### March 07, 2006

## The Falling Bullet #2

Now the time has come to complete my short treatise on the infamous free-fall bullet meme. Here are the data of our Kalashnikov bullet, and of the Winchester .22 short as a comparison: I think the .22 short is at the lower end of the lethality scale.

7.62 x 39

Weight: 7.97 g

Muzzle velocity: 710 m/s

Muzzle energy: 2010 J

Mainly steel

.22 short

1 gain = 64.8 mg

Weight: 29 gr. (1.88 g)

Muzzle velocity: 1095 ft/s (333.8 m/s)

Muzzle energy: 393 J

Lead

As I anticipated, the most difficult thing is to estimate the drag coefficient of a bullet falling freely in air. This website reports a

vt = sqrt( 2 m g / ro Sr

With:

And the results are:

(Ain't got time to perfect the formatting, chaps) So, even with a

I would say that this is not enough enough to kill an adult human, and neither a child. It would probably causes bruises, like heavy hail, and maybe somewhat more serious wounds in unfavourable circumstances.

7.62 x 39

Weight: 7.97 g

Muzzle velocity: 710 m/s

Muzzle energy: 2010 J

Mainly steel

.22 short

1 gain = 64.8 mg

Weight: 29 gr. (1.88 g)

Muzzle velocity: 1095 ft/s (333.8 m/s)

Muzzle energy: 393 J

Lead

As I anticipated, the most difficult thing is to estimate the drag coefficient of a bullet falling freely in air. This website reports a

*Cd*of 0.15 for subsonic 7.62 x 51 Nato bullets (fired from a gun); a projectile will probably fall butt first and thus be less aerodynamic; however I used Excel to calculate the terminal velocity for the*Cd*values of 0.1, 0.15, 0.2 and 0.3. The formula I used is:vt = sqrt( 2 m g / ro Sr

*Cd*)With:

**m**Bullet mass, kg**g**Gravitational constant, 9.81 m/s^{2}**ro**Air density, ~ 1 kg/m^{3}**Sr**Reference area: cross-section of the cylindrical part of the bullet, 4.56*10^{-5}m^{2}And the results are:

Cd | vt(m/s) | K(J) |
---|---|---|

0.1 | 185 | 137 |

0.15 | 151 | 91 |

0.2 | 131 | 68 |

0.3 | 107 | 46 |

(Ain't got time to perfect the formatting, chaps) So, even with a

*Cd*as low as 0.1 - which is rather unlikely - a free-falling Kalshnikov bullet would have roughly 1/3 of the energy of a .22 short, at a little more than half the speed.I would say that this is not enough enough to kill an adult human, and neither a child. It would probably causes bruises, like heavy hail, and maybe somewhat more serious wounds in unfavourable circumstances.

**Update 08/03:**This US Navy document places the minimum lethal energy at 100 J. I rest my case: free-falling Kalashnikov bullets are unlikely to kill.**Update 12/03:**I asked Kevin to weigh in, and he pointed me to a series of posts at Wadcutter: the guy over there did much more sophisticated calculations, but arrived to similar results - confirming the truism that the right mathemathical model is the one that gives usable results without being overly complicated. Huckel managed to make certain molecular quantum mechanics calculations simple (as far as matrix algebra goes) and rather accurate, by discarding some features that give better results but at the expense of much greater difficulty.__Comments:__

Fascinating. I have always wondered about this. Next time I see guys shooting AK47s in the air on TV I will feel much happier not having to worry about them landing on some kid and killing them.

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