Given the orientation matrix, for example: 0.284, 0.921, -0.265, -0.720, 0.388, 0.575, 0.633, 0.027, 0.773, and the angles of grain boundary in the XY plane (α= 166o) and in the XZ plane (β = 67o), how to obtain the index of the grain boundary plane.
The answer is: -0.562, 0.599, 0.570. This is an example from paper: Randle, V. (1995). Crystallographic characterization of planes in the scanning electron microscope. Materials characterization, 34(1), 29-34.
And in the attached file, I show details of my calculations:
The principle you are describing in your doc file is (from my point of view) correct although I think you defined the angles different.
The definition of my angles results in a description for vector
V1: (-cos(166),-sin(166),0), for V2 (cos(67),0,-sin(67)) so that the cross product V1 x V2 gives the triple
(0.223, 0.893, 0.095),
and when I multiply this with the given matrix (vector*matrix) the result is
(-0.520, 0.554, 0.528).
It is at least "close" to the result given by Randle. :-) but if you normalize this vector (which is not really necessary) you will end in the values
(-0.562, 0.599, 0.570)...which are exactly the description given in the paper..
Dear Gert,
thank you for your reply, I am trying to get the definition of your vector, and I do not see why the component X of your V2 is positive, I put some drawings in the attached file, could you let me know if this description is correct?, where alfa is an angle that rotates around the Z axes and beta is an angle that rotates around the Y axes
Regards and again thank you.
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