Whomever Smelt it, Dealt it

[Necromateur]

Yeah, I know. . . the title has very little to do with this question, I just thought it was funny.

Anyway, I'm reading the wiki and have been trying to figure out this whole % thing with the ore. I did successfully create a bar of iron on my first go round but was left with 5-6 dross.

The wiki says any ore that couldn't be melted into another bar, or a bar, will be put towards the next smelting. . . but it also says ore that couldn't be turned into a bar will be left as dross.

So which is it? It turns to dross or it goes towards another bar? How do I minimize dross leftovers, aside from using 20 ore and 3 lime?

[lachlaan]

Every unit of ore turns into dross after smelting, except for the ore that completed the progress towards 100% of a bar. Imagine progress of a bar as being an invisible counter the smelter doesn't show you. Every ore you smelt without lime in adds 2% of a bar to that counter, every lime you add into a smelt will add 1% to each ore's effect. When the smelter does the math for its counter, i think it counts ores left to right, top to bottom, and when an ore goes over 100% it becomes a bar rather than a dross, and it keeps adding the rest of the ores from the start. If it doesn't reach a full 100% at the end of the smelt, the progress gets remembered and added up with the next smelt's progress.

The highest yield you can get in the long run are smelts of 13 ore 12 lime, or 14 ore 11 lime. Because you'd get either 13 ores at a potency of 14%, or 14 ores at the potency of 13%, both resulting in 182% of a bar worth of progress.

[Necromateur]

Every unit of ore turns into dross after smelting, except for the ore that completed the progress towards 100% of a bar. Imagine progress of a bar as being an invisible counter the smelter doesn't show you. Every ore you smelt without lime in adds 2% of a bar to that counter, every lime you add into a smelt will add 1% to each ore's effect. When the smelter does the math for its counter, i think it counts ores left to right, top to bottom, and when an ore goes over 100% it becomes a bar rather than a dross, and it keeps adding the rest of the ores from the start. If it doesn't reach a full 100% at the end of the smelt, the progress gets remembered and added up with the next smelt's progress.

The highest yield you can get in the long run are smelts of 13 ore 12 lime, or 14 ore 11 lime. Because you'd get either 13 ores at a potency of 14%, or 14 ores at the potency of 13%, both resulting in 182% of a bar worth of progress.

Okay, I think I understand. . . So every piece of ore does actually get counted towards a 100% mark -- even towards a second 100% mark. . But the ore that gets counted, but not used for a bar, turns to dross? But it's % is still remembered for the next smelt?

Maybe that's worded horribly. . . . it becomes dross because its iron is removed and used toward the next smelt, but since it wasn't actually turned into a bar, it gets left over as dross, yet the % remains in the ore smelter? Is that right?

[lachlaan]

Yes, either it's the ore that completes 1 bar's worth of progress and turns into the bar it completes, or otherwise it turns into dross. But it always counts towards the progress, and the % remains memorized by the smelter, or the server, wherever it's stored

[Necromateur]

Yes, either it's the ore that completes 1 bar's worth of progress and turns into the bar it completes, or otherwise it turns into dross. But it always counts towards the progress, and the % remains memorized by the smelter, or the server, wherever it's stored

Awesome. Thank you so much, Lach. Very much appreciated.