by satelital
Tags: bullet, earth, return
 P: 185 I note this thread started some two years ago, however it is an interesting question. Friction due to air resistance is a very tricky thing according to my research. A great many factors will affect this value. In the original post, there was not suffcient information to say, for example, if the bullet still had kinetic energy from being fired, or if all the speed that was left was its terminal velocity. For the latter case: Assumptions: 1. Take the ideal case (unlikely) that the bullet is falling in a consistently "nose-down" attitude. 2. Air pressure is 1.29Kg/m^3 3. Use a .45 Cal bullet, mass 300g, drag coefficient 0.228. The "nose down" attitude gives us a cross-sectional area of 0.0001026m^2. 4. Still day. 5. Ignoring humidty. Process: We can use the Quadratic drag formula. $$v_{terminal} = \sqrt{\frac{2mg}{CP_{air}A}}$$ As I right this, LaTeX isn't working, so that is saying terminal velocity = square root ((2mg)/(CPA)). m = mass g = acceleration (gravity) C = drag coefficient P = air pressure A = cross-sectional area That comes out to about 442m/s = 1,448ft/s = 987mph. Gun enthusiasts are probably thinking that number is higher than the muzzle velocity of most guns, which means that in this "ideal" scenario, a bullet fired straight up wouldn't get high enough to attain terminal velocity on the way down. Much more realistic would be to assume the bullet is more or less sideways on the way down. This will result in a drag coefficient more like 0.6 (a sphere is about 0.5), and a cross-sectional area very roughly twice the nose-on area, or say 0.0002052m^2 Plugging those numbers in gives: 192m/s = 630ft/s = 430mph. That should be good enough as a "useless information" tidbit at your next BBQ.
 P: 447 This is interesting, from "Hatcher's Notebook": http://www.recguns.com/Sources/XD1.html "Out of more than 500 shots fired after adjusting the gun--only four shots hit the platform. One of the shots was a service 30.06, 150 grain flat based bullet, which came down base first...it left a mark about 1/16 inch deep in the soft pine board... It was concluded from these tests that the return velocity was about 300 feet per second. With the 150 grain bullet, this corresponds to an energy of 30 foot pounds. Previously, the army had decided that on the average, an energy of 60 foot pounds is required to produce a disabling wound. Thus, service bullets returning from extreme heights cannot be considered lethal by this standard... This supports the observations of those who wrote during WW2, that after a heavy battle, a number of bullets were found slightly embedded in tar rooftops, all pointed towards the sky.
P: 15,319
 Quote by McNamara A bullet is said to travel 33,000 MpH
This is a wildly inaccurate number - it's at least an order of magnitude too high.
P: 185
 Quote by Farsight This supports the observations of those who wrote during WW2, that after a heavy battle, a number of bullets were found slightly embedded in tar rooftops, all pointed towards the sky.[/i]
Interesting. If a bullet fired vertically does not change attitude througout its flight, therefore falls back "butt-first", its drag would be pretty high. Assuming 0.6...

BTW, I just realised I was taking a .45 as 300 grams NOT grains as I should have. Correcting that now....

Plug that into my "formula" gives a terminal velocity of 69m/s = 226ft/s = 154mph.

A 30.06 will have different characteristics. The largest my research tells me is 220 grains. Plugging the relevant numbers in...

79m/s = 259ft/s = 177mph.

For 150 grain bullet...

65m/s = 213ft/s = 145mph.

If one takes the drag coefficient to be 0.3...

92m/s = 302ft/s = 206mph.

It would seem to be a little low for drag for a flat surface, but that is what it takes to get about 300ft/s terminal velocity for a 150 grain 30.06 bullet falling butt-first!

Guess my numbers are wrong. The part I'm not sure of is the drag coefficient. Oh well...
 P: 57 I didn't think drag coefficient depended on the mass..?
 P: 1 Ok, what everyone seems to be missing is trajectory. All bullets fired do not return to earth at terminal velocity. Gravity is a constant, and a bullet fired on a flat trajectory, ) degrees elevation, it will hit the ground at the same time as a bullet dropped next to the chamber at the same time. Now, this doesn'tmean the bullet fired will hit with the same force, it also has a consicerable amount of force behind it, and the mass is coniderably greater. So, the question waas, how did a bullet fired into the air penetarate your tennis court, simple, thats what bullets do. Has nu\othing to do with terminal velocity or gravity. I also saw the episode of myth busters, and they were going under the asumption that a round was fired streight up, which is nearly impossible. All artillary is fired on trajectory, in an arch, and they still have devistating impacts. These objects are not particularly fast, but are moving faster then terminal velocity. SO to restate, the bullet did what bullets fired on a trajectory do.
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P: 6,679
 Quote by Mongo Now, this doesn'tmean the bullet fired will hit with the same force, it also has a consicerable amount of force behind it, and the mass is coniderably greater.
I am not sure why you say the mass of the fired bullet is considerably greater. They must have the same mass.

 All artillary is fired on trajectory, in an arch, and they still have devistating impacts. These objects are not particularly fast, but are moving faster then terminal velocity.
They are fired on an arch trajectory in order to maximize range. If they are explosive shells that are being fired, the damage they do doesn't depend so much on their speed.

It seems that falling bullets can do alot of damage. I wonder what the experience in Gaza was when they celebrated the election of Hamas?

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P: 39,565
 Quote by Mongo Ok, what everyone seems to be missing is trajectory. All bullets fired do not return to earth at terminal velocity. Gravity is a constant, and a bullet fired on a flat trajectory, ) degrees elevation, it will hit the ground at the same time as a bullet dropped next to the chamber at the same time. Now, this doesn'tmean the bullet fired will hit with the same force, it also has a consicerable amount of force behind it, and the mass is coniderably greater.
No, what you are missing is that everyone else is taking into account air resistance. Without that, there is no "terminal velocity" and a bullet dropped will hit the ground at the same time as a bullet fired on some trajectory. With air resistance, however, that's not true- given enough time to reach terminal velocity, a bullet will, eventually, return to earth with terminal velocity straight down (since there is no horizontal force, the horizontal component of "terminal velocity" will be 0.
It is true that the fired bullet, in your "airless" scenario will hit with more force since it has a horizontal component of velocity also. But I am mystified as to why you think the mass of the fired bullet will be "considerably greater" than the mass of the same bullet dropped.
P: 15,319
 Quote by HallsofIvy But I am mystified as to why you think the mass of the fired bullet will be "considerably greater" than the mass of the same bullet dropped.
Relativistic effects?
P: 57
 Quote by phun I didn't think drag coefficient depended on the mass..?
oops it does
 PF Gold P: 8,964 [semi-rant]Firing into the air in almost any circumstance is incredibly stupid. I'm not going to get involved in the calculations aspect of it, because I don't know math. What I do know is that one should absolutely NEVER shoot without knowing that there's an acceptable backstop for the round. It's nearly impossible to fire straight up, so you can't know where it's going to land. As Krysith and Andrew pointed out, people are injured or killed by this practice all the time. Even if terminal speed is only 100mph or so, that can still be lethal if it hits someone in the head. It's far more likely, moreover, that midrange-trajectory ballistics will apply rather than straight gravitational effects. People, especially when alcohol is involved, don't pay a lot of attention to where they're pointing guns as long as it's significantly above horizontal. I also don't always trust the math and physics as presented, either, if it conflicts with experience. Some gun magazine article that I read decades ago laughed at the folly of trying to hunt at over 50 yards with a handgun. It went through the math showing that a .44 magnum could be snagged by a guy with a catcher's mitt at 100 yards. Maybe, but I've killed gophers at 150+ yards with mine (not often, mind you; I'm not that good a shot). Even an innocent little .22LR bullet can kill someone at over a mile away. I don't intend, as stated, to enter into the scientific analysis of the situation; I merely want to stress that one should treat firearms (and other projectile weapons) as potentially lethal under any conditions and use them accordingly.[/semi-rant]
P: 185
 Quote by Mongo Ok, what everyone seems to be missing is trajectory. All bullets fired do not return to earth at terminal velocity.
In my defence, I had a CYA clause
 In the original post, there was not suffcient information to say, for example, if the bullet still had kinetic energy from being fired, or if all the speed that was left was its terminal velocity. For the latter case:

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