Train load/unload balancing

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lynce
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Train load/unload balancing

Post by lynce » Thu Nov 28, 2019 6:23 pm

Hi.

I'm still kinda new to this game, and keep figouring things out.
I remember watching Tuplex's vids when he did som kinda logic setup to load the chest evenly.
Can't remember exactly how he did it, so i was hoping someone here can give me some advice.

This is my setup: https://imgur.com/a/NrNyLW5

DaleStan
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Re: Train load/unload balancing

Post by DaleStan » Thu Nov 28, 2019 6:59 pm

I would use a 1-to-6 balancer for each wagon rather than using circuits.


But if you want circuits, wire each bank of 6 chests together and send that to the input of an "iron ore / -6, output iron ore" combinator. Use the same wire color to connect the combinator output to all the inserters. The use the other wire color to connect each inserter directly to its chest[1]. Now set the inserters to enable when iron ore ≤ 0.

If that stutters too much, you can give up some perfection for better chest-loading throughput by dividing by -7, and/or by setting a larger number in the inserters.

[1] You should now have eight circuits for each wagon: Six in one color that have one chest and one inserter each, and two in the other color, one connecting all the chests to a combinator, and the other connecting the combinator to all the inserters.

plussantana
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Re: Train load/unload balancing

Post by plussantana » Thu Nov 28, 2019 7:07 pm

think a little bit. You don't have to look for a guide for something basic. Use the wiki and it will show you what you have on hand. do it conveniently for your expectations and further development

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Re: Train load/unload balancing

Post by plussantana » Thu Nov 28, 2019 7:24 pm

To avoid all the circuit problem we could go directly to the division by 8 of the total like this: x / 8 if you use 7 splitters you can divide the result by 8 in this way: the splitter 1 (S1) is placed at the output of S1.S2 and S3 are placed at the exit of S2. S4 and S5 are placed and at the exit of S3. S6 and S7 are placed (keep in mind that the splitters are the letters S)

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