Say you're playing on a map where the mexes all give 2.0 metal/s.
you have 10 metal extractors and 14 solar collectors, your com is alive, the 14 solar collectors are connected in a grid with 2 mexes and 8 mexes are not in grid.
you have 9 metal extractors and 14 solar collectors, your com is alive, with a much more complicated grid. 7 solars are connected in a grid with 2 mexes, 3 solars connect a grid of 2 mexes, 1 solar connects a grid of 2 close mexes, there are 2 cases of a mex connected to a single solar and a single lone mex.
Ignoring any threat of enemies and assuming you are spending all the metal you produce into a detriment assisted by 50 build power(or any number more than your income), and your energy is at max storage, which scenario will produce the detriment the fastest?
Solution:
[Spoiler]
It's a trick question!
9 mxes 14 solars vs 10 mexes 14 solars.
the grids I gave in the post are very confusing, but what it amounts to is in the 9 mex sinario 8 mexes have access to 1 energy in it's grid, and in the 10 mex scenario 2 mexes have access to 4 energy in their grids.
But wait! what you're saying is that both scenarios have only 8 excess energy, but when you look at the math it's actually 10 excess energy and 12 excess energy! The answer to that is confusing, isn't it? if all the excess energy were converted to metal, you would not have any energy to SPEND that metal. what this means is that the excess energy is separated into 2 sections, the energy spent to overdrive into metal, and the energy saved to spend that metal. And these values end up equivalent, where the metal value of the energy for overdrive equals the energy value left over.
Here's a good time to remind you the formula for overdrive.
overdrive% = (sqrt(energy into mex)/4)*100%
that's a lot to unpack, so here are some simpler values to remember:
1e =25% overdrive
4e = 50% overdrive
9e = 75% overdrive
16e = 100% overdrive
64e = 200% overdrive
256e = 300% overdrive
having the first 2 key values is important to us. so let's go back to the problem.
in case 9 mexes, you have 12 excess energy. take my word for it, split that into 8 overdrive energy and 4 spend energy. The grids are confusing, but the energy can be evenly distributed to all 8 mexes at 1e/mex, and that translates to 25% of 2 or 0.5 m/e. 8*0.5=4, equal to our spend energy, so 12 excess energy converts to 4 spendable metal in overdrive.
Note that the 2 caretakers and strider hub produce .9 metal and energy.
in case 10 mexes, you have 10 excess energy. only 2 mexes are in grid, so we can only overdrive those 2 mexes. split the energy into 8 overdrive energy and 2 spend energy. the 8 OD energy goes 4e per mex. that's 50% overdrive on a 2.0 mex, or 0.25m/e. .25*8= 2 metal, equal to our spend energy. In this set up, the 10 excess energy converted into 2 usable metal in overdrive.
so, calculating the sum total of everything:
9 mex scenario, you have 22 metal base and 34 energy, excess energy converts into 4 metal for 26 total metal.
10 mex scenario, you have 24 metal base and 34 energy, excess energy converts into 2 metal for 26 total metal.
So the answer is that it's a trick question, production is the same and the detriments come out at the same time! silly lobsters, building detriments. Why don't you build a paladin in half the time and win the game?
