0

The connectivity of the pore network in the SiC/SiCN material is assumed to be very high since pyrolysis leads to a shrinkage of the matrix therefore leading to cracks throughout the matrix. In order to prove this assumption, a labeling of the objects in a smaller region of interest (1 x 1 x 0.5) mm3 of the binary volume is performed.

What does "binary volume" mean?

I can't find this notion at all. Maybe someone has assumptions what it can be?

4
  • 2
    Can you provide some context? – legatrix Dec 4 '20 at 16:41
  • The connectivity of the pore network in the SiC/SiCN material is assumed to be very high since pyrolysis leads to a shrinkage of the matrix therefore leading to cracks througout the matrix. In order to prove this assumption, a labelling of the objects in a smaller region of interest (1 x 1 x 0.5) mm3 of the binary volume is performed. – Саша Гула Dec 4 '20 at 16:43
  • I would also like to ask about the second sentence as I've not really got what the author wanted to say. – Саша Гула Dec 4 '20 at 16:46
  • 1
    Here is my assumption: The author is interested in the real-world physical object shrinking. They label various points on the computational model of the object. They then run a simulation of whatever the relevant real-world processes are. They then check to see if the labelled points in the computational model are closer together. – legatrix Dec 4 '20 at 16:51
1

A binary volume is a 3D representation of something using binary digits. In this case, it sounds like a computational model of a real-world physical object, very common in neuroscience, materials science, engineering, and many other fields. Below I show you a (badly drawn) binary representation of a sphere in 2D. If you imagine this in 3D (so with 125 bits instead of 25), that would be a binary volume of the sphere.

00100
01110
01110
00100

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.