"Myth Busters" did a show

Re: Ding Dong the Witch is dead

And graphs for bullets fired at 1 degree to horizon.

Now, I think this will be a repeat, my point is, that a bullet fired UP will be dangerous only as much as gravity would make it be. Virtually no energy from the shot will be left (of course, energy needed to get the bullet into the sky was also provided by the shot, but, I think you got my meaning - it will be the same as just dropping it from the sky.)

As for the "tumbling", I guess, if the bullet would "tumble" depends upon the loss of the bullet's rotation. If it loses rotation, it tumbles, if it does not lose rotation, it does not "tumble". And if it loses rotation or not will depend upon the time of flight, initial rotation parameters, state of bullet's surface, but will NOT depend upon the angle to horizon the bullet was shot with. I am not very strong with the theory of gyroscopes though.

[EDIT] on the "trace" graphs scale on X axis and scale on Y axis are NOT the same, unfortunately, so pay attention. I could not understand, why there is no 1 degree angle where it should be on them, for 1 whole long minute

 

Sorry about the thread derailing and spliitting. I had no idea there were so many ballistics experts at Co8.

So, Sergi, what is the terminal velocity of a non-tumbling, now downward trajectory 5.56mm round? Imho, mass, cross-sectional area and drag coefficient are sufficient to model the free flight of a "spent" round.

A rifle round losing most of it's energy in 2 to 3 seconds seems plausible, as most rounds have struck by then.

 

Gaear, I've uploaded a .rar.

GA, it will be enough, except that one has to search hard for a drag coefficient of a bullet due to its complicated form. And that is why I used a spherical bullet in my calculations.

I am being lazy, just put 90 degrees in my spreadsheet, and it showed terminal velocity of 55 m/s and energy 6,2 Joules for a 5,56 NATO bullet.

It is interesting to note that rounding mistakes along with inaccuracy of "Eulerian" integration lead to the error of about 2,5 meters in determining bullet's drop place - calculations showed it would meet earth at 2,5 meters from firing point, while it should be zero. [EDIT] I've found the main source of this error to be initial inaccuracy of the PI number, I used 3,14/2 to set 90 degrees, making it 3,1416 greatly decreases the error. Kind of cool.

I am in no way a ballistics expert, just used general equations for a moving body.

 

I would have guessed maybe 25% more at least, say 70-80 m/s with correspondingly higher energy. A pistol round would have much less energy at terminal velocity and would be more likely to tumble.

So, how much energy does it take to penetrate a human skull from above? I can't seem to find it on Wiki.

 

OK, I told you I was being lazy.
For terminal velocity determination it is not necessary to solve the system of two differential equations, only one simple equation is needed:

Terminal velocity is achieved when force of resistance equals gravity force.

m x g = 0,5 x C x s x r x V^2

V = sqrt (2mg/Csr)
(Velocity is in m/s)

5,56 NATO
s=2,42 x 10^-5 m2
- cross-section area

r=1,3 kg/m3
- air density

C=0,4
- resistance coefficient for a spherical bullet, since we do not know for the real one

m=4 x 10^-3 kg
- mass of the bullet

q=9,81 m/s2
- free fall acceleration

V = sqrt[(2 x 4 x 10^-3 x 9,81) / (0,4 x 2,42 x 10^-5 x 1,3)] = 79 (m/s)

Energy = 0,5mV^2 = 0,5 x 4 x 10^-3 x 79^2 = 12,5 (Joules)

So, well, I warned my spreadsheet has an accuracy of -50% +100%, did not I ?

[EDIT] Now, if we assume the bullet has the BEST aerodynamic form (waterdrop-shaped) ( C=0,04), then V=250 m/s, Energy 125 Joules.

 

DO NOT FIRE INTO THE SKY.

I have found on some site selling protective gear, that in building industry head traumas happen, when energy of impact is 100 Joules or more.

So, for a grown man, 5,56 NATO bullet falling from the sky would be painful but not dangerous (energy 12,5 Joules).

On the other hand, if the bullet is heavier AND has aerodynamics better than a sphere has, it COULD have energy large enough to injure a man or kill a child.

DO NOT FIRE INTO THE SKY.

 

Gets caught by his Mrs as he stands in the back yard with his Desert Eagle charged, loaded and pointed straight up at the night sky.

sirchet says ... terminal velocity my butt, tell me this wont hurt!!

Feeling a sudden weight upon his right foot he looks down and beholds all five foot of "The Mrs" :chick:, drops the magazine, unchambers the round and quietly slinks back to the house in hopes of avoiding the terrible wrath that could and would follow his earlier determined action.

 

from all the calculations it would seem that a falling round doesn't have the required energy to cause significant injury, but there have been people killed by bullets shot up - on the myth busters show they even had a doctor or coroner talking about his first had experience in dealing with real life cases of people killed by falling bullets.

something I was reading which I found frustration was ...

but this has no mention of mass at all, a piece of sponge traveling at that spead will do a lot less damage than a bus.

I guess that most of the time what the theroy says will be true, and the rounds will be basicly harmless, but in practice some do fall at speeds high enough to injure or kill.:twitch:

on a side note, I noticed Sergio's use of decimal marks, if it confused you at all read http://en.wikipedia.org/wiki/Decimal_mark#Hindu-Arabic_numeral_system

 

Being hit by a falling projectile is going to to hurt, and it's likely to cause property damage. It's not hard to find out the mass of a given type of round. The heavier the bullet, the more likely it is to cause damage. The mass of the bullet is a factor in the energy it has at terminal velocity.