How much filler metal do I need?
Pick your joint type, weld size, and welding process. Get the total pounds of wire or rod, number of spools or boxes to order, and the estimated cost. Accounts for deposition efficiency and waste.
Updated September 2026
Quick answer
A 1/4-inch MIG fillet weld in mild steel needs about 0.15 lb of wire per foot (after efficiency and 10% extra waste). A 10 lb spool covers roughly 68 feet. Enter your joint below for exact pounds, spool count, and cost.
How to calculate filler metal for a weld joint
The formula is straightforward: figure the cross-section area of the finished weld, multiply by length and the filler's density to get deposited weight, then divide by deposition efficiency and add any extra waste to get the amount you actually need to purchase. The cross-section already covers every pass from root to cap, so the pass count never multiplies it.
For a fillet weld, the cross-section is a triangle: half the leg size squared. A flat-faced 1/4" fillet works out to 0.106 lb/ft of steel, which matches the 0.106 lb/ft Lincoln publishes. Real fillets carry some crown, so the calculator multiplies by 1.2. That 1.2 is our convexity allowance, not a published figure. For a square groove, the area is the plate thickness times the root gap plus a crown allowance of a tenth of the thickness. For a single-V, it adds the two bevels (thickness squared times the tangent of half the included angle, 60° by default) and a 0.0625" crown across the face.
Each filler uses its own density: 0.2836 lb/in³ for mild steel, 0.289 for stainless, and 0.0975 for aluminum. The same 1/4" fillet in ER4043 deposits 0.044 lb per foot against 0.128 lb in steel, about a third of the weight.
Not sure which wire classification to use? The wire & gas selection guide covers filler and shielding gas pairings by process and base metal.
Deposition efficiency by welding process
Values from ESAB Quick Weld Productivity Analyzer and Welding Answers deposition efficiency research. E7018 at 62% assumes a 14" electrode. Shorter electrodes (12") have a higher stub-to-rod ratio and drop to 59%.
Fillet welds vs groove welds: filler metal differences
A fillet weld fills a triangular gap between two perpendicular plates. The cross-section grows with the square of the leg size. Doubling the leg from 1/4" to 1/2" quadruples the metal required. This is why oversizing fillet welds costs so much: a 3/8" fillet uses 2.25× the material of a 1/4" fillet for proportionally less added strength.
A groove weld fills the space between two plates. A square groove only has the root gap to fill, so it stays small. A single-V adds both bevels, and that area grows with the square of the thickness. On 1/2" plate with a 1/8" gap, a 60° single-V holds 0.80 lb of steel per foot against 0.30 lb for a square groove. The single-V here assumes no root face, so a real prep with a land takes a little less.
Questions welders keep asking about filler metal
How much MIG wire for a 1/4 inch fillet weld?
A 1/4" equal-leg fillet weld in mild steel deposits about 0.128 lb per foot of weld (0.0106 lb/inch). At 95% deposition efficiency for GMAW solid wire plus 10% extra waste, you need about 0.148 lb per foot purchased. A 4-foot weld needs about 0.59 lb of wire. A 10 lb spool covers about 68 feet of 1/4" fillet.
How many pounds of 7018 per foot of weld?
For a 1/4" fillet weld with E7018 stick at 62% deposition efficiency, you need about 0.206 lb of electrodes per foot of weld. The 62% already counts the stubs, coating, and spatter, so no stub waste goes on top. A 50 lb box of 1/8" E7018 covers roughly 243 feet of 1/4" fillet weld.
What is deposition efficiency in welding?
Deposition efficiency is the ratio of weld metal deposited to filler metal purchased. MIG solid wire is about 95% (the rest is spatter). Flux-core is 78%–86% (flux and spatter). Stick is 55–62% (stub ends, coating, and spatter). TIG filler rod is about 98% (almost no loss). A lower efficiency means you buy more material per pound of weld deposited.
How do you calculate weld metal weight?
Cross-section area (in²) × weld length (inches) × filler density (lb/in³) = deposited weight. For a fillet weld, cross-section = 0.5 × leg² × 1.2 (convexity allowance). For a square groove, cross-section = thickness × (gap + 0.1 × thickness). For a single-V, it is thickness² × tan(half the included angle) + gap × thickness, plus a 0.0625" crown. Density is 0.2836 for steel, 0.289 for stainless, and 0.0975 for aluminum. The cross-section is the whole finished weld, so the number of passes does not go into it.
What size MIG wire spool should I buy?
Hobby and light-duty: 2 lb spools. Small fab shop on a 200A machine: 10 lb spools (most common). Production MIG on a larger machine: 33 lb or 44 lb spools. Larger spools have a lower per-pound cost and fewer changeovers. If you weld more than 10 hours per week, step up from 10 lb to 33 lb.
How much waste should I add for welding filler metal?
The extra waste field is for grinding, test beads, wire trim, and scrapped parts. It sits on top of deposition efficiency, which already covers spatter, slag, and stick stubs. The calculator starts wire processes at 10% and stick at 0%, because counting stubs again would buy the same rod twice. For field work or less experienced welders, add 5–10% more. For production MIG with good technique, you can drop to 5%.
Does multi-pass welding use more filler metal?
No. The leg size or groove sets the finished cross-section, and every pass is a slice of that same area. A 3/8" fillet run in 2 passes needs the same metal as one done in a single pass. The calculator shows about how many passes to plan for, but the weight comes from the cross-section alone. Grinding between passes is what the extra waste field is for.
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Deposition efficiency data from ESAB QWPA and Welding Answers. Fillet weld formula from TWI Job Knowledge 95. Flat-faced 1/4" fillet weight (0.106 lb/ft) from the Lincoln Procedure Handbook of Arc Welding; the 1.2× convexity allowance is ours. Other Tack Box tools.