Calculate eigenvectors and eigenvalues from ODF

[Matteo]

Dear all,

I'm trying to calculate eigenvectors and eigenvalues from ODFs, but can't seem to find a dedicated function for it (or maybe I just can't find it). My objective is to try to replicate the Woodcock (1977) eigenvalues diagram to quantify the texture in my samples. Is there any way to do this in MTEX?

Thanks for the help

Cheers,

Matteo

[Ralf Hielscher]

This is not yet directly implemented. However you can do the following

ori = odf.calcOrientation(100000)
[~,~,lambda,eigV] = mean(ori)

Then lambda gives you the eigen values.

Ralf.

[Matteo]

Thank you very much Ralf. I don't understand why the outcome is a matrix 4x4 for eigV and four values for lambda (eigenvalues). Shouldn't eigenvectors and eigenvalues be only three?

Cheers

Matteo

------------------------

Matteo Demurtas

Ph.D. student

University of Padova

Department of Geosciences

via G. Gradenigo, 6

35131, Padova (Italy)

Skype: matteodemurtas

Fault rocks!

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[ruediger Kilian]

Hi Matteo,
for directions your will get 3 eigenvectors/values but for orientations there are 4 values. Roughly, for uniformly distributed orientations, you’ll get a relatively small largest eigenvalue, identical second und third values and a non-zero smallest value. For a “perfect" unimodal distribution, the largest eigenvalue is 1 and all others are 0. For a “perfect” fibre the two smallest eigenvalues values are 0. For real data, values will only approach 1 and 0.

In case you wanted to calulate eigenvectors/values for directions, here’s an example:
v=vector3d.rand(100) % some input
v=[v.x v.y v.z];
M=v.'*v;
[evec,eval]=eig(M,'vector');


Hope this helps,

Cheers,
Rüdiger

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