How to calculate this crack from this image

How to calculate crack from this image, i have picture ceramic and i want to calculate from this image, there are any defect from this image you can see this below

http://i1164.photobucket.com/albums/q570/diodonaika/diodo.jpg

you can see, there are cracks in the image defects how can I calculate the length of cracks in the image. I have to ask where else confused please help

Answer 1

[Image Analyst]

You might want to skeletonize with the bwmorph() function. Then all you will have are the centerlines of the cracks.

binaryImage = bwmorph(binaryImage, 'skel', inf');

Then call regionprops and ask for the area which will be the length of the cracks in pixels.

measurements = regionprops(binaryImage, 'Area');

[Dio Donaika]

image analyst thanks for your attention. I have to enter the code that you have given to me. The first to use the method skeleton, what the picture as below

http://i1164.photobucket.com/albums/q570/diodonaika/skeleton.jpg

sorry if I make mistakes

[Image Analyst]

OK. Looks good so far. You might want to run bwareaopen first to get rid of very small regions. You can use regionprops (like I showed you) to get each crack's length or you can get the total crack length for the entire image just by summing the binary image

totalCrackLength = sum(binaryImage(:));

[Dio Donaika]

may I ask about the program, whether the program would be like this

binaryImage = bwmorph(im1,'skel','inf');
figure, imshow(binaryImage);
title('skeleton');
measurements = regionprops(binaryImage, 'Area');
totalCrackLength = sum(binaryImage(:));

measurements
totalCrackLength

[Image Analyst]

Looks about right. You might want to extract the line lengths into an array instead of leaving them in the structure:

allCrackLengths = [measurements.Area];

That will be an array with the length for each individual crack. sum(allCrackLengths) will equal totalCrackLength.

[Dio Donaika]

What if I want to get how many cracks in the amount not in the array, and what if I want to get the longest and shortest of these cracks.

thanks so much for your attention and help my problem. ^_^

[Image Analyst]

That can all be done by examining the measurements returned by regionprops(). Have you studied my image segmentation tutorial yet, where it does that kind of stuff? http://www.mathworks.com/matlabcentral/fileexchange/?term=authorid%3A31862

Answer 2

[Walter Roberson]

How does this question differ from your existing question at http://www.mathworks.co.uk/matlabcentral/answers/41089-crack-detection ?

[Dio Donaika]

This question about counting and http://www.mathworks.co.uk/matlabcentral/answers/41089-crack-detection questions about identifying

[Walter Roberson]

You are already discussing length calculations there: http://www.mathworks.co.uk/matlabcentral/answers/41089#answer_54251

[Dio Donaika]

yes but the question has not been received, so the question on specialize to detect. and I made the two questions are connected. if you think it's the same then I'll delete this question

[Image Analyst]

They seem the same to me. Walter or I can delete this question later today and you can follow up on your prior thread. But it looks like your professors are not letting you use any code given to you by senior developers (like me) - that's what it seemed like you said - so I guess we're at the point where you need to take over.

[Dio Donaika]

thanks for help, I might be any suggestions for this problem. but i just want to learn and do not expect the code if you feel I am expecting from your code, yes indeed I expected. it is to learn so that I can understand. because learning from examples is the one - the only way to learn.

previous question I asked about the best way to detect defects on the surface of the ceramic is not included for the calculation, because the code that I wrote for the detection using the same method as before. I am in order to replace it and I am confused to use the best method.

for this question to calculations and I think the first question representing this question. I am so sorry it please delete this question