Here it is for the people who don't want the math

Beyond this point is some NERD ♥♥♥♥. Precede at your own peril.

I am going to talk about each belt as a fraction of the number of items per second it carries divided by the number of items the original belt carries.

For example, If the input belt carries 10 items per second, and it is split into 2 belts that each carry 5 items per second, the 2 belts will each be 1/2. This is so I can generalize and not have to talk about specific numbers of items per second, as it makes the math easier.

The input belt is split into 2 1/2 belts, then 4 1/4 belts, 3 of those 1/4 belt continue on to whatever building awaits them. The 4th belt is put back into the input splitting it into 2 1/8 belts, then 4 1/16 belts 3 continue on, and so on.




Each time the forth belt loops back around it is added to the other 3 belts making an infinite sum of fractions.

Algebra time.

S = the infinite sum of fractions being added to one output belt.

S = 1/4 + 1/16 + 1/64 +...

the first step is multiplying both sides by 4

4S = 4(1/4 + 1/16 + 1/64 +...)

due to the distributive property everything in the parentheses is multiplied by 4

4S = 1 + 1/4 + 1/16 + 1/64 +...

the next step is subtracting 1 from both sides

4S - 1 = 1/4 + 1/16 + 1/64 +...

this step is substituting the infinite sum with S, because the right side of the equation is identical to the definition of S

4S - 1 = S

The final step is to solve for S

4S - 1 = S
-4S -4S

-1 = -3S
*-1/3 *-1/3

-(-1/3) = -(-3S/3)

1/3 = S

1/3 = 1/4 + 1/16 + 1/64 +...

The simple option seems obvious, "Why don't you just not use one of the output? There are 3 belts if you just ignore one." The problem with that is a balance can't split a belt if there isn't 2 belts to split onto, so it will output everything onto one belt instead of 2. It will produce 2 1/4 belts and 1 1/2 belt instead of 3 1/3 belts,


With this configuration some buildings will be getting too many items, and some not enough. Efficiency is the name of the game, you can't have something so simple cause so much inefficiency.

This same logic can be used to split a belt into any number of equal belts. All you have to do is find the smallest number that when multiplied by 2 a few times equals the number of belts you want( A simple way to find this is to take the number of belts you want and keep dividing it by 2 until it can't be divided evenly anymore.), split the belt into the next biggest power of 2 (2, 4, 8, 16, 32, etc.) and send the unneeded belts back to the input, then split the remaining belts until you have the number of belts you want.

For this example I will split a belt into 10 equal parts.

First divide 10 by 2 until it cant be divided evenly.

10/2 = 5

Then split a belt into the next smallest power of 2, 8 parts and send the unneeded 3 back into the input.


Then split all 5 remaining belts in half


Then send the output to where ever you desire.

I'm gonna quickly show how to set it up with splitters and mergers its a lot easier and requires less thinking.

First make a line of outputs.

Second add a splitter for every other output.


Third repeat step 2 untill there are only 2 splitters


Fourth reconnect the unused outputs.


Fifth use a balancer to reinsert unused outputs and input.


If you use a splitter and merger for this step instead of a balancer there will be more than the max items/second on the belt between the splitter and inserter causing a clog and loss of efficiency.


And thats it.

This guide was inspired by a comment on another guide who suggested the simple option and my passion for math and overthinking automation games.
And thank you for reading my first guide, please tell me if I made a mistake I wrote this at 2 am and have subsequently discovered that I hate this with a passion, so I will not be taking feed back and will be promptly forgetting about this guide entirely, as it's 3 am right now and I didn't proofread it very well. Ok for real this time I'm not coming back. I just had to add that part about splitters and mergers.

I'm back with new knowledge and a whiteboard. this is a general proof if you care to decipher my scribbles.

General Proof























n=5 case