Weld Heat Input Calculator

Work out welding heat input from volts, amps and travel speed with the ASME IX QW-409.1 formula, or from the energy a pulse machine reports. Compare it with the EN 1011-1 heat input, find the slowest travel speed or highest current for a limit, and check every pass of a procedure.

Calculator Welding Updated Oct 4, 2026
How to Use
  1. Pick what to work out: the heat input, the slowest travel speed or the highest current or voltage that keeps you under a limit, the heat input from a pulse machine’s energy reading, or a whole table of passes.
  2. Enter the arc voltage and welding current, and the travel speed with its unit (in/min, mm/min, mm/s, cm/min or m/min).
  3. Choose the process. ASME and AWS heat input is the arc energy itself; the process factor k (0.8 for MIG, stick and flux-core, 0.6 for TIG, 1.0 for submerged arc) is used only for the EN 1011-1 figure.
  4. For a waveform-controlled or pulsed procedure, choose Pulse / energy and enter the energy (or average power and arc time) the machine or meter shows, and the length of the bead.
  5. Read the answer in the highlighted field and the readouts, in kJ/in or kJ/mm. Show Work lists each step.
Input
EN 1011-1 factor k
V
A
volts amps speed, one pass per line
Presets
Arc Energy and Heat Input
Heat input (ASME)
—
Same in
—
EN 1011-1 heat input
—
Arc power
—

Worked Example

A spray-transfer MIG fillet. Lincoln Electric’s GMAW guide (C4.200) lists 0.045 in wire at 300 A, 29 V and 18 in/min for a 1/4 in fillet. Heat input = 29 × 300 × 60 ÷ 18 = 29,000 J/in = 29 kJ/in, or 29 ÷ 25.4 = 1.142 kJ/mm. The EN 1011-1 heat input is 0.8 × 29 = 23.2 kJ/in.

A pulsed weld from the machine’s energy reading. In Lincoln Electric’s example the power source shows 22.3 kJ at the end of a 3.6 in bead. Under QW-409.1(c) the heat input is 22.3 ÷ 3.6 = 6.19 kJ/in; no voltage, current or travel speed is needed.

The common mistake: mixing the ASME and EN numbers. A stick weld at 30 V, 120 A and 150 mm/min has 30 × 120 × 60 ÷ 150 = 1,440 J/mm, so 1.44 kJ/mm is its ASME/AWS heat input. The EN 1011-1 figure is 0.8 × 1.44 = 1.152 kJ/mm. Checking an ASME procedure limit against 1.152 lets the weld run 25% hotter than the limit allows, because 1.44 ÷ 1.152 = 1.25. Leaving out the 60 when the speed is per minute is the other slip: it gives 0.024 kJ/mm, sixty times too small.

Show Work

Enter values and calculate to see the step-by-step breakdown.

Formulas

ASME IX QW-409.1(a)
HI = V × A × 60 ÷ S
S in in/min gives J/in; S in mm/min gives J/mm
QW-409.1(c)(1)
HI = energy ÷ bead length
Energy in joules from a meter that follows the waveform
QW-409.1(c)(2)
HI = power × arc time ÷ bead length
Power in J/s (W), arc time in seconds
EN 1011-1 heat input
Q = k × arc energy
k = 1.0 SAW; 0.8 SMAW, GMAW, FCAW; 0.6 GTAW and plasma
Minimum travel speed
S = V × A × 60 ÷ HImax
Any slower and each length of weld gets more energy
Units
1 kJ/mm = 25.4 kJ/in
1 kJ/cm = 0.1 kJ/mm; 1 kJ/in = 0.03937 kJ/mm

Arc Energy, Heat Input and the 2010 Code Change

Heat input matters because it sets how fast the weld and the heat-affected zone cool, and the cooling rate decides the microstructure and so the toughness and hardness. ASME Section IX, the welding and brazing qualification standard first published in 1941, deals with it in variable QW-409.1. Method (a) is the familiar volts × amps × 60 ÷ travel speed; method (b) works from the volume of weld metal deposited.

Pulsed and other waveform-controlled power sources broke method (a). Ordinary meters show average voltage and average current (or RMS values for AC), and the product of two averages is not the average power when both swing thousands of times a second. As Teresa Melfi of Lincoln Electric explained in the Welding Journal in June 2010, the 2010 edition of Section IX added method (c): take the energy, or the power and the arc time, from a meter that samples the waveform about 10,000 times a second, and divide by the bead length. For waveform-controlled procedures only methods (b) and (c) are allowed.

European practice draws a further line. The older British standard BS 5135 used the term arc energy; EN 1011-1 defines heat input as arc energy multiplied by a thermal efficiency factor k for the process, measured relative to submerged arc welding, which TWI tabulates as 1.0 for submerged arc, 0.8 for stick, MIG/MAG and flux-cored welding and 0.6 for TIG and plasma. American codes still call the arc energy the heat input, so always say which one a number is.

About This Tool

This calculator works the ASME IX QW-409.1 heat input from volts, amps and travel speed, or from an energy or power reading for pulsed and waveform-controlled welding, and gives the EN 1011-1 heat input alongside it. It can also turn a limit round into the slowest travel speed or the highest current or voltage that stays under it, and check a list of passes against that limit. The formulas follow ASME IX as quoted in Lincoln Electric’s Welding Journal article; the k factors are TWI’s.

The results are calculations, not qualification records: the limits that apply to your job come from the welding procedure specification and the code it was qualified to. Everything runs in your browser; nothing you enter is sent anywhere.

Related tools: Carbon Equivalent & Preheat Calculator, MIG Wire Feed Speed & Deposition Rate Calculator, and Welding Amperage Chart.

Frequently Asked Questions

How do you calculate welding heat input?

Multiply volts by amps by 60 and divide by the travel speed per minute: ASME IX QW-409.1(a). 29 V and 300 A at 18 in/min gives 29 × 300 × 60 ÷ 18 = 29,000 J/in, which is 29 kJ/in or 1.142 kJ/mm. With the speed in mm/min the same sum gives joules per millimetre.

What is the difference between arc energy and heat input?

Arc energy is V × A ÷ travel speed. EN 1011-1 heat input multiplies it by a process factor k (TWI lists 1.0 for submerged arc, 0.8 for stick, MIG and flux-core, 0.6 for TIG). A stick weld at 30 V, 120 A and 150 mm/min is 1.44 kJ/mm of arc energy and 1.152 kJ/mm of EN heat input. ASME and AWS call the 1.44 figure the heat input.

How do I work out heat input for pulsed MIG?

Since the 2010 edition, ASME IX QW-409.1(c) has you use the energy (or power × arc time) from a meter that samples fast enough to follow the waveform, divided by the bead length. Lincoln Electric’s example: 22.3 kJ for a 3.6 in bead is 22.3 ÷ 3.6 = 6.19 kJ/in. Average volts and amps from ordinary meters can be well off for pulsed waveforms.

How do I convert kJ/in to kJ/mm?

Divide by 25.4, because there are 25.4 mm in an inch. 29 kJ/in is 29 ÷ 25.4 = 1.142 kJ/mm, and a 1.5 kJ/mm limit is 1.5 × 25.4 = 38.1 kJ/in.

How fast do I have to travel to stay under a heat input limit?

Turn the formula round: minimum speed = V × A × 60 ÷ the limit. At 29 V and 300 A, staying at or under 40 kJ/in needs 29 × 300 × 60 ÷ 40,000 = 13.05 in/min (331.5 mm/min) or faster. Going slower puts more energy into each inch of weld.

How do I use the Weld Heat Input Calculator?

Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.

Is it free? Does it work without internet?

Yes to both. It is free with no sign-up, and once the page has loaded it keeps working even with no internet.

Where does my data go?

Nowhere — every calculation runs on your own device. Nothing you enter is uploaded, logged, or stored.

Common Use Cases

Checking a WPS

Enter each pass of a procedure qualification and a 45 kJ/in limit: passes at 33.6, 29 and 46.5 kJ/in show the third one over, so it needs a faster travel speed or less current.

Stainless steel

A TIG root at 12 V and 150 A on 300-series stainless held under 1.5 kJ/mm (EN, k = 0.6) needs a travel speed of at least 43.2 mm/min.

TMCP steel limits

TWI quotes limits such as 2.5 kJ/mm for 15 mm TMCP plate. A MIG pass at 29 V and 300 A meets it (EN, k = 0.8) at 167 mm/min, about 6.58 in/min, or faster.

Pulse and waveform machines

Divide the energy the machine reports by the bead length: 5.2 kW for 30 s over 150 mm is 156 kJ ÷ 150 mm = 1.04 kJ/mm.

Thin sheet

Short-circuit MIG on 16 gauge at 19 V, 120 A and 20 in/min (Lincoln’s starting point) is 6.84 kJ/in, about a quarter of a spray-transfer fillet.

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