## Accumulator / Solar panel ratio

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### Re: Accumulator / Solar panel ratio

WarDaft wrote:Then there are 125 seconds of darkness to deal with. Thus 42 KW * 125 s = 5.25 MJ to buffer the panel's overnight power. But that's a ratio of 20:21 panels:accumulators, not ~21:25, and something is wrong, and it has to be something in this paragraph.

I actually made the exact same mistake the first time I tried running through it in my head, too.

While dusk / dawn do last 83.33 seconds, the amount of time it takes for your solar panels to transition to/from accumulators is actually much shorter, because it takes your solar panels 83.33 seconds to transition to their max capacity, not to your current consumption. They hit your consumption level much more quickly.

Another way to word what Koub says is basically this: You'll switch between solar panels and accumulators as your solar panel output crosses the threshold of the power you actually need. Also the amount of power your solar panels can produce is of course proportional to the number of solar panels you have, but the day <-> night transition takes the same amount of time, it's constant. Therefore, the more solar panels you have, the earlier in the dawn the accumulators will switch off (because more panels = less light required to power your factory, so the panels get there more quickly), and the later in the dusk they'll switch on (same deal, the panels will take you longer into dusk).

If you're better at thinking about it purely geometrically, your (incorrect) thinking is that you have two overlapping trapezoids of the same height (one for accumulator capacity, one for solar panel capacity) that cross at exactly halfway through dawn or dusk. But in reality, the more solar panels you have, the steeper the sides get, and they don't "rotate" around the center point of dawn/dusk, they stay fixed to the start of dawn and the end of dusk.

Yet one more way of thinking about it, if you're better with conceptual extremes: Imagine you have infinite solar panels, so the most infinitely small amount of light will generate enough power for your factory. In this case your accumulators will switch off as soon as dawn starts, and switch back on as soon as dusk ends. So you can see in this extreme case you actually have 41.67 seconds of accumulator time (night) and 375 seconds of pure solar panel time (dawn, day, and dusk). This directly contradicts your notion that it's always a 50/50 split at dusk and dawn, and therefore your hypothesis must be false, by contradiction. The more solar panels you have, the more your effective power generation times approach this limit, and so you have to account for this when doing the math. The OP's geometric proof is one good way to break it down.
JasonC
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### Re: Accumulator / Solar panel ratio

That is exactly not how math works. Especially the part about infinities. When dealing with infinities you have to be very careful about how you approach them in the limit. If you have the exact number of solar panels to sustain your base on average, then it should always switch to drawing accumulator power at the same time during the day, no matter how many of each you need.

If you have 1 solar panel powering a base that draws 42 kilowatts, you will begin draining from your accumlators when the solar panel output falls to 42 KW/60 KW = 70% light level.

If you have a quintillion solar panels powering a base that draws 42 zettawatts, you will begin draining from your accumulators when when the solar panel output falls to 42 ZW/ 60 ZW = still 70% light level, thus the same point in time.

I'm not going to argue with the ultimate conclusion because I'm in no place to test it, but this reasoning for it to be the case is definitely not correct. If this reasoning were correct, there would be no one correct ratio as your total average power draw increases.

I'm more inclined to think that a different ratio is based on the part cost of the items, which I didn't factor in, so having a different ratio would make sense as the cheapest way to obtain a specific average power capacity.
WarDaft
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### Re: Accumulator / Solar panel ratio

WarDaft wrote:That is exactly not how math works. Especially the part about infinities. When dealing with infinities you have to be very careful about how you approach them in the limit. If you have the exact number of solar panels to sustain your base on average, then it should always switch to drawing accumulator power at the same time during the day, no matter how many of each you need.

If you have 1 solar panel powering a base that draws 42 kilowatts, you will begin draining from your accumlators when the solar panel output falls to 42 KW/60 KW = 70% light level.

If you have a quintillion solar panels powering a base that draws 42 zettawatts, you will begin draining from your accumulators when when the solar panel output falls to 42 ZW/ 60 ZW = still 70% light level, thus the same point in time.

I'm not going to argue with the ultimate conclusion because I'm in no place to test it, but this reasoning for it to be the case is definitely not correct. If this reasoning were correct, there would be no one correct ratio as your total average power draw increases.

I'm more inclined to think that a different ratio is based on the part cost of the items, which I didn't factor in, so having a different ratio would make sense as the cheapest way to obtain a specific average power capacity.

Yes it shifts the light level from 50% to 70%. The OP's reasoning is spot on, and so is mine. Either you didn't understand or I explained it poorly, but in any case, eventually it'll click for you.

(Also no, it's not related to item cost. Just read the original post before you get inclined to think anything else.)
Last edited by JasonC on Sat Apr 16, 2016 2:59 pm, edited 1 time in total.
JasonC
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### Re: Accumulator / Solar panel ratio

WarDaft wrote:If you have 1 solar panel powering a base that draws 42 kilowatts, you will begin draining from your accumlators when the solar panel output falls to 42 KW/60 KW = 70% light level.

Correct. 166.66 seconds is half and half, 70% of that is 116.66 seconds, and the total "accumulator darkness" from that is 58.33 seconds.

DaveMcW
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### Re: Accumulator / Solar panel ratio

DaveMcW wrote:
WarDaft wrote:If you have 1 solar panel powering a base that draws 42 kilowatts, you will begin draining from your accumlators when the solar panel output falls to 42 KW/60 KW = 70% light level.

Correct. 166.66 seconds is half and half, 70% of that is 116.66 seconds, and the total "accumulator darkness" from that is 58.33 seconds.

Aha! That was the missing step, thanks.
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### Re: Accumulator / Solar panel ratio

WarDaft wrote:I'm not going to argue with the ultimate conclusion because I'm in no place to test it, but this reasoning for it to be the case is definitely not correct. If this reasoning were correct, there would be no one correct ratio as your total average power draw increases.

You are correct, in that without better problem constraints, there is no single optimal ratio, but rather a range of viable ratios. But, there is the most obvious point, and that's what we calculate.

I found this in my notes file: "With minimal panel count, 0.84 accumulators are needed for each panel. Of course, arbitrarily more panels may be added, allowing the ratio of accumulators per panel in the limit of infinite panels to approach 0.35. But why do this when panels are more expensive and require more space than accumulators?"

I had 1.05, but I recalculated, it's 0.84. 3750 kJ excess during daylight, 450 kJ excess from dawn / dusk, 3750+450=4.2 MJ storage needed, 4.2/5.0 = 0.84. (Likewise, 1750 kJ at night, 2450 kJ transition, 2450+1750=4.2 MJ.) Now I have to make a new solar blueprint.

I'll come back with a formula for accumulators, that allows both panels and always-active steam engines! Because I think I saw a few people asking for it. Aaaand it's a lot more boring and obvious than I was hoping for. A steam engine is worth 510/42=85/7~=12.14 panels, essentially, and the 0.84 ratio is the same. But, it should be more interesting when steam is used as a backup...

edit: The 450 kJ, is from: 5000 ticks * 1/60 seconds/tick * (60kW - 42kW) * (1 - 42 kW/60 kW)/2*2
Last edited by Aru on Wed May 25, 2016 10:36 pm, edited 5 times in total.
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### Re: Accumulator / Solar panel ratio

Aru wrote:I'll come back with a formula for accumulators, that allows both panels and always-active steam engines! Because I think I saw a few people asking for it.

viewtopic.php?f=5&t=5594&start=20#p143317

Added to the wiki, too: https://wiki.factorio.com/index.php?tit ... imal_ratio

It's approximately 20:24:1 (21:25:1 is also reasonable) accumulators:panels:megawatts required (minimum accumulators and panels for a consistent 1 MW draw thru day and night). Precisely, it's 21:25:1.05 or 19.99:23.8:1.
JasonC
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### Re: Accumulator / Solar panel ratio

I'm not quite sure what you're saying, why the quote?

Update, I'm really, really slow with my analysis and math stuff... but, basically, if you build a steam array, then you were to transition to panels... you'd start by adding panels and accumulators in the normal ratio, as-needed to meet increasing power demands, instead of expanding the steam array. Then say, you reach a point where you want to start dialing down the steam array usage to switch to full solar. At this transition period, your hypothetical power demand stays the same. If you were to continue phasing engines out, they'd reach 0 usage and you'd have the normal 0.84 panel ratio again. But in between full steam usage and no steam usage, the optimal accumulator ratio varies! You don't need as many per panel at certain moments. This is what I'd like to calculate, and I'm working on it! I will get there.
Aru
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### Re: Accumulator / Solar panel ratio

hmmm
Fayez
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### Re: Accumulator / Solar panel ratio

Hi guys.

After many hours spending time making calculations on my own.

I finally undestood how this ratio was calculated.

Basically you solar panels starts providing enough power a little bit before full day light and stops providing a little bit later after full day light is gone.

Thats why the accumulator ratio is 20 and not 25. You dont need accumulated energy for all dawn/dusk/night period, just a fraction of it.

It depends of de number of solar panels you have.

For language limitation reasons i cant explain here all the math. But the graph on attachment will do.

I have made, a file in excel with customizable data input and a very detailed GRAPH that in my opinion helps a lot.

I hope you guys like it. (It took a while to do). The file is in english also. (i guess =)
Attachments
Solar Power Production Analysis.xlsx
Battutta
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### Re: Accumulator / Solar panel ratio

Maybe this is already old, but I haven't seen anything similar around so I'll post it...

If the base for the design grid is the Roboport (I have seen designs based on the power poles), assuming you want a small gap around the area, and that you want some light to see what you are doing... I came up with this simple solution:

180 Solar
150 Acc (ratio 0.833, very close to the perfect 0.84)
1 roboport
4 lights
16 substations

add lights on the outside wherever you like because they don't block passage.
Last edited by JoaoBernardino on Tue Jun 21, 2016 11:29 pm, edited 1 time in total.
Nom Nom Nom

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### Re: Accumulator / Solar panel ratio

In my opinion a bit less catchy for the eye but way more manual-placing friendly. Nice job!
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sillyfly wrote:kovarex just posted the thread... but with #118 in the title. I think they had too much beer

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### Re: Accumulator / Solar panel ratio

A couple weeks ago I embarked on a quest to find the lowest number of accumulators and solar panels necessary to keep a given circuit operational continuously. Long story short:

Rc = Rate of energy consumption by the circuit in kJ
Rp = Rate of energy production by solar panels in kJ = Ceiling( Rc / 42 ) * 60

Min. solar panels for a circuit: Ceiling (Rc / 42)

Min. accumulators for a circuit = Ceiling (((Rc^2) / (3600 * Rp)) + (Rc / 120))

The ratio does appear to be 1.2 panels per accumulators where possible.
organism31
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### Re: Accumulator / Solar panel ratio

Hint to all people who feel the urge to do the math by themselves (which I strongly encourage) : do whatever math you want. If the result is not a 0.84 accu per panel (or close to 1.19 panel per accumulator), then your math is wrong
Koub - Please consider English is not my native language.
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### Re: Accumulator / Solar panel ratio

Why not simplificate the whole solar panel and accu system to 1:1 :p

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steinio
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### Re: Accumulator / Solar panel ratio

steinio wrote:Why not simplificate the whole solar panel and accu system to 1:1 :p

That would take the fun away of all the calculating ^^
And the search for a practial -perfectratio- set-up
Choumiko wrote:
sillyfly wrote:kovarex just posted the thread... but with #118 in the title. I think they had too much beer

It's a wonder how good the game is, if you consider how bad they are with the FFF numbers
Ojelle
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### Re: Accumulator / Solar panel ratio

You do need an extra amount of accumulators to let laser turrets do their thing during the night. If you get attacked rarely then 1:1 will work fine. If you get attacked constantly then it's better to stick to the ratio.
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### Re: Accumulator / Solar panel ratio

You also need an extra amount of solar panels to power your extra amount of accumulators. In fact, it needs to be 25:21 solar panels:accumulators.

DaveMcW
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### Re: Accumulator / Solar panel ratio

DaveMcW wrote:You also need an extra amount of solar panels to power your extra amount of accumulators. In fact, it needs to be 25:21 solar panels:accumulators.

Only if you get a constant use of those extra accumulators. You need more accumulators than solar panels for that extra fluctuating bit than the ratio tells you. Since you don't get constant attacks you don't need as many solar panels as accumulators to power laser turrets.
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### Re: Accumulator / Solar panel ratio

Ratio is between 0,8469 (only one tile) and 0.8367 (infinite amount of tiles)

196 panels and 164 - 166 accumulators

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