Thursday, June 30, 2011
This is the second equipment post today, and probably not the last. So if you are just now stumbling across it, you may note that I also added an artillerist's equipment post earlier.
I really like this table, as it helps solve a problem that has been around D&D forever. The various prices are weighted so that the price is for the metallic value within the total amount of ore (1 ton) that is mined. The price is equally weighted to describe the richest likely ore that the players are most likely to find.
Thus, if the players remove 1 ton of stone from the interior of a mountain (which is not as much volume as you might think), and that stone is positively dripping with the metal indicated, this is the price the shop would pay (the green column) the players for bringing the stone ore in. Specifically, the shop would not require the players to smelt the ore (that is a different problem completely). All the party would have to do to get this price would be to load the ore onto a cart and get it to the shop.
Having a price for ore means that a random number can be generated to determine the richness of the ore. That random number could be a simple percentile die, as a percentage of what is rich ore and what is not. Or it could be a little more complicated, as follows:
If this simple table is used first, there is a much greater chance that a particular bit of earth will contain traces of the metal, but not enough to make it worthwhile mining. Thus, the players would first prospect for an element in a place where that element was known to exist. If no such availability had been specified by you, then the metal content in the ore is always zero.
6d6 are rolled, and then the % die. The idea would be to make finding a really rich vein of any material an unusual thing. Thus, if the party were prospecting for gold, and rolled an 18 on 6d6, and then a 32 on the d100, the value of the ore would be 0.716 g.p. per ton (36664/4096*0.32). In my world, I have 12 c.p. per s.p. and 16 s.p. per g.p., so that's a total of about 137 c.p. per ton. Probably not rich enough to mine.
Even at this, I am probably being too generous, and making the 'motherlode' too easy to find. The chances of rolling either a '6' or a '36' is only one in 23,328. If the player were allowed to roll once per week, on average that would take 448.62 years to have an average chance of hitting the motherlode.
Wouldn't it break your heart to roll six ones on the dice, then roll a '02' on the percentile?
Of course, when the overhead is considered, a mining operation must pay for the food, the timber necessary to shore up the operation, the broken tools, the danger and so on ... even a payback of 5 or 10 g.p. per ton might not be worth it.
And none of this determines how far the paydirt might go. I'd say probably another roll on the table above (6d6) to determine the number of tons, again modified by a percentage, so that the odds would be the paydirt would go on a long time.
Incidentally, panning would work the same way. I have some comments about panning for gold that can be read here.
I have intentionally listed the various ores without specifying the metal itself. I feel that if a player wants to get involved in mining, he or she ought to learn what the ores describe. Incidentally, some of the ores (such as galena) have the benefit of producing more than one kind of metal.
I had almost meant to mention that the material bought from the assayer is also in the form of an ore. So if the players wanted to smelt their own metal, the price for what was available is here.