November 7, 2018

Resources XIII: Placing Silver Ore: Redux

I made an error in the previous post. To enshrine my mistake, I'll fix it here.

I forgot to multiply by the hex area factor. If I do so, I get an occurrence of 700%! That implies that there are 7 deposits of silver in each hex. That has to be wrong, otherwise silver would be essentially worthless. So I made a bad assumption.

Instead, I will use the Mindat method (discussed in the future copper post). This yields:
\[{5,484\textrm{ deposits}\over 57,000,000\textrm{ mi}^2}\cdot {347\textrm{ mi}^2 \over 1 \textrm{ hex}} = 3.3\%\]
That's a bit more reasonable. Again, keep in mind that 347 square miles is a massive area, and that's just one hex!

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