solar day night cycle info needed

[inzain]

how long is the night cycle, im setting up a solar setup, to power my factories this game, and each accumulator outputs 300w, and stores 5kj(which is basically the same as saying 5kw is what it has to deplete i think yes?) so each accumulator would output 300w for 16.6 seconds before depleting its 5kj reserve, if my assumption of 5kj=5kw only time dependant. so if i had a 5kw using factory, and night was 120 seconds(just for calculations) i would need 17 accumulators per 16.6 seconds of darkness. so in total to cover 120 seconds of darkness i would need, 123 accumulators. or so my math states, so long as im correct about the way kj convert to kw in math. its been years since i messed with electrical stuff hehe so anyone able to help if im incorrect, and anyone know how long the cycles are? thanks in advance. and yes i know this will not cover turrets, or any abnormal stuff, i plan to build a ton to cover them also, just need the basics covered first, thanks again for any info.

note* i used the formula E(kJ) = P(kW) × t(s) and it showed i needed 600kj of energy consumption for the 120 seconds, for 5 kw sustained consumption. so yeah this is getting way too complicated lol, a simple, you need x amount of accumulators to sustain x amount of kw though the night would be awesome!!! lol

in game it seems 5kj =50 seconds of 100kw power. according to this, to supply 5kw i would need 50 accumulators per 50 seconds. so i still am lost lol. ill assume the bleed in effect of night and day will cover that 10 seconds, with solar covering half, and accumulators the other, ill try it.

[Dysoch]

From my own experience i have got to the assumption that the night cycle is between 30 to 40 seconds (this is the time the solar panels are off)

Ofcourse you have the minute before and after that the solar output is low.

[Tami]

You can use that statistics of your network at 10 minutes. You will see something like this picture.
A full day cycle is around 416 seconds, full night is 44 seconds, now use the statistics to see, if you need more panels or accus.

The accurefilltime is around 146 seconds with my deployments.

[inzain]

very awesome information thanks a lot, my question is how do you know the length of time on yer chart, i cant find any thing on mine to help determine it? is there a way? or are you just guestimating, how long each thing is in your picture, based on measuring its length in relation to the full days length of 416 seconds? or did you do like i did lasntight and break down using a stopwatch and watching the display lol ?

would be awesome if they added a timeline in seconds, and a sidebar of a numerical amount of energy used along the edges.

[Tami]

ive used ms paint, the full length of the window with graphs as a littlebit less then 600, around 593, so 1 pixel is around 1 second, if you have choosen the 10 minutes graph. The 416 seconds is the ime, when the sun reachs its 100% state, every 416 the suns reachs its 100% state.

But you have keep in mind that some structures have low power usage while idle like laser turrets, smelter, etc ...
You can use the green book level 1 2x for -80% energyusage and pollution.

[Dysoch]

very awesome information thanks a lot, my question is how do you know the length of time on yer chart, i cant find any thing on mine to help determine it? is there a way? or are you just guestimating, how long each thing is in your picture, based on measuring its length in relation to the full days length of 416 seconds? or did you do like i did lasntight and break down using a stopwatch and watching the display lol ?

would be awesome if they added a timeline in seconds, and a sidebar of a numerical amount of energy used along the edges.

He isnt quessing. Its really simple.
Take your electric network info, and put it on 10 minutes. Make a screenshot.
Then simple add it in paint, and divide the bar in 10 pieces (1 minute each). You then already have more exact info.

For even more exact info, divide the 10 bar in 20 pieces (30 seconds) or even 40 pieces (15 seconds each)

This is of course a crude way to find out, but it works.

[n9103]

Once you have an absolute length that matches an absolute time, then the rest is simple math.
The absolute length was given by the size of the graph being the same for all measurements.
Absolute time of that length was given by the scale (10m=600s).
You could make it easier to come up with numbers, and have your absolute length become a maximum length of 600 units, adjusting all lengths by the same ratio you used to get to 600 units. Using pixels for your unit is simplest. Simple way to do this is to just have your image editing software shrink/stretch to an absolute length. (Paint.NET is good.) At this point, the unit length of any line is the actual duration in seconds. You incur some inaccuracy in your measurements, as you are now measuring in a scale that isn't the same scale used to generate the raw data; However, for gaming purposes, it will be more than accurate enough.

If you wanted a more accurate method, you would measure the main length, and then the sub-lengths, and find your ratio, and then multiply the absolute time by that ratio to find the amount of time that sub-length corresponds to. This method can be the most accurate, but relies on the accuracy of your measurement and measuring tool, rather than simple counting of units.

[wrtlprnft]

If you want to know the exact timings, you can also look at these two posts by kovarex: length of a game day in real time and length of the night.

According to these, an in-game day is 416.6 seconds (so the pixel arithmetic seems to be pretty accurate), and assuming a day has 24 hours, a night takes 138 seconds (I guess this refers to the period where the sun is not shining with 100% brightness). The completely-dark part is then probably 2.5 in-game hours or 43.3958333 real seconds.

[rk84]

I got 43,055 s of "zero solar power" -time.
In testing, I used game.daytime in console and monitored power output of solarpanel.

Difference to "43.3958333" is probably because output did not die as soon as darkness was max.

[inzain]

Thank you all very much, exactly the info I needed, now there won't be any guesswork, I can mathematically figure the maximum and minimum electric storage needed in accumulators to make it through the night by simply adding my total usage, to the possible max usage( by adding everything I lay down to a tally) and making sure my accumulators store that much or close for 138 seconds, which can also be mathematically calculated. Rather be precise, than guesswork, of adding in one accumulator, then watch see what happens etc.. This way I can be sure, I'm good. Thanks again for all the information. It is much appreciated.

Ps the whole point is to run a green setup, no steam engines at all. here goes nothing.

[Psycho0124]

Sounds risky. One nighttime biter swarm could leave you without power and helpless.
I'd really consider keeping at least a few steam engines on standby in case things get dicey.

[ssilk]

This is about the same I want to answer.

Works only, if played in peaceful mode. Otherwise a horde of constant attacks can easily drop you out. But yes, if there is a formula, how much cells needed, I would appreciate that.

[inzain]

yeah it can get dicey, what the plan is, is to monitor consumption max's over a one night period, multiple nights on my current game (which i plan to use the same laser configuration of 14 turrets per section), and get an average of how much energy, my turrets use, to defend over a night, then multiply that by a minimum of 2, build a seperate power grid, of accumulators, and solar panels, and use it separated for defense to power them. ill have a pretty close estimate of how much accumulator stored energy itll take for a night, and being separate from my main factory, it will always have ample power as nothing will use it except when firing. but again we will see, i plan to put all this to the test come friday when the new patch releases. so cross some fingers and wish me luck


edit***
ps the math works out to be 100j/second supplies 100w this is only accumulator energy, it is not counting secondary power for the time the solar panels are supplying partial energy.

time wise, one accumulator supplies 50 seconds of power to 100w usage. this is way short of the needed time to supply this overnight.

with the ramp up time of approximately 47 seconds of partial solar, then 43 seconds of 0 solar, followed by another 47 seconds of partial solar, then 100% solar you need to supply 100j/s for 138 seconds as 1 accumulator under these conditions supplies 50 seconds then to fully support 100w for the duration of the night it would require approximately 2 accumulators =100 seconds of solid 100j/s power output, allowing for the solar drop, of 20 seconds prior to 0 solar, and 20 seconds after 0 solar time, where solar output would be at less than 50%, which leaves 7 seconds of more than 50% solar at both intervals totaling 14 seconds of >50% solar. this of course goes into account of having approx 5 solar panels per 1 accumulator(atleast thats what i used in my test)

so using 5 solar panels per accumulator, i will need 2 accumulators per 100w of power consumption, per night so a grand total of 20 per KW, or 100 per 5KW and 500 solar panels, so a typical 15Kw consuming setup (which is what mine uses now no counting laser fire) would require 3 times that, so 300 accumulators and a whopping 1500 solarpanels , remember though, using more accumulators would relieve the need for so many solar panels, the above 1500 produces 90kw at peak time, but IS required in order to keep the accumulators from emptying prior to sunup during the total of 95 seconds or so of partial solar. again doubling or even tripleing accumulator total would reduce this a great deal. goodness this is getting ridiculous, hehe half my map would be accumulators and solar panels according to this math. seeing as to how pollution cleans itself up over time, i think i might be using some fill in steam engines at night, with enough solar to cover me during the days, and accumulators to help take some of the bite off of it at night hehe. ok i surrender. i will be using some steam engines!

for the very basic formula, 3 FULL accumulators totaling 15KJ storage, would support 100w consumption over a full night of 134 seconds. regardless of solar output and partial solar output, so following this, i would only need 250 solar panels to supply the needed 15 KW power to my 15 KW demand during the day, and 450 accumulators. adding in enough extra solar panels to also recharge the accumulators (lets guess at 100 for safe keeping) totaling 350 solar panels, and 450 accumulators to supply ample power production.

so tldr: 30 accumulators + 16 solar panels, +additional panels to recharge said 30 accumulators, would be all thats needed per 1KW power consumption.

the above was derived from ingame testing, and seems to be correct, please keep in mind i am no mathematician and it is very possible i am incorrect somewhere in the above statements. lol but it seems to work on my game, so im happy with it.

[rk84]

This is about the same I want to answer.

Works only, if played in peaceful mode. Otherwise a horde of constant attacks can easily drop you out. But yes, if there is a formula, how much cells needed, I would appreciate that.

Well OP does give the formula.

E(kJ) = P(kW) × t(s)

Divide it with accumulator capasity to get count of needed accumulators.

P = your average power consumption (in kW)
t = 126.5s = ~43 s + ~167s / 2 (that is midnight time + the time that solarpower is lineary decreasing/increasing between 0 and 60 divided with 2)
Ca = 5 kJ (accumulator size)

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(P*t)/Ca = count of needed accumulators

P*25,3 = count of needed accumulators (little over 25 accumulator per 1kW)

You just need safe estimate for average power consumption. It can be tricky, because random attacks. I haven't played around alot to know where the safe line is, but I can make some experimental suggestions:
Here P is the average that you factory uses without attacks.

• Roughly picking safe factor. example P*1.5 for 50% extra energy in store.
  Or
• Calculating the energy need by estimating how many shots you want for laser turrets.
  Easy way: Roughly pick some ammo count you want to reserve per laser turret (One accumulator is like laser turret magazine of 25 shots)
  Hardway: Estimate size and composition of attacking biter force, calculate the needed damage to kill the wave (small_count*15 + medium_count*75 + big_count*375), calculate needed shots (damage/(turret damage*laser damage modifier)) and finally calculate needed accumulators.

  Code: Select all

  (P*t+shots*0,2kJ)/Ca = count of needed accumulators

[ssilk]

Some days ago: created enough solar panels/accu to have 10 factories online. Loading all accus took about 20 seconds. But !

I had a fight. Mother of fights. A big concentration of worms/spawners and it eats my power reserves down to only 10%. Never thought that, and I had been warned. But they did.

Conclusion: cannot be estimated. 50% reserve is too low. And if some says he has a formula, he lies. sorry rk84, no offencement. But I really think the estimations are only estimations in the ideal case, if nothing happens. But what game would that be?

[rk84]

Some days ago: created enough solar panels/accu to have 10 factories online. Loading all accus took about 20 seconds. But !

I had a fight. Mother of fights. A big concentration of worms/spawners and it eats my power reserves down to only 10%. Never thought that, and I had been warned. But they did.

Conclusion: cannot be estimated. 50% reserve is too low. And if some says he has a formula, he lies. sorry rk84, no offencement. But I really think the estimations are only estimations in the ideal case, if nothing happens. But what game would that be?

None taken. You are right. long fights can really dry your power grid. I think I will make separated power plant for my offensive turrets. And yea that 50% was bit of random selection, its relative on how big your factory is and if you are using lasers in offensive you need bigger. Mayby I can make formula for battles. Hmmm spawner count, spawn rate... Nah I better stop and go to sleep.

btw in my last fight I used only capsules. Distractor and poison. Against 8 spawners.

[ssilk]

Distractor is really powerful, but although I tried it now many dozen time I could not manage to use more than two weapon at a time.

What I wish is a primary and a secondary weapon, like in every shooter. It is not realistic to have more than a dozen weapons.
Primary and secondary weapons can be configured with the stacks on right side, about like now. You can configure two primary weapons (pistol and shotgun for example) and up to 5 secondaries. Key 1 switches between primary weapons, 2-6 between secondaries.

[rk84]

I'm back with fool proof formula. I dunno why I'm trying to be a miser with power reserves

Pf = avarage power consumption of factories
Pl = 0,6kW * laser turret count (power consumption of lasers when all laser are shooting non-stop)
t = 126.5s (low/no solarpower time)
Ca = 5 kJ (accumulator size)

needed accumulators for night

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(Pf+Pl)*t / Ca = count of needed accumulators

And solar panel count. And again lasers shooting non-stop day and night.
Ps = 0.06kW (solar panel output)
r = 126.5s / 416,7s = 0.3036 (night/day -ration)

Code: Select all

(Pf+Pl)*(1+r) / Ps = count of needed solars

hmm I have this one factory that uses 22kW currently and I have 653 laser turrets so...
(22kW + 391.8kW)*126.5 / 5kJ = 10469.14 count of accumulators. nice
(22kW + 391.8kW)*(1.3036) / 0.06kW = 8990.49 count of solars. cool

It seems I'm lacking alot of stuff if I want to organize non-stop laser party.

[ssilk]

Erhm.. 539 kilowatt?

Btw. Wasn't this planned, to replace watt with kilowatt?

https://forums.factorio.com/forum/vie ... f=6&t=1109

[rk84]

Erhm.. 539 kilowatt?

Btw. Wasn't this planned, to replace watt with kilowatt?

https://forums.factorio.com/forum/vie ... f=6&t=1109

I dunno. mayby they will revisit values in future, if they make new system for pipes/steam engines?
https://forums.factorio.com/forum/vie ... f=5&t=1515