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Material property lists, such as springermaterials.com, has lists of metal alloys and their properties. Is there a similar website that lists alloys along with their Mohs scale hardness? For springermaterials I can't seem specifically find hardness values. For example, when I search "hardness" or "Moh" in the search box nothing comes up. enter image description here

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    $\begingroup$ The name's Mohs, not Moh or Moh's. $\endgroup$
    – andselisk
    Jan 22 at 23:11

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You can find Mohs hardness values for various pure metals, but it sounds like you want something more specialized: Mols hardness values for specific alloys.

That's probably harder to find. But there is an alternative: One could create a formula to convert from the types of hardness values readily availble for alloys (e.g., Vickers hardness) to Mohs. That's what I do here.

I need to add the caveat that this is just an approximation: Mohs hardness is based on a scratch test, while Vickers hardness is based on indentation, so these two scales aren't fully comparable.

To start, I made use of a table of Vickers values for the ten Mohs reference substances, found here: https://geology.com/minerals/mohs-hardness-scale.shtml

Then I plotted Mohs hardness vs. Vickers hardness:

enter image description here

You can see from the above plot that, because of the highly non-linear nature of the Mohs hardness scale, it would be difficult to create a good fit that includes diamond (the point at the far right). Since metals probably top out at a Mohs hardness of ~$9$, I decided we could eliminate that value, and replot. This gives a curve that would be much easier to fit: enter image description here

A log-log plot nearly linearizes it, indicating it follows a power law:

enter image description here

This tells me I should try to fit it with a power law function. After some testing, I found that a nice fit can be obtained with a function of the form $a + b x^c$:

enter image description here

The above fit is:

$$\text{Mohs hardness} = -0.5853 + 0.4829 V^{0.3912}$$

where V is the Vickers hardness value.

To use it, just plug and play. For instance, the Vickers hardness of 316 stainless steel is 152. Substituting into the above, the estimated Mohs hardness of stainless steel is 5.0 (because of the problematic nature of comparing two different types of hardness scales, I wouldn't express the predicted Mohs values to anything beyond two significant figures; I included two guard digits beyond this in the above formula, to avoid round-off error).

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  • $\begingroup$ Should I be amazed that you managed to fit the two properties using such a simple analytical expression, that is, that it should follow a power law given a basis in a qualitative notion of hardness with Mohs scale really a ranking? Do you have an explanation for the high quality? $\endgroup$
    – Buck Thorn
    Jan 23 at 10:48
  • $\begingroup$ No idea. But I can speculate by considering the fit of the inverse function. Without diamond, it's: $V = -1.631 + 13.95 \text{ Mohs}^{2.270}$. I.e., actual hardness goes as $\pu{Mohs}^{2.270}$. Now let's suppose the reference minerals were chosen so that (a) the scale could go from talc to corundum in eight steps; and (b) each next step was hard enough to clearly scratch the one below. It may be that, physically, this happened by choosing minerals whose hardnesses increases as a power law (which made for an easy fit with power law). Again this just speculation.... $\endgroup$
    – theorist
    Jan 23 at 17:02
  • $\begingroup$ ...A materials scientist might be able to give a better answer. Of course, if my model had four or five free parameters rather than three, I could have cited Fermi's recounting of a quote by vonNeuman: "I remember my friend Johnny von Neumann used to say 'With four parameters I can fit an elephant, and with five I can make him wiggle his trunk.'" :) $\endgroup$
    – theorist
    Jan 23 at 17:03
  • $\begingroup$ Your last explanation is very funny, but I don't find it plausible. This is not a case of overfitting, you don't see weird loops in the fitted function between data points or other features that would indicate you used eg a polynomial of too high order. I speculate wildly but could it be that Vickers hardness was defined so as to exhibit the tidy quadratic dependence on Mohs that you revealed? That sounds farfetched, but the fit is truly uncanny. I suppose you may be right and Mohs did not pick the materials randomly out of a hat. $\endgroup$
    – Buck Thorn
    Jan 23 at 17:03
  • $\begingroup$ The last comment wasn't an explanation, it was just a joke. So of course it's not plausible, because it wasn't meant to be plausible.;) I wasn't implying that my three-parameter model is overfitting. As far as your speculation on Vickers goes, I wouldn't expect Vickers was defined to correspond in some way to Mohs, since Vickers is a quantitative scale based on measured resistance to plastic deformation. It wouldn't make sense to me to compromise Vickers' quantitative information by somehow rescaling it to correspond to a qualitative scale like Mohs. But I'm not an expert on such things. $\endgroup$
    – theorist
    Jan 23 at 17:13

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