I'm obsessed with trivial (and useless) information and i have a question thats been on my mind lately. I was watching Cops or something similiar and they were commenting on injuries and fatalities resulting from people firing guns into the air in celebration. This got mewondering about .223/5.56 rounds. What i want to know is: if you fired an AR straight up into the air 1)what would be the maximum altitude reached and 2)what would it's velocity be when it reached the ground (and more specifically, what is the potential for human harm) and 3) what would be the total time elapsed before impact?
for calculations assume "least effective" variables. i.e., short 14.5" barrel (lower pressure) and a "less-than-hot" round like a USA223R1, or even worse, Wolf.
Doesnt have to be super scientific, just rough estimates. this is just to satiate my curiosity.
Bullets coming back down are pretty dangerous when they hit someone. There was one story a couple of years ago in Germany where a police officer fired his .357mag in a new years night (what a prick)... the bullet came down about a mile from where he fired and hit a woman in her leg. It entered a few of inches, injuring her severly.
Another instance I know of resulted in the death of a guy, a 9mm or .45 (dont remember) came down straight, penetrated the upper skull, killing the man.
Unfortunately I dont have my books handy right now, but full calibre rifle bullets should easily go up nearly a mile. IIRC downspeed is in the region of 300-400fps. I will check later.
I've never understood the idea of firing a gun up in the air other than it's some dumb ass mimicing what he saw on TV.
Shoot a gun into the ground and it makes the same freaking amount of noise.
I can't help but wonder if there is some psychological macho idiot thing that these people are thinking when they goto shoot a gun in the air.
It comes back down at a speed that can be determined by the following information.
An object which is falling through the atmosphere is subjected to two external forces. One force is the gravitational force, expressed as the weight of the object. The other force is the air resistance, or drag of the object. The motion of any object can be described by Newton's second law of motion, force F equals mass m times acceleration a:
F = m * a
which can be solved for the acceleration of the object in terms of the net external force and the mass of the object:
a = F / m
Weight and drag are forces which are vector quantities. The net external force F is then equal to the difference of the weight W and the drag D
F = W - D
The acceleration of a falling object then becomes:
a = (W - D) / m
The drag force depends on the square of the velocity. So as the body accelerates its velocity and the drag increase. It quickly reaches a point where the drag is exactly equal to the weight. When drag is equal to weight, there is no net external force on the object and the object falls at a constant velocity as described by Newton's first law of motion. The constant velocity is called the terminal velocity .
We can determine the value of the terminal velocity by doing a little algebra and using the drag equation. Drag depends on a drag coefficient, Cd the air density, r the square of the velocity V and some reference area A of the object:
D = Cd * r * V ^2 * A / 2
At terminal velocity, D = W. Solving for the velocity, we obtain the equation
V = sqrt ( (2 * W) / (Cd * r * A) )
where sqrt denotes the square root function. Typical values of the drag coefficient are given on a separate slide.
The terminal velocity equation tells us that an object with a large cross-sectional area or a high drag coefficient falls slower than an object with a small area or low drag coefficient. A large flat plate falls slower than a small ball with the same weight. If we have two objects with the same area and drag coefficient, like two identically sized spheres, the lighter object falls slower. This seems to contradict the findings of Galileo that all free falling objects fall at the same rate with equal air resistance. But Galileo's principle only applies in a vacuum, where there is NO air resistance and drag is equal to zero.
This question is hard to answer in general. The best I can give is a "worst-case" estimation.
When a gun is fired vertically, the bullet after some time reaches a summit where the velocity is zero, and then falls back. I can estimate the velocity if it would fall nose first, that is the normal flying position for which drag is well known - so the real terminal velocity will actually be smaller than the following prediction.
* For a .22 lr bullet (m=40 grain, v0 = 1150 ft/s)
the summit will be at 1164 ft, the total flight time 30 seconds and the terminal velocity 270 ft/s
* For a SS109 military bullet (m= 55 grain, v0=3200 ft/s)
the summit will be at 2650 ft, the total flight time 44 seconds and the terminal velocity 404 ft/s.
For this bullet are indications that it will become unstable. This will further reduce summit height and terminal velocity considerably.
Most projectiles will not have any spin imparted upon them and will fall randomly, ie not nose first, not base first but a hap hazard tumble, which will increase drag and slow projectile. This drag cannot be calculated easily if at all. The worst case scenario is if the bullet is falling nose first using the minimal amount of drag.
Hmmm I would guess the projectile would still be spinning pretty fast when reaching the peak of the curve, even if it doesnt have any translatoric movement left. Drag along the rotational axis should be pretty low compared to translatoric drag.
For a vertical shot my money would be base-first when it begins to fall.
My understanding was that bullets return to earth base first, as it is the heavier end. They also are travelling at appx 400fps.
I nerd am I? Not convincing me to help.......
Yes, if the bullet was shot STRAIGHT up and came STRAIGHT down, with no wind affecting it, no breeze, and the bullet kept spinning to prevent it from yawing, yea, it would come base down first.
Hmmm, me thinks this not happen. Most likely it will arc and loose its rotation, then base goes first, then nose then base, etc....
Thanks guys. These calcs are close enough to satiate my curiosity. Even though it's for the most part useless, I at least now know that if I'm standing next to a mouthbreather who fires a rifle skyward, I've got about 30-45 seconds to get under something heavy. Not a likely scenario but nice to know none the less.