Free Tool
Welding Heat Input Calculator
Enter amps, arc voltage, and travel speed to calculate weld heat input in kilojoules per inch (kJ/in) and per millimeter (kJ/mm) instantly.
Heat Input Formula
kJ/in = (A × V × 60) ÷ (IPM × 1000)Arc energy per inch of weldkJ/mm = kJ/in ÷ 25.4Convert to per-millimeterUses arc-energy heat input. Some codes apply a thermal-efficiency factor for the process (e.g. 0.8 for SMAW/GMAW) — check your WPS or code.
How Welding Heat Input Is Calculated
Heat input is the amount of arc energy delivered to the weld per unit of length, and it is one of the most important variables a welding inspector monitors. It is calculated from three parameters: welding current (amps), arc voltage (volts), and travel speed. The formula multiplies amps by volts and by 60, then divides by travel speed in inches per minute times 1000, giving heat input in kilojoules per inch (kJ/in). The factor of 60 converts per-minute travel to a per-second energy basis, and dividing by 1000 converts joules to kilojoules.
For example, welding at 200 amps and 25 volts with a travel speed of 10 inches per minute gives (200 × 25 × 60) ÷ (10 × 1000) = 30 kJ/in. To express the same value in metric units, divide by 25.4 to get about 1.181 kJ/mm. Notice that faster travel lowers heat input while higher amperage or voltage raises it — which is why a welding procedure specification (WPS) fixes ranges for all three parameters. Travel speed sits in the denominator, so it must be greater than zero for the result to be defined.
Heat input directly affects cooling rate, the size of the heat-affected zone, distortion, and final mechanical properties, so CWI and welding-inspector exams test it heavily. Keep in mind this tool reports arc-energy heat input from raw voltage and current; some codes apply a process-dependent thermal-efficiency factor (for instance about 0.8 for SMAW and GMAW) to obtain "true" heat input, so always confirm which definition your code or WPS requires.
Frequently Asked Questions
How do you calculate welding heat input?
Heat input equals amps times volts times 60, divided by travel speed times 1000. Amps and volts are the arc current and voltage, travel speed is in inches per minute, and the result is in kilojoules per inch (kJ/in). For example, 200 A at 25 V traveling 10 in/min gives (200 × 25 × 60) ÷ (10 × 1000) = 30 kJ/in.
How do I convert kJ/in to kJ/mm?
Divide kJ/in by 25.4, since there are 25.4 millimeters in an inch. A heat input of 30 kJ/in equals about 1.181 kJ/mm. To convert the other way, multiply kJ/mm by 25.4. Many welding procedure specifications written to ISO standards list heat input in kJ/mm.
Why does travel speed have to be greater than zero?
Travel speed is in the denominator of the heat input formula, so a value of zero would mean dividing by zero — an undefined result. Physically, a travel speed of zero means the arc isn't moving, so there is no defined heat input per unit length of weld. The calculator requires a positive travel speed before it returns a result.
What is a typical heat input range for welding?
Typical heat input runs from roughly 15 to 60 kJ/in (about 0.6 to 2.4 kJ/mm) depending on process, material, and thickness. Welding procedure specifications set minimum and maximum limits to control cooling rate, mechanical properties, and the size of the heat-affected zone. Always follow the WPS for the job.
Does this include arc efficiency?
No. This calculator reports arc-energy heat input, which uses the raw voltage and current at the arc. Some codes multiply by a thermal-efficiency (arc transfer) factor that depends on the process — for example about 0.8 for SMAW and GMAW, or 1.0 for submerged arc — to get 'true' heat input. Check your governing code or WPS to see which definition applies.
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