What is a belt balancer?
A belt balancer splits saturated inputs into outputs that all carry exactly the same rate. It is what you want when N identical machines must be fed identically from M full belts, and no machine may be favoured over another.
Saturated means the input belts are running full: the rate to divide is the belt tier rate, not whatever a mine happens to be producing. The calculator returns the buildings to place — Splitters, Mergers, and a loopback belt when the output count demands one — with the exact rate written on every belt.
Why do five outputs need a loopback?
Because an acyclic network of Splitters and Mergers can only produce fractions of the form a / (2^x · 3^y), and one fifth is not one of them. A Splitter divides by 2 or by 3, so a chain of them can only build denominators made of 2s and 3s: no arrangement, however clever, yields exactly 1/5 without sending some flow back upstream.
This is not a limitation of this calculator. It holds for any splitter-and-merger network, including the butterfly layouts people build by hand — the loopback is where the missing prime factor is paid for.
The output counts that need no loopback are exactly the 3-smooth ones: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32. Every other count pays for a return belt, and the price is steep. Measured on one saturated Conveyor Belt Mk.6 carrying 1200 items/min:
| Equal outputs | Rate per output | Loopback | Buildings |
|---|
| 4 | 300 items/min | no | 3 |
| 5 | 240 items/min | yes, 2 of them | 14 |
| 6 | 200 items/min | no | 3 |
| 7 | 171.4 items/min (1200/7) | yes, 2 of them | 24 |
| 8 | 150 items/min | no | 7 |
| 9 | 133.3 items/min (400/3) | no | 4 |
Nine outputs cost four buildings; five outputs cost fourteen. If the number of machines is still yours to choose, that table is the cheapest decision on this page.
Why does the order of factors matter?
Because a tree of factors costs the sum of its prefix products, so the small factors belong at the top. Dividing into 12 as 2 × 2 × 3 costs 7 Splitters; the same 12 as 3 × 2 × 2 costs 10.
| Order of factors | Splitters per level | Total |
|---|
| 2, 2, 3 | 1 + 2 + 4 | 7 Splitters |
| 3, 2, 2 | 1 + 3 + 6 | 10 Splitters |
The depth is identical either way, because it is fixed by the factorisation itself, so a flatter tree is not the argument. Swapping two adjacent factors changes exactly one prefix product, which is why the cheapest order is simply the increasing one — and why this calculator always puts the 2s above the 3s.
When is a manifold better?
A manifold is a single line of splitters with the machines hanging off it, balanced by back pressure rather than by the Splitter round-robin. It is the better answer whenever its three conditions hold, because it needs no Merger, no loopback and no inflated trunk belt.
- The machines must be identical, so that they all pull the same rate.
- The supply must cover the demand. Under-fed, a manifold does not settle down — the machines at the far end of the line starve permanently.
- The total flow must fit on one belt, because the head of the line carries all of it. That is the same weakness as a single trunk belt, and it is what rules the manifold out above 1200 items/min.
It is not cheaper in buildings, and that is worth knowing before choosing: a manifold for M machines costs M − 1 splitters, while an exact tree for 27 outputs costs only 13. The manifold wins on simplicity and on having no return belt to route, not on the count. This page grades the three conditions for your numbers and greys the option out when one of them fails.
What happens if the belt would overflow?
Nothing is drawn, on purpose. Belt capacity is a hard constraint here: no plan is ever returned with one belt over its tier rate — not one belt out of forty — and a topology forced by hand is held to exactly the same rule, because it is an analysis tool and not a way around the ceiling.
Instead the tool refuses and names the absolute floor: the highest rate that any topology whatsoever would have to carry, which is the larger of the biggest single input and the total divided by the number of outputs. Below that floor no layout exists, so the message names the belt tier that clears it.
Two saturated Mk6 inputs of 1200 items/min each, split into 5 equal outputs, is refused on a Mk4: the best topology would put 1440 items/min on one belt, the floor is 1200 items/min, and the answer is a Mk6 — where the same problem builds in 28 buildings.
Frequently asked questions
How many outputs can a splitter have?
Three. A Splitter has one input and at most three outputs; a Merger is its mirror image, at most three inputs and one output. Every plan respects both limits, which is why dividing by 5 costs a loopback instead of a five-way building.
Why does 7 outputs cost so much more than 8?
Because 7 has a prime factor above 3 and 8 does not. Measured on one saturated Mk6 belt at 1200 items/min: 7 equal outputs cost 24 buildings and two loopbacks, 8 outputs cost 7 buildings and no loopback, and 6 outputs cost 3 buildings. If the machine count is negotiable, moving from 7 to 6 or 8 is the cheapest change available.
What is a 3-smooth number?
A number whose only prime factors are 2 and 3, so that it can be written 2^x · 3^y: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32 and so on. These are exactly the output counts a splitter tree can hit with no loopback, because a Splitter divides by 2 or by 3 and by nothing else.
Does the loopback lose throughput?
No. A loopback carries flow back upstream to be divided again, and the plan accounts for it exactly, so the outputs still sum to the inputs. What it costs is buildings, belt length, and extra load on the belts it rejoins — never items.
