I am trying to interpolate a table of the following form.
{{x1,y1,z1},{x2,y2,z2}...}
Where z depends on x and y, and x ranges from 0.01 to 1 in steps of 0.01, and y ranges from 0.01 to 15 in steps of 0.1 . It is a huge table so I cannot post the elements.
This is what the surface looks like.
I want to interpolate the table, however I am getting the following error regarding the mesh,
Interpolation::fememtlq: The quality 0.` of the underlying mesh is too low.
The quality needs to be larger than 0.`.
I read that this error can be caused by collinear points, so I performed the following perturbation. I tried this at multiple values of epsilon, and also without adding a perturbation to z and the results are more or less the same.
epsilon = 10^-5;
perturbData = {#1 + RandomReal[epsilon {-1, 1}], #2 +
RandomReal[epsilon {-1, 1}], #3 +
RandomReal[epsilon {-1, 1}]} & @@@ intrct;
reg = ConvexHullMesh[perturbData[[All, 1 ;; 2]]];
f = Interpolation[perturbData, InterpolationOrder -> 1]
However when I plot this to check how it looks I get this surface which does not seem right at all.
I am trying to find the roots of this function, but I cannot do this properly with this setup.
So I would like to know if there are any alternative ways to solve the interpolation error, or to refine the one I have.
Interpolation[...,InterpolationOrder->1]
? $\endgroup$