Effacer les filtres
Effacer les filtres

Interpolation of the values of points on the surface created by convex hull

12 vues (au cours des 30 derniers jours)
Nima
Nima le 22 Oct 2020
Modifié(e) : Matt J le 5 Juil 2022
I have some 3D points like and their values like . The convex hull of these points are generated by convhull or convexHull (see figure below)
Now I want to interpolate the values of triangle vertices for N new points on the convex hull. What are the possible methods to do this?
  7 commentaires
Jose Carrasco
Jose Carrasco le 3 Juil 2022
Modifié(e) : Matt J le 5 Juil 2022
Hi Torsten, I am sorry my point is not clear enough. With "tangent to a vertex" I mean that I want to produce the minimum surface that is "bubbly" (derivable at any point) and tangent to all the vertices of the convex hull made of all the triangular surfaces that connect the boundaries (most external or "extreme" points) of the scatter plot attached to this comment.
open untitled.fig
Matt J
Matt J le 5 Juil 2022
Modifié(e) : Matt J le 5 Juil 2022
@Jose Carrasco The surface you describe does not have a minimal form. It wll always be possible to make a "bubbly" surface smaller and smaller in a manner that converges to the convex hull. Think of a straight line segment in 2D and ask yourself, what is the smallest ellipse that can enclose the line. Clearly the solution must choose the major axis of the ellipse to be the line segment itself. However, the minor axis can be chosen to be any arbitrarily small eps>0.

Connectez-vous pour commenter.

Réponse acceptée

Nima
Nima le 1 Juil 2022
Hi everyone,
The answer I have already found is: Barycentric coordinate system.
In the section ' Interpolation on a triangular unstructured grid ' I have the solution.

Plus de réponses (2)

David Goodmanson
David Goodmanson le 30 Juin 2022
Modifié(e) : David Goodmanson le 30 Juin 2022
Hi Jose/Nima,
Suppose you have a triangle with vertex points p1, p2, p3, each of those defined by a column vector of their x,y,z coordinates. You can form the 3x3 matrix
M = [p1 p2 p3]
which is nonsingular if the vertex points are not colinear, which they are not in this case. Any point p0 (also defined by a column vector) is a linear combination of p1,p2,p3 with coefficients c = [c1;c2;c3]. In matrix notation that is M*c = p0
with solution
c = M\p0
For a quantity A with values A1,A2,A3 at points p1,p2,p3 respectively, the obvious way to interpolate and find the value at p0 is to use c1A1 + c2A2 + c3A3, or in matrix notation
Ainterp = c'*[A1;A2;A3]
so the three highlighted lines provide a solution. This is true whether p0 lies in the plane of the triangle or not, although it might not be a useful solution if p0 is far away from the triangle. If the point lies inside the triangle on the surface, I believe that 0<=c1<=1, 0<=c2<=1, 0<=c3<=1, c1+c2+c3 = 1 which is appropriate for intepolation.
  2 commentaires
John D'Errico
John D'Errico le 30 Juin 2022
Modifié(e) : John D'Errico le 30 Juin 2022
Yes. As long as the weights, as applied to the three corners of a triangle sum to 1, and all are between 0 and 1, this forms a convex combination. As such, a new point formed from such a linear combination of the corner vertices will lie strictly inside the triangle, and will lie in the plane of the triangle. If any of those weights were exactly 1, then the interpolated point lies exactly on the corresponding vertex. And if any of the weights were exactly 0, this corresponds to a point along the line opposite one of the vertices.
How to "interpolate" such a polytope is undefined, since there was never any direction as to what that interpolation would mean to either @Nima or @Jose Carrasco. For example, was the question targeted at sampling new points on the surface for some unspecified reason and based on some unspecified criteria? Or was the real question how to interpolate values at the corner nodes of the polytope, then trying to evaluate some function f(x,y,z) using interpolation on the surface of the polytope? Give me some time, and I might even guess some other reason behind the questions here, why I would not have touched this post when it originally was made.
Torsten
Torsten le 30 Juin 2022
The interpolation suggested assumes that the function can be linearly approximated over the triangle:
f(p0) = f(c1*p1+c2*p2+c3*p3) =(*) c1*f(p1) + c2*f(p2) + c3*f(p3) = c1*A1 + c2*A2 + c3*A3
where at the =(*), the linearity of the function f is used.
This might or might not be a good choice.

Connectez-vous pour commenter.


Matt J
Matt J le 30 Juin 2022
Modifié(e) : Matt J le 30 Juin 2022
For example, was the question targeted at sampling new points on the surface...?
Assuming that's the correct interpretation, then,
N=10;
P=rand(10,3);
k=convhull(P);
V=P(unique(k),:); %Generate vertices V
k=num2cell(convhull(V),1);
res=@(z) reshape(z,[],1,3); %Generate affine frame for each triangle
V0=res( V(k{1},:));
dV1=res( V(k{2},:)) -V0 ;
dV2=res( V(k{3},:)) -V0;
[a,b]=ndgrid(linspace(0,1,10)); %Sweep each triangle with extra samples
idx=a+b<=1;
a=a(idx)'; b=b(idx)';
XYZ=num2cell( reshape(V0+dV1.*a+dV2.*b ,[],3) ,1);
trisurf(cell2mat(k),V(:,1), V(:,2),V(:,3)); hold on %Visualize
scatter3(XYZ{:},'filled','r'); hold off

Catégories

En savoir plus sur Interpolation dans Help Center et File Exchange

Produits


Version

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by