Accumulator / Solar panel ratio

[featherwinglove]

I have more accumulators. I always have something which throws it off. If nothing mod-related, it's laser turrets and biters following the advice of Lionel Richie

[Hansi123]

I've been working on making a different kind of solar build -- one optimized for "fun", so to speak. Instead of min-maxing material and spatial costs, my goal is to optimize around playability concerns:

First, I hate the sound of accumulators so I have pushed them all into the middle of the build where they will be seen but not heard (as much).

Second, I don't like having to predict the future. So my build leaves space for a train track with signals in-between the tiles, in both the vertical and horizontal directions, and even enough room (well, not quite -- a small amount of futzing around with pieces is necessary) for a train to turn a corner within my solar farm.

Third, if I want to build some solar, I want it NOW. And, chances are, I want a LOT of it, more than I could carry in my inventory and anyhow to big to stand around waiting for personal roboport-bots to get it done. So my build greatly increases roboport density compared to, i.e., Madzuri's solar design, and leaves room to add even more roboports in the corners if necessary, so I can just shove everything required by my build in a provider chest, stamp down my blueprint, spray some bots, and leave.

Finally, I wanted something that was four-way rotationally symmetric, so that if I had to patch up a botched deployment, I wouldn't have to use the zoom command + squinting system to figure out if I had the right orientation when re-blueprinting.

Here is a picture

The attachment SolarBuild.png is no longer available

Note that, although I'm using mods in the picture, none are required, this is a vanilla thing. I am leaving two tiles of space between the "stamps" but I could think of reasons to use different spacings ranging from no gap up to six tiles, depending on what I was trying to accomplish. Since there is ample roboport coverage overhanging the build, any reasonable spacing arrangement should work just fine.

And here is a blueprint

Code: Select all

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IIRC the blueprint includes some rail tiles for alignment. If not, you may wish to add some -- otherwise, your odds of randomly placing your solar farm in a position that will prevent you from laying rail between the tiles is 75% since you run a 50% chance of missing your mark in both the horizontal and the vertical dimension (unless you leave even more than two tiles of space between the builds, but then when you lay your rails they'll be asymmetrical and ugly).

I think the optimal way to deploy them is to have an existing robot network in place, and the required solar panels and accumulators in logistic storage, preferably nearby, but not too nearby, but be holding the required roboports and substations. This way, your personal roboport bots can quickly place the substations and roboports, leaving your network with the task of laying down the solar panels and accumulators. In a large build, with sufficient construction bots in the network, this approach will prevent you from being nagged about "missing robots" or "missing material for construction" and the build will go much faster as you won't be waiting for robots to build out roboports to bring portions of your build into the network. By leaving a bit of distance between your storage and you build, you can also prevent your network construction bots from sealing you into the build forcing you to pull it to pieces to get out.

The build has 84 accumulators and 100 solars (plus twelve substations and one roboport)

Soo i was was building your setup and realized it wasn't exactly what i wanted, even though it is a cool setup.

I took some inspiration from you and came up with this:

My suggestion

20171015204727_1.jpg (1.06 MiB) Viewed 13568 times

It has
164 Solar panel
141 Accumulators
Which gives an ratio of 0.86, a bit higher than the optimal but in my opinion just perfect.

One of the biggest improvement of this design is the efficient use of space and substations.

If anybody has some suggestions to make this design even better please let me know

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[SuicideJunkie]

I found a 1:1 ratio is a great fit for my needs; that allows me to reserve a hearty 10% margin before the emergency power logic kicks in.

Between laser turrets and 30 MW of partial duty cycle factories, power use is quite spiky.

[Pascali]

What is the optimal ratio for solar per accumulator?

[steinio]

What is the optimal ratio for solar per accumulator?

Nobody knows so far.

[golfmiketango]

What is the optimal ratio for solar per accumulator?

Nobody knows so far.

I think it's pretty funny how at night all your lights come on. If you've been spamming lights everywhere, they can easily add up to become quite power hungry which absolutely will throw off the optimal solar-accumulator ratio if you'd like to not run out of power with minimum solar deployment.

Also, is it just me or do biters tend to get more aggressive at night? Maybe I'm just imagining that because I let power crap out too often: they might look more aggressive to me because they are breaking all my toys, when the actual reason is I turned all the turrets off and they simply took advantage of my hospitality.

[Pascali]

Nobody knows? Why? I read many different things like 25 solars on 21 accumulators. There will be a perfect ratio!?

[Acacel]

When you do the math in 0.15 was it the 25 to 21 but:

it depents on what you want to have? best solar use or some only over days reloadable backup for peaks in engery consumption.
So in the end everyone wants to achive a different thing and therefore thers never the perfect ratio to find

[Pascali]

o.k., so the optimal ratio without energy using is 25:21 solar:accu?

[Koub]

https://wiki.factorio.com/Power_product ... imal_ratio

[golfmiketango]

https://wiki.factorio.com/Power_product ... imal_ratio

I'm not sure if steinio was kidding, but FTR that's based on a pretty crude algebraic model of power consumption with no spikes and no other power sources. If your game is more complex than that (it is, trust me) then it becomes murkier. To really model this in terms of risk-of-ruin, I think you'd need to employ sort of variance analysis...

However, the good news is, the presence of secondary sources of power, I believe, would tend to prop up the validity of the simplistic model in the wiki. If your didn't use switches or something to prioritize accumulators over backup sources, however... yeah, you probably need an economics textbook to figure out your true optimal solar ratio

In practice, of course, this is just my sick idea of fun mental masturbation... the wiki ratio is a fine starting point and empirical tweaking from there is almost always going to be the way to go if you really want to fine tune it.

[Frightning]

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.

[Koub]

Actually the variability harms also when it occurs during daytime, especially if your power generation is very close to your average consumption : it will prevent your accus from filling, and you risk blackout the following night.

Always tailor your power generation for known (or expected) peak consumption, add an emergency backup with an alert as soon as it kicks in (so that you know your power production has become insufficient and you should add some more panels/accus asap) and you should be 100% safe.

The solar/accus ratio always assumes a constant average consumtion.
Lack of solar panels, and your accus might not be filled enough to ensure no blackout next night.
Lack of accus, and you might not have enough energy stored to prevent blackout.
And have excess of both panels and accus to account for variability, but keeping the same ratio is OK.

[creepercrack]

This is wrong, do not even bother to read it

The calculations for 0.16 (and the wiki) are wrong and there are more solar panles than you need.

=-=-=-=-=-=-=-=-=

Day time: 291.67 (1250/3) seconds
Night time: 125 seconds

Solar panel max power output: 60 kW
Accumulator capacity: 5000 kJ


- Now we need to know how much an accumulator needs to produce to last a full night and completely discharge:
5000 / 125 (accumulator capacity / night time) = 40 kW
- Then we need to know how much power we need to fill that accumulator over the duration of a day:
5000 / 291.67 (accumulator capacity / day time) = 17.14 kW
- Then we need to know how much energy need to produce over day time to keep that system stable:
That is because if a accumulator can produce over night 40 kW, we need to match that number and add how much we need to charge it over the day (hope this makes sense)
40 + 17.14 = 57.14 kW
- Now we only need to adjust that number to the production of a solar panel, and it gives us the ratio:
60 / 57.14 = 1.05
- So the ratio is 1.05 accumulators per solar panel
- 21 accumulators per 20 solar panels (simplified) in 0.16

=-=-=-=-=-=-=-=-=

I have an excel to automaticly calculate the ratio for any release of the game (you only need to set the correct numbers): "https://mega.nz/#!NFpDCKJY!e1NquCd5SJ4r ... Tl52QiGA2k"

[TruePikachu]

You are forgetting that the day/night power graph isn't a square wave

Code: Select all

100%---      ########      ########
             #      #      #      #
             #      #      #      #
             #      #      #      #
  0%---#######      ########      #######

but rather has nonzero time to transition between day and night

Code: Select all

100%---        ####          ####  
              #    #        #    # 
             #      #      #      #
       #    #        #    #        #    #
  0%--- ####          ####          ####

(graphs not to scale at all, the transition period for the second graph actually extends into what's generally considered nighttime, while daytime is at full power)

I've experimentially verified that the 0.84 ratio applies in 0.16 still, by successfully migrating my solar-powered radar outpost over.

[TruePikachu]

To elaborate, one method that can be used to compute the ratio is as follows:

Using numbers from here, we can derive a function that outputs the amount solar power we get from the tick in the day (using τ as a symbol for ticks):

Code: Select all

     ⎧       1  kJ/τ            0τ≤t<6250τ
     ⎪(11250-t)/5000  kJ/τ   6250τ≤t<11250τ
S(t)=⎨       0  kJ/τ        11250τ≤t<13750τ
     ⎪(t-13750)/5000  kJ/τ  13750τ≤t<18750τ
     ⎩       1  kJ/τ        18750τ≤t<25000τ

Integrating over the length of a day yields the power generated over a day/night cycle:

Code: Select all

⌠25000 τ
│  S(t) dt = 17.5 MJ
⌡0 τ

Dividing this by 25000τ yields 700J/τ continuous output possible. We can now create a function for how much power needs to flow out of our accumulators for this output:

Code: Select all

A(t)=700J/τ - S(t)

Solving for A(t)=0 yields the two brief moments where the accumulators aren't being charged or discharged; these are t=7750τ and t=17250τ.
Now, we check how much power the accumulators need to be able to store:

Code: Select all

⌠7750 τ     ⌠25000 τ
│ A(t) dt + │ A(t) dt = -4.2 MJ
⌡0 τ        ⌡17250 τ

⌠17250 τ
│ A(t) dt = 4.2MJ
⌡7750 τ

During the time where the solar panel has more power than the array outputs, it can charge the accumulators with 4.2MJ; during the time where the solar panel isn't sufficient to provide the 700J/τ, the accumulators discharge that 4.2MJ. Hence, you need at least 0.84 accumulators per solar panel. ∎

Here are graphs of S(t) and A(t):

06-30-2018 Image002.png (18.63 KiB) Viewed 12190 times

[bobucles]

Indeed the shape of the power generation is VERY important to the math. Consider 3 types of light conditions:
1) A massive flare that generates all the daily energy in one burst, and is dark the rest of the time.
2) The current vanilla one
3) A perpetual dusk that generates 42kW all the time

With the first option you need a huge number of accumulators to store the energy burst and distribute it the rest of the day. It's also pretty clear that with a permanent steady flow of power you don't need ANY accumulators at all. The amount of energy storage you need is based on how far the production deviates from the average. Dawn and dusk solar power is much closer to the ideal 42kW, so the amount of needed energy storage goes down when compared to a square wave day cycle. That's why the ratio is closer to 1.2 solar panels instead of being nearly 1:1.

[MisterSpock]

25 solar to 21 accu is perfect ratio for a 1mw plant. And its calculatable. Its not that difficult.
I use a 25 to 21,875 ratio: (featuring anti OCD-Measures)

[creepercrack]

This is wrong, do not even bother to read it

You are forgetting that the day/night power graph isn't a square wave

Code: Select all

100%---      ########      ########
             #      #      #      #
             #      #      #      #
             #      #      #      #
  0%---#######      ########      #######

but rather has nonzero time to transition between day and night

Code: Select all

100%---        ####          ####  
              #    #        #    # 
             #      #      #      #
       #    #        #    #        #    #
  0%--- ####          ####          ####

(graphs not to scale at all, the transition period for the second graph actually extends into what's generally considered nighttime, while daytime is at full power)

I've experimentially verified that the 0.84 ratio applies in 0.16 still, by successfully migrating my solar-powered radar outpost over.

I forgot to explain that I didn't forget about it not beign a square wave, what I did is because the wiki says that the light linearly increases from no light to light (and from light to no light), so I did this to simplify the math:


I have tested it with radars, and it works perfectly, but with 25 panles, it's too much panel heavy

Tha accumulator charge goes all the way up and all the way down perfectly

[TruePikachu]

Well then consider what bobucles said, the shape of the curve is extremely important. Note my rendered graph; the amount of time that the accumulators need to be responsible for powering the base is shorter in reality than if you model everything with a square wave.

EDIT: My solar outpost uses seven solar panels and five accumulators for a single radar (and only that single radar), which gives it enough power to continuously show the nearby area (though it browns out enough in the morning that it slows down the long-range charting). You can keep it 100% powered with eight solar panels and six accumulators, which is at a 0.75 actual ratio (more solar panels than needed). If your math were correct, then the 300kW needed for full power on a radar needs 7.142857->8 solar panels and (by your ratio) 7.5->8 accumulators. However, only 6 accumulators are needed...and exactly six accumulators (oddly enough), for a single radar. If you make an electric network with six accumulators, one radar, and 8 solar panels, you should notice that the accumulators run out of power at the same time that the solar panels can provide enough energy to run the radar. Fewer solar panels, and you either can't run the radar in the day, or it won't last the entire dusk/night/dawn; fewer accumulators, and it won't last the dusk/night/dawn; more accumulators, and the extra capacity will be wasted. This experiment will prove that the ratio is somewhere between 6:8 (0.75) and 6:7 (0.857143), and that your proposed ratio of 21:20 (1.05) is far too accumulator-heavy.