Day-Night cycle times in Space Age and Solar Power

[Frightning]

So I decided I wanted to calculate the correct solar-accumulator ratios for the other worlds (specifically interested in Vulcanus atm because I am in the process of establishing myself there), and I realized that there seems to be a small (but relevant) detail others have overlooked: Nauvis's day-night cycle was 416.(6) seconds long in vanilla Factorio, according to space age in game info, it's 7 minutes (which is *420* seconds)? This would actually *slightly* change the optimal ratio on Nauvis to not be exactly 25-21 solar-accum.

So my first question is: how long actually is Nauvis' day? 420 or 416.(6)?

Given that, how long *exactly* is Vulcanus' day (game says 1:30=90 seconds).

[Frightning]

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).

[Xenothar]

Hey,

I repacked everything in neat tables, including your numbers, but also a boatload more for all quality combinations!

viewtopic.php?f=18&t=119040

[fadden]

Thanks for doing this!

Intuitively, Vulcanus should require 1/4 as many solar panels. If the days have the same "shape", then the length of the day should only affect the amounts, not the ratios. (Right?)

One thing that wasn't clear to me in the math is why the "energy storage required" value is divided by 5000 on every planet. Can you explain what this value represents? (Edit: found it, it's the 5 MJ accumulator capacity, in kJ.)

Edit: I set up some tileable arrangements around a substation. With a 3.1 MW load, 47:14 acc:solar goes dead before the sun sets because the accumulators don't charge above about 2.5% (it's only generating 3.36 MW), while 22:25 fills completely and only dips down to about 45% at night. Upping the load to 4.2 MW, only 20:26 (0.769) runs all day and night (dips to 10%). This seems to agree with the math.

[Xenothar]

Thanks for doing this!

Intuitively, Vulcanus should require 1/4 as many solar panels. If the days have the same "shape", then the length of the day should only affect the amounts, not the ratios. (Right?)

One thing that wasn't clear to me in the math is why the "energy storage required" value is divided by 5000 on every planet. Can you explain what this value represents? (Edit: found it, it's the 5 MJ accumulator capacity, in kJ.)

Edit: I set up some tileable arrangements around a substation. With a 3.1 MW load, 47:14 acc:solar goes dead before the sun sets because the accumulators don't charge above about 2.5% (it's only generating 3.36 MW), while 22:25 fills completely and only dips down to about 45% at night. Upping the load to 4.2 MW, only 20:26 (0.769) runs all day and night (dips to 10%). This seems to agree with the math.

For a factory with the same power draw, you need 1/4 of the solar panels on vulcanus, thats correct, as they are still utilized to 70% on Vulcanus (or on any planet in SA). However, the second part holds not true. Vulcanus has the same shape of the day, just a shorter duration, but ratios are affected, even if we ignore the power factor difference. The reason is that the energy stored depends on the seconds of the day/dusk/dawn, not only the shape = ratios of the day. So a longer day leads to a longer time storing energy away in accumulators for more seconds in the night. As the capacity per accumulator did not change, you need more accumulators.

Tl;dr: Longer day-night-cycles with the same shape and same power factor need more acc per solar panel.

This is linear by the way. So twice the time of day, you need twice as many accumulators.

In general: The ratios acc/solar are a function of: day length, day-night-cycle-shape, power factor, solar panel output power, accumulator capacity (and accumulator in and out power limits, of which only out plays a role for Vulcanus).

Concrete example:

Nauvis: 1000 panels and 847 acc (normal qualities both) to power your facotry.
Vulcanus: 250 panels and 182 acc to power the same factory.

This is less than the ~212, which would be 1/4 of Nauvis acc. This is due to the day-night-cylce on Vulcanus being shorter than 1/4 of Nauvis day, which lowers the required amount below 1/4 of Nauvis.

If vulcanus day-night-cycle was 1/4 of nauvis, the ratio would stay the same, as power factor and day-night-cycle-length would cancel each other out.

If you factor in qualities, you get a lot of ratios, this is why you can look them up here: viewtopic.php?f=18&t=119040

[fadden]

Thanks, this helped.

Converting it to a simple thought experiment:

• Assume you have an acc+solar setup where all power is fully consumed, i.e. accumulators just barely drain fully overnight and just barely reach full charge at the end of the day.
• If solar power yield increases 2x, and the day length doesn't change, you only need half of the solar panels. acc:solar ratio increases.
• If solar power stays the same, but the day is half as long, then accumulators will only charge to 50% (and only need 50% charge). You can cut the number of accumulators in half. acc:solar decreases.

Vulcanus changes both, and on balance is pretty close to Nauvis.

[RunFact]

Hello! Thank you for this information! Super useful.

How would I calculate daily "usable" power from these numbers? With the old 416.6 second day/night cycle, it took 238.095 solar panels and 200 accumulators to get 10 MW of daily usable power. So (total solar panel output) x 0.7 = daily usable power.

With the new 420 day/night cycle, that 0.7 should go to 0.756... I think. Where I'm lost is, how do I calculate that daily usable power for the other planets?

[fadden]

How would I calculate daily "usable" power from these numbers? With the old 416.6 second day/night cycle, it took 238.095 solar panels and 200 accumulators to get 10 MW of daily usable power. So (total solar panel output) x 0.7 = daily usable power.

From the math post above:

Solar panels produce an average of 70% of their max power over the full day-night cycle.

Since the shape of the day doesn't change, solar panels should still be 70% effective. No energy is lost when charging/discharging accumulators, so all energy generated by panels will be available to you eventually if you match the correct acc:solar ratio.

(If you have more panels than prescribed, your estimate of power available will be too high. If you have more accumulators, your estimate will be accurate, but your accumulators will have unused capacity.)

I think this means that the number of solar panels required to hit 10MW usable power on Nauvis doesn't change. The number of accumulators does (by about +1.5). Intuitively (uh oh), what you're calculating is not the power generated per day (MJ), but rather the power available at any given time (MW), so the day length doesn't affect it.

Either I'm getting the hang of this stuff, or I'm about to get schooled.

[RunFact]

Thank you! So if we are trying to get 10 MW of daily usable power we'd need:

Planet, Solar Panels, Accumulators
Nauvis, 238.0952381, 201.6
Vulcanus, 59.52380952, 43.2
Gleba, 476.1904762, 288
Fulgora, 1190.47619, 86.4
Aquilo, 23809.52381, 576

[Xenothar]

That's correct, the formula being:

Necessary Panels = Required_continuous_power / (solar_power_per_panel * daily_planet_utilization * planet_power_factor)

Daily Planet utilization = 0.7 for all planets in SA
Planet Power Factor = 1 for Nauvis, 4 for Vulcanus and so on
Solar Power per Panel = 60 kW for normal quality, 78 kW for uncommon and so on

So for Nauvis, normal quality, 10MW draw by factory (not factory + acc, only factory = total power draw during nights):

Necessary_panels = 10000 kW / (60 kW * 0.7 * 1) = 238.095 panels

Acc/panel ratio on Nauvis (both normal quality) = 0.847 --> 201.67 acc

All ratios for all quality combinations for all planets can be found here: viewtopic.php?f=18&t=119040

Just be cautious for Vulcanus, as acc/panel ratio is throughput-limited unlike on all other planets.

Fadden is also on point: We are talking power (MW) here, not energy (MJ), so day length has no bearing for the panel calculation, only the average utilization of a panel, which is always 0.7 in SA. For accs, yes, the numbers are affected by day length changes.

[Seawolf159]

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

This is incorrect according to my Excel check. It says Solar: 500 and Accumulator: 363 is ideal. Your ratio is actually 0.72576, while this one is exactly 0.726. Am I doing something wrong?

[schorsch_76]

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

This is incorrect according to my Excel check. It says Solar: 500 and Accumulator: 363 is ideal. Your ratio is actually 0.72576, while this one is exactly 0.726. Am I doing something wrong?

Its just a rounding error.