Weld Heat Input Calculator

Arc power and heat input from voltage, current, and travel speed, in metric or imperial units.

Physics and EnergyUpdated 2026-10-193 min read
Weld Heat Input Calculator
Arc power
...
Heat input...
Heat input in the other unit...

What it does

Heat input is the amount of arc energy delivered per unit length of weld. It shapes cooling rate, grain structure, distortion, and the risk of cracking, which is why welding procedures often set a maximum or a range. The calculation takes the electrical power of the arc and divides it by how fast the torch or electrode travels.

Arc power is straightforward. Voltage in volts times current in amps gives watts, and dividing by 1,000 gives kilowatts. A 25 volt arc at 200 amps is 5.00 kW. That power enters the work over the length the arc travels in a minute, so heat input follows as 60 times volts times amps divided by travel speed.

Unit systems change the constant. In imperial units, travel speed is inches per minute and heat input comes out in kilojoules per inch. A 5.00 kW arc moving at 12 inches per minute deposits 25.0 kJ/in. In metric units, the same arc moving at 304.8 mm per minute deposits 0.98 kJ/mm, and the two figures describe the same weld.

The calculator reports both units at once so a procedure written in kJ/in can be checked against a metric specification without a separate conversion. The relationship is 1 kJ/in = 0.03937 kJ/mm, or equivalently 1 kJ/mm = 25.4 kJ/in.

The result is a lower bound on the energy that reaches the metal. Real processes lose energy to spatter, radiation, and the arc atmosphere, so standards sometimes apply an efficiency factor before comparing heat input with a limit. This tool follows the common practice of reporting the nominal arc energy with no efficiency applied.

How to use

  1. Pick imperial or metric units.
  2. Enter the arc voltage from the welding procedure.
  3. Enter the welding current in amps.
  4. Enter the travel speed in in/min or mm/min.
  5. Read the arc power and heat input in the selected system.
  6. Compare the other-unit figure with any specification written in the opposite system.

Formula

Arc power (kW) = voltage x current / 1,000. Heat input (kJ/in) = 60 x voltage x current / (1,000 x travel in in/min). Heat input (kJ/mm) = 60 x voltage x current / (1,000 x travel in mm/min).
V
Arc voltage in volts.
I
Welding current in amps.
Travel
Travel speed in inches per minute or millimeters per minute.
Heat input
Arc energy per unit length of weld, in kJ/in or kJ/mm.

The 60 converts minutes to seconds so the result lands in kilojoules per unit length. The formula uses nominal arc voltage and current and assumes all arc energy enters the weld. Process efficiency, joint geometry, and preheat are not included. For code work, apply the efficiency factor or the measuring method named in the applicable standard.

Worked example

Example: 25 V at 200 A traveling 12 in/min

A welding procedure uses 25 volts, 200 amps, and a travel speed of 12 inches per minute.

  1. Arc power: 25 x 200 = 5,000 W, or 5.00 kW.
  2. Imperial heat input: 60 x 25 x 200 / (1,000 x 12) = 300,000 / 12,000 = 25.0 kJ/in.
  3. Convert to metric: 25.0 / 25.4 = 0.98 kJ/mm.
  4. Metric check with 304.8 mm/min: 60 x 25 x 200 / (1,000 x 304.8) = 300,000 / 304,800 = 0.98 kJ/mm.
  5. The two systems agree because 304.8 mm/min is the same speed as 12 in/min.
  6. Faster travel would lower the heat input while leaving arc power unchanged.

Reference table

Heat input at 25 V and 200 A for common travel speeds
Travel speedHeat inputIn the other unit
6 in/min50.0 kJ/in1.97 kJ/mm
12 in/min25.0 kJ/in0.98 kJ/mm
18 in/min16.7 kJ/in0.66 kJ/mm
24 in/min12.5 kJ/in0.49 kJ/mm

Limits

  • Assumes all arc energy enters the weld, with no efficiency factor applied.
  • Uses nominal voltage and current, not measured waveforms.
  • Does not account for preheat, interpass temperature, or joint geometry.
  • Applies to a steady travel speed, not to weaving or pulsed techniques with varying speed.

Common mistakes

  • Using inches per minute in the metric formula or millimeters per minute in the imperial formula.
  • Forgetting the factor of 60 that converts per-minute travel into per-second energy.
  • Treating heat input as the total heat in the part rather than energy per unit length.
  • Comparing a nominal heat input with a code limit that requires an efficiency factor.
  • Changing current without noting that arc power and heat input both move with it.

Try it yourself

Quick checks. Answers sit tucked under each question.

30 V, 250 A, 15 in/min. What is the heat input?

30.0 kJ/in, which is 1.18 kJ/mm.

25 V, 200 A, 304.8 mm/min. What is the heat input?

0.98 kJ/mm, which is 25.0 kJ/in.

25 V, 200 A, 24 in/min. What is the heat input?

12.5 kJ/in, which is 0.49 kJ/mm.

Questions

What is weld heat input?

The arc energy delivered per unit length of weld, usually expressed in kJ/in or kJ/mm.

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

Divide by 25.4. For example, 25.0 kJ/in is about 0.98 kJ/mm.

Does faster travel increase heat input?

No. Faster travel spreads the same power over more length, so heat input falls.

Why does my code value differ from this result?

Codes may require an efficiency factor or a measured energy input rather than nominal arc power.

Does heat input include preheat?

No. Preheat and interpass temperature are separate variables that affect cooling but are not part of this calculation.

Which unit system should I use?

Use the one in your procedure. The tool reports both so you can cross-check a specification written in the other system.

Sources

How this page was checked. The formula above was matched against the cited sources, and the worked example was run through the tool on this page before publishing. If you spot an error, contact us so we can fix it and note the correction.