Maximal thickness of 3D triangulated part

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raymeout
raymeout 2020 年 5 月 11 日
編集済み: Matt J 2022 年 10 月 26 日
I am trying to find the maximal thickness of a triangulated part in 3D (with tetrahedrons; concave).
The thickness I am trying to calculate is shown in the simplified 2D-image below (green circle; the red ones are only of secondary interest):
Simple said: I need the biggest sphere that can stay completely inside the triangulation.
A part of the triangulation is shown below as an example (what you see is a cut through the inside of the triangulated part; the triangulation comes from a third-party software):
What I've thought about so far was an approach with using a kind of medial "axis" (I think this would be surfaces in 3D (?)) and then trying to find the biggest sphere with center on those points. But I could not think about a good way to do this so far.
Another approach was to use the function
bwdist
– which generaly would actually exactly do what I try to do. What I tried here was that I built up the triangulated volume with a kind of voxels and work with this function.
However, since the model lives in 3D and can get quite complicated (big parts in space, but mainly "thin" wall thicknesses (plastic parts)) there were many of those voxels needed. If I reduced the number of those voxels, I could do the calculation. But if I use a reasonable resolution of – let's say 0.01 mm (for accuracy) – I already got a matrix with 95 Gb data, which I cannot handle (while even only containing the voxels that really lie inside the part).
  • Is there another approach or even already a solution without the use of this kind of function?
  • Could such an approach as well calculate the minimal sphere-radius in analogy to the biggest one? This would mean, that localy such a sphere would be maximal, but seen over the whole part, it would be the smallest one. In the example above, it would be the right red circle.
Thank you for your help in advance
  1 件のコメント
Vijai Kumar Suriyababu
Vijai Kumar Suriyababu 2022 年 10 月 26 日
編集済み: Vijai Kumar Suriyababu 2022 年 10 月 26 日
This might be a little too late. But, what might help you here is the shape diameter function.
Check out this paper

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回答 (1 件)

Matt J
Matt J 2022 年 10 月 26 日
編集済み: Matt J 2022 年 10 月 26 日
Use alphaShape on the boundary points of the region. Basically, you would apply alphaShape() with increasingly larger alpha radii until you no longer have any holes, or equivalently when the volume of the alpha shape stops increasing. That critical radius is the radius you are looking for.

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