Internal porosity of 3D Volume object - part (A)

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yu sh
yu sh el 10 de Jun. de 2016
Editada: yu sh el 4 de Ag. de 2016
Hi everyone,
I want to find out the internal pores of a volume object. This 3D object object is resulted from binary images of X-ray tomography and then its voxel information is processed to obtain this 3D volume object. Any suggestion or guidance would be highly appreciated. Thanks
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afaf telbani
afaf telbani el 3 de Ag. de 2016
how did you get the 3D volume object please ? did you have multiple slices ??
yu sh
yu sh el 4 de Ag. de 2016
Editada: yu sh el 4 de Ag. de 2016
Yes I have multiple 2D microtomographic images. I write a MATLAB code using patch and isosurface functions to obtain the 3D volume object. But first I obtained the array containing the grey values of voxel intensities of the the complete stack of 2D slices.

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Image Analyst
Image Analyst el 10 de Jun. de 2016
Threshold to find material and sum the image. That's the material volume. Then invert that threshold so that now you have non-material/air. Now call imclearborder() to get rid of the surrounding air region. Then sum the volume, which will give you the volume of the internal void spaces. Then divide those two to get the volume fraction = (volume of material) / (volume of void spaces).
  5 comentarios
yu sh
yu sh el 22 de Jun. de 2016
@Image Analyst: At first I thought it is simple to apply but now I am facing difficulties.
Firstly please tell me what is the purpose of thresholding? in my case material is fixed (Maltodextrin) if necessary then please guide how should I perform thresholding.
Secondly my input is a complete 3D volume object. Please check it in this 'fig' format of MATLAB but it is resulting from a set of binary images obtained from X-ray tomography. So should I use my binary images as an input over here?
Please guide me I am stuck over here from last week not getting any idea to move forward.
Image Analyst
Image Analyst el 23 de Jun. de 2016
Thresholding tells your program what is material, and what is air. In my code, "material" is a binary image. Not sure what else to say.

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