Optimized Steam Engine Setup

[Patric20878]

Burner inserters do not use much energy in a steam engine setup! Like I said, they only insert at the rate boilers burn coal, which at completely maxed out power consumption from the steam engines, is only once per ~10 seconds, and a burner inserter can pull 57 times per coal! Since this means it's literally taking around 570 seconds for a burner inserter to consume ONE coal, the additional consumption is TINY. Even if you were to convert this into an energy calculation when it's not even energy, it's MASS, potential energy, I can calculate energy consumption from each to be ~31.66 KW, NOT 188 KW. It takes 96 seconds for an inserter to pull 57 items at max rate, and 96/570 * 188 = 31.66.

And all this talk about energy consumption of burner inserters is totally irrelevant, because it actually consumes ZERO energy. It is not electric, nor is it affected in any way by electricity - it consumes coal, so calculation needs to factor coal input rates, not electricity output rates! Mass is not energy until you convert it to energy! Since burner inserters consume 1 coal every 570 seconds and boilers consume one every 10 seconds, [(1/10) + (1/570)] / (1/10) gives just 1.75% more coal consumed. Even if you ignore the major advantage burner inserters give, i.e. NOT needing to manually disconnect power grids and inserters every coal field you deplete (the entire point here, automation), such a tiny amount of coal used is nothing compared to needing to craft massive amounts of circuits to make all those normal inserters instead. The only calculation you'd even have to do at this point to figure if the belt can still handle 10 steam engine rows, is divide the ~10.504 max supported engine rows you got by ~1.0175, = ~10.35 engine rows. With that calculation, you can be sure it can still handle 10 rows, with emphasis on the increased coal intake, which since they travel on belts unlike electricity, factors in possible belt related bottlenecks.

[Aru]

edit: Some of this stuff is wrong, it was assuming that inserters use their indicated power while working. But really their maximum is much less.

You are right about each boiler using about one coal every 10 seconds (~10.98s=14/(1.275/s)). And being used for steam, 14 burner inserters to 10 engines, the inserters are each using ~28.31236 kW in coal on average, which is not far from your 31.66 kW. But, coal is 8000 kJ of energy each. So a single burner inserter, assigned to this particular task, takes 8000 kJ / 28.31 kW ~= 282.56 seconds to consume one coal, not 570 seconds.

I got a factor of 2 mixed up (repeatedly at that, I kept going back and forth in the post with formulas) because of boiler inefficiency, but all it affected was the 7.78% of coal number. It's actually half that, 3.89%, I apologize for this error. I'll fix that post, and the one other post it's in. The power number (396.373 kW of coal), the total coal throughput for 100 engines with all burners (~13.245 per second, under the 13.392 maximum), everything with normal inserters, the thresholds for dropping to 0 power, those numbers are all still right. In fact the original assertion was that they use 7.78% of the power produced by the engines, which is still true. (And by that I meant, power_from_engines * 7.78%.) But only 3.89% of the coal. Which is still not 1.75%.

Even if you ignore the major advantage burner inserters give, i.e. NOT needing to manually disconnect power grids and inserters every coal field you deplete (the entire point here, automation)

You do have to do that, that's what I was saying. If the engines run out of coal, and there's a lot of them, you have to connect them directly to the coal mining drills, and feed them coal manually so they can power the mining drills to get more coal. Even if all of the boilers are fed by burner inserters.

such a tiny amount of coal used is nothing compared to needing to craft massive amounts of circuits to make all those normal inserters instead.

In resource weighting schemes, raw materials are usually valued at 1 each. Burner is 3, normal is 5.5, which is 1.5 or ~83% extra plates. Then, a normal inserter takes under 10 minutes to make up the price difference in saved coal. 20 minutes if coal is worth 1/2 resource. I didn't calculate the exact times. Far enough into the game, even coal has meaningful value, you need it for plastic. And it takes just as much resource in the way of mining drills to harvest, which is why they're usually all 1.

I would say a logical compromise is to start with burner inserters on boilers because they're cheaper, and at some point use normal inserters on new boilers instead. Perhaps at an even later point, go back and replace the burners, but that's not that important.

[Patric20878]

You are right about each boiler using about one coal every 10 seconds (~10.98s=14/(1.275/s)). And being used for steam, 14 burner inserters to 10 engines, the inserters are each using ~28.31236 kW in coal on average, which is not far from your 31.66 kW. But, coal is 8000 kJ of energy each. So a single burner inserter, assigned to this particular task, takes 8000 kJ / 28.31 kW ~= 282.56 seconds to consume one coal, not 570 seconds.

I got a factor of 2 mixed up (repeatedly at that, I kept going back and forth in the post with formulas) because of boiler inefficiency, but all it affected was the 7.78% of coal number. It's actually half that, 3.89%, I apologize for this error. I'll fix that post, and the one other post it's in. The power number (396.373 kW of coal), the total coal throughput for 100 engines with all burners (~13.245 per second, under the 13.392 maximum), everything with normal inserters, the thresholds for dropping to 0 power, those numbers are all still right. In fact the original assertion was that they use 7.78% of the power produced by the engines, which is still true. (And by that I meant, power_from_engines * 7.78%.) But only 3.89% of the coal. Which is still not 1.75%.

I just tested for just how long it really takes for a burner inserter to use 1 coal. From the same boiler I measured ~10 seconds from, I measured its burner, and both times, I got 565 seconds. You must be missing something in your calculations, because when a burner inserter pulls 57 times per coal, and each coal takes 10 seconds thanks to burner consumption to pull, I can't see how it'd be logically possible for it to somehow take 280 seconds instead of in the 570 second range. You can test this yourself. And I'm not sure if you're even comparing the same %, but for clarification, if a burner consumes 1 coal in 10 seconds, it consumes 0.1 coal per second. If you have burner inserters do the work and they consume 1 coal in around 570 seconds, they pull ~0.00175 coal per second. 0.1 + 0.00175 = 0.10175 coal per second, which is 1.0175x as much as usual, meaning it consumes an additional +1.75% coal per second. How much more coal it consumes compared to usual is what the number is, not % coal, power %, or what you were comparing it to.

Edit: Just realized something.

takes 8000 kJ / 28.31 kW ~= 282.56 seconds to consume one coal, not 570 seconds.

I got a factor of 2 mixed up (repeatedly at that, I kept going back and forth in the post with formulas)

Are you getting a factor of 2 mixed up here too? 282.56 * 2 = 565.12 is awfully close to the 570 and 565 seconds I got from calculations and testing.

You do have to do that, that's what I was saying. If the engines run out of coal, and there's a lot of them, you have to connect them directly to the coal mining drills, and feed them coal manually so they can power the mining drills to get more coal. Even if all of the boilers are fed by burner inserters.

That's on the mining drills. I was talking about the inserters. With electric inserters, you'll have to not only worry about starting the mining drills back up, but also restarting up the inserters, which you'd have to disconnect steam power too. And no one ever said you needed to power the mining drills with the same steam engines they're feeding, especially if it's very far away. Running out of electricity from steam engines does not logically require mining drills to stop also unless you're powering it from the same network, unlike with electric inserters. With burner inserters, I can stop and resume steam power simply by feeding or not feeding it coal, no additional power disconnects or manual setup required, and this was a point of emphasis I made originally with my first design, having the water tanks so there always is SOME power to keep the inserters powered, at least for much longer than usual. Being able to use burner inserters makes the steam engine setup itself fully automatic even with lapses in coal input, something electric inserters could never do, unless they're all powered by a separate power grid. Which of course seems rather silly.

In resource weighting schemes, raw materials are usually valued at 1 each. Burner is 3, normal is 5.5, which is 1.5 or ~83% extra plates. Then, a normal inserter takes under 10 minutes to make up the price difference in saved coal. 20 minutes if coal is worth 1/2 resource. I didn't calculate the exact times. Far enough into the game, even coal has meaningful value, you need it for plastic. And it takes just as much resource in the way of mining drills to harvest, which is why they're usually all 1.

I would say a logical compromise is to start with burner inserters on boilers because they're cheaper, and at some point use normal inserters on new boilers instead. Perhaps at an even later point, go back and replace the burners, but that's not that important.

As I've tested that burner inserters indeed use coal at once per ~570 seconds, specifically 565s from tests, it would take 2.5*565 = 1412 = ~23.5 minutes to catch up. But raw material resource isn't even the major thing you lose. Time is the major resource. Whatever extra time it takes you to craft or produce normal inserters instead of burner ones is the time you lose to do more useful things, like spending the time to just find a new coal field.

While theoretically I can perfectly agree that replacing the burner inserters with normal ones would save resources in the long run, assuming time is not a resource (um...never), practicality comes into play here. I can imagine that eventually, it would be worth it to replace the burner inserters, but it seems like to me the only time it'd ever be worth spending the effort on such a low yield gain is either if you're playing on a low-resource map, which in that case, you shouldn't even start with burner inserters, no less upgrading to them, or when you totally run out of more useful things to do in Factorio somehow, which seems like almost never. And even if there are other reasons to upgrade, do they come before one shifts to solar power as their primary source of power, rendering upgrading the steam engine setup pointless as they shift all coal mined to plastic production?

On using inserters for new steam engine rows' boilers though, I might be agree, since normally, one doesn't just build burner inserter factories. At the same time though, they're so easy to just handcraft with only iron required that it hardly uses any time to make them, but since time spent crafting inserters is time not spent crafting something else, it's unclear if it's worth using burner or normal inserters on new boilers. But more importantly, it's considering saving a tiny bit a coal vs having completely automated steam engine, which the latter wins for me, without a shadow of a doubt. But since saving coal can actually be more useful in certain cases too, I'd think I'll conclude that whether you should use burner or normal inserters on new engine rows should be a combination of preference and the specific case. As even finding one new coal field would make up for the 1.75% more coal used several times over, I think my preferred design will continue to use burner inserters for every new row, so I don't have to repeatedly manually starting up the steam engine every single time. I'd much rather minimize the time my factories are producing nothing thanks to having no electricity, thus improving overall production rate, than minimize coal "wasted". In the end, the person ahead is the one who produced more, not saved more.

[Aru]

You can make electronic circuits using assemblers, that's what I do. And I don't know how I can explain the thing with the mining drills any more clearly, it seems like you're deliberately misunderstanding.

Even if you ignore the major advantage burner inserters give, i.e. NOT needing to manually disconnect power grids and inserters every coal field you deplete (the entire point here, automation)

You do have to do that, that's what I was saying. If the engines run out of coal, and there's a lot of them, you have to connect them directly to the coal mining drills, and feed them coal manually so they can power the mining drills to get more coal. Even if all of the boilers are fed by burner inserters.

That's on the mining drills. I was talking about the inserters. With electric inserters, you'll have to not only worry about starting the mining drills back up, but also restarting up the inserters, which you'd have to disconnect steam power too. And no one ever said you needed to power the mining drills with the same steam engines they're feeding, especially if it's very far away. Running out of electricity from steam engines does not logically require mining drills to stop also unless you're powering it from the same network, unlike with electric inserters.

You do not have to disconnect the inserters, as they are powered by the engines. And you do have to connect the drills to the engines, if you want them to perform mining they need electricity. And like I said before, you only need 3 2 burner inserters if you want to completely avoid all possibility of this being an issue, not all 14. Why wouldn't you be powering them from the same network? Why would someone have multiple separate steam arrays?

Yes, I know it's on the mining drills, my original statement that you were referring to was also specifically about mining drills. And no, you wouldn't have to worry so much about the inserters, that was the point of the post. If the drills were all burner drills, and used burner inserters to fuel themselves, then you would also need (3 2) burner inserters at the boilers to allow everything to power back up from nothing. Without burner drills, the inserters don't matter so much. And this has only ever happened to me once, ever, and it was early enough in the game that I didn't have to disconnect anything. I learned, and found it very easy to prevent. The one time it did happen, had nothing to do with the coal field running out, and nothing to do with the inserters at the boilers, but with accidentally disconnecting the coal miners from electricity, cutting off coal supply.

The number discrepancy, is because inserters apparently don't use the indicated amount of power, even when transferring continuously. That's the problem. It's a jagged and peculiar power graph for all of them, and it's not possible for any inserter to consume nearly the amount of power that it indicates. I don't care to do any more math, the whole thing has become rather distasteful. Normal inserter is 6.2 kW here, less than half of the listed 13 kW.

[Patric20878]

Hm, 3 burner inserters and 11 normal inserters per engine row? So you were suggesting a hybrid setup. Looks like I read that last post from page 4 the wrong way then. I'll play around with this and see. What about when 13 boilers and 9 steam engines are used? And good to know that power behavior. Seems like it just uses half of its stated power consumption.

[Aru]

It uses 6.2 kW in that example, chest-to-chest, which is less than half of it's listed rate. Half would be 6.5 kW.
And by the way you really only need 2 burner inserters to refuel boilers from no power to full power, not 3 (I overestimated). Or more precisely 1.365 burners with (14-1.365) electric, needed quadratic formula for that number. 2 burners per row will fuel a 2*390/(510*10) or higher power ratio (assuming that base demand is matched to or below the maximum steam output), *12 electric inserters *.866 transfer rate is 1.589 coal/s from the 12 electrics, which is enough to fully fuel 12 boilers. 13 boilers and 9 engines also needs 2 burner inserters to refuel with 0 power.

The quadratic I mentioned,
a = -390/510./10.*0.866
b = (390/510./10.*0.866*14 + 390/8000.*2)
c = -510*10/8000.*2
x = 1.364727

Guess I did some more math, but I'm not trying as hard to spell it (the math) out now.

The 2 burner inserters only matter when you have enough coal on the coal belt for them to pick up, of course. And they only allow the engines to maintain that last ~11% power output from an 11% coal supply instead of dropping the rest of the way from 11% to 0. The electric drills are the more important component in the cycle of keeping the engines going. But should you goof up and allow coal supply to drop, and then you fix the problem, 2 burner inserters on each row will save you the subsequent task of manually inserting coal into 2 boilers on each to jump-start the engines. Instead, inserting coal to the coal belt is adequate, and the burner inserters can perform the jump-start for you. Either way, you have to add enough coal to last until the mining drills deliver coal to the engines, and they won't produce much coal at such low power. The fastest way to get out of that mess with a large engine array, is what I described before, disconnect the rest of the base from the engines and drills, so the drills can get more power. (Whether or not there are burner inserters.)

[Patric20878]

Oh, it actually consistently shows 6.2 or 6.3 KW used when pulling. So it isn't a rough average. Interesting.

And yeah, seeing I'm not too familiar with the calculations you use, unless it's simple and labeled, it wouldn't do me much good. I can only mainly rely on logical proof and in-game testing rather than mathematical proof at this point.

Math-wise, would there be any major difference in startup time between placing burner inserters in the beginning vs the end? I can imagine maybe a flat delay thanks to burner inserters being farther away, and maybe an additional delay of normal inserters on the way picking up coal but still being too underpowered to insert them, thus giving a possible additional delay.

Also, by your graph, what if I were to have 2/1/2 burner inserters on each row of 3 steam engines, which is 10/9/10? Would the 2 burner inserters on 2 of the rows make up for the 1 in the middle? Can this be simplified to ask if (2+1+2) / 3 = ~1.66 burner inserters per engine row of ~9.66 steam engines? I think the 1.365 number you got means the minimum number of burner inserters per row, but I'm not sure how this works with averages or if there's any additional factor when averaging them.

[Aru]

You can click the times, 5 seconds, 1 minute, etc. to show average values over that time period. The screenshot is an average over 1 minute (the 6.2), and as the graph shows it includes a lot of cycles.

Time delay for having burners in the back, is maybe 6.4s on average because yellow belt speed is 1.875. And slow orange could cause an extra delay, how long depends how much coal, because each orange inserter in front has to be occupied by one coal each if they're moving very slow, so 12 could be up to a 12 coal delay. The faster they are the less it matters, because they deposit one for every one they pick up. If power is actually 0, or extremely low, they won't delay at all. They might not even be fast enough to pick up coal at under 11% power (before the belt fills up), that would need a test.

And yes, it should work with 2-1-2 if the engines are all in the same electrical network. You could put them all in the first row, how many depends how many rows, but I just said 2 per row for simpler tiling. For other engine counts, substitute average per row where you see the 10 (as in 510*10 and /510/10), and use quadratic formula. This is what I used to get a/b/c:
510*10/8000*2 = 390*x/510/10*0.866*(14-x) + 390*x/8000*2
You could also tell whether a burner count is good by plugging it into other formulas, trial and error, but I went for a little overkill by solving the exact count. I could have made another error, but this time the result makes sense so I'm less doubtful.

[Patric20878]

And yes, it should work with 2-1-2 if the engines are all in the same electrical network. You could put them all in the first row, how many depends how many rows, but I just said 2 per row for simpler tiling. For other engine counts, substitute average per row where you see the 10 (as in 510*10 and /510/10), and use quadratic formula. This is what I used to get a/b/c:
510*10/8000*2 = 390*x/510/10*0.866*(14-x) + 390*x/8000*2
You could also tell whether a burner count is good by plugging it into other formulas, trial and error, but I went for a little overkill by solving the exact count. I could have made another error, but this time the result makes sense so I'm less doubtful.

Plugged in the a,b,c values you posted earlier, switching 14 out with 13. Got 1.48, an increase from 1.36. Assuming your model is correct and is at least 2 significant digits accurate (wouldn't know how to verify your method), that should mean each 10/9/10 row needs (1.36 + 1.48 + 1.36) / 3 = 1.4 burner inserters per row. Looks like I can't do something like 1/2/1 then, but should easily be able to do 2/1/2. Though uh, the way it's setup, the bottom 1/2 boilers would be fed by normal inserter, then burner inserter, in a belt - inserter - boiler - inserter - boiler - inserter - belt - burner inserter chain. Not sure if that'd work. Would the first 3 inserters last long enough to give coal to the burner inserters to prevent a downward spiral? And did you also factor in that inserters only need to be as fast as once per 10 seconds to avoid a downward spiral?

[Aru]

So long as the burner inserters can get coal, it should be okay, but I'm not sure what you're describing. And yes, though "10 seconds" isn't in the formulas, I didn't put any intermediate calculations in them. All the numbers are exact except for the 0.866 inserter speed, which is from the wiki, a result of someone else's testing (far more precise testing than I'd ever do).

510*10/8000*2 = 390*x/510/10*0.866*(14-x) + 390*x/8000*2

Left side is coal/second used by a row at full load. Next term is (power ratio produced by burner inserters) * (orange speed) * (orange count) which is the total transfer rate of the orange inserters as fueled by only the burner inserters, last is the total transfer rate of the burner inserters. The power ratio is a minimum, assuming that peak power from the engines can fully power the base. And x is the number of burner inserters of course, and this is a quadratic equation so x can be solved with quadratic formula.

[Patric20878]

I'm going to have to ask you again though. Did you actually verify any of your calculations in-game? I just setup 2 rows of 10/9 steam engines, with 14/13 boilers. I then attached 600 radars to it, so 19*510 = 9690, 600*300 = 180000, so 9690/180000 = ~5.3833% production available for consumption.

And the result is, 3 burner inserters couldn't keep up. Neither could 10. Hell, not even 26/27 burner inserters could even make the lone normal inserter keep up, and this is just 5.3833%. Unless my eyes are deceiving me, I'm going to have to conclude you're neglecting factors in your calculations again. From looking at it, power priority seems like a major factor here. Power graph after changing 24/27 burner inserters to normal ones:

[Aru]

I said several times, that certain calculations assume that the peak engine output is enough to fully power the attached load. With that kind of overload the inserters won't really do anything, even if all the boilers are full. If you plan on deliberately running a base with insufficient power, you should just use all burner inserters. Try matching the load to the engine output, without going over. That is, engine_count*510kW. And no, I didn't test anything other than what I said I did.

[Patric20878]

As for that downward spiral I mentioned with (14) electric inserters into (10) boilers, the threshold is ~10.96% (510*10/8000*2/.831/14). That is, if power supply drops under 10.96% of power demand, it can drop the rest of the way to 0 because the inserters don't get enough power.

What does this statement mean then? 5% is under that threshold isn't it? What's the meaning of 1-2 burner inserters with thresholds and power if not this?

And with that test comes another advantage of all burner inserters. This was an extreme test, where production is 5% of consumption since that's what I thought you meant by threshold, but in practice, adding more consumption than production wouldn't actually DROP production. The instant thing that comes to mind already is when pushing with laser turrets on a large camp - you don't have to worry about tearing down the last row of turrets before building the next, which normally would not only spread power of each turret thin, but would also drop electricity production too, further slowing them down. With burner inserters, even at 5% production to consumption, electricity production remains constant. You wouldn't need to worry about crippling your electricity production just by placing more laser turrets than you should've.

[Aru]

That makes the same assumption though, about peak engine output being enough to power the base. But if coal supply drops enough that the engine output goes under that threshold, then it will drop the rest of the way to 0, and won't be able to turn back on unless you put coal into a few of the boilers manually. Or you have a few burner inserters to do it for you. Turrets aren't an issue because they only use a lot of power while they're firing.

[Patric20878]

What assumption? I was still editing the post, but as said, not having enough production doesn't literally DROP production, unlike with burner inserters. Double whammy is bad. If coal supply drops, why would it just stop at 10% anyways, it's almost no difference from no production and 10% production, both require immediate attention. I hope I understand you correctly this time, that the % you meant is current production to max production instead of the max production to consumption I thought you meant.

And turrets are an issue, because it is when they are firing. A turret push on a large camp is typically a sustained effort, and with production being dependent on production:consumption ratio with electric inserters, it's much too easy to screw up and forget how many laser turrets you should place at max. And the part where you'd have to calculate this just to figure the number, such a pain.

[Aru]

See the rest of that post, "So really, the problem is not the decrease from this threshold to 0%, because it's a very small drop (11%), not far from 0% to begin with. The only real problem, then, is getting a huge engine array running again after it has run out of fuel, which might happen if your coal field unexpectedly empties."

"So, while that 10.96% power threshold is already pretty low, it can be reduced to 0% by swapping 3 orange inserters for burner inserters on each row of engines." (2, not 3)

Not having enough production will drop production, if it's under the 11%. That's the "not enough production". Then it can drop the rest of the way to 0 due to inserters on the engines being too slow, and then won't start again unless enough coal is inserted to meet the 11% power ratio.

I say the turrets aren't an issue because it should take a while to finish the 5 coal in each boiler. That's a long fight. And even if they do finish those 5 coal in a single fight, the 2 burner inserters would cause the engines to recover after the fight ends.

As for that downward spiral I mentioned with (14) electric inserters into (10) boilers, the threshold is ~10.96% (510*10/8000*2/.831/14). That is, if power supply drops under 10.96% of power demand, it can drop the rest of the way to 0 because the inserters don't get enough power.

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.

So really, the problem is not the decrease from this threshold to 0%, because it's a very small drop (11%), not far from 0% to begin with. The only real problem, then, is getting a huge engine array running again after it has run out of fuel, which might happen if your coal field unexpectedly empties. The solution in this situation, would be to first sever the engines from the rest of the grid. You can wire them to the mining drills on a new coal patch, and manually add some fuel, and re-attach the engines when the coal arrives. (This has only happened to me once, and it was early enough in the game that I didn't have to sever to restart the engines. I accidentally removed a pole that powered the coal miners, then later the engines ran out.)

So, while that 10.96% power threshold is already pretty low, it can be reduced to 0% by swapping 3 orange inserters for burner inserters on each row of engines. Because burner inserters use so much energy, make them the last three so they are only used if needed. And because 11% is already low, there's not much point doing this with just one or two burners, it should be three. But even if the inserters are adequate to fully fuel the engines, they need adequate fuel, and burner inserters won't help keep the drills mining the coal powered. They could hurt, by taking a cut of the coal supply (3.7% with all burners).

[Patric20878]

I read that part, and I'm still not clear what part you're actually addressing. Say the thesis in simple terms, then go into detail, so I can get what the point actually is. I'm assuming at least 3 burner inserters for my 19 engine setup, since 2 rows of 1.36 + 1.48 = 2.84. That doesn't work for 5% production:consumption ratio. And I only needed 1 burner inserter to get everything started at near 1:1 ratio. So what's the 2.84 number for?

"So, while that 10.96% power threshold is already pretty low, it can be reduced to 0% by swapping 3 orange inserters for burner inserters on each row of engines." (2, not 3)

Not having enough production will drop production, if it's under the 11%. That's the "not enough production".

Still vague, not having enough production will drop production? Current production? Max production? Production:Consumption ratio? WHAT production? And what power threshold? I apologize if it seems like I'm intentionally not understanding what you mean, but your wording is way too ambiguous to interpret it in just one way.

And haven't you turret pushed against a large camp before? It easily takes double digit minutes, unless you want to lose 80% your laser turrets trying. If you add more laser turrets than your production can support, production drops. And of course it's likely one will, no one has such time to be counting how many turrets they placed while trying to push a large camp.

[Aru]

I read that part, and I'm still not clear what part you're actually addressing. Say the thesis in simple terms, then go into detail, so I can get what the point actually is. I'm assuming at least 3 burner inserters for my 19 engine setup, since 2 rows of 1.36 + 1.48 = 2.84. That doesn't work for 5% production:consumption ratio. And I only needed 1 burner inserter to get everything started at near 1:1 ratio. So what's the 2.84 number for?

Okay... when you click a power pole, and see the graphs and such. If the bar is yellow, if it's less than full, that means that the base is asking for more power than it's getting. So everything runs slower. If that bar drops under 11%, and the base is engine powered, and peak engine output (that is, number_of_engines*510kW) should be enough to power the base, then it can drop the rest of the way to 0. (edit: Even if it doesn't, recovery will be delayed until it reaches ~11%.) If you have excess engines, more than enough to power the load at peak output, then that threshold should be lower (I think). But if you don't even have enough engines to power the base, it will be higher.

I don't know about 2.84, you tell me, hah.

I know it's vague, it was me quoting you. I did my best to interpret it anyway. You: "not having enough production doesn't literally DROP production"

Even with a long invasion, I doubt that the turrets will drain such a huge amount of power for all those minutes *continuously*. When I'm doing that, I'll drop a pattern with almost 100 turrets, and they will all be active for a few seconds, but no more. Then they only face a small trickle from the rest of the enemy base. Then I place it again, and they all shoot again, but it's not hundreds of turrets firing continuously for several minutes. At that point in the game, I'm using solar and accumulators anyway. It's possible that even if the entire base shuts down to divert power to the lasers, they might not get enough from engines to all shoot at once.

[Patric20878]

Okay... when you click a power pole, and see the graphs and such. If the bar is yellow, if it's less than full, that means that the base is asking for more power than it's getting. So everything runs slower. If that bar drops under 11%, and the base is engine powered, and peak engine output (that is, number_of_engines*510kW) should be enough to power the base, then it can drop the rest of the way to 0. If you have excess engines, more than enough to power the load at peak output, then that threshold should be lower (I think). But if you don't even have enough engines to power the base, it will be higher.

I know it's vague, it was me quoting you. I did my best to interpret it anyway.

I don't know about 2.84, you tell me, hah.

Which bar, the consumption bar? If that's the bar you're talking about, that's the same bar I'm talking about when saying 1.36 burner inserters isn't enough to prevent production from dropping further. Unless all inserters are burner ones, inserters can't keep up with pulling coal like in my 600 radar test. If I'm correct, you've only found the number to prevent production from dropping all the way to 0, but it doesn't change the fact that it still drops immensely, to under half of what it should be producing.

On the laser turret push, only if you're pushing from one direction, When the camp is large enough, you need several hundred turrets, and each have to be firing constantly at the not-really-a-trickle that is all those evolved mobs. Otherwise, you risk them attacking from the sides. And while I keep those as a safe zone of sorts, pushing with laser turrets to advance then risks this issue. Boilers only keep 60 seconds worth of coal, it's incredibly easy for them to run out when inserters are barely working, and this forces you to take into account the production drop when you place more laser turrets than production can handle. And then you have to calculate somehow just how many turrets you can place before placing more actually DECREASES their combined power. Not a thing anyone would want to do when pushing a large camp.

[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.

You must be using mods or some crazy settings, if you actually have several hundred turrets firing, all of them continuously for several minutes. When I do it, there's 2 or 3 hundred turrets out at a time, from 9 personal roboports. But not every one fires non stop for several minutes. When enemies come, enough turn on to kill the enemies, but the rest are off except for shortly after placement, and perhaps during a wave of enemies. The only reason I drop so many is so that they can defend each other, to cut losses, so I don't lose more turrets than I have to. If some turrets let up even a little, the boilers can get coal. The only way they can cause the engines to stop, is if you don't have a couple burner inserters at each row of engines, and the turrets drain all or nearly all of your available power, and they do so without even a 1 second break for a long enough interval for the boilers to deplete their 5 coal. All three conditions would have to be met for lasers to turn off the engines. And if they're drawing all of your power, some of them aren't even doing anything, you need more power.

Maybe you're unaware, that laser turrets draw a lot of power only when they're firing? When they're not firing, they don't draw power, except for a measly 6 kW. Only while actively shooting at an enemy do they draw 2400 kW. You would need 471 steam engines to use 100 turrets at the same time anyway, which is a lot more than I've ever had. I rely on accumulators to power them.