Interpolation of the values of points on the surface created by convex hull
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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 Comments
Matt J
on 5 Jul 2022
Edited: Matt J
on 5 Jul 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.
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More Answers (2)
David Goodmanson
on 30 Jun 2022
Edited: David Goodmanson
on 30 Jun 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 Comments
Torsten
on 30 Jun 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.
Matt J
on 30 Jun 2022
Edited: Matt J
on 30 Jun 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
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