Parametric form of conical helix
Given a cone as
\begin{array}{l} x=s\cos t\\ y=s\sin t\\ z=s\cot\alpha\end{array}
How could I get parametric form of any conical helix?
Thank you.
$\endgroup$1 Answer
$\begingroup$Parametrized by arc-length
\begin{align*} \mathbf{r}(s) &= \begin{pmatrix} \dfrac{as}{\sqrt{1+b^2}} \, \cos (b \ln s) \\[2pt] \dfrac{as}{\sqrt{1+b^2}} \, \sin (b \ln s) \\[2pt] s\sqrt{1-a^2} \end{pmatrix} \\[5pt] \kappa &= \frac{ab}{s} \\[5pt] \tau &= \frac{b\sqrt{1-a^2}}{s} \\[5pt] \frac{\tau}{\kappa} &= \frac{\sqrt{1-a^2}}{a} \\[5pt] \left( \frac{\tau}{\kappa} \right)' &= 0 \end{align*}
where $\cot \alpha=\dfrac{\sqrt{(1-a^2)(1+b^2)}}{a}$ for $a\in (0,1)$ and $b\in \mathbb{R} \backslash \{ 0 \}$.
See also the cone geodesic in another answer for which $\dfrac{\tau}{\kappa}=\dfrac{s}{a}$ and so it's a conical spiral instead of helix.
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