# Why aren't guns impossible? (physics, not a debate) **URL:** \ **Category:** Factual Questions\ **Created:** [June 28, 2001, 6:20am UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433 "2001-06-28T06:20:28Z")\ **Posts on this page:** 13\ **Page:** 3
**Author:** ![Tranquilis](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tranquilis/32/3639_2.png) [@Tranquilis](https://boards.straightdope.com/u/Tranquilis)\ **Post date:** [July 1, 2001, 4:59am UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/41 "2001-07-01T04:59:43Z")
> [@](#): > > \*Originally posted by Joe\_Cool \* > \*\*Last I heard, _nobody_ could prove with certainty whether it’s momentum or K.E. that kills, or which has greater effect on the severity of the wound. \*\* Go read Fackler (links are found in the other thread I linked to, above). What enables a bullet to kill is the amount of energy possesed by the bullet that is converted into work producing the perminant wound channel. High-velocity bullets tend to waste a lot of energy creating temporary wound channels, while low-velocity bullets tend to produce limited temporary wound channels, but also tend to produce limited perminant wound channels. Mid-to-high velocity bullets of moderate weights tend to do the best job of creating large perminant wound channels, unless fragmentation is involved, in which case the higher the velocity and weight, the bigger the perminant wound channel, and the deadlier the bullet (Example: German-issue 7.62 NATO Ball). ---
**Author:** ![Father\_Pacelli](https://avatars.discourse-cdn.com/v4/letter/f/c37758/32.png) [@Father\_Pacelli](https://boards.straightdope.com/u/Father_Pacelli)\ **Post date:** [July 1, 2001, 12:15pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/42 "2001-07-01T12:15:58Z")
Speaking of physics, why does the AK-47 climb up and to the right when it’s fired at full-auto? [http://kalashnikov.guns.ru](http://kalashnikov.guns.ru) ---
**Author:** ![LabRat](https://avatars.discourse-cdn.com/v4/letter/l/898d66/32.png) [@LabRat](https://boards.straightdope.com/u/LabRat)\ **Post date:** [July 1, 2001, 5:07pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/43 "2001-07-01T17:07:35Z")
Golden Child, If you actually think that “grains” on the ammo box refers to grains of powder then you don’t know enough about guns to comment. A grain is a unit of measure from the apothecary system. IIRC 1 grain = 60 mg. In ammunition, this refers to the weight of the bullet, not the amount of powder. RE: your second response, 1. Spin is imparted for stability only. 2. “air pocket” in front of the bullet? Nonsense. 3. Exit wounds have more to do with the bullet mushrooming in the target than anything else. -LabRat ---
**Author:** ![DougC](https://avatars.discourse-cdn.com/v4/letter/d/7feea3/32.png) [@DougC](https://boards.straightdope.com/u/DougC)\ **Post date:** [July 1, 2001, 5:49pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/44 "2001-07-01T17:49:37Z")
> [@](#): > > …If you actually think that “grains” on the ammo box refers to grains of powder then you don’t know enough about guns to comment… - Labrat - - - 45-70? -I think that’s for the black-powder version, but that counts. There’s others too. It _is_ an older system of classification though, and doesn’t apply to the .223. - Also cilivian .223’s are almost always open-nose jacketed and do tend to go straight through/in. Mil FMJ’s go in straight, but come out every which way. - MC ---
**Author:** ![Kalashnikov](https://avatars.discourse-cdn.com/v4/letter/k/2bfe46/32.png) [@Kalashnikov](https://boards.straightdope.com/u/Kalashnikov)\ **Post date:** [July 1, 2001, 6:30pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/45 "2001-07-01T18:30:50Z")
The muzzle of any gun will tend to climb because the barrel is above the grip. the backwards force applied above the pivot point makes the gun rotate. This is why many guns that fire full auto (and semiautos derived from them) have straight stocks and pistol grips. Why it would go to the right I don’t know. I’ve never noticed any tendency to go left or right. If anything, the ejecting shells would push it to the left. Perhaps it has something to do with how a righthander holds the gun, or with a particular muzzle break. ---
**Author:** ![Turbo\_Dog](https://avatars.discourse-cdn.com/v4/letter/t/ce73a5/32.png) [@Turbo\_Dog](https://boards.straightdope.com/u/Turbo_Dog)\ **Post date:** [July 1, 2001, 7:19pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/46 "2001-07-01T19:19:45Z")
kalishnikov, most light to medium weight full-auto’s will veer to the right (shoulder fired weapons) simply because the shooters body tends to pivot in that direction from the recoil. Left hand shooters will notice the opposite. With the AK, it is more pronounced, so in later years, they had a muzzle compensater shaped to correct this for right handed shooters. That’s why it takes a lot more training to fire a full-auto weapon with any practicality, to learn how to properly hold and fire to help correct this. ---
**Author:** ![Kalashnikov](https://avatars.discourse-cdn.com/v4/letter/k/2bfe46/32.png) [@Kalashnikov](https://boards.straightdope.com/u/Kalashnikov)\ **Post date:** [July 1, 2001, 7:45pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/47 "2001-07-01T19:45:43Z")
Oh ok. Actually the only full autos I’ve fired recently are an mp5 (low recoil because its only 9mm) and a Thompson (not too much recoil because its fairly heavy for a submachine gun). ---
**Author:** ![sewalk](https://avatars.discourse-cdn.com/v4/letter/s/b3f665/32.png) [@sewalk](https://boards.straightdope.com/u/sewalk)\ **Post date:** [July 1, 2001, 8:16pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/48 "2001-07-01T20:16:52Z")
I’ve seen it in guns as large as the M60 and MG3, when fired from the hip. In the old M3 submachine gun (.45ACP), it is incredibly pronounced. Keeping one of those anywhere near the target after the first round leaves the muzzle is nearly impossible. They sure look cool, though. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [July 1, 2001, 10:16pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/49 "2001-07-01T22:16:14Z")
**Joe\_Cool** , your calculations for the guy in armor are a bit off. Again, this is a situation where you have to use momentum, not energy. Using your figures of v = 3240 ft/s (987.6 m/s) and KE = 1740 J, I get that the mass of the bullet is 3.568 grams, correct? At m = 3.568_10[sup]-3[/sup] kg and v = 987.6 m/s, we’ve got an initial momentum of 3.524 Kirks. If we assume that the bullet gets embedded in the armor, then the man+armor after the collision has that same amount of momentum, which would amount to a velocity of .03524 m/s, or 2.19_10[sup]-5[/sup] MPH. If the bullet rebounds instead of sticking, then the velocity of the target will be twice that, and if the bullet penetrates, then the target velocity will be smaller. This is a lot less than the speed you get from conservation of kinetic energy, because _kinetic_ energy is not conserved. Most of the initial kinetic energy of the bullet is converted to heat. ---
**Author:** ![Golden\_Child](https://avatars.discourse-cdn.com/v4/letter/g/7ea924/32.png) [@Golden\_Child](https://boards.straightdope.com/u/Golden_Child)\ **Post date:** [July 2, 2001, 4:56pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/50 "2001-07-02T16:56:47Z")
> [@](#): > > \*Originally posted by sewalk \* > \*\* > > > [@](#): > > > > \*Originally posted by Golden Child \* > > \*\*Recoil in rifles is softened by a rubber stopper like thing which is called a plunger. \*\* > > In most (semi)automatic rifles, the heavy moving part that soaks up the energy of the recoil is called, appropriately, the _buffer_. In the M16, that’s the thing between the rear of the bolt carrier and the recoil spring. In general, the bigger the cartridge being fired, the bigger the buffer. You oughta see the buffer in the .55 Boys Rifle (A very, very large rifle used very ineffectively against German tanks in WWII). \*\* I knew it was something along those lines. I just had my moment a forgot what it was called ---
**Author:** ![BF](https://avatars.discourse-cdn.com/v4/letter/b/b5ac83/32.png) [@BF](https://boards.straightdope.com/u/BF)\ **Post date:** [July 2, 2001, 5:28pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/51 "2001-07-02T17:28:59Z")
> [@](#): > > \*Originally posted by LabRat \* > \*\*Golden Child, > > If you actually think that “grains” on the ammo box refers to grains of powder then you don’t know enough about guns to comment. A grain is a unit of measure from the apothecary system. IIRC 1 grain = 60 mg. In ammunition, this refers to the weight of the bullet, not the amount of powder. > > -LabRat \*\* Correct, however, the amount of powder is measured in grains too. When loading shotshell for instance, I load 1 and 1/8 ounce of shot (approx. 500 grains) and 18.3 grains of powder for the type of load I’m shooting. Handloading is similar, for instance a .357 Winchester round could use a 110gr. projectile propelled by 9.5 grains of powder. We now return you to your previously scheduled debate… ---
**Author:** ![Joe\_Cool](https://avatars.discourse-cdn.com/v4/letter/j/ecb155/32.png) [@Joe\_Cool](https://boards.straightdope.com/u/Joe_Cool)\ **Post date:** [July 2, 2001, 8:03pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/52 "2001-07-02T20:03:22Z")
> [@](#): > > \*Originally posted by Chronos \* > \*\ ***Joe\_Cool** , your calculations for the guy in armor are a bit off. > > … > > This is a lot less than the speed you get from conservation of kinetic energy, because _kinetic_ energy is not conserved. Most of the initial kinetic energy of the bullet is converted to heat. \*\* Thanks for the correction. I always forget that KE isn’t always conserved. I hate being wrong though. 🙂 ---
**Author:** ![LabRat](https://avatars.discourse-cdn.com/v4/letter/l/898d66/32.png) [@LabRat](https://boards.straightdope.com/u/LabRat)\ **Post date:** [August 21, 2001, 7:06pm UTC](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433/53 "2001-08-21T19:06:06Z")
Thanks for the clarification BF. I should have added that. [Previous page](https://boards.straightdope.com/t/why-arent-guns-impossible-physics-not-a-debate/70433.md?page=2)