HELP : evenly feed a full belt onto one side (side insert)

[DrNoid]

I take my hat off to thee and bow!
That's some nice wiring!

[DrNoid]

Actually, you where very close with your first design!

Factorio_even_sided_insert.png (337.4 KiB) Viewed 50225 times

The change I made there was to make the final inserter a smart one that only works if the first two boxes contain 1 item.

• Inserters 1 and 2 only insert when their first box contains no items.
• Inserters 3 and 4 only insert when the last box is empty.
• Inserter 5 only inserts when the first two contain 1 item.

That last rule makes sure that the system can not run empty on one side.

[just_dont]

Huh, so any kind of separate circuits weren't even needed Good to know, that makes things much more simple.

Any kind of wiring solution tend to grow in complexity up until the point you realize that half of it isn't even needed.

EDIT: Ah, no, I'm pretty sure that your system produces the same errors as my #1 schematic -- i.e. 2-1 splits when the system recovers from 1-side input shortage.
1) A state of shortage: 1 item in one of the first boxes, 0 in other one, only 1 item in the last box, output inserter stopped.
2) Recovering: 1 other item comes in the first box, output restarts and removes that 1 leftover item from the last box.
3) Then 2 middle inserters put their items in the last box.
Question: which one will be picked next by the output inserter?

[DrNoid]

Any kind of wiring solution tend to grow in complexity up until the point you realize that half of it isn't even needed.

I have that a lot too

EDIT: Ah, no, I'm pretty sure that your system produces the same errors as my #1 schematic -- i.e. 2-1 splits when the system recovers from 1-side input shortage.
1) A state of shortage: 1 item in one of the first boxes, 0 in other one, only 1 item in the last box, output inserter stopped.

I've tried it quite a bit, and have not had that happen. The thing is, when one side runs dry, the system stops with *two* items in the last box, because the last inserter can not work unless there is 1 item in each of the first two boxes. So the state you suggest there can not happen.

1) Inserter 5 picks the last item from box 3 -> inserter 3&4 start to move their item from box 1&2 to box 3
2) Box 1&2 are empty, so inserter 1&2 try to move an item, but one of them can't
3) Inserter 5, 3&4 finish the move of their item. And one of the inserters 1&2 finish its move
Now there are 2 items in box 3, and 1 item in box 1 or 2. The other box is empty.
Inserter 5 can not start on those last two items, inserter 3&4 stop because box 3 is not empty.

Question: which one will be picked next by the output inserter?

I have no clue how that output inserter decides what to pick, but for some reason the order is always the same...

[just_dont]

That's even more nice! I'll be running some tests, then.

UPD: You are amazing It works properly in all cases.

So, the prize for a perfect most compact and fast deterministic solution goes to DrNoid:

Code: Select all

> +O
 +P+P
 O+O+>

O - smart chest;
P - pole;
+ - smart inserter
> - input and output.

Wire two intermediate chests to one pole, the last one to another pole, both wires. Wire last inserter to pick only when intermediate chests have =1 of both items. Wire intermediate inserters to pick only when last chest doesn't (<1) have both items. Wire first two inserters to filter-pick each of the two input items, and pick only when intermediate chests don't have their corresponding item (item1 <1 if inserter picks item1, item2 for other inserter).

And it'll work!

PS: There's also the matter of mixing stuff with different ratios...

[mohreb]

I made an alternator which is far from perfect but is kind-of deterministic and works for any number of incoming materials (here it is tested with 6) :


the upper chests giving the tick for inverters by moving the fuel (anything) by one if there is an output possibility in the chest, are linked to each of the 6 chest by green cables. the inputs are limited to 1 (in green circuit) of input and they take it when there is a coal in the next tick. the out puts put it out when it is their tick.
the order of output is dependent on how the belts are put together and need a bit of testing for each configuration (so in max speed they will still be in order). once set up this is slow but reliable way of putting things in desired order ,just need to be sure there wont be an overload, by assuring the output at the end. I have tested the case where one material is not coming, it should work (the split will be stopped until all lanes are providing). if you want to make more then one lane of the same material you should limit the ticks for more then 1 inserted or put the green wires more separately.

to speed things up you could put the tick-chests in a circle. or if you have many lanes i guess you can slow them down a bit by additional slow-inserter steps => more time between outputs so they will be easier to co-ordonate in the arriving lane.

[ironchefpython]

I use this layout. Without the stack limit on the bottom left smart chest, it can get in a state where it starts moving only a single color. This layout is very forgiving of wiring, you can wire all colors to all inserters and chests, and as long as you set the conditions right, it will work. The underground belts aren't needed, but I use them to keep everything linear, and be able to run series of belts across the map, and split out the desired combination at any point without interrupting belt flow



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