For a spherical spiral curve, parametric representation is given as:
$x=r \sin(t) \cos(ct)$, $y=r \sin(t) \sin(ct)$, $z=r \cos(t)$ with $t=[0,\pi]$ and $c$ a constant.
How can I translate this to a surface parameterization of a 3D spherical spiral, having its own minor radius ?
The surface should look like something similar to this: Spherical Helix
Best Answer
Converting your original equation into a vector gets $$\overrightarrow C=\langle r\sin(t)\cos(ct) , r\sin(t)\sin(ct), r\cos(t)\rangle$$ Then, the tangent vector is $$\overrightarrow T=\langle r\cos(t)\cos(ct)-rc\sin(t)\sin(ct) , r\cos(t)\sin(ct)+rc\sin(t)\cos(ct), -r\sin(t)\rangle$$ A normal vector is $$\overrightarrow N=\langle r\cos(t)\sin(ct)+rc\sin(t)\cos(ct) , rc\sin(t)\sin(ct)-r\cos(t)\cos(ct), 0\rangle$$ Then the binormal vector is the cross product which is $$\overrightarrow B=\langle-r^2 \cos(t) \cos(c t) \sin(t) + c r^2 \sin^2(t) \sin(c t), -c r^2 \cos(c t) \sin^2(t) - r^2 \cos(t) \sin(t) \sin(c t), -r^2 \cos^2(t) \cos^2(c t) - c^2 r^2 \cos^2(c t) \sin^2(t) - r^2 \cos^2(t) \sin^2(c t) - c^2 r^2 \sin^2(t) \sin^2(c t)\rangle$$ (This is getting out of hand quickly) Using $u$ as the second parameter, we get the final equation for the helix is $$\overrightarrow C+\frac{\overrightarrow N}{\|\overrightarrow N\|} R\cos(u)+\frac{\overrightarrow B}{\|\overrightarrow B\|} R\sin(u)$$ which I do not have the guts to write out fully.