Two of the fastest questions you can ask a metal. The first is an indent. Press a square-based diamond pyramid into a polished face, measure the two diagonals of the mark, and divide the load by the SLOPING area of that mark: $HV = \dfrac{1.8544\,F}{d^{2}}$ — read aloud, H V equals one point eight five four four F over d squared. $F$ is the test load in kilograms-force (kgf — the hardness world never converted, and HV30 means thirty of them), $d$ is the mean of the two measured diagonals in millimetres, and $HV$ is the hardness that comes out, quoted bare: 220 HV. The 1.8544 is not a fudge — it is $2\sin 68^\circ$, pure geometry from the indenter's 136° faces. And because a pyramid is self-similar, the number barely moves with the load, which no other common hardness scale can claim.

The second question is about grain size, and Hall measured its answer on mild steel in 1951: $\sigma_y = \sigma_0 + \dfrac{k_y}{\sqrt{d}}$ — sigma-y equals sigma-nought plus k-y over root d. $\sigma_y$ is the yield strength in MPa, the thing you want. $\sigma_0$ is the friction stress in MPa — what a single crystal with no boundaries at all would yield at. $k_y$ is the Hall–Petch slope in MPa·√m, the price the boundaries charge. And $d$ is the mean grain diameter. Careful with that letter: the $d$ up in the Vickers formula is an indent diagonal, this $d$ is a grain — same symbol, two entirely different jobs, one page.

One unit discipline before you compute. $k_y$ carries MPa·√m, so $d$ belongs in metres under that root. A grain quoted at 4 µm is $4 \times 10^{-6}$ m, whose square root is $2 \times 10^{-3}$ √m — which is where the factor of a thousand in every worked line comes from. Finer grain, stronger metal, and it is the one strengthening route that costs nothing in toughness. But the inverse square root is a lazy curve: halving the grain does not double the boundary term, it multiplies it by $\sqrt{2}$. The last factor of two is far dearer than the first.