Alternatives to Delaunay triangulation

Hello,

I am in need of calculating the volume of a shape that is given as a cloud of points. I thought a first step would be to triangulate using delaunay but the results are not satisfactory. I have attached some images to show what is unsatisfactory. I zoomed in to look at a part of the cloud that is kind of shaped like the cone of an airplane. I have provided two angles so a sense can be gotten. The super elongated triangles do not seem like they will be useful in getting me surface area and volume.

Does anyone know of alternatives that would produce better results?

Thanks.

[Richard Brown]

Looks to me like the cone is "open" - like an ice-cream cone with no ice-cream. Should there be points over the disc-shaped part of it?

[David Wilson]

You could try alphashape. Further details on how to extract its volume are given in:

https://au.mathworks.com/matlabcentral/answers/152189-volume-of-3d-polyhedron

or you could try convhulln.

I'll try to regenerate your cone shape. One could, if they bothered, compute the exact volume for comparison.

%% Make a nose cone 
npts = 1e3; 
a = 1; % change this to pi/2
l = a*rand(npts, 1); 
phi = rand(npts,1)*2*pi; 

p = deg2rad(3); 
r = sin(l); 
x = cos(l); 
y = r.*sin(phi); z = r.*cos(phi); 

plot3(x,y,z,'.');
axis equal
grid on 

Now compute the convex hull and the volume.

pts = [x,y,z]; 
[k,vol] = convhulln(pts);

For a sanity check, if you re-run the code above with

a = pi/2 % half a sphere

I get vol = 2.0593 which is not too far off the exact 4/6 pi = 2.0944.

[David Winthrop]

Richard, it is just the zoom of the MATLAB plotter that cuts off part of the figure when you zoom in. The shape is as closed as a cloud of points can be. I will make an "answer" that has the data in a text file and provide the code I am using to plot it.

[David Winthrop]

David,

Thanks for the input on how to calculate the volume, but won't convex hull exclude areas of concavity?

Answer 1

[Richard Brown]

As David Wilson suggested, use alphashape:

U = csvread('Coords.txt');
shp = alphashape(U(:, 1), U(:, 2), U(:, 3));
plot(shp)
disp(shp.vol)

I get a volume of 3.59e-8. The picture is here:

Or, I just noticed that the commands are different in R2019a (I did the previous with R2018a by accident):

U = csvread('Coords.txt');
shp = alphaShape(U(:, 1), U(:, 2), U(:, 3));
plot(shp)
disp(shp.volume)

Answer 2

[David Winthrop]

I have attached the point cloud as a file Coords.txt. Below is the code I use to plot it. I have found that the boundary code works well but still has some very large and skew triangles in it. It is an improvement. Any further suggestions would be welcome.

fl = fopen('Coords.txt');
T = textscan(fl,'%f %f %f','Delimiter',',');
X = T{1};
Y = T{2};
Z = T{3};
B1 = boundary(X,Y,Z,.8);
figure
plot3(X,Y,Z,'*')
Ax1 = gca;
xlabel('X')
ylabel('Y')
zlabel('Z')
figure
trisurf(B1,X,Y,Z)
Ax2 = gca;
set(findall(Ax2,'type','patch'),'edgecolor',[.2,.2,.2])
hlink = linkprop([Ax1;Ax2],{'xlim','ylim','zlim','cameraposition'});
xlabel('X')
ylabel('Y')
zlabel('Z')

Thanks

Answer 3

[darova]

Volume can be found as summation volume of pyramids. Here is an idea:

clc,clear

load Coords.txt
x = Coords(:,1);
y = Coords(:,2);
z = Coords(:,3);
    % move to origin
x = x-mean(x);
y = y-mean(y);
z = z-mean(z);
    % convert to spherical system
[t,p,r] = cart2sph(x,y,z);
tri = delaunay(t,p);
trisurf(tri,x,y,z)
    % indices of triangle
i1 = tri(:,1);
i2 = tri(:,2);
i3 = tri(:,3);
    % vectors of triangle base
v1 = [x(i1)-x(i2) y(i1)-y(i2) z(i1)-z(i2)];
v2 = [x(i1)-x(i3) y(i1)-y(i3) z(i1)-z(i3)];
A = 1/2*cross(v1,v2,2);                     % surface of a triangle
V = 1/3*dot(A,[x(i1) y(i1) z(i1)],2);       % volume of a triangle
V = sum(abs(V));

[David Winthrop]

Something is wrong with your volume calculations. If I do this:

fl = fopen('Coords.txt');
T = textscan(fl,'%f %f %f','Delimiter',',');
X = T{1};
Y = T{2};
Z = T{3};
[B1,V1] = boundary(X,Y,Z,.8);
V1
V2 = VolOfTriangulation(B1,X,Y,Z)

The function VolOfTriangulation is simply your code for calculating the volume based on triangulation.

V1 = 3.746596463335299e-08

V2 = 5.908642403710292e-08

[darova]

Maybe it's because of wrong triangulation in this place:

alphashape should be cool in this case then

[darova]

It's because of converting cartesian coordinates to spherical

[t,p,r] = cart2sph(x,y,z);

Geometrically -pi and +pi are the same, but delaunay interprets it as different numbers.

That is why there is a gap between -pi and +pi

Don't know how to resolve it

Answer 4

[David Winthrop]

Thanks everyone!

I found a file exchange function that does a very nice job of displaying the mesh so that it matches the mesh exported from a Finite Element suite. However, I am still in need of calcuating the volume accurately. If I plot my points using that function and use the volume calculated by dorova I do not get the same value that the FEM code gets.

I get 5.901791412991205e-08 whereas the software says the volume is 3.739248319033497e-08. Since I am trying to use MATLAB to track the volumes I need to match the software volume as closely as possible. Does anyone else know of a good method for this?

[Richard Brown]

Did you see my solution using alphaShape above? It produces a mesh that looks very similar, and a volume of 3.59e-8 with default parameters ...

[darova]

I get 3.7336e-08 using my script. How is this possible?

[David Winthrop]

Hello Richard. I did see yours and I had tried it before. If you will notice, there are some large triangles near the circular opening as the method you used is covering a convex section. If you feel like it, please download the code I linked to and check it out. It is very nice.

[David Winthrop]

darova, I am not sure. I will double check .

[M.S. Khan]

David, could you plz explain how you applied trgingulations.

regards,