Feed Three Iron Smelters with a Simple Manifold

Feed three Smelters from a single line carrying at least 90 Iron Ore per minute. This unmodded PC example uses the standard Iron Ingot recipe, ordinary splitters and unlocked Mk.2 belts.

Each Smelter at 100% clock uses 30 ore per minute and makes 30 ingots. Three therefore need 90 ore and produce 90 ingots per minute. A belt’s capacity does not guarantee that your Miner supplies that amount. Iron Ingot recipe

Connect the row

Place three Smelters, select Iron Ingot on each, and connect all three to power.

1. Run a Mk.2 ore belt into ordinary splitter A.
2. Branch from A to Smelter 1 with a Mk.1 belt.
3. Continue the main Mk.2 belt from A into splitter B.
4. Branch from B to Smelter 2 with a Mk.1 belt.
5. Run the remaining main line directly into Smelter 3.

Two splitters are enough; the final machine takes the tail of the line. A Mk.1 branch can carry 60 items per minute, enough for one Smelter’s 30. The Mk.2 main belt can carry 120. Keeping the main line at Mk.2 avoids an accidental 60-item limit before the first split. Conveyor capacities

Connect each ingot output to separate storage or a consumer that has room. If you combine the outputs, the shared belt needs capacity for 90 ingots per minute; follow the merger capacity guide.

Let the manifold fill

Earlier machines fill before later ones. Once their inputs back up, the splitters send items toward the remaining available branches. Allow that startup period before judging the last Smelter. Manifold behavior

The setup is working when all three keep receiving ore and producing ingots. Persistent starvation calls for an actual supply-rate check, then a search for a slow main-belt segment, reversed connection, wrong recipe or clock setting. A blocked ingot output also stops a machine. For storage that regularly fills, add Smart Splitter overflow.