How to calculate spray penetration length and spray cone angle?
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In order to continue processing data for the experiment results, I have to calculate the spray penetration length based on the spray image. Most of the cases In this case, the spray is mostly a single beam, but to be more realistic, how to calculate the spray penetration distance and spray cone Angle if have more sprays per image? If you guys have any ideas to solve this problem, please help.
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William Rose
il 1 Nov 2024
It depends partly on how much you want to automate the proesss. If you have more user inut on each image then the amount of fancy image processing will be less, and vice versa.
I dont understand your question "In this case, the spray is mostly a single beam, but to be more realistic, how to calculate the spray penetration distance and spray cone Angle if have more sprays per image?" Are a beam and a cone the same thing? The six images you provided appear to be a time lapse set , and each shows nine spray cones. By angle, do you mean the angle between cones (40 degrees, since there are 9 equally spaced) or the angle of each cone?
I recommend you download this image segmentation tutorial by @Image Analyst from the File Exchange. And maybe some of @Image Analyst 's other contributions - see a list here.
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Image Analyst
il 3 Nov 2024
It looks pretty easy. The steps would be
- Threshold the first image and use regionprops to find the centroid of the center spot (xCenter, yCenter)
- Use imabsdiff to compute the difference of the current image from the first image.
- Threshold it to find the spray "arms".
- Use convhull() to find the locations of the convex hull vertices, which will be the tips of the arms.
- For each arm, compute the distance of its tip to the center (yCenter, yCenter).
- Zero out the center of the mask to isolate the 9 arms.
- Label the image with bwlabel.
- Loop over all labels, using ismember() to extract each spray arm mask one at a time.
- For each arm, determine the that spray's cone angle. I attach an old example that you can adapt.
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