takisama wrote: Sun Mar 26, 2023 3:46 am
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
