![]() ![]() Basic_cycletime for yield 24 is 97s, so Efficiency is only 0,99. Ore mine M on a 25 roid does produce 6 Ore in 72 seconds. If we do try some numbers, we notice that yield 25 Ore roid has Efficiency=1,0 and yields 50,51,52,53 all have Efficiency=2,0. Crystal Fab M consumes silicon at Fab_equiv=2, so one Mine M with Efficiency=1,0 is sufficient to support it. Continuing from the equations above:Ĭode: Select all Efficiency = Fab_unit / Basic_cycletimeįrom this we get that if the Efficiency is 1,00 then the Mine M has Fab_equiv of 2,0 and the Mine L has Fab_equiv of 5,0. Say Fab_unit=96s for Silicon and Fab_unit=24s for Ore. How about presenting mine yield as how many factories it can support? Thus, yield 24 is slightly too slow but yield 25 is already faster than the standard rate of wafer (and ore) consuming factories. However, yield 25 mine produces that same wafer in 1:33, three seconds faster than necessary. But a M mine would still produce the two silicons required by M crystal fab one second too slow. With the equation above we see that it takes 1:37 from yield 24 mine to produce that one wafer. Lets go back to that "how much is much?" question. One unit of mineral costs always the same amount of energy. It does not matter what the yield and cycle times are. It takes the common 900 ECells to mine the Ore consumed by a weapon factory in one hour. Ore mine consumes 6 ECells per 1 Ore produced. One Silicon Wafer costs 24 ECells to mine. Four yield 8 mines produce almost as much as one yield 35 mine. For example, yields 12 and 14 produce almost as much as one yield 27. ![]() However, this has a side-effect: when you have several asteroids, their total production is slightly more than just the sum of their yields. Put a mine on a roid of wrong type, and you get yield of 0. Note too that the division for Basic_cycletime has (Yield + 1) and not just Yield. The purpose of the Multiplier is to prevent the cycle time to be shorter than 60 seconds. Products per cycle = Multiplier * Fab_size Multiplier = rounddown( 59.9 / Basic_cycletime ) + 1 ![]() Fab_size is 2 for M, and 5 for L.Ĭode: Select all Basic_cycletime = rounddown( BASETIME / (Yield + 1) ) + 1 seconds There is a BASETIME, which is 2400 seconds (40 minutes) for Silicon and 600 seconds (10 minutes) for Ore. The production rate of mines is calculated from the yield of the asteroid exactly as it was done in X2. The question is how much more? And are the low-yield asteroids worth any? One can save on station costs by placing the mine on a higher-yield asteroid, because then that mine can support more factories. It is a safe bet to say that a mine on an asteroid with yield 26 is sufficient. However, there is an another "non-conforming" factory and that is the mine. The behaviour of solar power plants and how they differ from the rest of the factories has been explained too. The matching of production rates of different factories have been described quite extensively. I've looked through many of the excellent guides that describe station and complex building. ![]()
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