Welding energy comparison and calculation.
I have to admit that it may look strange to you when you start reading this welding article and all of a sudden i start talking about bullets. The reason for it is rather simple, i am a firm believer that we understand the reality of the natural world the best if we have the opportunity to compare new phenomena to our existing experiences. We all know that it is possible to weld pieces by use of explosives and also by drop forging, and some of us may argue that in principle, both of these welding processes are rather similar to each other despite the difference in the energetic values of the causing forces. So let’s have a look at the firearm component of this equation.
Energy of Bullets
The energy of a bullet, also known as muzzle energy, is the kinetic energy that a bullet will have as it is pushed out of the muzzle of a firearm. This is often used as a rough indication of how much destructive potential a given firearm or bullet will have. The heavier the bullet and the faster it moves, the higher its muzzle energy will be and the more damage it will potentially do.
Every firearm shoots a different size of bullets with different weights, diameters, and speeds, so every single one will be somewhat different than another. Let’s look at four of some of the most popular firearm cartridges in the world today:
- 5.56x45mm NATO (standard issue cartridge for both US and UK troops)
- 7.62x39mm (standard caliber for the world-famous AK47)
- 308 Winchester (most popular sniper cartridge)
- 12.7x108mm (high powered sniper rifle caliber)
5.56x45mm NATO
The standard weight of the 5.56 round is 62 grains. The diameter of the bullet is 5.70 mm (0.224 in). Bullet energy at the muzzle is 1796 joules, and this round will remain lethal out to 2000 to 2500 meters While it would still be deadly past these ranges, at this point, it loses much of its energy and you might be better served with a larger caliber.
7.62x39mm
The standard weight of the 7.62 round is 122 grains. The diameter of the bullet is 7.9 mm (0.311″). Bullet energy at the muzzle is 2108 joules, and this round will remain lethal out to 2000 meters and beyond.
308 Winchester
The standard weight of the 308 round is 150 grains. The diameter of the bullet is 0.308 in (7.8 mm) hence the name. Bullet energy at the muzzle is 3590 joules, and this round will remain lethal out to 800 meters, although many snipers and shooters will claim that it still carries plenty of energy to be lethal out to 4000 meters and even more.
12.7x108mm
The standard weight of the 12.7 round is 855 grains. The diameter of the bullet is 12.98 mm (0.511 in). Bullet energy at the muzzle is 18,625 joules, and this round will remain lethal out to 1500 meters, although it can still be accurate and deadly even out to 6,000 meters.
Back to welding energy and heat input calculations.
Yes, we will compare the energy of the welding process to the energy of a bullet that is fired from the barrel of a gun. I am not talking about the possibility that a bullet or its parts may weld itself into any metal objects that it may hit, i am talking purely of the energy comparison. As a “weapon” of choice in this comparison, i will use a very common welding scenario of welding 69mm OD steel pipe with TIG welding method.
Here is an image of extract from one of the welding procedures that i have.
If you pay attention you will see that that in the very first root run our heat input energy value is 2000 joules which is right on par with an AK rifle bullet. Amazing thing is that when we weld, we release this amount of energy into every millimeter of welding.
I let this to think in, and if you consider how much energy will be released during the working day, month, or professional career.
It should also be clear to everyone involved in welding work that PPE is paramount in welding as being to such an intensive energy source without adequate personal protection equipment may have very dire consequences.
If you have a range of typical materials and thickness that you weld and you interested to know your own welding heat input, you can calculate it relatively easily. It may be a lot easier than you think.
If your welding machine has digital or analog meters you are all set. Simply take the reading of amperage and voltage while welding and divide it by the travel speed to get your heat input in joules per mm.
In case that the welding machine does not have any meters showing the voltage and amperage delivered to the arc, the readings are taken by using an amperage/voltage meter or Fluke Meter and are measured during the arc.
The formula is as follows:
Heat Input = (60 x Amps x Volts) / (1,000 x Travel Speed in mm/min) = KJ/mm
The 60 and the 1,000 are there to turn the final answers into Kilojoules per mm. The vast majority of fabricators are not typically concerned with heat input. For the most part, this is OK. But when you are welding on materials whose microstructure can be significantly affected by welding procedures it is important to know about heat input. The reason why heat input is critical in certain applications is that it has a huge bearing on the cooling rate. Typically, faster cooling rates are detrimental to a weldment because they cause embrittlement in the heat-affected zone. An example of this is when dealing with materials susceptible to hydrogen-induced cracking in which adequate heat input is critical.
Glad you found it to be interesting! Thank you for letting me know!