"Myth Busters" did a show that claimed that the bullets coming down wouldn't kill anyone. That same year, a child in Houston was killed that way. I've gone up on the steel roof of our karate school to patch a hole made by a falling bullet that punched all the way through and dented on our concrete floor below. While it is statistically unlikely that a falling bullet would hit anyone, there is no good way to be struck by a bullet from above. People who discharge firearms into the air like that are too irresponsible and stupid to be in possession of a firearm.
Re: Ding Dong the Witch is dead In that show they showed that if the buttle was fired straight up, and therfore feel straight down (and would tumble) it wouldn't have the speed to kill someone, but if on even the slightest angle it wouldn't tumble and still have the energy to seriously harm someone
Re: Ding Dong the Witch is dead While firing into the sky is certainly reckless, a bullet that is simply falling from the sky would not be moving at the same FPS as a normally discharged round. I assume it would be like a bullet-sized rock falling from the sky. I wouldn't want to get hit by one, but I'd rather that than get shot the normal way. I suppose firing toward the horizon would be different ... when the bullet reaches the ground it might still have some of its discharge velocity, depending on the distance traveled ...? Cujo :ninja:
Re: Ding Dong the Witch is dead One of the first things you learn in the military… Correct. A straight 90 degree shot up would tumble, but the slightest ballistic angle will keep the round lethal.
Re: Ding Dong the Witch is dead The speed the bullet would be at would be it's terminal velocity: where the wind resistance is equal to the force of gravity. This would depend on the bullet's geometry and whether it was tumbling or not. The bullet that penetrated our roof must not have been tumbling.
Re: Ding Dong the Witch is dead from what mythbusters said, basicly a round fired straight up will stall and wind resistance will cause it to tumble, but a round fired almost vertically will still travel in a parabolic curve so it wont tumble and will retain much of its muzzle energy. IRL its almost impossible to fire a round at 90 dergrees to the horison, so most will retain an element of lethality.
Re: Ding Dong the Witch is dead What if you fired a round at say an 80 degree angle to the horizon (or anywhere from 45 degrees to 135 degrees)? It would still go a lot further up than over, and wouldn't it lose most of its lethal energy going up, regardless of whether it tumbles on the way down? I'm no balistics expert, but I can't fathom how only a lack of tumble would account for a bullet's destructive power. I would hazard a guess that when most people recklessly fire their guns 'up,' it's between 45 and 135 degrees to the horizon. Otherwise it wouldn't really be 'up.' -up- -up-_over_\ :wyatt: /_over_
Re: Ding Dong the Witch is dead The longer a round is in flight, the more it's slowed by the air. The vertical component of of the velocity will decrease to terminal velocity. The horizontal component will decrease towards zero. Rounds fired at an angle closer to horizontal spend less time in the air, and so are slowed less before impact. All rounds travel in a ballistic arc. If we could neglect air friction, that arc would be a parabola. I suspect a non-tumbling round will have a terminal velocity of around 200 to 300 mph, depending on the slug's density and cross-sectional area. It's deadliness would increase with the mass, and depend a lot on random chance.
Bullet in the sky investigation I have made some simple calculations on the matter of bullets fired to the sky, feel free to check and correct me (surely, there are some mistakes... OK, a lot of them, actually...) It turned out, for most angles, which send the bullet into the sky, it loses almost all its energy provided by the shot very fast and only retains energy provided by the gravity when it reaches zero height, so it is as dangerous as a rock would be. There is a .pdf with my thoughts and a spreadsheet in OpenOffice format with calculations. You can change anything, which is on cyan background, and see what happens on graphs. If you have OpenOffice installed, that is. GA, please, do not punish me for such a lame application of math. [EDIT] I'll post some pictures tomorrow, now it is too late. [EDIT] Uploaded a .rar instead of .zip.
Re: Ding Dong the Witch is dead You should put that in a .rar, Sergio. For some reason .zips are corrupting here.
Re: Ding Dong the Witch is dead Using a spherical bullet in your calculations is bad, its balistic properties are different to a modern round. A sphere has uniform drag regardless of which direction it is "facing" - its basicly like a heavier hail stone. A "bullet shaped" bullet has far less drag when facing forwards then when facing sideways. What Mythbusters showed was that a bullet which has been droped from height (rather than fired) will naturally tumble and fall sideways, which given the shape of the round isn't very aerodynamic (low sectional density). To achieve this when fired from a gun the round basicly has to stall, altho some rounds will tumble anyway when their airspeed is slow enough. A round fired upwards (but not straight up) that doesn't stall will back down "bussines end first" so theres far less drag acting on it than if it was falling sideways, and therefore will fall at a higher speed. This isn't such an issue with smaller, lighter or shorter ammunition like pistol ammunition, but heavier or longer rifle rounds will still impact quite hard. To use an extreme example, in the case of the 9x39mm ammuntion which is VERY heavy round (approximately 16 g (250 gr)), and has also been designed to have a high ballistic coefficient and sectional density, I'm guessing that they would hurt a lot coming down. Of course if a rifle round did tumble it would end up with an extremely poor sectional density due to its length, and would slow down a lot more.
Re: Ding Dong the Witch is dead Do you mean the file will become corrupted after some time passes? I was able to open it just fine now, from a different PC. I assign MASS independently of DIAMETER (without regard to density of the bullet's material). So it is like the WORST case of "non-tumbling" bullet is always considered. As far as I know, long bullet ("non-tumbling") with a given section and mass will have approximately the same resistance (drag) coefficient as a spherical round of the same section and mass. If we could make a waterdrop-shaped bullet moving with a round end forward (>, it will have lesser coefficient (at high Re-s C=0,045 as opposed to C=0,4 for a sphere, reverse waterdrop <) would have C=0,1. If you, by chance, know C for the bullet in question, you can use it with my calculations because the assumption of a spherical bullet was made only to get the value of C. (Also S might have some other value, maybe the correct formula in case of an elongated body is Volume^(2/3) ), [EDIT] OK, I've made some calculations for 5,56 NATO bullet fired from an M16. Some pictures follow. I estimate their mistake as -50% +100%. Maybe this is too optimistic.