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Old March 10, 2007, 07:37 AM   #1
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Lethal distance of hi powered rifles?

Let's say I am out shooting my .243 or my larger 30.06 rifles. If I were to let a round go by accident into the open air, what is the maximum distance the bullet can travel and what is the maximum distance it could be lethal to a human?
I know the answer will depend on the calibers, loads, bullet weight, wind, etc....
Looking for an average.
Also, where can I find a calculator to better assist me in this question?
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Old March 10, 2007, 08:43 AM   #2
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Interesting and disturbing questions. You have guns for which you don't know their capabilities that you are worried about firing off by accident into the air (we call that a negligent discharge, by the way) and so you want to know the AVERAGE range at which they would be lethal to a human being without specific consideration of critical facets (caliber, weight, velocity, bullet type, angle, altitude, weather, etc.).

Asking for knowledge is a good thing, but you have gone about it in a bad manner. First, you don't want an AVERAGE. You want a MAXIMUM RANGE. You need to consider that for such a negligent discharge, anything within the maximum range it potentially at risk. Plus, you need the MAXIMUM range for harm, not lethality because if the bullet is traveling with enough velocity to harm someone, you run the risk it can kill them. Either way, if somebody is killed or injured, you are responsible for what happens as a result of that round you fired by ACCIDENT. Heck, you are responsible for property damages for that matter.

As far as a calculator, there are a bunch online. Use GOOGLE. Use your SEARCH button here to check for ones already posted on the forum.
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Old March 10, 2007, 09:07 AM   #3
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There should be a maximum range listed on the ammunition boxes you buy, the round will have lethal force all the way to it's maximum range.

If not then you can look up the ballistics for the ammo on the manufacturers websight.

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Old March 10, 2007, 09:36 AM   #4
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On a box of 22 it will say " caution range 1 1/2 miles.

A 180 grain 30-06 bullet fired at the right angle can travel over 3 miles.

Rifles should never be shot into the air like that, always know your backstop.
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Old March 10, 2007, 10:19 AM   #5
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I'm not a ballistic expert... but

I would say as long as the rifle is held at below a 45 degree angle then what ever the maximum distance of the round is...that is the lethal distance.

Several miles for anything over 22lr for sure.
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Old March 10, 2007, 11:03 AM   #6
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Providing I'm computer literate enough to make this work, here is a chart of the 308 Winchester at 3200'ps with a Sierra MK bullet. The ballistics would look like this:
JBM Maximum Distance Output
Input Data
Manufacturer: Sierra Description: .308 dia. 150 gr. HPBT MatchKing™

Muzzle Velocity: 3200.0 ft/s

Temperature: 59.00 °F Pressure: 29.92 in Hg
Humidity: 0.0 % Altitude: 0 ft

Std. Atmosphere at Altitude: No Corrected Pressure: Yes
Calculated Parameters
Atmospheric Density: 0.07647 lbs/ft³ Speed of Sound: 1116.5 ft/s

Initial Angle: 30.0 deg Terminal Angle: 62.9 deg
Terminal Range: 4476.7 yds Terminal Velocity: 413.9 ft/s
Terminal Time: 28.082 s Terminal Energy: 57.1 ft•lbs

This for the absolute maximum range of this bullet/load combination at the ideal initial angle as computed by the program. The bullet remains lethal as long as it is moving in air. There are some morons that think it isn't dangerous by firing guns up in the air for New Years and such. Even once the bullet reachs maximum arch, the weight of the bullet in free fall has killed folks. I think that was the real question here.
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Old March 10, 2007, 11:25 AM   #7
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Billy Dixon shot an Indian off a horse at 1538 yards with a Sharps 50-90. That ought to tell you something.
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Old March 10, 2007, 11:29 AM   #8
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Distance depends at what altitude the round is fired.....this should give you a good idea.

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Old March 10, 2007, 12:24 PM   #9
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Rembrandt, are those charts for if the gun is held horizontal, or if the gun is held 45 degrees? But thanks; very interesting...
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Old March 10, 2007, 12:32 PM   #10
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Quote:
Originally Posted by FirstFreedom
....are those charts for if the gun is held horizontal, or if the gun is held 45 degrees?
Not sure, but since the chart says "maximum", I'd guess it's at the optimum angle to achieve the greatest distance.
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Old March 10, 2007, 12:41 PM   #11
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That's what I'd guess, too. I think the optimum angle is between 39-45, somewhere in there. I want to say that, for some strange reason, it's a little less than 45, contrary to intuition.
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Old March 10, 2007, 01:04 PM   #12
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Around 30 or 35 years ago: I know of a case where a guy was working on his house, doing some painting, when he fell off the stepladder. Nobody in the neighborhood had heard a shot, but there was a 130-grain .270 bullet, unexpanded, inside his skull.

This was during deer season. Now, a .270 is pretty loud, but nobody heard any shot. The shot could have been fired anywhere from one to two miles away; probably over a mile. Odds are that the hunter didn't even know about it until he read the newspaper article.

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Old March 10, 2007, 02:05 PM   #13
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And don't forget...

The bullet can be lethal along the entire path. A few years back a person was killed by a .44 bullet that stuck them in the head as they sat on a park bench. Investigation indicated the only place the shot could have come from was over 2 miles away! ALways, always, ALWAYS have a safe backstop!
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Old March 10, 2007, 03:04 PM   #14
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Quote:
I think the optimum angle is between 39-45, somewhere in there. I want to say that, for some strange reason, it's a little less than 45, contrary to intuition.
Strange Reason=Air Resistance.

In a vacuum, 45 degrees would be optimal.
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Old March 10, 2007, 07:44 PM   #15
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Quote:
In a vacuum, 45 degrees would be optimal.
Quote:
Initial Angle: 30.0 deg Terminal Angle: 62.9 deg
If memory serves me in one of my older reloading manuals optimal angle is 30 degrees. Might be my older Hornady manual but I cannot swear to that.
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Old March 10, 2007, 08:40 PM   #16
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Optimal angle depends on the ballistic coefficient of the bullet. Higher B.C.; higher angle.
As John said (beat me to it ), in a vacuum or with an infinite ballistic coefficient, the optimal angle would be 45°.
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Old March 10, 2007, 09:30 PM   #17
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If you are in a vacuum, you have worse problems than stray bullets. The angle of departure for maximum range in air is something between 30 and 35 degrees.

Hatcher found that the maximum range of a .30-06 M2 150 grain flatbase was about 3500 yards; the M1 173 grain boattail would carry for 5400, two and three miles, respectively. They would carry a dangerous wallop most of the way.
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Old March 10, 2007, 11:52 PM   #18
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Quote:
what is the maximum distance the bullet can travel
Quote:
In a vacuum
So, if shot in a vacuum how far will the bullet travel?

If it is zero today and gonna be twice as cold tomorrow,,,,
How cold is it going to be?

Last edited by rem33; March 11, 2007 at 01:14 AM.
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Old March 11, 2007, 12:09 AM   #19
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The chart seems to give one a good idea, but I think it is developed for a particular round, since the 30-06 has more range than the 300 H&H? If anything, the one that retains a positive vector will leave the bullet free falling the furthest (generally speaking)

One might note that while any bullet falling would inflict injury, the 300-grain slug would be a disaster.
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Old March 11, 2007, 01:04 AM   #20
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It all depends. LOL

Rem33, if it was zero Kelvin today, there would be no tomorrow. You would be dead.
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Old March 11, 2007, 12:30 PM   #21
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While this is a minor detail, 45 Degrees will never be an optimal angle. If you are shooting in a vacuum which only effects bullet speed, you still have gravity to contend with. If you eliminate the gravity in a vacumm, then 45 degrees is far to inefficient as compared to a flat trajectory where the bullet has no limits on distance. Optimum angle is generally at 30 to 32.5 degrees and that includes the 16" guns on the Iowa class battleships.
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Old March 15, 2007, 01:47 PM   #22
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Quote:
the weight of the bullet in free fall has killed folks. I think that was the real question here.
It is imposable(unless it hits your eye, but you would have to be looking straight up) for a bullet to kill someone in true free fall. If it is fired with any angle then it can kill on the way down. But true free fall, by a bullet fired straight up in the air will not kill someone on the ground. The terminal velocity of a bullet will not kill you. The thing is it is very hard to actually fire a round straight into the air. So when people shoot at new years party's in the air, there bullets are coming down at an angle not straight down.

http://www.loadammo.com/Topics/March01.htm

also, myth buster's did it and received the same answers. again only if its fired straight up in the air.
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Old March 15, 2007, 01:54 PM   #23
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Sorry, Trapper, in a vacuum with no force on the bullet except gravity, and with a range short compared to the radius of the planet, a 45 degree launch angle will indeed give the longest range in a pure parabolic trajectory.

I agree with you, mindwip, a bullet dropping straight down at terminal velocity would be painful but not lethal, according to Hatcher's work. I doubt many New Year's drunks use a plumb bob to align their shots in the air, and if there is any angle at all, the bullet will come back down with some residual velocity, enough to be dangerous.
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Old March 15, 2007, 04:27 PM   #24
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Quote:
It is imposable(unless it hits your eye, but you would have to be looking straight up) for a bullet to kill someone in true free fall. If it is fired with any angle then it can kill on the way down. But true free fall, by a bullet fired straight up in the air will not kill someone on the ground. The terminal velocity of a bullet will not kill you.
Sorry, that is incorrect. It most certainly could kill a person, coming straight down, depending on a variety of different factors:

Did the bullet turn around due to wind or bullet contruction such that the nose rather than the base is heading down first on the way down? If nose first, then what is the BC of the bullet? BC applies in cutting the wind straight down too. The higher the BC, the higher the terminal velocity from gravity. Also, the higher the BC and the highly-correlated SD, the higher it's penetration ability through the skull, so again, depends on whether the bullet turns around, and to what extent it turns around, and the BC. Also, if it does NOT turn around, then what's the BC of the bullet's BASE (which at that point becomes its nose)? Is it a boattail bullet, which will then have a higher BC than a flatbase?

Next, what's the weight of the bullet? The heavier the bullet, the deadlier, due to sheer momentum, which can translate into more penetration and crushing of skull bone. Further, regardless of whether it penetrates, the sheer weight concussing the skull can cause brain trauma or death, due to the massive jolt, if the bullet is big enough. Is it a 700 gr 12 ga slug? 750 gr .50 bmg? Going terminal vel? Are you kidding - death would be more likely than not.

Next, what's the bone density and resiliency of the person in question? Is it a baby getting hit in the soft spot? Is it a young person with flexible bones but not very thick and hard bones (skull)? Is it an elderly person with osteoporosis?

Next, is there any downward wind shear force aiding the bullet's velocity with a "tailwind"? Wind goes in all directions, not just left and right on a 2-dimensional scale. Many factors come into play, and I assure you that a bullet in the brain can be deadly, and there is a high probability that a given bullet will penetrate a skull.

Lookit, there is no magical sharp distiction between a bullet shot directly up at 90 degrees with a plum, versus a bullet shot at 89.999995 or 90.000005 - that's got nothing to do with it. Any bullet in the *general* range of 80 to 100 degrees, give or take, as far as the horizontal vector goes, is going to have a VERY minimal amount of a horizontal energy/momentum, from that particular vector - the damage comes from the bullet more or less stopping at its apex, turning around and coming down at its terminal velocity (if it has enough time to reach terminal, which most any bullet would). There is no magic distinction on either side of 90 degrees. At SOME point along the scale, the horizontal vector is low enough that momentum/energy is carried horizontally, at least until that point near the very end of the trajectory, where the parabolic curve is such that the drop rate is much higher than horizontal momentum, due to energy bleed from air resistance.
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Old March 15, 2007, 06:06 PM   #25
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Theoretically at least, a bullet fired directly vertically would have the same velocity upon returning to earth as it had when fired minus some velocity loss to air resistance. The round is fired, the acceleration of gravity slows it down, it stops momentarily, then the acceleration of gravity accelerates it towards the earth again. Same as if you jump off of a trampoline, or throw a ball into the air, etc. In fact, the calculation is much easier when you talk about firing the bullet vertically, as the calculation gets less complicated.

Assume-
3,200 fps muzzle velocity
Acceleration= 32 ft/sec/sec (G)
Time= Velocity/G = 100 seconds to stop
Then conversely, upon returning to earth
Velocity= Acceleration (G) X Time= 32 ft/sec/sec X 100 sec= 3,200 fps

I will accept that there is velocity loss due to air resistance both on the ascent and descent, but even ifyou were to lose 1/2 of the initial velocity to air resistance, you can see how a bullet fired into the air is still very dangerous when returning to earth.
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