Integrated Composite Analyzer (ICAN/JAVA)
Particle Slicing Computation

Particle Slicing

Because the particles are assumed to be evenly spaced, it is sufficient to consider one particle within a volume that is repeated as required.
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The size of the cube containing the particle may be computed from the diameter of the particle, d_p, and the particle volume ratio, k_p. The particle volume ratio may be expressed as the ratio of the volume of the particle to the volume of the cube that may be repeated.
(1.1)   k_p = (/6) (d_p)³ / D³
where D is length of the side of the cube over which the particles are evenly spaced.

Solving for D, one obtains:
(1.2)   D = d_p [(/6) / k_p]^(1/3)

Consequently, one need consider a horizontal slice, possibly containing a piece of the particle, only within a cube over which the particles are evenly spaced. One such slice is shown below:
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The particle slice volume ratio may be expressed as the ratio of the volume of the particle within the horizontal slice to the volume of the horizontal slice. Further information is available about computing volume ratios for a slice of a spherical particle. The analysis below uses the volumes and volume ratios for the slice.

To analyze further, the horizontal slices may be sliced in the vertical direction; for example, along the indicated lines above. The round edges are approximated by square edges, as shown below:
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Particle Subslicing

1. If the subslice is entirely inside the sphere, i.e. the sphere contains the entire subslice, there is obviously an intersection. The entire subslice is inside the sphere if the farthest point of the subslice is within a radius from the center of the sphere, i.e. ||F - O|| < R.

2. If the subslice contains the entire sphere, i.e. the sphere is entirely inside the subslice, there is obviously an intersection.
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   For each axis, let T be a unit vector defining the direction to test. The sphere is within the subslice if, for all such directions, T, |(C-O)·T| < ½ |D·T| - R.

3. If the subslice is entirely outside the sphere, i.e. the sphere is entirely outside the subslice, there is obviously no intersection.
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   For each axis, let T be unit vector defining the direction to test. The sphere is outside the subslice if, for any such direction, T, |(C-O)·T| > ½ |D·T| + R. This test determines whether the sphere is outside of the box enlarged by the radius of the sphere on each side. The "green" areas may be described as the area outside the locus of points made by the center of the sphere rolling around on the outside of the box and inside a box enlarged by the radius of the sphere on each side. The "green" areas in the figure, near each corner or edge, are left for the following analysis.

4. If the center of the sphere is outside the subslice by an amount greater than R, the sphere does not intersect the subslice. For each axis, let T be unit vector defining the direction to test. The distance that the center of the sphere is outside the subslice in the direction specified by T, is given by:
   d_T = |(C-O)·T| - ½ |D·T|
   If d_T < 0, use zero for this term in the next equation, because one only needs the distance outside the corresponding face of the subslice.
   Let d = SQRT( d_Tx² + d_Ty² + d_Tz² ).
   There is no intersection when d > R.

5. Otherwise, there is an intersection.

The above analysis may be used as follows:

• Use (1) for R = ½ d_p, to determine whether the subslice is particle only. In that case, k_p = 1.
• Use (1) for R = ½ d_i , and (3) and (4) for R = ½ d_p, to determine whether the subslice is interphase only. In that case, k_i = 1.
• Use (2) for R = ½ d_i , to determine whether the subslice contains the entire particle and interphase to simplify the computation.
• Use (3) and (4) for R = ½ d_i , to determine whether the subslice is matrix/binder only. In that case, k_m = 1.

The volume of the intersection may be computed by integration or approximation. The problem with integration is computing the limits of integration, which occur at either the spherical interface or some edge of the subslice, and integrating differently within the different regions (difficult). Approximation allows evaluating the limits along a single line, computing the length of the line within the intersection, and using the weighted sum of many such lengths to represent the integral (easy).

To approximate the integral, sum the vertical lengths over all x (1) and y (2) values. The z location(s) where the vertical line at (x,y) intersects the sphere may be computed as follows:
z = O_Z ± SQRT[ R² - (x - O_X)²  - (y - O_Y)² ]
If the computation gives imaginary results, there is no intersection. If real z value(s) can be computed, there will be two solutions z₁ and z₂, corresponding to the negative and positive signs, respectively. The length inside the sphere at (x,y) is the length of the line from (x,y,z₁) to (x,y,z₂). The amount of this line to use, i.e. the amount inside the subslice, is computed by limiting the (maximum) z₂ value to the maximum for the slice and limiting the (minimum) z₁ value to the minimum for the slice. If the resulting (z₂ - z₁) is positive, sum the corresponding volume: (z₂ - z₁) * dA.

Normal Moduli

Shear Moduli

Poisson's Ratio

Thermal Expansion Coefficients

Thermal Heat Conductivities

Heat Capacity

Moisture Expansion Coefficient

Moisture Diffusivities