Balancing with Krastorio 2 #41
Replies: 2 comments
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Sorry for the late response. But I love the suggestions you gave me, that's something I can probably work with. |
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No worries at all, I appreciate you taking the time to read this. No rush on any of this. A few more ideas:
As much as I want more U235, the Thorium processing recipe seems very overpowered: What do you think? |
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Hey there,
I was wondering if you have given any thought to balancing with k2?
I am in the middle of a K2SE with atomic overhaul and nuclear fuel (for the breeder reaction).
I understand that this might make things a little more complex to balance but am hoping a discussion might help come up with something.
First off would you consider adding the breeder reaction for the full configuration of the mod? Would be able to remove NF then.
Would you be able to detail your thoughts behind the current balancing of AO?
Looking at the current ratios, excluding reprocessing and ease of obtaining the elements:
Uranium cells:
1 U235 is worth 80GJ (same as vanilla)
Calculation:
1xU235 makes 10 cells @ 8GJ.
Plutonium Cell:
1 Pu is worth 5.3GJ
Calculation:
19xPu makes 10 cells @ 10GJ.
MOX Cell:
each Pu is worth 0.66GJ
each U235 is worth 10.5GJ
Calculation:
45xU235 + 5xPu makes 50 cells @ 12GJ = 480GJ
base on previous cells 1 PU is worth 1/16 the energy of U235.
This one feels a little underwhelming considering we get so much less energy per element. Is it supposed to be obtained from the waste instead of crafted?
Breeder cells from NF:
1 Pu is worth 40GJ.
Calculation:
1xPu makes 10 cells @ 4GJ.
Anyways the other elements follow the same principals so I will leave them for now.
At this point it would seem like the it would best to use breeder cells and Uranium cells to get the most energy and as a side benefit, lots of extra U235 and PU from the breeding.
In base k2:
Uranium cells:
1 U235 is worth 25GJ.
Calculation:
2xU235 makes 1 cell @ 50GJ.
Now with K2 AO NF:
Uranium cells:
1 U235 is worth 50GJ.
Calculation:
1xU235 makes 1 cell @ 50GJ.
This also makes the modified recipe very odd:
1 xU235 + 19 Graphite makes 10 rods -> 10 rods + 10 empty cells make 1 U Cell
The other change is to breeder cells from NF:
1 PU is worth 50 GJ.
Calculation:
1 Pu makes 2 breeder cells @ 25GJ.
Because the other cells are not adjusted it again seems to make Uranium and Breeder cells the most efficient energy per element.
Let me know if there is anything wrong with my math.
I would love to hear ideas
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