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Calculate number of fringes

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Does anyone know the feasability of using Machine Vision or LabView to calculate the number of fringes in each region? (Image attached). I want to count how many white line on each region. For example, on attached image, there is about 6 regions, and each region will display the number of white line of their own region. For the information, I have LabView V8.5 and NI Vision Assistant V8.5. Please, give me some ideas on how to solve it.

 

Thank you.

 

Farid

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You can use a simple edge detection like this

 

edge.PNG

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The attached vi will give you the particles that for which the the whitelines can be counted. Feed the output of this particle analysis bounding box (Top,Left,Bottom,Right) to the edge detection region of intrest and adjust the settings accordingly
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helped?solved?
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Hi muks,

 

1) Thanks for the quick reply. I've done the simple edge detection. Yes, I  am able to count the number of white line by applying edge detection. But, I want to determine the number of line for all the region automatically. Is there any way that I can do that without draw a single line?

 

2) Thanks for the fringes.vi file. I can run this VI, and it produces 7 particles from the image that I posted before. That is good. Btw, how to count the white line of each particle? The output that I'm looking for such as : Particle 1 = 10 lines. Particle 2 = 12 lines. Particle 3 = 7 lines. and so on.

 

3) I am not understand what do you mean by 'Feed the output of this particle analysis bounding box (Top,Left,Bottom,Right) to the edge detection region of intrest and adjust the settings accordingly'. Could you elaborate more detail please.

 

Thanks

Farid   

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Solution
Accepted by topic author MFarid
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Thank you so much.

 

This VI is really help.. I think I've got the idea to solve my problem.

 

 

Thanks

 

Farid

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You can also use a "roi profile and edge detection vi". The roi profile will have a ROI descriptor input. You can unbundle this and input the particle analysis output (again top,left,bottom,right) .
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