Gnarly Gnu Posted June 2, 2015 Share Posted June 2, 2015 Gotta love Americans.... this is quite nuts! 2 1 Link to comment Share on other sites More sharing options...
bexrbetter Posted June 2, 2015 Share Posted June 2, 2015 Well he's raised the bar. Link to comment Share on other sites More sharing options...
SDQDI Posted June 2, 2015 Share Posted June 2, 2015 Lol she'll be apples boys. Ok so we aren't allowed bullet proof vests but are these technically legal here? Link to comment Share on other sites More sharing options...
old man emu Posted June 2, 2015 Share Posted June 2, 2015 Reminds me of that great piece of classical music - The William Tell Overkill. Hi Ho Silver. Away! OME Link to comment Share on other sites More sharing options...
rgmwa Posted June 2, 2015 Share Posted June 2, 2015 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 Link to comment Share on other sites More sharing options...
Gnarly Gnu Posted June 2, 2015 Author Share Posted June 2, 2015 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. Link to comment Share on other sites More sharing options...
SDQDI Posted June 2, 2015 Share Posted June 2, 2015 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' Link to comment Share on other sites More sharing options...
rgmwa Posted June 2, 2015 Share Posted June 2, 2015 Only a 22lr? I'm disappointed. He obviously didn't have the balls for the job. rgmwa Link to comment Share on other sites More sharing options...
old man emu Posted June 2, 2015 Share Posted June 2, 2015 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 Link to comment Share on other sites More sharing options...
rgmwa Posted June 2, 2015 Share Posted June 2, 2015 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 1 Link to comment Share on other sites More sharing options...
pmccarthy Posted June 2, 2015 Share Posted June 2, 2015 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! Link to comment Share on other sites More sharing options...
fly_tornado Posted June 2, 2015 Share Posted June 2, 2015 Good thing he was wearing a cup Link to comment Share on other sites More sharing options...
Marty_d Posted June 3, 2015 Share Posted June 3, 2015 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. Link to comment Share on other sites More sharing options...
old man emu Posted June 3, 2015 Share Posted June 3, 2015 Good thing he was wearing a cup[ATTACH=full]36177[/ATTACH] I spoke to him. He said, " It was ........... murder." OME 1 Link to comment Share on other sites More sharing options...
old man emu Posted June 3, 2015 Share Posted June 3, 2015 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 Link to comment Share on other sites More sharing options...
spacesailor Posted June 3, 2015 Share Posted June 3, 2015 AND all the bullets COULD have a reduced charge load! spacesailor Link to comment Share on other sites More sharing options...
bexrbetter Posted June 3, 2015 Share Posted June 3, 2015 AND all the bullets COULD have a reduced charge load! I'm sure they would reduce charge load. 1 Link to comment Share on other sites More sharing options...
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