Find the coordinates of a triangle in a mesh

I am working with a 2D mesh generated by Gmsh and here I'd like to compute the area of the triangles of in my mesh. What I don't understand is how can I find the coordinates of a given triangle?

In my .m file, I have: msh.nbNod, msh.POS, msh.Max and msh.MIN, msh.LINES, msh.TRIANGLES and msh.PNT.

I think I should somehow use msh.TRIANGLES but I don't get what the numbers listed refer to. If needed, I can post my .m file. Thanks for the help.

[Turlough Hughes]

Can you attach a msh.mat?

[Florian Spicher]

Done!

[Florian Spicher]

Or maybe I misunderstood what you meant since you edited several times your comment, sorry.

Answer 1

[Dave B]

It looks to me like msh.TRIANGLES is referring to indices in msh.POS, which like a reasonable way to describe a mesh.

I'd grab the co-ordinates of a triangle's vertices like this:

squareRing;
triangle_ind=500;
msh.POS(msh.TRIANGLES(triangle_ind,1:3)',:)

ans = 3×3

-1.1902 -0.8650 0 -1.1905 -1.0510 0 -1.3512 -0.9948 0

(Here each row is a vertex, and the columns are x,y,z)

I say it's reasonable, because this is similar to what patch would use (and because max(msh.TRIANGLES(:))==size(msh.POS,1)).

patch('Faces',msh.TRIANGLES(:,1:3),'Vertices',msh.POS,'FaceColor','none','EdgeColor','r')

Note that in the data file you provided, the third column of msh.POS is all zeros, which I assume means that they are all in a 2-D plane.

[Florian Spicher]

That's the case, it's in the Oxy-plane. Thank you very much for the help :)

[Florian Spicher]

I'm not quite sure to fully understand why you set triangle_ind to be equal to 500 though.

[Dave B]

Oh I just picked some particular triangle off the top of my head because you said 'given triangle'...

You can get all the triangles with (I dropped the z-coord because they're all 0):

alltri=msh.POS(msh.TRIANGLES(:,1:3)',1:2);

Now the first 3 rows of alltri are the first triangle, rows 4-6 are the second triangle, etc.

You could reshape that into a 3-D matrix with some gymnastics (maybe there's an easier incantation of reshape/permute, this is what I came up with):

alltri3=permute(reshape(alltri',2,3,[]), [2 1 3]);

So now alltri3(:,:,1) is the first triangle, alltri3(:,:,2) is the second triangle, etc.

I think that all of the areas would be:

ar=squeeze(polyarea(alltri3(:,1,:),alltri3(:,2,:)));

[Florian Spicher]

Aha! Now it’s 100% clear! Well, thank you very much :) you really helped

[Camille Grimaldi]

Hi

How did you get the xy for each triangle in the first place?

Cheers

Camille