Angle between a scratch and mean orientation of the grain where the scratch resides

[Mustafa-Rifat]

Hello mtex community,

The EBSD plot that I have shown here have 6 scratches, made intentionally. Let's consider the 3rd scratch from right. The green arrow shows the scratch direction. The scratch passes through 3 grains. I want to find the angle between the mean orientation of each of the grains with the scratch direction (i.e. angle between the mean orientation of grain 1 and scratch direction; angle between the mean orientation of grain 2 and scratch direction;angle between the mean orientation of grain 3 and scratch direction).
I know how to get the mean orientation of a grain. I use:
grains (x,y).meanOrientation
Here x and y are any point coordinate which is inside a grain. This gives me euler angle. If I find the mean orientation of another grain, I can use this line of code to find the misorientation between 2 grains using this code:
angle(grains (p1(1),p1(2)).meanOrientation,grains (a,b).meanOrientation)/degree
Where (a,b) is a point coordinate inside the second grain. Can anyone help me how to find the angle between the scratch and mean orientation of a grain here?

[kilir]

Hi,
according to the title you are looking for the angle between a direction and an orientation. I don't think this can be defined. Can you give an example what this should be?

Asking for the direction between a direction ("scratch") and an orientation is equally confusing to me. Maybe you mean the crystal direction parallel to the scratch? Maybe you need to explain a little more in detail.

Cheers,
Rüdiger

[kilir]

Hi Mustafa,
orientation .* crystal direction(s) will give you the direction(s) in specimen coordinates and inv(orientation).* specimen direction ref will give you the crystal direction parallel to ref. Then you can use angle.
See also here: https://mtex-toolbox.github.io/DefinitionAsCoordinateTransform.html

Hope that helps.

Cheers,
Rüdiger

[Mustafa-Rifat]

Let's say I am considering r = ori * h from the link: https://mtex-toolbox.github.io/DefinitionAsCoordinateTransform.html
These are the operations I am performing:

%defining the crystal symmetry similar to the way I defined in the EBSD plot
cs = crystalSymmetry('m-3m',[3.57 3.57 3.57],'mineral', 'Ni-superalloy', 'color', 'light blue')
%defining the [100] direction
h = Miller(1,0,0,cs,'uvw')
%finding the mean orientation of the grain I am interested
ori=grains(151,25).meanOrientation
%transferring h to specimen coordinate
r=ori*h
%finding the angle
angle(r,vector3d.Y,'noSymmetry') / degree

I have got an answer about 65 degrees.
However, You will see that in I have taken vector3d.Y at last line with r. This is (i.e. [0 1 0]) what I assumed to be the direction of the scratch in the specimen. Is it alright?

[kilir]

Hi Mustafa,
vector3d.Y is the same as vector3d(0,1,0), however, whether this direction coincides with the direction of your scratch as you see it on the screen depends on how you set the plotting conventions. Maybe just use xy = ginput(2) to get the start and end coordinates of your scratch when in doubt where your x-axis points to.

Another note: you may want to use r=ori*h.symmetrise('unique') since I guess you can't distinguish between symmetrically equivalent crystal directions.

Cheers,
Rüdiger

[Mustafa-Rifat]

Hello Rüdiger,

Thanks a lot for your quick replies. I did find the start and end point of the arrow and saw there is an increase of y-coordinate. So, I have chosen the vector3d.Y.

When I use

 r=ori*h.symmetrise('unique')
angle(r,vector3d.Y,'noSymmetry') / degree

as you suggested, I get 6 angles. Moreover, I found h.symmetrise to be

u 1  0  0 -1  0  0
v 0  1  0  0 -1  0
w 0  0  1  0  0 -1

So, I guess the 6 angles that I get are the angles between scratch direction [0 1 0] and the 6 [u v w] here respectively, right?

[kilir]

Exactly.
Adding 'antipodal' should leave you with three unique values.
Rüdiger