Talk:Compression Furnace

Details of Furnace Operation

If you want to find the most efficient recipes, it's not enough to know the calculations involved, but also to know how those calculations are rounded. So here are the details I've figured out from experimenting:

Basic Details

Charcoal is used by the furnace at the rate of 1 cc / minute. But more specifically, charcoal is used at the 30 second mark. The furnace runs for 30 seconds without using any charcoal, and uses its last charcoal at 19 minutes, 30 seconds.

Without rounding, one metal is produced by the furnace every (40000 / ore) seconds. (So with the maximum load of 500 ore, one metal takes 80 seconds to enter the reaction chamber.)

Even without rounding, there are two factors that affect this:

• The first metal is created in half the time.
• The furnace gives you a head start of 20 seconds. So, if the furnace contains 500 ore, the first metal shows up after 20 seconds instead of 40, and the next metal shows up at 100 seconds, rather than 120.

But this is only theoretical, since the exact time for metal to appear depends on rounding (see next section).

Rounding and Calculations

So, now for the part that lets us find more efficient and less efficient recipes: rounding. The furnace first checks whether a metal has been smelted at 20 seconds. After that, it checks every 40 seconds. If the internal calculation shows that a metal would have been created by the time of the check, it will be added to the reaction chamber.

So, theoretically, the time to generate x metal from y ore is:

time in seconds = ((x - 0.5) * (40000/y)) - 20

In practice, you need to first round up ((x-0.5) * (40000/y)) to the next 40 second mark, then subtract 20 seconds.

On the other hand, if you want to calculate how much metal you can generate in a given amount of time from y ore, you first want to massage the time to account for the fact that checking occurs 40 seconds apart:

rounded time = -20 + ((theoretical time + 20) rounded down to the nearest 40)

So, if you want to calculate how much metal will be generated in the full 20 minute run, you will use 1180 seconds for the following calculation rather than the full 1200 seconds.

Finally, to calculate the metal produced in that amount of time

metal produced = ((rounded time + 20)* y / 40000) + 0.5

Then round that down to the next whole integer.

Excel Formulas

To calculate the time at which x metal will appear in the reaction chamber from y ore, and how much charcoal that will require:

RC[-2] contains x
RC[-1] contains y
RC (time) contains =CEILING((40000/RC[-1])*(RC[-2]-0.5),40) - 20
RC[1] (cc) contains =IF(ROUNDUP((RC[1]-30)/60,0)>=0, ROUNDUP((RC[1]-30)/60,0), 0)

Conversely, to calculate the amount of ore needed to get x metal during z seconds:

RC[-2] contains x
RC[-1] contains z
RC (ore) contains =CEILING(40000*(RC[-2]-0.5)/(FLOOR(RC[-1] + 20,40)),1)

To calculate how much metal will appear from y ore after z seconds:

RC[-2] contains z
RC[-1] contains y
RC contains =FLOOR((FLOOR(RC[-2]+20,40) * RC[-1] / 40000) + 0.5, 1)

Finding Efficient Recipes

Now for the point of this entire exercise: how to find the recipes most efficient for charcoal per metal or metal per ore.

To find recipes that use charcoal most efficiently, look for recipes that yield the metal you want at any 20 second mark, such as 5:20, 19:20, and so on. (This is just before the 30 second mark where the next charcoal is used.) So, 500 ore will give you 1 metal at :20, which will use 0 charcoal! (Of course, you have to add charcoal to start the furnace, but if you open the chamber between :20 and :29, you will not lose that charcoal.)

To find recipes that use ore most efficiently, you will want to run the furnace as long as possible, up to 20 minutes / 1200 seconds. Then the amount of ore needed is the following, rounded down to the next whole integer:

40000 * (desired metal - 0.5) / 1200

Desired Metal	Minimum Ore		Desired Metal	Minimum Ore		Desired Metal	Minimum Ore
1	17		6	184		11	350
2	50		7	217		12	384
3	84		8	250		13	417
4	117		9	284		14	450
5	150		10	317		15	484

Of course, if you have access to a Gyration Cell, this whole exercise is moot, since they are far more efficient in metal per ore than compression furnaces, and are more efficient in charcoal per metal in most cases.

The woman of too many details,

-Talibeh 12:28, 21 October 2010 (EST)