Welding Rod Consumption Calculator
Work out how many stick electrodes a fillet weld will burn from the weld length, leg size and deposition efficiency.
How it works
Deposited weight = (leg² ÷ 2) × length × 0.284 lb/in³ · Electrode consumed = deposited ÷ efficiency · Rods = deposited ÷ deposit per rod (rounded up)
An equal-leg fillet has a triangular cross-section equal to leg² ÷ 2, so the deposited weld-metal volume is that area times the weld length, and steel weighs 0.284 lb per cubic inch. Deposition efficiency accounts for spatter, slag and the stub you throw away — stick sits around 60–65%, which is why the electrode you buy always weighs more than the metal that ends up in the joint. Buy a little extra: real jobs lose rods to arc strikes, restarts and short stubs.
Worked example
A 20 ft (240 in) run of 1/4-inch fillet: cross-section = 0.25² ÷ 2 = 0.03125 in²; volume = 0.03125 × 240 = 7.5 in³; deposited = 7.5 × 0.284 = 2.13 lb. At 0.035 lb per rod that is about 61 rods, consuming roughly 3.28 lb of electrode at 65% efficiency.
Frequently asked questions
Why is electrode consumed more than metal deposited?
Deposition efficiency is always below 100%: some of the rod becomes spatter and slag, and every stick leaves an unusable stub. At 65% efficiency you burn about 1.5 lb of electrode for every 1 lb that stays in the weld.
How much does one stick electrode deposit?
It depends on diameter and length, but a common 1/8-inch × 14-inch E7018 deposits roughly 0.03–0.04 lb of weld metal after stub loss. Larger 5/32 and 3/16 rods deposit more per stick.
Does this cover multi-pass welds?
Yes — enter the finished leg size and the fillet-area formula captures the total metal, whether you lay it in one pass or several. For grooves or very large fillets, add a little for reinforcement (crown) above the theoretical throat.
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This calculator is for educational purposes. Double-check important results before acting on them.