Offsetting the longer row instead would not gain the extra reactor any bonus, while the reactor on the other end of the same row would lose its bonus as well. Odd numbers of reactors are inefficient in maximizing the bonus, but, if needed, the odd reactor should be aligned with one of the rows. Splitting the row, while possibly logistically beneficial, reduces total power output by 160 MW per split. For even numbers of reactors, the total output of the array is 160n − 160 MW (where n = total number of reactors, and assuming all are fueled). The most efficient practical layout is an aligned double row of arbitrary length (number of reactors as needed).
These fuel cells are then burned in a nuclear reactor to create heat. In general, nuclear power is produced by the following production chain: Uranium ore is mined and processed to uranium-235 and uranium-238, then uranium fuel cells are created from the two. Nuclear power See also: Tutorial:Nuclear power 1 solar panel produces 42KW after factoring in the night (Factorio forums).Perfectly optimal solar network (Factorio forums).Where game_day is the number of seconds in the game day which is 25000/60 s by default.
FACTORIO STEAM POWER PLANT MOD
If the only effect the mod has on the game is it changes the total length of one day, without changing the ratio of dusk : day : dawn : night, then the equation can be simplified asĪccumulators / Solar_panels = 0.002016 /s × game_day This equation could also be used to remember the vanilla optimal ratio given its simplicity.
FACTORIO STEAM POWER PLANT MODS
If the player uses mods which change the power generation of solar panels, or the energy storage of accumulators, but not the length of days, a simplified version of this equation can be used.Īccumulators / Solar_panels = 70 s × Solar_power / Accumulator_energy Which, given the default time lengths of: day = 12500/60 s dawn or dusk = 5000/60 s night = 2500/60 s, and the default: Solar_power = 60 kW Accumulator_energy = 5 MJ = 5000 kJ, gives the optimal ratio of 0.84 accumulators per solar panel. (day + dawn) × (night + dawn × (day + dawn) / game_day) / game_day In vanilla factorio, without mods which change any of these values, the optimal ratio will be the same. There are also times between day and night called dusk and dawn which complicate the calculations. These include the power generation of a solar panel, the energy storage of an accumulator, the length of a day, and the length of a night. The optimal ratio of accumulators per solar panel relies on many values in the game. This is taken from Accumulator / Solar Panel Ratio (which calculates this in an impressive mathematical way!) and another post in that thread (which calculates the solar panel to megawatt ratio in a different way).Ī small 9x9 setup demonstrating the 20:24 "close enough" ratio above. This means that you need 1.428 MW of production (of solar panels) and 100MJ of storage to provide 1 MW of power over one day-night cycle.Ī "close enough" ratio is 20:24:1 accumulators to solar panels to megawatts required (for example, a factory requiring 10 MW can be approximately entirely powered, day and night, by 200 accumulators and 240 solar panels - this approximation differs from optimal only in that it calls for 20 extra solar panels, which is negligible but remember that the difference between the "close enough" ratio and the optimal ratio increases as you add more solar panels). The optimal ratio is 0.84 (21:25) accumulators per solar panel, and 23.8 solar panels per megawatt required by your factory (this ratio accounts for solar panels needed to charge the accumulators). A possible setup Solar panels and accumulators Optimal ratio
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