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Posted

Lol she'll be apples boys. Ok so we aren't allowed bullet proof vests but are these technically legal here?

 

 

Posted

Reminds me of that great piece of classical music - The William Tell Overkill. Hi Ho Silver. Away!

 

OME

 

 

Posted

Really? Surely the energy of a bullet fired at such close range from a weapon like that should at least have made him stagger back a pace or two? Even with his protector firmly in place the jolt should have been severe enough to knock the grin off his face, even if it didn't knock anything else off. Don't try this at home is all I can say.

 

rgmwa

 

 

Posted
Really? Surely the energy of a bullet fired at such close range from a weapon like that should at least have made him stagger back a pace or two?

Nah, not really (below) unless it was some insane calibre - or in a movie. But the fact there was no slo-mo or close up of the impact damage makes it seem a bit sus. Risk of ricochet due to the convex shape, risk of a dud round.... quite close to some major blood vessels there so it might be faked but I still thought it was funny. There are plenty of test clips about, it's a pretty common demo method but not in the groin from a distance shooting at a curved plate.

 

 

 

Posted

It was a 22lr so not much energy involved.

 

Rgmwa surely the 'pace or two backwards' was a bit tongue in cheek? It's basic physics, for the target to be knocked back a pace or two the shooter would also have to be knocked back a pace or two. (Won't happen unless a HUGE caliber is used) The movies make it look so glamorous but it is really eerily mundane.

 

For more interesting vids try googling 'the armoured Armani'

 

 

Posted

Only a 22lr? I'm disappointed. He obviously didn't have the balls for the job.

 

rgmwa

 

 

Posted

Using a .22 Magnum loaded to give a muzzle velocity of 610 m/s, we can calculate the Force applied to the target using the Law of Conservation of Momentum.

 

Bullet weight = 40 grains = 2.6 gm - 0.0026 kg (m1)

 

Muzzle velocity = 610 m/s (v1)

 

Target = 100 kg (m2)

 

Target initiral velocity = 0 m/s (v2)

 

m3.v3 = momentum of bullet after impact

 

m4.v4 = momentum of target after impact

 

m1.v1 + m2.v2 = m3.v3 + m4.v4

 

(0.0026)* 610) + (100 * 0) = (0.0026 * 0) + (100 * v4)

 

1.56 kg m/s + 0 = 0 + 100v4

 

(1.56 / 100) = v4

 

0.0156 m/s = v4

 

0.0156 m/s = 0.056 Kph

 

If we knew how long it took for the bullet to stop, we could calculate the deceleration (v1 - v3)/t and from that work out the force exerted on the target by the bullet, F = ma.

 

OME

 

 

Posted

Let's guess the time for the bullet to stop is about 1/25th of a second (the proverbial `blink of an eye'). That makes the deceleration 610/.04=15,250 m/sec2. For a bullet with a mass of .0026kg, we get an impact force of .0026x15,250 or about 40kg. That's a fair punch in the nether regions.

 

rgmwa

 

 

  • Like 1
Posted
Let's guess the time for the bullet to stop is about 1/25th of a second (the proverbial `blink of an eye'). That makes the deceleration 610/.04=15,250 m/sec2. For a bullet with a mass of .0026kg, we get an impact force of .0026x15,250 or about 40kg. That's a fair punch in the nether regions.rgmwa

You would not be smiling!

 

 

Posted

If you're silly enough to stand there while someone shoots you in the nuts, you should remove the cup and remove yourself from the gene pool.

 

 

Posted
Good thing he was wearing a cup[ATTACH=full]36177[/ATTACH]

I spoke to him. He said, " It was ........... murder."

 

OME

 

 

  • Haha 1
Posted
Let's guess the time for the bullet to stop is about 1/25th of a second (the proverbial `blink of an eye'). rgmwa

I was scratching around for a suitable figure to use to calculate the acceleration in this example so I could work out the Force. Your guess seems reasonable to me.

 

OME

 

 

Posted
AND all the bullets COULD have a reduced charge load!

I'm sure they would reduce charge load.

 

 

  • Haha 1

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