Welding math

Deposition Rate

Deposition rate measures how fast a process puts metal in the joint while the arc is actually burning, which is only half the productivity picture and often the less important half.

The formula

DR = W_deposited / t_arc, or DR = melt-off rate x efficiency

Deposition rate in pounds per hour is the weight of metal deposited divided by the arc-on hours it took, or equivalently the rate at which electrode is consumed multiplied by the deposition efficiency.

What each symbol means

SymbolMeaningUnits
DRDeposition rate, weld metal laid down per hour of arc timelb/hr
WWeight of weld metal actually deposited in the jointlb
t_arcArc-on time, not clock timehours
MRMelt-off rate, the rate the electrode itself is consumedlb/hr
eDeposition efficiency, deposited weight per pound of electrode consumedfraction

Why it works

The definition is deliberately narrow. Deposition rate counts only arc-on time, so a process that lays down twenty pounds an hour while burning does exactly that regardless of how long the welder spends fitting, chipping, or waiting. Keeping the definition narrow is what makes the number comparable between processes; combining it with operating factor is what makes it comparable between shops.

There are two ways to arrive at it. Weighing a test coupon before and after and dividing by the timed arc period is the direct measurement, and it is the one that settles arguments. The indirect route multiplies the melt-off rate by deposition efficiency, which is useful because melt-off is easy to compute from wire feed speed and wire diameter, and because it makes the efficiency loss explicit.

Deposition rate rises with current, which is why processes and positions that tolerate high current are the productive ones. A flat-position submerged arc weld runs currents no overhead stick welder could hold. That relationship also sets a limit: pushing current for deposition raises heat input at the same time, so on a heat-input-limited procedure the productive setting and the permissible setting can be different numbers.

Worked examples

Weighing a timed test coupon

Weld metal deposited
6.4 lb
Arc-on time
48 minutes
  1. Convert arc time to hours: 48 / 60 = 0.8 hours.
  2. Divide deposited weight by arc hours: 6.4 / 0.8.

Result: 8 lb/hr
Eight pounds an hour of arc time is a respectable semi-automatic figure. What it does not say is how many of the shift's hours were arc-on, which is where most of the real difference between shops lives.

Deriving it from melt-off and efficiency

Electrode melt-off rate
12 lb/hr
Deposition efficiency
0.85
  1. Recognise that only the efficient fraction of consumed electrode stays in the joint.
  2. Multiply: 12 x 0.85.

Result: 10.2 lb/hr
Nearly two pounds an hour of electrode never becomes weld metal, leaving instead as slag, spatter, and fume. That gap is what separates the consumable purchase from the deposit.

In practice

  • Arc-on time is not shift time. A deposition rate quoted against clock hours is roughly three times too low and describes something else entirely.
  • Melt-off rate can be computed from wire feed speed and wire cross-section without weighing anything, which makes the indirect route practical for wire processes.
  • Position dominates deposition rate on manual work. The same welder, consumable, and machine will deposit far less overhead than flat, because the puddle sets the ceiling on current.
  • Chasing deposition rate on a heat-input-limited joint is self-defeating; the limit will bind before the process runs out of capability.

Related topics

Practise the arithmetic under pressure

The welding-math domain in the question bank is where these formulas get tested the way they get tested on paper — with distractors built from the mistakes people actually make.

Welding math questions