Posts by Uristqwerty

    Doesn't help that placing cable with a deployer seems to NPE...
    (at xd.a(
    at ic2.common.ItemCable.a(
    at eloraam.machine.TileDeployBase.tryUseItemStack(

    So, moving the massfab is out, using cables won't work either... Leaving the only reasonable strategy to create a new massfab, stock up on UUM, remove the old massfab, then place as many panels as possible, using a line of MFSU to transfer power, then once the UUM runs out, place the (already created) new massfab and start over? Expensive...

    So, each ore can become 8 near-depleted cells. Each of those, once processed and used, has a 25% chance to become a near-depleted cell. Of those 25%, 1/4 will become a near-depleted cell when used, again. It continues on like this, for a total of 100% + 25% + 6.25% + 1.5625% + 0.390625% (and so on, though I won't bother with them). With just those 5 cycles, you get a total of 1.33203125%, and this continues to approach 1 + 1/3 the longer it goes.

    So, each uranium ore can potentially provide 8 + (8/3) = 10 + 2/3 cells, with processing.

    Now, assuming that one uranium cell can recharge 4 near-depleted cells, it would take (10 + 2/3)/4, or 32/12, or 8/3, or 2 + 2/3 uranium cells to recharge the full output of a single uranium ore into 10 + 2/3 uranium cells. Now, if you were to use 2 + 2/3 of those in order to recharge the next ore's cells, that leaves exactly eight uranium cells that can be used elsewhere.

    So, as long as you aren't stupid, and use recharged cells to recharge later near-depleted cells (rather than making a fresh uranium cell each time), each uranium ore you gather can produce a full 8 cells to be used in whatever non-breeder you choose. As mentioned by people previously, that could be as high as 4.44, giving 35.52 for each uranium ore you mine.

    Oh, and that's *completely* ignoring the energy produced by the breeder reactor.

    If you use a 9k temperature breeder (if I understand correctly halving the recharge time, so a single uranium cell can charge two sets of four near-depleted cells), you get an extra 1 + 1/3 cells output, since you only need half as many.

    So, with the extra output, and then assuming that each uranium cell in the breeder gets at least an efficiency of 1, you get ((9 + 1/3) * 4.44) + ((1 + 1/3) * 1) = 41.44 + 1.3333... = 42.7733 per ore of input.

    Now, the very first time you run the breeder, you'll need to supply a uranium cell to bootstrap the whole process, but since you can use the output of that very first cycle to provide the next uranium cells for later cycles, all other uranium you use can have that nice 8x multiplier.