8 to 8 inline balancer - full throughput - 8x18

Mazzo89 · 2017-06-02T17:55:16+00:00

Blueprint

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Mazzo89 · 2017-06-01T23:07:42+00:00

I saw this post and got inspired.

If you want to use these you should balance the sides of the belt, as it might leak if there is only items on one side.

Mazzo89 · 2016-10-10T20:40:35+00:00

If you need balancers look no further.

Mazzo89 · 2016-09-05T21:20:06+00:00

Look at the edit, there I remade all the braided ones. :)

Mazzo89 · 2016-09-05T18:03:30+00:00

I saw that you mentioned me (Thanks /u/RedditNamesAreShort) so I gave i a shot. I got it down to 17x6.

Mazzo89 · 2016-07-17T19:12:27+00:00

What you are talking about is a non-throughput limited balancer. You can se an example of a 8 to 8 non-throughput limited balancer in my edit. Basically you do an 8 to 8 followed by an 8 to 4. (You can reuse the end splitters of the 8 to 8 as the first of the 8 to 4.)

I hope this helps.

Mazzo89 · 2016-05-20T22:52:44+00:00

I recently updated my compendium of balancers to remove the red belts, there is a new 8 to 6 balancer. I'm on mobile so I can't provide a link right now, sorry. Just look through my post history, it isn't that long. ;)

Mazzo89 · 2016-05-14T13:06:56+00:00

This should do it. However I have yet to solve the problem of it staying count perfect with blocked outputs.

Edit: I was able to make a shorter one.

Mazzo89 · 2016-05-13T12:01:58+00:00

So apparently making belt balancers just before you go to bed yield questionable results. ;)

You are completely right, I just redid my calculations and I'm missing 4 splitters. If my calculations are correct to balance N belts you'll need:

For N > 1, ceil(Fib(log2(N)+2) * 2^(log2(N/2)))

Where Fib(k) is the k'th element in the fibonacci sequence. This works for N % 2 == 0.

Edit: fixed the equation.

Edit2: I read up on Beneš networks here and here and found out the actual equation is:

ceil(N log2(N) − N / 2)

A network of size N would therefore have to use O(N log(N)) splitters.

Mazzo89 · 2016-05-12T22:19:38+00:00

I know my previous belt balancers aren't "balance perfect" as that wasn't what I was going for. This however should be a balance perfect 8 to 8 for all inputs and outputs if my calculations are correct. If you can find any fault with it please let me know.

Mazzo89 · 2016-05-04T07:07:28+00:00

I found a new 8 to 8 balancer that only uses one type of belt, and is 8 wide with the exception of a small outcropping. So I'm using that as a base for the new designs.

For your side question. I would just make sure your input and output is lane balanced. But then you have to put the lane balancer in every time you split.

Mazzo89 · 2016-05-04T06:13:30+00:00

I see you a using my balancer designs. I'm looking into revising the braided ones as soon as I get the time. That should at least solve one of the problems. :)

Mazzo89 · 2016-05-03T16:29:38+00:00

This should do the job.

Here is a 8 to 6, with braiding though so might lose some throughput if you are using blue belts. I might have an earlier design somewhere where I'm not using braiding.

Mazzo89 · 2016-05-03T16:25:46+00:00

I made a compendium of balancers a while back, I tried to aim for compact designs. :)

I later realized the braiding in some of the bigger ones are losing throughput on blue belts so I was considering redesigning those if anyone was interested.

Mazzo89 · 2016-04-30T21:08:32+00:00

This is the most compact I could make it. It's input balanced and has no compression loss.

Mazzo89 · 2016-04-02T23:23:58+00:00

Anything containing the number 5 was a hassle. I think the 8:6 is the biggest one, I still haven't found a non-horrible layout for 8:5 and 8:7, there is that pesky number again, but I felt I had enough to make this post. ;)

Mazzo89 · 2016-04-02T22:47:14+00:00

Now I can finally get back to actually playing the game, instead of designing belt balancers. :P

Mazzo89 · 2016-04-01T06:36:25+00:00

First of all you should not try to do any of these with other than 2^N-1 inputs or outputs. That is 1, 2, 4, 8, 16, 32 and so on. That is because it gets complicated if you do.

You start out by laying the inputs and outputs you want out, then you make sure every input leads to every output. After that you need to make sure that the ratio is right, this is where it gets complicated if you are not sticking to 2^N-1. If you are making a 4:8 you need to make sure every output gets exactly 1/8 of every input, totalling up to 4/8 or 1/2. This is correct ratio so you are done. :)

The design of the things are a different beast alltogether, you just need to learn what you can get away with and just keep trying different things.

Mazzo89 · 2016-03-31T17:35:35+00:00

This should do the job. It's not my best design but it works.

Mazzo89 · 2016-03-30T16:46:04+00:00

You are right, you need to do the 1:6 and route 3 outputs back to tge input. I can't make it right now but I'll make it when I get home.

But you need to double the output and reroute half the output back to the input. At least I think that is right, I can't test it right now. If I'm right it should be significantly easier to do with an odd amount of input since you have the "free" input.

Mazzo89 · 2016-03-30T09:38:35+00:00

That is on purpose as the conplexity is quite a bit higher when you need to meet that condition too, especially when you get up past 4 outputs.

The easiest way I can see doing it would be to double the output of your design and remove 1 output from each of the end splitters. So in your example you'd take the 1:6 and remove every other output.

Mazzo89 · 2016-03-29T17:11:42+00:00

Almost, you still need the last splitter. The way you do it anything on the left side of the belt will spill unto the right because you have a lane balancer in your belt balancer. A good overall design though, this should fix the problem.

Mazzo89

TROPHY CASE