# Bullets kill

1. May 18, 2006

### wolram

I all ways thought films dramatized the act of some one being shot, like they
fly backwards or jump in the air, but according to this it is true.

http://www.bobtuley.com/terminal.htm

Last edited by a moderator: Apr 22, 2017 at 10:21 AM
2. May 18, 2006

### Pengwuino

Who's claiming you fly backwards???

3. May 18, 2006

### Cyrus

It's false. They had a whole episode of this on mythbusters.

Action = reaction

If the gun kicks you back a few inches at most, how will the bullet make you fly? It cant get all this extra energy.

4. May 18, 2006

### wolram

Upon entering a fleshy target, the 7.62 bullet travels strait nearly six inches before the massive shock wave ahead of the bullet transfers incredible energy into the target as the bullet begins to tumble. Thus the bullet can exit before the maximum shock wave expansion can occur. 30 Caliber rifle bullets of this type are known to knock men down, and throw them off their feet back some distance. The cartridge is powerful, accurate, and humane in it's ability to kill quickly. The permanent cavity produced remains after the bullet exits the body. The temporary cavity causes tearing of tissues and muscle damage. The temporary cavitation (shock wave) causes death when it impacts the heart or liver but not necessarily in other areas of the torso.

5. May 18, 2006

### franznietzsche

You cant cheat the laws of physics, conservation of momentum in this case. Any knocking people down or knocking them back is going to be a physical reaction of the person to sudden pain, not directly from the force of the bullet.

6. May 18, 2006

### Cyrus

I was looking at wiki, they say that the bullet has 2010 J of energy. A 180lb man only has a force of 440N. That would be enough energy to move him 5 meters.......that does not sound right, because the guy firing the gun would also get knocked back 5 meters........

Hmmmmm? Any ideas franz?

7. May 18, 2006

### dav2008

Is that a joke post? What do you mean the man has a force of 440N?

Momentum is conserved.

(I just googled for "mass of bullet" so I'm using those numbers) http://hypertextbook.com/facts/2000/ShantayArmstrong.shtml

$$m_{bullet}=0.0042\ kg$$
$$v_{bullet}=900\ m/s$$
$$m_{man}=70 \kg$$

$$(.0042\ kg)(900\ m/s)+0\ kg\cdot m/s=(v_f)(70.0042\ kg)$$
$$v_f \approx .054\ m/s \approx .12\ mph$$
(Of course ignoring any friction)

In all honesty I have never seen anyone blown backwards from a gunshot in a movie. I've seen people shake or something but that could be probably explained by muscles twitching or something.

Last edited: May 18, 2006
8. May 18, 2006

### Rach3

Newtonian mechanics is simply not valid at two o'clock in the morning. :zzz:

Last edited by a moderator: May 18, 2006
9. May 18, 2006

### Pengwuino

And thats only a kinetic energy of 1700J. And lets even be realistic, some of the energy has to be lost as thermal energy inside the gun and you cant consider the gun really "part" of the person so you're not going to have 100% of that momentum transfered to the person.

10. May 18, 2006

### Pengwuino

You mean except in a movie?

And if you've ever seen someone get shot, they just get dropped like a sack of a bricks. I mean MAAAAAYBE multiple automatic rifles can do it... but i don't think anything less can knock someone any considerable distance backwards. I've seen people hit by multiple .45's (cop showdown) adn they dropped like a brick.

11. May 18, 2006

### wolram

The way i read this is that, the bullet is not the cause, it is the (shock wave) as the gun does not produce the shock wave your calculations are
invalid.

12. May 18, 2006

### Pengwuino

But the shockwave can't just be created out of nowhere. It has to have a source; the bullet.

13. May 18, 2006

### franznietzsche

Im sure by force he meant weight.

Ok, well lets see what I can do.

A 175 grain bullet (such as wolrams referenced 7.62) weighs in at 11.3 grams (took me a while to find this, and it may be wrong. But if you check it make sure youŕe looking for the right bullet, theyre not all the same). Lets say youve got a man at 80 kg (heck even I weigh more than that, and people still think I am malnourished). Now, 2010 joules doesn seem unreasonable for the amount of chemical energy released in the gunpowder charge (dav2008s numbers give about 1700 J). But not all of that energy goes into the bullet, some of it goes into recoil. So, lets assume a simple one-dimensional model, with conservation of energy and momentum.

The exit velocity of the bullet can be determined by

$$m_b v_b = m_1 v_1$$
$$\frac{1}{2}m_b v_b^2 + \frac{1}{2}m_1 v_1^2 = 2010 J$$

Where
$$m_b$$ Mass of bullet, 11.3 g
$$m_1$$ Mass of shooter, 80 kg.
$$v_b$$ Bullet exit velocity
$$v_1$$ One dimensional recoil velocity

We get

$$v_b = 596 m/s$$

Not 900 m/s. So either the bullet has a lot more energy in that charge(and Im no munitions expert), or something funny is going on (like momentum not being conserved funny). Incidently at this speed, the bullet has 2006.97 J of kinetic energy.

Now, on impact, the bullet does not impart all of its energy into the target. Nor is it an elastic collision either though. That said, lets solve for each case to get an idea of the limits.

For an elastic collision, we solve again:

$$m_b v_b = m_2 v_2$$ (where m2 =m1, just a different person)
$$\frac{1}{2} m_b v_b^2 + \frac{1}{2} m_2 v_2^2 = 2006.97$$

This gives

$$v_2 = 8.417 \times 10^-2 m/s$$

Or roughly 3.3 inches per second, 0.188 miles per hours, etc.

And if the bullet were to deposit all of its energy into the target in an inelastic collision (and not exit the target)

$$m_b v_b = (m_b+m_2) v_2$$

gives

$$v_2 = 8.423 \times 10¯2 m/s$$

Again a tiny 3.3 inches per second. The difference between elastic and inelastic is quite small, but the bullet cannot impart more speed than this to the person.

Now, certainly getting shot by such a large bullet is going to cause all sorts of problems fo the person, that will cause them to fall or stagger, even in the direction the bullet was traveling, but theyŕe not going to go flying or get floored, not by the impact anyway.

14. May 18, 2006

### wolram

This is a good site, if you read down to recoil i think you will be interested.

http://www.xmission.com/~fractil/math/kp.html

Also on this site.

The formula for momentum of a bullet in motion is momentum is equal to
mass times velocity and it's unit of measure is the (slug ft/sec) or
symbolically:

momentum = ( m * v ) / 225218 slug ft/sec.

225218 resolves mass in terms of grains which is the common unit of mass for a
bullet. There are 7000 grains in a pound. Or 7000 grains = 1 lb of mass or .0310815 slug.

The force a bullet exerts on impact is a ratio to the time it takes to
stop the bullet divided into the momentum or symbolically:

Force = momentum / (time to stop) or F = mv/t

If a bullet hits an object and it stops in 1/10000 of a second the force
is for a bullet traveling a 2000 ft/s and weighing 180 gr is for example

Force = ( ( 180 gr * 2000 ft/s ) / 225218 ) / .0001 = 15984.6 lbs .

That is if it stops in 2.4 inches of medium 15984.6 lbs is applied.

If a bullet hits an object and it stops in 1/1000 of a second the force
is for a bullet traveling a 2000 ft/s and weighing 180 gr is for example

Force = ( ( 180 gr * 2000 ft/s ) / 225218 ) / .001 = 1598.46 lbs .

That is if it stops in 24 inches of medium 1598.476 lbs of force is applied.

Last edited: May 18, 2006
15. May 18, 2006

### wolram

From the same site.

The great invention of the ballistics is that 99.99...% of the energy (also kp)
goes with the bullet and not to the shoulder of the marksman.

16. May 18, 2006

### franznietzsche

Did you ignore what I posted? Thats exactly what I found. And its no great invention, its simple high school physics.

Force is irrelevant. Impulse is what matters. Thatś $$F \Delta t$$.

Incidently 7000 grains per pound gives the 11.33 grams for the 175 grain 7.62 bullet. So unless you can find a more accurate number for the chemical energy in the charge, I stand by numbers. Note however that doubling the energy still only increases the velocities by a factor of $$\sqrt{2}$$. So even if its 4020 J per charge (giving an exit velocity of about 842 m/s, the target is only going to be given a maximum velocity from the impact of 0.119 m/s, or a whopping 4.6 inches per second. He is not going anywhere.

17. May 18, 2006

### wolram

I am not disputing any thing franznietzsche, just trying to understand how theese so called experts come out with statments like this.

18. May 18, 2006

### Cyrus

:rofl: I am stupid. Why did I conserve energy in a collision.

<Goes and hangs myself>

19. May 18, 2006

### Staff: Mentor

I actually saw a man get shot with a rifle. He just staggered. Well, until he fell dead.

20. May 18, 2006

### wolram

Have you been rustling again Evo?