### Using Marching Cubes to get the dimension of a surface

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#### Using Marching Cubes to get the dimension of a surface

« on: October 26, 2019, 02:34:27 AM »
I wrote a paper about fractal dimension. I would have attached it here, but PDF is not an allowed file type.

The paper is at: http://vixra.org/abs/1812.0423

I'm looking for criticism, or whatnot. Thanks for your time.

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #1 on: October 26, 2019, 03:33:21 AM »
You could post text here from your article/paper if you like
this is the first (I think) request for PDF filetype, we don't use it because PDFs can get quite large and most of the allowed filetypes are intended for different program parameters and settings along with pictures.
Feel free to zip the pdf and attach that to a post but it will still be subject to the maximum allowable size for attachments.

Welcome to FF btw

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #2 on: October 26, 2019, 04:58:21 AM »

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #3 on: October 26, 2019, 07:10:54 PM »
Does this "curvature dimension" coincide with "box counting dimension" or other dimension measures?

That doesn't mean it's not a useful metric, as often its dimension is the least interesting thing about a fractal.

Being based on meshes seems a bit ungeometrical - maybe different ways of meshing the same surface would give different results?

I calculated* this curvature metric for the Koch snowflake curve, I think it comes out as 7/6 = 1 + lim n→infinity (2/3 (4^n - 1) cos(60deg) + 1/3 (4^n - 1) cos(120deg)) / 4^n, which is different from the similarity dimension of log(4)/log(3).

*by averaging over corners instead of edges, in topological 1D it should work out the same (I think? edges are interspersed between the corners). surfaces are more complicated...
« Last Edit: October 26, 2019, 07:27:17 PM by claude, Reason: typo »

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #4 on: October 26, 2019, 10:38:23 PM »
This measure of dimension does not coincide with other measures of dimension, as far as I know. It cannot be said enough times... I'm an amateur. That said, getting the box counting dimension is simple enough when using Marching Cubes (or Marching Squares) -- boxes containing triangles (or line segments for Marching Squares) are easily discernible from those boxes without triangles (or line segments). I'll look into it further. I hadn't thought about it too much before you brought it up; I didn't realize how simple it is.

Yes, you are right, it does matter how the shape is meshed. I updated the source code about a month ago to handle sliver triangles and whatnot. I left that out of the paper. Perhaps I should add it into the paper?

PS The source code can be found at: https://github.com/sjhalayka/meshdim
« Last Edit: October 26, 2019, 11:22:32 PM by sjhalayka »

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #5 on: October 27, 2019, 02:08:54 AM »
I implemented the box-counting dimension code and put it here: https://github.com/sjhalayka/marching_cubes

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #6 on: October 28, 2019, 04:59:08 AM »
I'm making the acknowledgements section, and I'm wondering if anyone wants me to put their initials in the thanks?

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#### Re: Using Marching Cubes to get the dimension of a surface

« Reply #7 on: November 29, 2019, 11:57:11 PM »
I changed the topic from Geometry to Quantum Gravity:

https://github.com/sjhalayka/meshdim/raw/master/bezier_escape.pdf

The only sections that are absolutely necessary are Section 1 and Section 4.3.

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