Units when measuring grain parameters / m-index

[Filippe Ferreira]

Hello, everyone. First of all thanks for all  the developers of MTEX. It is such a great tool with infinite possibilities.

Actually, i have two questions:

1-By measuring grain parameteres such as area, for example:

{

%% Grain-size Analysis

% Lets go back to the grain size and analyze its distribution. TO this end

% we consider the complete data set.

mtexdata forsterite

grains = calcGrains(ebsd)

%%

% and compute the vector of grain areas

ar = grains.area;

%%

% The following script compute the distribution of grain areas within the

% data set.

% define the binning of the areas

bins = linspace(0,max(ar)+eps,15);

% find for each area the binId

[tmp,binId] = histc(ar,bins);

% compute the sum of areas belonging to the same bin

cumArea = zeros(1,numel(bins)-1);

for i = 1:numel(bins)-1

cumArea(i) = sum(ar(binId == i)) ./ sum(ar);

end

% plot the result as a bar plot

binCenter = 0.5*(bins(1:end-1)+bins(2:end));

bar(binCenter,cumArea,'BarWidth',1)

xlim([bins(1),bins(end)])

xlabel('Grain Area')

ylabel('relative volume') } The x axis gives values in pixels, right? Is there a possibility that this value could be arranged accordingly with the step size for EBSD data, giving the results in µm², for example?  2-Does MTEX computes M-index(Skemer 2005)? Thanks!

[David Mainprice]

The units used by MTEX for the EBSD map will be the same as used for grains. For Oxford/HKL data it is in microns, for other sources of EBSD data you should check the scale used.

You can calculate the equivalent to the Skemer M-index

%**************************************************************************

% Misorientation angle distribution for a uniform distribution

%**************************************************************************

% misorientation symmetry for the crystal symmetry of your data

CS_Orthorhombic = symmetry(‘mmm');

[density_uniform,~] = angleDistribution(CS_Orthorhombic);

density_uniform = density_uniform/sum(density_uniform);

% uncorrelated mdf from odf

mdf = calcMDF(odf);

%**************************************************************************

% Misoreintation angle distribution of a misorientation distribution function

%**************************************************************************

[density_MDF,~] = calcAngleDistribution(mdf,'resolution',1*degree);

density_MDF = density_MDF/sum(density_MDF);

% M index

M_index = (sum((abs(density_MDF-density_uniform))/2));

all the best David

ylabel('relative volume') } The x axis gives values in pixels, right? Is there a possibility that this value could be arranged accordingly with the step size for EBSD data, giving the results in um^2, for example?  2-Does MTEX computates M-index(Skemer 2005)? Thanks!

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David Mainprice

15 Les Romarins

34270 Les Matelles

Tel. +33 4 67 67 03 30

(In France 04 67 67 03 30)

E-mail: davidma...@icloud.com

[Filippe Ferreira]

Thanks very much for the help, Mr. Mainprice. I made a comparison of the values of different texture indices i.e. entropy,J and M index, with and without calculating the psi function for optimal kernel function, as follows:

 

With kernel estimation: 

grains = calcGrains(ebsd)
psi = calcKernel(grains('Hematite'))
odf = calcODF(ebsd('Hematite'),'kernel',psi)

CS_Trigonal = symmetry('-3m');
[density_uniform,~] = angleDistribution(CS_Trigonal);
density_uniform = density_uniform/sum(density_uniform);
mdf = calcMDF(odf);

[density_MDF,~] = calcAngleDistribution(mdf,'resolution',1*degree);

density_MDF = density_MDF/sum(density_MDF);

M_index = (sum((abs(density_MDF-density_uniform))/2));

M_index
textureindex(odf)
s = real(entropy(odf))

Without kernel estimation:

odf = calcODF(ebsd('Hematite'))

CS_Trigonal = symmetry('-3m');

[density_uniform,~] = angleDistribution(CS_Trigonal);

density_uniform = density_uniform/sum(density_uniform);

mdf = calcMDF(odf);

[density_MDF,~] = calcAngleDistribution(mdf,'resolution',1*degree);

density_MDF = density_MDF/sum(density_MDF);

M_index = (sum((abs(density_MDF-density_uniform))/2));

M_index

textureindex(odf)

s = real(entropy(odf))

giving the results (with kernel estimation / without kernel estimation):

M index:    0.0796 / 0.0741

J index =    4.6087/ 2.5176

Entropy =   -1.0167/ -0.5461

Maybe apart from the M index, the values showed a considerable discrepancy. For this type of quantification analysis, does the kernel estimation adds more reliability to the data?