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Physics of a bullet

Kinetic Energy Released


The formula for kinetic energy is

KE = ½mass x velocity2

So for example, a small .45 caliber bullet weighing 15 grams and traveling at 288 meters per second yields is 619 joules of energy. Further explained, if a man weighing 139 lbs (63.2 kg) were to fall off of a bed, it would yield roughly the same energy as being shot by that bullet; the difference being with a fall the energy is dispersed through the entire surface area of the man's body versus a bullet where the focal point is a tiny circle.

KE = ½mass x velocity2
KE = (.015kg / 2) x (288 m/s x 288 m/s)
KE = 619 joules of energy

PE = mass x gravity x height
PE = 63.2kg x 9.81 m/s x 1 meter
PE = 619 joules of energy

So taking this information, let's plug in the numbers of the Apache's M230 automatic gun ammunition. We have each 30mm round weighing 350 grams and traveling at 800 meters per second.

KE = (.3505kg / 2) x (800 m/s x 800 m/s)
KE = .175 x 640,000
KE = 112,160 joules of ass whoopin

Now that's a little hard to wrap your army around... I mean just how much energy is 112,000 joules? Well, for starters it's 180 times the energy of the .45 caliber handgun bullet. So imagine 180 people all pointing guns at this guy's body and everyone pulling the trigger all at the same time. Hmmm, yes...messy.

Furthermore, we can calculate just how high up this guy would have to plunge in order to release the same amount of energy as was released when he caught one of the Apache's 30mm rounds square in the chest...

112,160 = 63.2kg x 9.81 x height
height = 112,160 / (63.2 x 9.81)
height = 112,160 / 619.99
height = 180.9 meters (or 593 feet)

Now, taking our queue from the evolution of skyscrapers, I found an average 4.26 meters (13.96 feet) per foor. Thus this poor bastard you see splattered all over Main Street in downtown Baghdad? He looks the same as if someone tossed his happy ass off a 42 story building. This would have to happen in a vacuum to account for terminal velocity and wind resistance. But you get the point.

And the best part? The Apache's 30mm gun is really a popgun compared to the 30mm gun of an A-10 -- same diameter slugs but they're much heavier and travel much faster. So should you be unlucky enough to eat one of the Warthog's tank killing depleted uranium slugs...

KE = (.91kg / 2) x (1500 m/s x 1500 m/s) = 1,023,750 joules of smackdown

1,023,750 joules / 619 joules per .45 cal bullet = 1,626 people shooting you at once

1,023,750 joules = 63.2kg x 9.81 x height
height = 1,651 meters or 5,417 feet or a 1.02 mile freefall

But at a fire rate of 3,900 rounds per minute, the A-10's bullets will be more like Lays potato chips -- nobody can gonna eat just one. All you fuckers in Iran better keep that in mind when you hear the whoop-whoop-whoop of helicopter blades, eh?


Added: Apr-28-2009 
By: wilkinsonm
In:
Iran, Arts and Entertainment
Tags: Physics, of, a-10, and, apache, bullets
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