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Non-linear Transfer Functions for Accurately Estimating 3D Particle Size, Distribution, and Expected Error from 2D Cross Sections of a Lognormal Distribution of Spherical Particles

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A Correction to this article was published on 19 August 2021

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Abstract

Accurately estimating the mean size of features within an opaque material using only 2D observations is a common requirement in the materials and medical communities. Attempting to employ numerical methods to obtain an accurate estimate of the full 3D size distribution, which is often important for calculating structure-dependent material properties, is substantially more challenging. This paper circumvents the error propagation issues observed with classical numerical approaches, such as those formulated by Saltikov, Johnson, and Schwartz, through comparison of two calculation methods: (i) the direct simulation of synthetic particles and (ii) inversion of the conventional Saltikov analysis. Each method was used to generate dispersions whose 3D particle diameters followed lognormal distributions of varying shape parameters; the mean and variance were then predicted for 2D cross-sectional measurements. Parameters from both the 2D and 3D distributions were analytically coupled to produce a set of equations for accurately transforming cross-sectional observations into estimates of the full 3D distributions. Accuracy of the 2D-to-3D transformation is, for the first time, characterized as a function of the population size and measurement resolution. Specifically, included figures allow for rapid conversion from 2D to 3D metrics with the corresponding margin of error. The Inverse Saltikov method is shown to provide superior transformation accuracy for the idealized cases of lognormally distributed and randomly dispersed spherical particles when compared to either the Direct Simulation method or classical numerical approaches.

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Correspondence to A. R. C. Gerlt.

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The original online version of this article has been corrected. There were errors in Eq. 26 and Figure 9. In addition, several table and equation citations were corrected in the text.

Manuscript submitted May 4, 2020; accepted October 17, 2020.

Appendix

Appendix

Similar to how mean and standard deviation may be used to characterize a population sample, expectation and variance can be calculated to characterize a continuous function. The equations for doing so are as follows[11]:

$$ E\left( x \right) = \mathop \smallint \limits_{ - \infty }^{\infty } x*f\left( x \right)dx $$
(A1)
$$ Var\left( x \right) = \left( {\mathop \smallint \limits_{ - \infty }^{\infty } x^{2} f\left( x \right)dx} \right) - \bar{x}^{2}. $$
(A2)

Using these two equations as well as Eqs. [8] and [9] above, it is fairly straightforward to calculate \( K_{\ell } \), \( K_{d} \), \( S\left( {\ell /\bar{\ell }} \right) \), and \( S\left( {d /\bar{d}} \right) \) for a single sphere, which is equivalent to a population of spheres where \( \sigma = 0 \).

$$ \bar{\ell } = E\left( \ell \right) = \mathop \smallint \limits_{ - \infty }^{\infty } \ell f\left( \ell \right){\text{d}}\ell = \mathop \smallint \limits_{0}^{1} 2\ell^{2} {\text{d}}\ell = \frac{2}{3} \to K_{\ell } = \frac{D}{{\bar{\ell }}} = 1.5 $$
(A3)
$$ S\left( {\ell /\bar{\ell }} \right) = \frac{{\left( {Var\left( \ell \right)} \right)^{1/2} }}{{\overline{\ell } }} = \frac{{\left[ {\left( {\mathop \smallint \nolimits_{ - \infty }^{\infty } \ell^{2} f\left( \ell \right){\text{d}}\ell } \right) - \overline{\ell }^{2} } \right]^{1/2} }}{2/3} = \frac{3}{2}*\left[ {\frac{1}{2} - \frac{4}{9}} \right]^{{\frac{1}{2}}} = \left( {\frac{1}{8}} \right)^{{\frac{1}{2}}} \to S\left( {\ell /\bar{\ell }} \right) = 0.35355 $$
(A4)
$$ \overline{d} = E\left( d \right) = \mathop \smallint \limits_{ - \infty }^{\infty } df\left( d \right){\text{d}}d = \mathop \smallint \limits_{0}^{1} \frac{{2d^{2} }}{{\sqrt {1 - d^{2} } }}{\text{d}}d = \frac{\pi }{4} \to K_{d} = \frac{D}{{\bar{d}}} = \frac{4}{\pi } $$
(A5)
$$ S\left( {d /\bar{d}} \right) = \frac{{\left( {Var\left( d \right)} \right)^{1/2} }}{{\overline{d} }} = \frac{{\left[ {\left( {\mathop \smallint \nolimits_{ - \infty }^{\infty } d^{2} f\left( d \right){\text{d}}d} \right) - \overline{d}^{2} } \right]^{1/2} }}{\pi /4} = \frac{4}{\pi }*\left[ {\frac{2}{3} - \frac{{\pi^{2} }}{16}} \right]^{{\frac{1}{2}}} \to S\left( {d /\bar{d}} \right) = 0.28418. $$
(A6)

As can be seen, the calculated values agree almost perfectly with the values seen at σ = 0 in Figures 3(c) through (f).

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Gerlt, A.R.C., Criner, A.K., Semiatin, S.L. et al. Non-linear Transfer Functions for Accurately Estimating 3D Particle Size, Distribution, and Expected Error from 2D Cross Sections of a Lognormal Distribution of Spherical Particles. Metall Mater Trans A 52, 228–241 (2021). https://doi.org/10.1007/s11661-020-06072-w

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