I have function surface given by f(x,y) values (plus x and y). I know normal at each point and neighbourhood points as well. Is it possible to use this information to calculate Gaussian and Mean curvature? I have almost no knowlegde of this kind of math, so some kind of "5-years old" explanation would be good.

I have found, that Gaussian map can be used to calculate what I want - http://mathworld.wolfram.com/GaussMap.html. But again, I have not found, how to construct this or manipulate it.

  • $\begingroup$ Do you have $f(x, y)$ itself, or just its value at some finite number of points? In the former case, this is just working out the Gaussian and mean curvature for a graph surface; in the latter case, since these are local properties, you can't say anything at all without some more information. $\endgroup$ Nov 4, 2014 at 12:43
  • $\begingroup$ @Travis I have only points, f(x,y) is unknown. What more info do I need to calculate curvature. I could also use only estimate value, not 100% accurate. In my case, speed is important, precision is second. $\endgroup$ Nov 4, 2014 at 13:31
  • $\begingroup$ The curvatures will depend on the derivatives of $f$, so you need some information about how much the derivatives vary. Otherwise, the function $f$ could be very wild in between your sample points. $\endgroup$ Nov 4, 2014 at 13:33
  • $\begingroup$ @Travis I am taking it as the sampling points are direct neighbouring values, eg. my sampling is detailed enough. There is no wildness in non-sampled points. $\endgroup$ Nov 4, 2014 at 13:42

1 Answer 1


You can find formulae for the curvatures of a graph here. These are in terms of the first and second derivatives of the function, so you just need to replace these smooth derivatives with approximations using your data set. If your data points are laid out on a regular lattice then you can just use the standard finite difference formulae, while if they are not I guess you will need to use some form of interpolation - this should be a very common problem in finite element analysis, which you probably know more about than I do.


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