Optimized Steam Engine Setup

[Patric20878]

I don't remember the names of the bars, it's whichever bar meets the description. It changes color when it drops, and it corresponds to how much of the demand is met, and when it drops everything slows down. Again, those engines would never have been able to fully power 600 radar. Not even operating at their peak, 510 kW each, not even remotely, distantly, vaguely close to being able to power 600 radars.

That'd be the consumption bar then. And like said, the only reason I used 600 radars is because of the 1-11% threshold numbers you gave with burner inserters. So to test those numbers, I made the consumption bar 1-11% of full also. From tests, my closest guess as to what you mean by threshold is that without burner inserters, production drops to 0 since there isn't enough burner inserters to feed enough coal to even power the normal inserters. And with enough, it no longer drops to 0, but the production is still awful. Only with all burner inserters, does production not drop AT ALL. Yes, you will never attempt to power 600 radars with 19 steam engines. But can you tell me you'll never have a single instance where that consumption bar drops for 60 seconds? I can think of many reasons why it'd drop longer than that. Laser turret pushing was just one of them.

And yeah, you ever toggle enemy base setting to max? That's what it's like. Some those mobs take an absurd number of hits to die. Inserters don't have a buffer like boilers do, so they slow down soon as turrets start firing.

Again, you need to clarify what that 11% number really means. Was this just out of academic interest or what? If that number is about the consumption bar, it would never drop that low, even if not careful with production. If there's any number important, it'd be how low that consumption bar can go before inserters can't keep up, i.e once per 10 seconds.

[Aru]

I don't remember the names of the bars, it's whichever bar meets the description. It changes color when it drops, and it corresponds to how much of the demand is met, and when it drops everything slows down. Again, those engines would never have been able to fully power 600 radar. Not even operating at their peak, 510 kW each, not even remotely, distantly, vaguely close to being able to power 600 radars.

That'd be the consumption bar then. And like said, the only reason I used 600 radars is because of the 1-11% threshold numbers you gave with burner inserters. So to test those numbers, I made the consumption bar 1-11% of full also. From tests, my closest guess as to what you mean by threshold is that without burner inserters, production drops to 0 since there isn't enough burner inserters to feed enough coal to even power the normal inserters. And with enough, it no longer drops to 0, but the production is still awful. Only with all burner inserters, does production not drop AT ALL. Yes, you will never attempt to power 600 radars with 19 steam engines. But can you tell me you'll never have a single instance where that consumption bar drops for 60 seconds? I can think of many reasons why it'd drop longer than that. Laser turret pushing was just one of them.

And yeah, you ever toggle enemy base setting to max? That's what it's like.

It has to not only drop, but drop below 11%, and it has to stay there for a sustained period. For example, because there's not enough coal. In such a circumstance, yes indeed, it will drop for a sustained period. It will probably drop to 0. That was the entire point, of the whole calculation, in the first place. And I have said this so many times, but I'll say it again, the calculations assume that the engines are capable of fully powering the base, when operating at their peak. That is, number_of_engines*510kW. I've been saying that from the beginning, as clearly as possible, as many different ways as I could think of. I have clarified it so many times. I don't know how else to explain it to you. If the engines can't power the base fully, then that's when I build more engines. Make sure that when you're playing, you don't accidentally drop 600 radars with less than 20 engines and then forget to add any additional power production.

And again... you need over 470 engines to power 100 laser turrets simultaneously anyway. I've never had that many engines, I use accumulators. There would be no point in dropping that many turrets at the same time, if you can't power them.

[Patric20878]

the calculations assume that the engines are capable of fully powering the base, when operating at their peak

So normally, the consumption bar is max, but in certain circumstances like running out of coal, it drops? Well okay. Running out of coal would cause it to drop to 0, no matter the number of burner inserters, and otherwise, I can only think of the bar not being full when the engines aren't capable of fully powering the base.

But anyways, I'm going to test just how low that bar can go before production drops. And I'm talking about permanent sustainability here. The lowest it can go before inserter rate drops to below the once per 10s rate it needs to pull.

[Aru]

the calculations assume that the engines are capable of fully powering the base, when operating at their peak

So normally, the consumption bar is max, but in certain circumstances like running out of coal, it drops? Well okay. Running out of coal would cause it to drop to 0, no matter the number of burner inserters, and otherwise, I can only think of the bar not being full when the engines aren't capable of fully powering the base.

YES! WOW! You see now, the point of the posts from 2 or 3 pages ago, where I was emphatically saying that powering the drills on coal are a bigger issue than the inserters on the boilers. And also see, perhaps, that I said over and over, that the calculations assume that you build enough power supply to fully power the base. The main significance of this threshold, which I only bothered to be concerned with and to calculate because you seemed to think it was more important than it really was, is that in order to get the engines going after they stop (without delay), they have to be able to meet a certain minimum ratio of supply to demand. (They also need to have access to coal.) If there's only burner inserters on the boilers, or if there's at least 2 burner inserters per 10 engines, then that minimum ratio is 0. If it's all orange inserters, it's 11%.

[Patric20878]

Well hey, if you said that first in plain english at the beginning of the post, without all the calculations in the way, and also mentioned that this kind of thing will basically never happen in game, it'd be understandable immediately, wouldn't it? And I could've sworn I said before, my engine setup is concerned with the steam engine setup ONLY, not with coal supply. Also, why wouldn't 1 burner inserter have 0 minimum threshold? If that causes any power to be generated, inserters would start up eventually, wouldn't it?

Either way, if that's what the number means, it has much less use than what l was thinking. I need to test for how low the bar can get before ANY production drop, not ALL production drop.

[Aru]

Well hey, if you said that first in plain english at the beginning of the post, without all the calculations in the way, and also mentioned that this kind of thing will basically never happen in game, it'd be understandable immediately, wouldn't it?

Either way, if that's what the number means, it has much less use than what l was thinking. I need to test for how low the bar can get before ANY production drop, not ALL production drop.

I did say that. Then I said it some more. If the drop is due to insufficient coal, which is the only reason why it might suddenly drop, then you can get it down to 11% (or lower) (with orange inserters). If you accidentally drop 600 radars or something like that, that would also drop it to 11% (or lower). The orange inserters won't do much because the power ratio stays so low, no matter how well the engines are performing, they're just insufficient to make up for the fact that you don't have enough engines to power the base.

As for how low the bar can get before there is a further drop of power production, when all boilers are fed by orange inserters... about 11%.

Yes, one burner inserter for 100 engines would eventually allow them to come back to peak, that should be obvious. But it could be very slow, if all the other inserters and mining drills are getting low power. So slow that it might as well not be there. The 2 burner inserters per row number, was how many it takes to raise the power ratio enough that the other inserters can fully fuel the steam engines, without further delay. I also said this before... repeatedly!! Also assuming that the engines are capable of fully powering the base, and assuming that there is adequate coal. If there's not enough coal, you can't fully fuel the steam engines. But as soon as there is, the 2 burner inserters will ensure that the steam engines will come back to peak power, as quickly as they can with the available coal.

[Patric20878]

Edited last post to ask why 1 burner inserter wouldn't have a minimum threshold of 0.

With one burner inserter and 13 electric, it's 6.44%. With two burners, it's 1.17%. Three burners, it's impossible. You only need 2 to make this impossible.

How does this not heavily imply that with 1, it's still possible? Now you're saying it's impossible, so you actually only need 1, but it's slow. You gave the number, 1.36, for the minimum number to make power dropping impossible, yet 1 burner inserter makes it impossible too. If you said it repeatedly, you need to explain this line. Clearly it's impossible for engines to drop to 0 if you have at least one burner inserter in the setup.

[Aru]

More specifically, it makes it impossible for a lack of power at one moment to impede the process of steam engines coming back to full power. The calculations weren't really emphasizing what happens if you have some burner inserters, but not enough to get the engines back to full power at full speed, but rather what happens if you have no burner inserters, and how many burner inserters you need to restart the engines from 0 without an arbitrary delay.

If you have one burner inserter on 100 engines, and the power drops low (not necessarily to 0), it will take a very long time to come back to full power. You might as well do it manually, rather than wait.

"And because 11% is already low, there's not much point doing this with just one or two burners, it should be three two."

[Patric20878]

And whether it's for 1 or 2 burner inserters, production only returns to max if the consumption bar is at least 11%, correct? Burner inserter count affects recovery speed, while consumption % affects how much % the production will be recovered, right?

Then first, if it's true that boilers should consume coal at a max of 8000/390*2 = ~10.2564 seconds, 11% is the minimum consumption % to keep inserters operating at that rate. Now say I need inserters to operate at double that rate, or once per ~5.13s. Does that mean consumption % must be at least 22% then?

With my design, I will actually need faster than once per 10.25 seconds, thanks to boiler chaining. There's currently 4 boilers in the biggest chain. Then I think 10.96%*4 = 43.84% of consumption would be the lowest it could go before dropping production.

[Aru]

Yes, one burner inserter for 100 engines would eventually allow them to come back to peak, that should be obvious. But it could be very slow, if all the other inserters and mining drills are getting low power. So slow that it might as well not be there. The 2 burner inserters per row number, was how many it takes to raise the power ratio enough that the other inserters can fully fuel the steam engines, without further delay. I also said this before... repeatedly!! Also assuming that the engines are capable of fully powering the base, and assuming that there is adequate coal. If there's not enough coal, you can't fully fuel the steam engines. But as soon as there is, the 2 burner inserters will ensure that the steam engines will come back to peak power, as quickly as they can with the available coal.

The only reason to use burner inserters, is that they make this restart process entirely automatic when coal supply is restored after being cut off. So you don't have to add coal yourself to restart without delay. Originally, you were talking of all burner inserters, so I calculated how many you need to meet this reason. My designs don't have any burner inserters.

Yes, they affect recovery speed, ranging from impossible to full speed with no delay. The 2 per row, technically is the minimum needed for no delay. Or rather 1.3 ish or whatever the number was.

And, the 10.96% was not quite what's needed to fully power the boilers, but what's needed to fully power the steam engines, which is close but slightly less. The last boiler doesn't burn at the full rate.

[Patric20878]

If I can find a way to minimize boilers per chain in this setup, production dropping due to low consumption bar will no longer be an issue. I'm starting to understand why I had the problem with laser turrets dropping production, where you didn't. You didn't chain boilers. I did with the old setup, often in chains of 9, so the minimum % before production drops was much higher for it than yours, potentially as high as 100% depending on how many times the old setup is stacked. I'll have to see again if I can compact the belts the same way you did in that pic, in groups of 2 with no boiler chains, as opposed to my current design, with the topside of the belt not being chained and the bottomside being chained in groups of 2 and a few at 3 or 4.

Oh and since I now know that threshold is about recovery speed rather than ability to recover, logically then, the inserters must be available immediately from the belt rather than delayed, from being in a boiler chain somewhere, otherwise it causes delay.

And since this is all about delay, I wonder if it's just better to save the 6s or whatever delay it is and just place them at the beginning of the belt. If 14 boilers use 1.75% more coal with 14 burner inserters (not factoring the 14th boiler's slower consumption rate if that makes a slight difference), then 2 burner inserters would use like, 0.25% more coal. Leaning towards this idea, surely you can produce way more in 6s than you can save in coal in a long period, no? Like, in the same time it takes to mine 10000 coal, you either save 25 coal, or you produce some amount in 6 seconds. I think in virtually all stages of the game past early game, you'd produce more than 4 items a second (not to mention each item costs multiple raw materials), heh.

Hmm, actually no, I'd need to know how big the average coal field is. 10000 seems a bit too small for a coal field. With help of the map editor's brush tool to remove ore by area (or just straight up adding all the tiles, but that's tedious), I'm estimating somewhere in the 100k to 1m range. Then if I use 500,000 as the average, it's 1250 coal saved vs some amount produced in 6 seconds. In late-game, should easily still produce more than 1250 raw items worth of stuff per 6 seconds. But not sure about mid-game. If 3 raw items go into crafting something at the basic crafting level, that's 70 items/s. That's like what, 30-40 assemblers?

[Aru]

I see. Because the first inserter has to transfer enough coal to fuel several boilers, it is inhibited at a higher power ratio. You could switch it for a fast inserter. I guess you already know that though.

Oh and since I now know that threshold is about recovery speed rather than ability to recover, logically then, the inserters must be available immediately from the belt rather than delayed, from being in a boiler chain somewhere, otherwise it causes delay.

If I'd known that distinction was a factor in your current designs, I would have considered more carefully that the "impossible" condition is met with any number of burner inserters. But I was focusing on the minimum needed to prevent delay in restarting.

And since this is all about delay, I wonder if it's just better to save the 6s or whatever delay it is and just place them at the beginning of the belt. If 14 boilers use 1.75% more coal with 14 burner inserters (not factoring the 14th boiler's slower consumption rate if that makes a slight difference), then 2 burner inserters would use like, 0.25% more coal. Leaning towards this idea, surely you can produce way more in 6s than you can save in coal in a long period, no? Like, in the same time it takes to mine 10000 coal, you either save 25 coal, or you produce some amount in 6 seconds. I think in virtually all stages of the game past early game, you'd produce more than 4 items a second, heh.

Hmm, actually no, I'd need to know how big the average coal field is. 10000 seems a bit too small for a coal field. Happen to know?

I don't, but you can use the YARM mod to find out quickly. It introduces a research, which allows you to craft an item, which you can use to track the size, drain rate and remaining fraction of contiguous resource patches. You click on a resource patch and it tells you the total. If putting the burner inserter at the front improves the design, do it, the difference in coal consumption isn't enough to warrant hindering the design.

[Patric20878]

Yeah, that's why the old design used fast inserters at the belt. Since it then uses a long inserter to pull after that, there was no point making the 3rd inserter row also fast inserters.

But since I am placing an emphasis on tech/resource efficiency this time around, a much better solution as mentioned is just keeping boiler chains as small as possible. Makes for a slightly funny setup though, with steam engine rows in groups of 3 and inserter rows in groups of 2, but that's probably an aesthetic concern at best, so isn't all that important.

And cool. I'll be trying that mod then. With its help, it'd be easy to compare coal saved vs production hindered. And yeah, by my intuition, there's no way 0.25% more coal used could ever beat 6 more seconds of production in mid-late game, but I want to make sure heh.

And I should mention, these 4 pages of discussion offered great insight on how to improve steam engine setups on a deep level. So, thanks. Didn't expect there'd be so much more to improve besides the initial stackable, unchained belt feeder design.

[Patric20878]

And it's done! One thing led to another, leading me to finding how to compact 19 steam engines in a 50 by 6 area, and with only small poles! So now just like your design, it takes no research beyond logistics 1, heh. Would've been 29 engines in a 50x9 area too like how it used to be, but the new belt feeding system makes that basically impossible without losing compactness or adding not-so-neat inserter and boiler chains. And fortunately, the lone boiler chain on top left of each 2 rows goes burner inserter to boiler to basic inserter to boiler, and since burner inserters aren't affected by low electricity, this would not count as an actual boiler chain, that would raise minimum consumption bar to operate! I really should just say basic inserter powered boiler chains instead, heh.

Realizing that with this setup, having small pumps to the right of the boilers would no longer save space anymore, I can now put them to the left for the battery setup, and start every row with 2 unspaced engines. And then only from that does the spaced rows barely fit in 50 tiles wide. All that and still kept the design to only 1 row of small poles below the design, with no belts, thanks to the alternating 10/9 engine rows, heh. Pretty happy with the design now - for more info, see the updated original post and the new pictures.

Oh and I tried that YARM mod. Really handy tool. And so measuring sets of 30 coal fields each, I got 200k to 250k being the average coal per field. Then from that, 250,000*0.25% = 625 coal. Then if it takes 3 materials per craft early game, it's saving 625 coal vs producing at least 200 items in the time it takes to deplete the average coal field. Welp, even assuming you notice the coal field running out most the time to prevent steam engine production from stopping, I think it's a certainty that you'll produce FAR more than a couple hundred items per depleted coal field, and this doesn't even include smelting raw materials. Then good, my setup does need the burner inserter for that one boiler chain in the beginning, putting all 3 there without loss is just great!

I wonder if there's any other major optimizations possible beyond this point. Now, compactness, required tech, and resources required are all pretty much optimized. Can't think of much other aspects to improve on now heh.

[Aru]

That is nice, it's compact, 3 tiles shorter. (4 shorter if you count a stretched, movable row of small poles, which can be walked through, so maybe 3.25 or 3.5 shorter)

2.89% less power per raw resource (with same burner inserter count)
4.28% - 5.63% less area
0.04% - 0.39% less power per area (with sparse movable pole row counting as 0.5 - 0.25 width)

Power per area is pretty much the same, but it's smaller overall, and a little costlier per resource. Which is about what I'd conclude from looking at them - 9.5 engines vs. 10, added underground belts. So at least, I wouldn't say that resource is strictly better.

You could probably replace the underground pipes with normal ones, and it would be more resource efficient, but then you couldn't walk through that side. Then it's 0.36% higher power per resource.

Calculation notes

3 poles per width of 7 engines at the right edge
6/7 poles per 2 engines stretched

10*5+2+14+4+(0 to 1) = 70-71
10*5+14+3 = 67

pump: 5 iron, 3 copper
engine: 20 iron
boiler: 1 iron, 5 stone
pipe: 1 iron
inserter: 4 iron, 1.5 copper
b. ins.: 3 iron
belt: 1.5 iron
split: 16 iron, 7.5 copper
small pole: 0.5 copper, 0.5 wood
und. belt: 8.75 iron
und.pipe: 7.5 iron

Patric:
2 pump
19 engine
27 boiler
7 pipe
24 insert
3 b. insert
16 belt
1 split
8 small pole
2 und. belt
4 und. pipe

Aru:
2 pump
20 engine
28 boiler
5 pipe
25 insert
3 b. insert
20 belt
1 split
7+6/7 small pole (~= 7.857)
0 und .belt
4 und. pipe

Code: Select all

# Python
iron_cost = [5,20,1,1,4,3,1.5,16,0,8.75,7.5]
copper_cost = [3,0,0,0,1.5,0,0,7.5,0,0,0]
# This one is no underground pipes
patric_count = [2,19,27,11,24,3,16,1,8,2,0]
# This one is underground pipes
#patric_count = [2,19,27,7,24,3,16,1,8,2,4]
aru_count = [2,20,28,5,25,3,20,1,7+6./7,0,4]

patric_iron = sum([x*y for x,y in zip(iron_cost, patric_count)])
patric_copper = sum([x*y for x,y in zip(copper_cost, patric_count)])
aru_iron = sum([x*y for x,y in zip(iron_cost, aru_count)])
aru_copper = sum([x*y for x,y in zip(copper_cost, aru_count)])

aru_power = 510*20
patric_power = 510*19

aru_power_per_resource = float(aru_power) / (aru_iron + aru_copper + 5*aru_count[2] + 0.5*aru_count[8])
patric_power_per_resource = float(patric_power) / (patric_iron + patric_copper + 5*patric_count[2] + 0.5*patric_count[8])

[Patric20878]

That is nice, it's compact, 3 tiles shorter. (4 shorter if you count a stretched, movable row of small poles, which can be walked through, so maybe 3.25 or 3.5 shorter)

Eh, the steam engines themselves are 3 tiles narrower, and the belt/pipes are 1 tile narrower, compared to yours, so it's 4 tiles shorter. 67 tiles wide, vs your 71. The 71 includes the poles on the right in your design.

And so, how are you calculating the power by area for my design to actually be lower? I'm understanding that to mean steam engines over rectangular area, so if that wasn't what you were referring to, you can ignore everything below. As the sparse power pole row on the bottom is a flat +1 to total rows in a setup, the height would be 3x+1, since it is literally 3 tiles high per engine row. And like said, you can't just reduce the width just because a row is less occupied than the rest. Why would a distinction be made for a less occupied edge column, vs a less occupied middle column? You can route wires, belts, and even buildings right through the middle of a design if the tiles there aren't occupied too. If its compactness % (occupied tiles / rectangular area) affects its width, I can reduce my design's width instantly by removing a random inserter, an absurd proposal.

On 2 engine rows:
Mine: 67 * [3(2) + 1] = 469 tiles. 469 / 19 = ~24.68 tiles per steam engine.
Yours: 71 * [3(2) + 1] = 497 tiles. 497 / 20 = 24.85 tiles per steam engine.
As that's area over power, not power over area, you can take the inverse of both numbers, or simply divide 24.85/~24.68 instead of the other way around, and the result is, mine has ~0.67% higher power by area, for 2 engine rows.

On 10 engine rows:
Mine: 67 * [3(10) + 1] = 2077. 2077 / 95 = ~21.86 tiles per steam engine.
Yours: 71 * [3(10) + 1] = 2201. 2201 / 100 = 22.01 tiles per steam engine.
Result, mine still has ~0.67% higher power by area.

In any number of steam engine rows, my design's power by area is always higher, and that was a major point of emphasis upgrading the design, to keep it higher or at least equal to the compact power plant and your design. Unfortunately, with this upgrade, since it's no longer arranged in triple rows, power by area did drop slightly, but is still overall higher.

Note: For readers, the numbers in this post are correct and take precedence over the incorrect numbers I posted later, as I edit this post with the latest information to reflect accuracy.

[Aru]

Mine doesn't need a row above. And I was using the tileable parts for area. If you count it as 3 shorter as you said, 10 engines per 70 length, vs. 9.5 engines per 67 length, (9.5/67)/(10/70) = 0.99253. But in my post I was counting it as 3.25 and also 3.5 shorter, for the small poles on the right side. Your latest design is a hard 67, I don't know where the 66 comes from. Maybe you counted the offshore pump in one but not the other? The area efficiency comes from this part:
"10*5+2+14+4+(0 to 1) = 70-71
10*5+14+3 = 67"

Page 2:

Also, the poles at the top can easily be moved down to reduce vertical size by one. Right now they're positioned for an additional row.

And I don't know what you're calling a "less occupied middle column", since you can't put buildings in the middle, and you can't reposition things in the middle.

[Patric20878]

Mine doesn't need a row above. And I was using the tileable parts for area. If you count it as 3 shorter as you said, 10 engines per 70 length, vs. 9.5 engines per 67 length, (9.5/67)/(10/70) = 0.99253. But in my post I was counting it as 3.25 and also 3.5 shorter, for the small poles on the right side. Your latest design is a hard 67, I don't know where the 66 comes from. Maybe you counted the offshore pump in one but not the other? The area efficiency comes from this part:
"10*5+2+14+4+(0 to 1) = 70-71
10*5+14+3 = 67"

Page 2:

Also, the poles at the top can easily be moved down to reduce vertical size by one. Right now they're positioned for an additional row.

Edited last post to reflect yours not needing 2 flat rows of poles. Read again for why it's 66 hard tiles wide, not 67. It's 10*5+14+2. And yours is 70 soft tiles wide, as in including the small poles.

On the example about middle columns and such, I'm saying that it's an arbitrary definition to just count your design as less than 70 tiles wide just because the last column is a soft 1 tile column. If the middle columns were soft enough, you CAN fit buildings there. No one counts width as less than what it is based on what's in it, the fact that your design's 70th tile is soft doesn't change the fact that it won't fit in a 69 tile wide land mass.

[Aru]

It doesn't explain why. Maybe you're not counting the offshore pumps? And I don't think a 69 tile land mass will ever be a problem. The poles can be repositioned, they provide power, they can be (and are) stretched as far apart as possible, you could for example put a row of adjacent assemblers against the side of the engines with no gaps, and move the poles out, and a row of poles doesn't block movement (like the underground pipes).

I still have no idea what you mean by "soft enough". The middle, is steam engines, and boilers, and an offshore pump, and a couple underground pipes. You can't put anything in there but power poles, and it doesn't need extra poles.

My contention, was that they are about the same in power per area, not a 15% difference.

[Patric20878]

Yes, I didn't include the offshore pumps for either design, since that's built on water and has nothing to do with occupying land.

And the same can be said of many designs in redstone and circuitry, the edge rows/columns often have wiring that don't need to be repeated when stacking, but it doesn't change the fact that it counts as column. I can use the rightmost side of a steam engine to output into storage tanks too, but you don't see me counting steam engines as 4.5 tiles do you? You counting it as somewhere between 0 and 1 is way too arbitrary, tell me when those poles can allow a solid wall of labs or assemblers to occupy the same space it is, then you can count it as something less. Otherwise, literally, it counts as 1 tile wide. And so does that row of poles needed, it counts as 1 tile high.