How to divide full belt into precision number

[takisama]

Hello guys. I will go straight to the question.
Right now I'm using Factorio Calculator to setup my base. The problem is instead of paying full belt or half belt for 1 line of assembling machine, I need to divide the belt into float number. Below I posted a picture for an example.



Have anyone had the same problem? How do you guys solve this? Thank you!

[robot256]

There is no problem, you just need regular splitters. If you provide enough to begin with, the machines will consume what they need and the rest will back up on the belt, forcing the unneeded items to take a different path towards another machine that needs them.

[FuryoftheStars]

Your image does not load for me. Try attaching it directly to your post (see "Attachments" tab below where the submit button is while composing/editing your post) vs using a third party host that you link to.

[takisama]

Your image does not load for me. Try attaching it directly to your post (see "Attachments" tab below where the submit button is while composing/editing your post) vs using a third party host that you link to.

Thank you for noticing me that. It's the first time I post in forum

[takisama]

There is no problem, you just need regular splitters. If you provide enough to begin with, the machines will consume what they need and the rest will back up on the belt, forcing the unneeded items to take a different path towards another machine that needs them.

Sooner or later my base will become spaghetti if I do that because I will need to connect the rest to main stream and it will look terrible. Do you have a better way?

[takisama]

There is no problem, you just need regular splitters. If you provide enough to begin with, the machines will consume what they need and the rest will back up on the belt, forcing the unneeded items to take a different path towards another machine that needs them.

Oh. I just realize I could set a inserter to collect the rest and use bot to send it to main bus. It's not a good solution but it works

[jodokus31]

So, instead of not belting it away from the main bus, you use bots to bring it back to the main bus.
Sounds funny.

[mmmPI]

Sooner or later my base will become spaghetti if I do that because I will need to connect the rest to main stream and it will look terrible. Do you have a better way?

Not all spaghetti looks terrible ! I think some looks even terrific

But also,It doesn't necessarily tranform into spaghetti with some set of rules : if 1) you do all your branching for similar reasons with similar method, 2) you don't connect things "back".

Make the "main stream" a straight lane, make all splitters create 90° angle branch off the main stream on the same side of it. And that's one definition of a what is refered to as "main bus" which is a concept that solve the problem you describe in a way i'll try to explain

If you give priority with the splitter to the branch going to the side of the main stream, the material will try to fill in the belt as mentionned, if you need 2.23 belt, you first need to have at least that amount in the main stream, let say you have 4 belts worth or green circuit running parralel, you could use 3 splitters to force a doubling of 3 of those 4 lanes, and those 3 extra lane created from the process ,you make them turn 90° immediatly while apart from that the main stream is left untouched.

Any machines attached to the branch can share up to 3 belts worth of green circuit. No need to connect things back, if machines on the branch do not consume (all) the material, eventually the (input) material will go the other way at the earlier splitter because one of the 2 choices will be full. (with no output priority a splitter tries a 50%-50% split, with priority on one output lane, it will try to give 100% of the incoming material to that output lane, and if that output lane is full or if there is too much input becomes it comes from 2 lanes, only then the extra material are placed on the non-priority output lane)

Depending on the number of machines, maybe it will consume 2.33 or 2.23 or 1.651423 on average overtime. No more than 3 that's for sure. The exact consumption depend on the ouput of those machine, if it's full /blocked those machines will stop their production untill it's not anymore; this can passively prevent the machines to eat all 3 belts at all time if the later step on the production chain is balanced.

Now if you take 2.23 belts out of 4 belts from the main stream using this branch method , and you plan than branch to be running at all time like science assembly machines, you can make the main stream thinner after the splitters since there will be at most 1.77 belts left you can make it 2 belts wide instead of the inital 4.

Or instead of making it thinner by 2 lanes, you can use 1 of those now available lane to carry the ouput of the machine as a method of organisation.

Or you can "refill" the 4 lane wide but depleted main stream from the side. Where you combine 2 input lane from a splitter, but only use 1 output lane.
Any which you find the most pleasing to the eye

TL DR : if not all output can go out, not all input is processed, the 2.23 number you could see as 2.5 belt, or 3 belt is the minimum you need to fully supply the production step, if the expected output is 1 lane, and you need 2.23 lane of input, you can give it 2.5 lanes with a bypass made from splitters at the start to let the overflow go somewhere else knowing it's then (2.5 minus 2.23) belts worth of material minimum.

There are others way to do, if you are interested into more bizarre machinery but i felt the previously described approach is more common than attempting to count items precisely, or count time to stop belts and is a solid base on which one can try improve the look of it rather than spending more time on making it work in the first place

[robot256]

There is no problem, you just need regular splitters. If you provide enough to begin with, the machines will consume what they need and the rest will back up on the belt, forcing the unneeded items to take a different path towards another machine that needs them.

Oh. I just realize I could set a inserter to collect the rest and use bot to send it to main bus. It's not a good solution but it works

Items don't fall off the ends of belts! They just stop, and make a traffic jam instead (the belt is "backed up" when it's full of non-moving items), that moves a little whenever an item is picked up. You just make it dead-end after each row of assemblers and let it fill up. Once full, the assemblers will pick up items at whatever rate they please, and incoming items will fill in the gaps created at the beginning of the queue.

[FuryoftheStars]

Items don't fall off the ends of belts!

Maybe they're using this mod?

https://mods.factorio.com/mod/belt-overflow

[Tertius]

@takisama It seems you ordered the calculator to create the numbers for an arbitrary number of items to output (blue=300, red=1900). However, you will usually not consume exactly this quantity. Instead, you will build production lines that will create a full belt of yellow/red/blue belts. Reasoning behind this: Producing the full capacity of a blue belt plus some more item makes no sense - you want to fully use the output belts.

Or you want production lines that will consume full belts of yellow/red/blue belts. Often, the capacity of some belt will inherently limit the throughput of some production line part, and the items around it have to adapt to this. Or you have a train station that isn't able to provide more than 4 belts, and you don't want to add more stations.

And then there is the internal setup of a production line. You want your assembling machines running at all time with no interruption - would be a waste otherwise. So if you have a assembler type 3 for green circuits, it will produce 1 circuit in 0.5/1.25 = 0.4s or 2.5 circuits per second. 6 assemblers will produce a full yellow belt (2.5/s * 6 = 15/s). This is probably the smallest production line you want to have. Everything around it would be a multiple of this to keep numbers integer and not have fractional values.

So, people usually order these calculators to produce X yellow/red/blue belts and see how much input this will take. Then look if this input is near a multiple of yellow/red/blue belts, round down, then calculate again. The most common production ratios (1/2, 1/3, 1/4, 2/3, 3/4) will lead often to production lines that will consume full belts and at the same time produce full belts.

If it comes to belt capacity, mentally my smallest unit is "one yellow belt lane". 1 yellow belt has 2, 1 red belt has 4, 1 blue belt has 6. If one production line needs 3 of these, I either use a red belt and have a surplus of 1 or use a blue belt and have 3 surplus and can double that production line to produce more and fully consume that belt.