Accumulator / Solar panel ratio

[DaveMcW]

I did this to simplify the math

You simplified it too much. The first 30% of dawn and dusk do not require accumulators, because they are producing more than 42kW.

accumulator.png (12.42 KiB) Viewed 11175 times

If you cut out the green regions, it reduces the accumulator power needed to only 100 seconds, not 125.

[creepercrack]

I did this to simplify the math

You simplified it too much. The first 30% of dawn and dusk do not require accumulators, because they are producing more than 42kW.

accumulator.png

If you cut out the green regions, it reduces the accumulator power needed to only 100 seconds, not 125.

Now I see it, I was wrong, thanks!

[AJaromir]

Looks like in this thread will be someone who will help me.

I am just testing Solar/accumulator ratio alchemy with 20 radars. Radars have constant power usage so they are good for testing.

20 radars have 6 MW power usage. That means I need 100KJ for every game tick.
If my calculations are right, then I need 150 accumulators to keep it working in night. (100 KJ x 7500 ticks = 750 MJ -> 750 MJ /5 MJ = 150 accumulators)

To test if it's the right value, I pre-charged them with 100 solar panels which generates just 6MW -> same power like needed by 20 radars. After the cells were fully charged, I connected radars to the circuit with 100 solar panels and 150 pre-charged accumulators in mid-day.
Then something strange happened. Even the everything was balanced, about first 4 seconds the radars used the energy stored in accumulators instead of solar panels, so I had to add one solar panel for a short time to refill the accumulators to full capacity again.

The first test passed. The radars worked on 100% for the whole night and the stored power at the end of dawn was 0J. The second test was without re-charging accumulators after connecting radars. This one almost passed. There was just for less than 1 second 5.98 MW power usage.

So, let's continue.
I just need to generate the 2500 MJ during the whole daytime. -> 2500 MJ / 17500 game ticks = 142+(6/7) KJ per game tick. (142+(6/7)) KJ * 60 game ticks = 8571+(3/7) KW -> / 60 = 142+(6/7) solar panels which is just rounded 143 solar panels.

Now the practice test which have the different result:
Looks like the 143 solar panels is good number because the accumulators are accumulating or discharging almost continuously for almost 100% of daytime and the accumulators are always charged to 100%. BUT I've found that I have 150MJ of useless reverse which is just 30 accumulators. So In the end I just need only 120 accumulators instead of 150. And this is where I am getting lost in my math.

Edit. I think I am getting it. Because there is more solar panels, the energy production during dawn and dusk is higher so less accumulators is needed. Easy to calculate.

Edit 2 - the final result (yeah, finally! ):

For 6 MW power consumption we need exactly:
8571+(3/7) KW solar panels
600 MJ capacity of accumulators

So, the result is we need 25/21 (= 1.190476^_ (period 6)) more solar panels than accumulators. Or we can say we need 21/25 (=0,84) accumulators per solar panel. Or we can say we need 25 solar panels per 21 accumulators. So the OP post is 100% accurate.

[BurritoBotV3000]

Greetings, AI's of factorio, I have something for you, this is my first post. this post is a blueprint for the solar farm in the picture. Enjoy!

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

[ownlyme]

from my ingame tests (0.18) it seems to be slightly more than 0.84, more like 0.8407 ...
and your base may only consume 69.99% of your solar panel's production

[leadraven]

from my ingame tests (0.18) it seems to be slightly more than 0.84, more like 0.8407 ...
and your base may only consume 69.99% of your solar panel's production

Tests vs Math... Good luck.

[Marco3104]

Sry if i dont get this correct but i have one question. The equation from the first post:

n / m = [(0.5 + 0.2) * [0.1 + 0.2*((0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

Iam i stupid or cant i count? There are more open brackets than close brackets right ?? I dont say there is an error in the calculation, but the representation of the equation is incorrect. If iam wrong please tell me what i did wrong. (sry for my english)

[Koub]

Sry if i dont get this correct but i have one question. The equation from the first post:

n / m = [(0.5 + 0.2) * [0.1 + 0.2*((0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

Iam i stupid or cant i count? There are more open brackets than close brackets right ?? I dont say there is an error in the calculation, but the representation of the equation is incorrect. If iam wrong please tell me what i did wrong. (sry for my english)

n / m = [(0.5 + 0.2) * [0.1 + 0.2*((0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

should be read :
[Edited]
n / m = [(0.5 + 0.2) * [0.1 + 0.2*(0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

[Qon]

n / m = [(0.5 + 0.2) * [0.1 + 0.2*((0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

should be read :

n / m = [(0.5 + 0.2) * [0.1 + 0.2*((0.5+0.2) / 1.0) / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

Still wrong . Now you have [[ but only one ]

[mrvn]

Has anyone updated their Accumulator / Solar panel blueprints to have a global grid setting so you can just put them down anywhere and it aligns right?

[mrvn]

If your power draw is constant, then the optimal ratio is 25 Solar panels to 21 Accumulators. If there is variance, than you need to know what the greatest possible drain over the timespan from beginning of dusk to end of dawn is (i.e. entire timespan that Solar panels are not producing 100% energy) will be. In practice, running a bit accumulator heavy would ensure that small amounts of variance won't leave you starving for energy in the morning. However, I never had issues with volatility in power drain with solar because I usually had a least a little bit more average power production than I had average demand, and that excess+accompanying storage capacity from matching accumulators was always enough to make to the next day.

Actually let me challange that statement.

The 25:21 ratio is calculated so that at a constant power draw you have just enough accumulators to last the night and just enough solar cells to recharge them fully during the day. It maximizes the number of accumulators while minimizing the number of solar cells. Since accumulators are also smaller it probably also minimizes space. Or maximizes the Watt / m^2.

But maybe that isn't what we should be optimizing? If you build more accumulators then they simply won't get charged since 25 solar pannels - constant consumption only leaves so much to charging. So 21 is the maximum number of accumulators you can use with 25 solar cells. But maybe it makes sense to use less?

If you use more solar cells then they will provide more power earlier in the day and last longer in the evening. That means less power will be drawn from the accumulators over night and you can use less of them. They will also be recharged faster during the day and then there will be excess power that will simply go to waste (assuming our constant consumption of power). But to some accumulators are expensive and solar cells are cheap. Space is also often not a problem or you probably wouldn't be using solar. So wouldn't a 30:18 (or whatever ratio makes it work again) be better if you assign accumulators a higher price than solar cells?

So what we really should have is a table with solar cells at one axis and accumulators at the other axis and the garanteed constant power as values in each cell. There would be a region in the table with sensible combinations of solar cells and accumulators. But also regions that don't make sense. Any cell that doesn't improve on the one with one less accumulator is out too. That's just wasted. On the other side though, asides from having 0 accumulators there is nothing to obviously exclude. Every added solar cell will increase the power even one accumulator will alow as constant draw. But it quickly becomes near 0.

There would be a line starting at 1,1 and go through 25:21 that reflects the "optimal" 0.84 ratio. The line of maximum accumulators. How many additional W/solar cell does that line give? I think it's something around 40W/cell.

The question I would want answered is: Given a cost ration X:Y for solar cells to accumulators what ratio gives you the most W per cost and how much is it?

[Marco3104]

n / m = [(0.5 + 0.2) * [0.1 + 0.2*((0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

should be read :
[Edited]
n / m = [(0.5 + 0.2) * [0.1 + 0.2*(0.5+0.2) / 1.0] / 1.0] * (25 000 / 60) * (60 / 5000) = 0.84

Iam sry i dont want to be inpolite or proof something wrong. I just want to learn why it is that way. Can you tell me why that double open bracket should be read as a single one? I know that for every open bracket there has to be one that closes the bracket, i wrote that equation now several time on a piece of paper to see where i made a mistake but i cant find one. (sry for my english)

[Koub]

a*(b+c)/d/d = a*((b+c)/d)/d.
You can see it here written in another manner : https://wiki.factorio.com/Power_product ... imal_ratio

[Qon]

Has anyone updated their Accumulator / Solar panel blueprints to have a global grid setting so you can just put them down anywhere and it aligns right?

Since it is done trivially in seconds at most, yes, obviously.
If you are asking for examples, you can have mine:
Your message contains 61426 characters.
The maximum number of allowed characters is 60000.
Well maybe you can't have it as a BP tag here....
Link: https://pastebin.com/A1tweXz3

Actually let me challange that statement.

It's rare that I see someone actually thinking critically about this topic. Not many seem to realise this and assume that the ratio actally means something instead of the count of solar panels and accumulators compared to your power draw is what matters.

The answer happens to be 42.
That is, kW constant draw with maximised accumulators.

With infinte solar panels you (ratio 1.0) you just need enough accumulators to last through the times when solar outputs 0, which will reduce your need of accumulators since you no longer need them to power your base at dusk and dawn. That isn't very cost effective though, even though accumulators are expensive compared to solar panels. I'm also interested in the table. Depends on how you value oil compared to metal ores.

[JimBarracus]

Not many seem to realise this and assume that the ratio actally means something instead of the count of solar panels and accumulators compared to your power draw is what matters.

The answer happens to be 42.
That is, kW constant draw with maximised accumulators.

every solar panel can provide 42kw constantly with the correct amount of accumulators.
What about a safety margin?
How much would be acceptable without being considered wastefull?

Or is "better" to have some boilers left as a backup generator. How big should this be?

[Impatient]

I made an excel sheet once to calc the correct numbers precisely.
And of course there is no perfect ratio. Because of the two periods when there is a gradient of sunlight, it is possible to vary the ratio within some boundaries. And the larger the complex, the more the two numbers can vary obviously.

[Hannu]

every solar panel can provide 42kw constantly with the correct amount of accumulators.
What about a safety margin?
How much would be acceptable without being considered wastefull?

It depends on your personal opinions. Many Factorio players want to have exact ratios everywhere, but practically power consumption varies tens of percents. 10-30 % is common compromises in most engineering work. I use large numbers of accumulators rarely, because I like to have more complex power system, but when I use i have usually 10-20 % excess to give reserve.

Or is "better" to have some boilers left as a backup generator. How big should this be?

It also depend on situation and your personal preferences. If you have high biter settings and laser weighted defense blackouts may be more severe than with normal biters and defenses relying on machine guns or flame throwers. Also if you are careful player and check power often you need less reserves than if you are careless or easily distracted.

If I use solar energy with accumulators, I keep by first coal powerplant as a reserve. It is usually 60 boilers. At very late game I have usually around 1:1:1 ratio between coal, nuclear and solar and accumulators only for controlling coal plants. I consider nuclear fuel as free and do not care if it is wasted during low power situations.

[Hannu]

I made an excel sheet once to calc the correct numbers precisely.
And of course there is no perfect ratio. Because of the two periods when there is a gradient of sunlight, it is possible to vary the ratio within some boundaries. And the larger the complex, the more the two numbers can vary obviously.

Power output during morning and evening are exactly defined as linear functions and 25:21 ratio takes it into account. Accumulators deplete in morning at exactly the moment when solar panels give average power. However, consumption of the base varies in quite large range which makes exact ratio for constant power as academic nitpicking (or Factorio community calls it OCD).

[mrvn]

Not many seem to realise this and assume that the ratio actally means something instead of the count of solar panels and accumulators compared to your power draw is what matters.

The answer happens to be 42.
That is, kW constant draw with maximised accumulators.

every solar panel can provide 42kw constantly with the correct amount of accumulators.
What about a safety margin?
How much would be acceptable without being considered wastefull?

Or is "better" to have some boilers left as a backup generator. How big should this be?

Backup is always good. The power consumption of a factory is not constant. The biggest variance are laser turrets. Because whenever aliens attack you get HUGE spikes. And it seems aliens like to attack at night and then your accumulators will drain before morning and you have no power at all. To be safe have enough boilers to power your lasers.

But practically aren't you constantly expanding, constantly need more power, constantly adding more solar cells and accumulators? Just build ahead on power generation till you have a good margin of available vs. consumed and keep an eye on the ratio. Some people stick with 50% power consumed. Any more and they add more solar.

[mudcrabempire]

Meh, ratios in the honor they deserve but on the solar panel/accumulator ratio I've kinda given up.
Space optimization is kind of whatever, since if you need high power per space, you simply don't use solar.
I'm also fairly certain that solar is not the way to go for a good resource to power ratio.
There are some considerations about the ratio, like how many accumulators you need to last the night and how many solar panels you need to recharge them while powering your factory as well... but honestly, all those considerations presume that you want to build just enough power to drive your factory.
I don't care for that. If I've learned one thing about power, it's that you don't build just enough of it. With power, your're generous. Give me twice the average consumtion of my factory and I'll call it decent.

So as far as I'm concerned: Average output of solar panels > average consumption of factory+margin. Charge of accumulators > what's needed for the night+margin.

To put some actual argument to the above conclusion so it's not just some guy saying "meh":
As far as I know, most discussions about the solar panel/accumulator ratio work with average factory consumption. Factory consumption is not constant. Far from it. Large biter waves versus laser turrets can make large dents in your power budget. Even if your solar farm is enough to handle "average" consumption, these spikes can easily exceed your maximum production, especially at night, leaving your poor defenses dangerously underpowered. Depending on the setup of the factory, several trains delivering ore to the furnaces at once or other comparable scenarios can cause similar issues.

So you need to plan above your average. And since in the long run these spikes will cancel out with some lows (otherwise the average would be higher), I prefer to err on the side of more accumulators to have a generous buffer for when things get hairy.