I'm trying to calculate the surface area of the particles that are attached to the bubble in the image. But I'm not too sure how. Edge Detection doesn't work accurately.
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Syamsundar Menon Murali Mohan
on 20 Sep 2021
Commented: Syamsundar Menon Murali Mohan
on 27 Sep 2021
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/743714/image.jpeg)
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Accepted Answer
yanqi liu
on 26 Sep 2021
sir, may be use the image seg to locate the target.
clc;
clear all;
close all;
im = imread('https://www.mathworks.com/matlabcentral/answers/uploaded_files/743714/image.jpeg');
bw = imbinarize(im,'adaptive','ForegroundPolarity','dark','Sensitivity',0.2);
bw = ~bw;
bw = imfill(bw, 'holes');
bw = imclearborder(bw);
bw = bwareafilt(bw, 1);
be = imdilate(bwperim(bw), strel('disk', 3));
im2 = im;
im2(be) = 255;
figure; imshow(im2);
2 Comments
Image Analyst
on 26 Sep 2021
Yes, that can identify the bubble plus the particles attached to it, but it does not give the surface area of the particles. I think it's basically impossible - you'd need a 3-D image. I think @Syamsundar Menon Murali Mohan should work on a different metric that might correlate with something else. For example, maybe the perimeter or standard deviation of the blob correlates with temperature or pH or viscosity or whatever of the solution he's looking at.
More Answers (1)
Image Analyst
on 26 Sep 2021
What I'd do it to find the blob (bubble plus particles) like @yanqi liu did. But now you need to determine what is bubble and what are particles. So I'd use stdfilt() to find high standard devation areas. The idea is that the bubble itself will be rather smooth. You could even synthesis a bubble. Just assume some radial intensity profile using the equivalent circular diameter and subtract it from your blub with particles on it, leaving just the particles themselves.
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