The Reality of Terminal Ballistics: Is it Really Like the Movies?

In summary: So if the man has a weight of 70 kg, and the bullet has a weight of 11.3 grams, then the man would be moved (by the bullet) with a speed of .054 m/s. That's about 12 mph.
  • #1
wolram
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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 [Broken]
 
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  • #2
Who's claiming you fly backwards?
 
  • #3
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 can't get all this extra energy.
 
  • #4
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
wolram said:
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.
You can`t 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
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
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

[tex]m_{bullet}=0.0042\ kg[/tex]
[tex]v_{bullet}=900\ m/s[/tex]
[tex]m_{man}=70 \kg[/tex]

[tex](.0042\ kg)(900\ m/s)+0\ kg\cdot m/s=(v_f)(70.0042\ kg)[/tex]
[tex]v_f \approx .054\ m/s \approx .12\ mph[/tex]
(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.
 
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  • #8
Is that a joke post?
Newtonian mechanics is simply not valid at two o'clock in the morning. :zzz:
 
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  • #9
And that's only a kinetic energy of 1700J. And let's even be realistic, some of the energy has to be lost as thermal energy inside the gun and you can't consider the gun really "part" of the person so you're not going to have 100% of that momentum transferred to the person.
 
  • #10
dav2008 said:
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.

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
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
But the shockwave can't just be created out of nowhere. It has to have a source; the bullet.
 
  • #13
dav2008 said:
Is that a joke post? What do you mean the man has a force of 440N?

I`m sure by force he meant weight.

Momentum is conserved.

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

[tex]m_{bullet}=0.0042\ kg[/tex]
[tex]v_{bullet}=900\ m/s[/tex]
[tex]m_{man}=70 \kg[/tex]

[tex](.0042\ kg)(900\ m/s)+0\ kg\cdot m/s=(v_f)(70.0042\ kg)[/tex]
[tex]v_f=.054\ m/s = .12\ mph[/tex]
Ok, well let's 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, they`re not all the same). Let's say you`ve 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, let's assume a simple one-dimensional model, with conservation of energy and momentum.

The exit velocity of the bullet can be determined by

[tex]
m_b v_b = m_1 v_1
[/tex]
[tex]
\frac{1}{2}m_b v_b^2 + \frac{1}{2}m_1 v_1^2 = 2010 J
[/tex]

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

We get

[tex]v_b = 596 m/s[/tex]

Not 900 m/s. So either the bullet has a lot more energy in that charge(and I am 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, let's solve for each case to get an idea of the limits.

For an elastic collision, we solve again:

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

This gives

[tex]v_2 = 8.417 \times 10^-2 m/s[/tex]

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)

[tex]m_b v_b = (m_b+m_2) v_2 [/tex]

gives

[tex]v_2 = 8.423 \times 10¯2 m/s[/tex]

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
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.
 
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  • #15
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
wolram said:
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.


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

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

Force is irrelevant. Impulse is what matters. Thatś [tex] F \Delta t[/tex].

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 [tex]\sqrt{2}[/tex]. 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
I am not disputing any thing franznietzsche, just trying to understand how theese so called experts come out with statements like this.
 
  • #18
:rofl: I am stupid. Why did I conserve energy in a collision.

<Goes and hangs myself>
 
  • #19
I actually saw a man get shot with a rifle. He just staggered. Well, until he fell dead.
 
  • #20
Evo said:
I actually saw a man get shot with a rifle. He just staggered. Well, until he fell dead.

Have you been rustling again Evo?
 
  • #21
wolram said:
Have you been rustling again Evo?
Hey I have all the cattle I want living behind my house. Longhorn steer.
 
  • #22
Evo said:
Hey I have all the cattle I want living behind my house. Longhorn steer.

You are not supposed to tell the world where you hide them :uhh:

when is ox roast night ? :smile:
 
  • #23
wolram said:
You are not supposed to tell the world where you hide them :uhh:

when is ox roast night ? :smile:
Steaks and burgers on the hoof. :tongue2:

I think there my be a gap in the fence in the next few days, can I help it if one wanders onto my property? o:)
 
  • #24
Evo said:
Steaks and burgers on the hoof. :tongue2:

I think there my be a gap in the fence in the next few days, can I help it if one wanders onto my property? o:)

It would not be your fault, nor if it got accidentaly skined and butchered.
 
  • #25
wolram said:
I am not disputing any thing franznietzsche, just trying to understand how theese so called experts come out with statements like this.
The simple answer is either a) they are not experts, or b) if they have actually seen such a thing, they fail to disinguish between the reaction of the person to the sudden pain (such as stumbling and falling backwards) from the actual force of the bullet.
 

1. What is terminal ballistics?

Terminal ballistics is the study of the behavior and effects of projectiles when they strike a target, specifically focusing on the damage and penetration caused by the projectile.

2. How accurate are depictions of terminal ballistics in movies?

The reality of terminal ballistics is often exaggerated or misrepresented in movies for dramatic effect. While some aspects may be accurate, such as the general concept of a projectile causing damage upon impact, the specific details and behaviors are often exaggerated or simplified for entertainment purposes.

3. Are there any common misconceptions about terminal ballistics?

One common misconception is that bigger or louder guns automatically cause more damage. In reality, factors such as bullet design, velocity, and target composition have a greater impact on terminal ballistics than the size or loudness of the gun.

4. How do scientists study terminal ballistics?

Scientists use a combination of experimental studies and mathematical models to understand the behavior of projectiles and their effects on different targets. This may include testing different types of ammunition on various materials and analyzing the results to determine patterns and principles.

5. How does terminal ballistics impact real-life scenarios?

Understanding terminal ballistics is important for various fields such as forensic science, military strategy, and law enforcement. It can help determine the type of weapon used in a crime, the effectiveness of body armor, and the potential injuries caused by different types of ammunition. It is also crucial in developing and improving weapons and protective gear.

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