Texture Fiber HCP Prismatic and Pyramidal

[Camilo Garzón]

Hello,

I've been trying to make prismatic and pyramidal fiber of an hcp system without success, next, my procedure:

First I try with the basal fiber doing this:

1. Define the symmetry and an orientation of a basal pole

csa = crystalsymmetry('hexagonal');
o_basal = orientation.byEuler(0,0,0,csa);

2. Define the fiber with the orientation and a vector 3d to align every direction parallel to the normal to the orientation defined.

f_bsl = fibre(o_basal,vector3d.Z)

---
 size: 1 x 1
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)
 o1: (0°,0°,0°)
 h: (0001)

---
plotPDF(f_bsl,h,csa,'lineColor','r','linewidth',2)

Luckily, I built the basal fiber right, as you can see in the plot attached. However, I couldn't get the same results for prismatic and pyramidal ones. 

For that reason, I try to build the prismatic and pyramidal fibers, with two orientations.

For prismatic I take as reference the orientation (1 1 -2 0) (1 -1 0 0),

o_pr = orientation.byMiller([1 1 0],[1 -1 0],csa);

--

 Bunge Euler angles in degree

  phi1  Phi phi2 Inv.

   360   90   30    0

 

The idea is to complete the fiber since phi1 = 0*degree to phi2 = 360*degree

o_pr1 = orientation.byEuler(0*degree,90*degree,30*degree,csa);
f_pr = fibre(o_pr1,o_pr);

---

 o1: (0°,90°,30°)

 h: (2-1-10)

>> 

It supposed that it must did the prismatic fiber, but the result is other as you can see in the attached pole figure.

How can I solve it? And also I would like that the point in the center of pole figure appear when it supossed to be there?

Best Regards

Camilo Garzón

[Rüdiger Kilian]

Hi,
how about

fp = fibre(Miller(1,0,-1,1,cs),zvector)

?
That should be a fibre with (1,0,-1,1) as axis parallel to 001 in specimen coordiantes.

Hope that helps.
Cheers,
Rüdiger
________________________________________
From: mtex...@googlegroups.com <mtex...@googlegroups.com> on behalf of Camilo Garzón <scga...@gmail.com>
Sent: Saturday, September 21, 2019 7:32:59 PM
To: MTEX
Subject: {MTEX} Texture Fiber HCP Prismatic and Pyramidal

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[Camilo Garzón]

Hi Rüdiger,

Thanks for your answer, I tried how you tell me, however, I didn't get the expected results.

f_prismatic = fibre(Miller(1, 1, -2, 0,csa),zvector))
--

 size: 1 x 1
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)

 o1: (330°,90°,30°)
 h: (11-20)


f_pyramidal = fibre(Miller(1,0,0,0,csa),zvector)


--

size: 1 x 1
 mineral: Titanium (Alpha) (622, X||a, Y||b*, Z||c)

 o1: (300°,90°,60°)
 h: (10-10)

When I plotted, I expected different results in the PDF of what I get, In the attached image the I put the pole figures for prismatic and pyramidal fibre according to with the next article.

Texture Analysis in hexagonal materials /DOI: 

• 10.1016/S0254-0584(03)00168-8
• 
  
• What do you think?

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[ruediger Kilian]

Hi,
if you look at the pole figures for the {11-20} fibre, you see that (10-10) makes the small circle ring (such as (01-10) would do) while you get the equatorial great circle for (1-100) and (-1100). So the fibre is fine so far but I guess you need to superpose those pole figures to see the same as in that publication for now.

Hope that helps.

Cheers,
Rüdiger

cs = crystalSymmetry('622','X||a')
f = fibre(Miller(1,1,-2,0,cs),zvector)
h = Miller(1,0,-1,0,cs)
plotPDF(f,h.symmetrise,'linewidth',2,'linecolor','r')

[João Rodrigues]

Hi Camilo,

You can calculate the odf and then plot pole figure.

f1 = fibre(Miller(1,1,-2,0,cshex),zvector);

odf = fibreODF(f1,'halfwidth',5*degree);

plotPDF(odf,Miller(-2,1,1,0,cshex));

Best regards,

[Camilo Garzón]

Hi

Thanks, Rüediger and João for your help, the solution of João, confirms clearly what you said Rüediger.

 

Best Regards

Camilo