calculate volume from iso-surface coordinates (x,y,z).

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Hello,
I have coordinates (x,y,z) of an isosurface. How can I calculate volume of that isosurface? I have attached an image of iso-surface and coordinates file here.
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Fifteen12
Fifteen12 2023-9-21
编辑:Fifteen12 2023-9-21
Do isosurfaces necessarily have a volume? Are these completely closed surfaces? If you just need to calculate the surface area you could check out this approach (I haven't looked at it myself): https://www.mathworks.com/matlabcentral/fileexchange/25415-isosurface-area-calculation?s_tid=answers_rc2-2_p5_MLT

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回答(3 个)

Walter Roberson
Walter Roberson 2023-9-21
If you use boundary then it has an optional second output which is the volume.
However, I would not expect boundary() to be able to deal with disconnected components, so you would need to separate out the different components based on the vertices returned by isosurface().
You could potentially turn the vertices into a connection table and create a graph object from it, and use something like conncomp to determine which sections are disconnected from the others.
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Bruno Luong
Bruno Luong 2023-9-22
@Walter Roberson "I'am not sure at the moment why one of the volumes comes out as 0.... "
The text is clipped.

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William Rose
William Rose 2023-9-21
Find the delaunay triangulation of the 3D points with
DT=delaunay(x,y,z);
This gives a set of tetrahedrons which fill the volume. Then compute and add up the volumes of the tetrahedrons.
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Walter Roberson
Walter Roberson 2023-9-22
When we have values for each point but no connectivity information for the vertices, then the only possibility is to treat the points as being scattered samplings of a continuous function, and to interpolate those scattered positions over a grid and construct isosurfaces of the result.
... It doesn't look very good.
data = readmatrix('Q.txt');
x = data(:,2);
y = data(:,3);
z = data(:,4);
q = data(:,5);
F = scatteredInterpolant(x, y, z, q);
N = 50;
[minx, maxx] = bounds(x);
[miny, maxy] = bounds(y);
[minz, maxz] = bounds(z);
[qX, qY, qZ] = meshgrid(linspace(minx, maxx, N), linspace(miny, maxy, N), linspace(minz, maxz, N));
qQ = F(qX, qY, qZ);
[minq, maxq] = bounds(qQ(:));
isolevels = linspace(minq, maxq, 6);
isolevels([1 end]) = [];
for V = isolevels
isosurface(qX, qY, qZ, qQ, V);
end
view(3)
legend("q = " + isolevels);
figure()
h = scatter3(x, y, z, [], q);
%h.AlphaData = 0.3;
h.MarkerEdgeAlpha = 0.1;
h.MarkerFaceAlpha = 0.1;
Raju
Raju 2023-9-24
@Walter Roberson I appreciate your effort. I will have a look if I can get some connection data.

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Bruno Luong
Bruno Luong 2023-9-21
编辑:Bruno Luong 2023-9-21
Do you have connectivity face of these points coordinates?
If you use the command isosurface https://www.mathworks.com/help/matlab/ref/isosurface.html you should have. Please share the outputs faces and verts or structure s (save in matfile and attach here).
Or try this formula:
[x,y,z] = meshgrid([-1.1:0.05:1.1]);
V = x.^2 + y.^2 + z.^2;
s = isosurface(x,y,z,V,1) % replace this command using your data
s = struct with fields:
vertices: [7470×3 double] faces: [14936×3 double]
VF = permute(reshape(s.vertices(s.faces,:),[size(s.faces) 3]),[3 1 2]);
Vol = 1/6*sum(dot(cross(VF(:,:,1),VF(:,:,2),1),VF(:,:,3),1)) % close to 4/3*pi volume of the sphere of raduius 1
Vol = 4.1809
4/3*pi
ans = 4.1888
This formula works for non-convex volume enclosed by the surface given by triangular connectivity (correctly oriented).
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Bruno Luong
Bruno Luong 2023-9-22
编辑:Bruno Luong 2023-9-22
@Raju Sigh, I still don't see any connectivity data. Can't help you more.
[X, Y, Z] = meshgrid(linspace(-2*pi, 2*pi, 200));
iR2 = 1./(X.^2+Y.^2+Z.^2);
C = iR2 .* (sin(X).*cos(Y) + sin(Y).*cos(Z) + sin(Z).*cos(X));
s = isosurface(X, Y, Z, C, 0.05); % replace this command using your data
% the connectivity mooke like this
s.faces(1:10,:),
ans = 10×3
1 2 3 1 4 2 2 4 5 4 6 5 5 6 7 6 8 7 9 10 11 9 12 10 10 12 1 10 1 3
The connectivity tells the mesh triangles of the surface connect which vertexes. As above the last line tell the 10th triangle is composed of of three vertices (#10, #1, #3)
Raju
Raju 2023-9-24
@Bruno Luong Thanks for your effort. But these are the data I have right now. I will have a look if get some additional data (connection data you mean).

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