I've got a discrete grayscale image (a.k.a. 2D rectangular array of floats). Its continuous representation (whether received via reconstruction with sinc, with a cubic kernel, with a triangle kernel...) is a continous 2D function, so it has isolines. I'm interested in calculating the magnitude of the derivative of the isoline at a given point.



Red line is part of an isoline, and green arrows are example derivatives of that isoline at 2 points.

Here's my approximate solution:

float getCurvature(vec2 p) {
    vec2 grad = getBilinear(gradients, p);
    if(grad = vec2(0, 0)) return 0;
    else grad = normalize(grad);

    vec2 gradP = perpendicularLeft(grad);

    float val = getBilinear(img, p);
    float valLeft = getBilinear(img, p+gradP);
    float valRight = getBilinear(img, p-gradP);
    return (val - (valLeft + valRight) * .5f);

It basically computes the second directional derivative in a direction perpendicular to the gradient. It works fine for my purposes, except that it's somewhat slow in the CPU implementation (and I have reasons not to port it to GPU).

Any faster (and/or more accurate) method?

Edit: I found a twice-faster method, still calculating the second derivative, but in a simpler way. It's equation (5) on this page. It works, but the result is more crude. I think I should switch to a higher-quality differentiation kernel than [+1, 0, -1].


Gradient magnitude and angle can be found using a Sobel operator. It's more sophisticated than the [1, 0, -1]. This is commonly used either directly for edge detection, or as part of a more complex algorithm.

Or, if I've misunderstood, and you want only the curvature of the 2D isoline, this answer on Math Overflow might be of use.

  • $\begingroup$ Thanks, I'm aware of the Sobel operator, but it slows my algorithm down too much. The MO answer assumes I have the isoline as a set of points. I don't, I have a 2D array of grayscale values. $\endgroup$ Nov 20 '17 at 12:29
  • $\begingroup$ In that case, you might consider posting your code to the CodeReview StackExchange and ask for performance help. $\endgroup$ Nov 20 '17 at 15:50
  • $\begingroup$ I could, but the folks over there are less likely to have the specialized expertise for doing this graphics thing efficiently. Thanks again. $\endgroup$ Nov 20 '17 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.