# How to measure the curvature of object boundary?

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sana3 sal on 20 Jun 2018
Hello, I have tried the solution : https://www.mathworks.com/matlabcentral/answers/164349-how-to-calculate-the-curvature-of-a-boundaries-in-binary-images and i tried it on my image. it works, but i have to analyse the curvature. let's be clear, the first photo attached is representation for normal case of some inter-vertebrae discs. the second is abnormal case, where we can notice the disc bulge makes the curve. now in order to classify each one, we need to look at the curve in the upper part of the object, so that it will classify them automatically based on this. How i can represent this curve mathematically in order to make this automatic classification? or what does the plot means in my cases?
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sana3 sal on 22 Dec 2018
I deleted the part that is related to my thesis

Anton Semechko on 21 Jun 2018
Thanks for the clarification. If I understand correctly, the features you seek should be extracted from an open-curve that represents dorsal portion of the interververtebral disk space. In this regard, I can think of a couple curvature-based features that can be used to discriminate between 'normal' and 'abnormal' cases. The first, and simplest, feature would be mean curvature of the curve (obtained by integrating curvature along the curve and dividing by the total arc length). When using this feature, one would expect that pathological cases would have higher mean curvature. Second feature would be a histogram of curvatures. This is a generalization of the first feature. One would expect that histograms of pathological samples would have thicker tails and modes shifted to the right. In any case, going forward, it would helpful if you posted some training data on here, so that I, and other people on here have something to work with.
sana3 sal on 25 Jun 2018
I understand you, thank you for your guidance.

Preetham Manjunatha on 2 Aug 2021
Here this function measures the curvature and other shape properties. In addition, it displays the curvature.