tau$, but is much nastier to write down.


Only when curvature is non-zero we can say that zero torsion space curve lies in a plane.

Surface theory parameterization to appreciate this in association with its surrounding surface is necessary here.

Solitary lines embedded in 3-space can be inadequately understood this way.

In the examples given below all lines are straight lines.

Take the case of the central asymptotic line in a helicoid:

$$ u \cos v, u \cos v, c\, v $$

Lines $u=0 $ twist around themselves just like a single thin straight twisted hair... or the central spine of a spiral staircase.

The line $ u=0 $ is twisted, although it has no curvature.

It is recommended you investigate generator lines of a hyperbolic paraboloid for twist curvature $ \partial^2 z/ \partial x\partial y. $

i.e., please read about how the surface is generated by $$ x = u+v, y = u-v , z = u v, $$ or simpler, the straight generators of a one-sheet. hyperboloid of revolution.

Ruled surface straight line generators are good examples. Without exception these are lines belong to negatively (Gauss) curvature surfaces.


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