Find the Radius of Curvature of a 3d Helical Sinusoidal Function

Question

I am creating a 3d Helical Sinusoidal Function as found here: Set of Equations to Helically Wrap a Sine Curve

However, my typical method for finding radius of curvature does not work (as intended) because it 'sees' a tiny radius, assumingly because I am not 'unmapping' the sine curve from it's cylindrical projection, giving very tight radii at the peaks of the sine curve. See image below for example.

For reference, I am calculating a radius of curvature (min) of ~0.003 units which is way too small… If you look at the 3d plot of the helically wrapped sine wave, it doesn't look as tight.

Note that both the directly plotted sine curve, $t,f(t)$, and the 3d plot are the same parameters except I reduced the range of $t$ for the 2d plot.

How would I go about extrapolating this Radius of Curvature equation to find the actual radius of curvature of the sine curve as if it were 'unwrapped' from the cylinder?

Answer 1

Given the functions $\overline{f},f : [0,T] \to \mathbb{R}$ of law:

$$ \overline{f}(t) = a\sin(\overline{\omega}\,t+\phi), \quad \quad \quad f(t) = a\sin(\omega\,t+\phi) $$

and given the curves of parametric equations:

$$ \begin{aligned} & \mathbf{r}_1(t) = \left(t,\;\overline{f}(t),\;0\right); \\ & \mathbf{r}_2(t) = \left(r\cos t + f(t)\sin\theta\sin t,\;r\sin t - f(t)\sin\theta\cos t,\;r\,t\tan\theta + f(t)\cos\theta\,\right); \\ \end{aligned} $$

the local maximum points are located by:

$$ t^* = \frac{\pi/2+2\,n\,\pi-\phi}{\omega^*}, \quad \quad \quad n \in \mathbb{N}\,. $$

So, all that remains is to impose the equality of the radii of curvature at such points:

$$ \rho_1(t^*) = \rho_2(t^*) \quad \quad \Leftrightarrow \quad \quad \frac{||\mathbf{r}_1'(t^*)||^3}{||\mathbf{r}_1'(t^*) \times \mathbf{r}_1''(t^*)||} = \frac{||\mathbf{r}_2'(t^*)||^3}{||\mathbf{r}_2'(t^*) \times \mathbf{r}_2''(t^*)||} $$

that is, by performing the calculations:

$$ \frac{1}{a\,\overline{\omega}^2} = \frac{a^2\left(1-\cos(4\theta)\right)+8\,r^2}{(2\cos\theta)^2\sqrt{a^2\left(4\,\omega^4+4\,\omega^2+2\right)-2\,a^2\left(2\,\omega^2+1\right)\cos(2\theta)+4\,r^2}} $$

from which:

$$ \boxed{\overline{\omega} = \frac{2\cos\theta\,\sqrt[4]{a^2\left(4\,\omega^4+4\,\omega^2+2\right)-2\,a^2\left(2\,\omega^2+1\right)\cos(2\theta)+4\,r^2}}{\sqrt{a}\,\sqrt{a^2\left(1-\cos(4\theta)\right)+8\,r^2}}} $$

or:

$$ \boxed{\omega = \sqrt{\frac{\sqrt{\left(a^2\left(1-\cos(4\theta)\right)+8\,r^2\right)\left(a^2\,\overline{\omega}^4\left(a^2\left(1-\cos(4\theta)\right)+8\,r^2\right)-8\,(\cos\theta)^4\right)}}{8\,a\,(\cos\theta)^2}-(\sin\theta)^2}}\,. $$

Note: to perform this comparison in Mathematica use AspectRatio -> Automatic.