ASTM E112 grain size number calculator

$$N=2^{G-1}{\text{grains per in}}^2\text{at }100\times$$$$N_{\mathrm{A}}=15.50\times N{\text{grains per mm}}^2\text{at }1\times$$$$G=3.321928\cdot {\log }_{10}(N_{\mathrm{A}})-2.954$$$$\begin{array}{rl}G& =-6.643856\cdot {\log }_{10}(\overline{\ell })-3.288\\ & =6.643856\cdot {\log }_{10}(N_{\mathrm{L}})-3.288\end{array}$$

The definition. ASTM E112, Standard Test Methods for Determining Average Grain Size, defines the grain size number $G$ through $N$, the number of grains per square inch at 100× magnification: $N=2^{G-1}$. Multiplying by 15.50 (${100}^2÷645.16$ mm² per square inch) gives $N_{\mathrm{A}}$, the grains per mm² at 1×, and taking logarithms gives the planimetric relation above, with $3.321928=1/{\log }_{10}2$. Each step of 1 in $G$ halves the mean grain area, so a higher $G$ means finer grains.

Planimetric method. Count the grains per unit area of the specimen: in Jeffries’ count, the grains completely inside a test area plus half of those cut by its boundary, divided by the true area. Image analysis gives the mean grain area $\overline{A}$ of the grains not cut by the frame directly, and for a single-phase structure $N_{\mathrm{A}}=1/\overline{A}$. A mean area of 39,340 nm² is $3.934\times {10}^{-8}{\mathrm{mm}}^2$, so $N_{\mathrm{A}}=2.542\times {10}^7$ grains per mm².

Intercept (Heyn) method. Lay test lines of known total length $L$ on the image and count $N$, the grains they intercept (or $P$, the grain boundaries they cross). The mean lineal intercept is $\overline{\ell }=L/N$, in mm at the specimen, and $N_{\mathrm{L}}=1/\overline{\ell }$. $G=0$ corresponds to $\overline{\ell }=0.32\mathrm{mm}$, and halving $\overline{\ell }$ adds 2 to $G$ ($6.643856=2/{\log }_{10}2$).

Converting back. For a given $G$ the calculator gives $N_{\mathrm{A}}$, the mean grain area $1/N_{\mathrm{A}}$, its square root (the average diameter of the E112 tables), the diameter of a circle with that area, $\sqrt{4\overline{A}/\pi }$, and the mean intercept from the intercept relation. The two relations agree only for uniform, equiaxed grains, for which they make $\overline{\ell }=0.891\cdot \sqrt{\overline{A}}$; a real structure measured both ways gives $G$ values that differ by a fraction of a unit.

Constants and edition. The relations are those printed in ASTM E112 (checked against the E112-13 text) and in Metkon’s guide to E112 (metkon.com). Some references print the intercept relation as $G=-6.6457\cdot {\log }_{10}\overline{\ell }-3.298$ (G. F. Vander Voort, vacaero.com), which differs by 0.01 $G$ or less for intercepts from 1 nm to 1 mm. The current edition is ASTM E112-25 (store.astm.org); consult the standard for the full procedure: specimen preparation, the comparison, planimetric and intercept procedures, how many fields to measure and the confidence interval. ASTM’s summary of E112-13 (reapproved 2021) states that a precision of ±0.25 grain size units can be reached with a reasonable amount of effort.

Assumptions and limits. E112 measures planar grain size, on a two-dimensional section, and applies chiefly to single-phase structures; the relations here are for equiaxed grains, and the standard treats non-equiaxed grains (elongated, partly recrystallised or cold-worked) separately, measuring in more than one direction. Duplex structures, with two grain size populations, are covered by ASTM E1181. Count enough grains, over several fields, for the mean to be representative. Nanocrystalline films give very high $G$: a mean grain area of 39,340 nm² (equivalent diameter 224 nm) is $G$ 21.6, where the mean grain area and equivalent diameter, with the number of grains measured, tell a reader more. A top-view image of a film is also not the random section the method assumes.

1. A perovskite film, from the mean grain area. The built-in thin-film sample in LabCanvas Image Analysis (a synthetic image with a scale in nm) gives 118 grains not cut by the frame, with a mean grain area of 39,340 nm² and an equivalent diameter of 224 nm.

1. $\overline{A}=\mathrm{39,340}{\mathrm{nm}}^2=3.934\times {10}^{-8}{\mathrm{mm}}^2$, so $N_{\mathrm{A}}=1÷\overline{A}=2.542\times {10}^7$ grains per mm².
2. $G=3.321928\times {\log }_{10}(2.542\times {10}^7)$ $-2.954$ $=3.321928\times 7.4052-2.954$ $=\mathbf{21.6}$.
3. Back from $G$: $\sqrt{\overline{A}}=198\mathrm{nm}$ and the mean intercept for the same $G$ is 177 nm.

$G=21.6$ is a correct number but not a useful one: it lies far beyond the table below, which reaches a mean intercept of 2.5 µm at $G$ 14. Report the mean grain area (39,340 nm²) and equivalent diameter (224 nm) with n = 118, and $G$ only if a reader asks for it.

2. A steel, from the mean intercept. Test lines with a total length of 2 mm on the specimen cross 100 grains, so $\overline{\ell }=2\mathrm{mm}÷100=0.020\mathrm{mm}$ (20 µm).

1. $G=-6.643856\times {\log }_{10}(0.020)-3.288$ $=-6.643856\times (-1.69897)-3.288$ $=\mathbf{8.0}$.
2. The same $G$ from the definition: $N=2^{8-1}=128$ grains per square inch at 100×, $N_{\mathrm{A}}=15.50\times 128=1984$ grains per mm², a mean grain area of 504 µm² and an equivalent diameter of 25.3 µm.

3. A planimetric count. 95 grains lie completely inside a 0.5 mm² field and 46 are cut by its edge: $N_{\mathrm{A}}=(95+46/2)÷0.5=236$ grains per mm², $G=3.321928\times {\log }_{10}(236)-2.954$ $=\mathbf{4.9}$.