Given a curve $r(s)$ with constant positive curvature and constant torsion, we consider the curve $r'(s)$. If $k_r$ and $\tau_r$ are the curvature and torsion of $r(s)$, what are the respective curvature and torsion of $r'(s)$ in terms of $k_r$ and $\tau_r$?

My answer isn't very rigorous but here goes:

We have a helix curve. The tangent curve of a helix curve is a circle. A circle has no torsion so the torsion of $r'(s)$ is $0$. A circle has constant curvature and since $r'(s)$ is tangent to $r(s)$ they share the same curvature $k_r$.

Does this make sense?


\begin{align*} \kappa &= \frac{|r'\times r''|}{|r'|^3} \\[5pt] &= \frac{|T\times \kappa N|}{1} \\[5pt] \kappa^* &= \frac{|r''\times r'''|}{|r''|^3} \\[5pt] &= \frac{|\kappa N\times (\kappa N)'|}{\kappa^3} \\[5pt] &= \frac{|\kappa N\times (\kappa'N-\kappa^2 T+\kappa \tau B)|}{\kappa^3} \\[5pt] &= \frac{|\kappa B+\tau T|}{\kappa} \\[5pt] &= \frac{\sqrt{\kappa^2+\tau^2}}{\kappa} \\[5pt] \tau &= \frac{r' \cdot r''\times r'''}{|r''|^2} \\[5pt] \tau^* &= \frac{r'' \cdot r'''\times r^{(4)}}{|r'''|^2} \\[5pt] &= \frac{ \kappa N \cdot (\kappa'N-\kappa^2 T+\kappa \tau B) \times (\kappa'N-\kappa^2 T+\kappa \tau B)'} {\kappa'^2+\kappa^4+\kappa^2 \tau^2} \\[5pt] &= \frac{ \begin{vmatrix} 0 & \kappa & 0 \\ -\kappa^2 & \kappa' & \kappa \tau \\ 3\kappa \kappa' & \kappa''-\kappa \tau^2-\kappa^3 & 2\kappa' \tau+\kappa \tau' \end{vmatrix}} {\kappa'^2+\kappa^4+\kappa^2 \tau^2} \\[5pt] &= \frac{\kappa^3(\kappa \tau'-\kappa' \tau)} {\kappa'^2+\kappa^4+\kappa^2 \tau^2} \\[5pt] &= 0 \tag{$\kappa'=\tau'=0$} \end{align*}

  • $\begingroup$ It's far easier to do directly from a parametrization of a circular helix (especially after realizing geometrically that the new curve is a circle), but sometimes a parametrization-free approach is elegant. This time, not so elegant :P $\endgroup$ – Ted Shifrin Nov 29 '16 at 22:29

You're right that the curve swept out by the unit tangent vector of the helix is a circle and therefore has zero torsion. However, you need to compute the radius of the circle to compute its curvature. You are incorrect in your guess. To see that your reasoning has no foundation, consider a circle of radius $a$; it has curvature $\kappa=1/a$, but its unit tangent vector curve sweeps out a circle of radius $1$, which in turn has curvature $1$.

  • $\begingroup$ Should it be $\dfrac{\sqrt{\kappa^2+\tau^2}}{\kappa}$? $\endgroup$ – Ng Chung Tak Nov 29 '16 at 20:33
  • $\begingroup$ Looks right to me. $\endgroup$ – Ted Shifrin Nov 29 '16 at 20:44

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