Plot Taylor Factor

[pab...@gmail.com]

Hi. How can I plot Taylor factor? I use these commands for Taylor factor calculation:

>> q = 0.5;

>> epsilon = strainTensor(diag([1 -q -(1-q)]))

>> ori= odf.calcOrientations(10000)

>> [M,b,mori] = calcTaylor(inv(ori)*epsilon,sS.symmetrise);

[Ralf Hielscher]

I guess you want to plot the Taylor factor in dependency of orientation. But how?

As a function of the three Euler angles in a 3d plot? 

In phi2 section? 

Ralf.

********************************************************************
Ralf Hielscher                   Tel: +371-531-38556
Fakultät für Mathematik               +371-531-22200 (Sekr.)
Technische Universität Chemnitz  Fax: +371-531-22109
Reichenhainer Str. 39            E-mail: ralf.hi...@mathematik.tu-chemnitz.de
D-09126 Chemnitz                 http://www.tu-chemnitz.de/~rahi
********************************************************************

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[ali hosseinabadi]

yes of course

[Paul black]

absolutely yes

[Ralf Hielscher]

Have you seen this side:

http://mtex-toolbox.github.io/files/doc/TaylorModel.html

Ralf.

********************************************************************
Ralf Hielscher                   Tel: +371-531-38556
Fakultät für Mathematik               +371-531-22200 (Sekr.)
Technische Universität Chemnitz  Fax: +371-531-22109
Reichenhainer Str. 39            E-mail: ralf.hi...@mathematik.tu-chemnitz.de
D-09126 Chemnitz                 http://www.tu-chemnitz.de/~rahi
********************************************************************

[Paul black]

yes, I've seen that but I have a problem with this. it doesn't work for me. I change my pole fig data but every time it plots the same figs.

[Ralf Hielscher]

Hi Paul,

please explain a bit more in detail what you want to do. According to what I guessed you wanted to plot the Taylor factor as a function of the orientation, e.g. in phi2 section. How this is explained in section "The orientation dependence of the Taylor factor" in the link I posted above. 

* If this does not work for you then their is probably an error message. Please post it.

* If this is not what you wanted to plot - try to explain your problem in more detail.

Ralf.

********************************************************************
Ralf Hielscher                   Tel: +371-531-38556
Fakultät für Mathematik               +371-531-22200 (Sekr.)
Technische Universität Chemnitz  Fax: +371-531-22109
Reichenhainer Str. 39            E-mail: ralf.hi...@mathematik.tu-chemnitz.de
D-09126 Chemnitz                 http://www.tu-chemnitz.de/~rahi
********************************************************************

[Paul black]

Dear Ralf,

Actually, I use "The orientation dependence of the Taylor factor" and it plots a Taylor factor that I attach it here. when I use this for another sample, it plots the same Taylor factor again without any changes. one Taylor factor for two different samples.!! 

[ruediger Kilian]

Hi Paul,
it might be helpful to attach the exact code you use.
Cheers,
Rüdiger

[Paul black]

Pole Figure

>> pf;

>> pf(pf.intensities<0)=0;

>> pf_back = correct(pf,'background');

>> pf_def = correct(pf_back,'def');

>> pf_norm = normalize(pf_def);

>> plot(pf_norm,'contourf')

>> CLim(gcm,'equal');

 

ODF

>> odf = calcODF(pf_norm);

>> plotSection(odf,'phi1','sections',9,'silent')

 

Slip systems of hcp

sSBasal = slipSystem.basal(CS);

sSBasalSym = sSBasal.symmetrise('antipodal')

sSPrismatic = slipSystem.prismaticA(CS)

sSPrismaticSym = sSPrismatic.symmetrise('antipodal')

sSPyramidal = slipSystem.pyramidalCA(CS)

sSPyramidalSym = sSPyramidal.symmetrise('antipodal')

sSPyramidal2 = slipSystem.pyramidal2CA(CS)

sSPyramidal2Sym = sSPyramidal2.symmetrise('antipodal')

sS= [sSBasalSym; sSPrismaticSym; sSPyramidalSym; sSPyramidal2Sym]

 

Taylor factor calculation (compression) 

>> q = 0.5;

>> epsilon = strainTensor(diag([1 -q -(1-q)]))

>> ori= odf.calcOrientations(10000)

>> sP = phi1Sections(CS,specimenSymmetry('mmm'));

>> sP.phi1 = (0:10:90)*degree;

>> oriGrid = sP.makeGrid('resolution',2.5*degree);

>> oriGrid.SS = specimenSymmetry;

tic

[M,~,mori] = calcTaylor(inv(oriGrid)*epsilon,sS.symmetrise);

toc

sP.plot(M,'smooth')

 

sP.plot(M,'smooth')

 

mtexColorbar

 

 

 

[ruediger Kilian]

Hi Paul,
I believe you must have something else wrong, using two different input pf gives two different results.
Are you sure the input you are using is distinct?
Cheers,
Rüdiger

[Paul black]

yes, I'm sure. could you please check this code that I use? 

[ruediger Kilian]

That's what I did. Your polefigures and sections look different?
Cheers,
Rüdiger

[Paul black]

Yes. They look different but Taylor factor is the same. 

[Ralf Hielscher]

Hi Paul,

I think there is an understanding problem. You need to distinguish two things:

1. For each orientation you have a certain Taylor factor. This dependency can be visualized - as you did in your code. This dependency has nothing to do with the texture of you specimen. It is a purely geometrical model about how certain orientation deform under strain. Hence, your pole figures have now influence on the result. If you look through your code you will observe that the simulated orientations ori are not used anywhere.

2. You might be interested in the Taylor factor of the whole specimen. This value depends on the actual texture and  hence on the pole figures. But this is only a single number - so plotting makes sense here.

You see how important it is to explain clearly what you want to do. Because otherwise we are guessing all the time, which is not very efficient. 

Ralf.

********************************************************************
Ralf Hielscher                   Tel: +371-531-38556
Fakultät für Mathematik               +371-531-22200 (Sekr.)
Technische Universität Chemnitz  Fax: +371-531-22109
Reichenhainer Str. 39            E-mail: ralf.hi...@mathematik.tu-chemnitz.de
D-09126 Chemnitz                 http://www.tu-chemnitz.de/~rahi
********************************************************************