For the purposes of modelling a fluid mechanics experiment, I'm dealing with a convex surface parametrized by the azimuth $\theta$ and an arc length $s$ along the surface. The points on the surface are described in a coordinate system of $r$ and $y$, as shown in the image below.

I would like to build an ordinary differential equation in terms of the derivatives of $y$ and $r$ which I will then solve numerically to verify the model. Since surface tension is involved, I need to find the mean curvature at any point on the surface.

Finding the curvature in the $y$-$r$ plane is straightforward: $\theta$ is fixed, and we're looking at a simple 2D parametric curve. But how can I get the curvature in the other principal direction, i.e. in the plane that lies normal to the tangent along the arc $s$? (I'm assuming those are the principal directions; maybe that's already a mistake …?)

I have looked at expressions that were worked out for cylindrical and spherical systems. Both are awfully complicated. My attempts to derive a formula from scratch haven't borne fruit so far—there are just too many terms to juggle. I'm also thankful for any suggestions that would allow me to recast the problem in a coordinate system that makes this easier to handle, or suggestions that will result in an approximate solution.


Update: if I find the curvature parallel to the $\theta$-$r$ plane, and multiply that value by the sine of the tangent angle … would that be a valid measure of the curvature in the other principal direction?

  • $\begingroup$ Is your surface a surface of revolution? $\endgroup$ – Xipan Xiao Mar 4 '15 at 3:24
  • $\begingroup$ Unfortunately not, it varies with $\theta$. It's symmetrical, so technically you only have to worry about $\theta \in [0, \pi]$, but that's not very helpful, is it? $\endgroup$ – sebastian_k Mar 4 '15 at 4:22
  • $\begingroup$ Then I didn't get from your post how do you define the surface? $\endgroup$ – Xipan Xiao Mar 4 '15 at 17:21
  • $\begingroup$ Or do you have a discrete set of coordinates of points of the surface? $\endgroup$ – Xipan Xiao Mar 4 '15 at 17:22
  • $\begingroup$ Yep, exactly. There are two functions $y(s, \theta)$ and $r(s, \theta)$, which give you the coordinates of all of the points on the surface. I don't know what those functions are, but I will find out numerically. Please see my update too. $\endgroup$ – sebastian_k Mar 4 '15 at 18:43

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