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Author Topic: Amount of metal per "ore"  (Read 1172 times)

Turgid Bolk

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Amount of metal per "ore"
« on: November 06, 2007, 05:09:00 pm »

We were talking about the ginormous gold nuggets over at http://www.bay12games.com/cgi-local/ultimatebb.cgi?ubb=get_topic&f=6&t=001758

The problem is, each unit of ore = one bar...but that would mean there's either huge pieces of metal, or each tile contains a huge amount of rock.

Perhaps Toady should make the mined ore contain only 1/2 or 1/3 or 1/10 the total gold for a bar. That way it would be a more reasonable chunk of rock, weighing less, but you'd need a few units of it to get a full-size solid gold bar. (Bars seem to be pretty big in DF, bigger than a typical real-life gold bar. It only takes 3 to make a barrel, for instance.)

In real life it takes a lot of digging to get enough gold for a bar, and it mostly comes out of the rock in little bitty nuggets and flakes. This could be applied to other metals as well, particularly precious metals like platinum which are far too plentiful.

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axus

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Re: Amount of metal per "ore"
« Reply #1 on: November 06, 2007, 05:23:00 pm »

It's a good idea, but right now there's not a good concept of fractional quantities of materials in DF.  The way around that is to task multiple ores to create one ingot.  Of course I like having more metal than less  ;)
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Turgid Bolk

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Re: Amount of metal per "ore"
« Reply #2 on: November 06, 2007, 10:39:00 pm »

Right, that's what I meant. It would take 10 ores to make a bar (or whatever amount Toady decides is appropriate for that metal). Sorry for not being clear.  :)

And the smelter already holds fractional quantities of metal when you melt things. This basically works the same way; you'd have to "melt" so many ores. So hopefully not too hard to implement?

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RPB

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Re: Amount of metal per "ore"
« Reply #3 on: November 06, 2007, 11:01:00 pm »

Actually, this suggestion is precisely backwards. The major problem with existing mechanics is that smelted ore does not produce enough bars. See  http://www.bay12games.com/cgi-local/ultimatebb.cgi?ubb=get_topic&f=5&t=001653 .

The short of it is, it takes 3 sets of gold bars to make a solid gold statue. But your mason can carve a gold statue out of a single block of ore from a vein of native gold, and the game treats the gold-containing stone (in terms of value and weight) as if it were made out of solid gold too, rather than being a block of stone peppered with little chunks. The main problem is with the ore object itself, not the bars. You are getting an unrealistic amount of gold when you smelt the ore into bars, but what's really outrageous is the amount of gold that is implied to be in the ore before you smelt it down.

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Turgid Bolk

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Re: Amount of metal per "ore"
« Reply #4 on: November 06, 2007, 11:44:00 pm »

quote:
Originally posted by RPB:
<STRONG>You are getting an unrealistic amount of gold when you smelt the ore into bars, but what's really outrageous is the amount of gold that is implied to be in the ore before you smelt it down.</STRONG>

...right. That's what I'm saying. Ore should have less metal. We're basically arguing for the same thing. You're saying more bars for the same amount of metal, I'm saying less metal per ore.

Boy, I guess I'm just not being very clear tonight. I'll change my post over at that thread to explain what I mean better.

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PTTG??

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Re: Amount of metal per "ore"
« Reply #5 on: November 07, 2007, 12:46:00 am »

OK!
PTTG is has it all figured out!
Gold, silver, and other metals come in TWO forms; more commonly in "ore"; that is, small flecks or oxidized metal that is normally not all that valuable. More valuable and rare is the "nugget" form, or a native metal, and this forms more like a gemstone than an ore or stone. It also seems that ore may need to be along the lines of "granite b/w gold", "Limestone and silver ore", and so on, the value of a chunk of gold ore being more dependent on the host stone. This could also be helped if there was an easier way to get experienced metalsmiths; for instance, if horses need shoes, or perhaps picks need repairs?
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RPB

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Re: Amount of metal per "ore"
« Reply #6 on: November 07, 2007, 12:59:00 am »

quote:
Originally posted by Turgid Bolk:
<STRONG>

...right. That's what I'm saying. Ore should have less metal. We're basically arguing for the same thing. You're saying more bars for the same amount of metal, I'm saying less metal per ore.</STRONG>


No, that is not what I am saying. Ore does not "have" metal. Ore is ore. Metal is metal. The two are unrelated in DF, except in that the smelter can magically transform chunks of ore into metal bars. The properties of one have no bearing on the properties of the other, and the big issue is that ore itself is worth an utterly unrealistic amount for the sheer amount of objects masons/stonecrafters can make from blocks of ore. It is even more unrealistic when compared to the number of metal bars it currently takes a blacksmith to produce the same objects, considering that the ore produces equally valuable objects.

It is possible to combine the two solutions: make ore worth less, and make smelters produce fewer metal bars. This becomes problematic, however, because if you decrease the number of bars produced per ore, or increase the number of bars consumed per object, then by necessity you must reduce the value of ore even further to keep things sane and quickly reach a point where ore must become less valuable than non-ore rock! Actually, the only way to reconcile this problem completely would be, in fact, to increase a smelter's metal production per unit of ore, unless you take the view that base metals such as copper and tin should in fact be worth less per unit volume than simple rock.

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Turgid Bolk

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Re: Amount of metal per "ore"
« Reply #7 on: November 07, 2007, 02:57:00 am »

Ok, I'll just continue the conversation over at the other thread ( http://www.bay12games.com/cgi-local/ultimatebb.cgi?ubb=get_topic&f=5&t=001653  ).

This two-threads-at-once thing doesn't work for me.

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