Welding math

Electrode Consumption and Deposition Efficiency

The metal that ends up in a joint is always less than the electrode bought to put it there, and deposition efficiency is the factor that converts between the two, which is why price per pound of rod is not price per pound of weld.

The formula

N = W_deposited / (w_electrode x e), W_gross = W_deposited / e

The number of electrodes a job needs is the deposited weight divided by the product of one electrode's weight and the deposition efficiency, and the gross electrode weight to purchase is the deposited weight divided by the efficiency alone.

What each symbol means

SymbolMeaningUnits
NNumber of electrodes requiredcount
W_depWeld metal needed in the jointlb
w_eWeight of one whole electrodelb
eDeposition efficiency after slag, spatter, fume, and stub lossfraction
W_grossElectrode weight that must actually be bought and consumedlb

Why it works

Start from the deposit. Geometry gives the weight of weld metal a joint needs, and that number is fixed by the design. What varies is how much consumable has to be bought to deliver it. Dividing the deposited weight by the deposition efficiency gives the gross electrode weight, and dividing that by the weight of one electrode gives the count to issue from the store.

Efficiency differs enormously by consumable and by process. A covered electrode loses metal to slag, to spatter, and to the stub the welder discards, and the discarded stub alone can be a substantial fraction of a short rod. A continuously fed solid wire under gas has no covering to become slag and no stub to throw away, and it sits far higher. A flux-cored wire sits between the two, paying for its slag in efficiency.

Comparing consumables therefore requires dividing price by efficiency before comparing. Two products at the same price per pound are not the same price per pound of weld metal if one delivers sixty-five percent and the other ninety-five. Nor is that the whole comparison, because efficiency says nothing about deposition rate or about the cleaning time a slag system imposes, which land in operating factor instead.

Worked examples

Stick electrodes for a structural job

Weld metal required
20 lb
Weight of one electrode
0.10 lb
Deposition efficiency
0.65
  1. Gross electrode weight: 20 / 0.65 = 30.77 lb.
  2. Divide by the weight of one rod to get the count: 30.77 / 0.10 = 307.7 rods.
  3. Round up, since a fraction of a rod is not issuable.

Result: 30.77 lb of electrode, about 308 rods
More than ten pounds of the purchase never reaches the joint. On covered electrodes that loss is normal, and it is the number that has to be carried into any cost comparison against a wire process.

Larger electrodes at better efficiency

Weld metal required
50 lb
Weight of one electrode
0.25 lb
Deposition efficiency
0.70
  1. Gross electrode weight: 50 / 0.70 = 71.43 lb.
  2. Electrodes needed: 71.43 / 0.25 = 285.7, so round up.
  3. Note that the waste is 71.43 - 50 = 21.43 lb.

Result: 71.43 lb of electrode, about 286 rods
The heavier electrode carries a proportionally smaller stub loss, which is part of why efficiency improved. Over twenty pounds still leaves as slag, spatter, and fume.

In practice

  • Stub loss is a real and controllable cost. A shop that habitually discards long stubs pays for it directly, and the effect is largest on the shortest electrodes.
  • Efficiency figures quoted for a consumable assume reasonable technique. Excessive spatter from wrong polarity or a long arc pushes the real figure well below the published one.
  • Buy on cost per pound of deposited metal, not cost per pound of consumable, and remember to include the shielding gas and the cleaning time in the comparison.
  • Estimates built from ideal joint geometry are already low, so apply a convexity and overwelding allowance before dividing by efficiency, not after.

Related topics

Practise the arithmetic under pressure

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Welding math questions