EENS 2110 
Mineralogy 
Tulane University 
Prof. Stephen A. Nelson 
Crystallographic Calculations 

Crystallographic Calculations Crystallographic calculations involve the following:
Most of what you will do in lab or on exams will involve crystallographic calculations in the orthorhombic, tetragonal, hexagonal, or isometric systems, where the axes angles are fixed. Note that you will always be given enough information to solve the problem. Some of the problems you might be expected to solve involve determination of Miller Indices from the r and f angles, to determine the r and f angles for faces of mineral with known axial ratios, or to determine axial ratios of minerals that have faces of known r and f angles. First let's review some of the things we know about Miller indices and r and f angles. In the orthorhombic, tetragonal, or isometric systems.
So, f = 0^{o} or f = 180^{o}
So, f = 90^{o} or f = 270^{o}
So, r = 90^{o} and f = 0^{o} or f = 180^{o}.
Now let's do some examples 

For the (110) we note that it does not intersect the c axis, so we can look at it in the two dimensional plane containing the a and b axes, as shown in the drawing below. From this face we should be able to determine the a/b axial ratio. Since the f angle is the angle between the normal to the face and the b axis, by similar triangles we know that the f angle also occurs between the a axis and the face.


Thus, we can write:
tan 45^{o} = 1b/1a then 1a/1b = 1/tan 45^{o}= 1 So, a/b: b/b = 1: 1 
We next note that the (011) intersects the b and c axes
only, so we can examine this face in the plane containing only b
and c, as shown below. From this drawing we can obtain the
c/b axial ratio. Since the r angle is
the angle between the pole to the face and the c axis, again by similar
triangles we know that the r angle also
occurs between the b axis and the (011) face. 

Thus, we can write:
tan 70^{o} = 1c/1b then c/b = arctan 70^{o} So, c/b = 2.7475 
so, a : b : c = 1 : 1 : 2.7475 and the mineral must be tetragonal, since a/b =1. 
The face (120) does not intersect the c axis, so we can look at this face in the plane containing only the a and b axes. We must also remember that Miller Indices represent the inverse of the intercepts, so the face (120) intersects the a axis at twice the number of unit lengths that it intersects the b axis. Since the f angle is the angle between the normal to the face and the b axis, by similar triangles we know that the f angle also occurs between the a axis and the (120) face. 

Then we can write: tan 70^{o} = 1b/2a a/b = 1/2tan70^{o} 1a/1b = 0.18199 
We next note that the (011) intersects the b and c axes only, so we can examine this face in the plane containing only b and c, as shown below. From this drawing we can obtain the c/b axial ratio. Since the r angle is the angle between the face and the c axis, again by similar triangles we know that the r angle also occurs between the b axis and the (011) face. 

For this face we can determine that tan 32^{o} = 1c/1b 1c/1b = 0.6248 

c. Faces r
f

We first attempt to draw a 3dimensional view of this face.
Notice that the f angle is
measured in the horizontal plane that includes the a and b
axes. The r angle is measured in a vertical
plane that includes the c axis and the line normal to the face in
the a  b horizontal plane, and is measured between the c
axis and a line normal to the face. 

We can determine the a/b part of the axial ratio by looking at the projection of this face in the a  b plane. 1b/(1/3)a = tan 33^{o
}1a/3b = 1/tan 33 
In order to determine the length of the c axis, we need to know the length of the line labeled t, because this line forms the base of the triangle in which the r angle is measured. The length of the line t is: t/b = cos 33^{o
}t/b = 0.8397 

We can now use this to determine the c/b axial ratio. 1c/t = tan 24^{o }1c = 0.8387 b tan 24^{o} c/b = 0.3747 
Thus, the axial ratio for this mineral is 4.6196 : 1 : 0.3747 
Now we'll look at an example where we are given the axial
ratio of the mineral and asked to calculate the r
and f
angles for the faces.

In this case the intercept on the all three axes is 1. 

Since the f angle for this face is measured in the horizontal a  b plane, we can draw the plane containing only the a and b axes to determine the angle. Since the axial ratio tells us that the relative lengths of the a and b axes are equal tan f = 1b/1b = 1 f = 45^{o} 
t/b = cos 45^{o
}t = 0.7071 b 

Now we can determine the angle by drawing the plane that includes the c axis and the line t. In this plane we can let the length of the c = 5b, from the axial ratio. Then: tan r = 5b/t tan r = 5b/0.7071b tan r = 5/0.7071 tan r = 7.071 r = arctan (7.071) = 81.95^{o} 
So for the (111) face in this crystal r =81.95^{o} and f = 45^{o}.

For the face (101) r = 70^{o} f = 90^{o}


tan r = 1c/1a so, tan 70^{o} = c/a = 2.7475 and the axial ratio is: 1 :1 : 2.7475 
tan (53.9478) = 2.7475x/1, and thus, x = 0.5 The parameters for this face are then: 1, ¥, 1/2, which can be inverted to give the Miller Index  (102).

Since the mineral is tetragonal, the face labeled (0hl) would have the Miller Index (012). Problems such as these could be asked on any laboratory exam. 