Did some digging around in game files, and it appears the day-night cycle on Nauvis is, in fact, 420 seconds (60 seconds per minute times 7 minutes; each second is 60 ticks ofc).
This means the old calculations for solar-accumulator ratios are now slightly off.
I'm going to re-derive them in a way where I can switch out the solar power and cycle length values for other worlds. Apparently, all worlds day-night cycles have the same relative shape: 50% of time is spent in full daylight, 10% at full night, remaining 40% is split into 2x 20% linear transition between them for power output of panels. (reference:
https://wiki.factorio.com/Time#Days)
Some useful observations for computational purposes are as follows:
-Solar panels produce an average of 70% of their max power over the full day-night cycle.
-Hence, the excess produced above 70% of max (which occurs during the last part of dawn, all of full daylight, and the first part of dusk) must be exactly the same amount of energy as the accumulators can store (excess in daylight = shortfall in darkness) because this is the energy needed to supplement what the panels provide for the rest of dawn and dusk+completely replace them during the night.
-Because the transition from full power to none and vice versa occurs linearly, we can easily compute the exact time when the solar panels are producing precisely their average power output, this is useful for calculating the area of the trapezoidal region that represents the excess power (and hence the required Accumulator charge and thus accumulator ratio).
First, we compute the 70% or more power time proportion (this gives us length of long side, the short side is: 0.5 * day-night cycle length)
The transitions each take 20% of the cycle, the 70% power mark is reached 30% of the way into dusk or from the end of dawn, hence 0.3*0.2=0.06. Since this happens twice (once for dawn, and for dusk), we have a total of 12% of the day-night cycle that isn't full daylight but the panels are still making 70% or more of their maximum output. Hence out long side of the trapezoidal region is 0.62 * day-night cycle length.
One way to compute the area of a trapezoid is the split it into 2 triangles (which are identical) and a rectangle. Doing this here gives a rectangle of width 0.5 (50% of day-night cycle length) and height 0.3 (30% of max solar power), the triangles are each have base 0.06 (6% of day-night cycle) and height 0.3, hence their areas are each half of that product, but there are 2 of them, so the 1/2 cancels with 2 giving 0.3*0.56=0.168 (this is the %, in decimal form of max solar power per day-night cycle that your accumulators need to be able to store; this is 18.6% of max solar energy times day-night cycle length, or in otherwords, 18.6% of the power your panels could theoretical produce in that time *if they provided max power the entire time*). This number represents a bit of odd concept, but it's really computationally useful as we will soon see.
Here we gather some universal constants for reference:
Normal (100%) max solar power: 60kW
Accumulator capacity: 5MJ (a Joule is 1 W for 1 second, so 60kW is 60kJ/sec)
Solar energy factor: 0.168=21/125 (explained above)
With these constants we can use the world's Solar Power modifier and day-night cycle length (convert to seconds) to compute how much accumulator energy storage is needed, this can then be divided by how much energy each can store to obtain the proportion of accumulators per solar panel (intentionally sticking with rational numbers will allow exact ratios to be recovered).
World:
Nauvis:
Cycle time: 420 seconds
Solar Power: 100%
Energy storage required: 60kW * 420sec * 0.168(factor) = 529,200/125 =21,168/5= 4,233.6 kJ per panel
Ratio: accumulators/panels: 4233.6/5000 = 42336/50000 = 2646/3125 = 0.84672 (this is 21 and 21/125 accumulators per 25 panels, slightly more accumulator heavy than old ratio, which is expected since old ratio was based on 416.(6) sec day-night cycle)
Exact ratio:
3125 Solar Panels
2646 Accumulators
World:
Vulcanus:
Cycle time: 90 seconds
Solar Power: 400%
Energy storage required: 240kW * 90sec * 0.168 = 453,600/125 = 18144/5 = 3,628.8 kJ per panel
Ratio Acc/Panel: 3628.8/5000 = 36288/50000 = 2268/3125 = 0.72576
Exact ratio:
3125 Solar Panels
2268 Accumulators
World:
Gleba:
Cycle time: 600 seconds
Solar Power: 50%
Energy storage required: 30kW * 600sec * 0.168 = 378,000/125 = 3024 kJ per panel
Ratio Acc/Panel: 3024/5000 = 378/625 = 0.6048
Exact ratio:
625 Solar Panels
378 Accumulators
World:
Fulgora:
Cycle time: 180 seconds
Solar Power: 20%
Energy storage required: 12kW * 180sec * 0.168 = 45,360/125 = 9072/25 = 362.88 kJ per panel
Ratio Acc/Panel: 362.88/5000 = 36288/500000 = 1134/15625 = 0.072576
Exact Ratio:
15,625 Solar Panels
1134 Accumulators
World:
Aquilo:
Cycle time: 1200 seconds
Solar Power: 1%
Energy Storage required: 0.6kW * 1200 sec * 0.168 = 15,120/125 = 3024/25 = 120.96 kJ per panel
Ratio Acc/Panel: 120.96/5000 = 12096/500000 = 378/15625 = 0.024192
Exact Ratio:
15625 Solar Panels
378 Accumulators
One might have naively expected worlds with less solar power output to need more accumulators, but the opposite is actually the case since the accumulators still store 5MJ each and it takes a lot more panels to make that energy, obviously longer day-night cycle translates to needing more accumulators, all other things being equal. Somewhat surprisingly, Nauvis is actually the most accumulator heavy to build solar on, but ofc a more important consideration is how much solar power you get on the planet's surface, since that has a much greater influence on resource and space cost per MW. The decimal values for accumulators per panel can be handy when building to a certain goal amount of power. You can take your desired power level and divide by:
70% times solar power on planet times 60kW (panel base value)
This gives the number of panels you need to produce enough energy (round up here), then multiply that result (after rounding) by the accumulator/panel ratio and again round up and you have the needed accumulators to go with it (using the ratio for the world in question).
