Accumulator / Solar panel ratio

[jockeril]

I'm looking for a good solar farm that uses the latest ratio relevant to 0.12.4 (should be 23:25 right ?)

Can anyone give me a link to a post or picture of one ?

[sinsiliux]

Where did you get that information from?

[jockeril]

Where did you get that information from?

If that question was for me, then read the post from the start...

[Ojelle]

Where did you get that information from?

If that question was for me, then read the post from the start...

Can I upvote this?

[jeroon]

I'm looking for a good solar farm that uses the latest ratio relevant to 0.12.4 (should be 23:25 right ?)

What has changed in 0.12.4? Isn't it still 21:25 ("It does mean that for 100 solar panels, we need 84 accumulators" from the first post)?

[PandaGOD]

I would love to know if this desing is still the best at the moment of this post and version of the game (i use beta version 0.12.16, but should be the same result for 0.11 i assume)

[Absolute]

Used this information to design mine own solar plant. First tried to make it from scrach around single roboport, but failed. Then more or less copied the scheme from the OP and began rebuilding it. Took me a while to understand that the main rule is that 4 panels form a squarer, identical to 9 accums, and 4 accums are as whide as 2 panels and 1 accumulator. Using this principles and the fact, that 4 panels exactly fit between two substations, I managed to crate this. Nice scalable sysytem with overlaping edges and none free space over the entire roboport logistical cover area! The ratio is a bit high in single block, but slightly lower than 0,84 when part of an "infinite" filled area.

[Jackalope_Gaming]

I got looking at the 25 to 21 ratio and wondered if something similar could be done. Eventually I found that 6 to 5 is 83.333, which is less than 1% relative and absolute to the 84% number. 6 solar to 5 accumulators can be done with a single medium power pole. Here it is:



Just make sure all of your tiles are oriented the same way and you're good to go. If they aren't positioned the same way then you run the risk of the poles not connecting up because too much distance is between them.

Due to the position issue it's not advisable to try and make walkways using that interior space.

[Hexicube]

This result only apply when solar energy is the only source of electricity production. Otherwise, other sources of electricity production will have the priority on accumulators. In this case, accumulator can provide a temporary increase in power when needed (for instance for laser turret during an alien attack) and their number is not calculated this way and generally much lower.

Actually, the result is always useful. If you're producing steam power, you can just subtract a flat amount of demand and do the math as normal. Might be worth mentioning that in case it's not obvious to anyone wanting to make a hybrid system.

[JasonC]

Btw, here's how to calculate the number of solar panels you need for a given power requirement including what you need to charge your accumulators, it's really straightforward. First, as a constant, here is how much energy a single solar panel can produce in a single cycle (recall trapezoid shape from OP's diagrams for explanation of why we halve dusk and dawn time):

• A solar panel outputs 60 kW power.
• Effective output time in game seconds is Tday + (Tdusk + Tdawn) / 2, where the times are in game seconds. So that's currently 208.33 + (83.33 + 83.33) / 2 = 291.67 seconds.
• Therefore a solar panel can produce 60 kW * 291.67 seconds = 17.5 MJ per cycle.

I don't think this was mentioned in the OP but Joules are the unit of measurement of the area of those trapezoids. 1 J = 1 watt * 1 second (Joules measure the same thing as kilowatt-hours). That's why accumulator capacities are measured in MJ.

Anyways, a solar panel can output 17.5 MJ per cycle. So to calculate how many solar panels you need, you first calculate how many Joules you need in a cycle. A cycle is 416.66 game seconds so:

Energy = Power * 416.66

Then divide by how much energy a solar panel can provide and that's it:

Panels = Energy / 17.5

So the full equation is:

Panels = Power * 416.66 / 17.5

Then you just use the 21:25 ratio to figure out accumulators.

For example, if you need 2.1 MW continuous power output:

2.1 MW * 416.66 seconds = 875 MJ energy required
875 MJ / 17.5 MJ = 50 solar panels

That means for 2.1 MW output you need 50 solar panels and 42 accumulators.

The idea here is that energy is energy, no matter when you generate it. If you need some amount of power for an entire cycle (day and night) then you need to generate that amount of energy, period. Since solar panels only generate during the day you have to have enough solar panels to generate that amount of energy while it's light out. That's what the above calculates. The accumulators are just a means to move that extra daytime energy into the night, but the total energy is always the same regardless of how many accumulators you have.

One more example for the road:

• You want to power a factory with a 60 MW continuous demand.
• You'll need 60 * 416.66 / 17.5 = 1429 solar panels.
• You'll need 1429 * 21 / 25 = 1200 accumulators.

If you don't want to do math or remember numbers just remember it's roughly 24 (23.81) solar panels per megawatt needed. If you really don't want to do math and don't mind making a few extra solar panels and accumulators you could just call it 25 panels per megawatt and have a nice easy 21:25:1 accumulator:panel:megawatt ratio to remember.

Alternatively, Jackielope's cool little square layout above has 6 panels : 5 accumulators, which is close enough, and works into the 24:1 panel:megawatt ratio quite nicely, so you could also have 4 of Jackielope's blueprints per megawatt, i.e. 20:24:1 accumulator:panel:megawatt.

[zytukin]

Either things have changed since topic started or the math is off (or I screwed up when monitoring my factory one day)
I had found that a ratio of 1.2 to 1.5 accumulators per solar panel worked out perfectly fine.
The ratios posted result in more solar panels then are needed to power factory and charge accumulators

I had posted this in a topic on the steam forum a week ago when somebody was asking about solar panels and another person brought up the 0.8/1 ratio.

If I may use my factory as an example. I know it's not 2:1 like I mentioned above, but it's still off from that 'ideal' ratio
9.1k solar planels, 11k accumulators, a ratio of about 1.2:1
9,100 solar panels vs 0.8 ratio = 7,280 accumulators

Would that power my factory overnight?
My factory uses 350MW. With 11k accumulators, I get about 75% drain.
7,280 accumulators is 66% of my current amount. But I can only afford to lose 25% of my accumulators based on current usage, which means that a ratio of 0.8 accumulators to 1 solar panel would not work, I would run out of power overnight.

So, go the other direction and build more solar panels to preserve the ratio?
Following that 'ideal' ratio, I would need 13,750 solar panels for my 11k accumulators
But, the 9.1k solar panels puts out 547MW while my factory only uses 350MW and they can fully charge the accumulators in 3/4 of the daytime.

So, that 'ideal' ratio is not so ideal.
Instead, it resaults in building far more solar panels then are needed.

[JasonC]

Either things have changed since topic started or the math is off (or I screwed up when monitoring my factory one day)

I can tell you that I just today came to this topic skeptical, tested with the ratios here on 0.12.29, and can confirm that they're still dead on.

I tested with 7 radars = 2.1 MW draw, 50 solar panels (as per my previous post), and 42 accumulators (as per OP), and it was so accurate that the accumulators hit full charge the moment they stopped charging and hit empty the moment they started recharging.

Maybe you ran into an issue with rounding or uneven power draw? I chose radars to test because they're ideal, always flat. I chose 7 because 2.1 MW requires 49.99 solar panels, so ideal for avoiding rounding.

[zytukin]

Either things have changed since topic started or the math is off (or I screwed up when monitoring my factory one day)

I can tell you that I just today came to this topic skeptical, tested with the ratios here on 0.12.29, and can confirm that they're still dead on.

I tested with 7 radars = 2.1 MW draw, 50 solar panels (as per my previous post), and 42 accumulators (as per OP), and it was so accurate that the accumulators hit full charge the moment they stopped charging and hit empty the moment they started recharging.

Maybe you ran into an issue with rounding or uneven power draw? I chose radars to test because they're ideal, always flat. I chose 7 because 2.1 MW requires 49.99 solar panels, so ideal for avoiding rounding.

hmm...it is possible. My factory's consumption can fluctuate quite a bit, I've seen it change by as much as 100MW depending on what assemblers/furnaces are running. Could have just been a higher usage time shortly before I typed that post.

[mooklepticon]

These calculations don't usually include laser spikes. Some days, the backup steamies are in full force because there are sooooo many biters.

[zytukin]

I also wonder how much surplus energy production it accounts for.

If you make enough to have a surplus (ie, make 2-3 times more then you are using), then you wont have to worry as much about having a proper ratio, you could easily have more accumulators because the excess solar plants can charge them with ease to offer better coverage for spikes.

[JasonC]

I also wonder how much surplus energy production it accounts for.

If you make enough to have a surplus (ie, make 2-3 times more then you are using), then you wont have to worry as much about having a proper ratio, you could easily have more accumulators because the excess solar plants can charge them with ease to offer better coverage for spikes.

Yeah, you'll just have to tweak it for laser spikes. If you wait for like a 1 or 10 minute period of heavy activity, you can look at the power graph and switch it to 10 minutes or whatever. Then add up the numbers in the production column for an average power need over that time (the little green bar at the top always shows current usage but the consumption / production numbers show the average over the selected time).

Then you can use the 5:6:1 accumulator:panel:megawatt ratio, and maybe throw in a couple more 5:6 accumulators:panels to taste if you still find you're hitting heavy spikes.

2x or 3x as much as you are using might be overkill, but it really depends on how often you get attacked - the real goal is to have enough energy stored in accumulators for heavy spikes, and also remember accumulator max discharge rate is 300kw - but never hurts to plan for future expansion.

[mooklepticon]

2x or 3x as much as you are using might be overkill, but it really depends on how often you get attacked - the real goal is to have enough energy stored in accumulators for heavy spikes, and also remember accumulator max discharge rate is 300kw - but never hurts to plan for future expansion.

I aim for 33% buffer. I find my current MW usage and then divide by .75 to find my assumed peak and build to that amount.

[zytukin]

Also depends on how much your planning on expanding.
I mean, build a massive setup once vs over and over as you upgrade.

[WarDaft]

Hmm, my math seems completely out of line with everyone else's on this, and I am still working on making my first battery.

Maths:
Day cycle is 416.66 seconds
Daylight is 208.33 seconds
Night is 41.66 seconds
The remaining 166.66 is half and half.

So the day is 291.66 s effective light and 125 s effective dark or equivalently 70% full brightness, 30% full darkness.

And so solar panels are 70% efficient overall and generate an average of exactly 42 kilowatts - this agrees with other folks' number of 23.81 panels/MW.

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 suppose I must start making batteries and find out.

[Koub]

So the day is 291.66 s effective light and 125 s effective dark or equivalently 70% full brightness, 30% full darkness.

This is where you get it wrong. It's correct to calculate the average output of your solar panels.
It's incorrect to calculate the solar panel / accumulator ratio, because what you assert (without saying it) is that at exactly half way between the last tick of full light and the first tick to full night, you switch from solar power to accu power. This is false, and it's what makes your calculations false.