Grinds my Gears: Inaccurate Hollywood Portrayals of Firearms and Firearm Accessories

This one's been brewing in the back of my head for a while, but I never really cared to post because 1) Jon and Chris already know the things I'm about to say 2) Everyone else doesn't really care. Well too bad, it has gone on long enough! I saw something that was the final straw. I've been watching Top Shot on Netflix and I consider that a show that knows everything about firearms. But look at it's promotion poster!
















Any of the contestants on the show could have told you what's wrong with it. The flying bullet still has it's case attached to it. Only the conical part (bullet) leaves the gun. The rectangular part houses the gunpowder and is ejected out the side of the gun. I've seen this is a couple movie posters with flying bullets and just laughed at it. But this is sad. I thought this show was the real deal. Why couldn't they get a photoshop guy who knew something that all the fans know?

Anyway, here's a couple more that I can list off the top of my head.

1. Never Needing to Reload and Not Aiming Down the Sights

Anyone who plays Call of Duty or Counter-Strike knows about how long you can hold down a trigger until the magazine is empty. It's about 3 seconds.

Check out Arnold in Commando. His first burst against the two guards alone is already one magazine, yet he doesn't have to re-load in the whole battle. Also he aims from the hip and is able to hit all his targets. God mode indeed.



2. Cocking back the hammer on a pistol that doesn't have a hammer.

One common thing in gun movies is when the bad guy points a gun at someone and uses his thumb to cock back the hammer. It makes a "click" sound. I suppose movies use that sound alot because it's the modern equivalent of hearing the executioner sharpen his axe. Most guns can shoot either from a hammer-decocked or hammer-cocked position. But cocking the hammer makes the trigger pull shorter, and thus making a more accurate shot. But it really doesn't matter from a distance of five feet, I guess movies do it for the "click" sound factor.

This guy makes a good point:




But the FAIL in movies and TV shows comes when they apply a click sound to a gun that has no external hammer, like that first revolver in the previous video. The most famous example is a Glock.


If you look at the back edge there is no hammer to cock back. The firing mechanism is internal. And yet when the guy points a Glock, there is a mysterious hammer cocking sound. Of course the biggest culprit in my head is the show 24. They've done this specific mistake at least five times, maybe more.




3. Gratuitous Shotgun pumping / Rifle cocking

You only need to load a gun by pulling on the slide or pump once. That loads one round from the magazine into the firing chamber, and it's ready to go. The sound effect editor or director likes that cocking sound so much that even on guns without hammers (rifles and shotguns) they will have the guy pumping the gun to make that sound. I'm a super 24 junkie and as much as I love the show, every gun battle has way too much gun cocking. In one scene, Jack Bauer has gone rogue and is inside a gas station. The cops surround him outside. Every time they have a cut scene to the cops outside (about 3 times), the cop with a shotgun pumps his shotgun. I face-palmed because unlike pistol hammer cocking, every time he does this a perfectly good round is ejected out of the gun, and he has one less round in the magazine.


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There's more examples, but I've made my point. Just know that this Grinds my Gears because I wish Hollywood would consult with technicians who know its better to give up cool "effects" for the sake of being legit. I know guns are powerful and dangerous, I don't need a stupid clicking sound to tell me.

On a positive note, I remember seeing this video from CNN. It's a firefight between rebels in Libya and the Gaddafi forces. The fun part starts at 20 seconds, when you realize you're not watching "Commando" but a real fight. And yet they hop into the line of fire and take a couple shots from the hip. They must be thinking "This is how the big American fights, I shall be like him, he always hits his target and never gets hit. Veery Naaiice." (no one gets hit, you can watch it)


If it at least turns our potential enemies into horrible infantry, then keep it up Hollywood. Keep making movies with Arnold and Chuck and Sylvester. For all we can guess, their hand-to-hand combat training involves only roundhouse kicks.

5 comments:

J.Frosty said...

LMAO, ALLAH ACKBAR *roundhouse*

I cannot wait till we meet in battle.

and stop hating on arnold's God Mode lol

Khorn said...

A man cock a 0.45 glock and fire a bullet with a of Mass 13 g having a muzzle velocity of 1050 ft/s. The bullet hits a target with a Mass 150 lb running at 5 miles/hr. What is the final speed of the bullet embedded target? Given that the target is on ice (which result in negligible friction) and the shell of the bullet casing is fully attached (which according to Mac Bryan is totally impossible). How far will the target fly if it is running toward a cliff.

Khorn said...

Please answer in terms of height of cliff for second question.

Mac OBryan said...

first off I will say that I know you didn't pull these numbers out of your ass because a .45 bullet is indeed around 230 grains(13 grams) and has a velocity of around 1000 fps. either this is a real question from somewhere or you looked up the stats.

anyway here's my guess to the question.

assuming there's no friction from the body and the ground and no bullet passthrough, the full momentum will be transferred to the target. the final speed of the bullet will be 0.

there must be conservation of momentum between the bullet and the target. (Mass)(Velocity)= (Mass)(Velocity) or MV=MV

bullet is traveling at 1050 fps which I converted to 350 meters/second. the target weighs 150 lbs or 68 kg

MV=MV
(350m/s)(.012kg)=(v)(68kg)

v=0.06 m/s

body is running toward cliff at 5 miles per hour which i converted to 2.5 m/s

if the running target was shot in the back, the two velocities are added together to make 2.56 meters/second. The bullet doesn't matter, his running start mattered much more -_-

The distance in the x-axis that the target will "fly" is the speed multiplied by the time spent in the air or x= vt = (2.56)(t)

because motion in the x and y axis are independent of each other, we can figure out distance he flies in x-axis by the time he takes to hit the ground in the y-axis.

the equation for freefall is
y= vt + (1/2)a(t^2)

y = the height of the cliff

since initial velocity in y-axis is zero, all we need is y=(1/2)at^2

a= gravity = 10 m/s

and so y=5(t^2)

t = .445y seconds = time until he hits ground in terms of height of the cliff

and so distance he "flies" in the x-axis is

x = 2.56(t) = 2.56(.445)(y) meters

So if the cliff was 100 meters high, he would fly for 113 meters in the x-axis. that's pretty far!

Khorn said...

Almost correct! Yes I didn't pull these numbers from the top of my head, had to look it up (quite interesting). But the problem, I made up. I'm glad that someone solved the problem. It was more of a joke post, but wanted to make the problem solvable. Anyway, here is my answer:

Solve Final Velocity of Bullet embedded Target...

Mass of Bullet = m
Mass of Body = M
Initial Velocity of Bullet = vb
Initial Velocity of Body = vB
Final Velocity = vf

m*vb + m*vB = (m+M)vf

(13g)(320m/s) + (68000g)(2.24m/s) = (13g+68000g)(vf)

vf = 2.30 m/s

Solve Distance (X & Y) in terms of Cliff Height (y)...

Derived projectile equation from Calculus:
a = g
v = vi + gt
y = yi + vt + 0.5gt^2
y = 0 + 0 + 0.5(9.8m/s)t^2
t = 0.452sqrt(y)

x = xi +vt + 0.5gt^2
x = 0 + (2.3m/s)t + 0
x = (2.3m/s)(0.452sqrt(y)) = 1.0396sqrt(y)

So if I use cliff height of 100 m, the distance of travel would only be equal to 10.396 m. You forgot to square root the height!