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.
[Math] Parametric form of conical helix
parametric
Related Solutions
Treat $x$ as free variable and solve for $y$ we get $y=(-3x-10)/2$, now you can let $x$ be whatever you like and get the corresponding $y$.
Let us denote your original curve by a vector function $\vec{p}$. Thus,
$$\vec{p}(t) = \begin{pmatrix} x(t)\\ y(t)\\ z(t)\\ \end{pmatrix} = \begin{pmatrix} r_1\cdot \cos(\omega_1\cdot t)+ r_2\cdot \cos(\omega_2\cdot t)\\ r_1\cdot \sin(\omega_1\cdot t)+ r_2\cdot \sin(\omega_2\cdot t)\\ 0 \\ \end{pmatrix}. $$ With this notation, the vector $\vec{p}'(t)$ obtained by differentiating componentwise $$\vec{p}'(t) = \begin{pmatrix} x'(t)\\ y'(t)\\ z'(t)\\ \end{pmatrix} $$ is tangent to the curve. Since your curve is contained in the plane, an easy way to obtain a vector perpendicular to the curve is to compute the cross product of $\vec{p}'$ with $\langle 0,0,1\rangle$. Using this, I obtained the unit normal vector $$ \vec{N}(t) = \begin{pmatrix} \frac{r_1 \omega _1 \cos \left(t \omega _1\right)+r_2 \omega _2 \cos \left(t \omega _2\right)}{\sqrt{2 r_1 r_2 \omega _2 \omega _1 \cos \left(t \left(\omega _1-\omega _2\right)\right)+r_1^2 \omega _1^2+r_2^2 \omega _2^2}}\\\frac{r_1 \omega _1 \sin \left(t \omega _1\right)+r_2 \omega _2 \sin \left(t \omega _2\right)}{\sqrt{2 r_1 r_2 \omega _2 \omega _1 \cos \left(t \left(\omega _1-\omega _2\right)\right)+r_1^2 \omega _1^2+r_2^2 \omega _2^2}}\\0 \end{pmatrix}. $$
At this point, your spiral (or, helix, really) can be parametrized by the vector function
$$\vec{p}(t) + r_3 \vec{N}(t)\cos(\omega_3 t) + r_3 \langle 0,0,1\rangle\sin(\omega_3 t).$$
Using your parameters, this produces the following pictures:
Best Answer
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.