What Is Heat Flux, and Why Is It Important?


Why do some cartridges burn barrels more than others? What makes a smaller-caliber, higher velocity round toastier to your bore, and how can you compare one caliber to another in this respect? While it’s not the whole answer to this question, today we’re going to talk about a major component to this equation, and that’s heat flux.

Note: For all the engineers out there, the units we’ll be working in are kJ/cm^2, which are not the normal units for heat flux in engineering (those would be Watts/unit area). The reason for this is that for the purposes of a gunshot, the missing unit of time is not important. In other words, we are measuring the energy dumped per gunshot. This is very useful, as we can combine this unit with a rate of fire figure to approximate the actual heat flux (in W/unit area) of a gun over a given course of fire.

Oh, and for those following my math, I am using approximate unit conversion factors of 25.4 mm per inch, 15.43 grains per gram, and 3.28 feet per meter.


It may be easy to tell which of the pair on the left will erode the barrel faster, but it’s much more ambiguous for the pair on the right. Having a way to compare cartridges in this respect is useful.


First, we need a couple of examples to compare. Let’s take the common 5.56x45mm M193 Ball loading, for starters. We know a few relevant things about this cartridge:

– It has a bore diameter of 5.56mm/0.219″, and a groove diameter of 5.69mm/0.224″. Either of these dimensions can be used for calculating heat flux, as long as which one is chosen is kept constant when comparing different calibers. Since groove diameter is the more commonly available figure – it is usually the same as the bullet diameter, we will use it.

– It has uses a charge of approximately 1.72g/26.5grs.

– It produces a muzzle velocity from a 50.8cm/20″ barrel of about 997 m/s or 3,270 ft/s.

– It has a bullet weight of 3.56 grams or 55 grains.

From these figures alone we can calculate heat flux. We should start by calculating the muzzle energy of the round, using the kinetic energy equation KE  = 0.5 * (mass of the bullet) * (muzzle velocity)^2. I prefer doing my calculations in metric, so that would be:

MKE = 0.5 * 3.56 g * (997 m/s)^2 = 1,769 Joules

Next, we need to calculate the total energy produced by the cartridge. This is not the same as the muzzle energy, as guns are not perfectly efficient engines, so to approximate this we need to take the powder charge in grams and find out how much energy it produces when it burns. There’s not an easy way to do this exactly, so instead we’ll take a shortcut by just using a fixed figure of 3,870 Joules produced per gram, which is converted from a figure of 185 ft-lbs per grain estimate for IMR brand powders. This step is where most of the error will be introduced as that figure may not be correct for every kind of propellant (though it shouldn’t be wildly off for nitrocellulose-based propellants), and because for some barrel length-cartridge combinations a lot of unburnt powder may be ejected out the muzzle (and therefore its energy won’t go into heating the barrel). However, for a quick rule-of-thumb estimate, it’s not bad.

For M193, we find:

TKE = 3,870 J/g * 1.72 grams = 6,656 Joules

Giving us a wasted energy figure of:

6,656 Joules – 1,769 Joules = 4,887 Joules

Or 4.887 kilojoules (kJ). Now, find the area of the bore, or in this case the approximated area from the groove diameter:

5.69 mm / 2 = 2.845 mm

π * (2.845 mm)^2 = 25.4 mm^2 = 0.254 cm^2

Now, finally, we can find the heat flux per shot by dividing the approximate wasted heat in kilojoules by the approximate area of the bore in square centimeters!

4.887 kJ / 0.254 cm^2 = 19.2 kJ/cm^2

And that’s the approximate heat flux for a single shot of M193 Ball ammunition, through a 20″ barrel. By calculating values for different calibers, you can compare them to get a better idea of which calibers burn barrels out faster than others.

You can also use the excellent Powley computer hosted at kwk.us to do this calculation as well, though it’s not necessary. To do so, fill in all the values for case capacity, case length, cartridge overall length, bullet length, etc, and adjust the pressure to the correct value by clicking the checkbox next to the “Pressure” section. Then use it to find the total kinetic energy value by dividing the muzzle energy by the efficiency (convert the efficiency from percent to decimals first), and pick up the instructions above from there! A screenshot of what a filled in Powley sheet looks like is below:


Nathaniel F

Nathaniel is a history enthusiast and firearms hobbyist whose primary interest lies in military small arms technological developments beginning with the smokeless powder era. In addition to contributing to The Firearm Blog, he runs 196,800 Revolutions Per Minute, a blog devoted to modern small arms design and theory. He can be reached via email at nathaniel.f@staff.thefirearmblog.com.


  • guest

    Heat is one thing, mechanical erosion is another. I’ve had a heated debate a long time ago about the topic, and it boils down to two things: the topic itself from a technical standpoint does not have a clear consensus even amongst firearms manufacturers and users, although there are mounds of data (as in detailed studies of barrel erosion of all kinds, from all kind of guns in a plethora of calibers). For this reason I am guessing that from the MFGs standpoint there is no urgent need to “get to the bottom of this”, because having an acceptable barrel wear beats putting X amount of hours into researching how exactly barrel wear occurs and how to prevent or limit it, which most likely would result in something similar to exotic ceramic barrel inserts (especially close to the chamber) etc, or in other words “just make a cold-hammered high strength chromed barrel and forget about the rest”.
    The second thing kinda comes back to the lack of a consensus issue: better leave the subject alone, unless one is directly working on research aimed at understanding barrel wear.

    • Jwedel1231

      Sounds like we need someone to look into this purely for the science of it. Can I get government grant money for this?

      • Nicks87

        Nope, the money has already been spent on researching climate change and why gun ownership causes racism.

        • Jwedel1231

          If mathematicians can be paid to discover and figure out how many types of infinities there are, and describe the shapes of those infinities, surely someone somewhere has some money that can be spent on discovering the exact mechanism(s) behind throat erosion.

          • LG

            It is well know. Please see the second edition of the excellent text “Ballistics: Theory and Design of Guns and Ammunition”. Remember, the actual physical, crystalline, structure of the metal in the throat is changed at it’s surface by intense energy absorption. That is one of the reasons for the alligator hide appearance of a “burned out

    • nova3930

      I think in most instances, mechanical wear is negligible. You’re talking either solid lead or copper/alloy jacketed projectiles running on a steel, chrome lined steel, nitride steel, etc etc barrel, which is soft running on relatively hard. I think the thermal erosion is orders of magnitude more important since no matter how hard a metal is, it will literally burn away at the surface with enough heat input…

      • Mechanical erosion is significant, but it’s more minor than heat. Here’s an Australian gov’t paper on barrel erosion:

        “Of the erosion processes, mechanical erosion perhaps receives the least a‘ention in the literature. It has been cited as most the most dominant erosion mechanism for low temperature firings [24], where there is insufficient heat to drive chemical reactions or cause thermal erosion. At higher temperatures, the three mechanisms act concurrently.”

        • nova3930

          Interesting paper. I neglected the thermal degradation of material properties which are going to be pretty significant for most steel alloys.

          Most interesting was the section talking about particulaten matter in the gas flow. Not hard to see how that could be a major source of mechanical erosion, being effectively sand blasting of a sort.

        • guest

          I did not elaborate clearly enough: mechanical erosion caused by the propellant. There are of course factors such as erosion caused by the bullet itself (does not matter if bullet metal is softer than the barrel, it will still cause erosion trough friction over time).
          Some people would like to say barrels “burn out”, especially close to the chamber and to some degree at the crown. Heat, especially coming from oxygen-poor combustion products of burning propellant never “burned away” anything, unless one considers that heat + mechanical erosion by the propellant (which is not a pure gas, but a cocktail of small solids and various gases) can indeed work together to cause that wear. My 0.02$ coming from someone with no scientific education in this field. But if a high-pressure washer can pierce wood and other materials with a fluid that has exacly ZERO hardness, as can water drops slowly make holes in any materials that they fall on over time especially in the same exact spot, I’d say it’s erosion by propellant particles and gases aided by heat.

  • Anonymous

    Sorry to be a pedant, you mean 3.28 ft per meter.

    • Ax

      …and if there are 15 grams to a grain then I need to go pull the bullets out of all of my handloaded ammo.

  • Ranger Rick

    I’m not a math whiz,but is there a way to take into account how tracers would effect the calculations?

    • Jwedel1231

      Not in the model used above. You would have to add in somewhere the heat generated by the tracer compound, and the model above does not have a place for it yet.

  • rayward

    Shouldn’t the area of the bore calculation include barrel length? It is entered on the Powley form…

    • tony

      I would presume only the cross section area is used for calculation.
      Others familiar with heat flux formulas please chime in

    • It’s better to keep the bore length calculation constant, as more error is introduced if you are comparing barrels of different lengths, due to differences in how much propellant is wasted for a given barrel length.

  • nova3930

    Love the deep dive tech articles. My only quibble is that 1 inch = 25.4mm isn’t an approximation. By definition 1 inch is exactly 25.4mm. :p

  • To any people out there who want to learn about the engineering behind ballistics, read “Ballistics: theory and design of guns and ammunition” excellent resource.

    • Smiddywesson

      thanks, I will.

    • Smiddywesson

      Ah, 160 pages are devoted to terminal ballistics, my kind of guy.

  • MPWS

    This is (muzzle vs. total energy) worse ratio that with gas combustion engine. Just that we know what is “thrown away”.
    It should be suggestion what to do about next generation propulsion system.

    • Huh? Don’t gasoline ICEs generally operate at about 25-30% efficiency? That’s right around how efficiency small arms are.

      • nova3930

        That’s pretty close. Max theoretical efficiency of typical gasoline engines is 37% based on carnot’s theorem iirc. Goes without saying real engines will be worse. Propulsion was a long time ago though so that number could be off a bit.

  • Vitor Roma

    Quick question: does the 7.62×39 erodes the barrel faster or slower than the 5.56? It is much considerably slower and lower pressure, but about 300 joules more powerful.

    • Trey

      The bullet construction is often the issue with the 7.62×39. Bimetal and steel jackets tend to erode a barrel faster than a Copper or Gilding metal jacket.

    • Jwedel1231

      On it’s face, I’d guess 5.56 erodes much faster than x39. You ought to follow the mathematical steps above and find out for yourself 🙂

    • iksnilol

      Given same case and bullet material 7.62×39 will last longer.

      Lower pressure that gets distributed across a larger surface.

  • tony

    Thank you for using metric units, it makes calculations much easier.

  • LG

    You have neglected the energy deposition in the barrel from metal plasticity, the barrel actually expands as the bullet traverses the bore. You have also neglected expelled energy in the propellant gases leaving the barrel and the ejected brass/steel/metal/whatever case. You have only very superficially touched upon the parameters and effects.

    • From the first paragraph of the post:

      “While it’s not the whole answer to this question, today we’re going to talk about a major component to this equation, and that’s heat flux.”

    • RocketScientist

      Was my first thought as well. Those factors MAY be insignificant compared to the ones the author focusing on, but they may NOT be. Only way to know is to actually run the numbers and get a fee for it. Otherwise these numbers are worthless.

      • LG

        Barrel heating secondary to elasticity of the steel tube during bullet passage and chamber stretch under pressure is quite significant and has been measured several times. Also as one progresses along the length of the barrel or progressively longer barrel heating from elastic deformation becomes one of the primary sources of barrel heat near the muzzle. Heat transfer to the bullet is quite significant. Very early literature from Newton, of the Newton Arms Co., demonstrated molten cores of jacketed rounds, even at target impact. Also the location of the powder burn in the chamber and tube has profound effects. The greater the bore diameter compared to the case diameter and the less sharp the case shoulder, the more powder burn will be in the throat and chamber than in the case. The Late Great Elmer Keith showed thin very early with his “duplex” loadings. And the surface effects of heating, especially in the throat region, effect the crystalline nature of the steel. This significantly changes it’s properties of hardness and surface finish. The alligator hide of an “eroded” throat is more from crystalline change and surface vaporization than mechanical abrasion. But do not believe me. Just read the textbooks. Again I reference “Ballistics: Theory and Design of Guns and Ammunition, Second Edition”.

  • Shocked_and_Amazed✓ᵛᵉʳᶦᶠᶦᵉᵈ

    Barrel length .750 inches ??

  • uisconfruzed

    Where’s the summary of popular rounds?
    I’d like to see #’s for 260 Rem & 6.5 Creedmore, same bullet & almost the same velocity, yet the CM has longer barrel life.

  • Wynter

    I was hoping for more of a “here’s what we learned” type article instead of “here’s the formula, figure it out yourself”. lol

    When I got out of the precision manufacturing QC business math was the first thing that left my brain. I didn’t ask it to leave and I don’t know where it went but it’s gone. Now when I look at math I imagine a monkey wearing a pope hat riding a unicorn to the Jeopardy music…I should get that checked out.

  • Mazryonh

    So will caseless rounds leave behind more heat because there’s no heat-conducting casing to take some away? How about rounds like the VOG-25 grenade?

    Plenty of kids think they want “blaster rifles.” For the sake of argument, would a laser weapon putting out the same energy on target as a 5.56mm M193 round at, say, 100 meters from an M4A1 be more or less efficient (resulting in more heat buildup in the weapon) than the M4A1?

  • georgesteele

    I keep staring at your calculations and cannot make sense of why you would use the cross-sectional area of the bore instead of the area of the internal barrel surface. You are dissipating the heat of combustion of the powder across the internal surface area of a 20″ long tube – so the equation wouldn’t be pi*r^2, which gives the area of the circle that is the bore, but rather pi*D*L – the first two giving the circumference and the latter being the barrel length.

    That provides the surface area of the cylinder that is the barrel internal bore area, through which the heat is dissipated (that is, the component that is transferred from the high temperature gas minus that which vents from the barrel in the form of hot gas and unburned powder). Also, the heat generated within the first few moments, at the highest chamber temperature and pressure, would be (partially) transferred to the throat and leade area, but it would diminish as the bullet traversed the barrel, the pressure and temperature diminished, and the last of the powder contributing to thrust against the bullet burned, so it’s not a linear condition – of course, barrels burn out from the chamber forward, that being one of the reasons.

    By “not linear”, I mean that once the bullet leaves the barrel, the temperature and pressure drop off very rapidly, so the muzzle is subjected to the lowest pressure and temperature, and for the shortest time. The area subjected to the highest combustion heat, for the longest time, is in the throat/leade area – both as the gases pass by the bullet as it jumps from the case neck across the leade into the first “engraving” portion of the rifling to create the barrel seal, and as the pressure and temperature skyrocket as the bulk of the powder ignites. As the bullet progresses down the barrel, the increasing volume behind the bullet drops the temperature and pressure by expansion, so heat transfer is reduced because the barrel gas temperature is reduced. But the leade area is still at the highest temperature, as the powder was burning in that area at the highest pressure level.

    Back to the equation, what is happening within the center of the “tube” of burning gas behind the bullet shouldn’t be as important as what is happening at the surface of that tube, which is in contact with the barrel grooves and rifling. Conductive (vs. radiative) heat transfer from the center of the cylinder of combustion gas isn’t instantaneous through the surrounding gas to the outer edge of that expanding cylinder, nor is the heat transferred instantaneously from the gas in contact with the barrel surface – but as you say, the idea is to compare one round/barrel combination with another, rather than to measure exactly the thermal flux value.

    So I’m just not seeing why the cross-sectional area equation pi*r^2 is used instead of the cylinder surface area equation, pi*D*L. Is the radiative component and combustion mass more predictive of the heat flux than the conductive/convective? Is there a source that might clarify this?