kinetic energy?

[ccsniper]

How much kinetic energy is needed for a bullet to be considered lethal? I have done several searches and cannot find anything other than what is considered lethal for deer. I need it for an argument against firing your shotgun in random directions to "scare" away home invaders.

 

[Sam1911]

You can easily kill a deer with a .22, and you could almost certainly do so with a very good pellet rifle under the right conditions.

That's not a very complete answer, but no answer based solely on kinetic energy could be.

 

[481]

There is no hard and fast number for the amount of KE required to be lethal.

Under the proper conditions, a pellet gun (3.5 - 6 fpe) can kill a person and a .375 H&H (4300 fpe) may not.

KE by itself means little. It is how it is used.

 

[ccsniper]

There is no hard and fast number for the amount of KE required to be lethal.

I am asking for a number that would allow a bullet to penetrate sufficiently to be considered lethal.

In other words, what is the exact amount of energy required to penetrate deeply enough into a person that could potentially be lethal.

 

[Blackstone]

If you're talking about eyes, then 4-6ft/lb from a pellet gun is enough

 

[1911Jeeper]

Simple answer, randomly discharging a firearm is stupid, reckless, and has the potential to main or kill innocent persons.

By definition, gunfire is considered lethal force. Kinetic energy has no relevance to the argument. "Warning shots" are not a legally recognized defense in court should the shots injure or kill someone innocent. Depending on state and local laws, they are probably illegal as well.

If a shooter cannot articulate how/why they were threatened, and also identify their target, assault with a deadly weapon, manslaughter, or murder charges may be the brought against them.









.

 

[481]

I am asking for a number that would allow a bullet to penetrate sufficiently to be considered lethal.

That is still too simplistic- KE alone doesn't dictate lethality. Lethality is determined by what is hit by the bullet, the function and importance of what is hit by the bullet to the survival of the person shot, and even to the mental/psychological determination of the person being hit.

What you are asking is very complex and there is no simple answer that can phrased in terms of a simple KE value.

In other words, what is the exact amount of energy required to penetrate deeply enough into a person that could potentially be lethal.

Any amount of KE can be potentially lethal. Just depends on how it is used.

 

[Deus Machina]

There's just no way to answer that.

Assuming ideal conditions, 20 ft/lb will be plenty enough to put a .17 round deep enough to pierce vital organs. A .45 carrying 20ft/lb would leave an aching bruise.

The body of an animal varies too much for any hard numbers--that .17 will get right to the heart, but a half-inch up or down and it will break a rib and stop--and even under ideal, controlled conditions, the numbers themselves will vary wildly between different diameters and even designs of the bullet.

Worry more about penetration and end size.

 

[Sam1911]

I am asking for a number that would allow a bullet to penetrate sufficiently to be considered lethal.

I'm really struggling to come up with a satisfactory answer to your question.

There is NO level of energy so low that a FIREARM is not considered using "lethal force." Any firearm, fired at another human is considered using deadly or lethal force in every jurisdiction in this country.

Every jurisdiction in the USA that has a law prohibiting discharging a firearm within city limits includes ALL firearms, every one, period. (And some include air rifles, too!)

So if the question is what firearm produces so little energy that it isn't considered "lethal force" then the answer is NONE.

In other words, what is the exact amount of energy required to penetrate deeply enough into a person that could potentially be lethal.

That's very much the same as asking, what's the exact horsepower required to make a car go 100 mph? There's so many things that will influence the question that you really can't answer the question at all -- and one suspects no answer given will get to the underlying point anyway. Even knowing a theoretical average quantify of FPE (if such a thing could be calculated) would really not answer what is seems like you're trying to ask -- i.e. the legal question(s) I answered above..

 

[Blackstone]

Here in the UK, a gun is only a firearm if it's power is greater than 1J as that's the medically defined energy required to break skin.

 

[hso]

No, what you need is the lethal range for a shotgun.

Consider buckshot lethal to 120-ft. and slug lethal to 300-ft.

 

[230RN]

Possibly helpful:

Hatcher's Notebook, pp 405-408, indicates that 100 foot-pounds is considered the minimum to inflict a wound on the human body.

However, in another section on bullets fired vertically, he says, "Previously the Army had decided that on the average, an energy of 60 foot-pounds is required to produce a disabling wound." (p. 514.)

Of more direct interest in terms of Mr. Biden's expert advice might be the maximum range a shotgun will shoot. We find this informaton on p. 547: "Ordnance Technical Manual 9-1990 gives the following distances as the maximum ranges attained by standard shot loads when fired in full-choke guns: No. 00 Buckshot, 600 yards, No.8 shot, 230 yards, and No.9 shot, 210 yards."



Terry, 230RN

 

[ccsniper]

However, in another section on bullets fired vertically, he says, "Previously the Army had decided that on the average, an energy of 60 foot-pounds is required to produce a disabling wound." (p. 514.)

Thank you, that is what I was asking. Sorry to all if my question was worded wrong,.

 

[2zulu1]

Possibly helpful:

Hatcher's Notebook, pp 405-408, indicates that 100 foot-pounds is considered the minimum to inflict a wound on the human body.

However, in another section on bullets fired vertically, he says, "Previously the Army had decided that on the average, an energy of 60 foot-pounds is required to produce a disabling wound." (p. 514.)

Of more direct interest in terms of Mr. Biden's expert advice might be the maximum range a shotgun will shoot. We find this informaton on p. 547: "Ordnance Technical Manual 9-1990 gives the following distances as the maximum ranges attained by standard shot loads when fired in full-choke guns: No. 00 Buckshot, 600 yards, No.8 shot, 230 yards, and No.9 shot, 210 yards."



Terry, 230RN

100fpe as the minimum KE# is inaccurate. There is 22lr ammunition with ~100fpe ME that cause wounds at distance, therefore, the minimum amount of KE needed to cause wounding is less than 100fpe. Actually it is much less, and as others have stated, more complex than what Hatcher knew at that time.

 

[JRH6856]

Here in the UK, a gun is only a firearm if it's power is greater than 1J as that's the medically defined energy required to break skin.

Well, 1 Joule = .737554 ft/lbs, but that seems pretty meaningless unless you know the area to which the energy is being applied as pressure. 1J applied to 1 sq ft. of skin is not likely to penetrate, whereas applied to .01 sq in it is.

 

[230RN]

Me:

Possibly helpful:

There are a lot of variables there. Hatcher, for example, did not say whether the target was in battle dress or stark nekkid, or whether the projectile was flying point first or base first (in the case of bullets coming down from the sky).

I can shoot a dart out of my blowgun which will penetrate clean through 3/8" particle board (which is pretty hard stuff) so that you need pliers and a lot of effort to pull it back out. (I've done it, so I know. It was the first thing I shot with the blowgun, figuring it would just make a little pinhole and stick there like a thumbtack. Boy, was I surprised!)

You can drop a broadhead arrow down on a beefsteak and it will go through and stick in the cutting board, but if you drop a .45 bullet from a height which gives the same calculated energy it will just plop down onto the surface of the steak.

So, while I recognize the disparity between Hatcher's report of 60 to100 foot-lbs for a (rifle) bullet, I would hope that is "possibly helpful" in terms of ball-parking a number.

But 1911Jeeper has the right answer anyhow:

Simple answer, randomly discharging a firearm is stupid, reckless, and has the potential to maim or kill innocent persons.

Terry, 230RN

 

[denton]

The Earth intercepts protons from deep space that hit with a surprising amount of KE. One of those won't kill you. A 2,000 pound car traveling 5 miles per hour is carrying just under 3,000 foot pounds of energy, and that won't generally kill you, either.

KE is simply not a useful predictor of lethality. So, as asked, the question cannot be answered.

Force, momentum, and kinetic energy are all derived from the same fundamental quantities. You can move from force, to momentum, to kinetic energy by performing integration, and move the other way by performing differentiation.

As others have pointed out, what really matters is force, or more precisely, force per unit area (pressure) exerted by a bullet on impact. If the pressure of the bullet is greater than the strength of the tissue in front of it, then the tissue will be crushed or torn.

The force, and hence the pressure exerted, is the rate at which the bullet is shedding momentum. When the bullet is no longer shedding momentum fast enough to crush or tear tissue, it comes to rest. Hopefully, that is just under the off-side hide of the most humoungous elk known to man.

 

[shootniron]

I think there are far too many variables to definitively answer your question.