Apache owns enemy soldiers... ( video )
- Thread starter Spidey
- Start date
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thats crazy...
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i have a video just like this one of a bomber flying over head and they are dropping lazer guided missles... its crazy to see some of this stuff.
i have another thats the footage from a smart missle, you can see the poor victim running from its target right before it hits - movie is called "fastest man in serbia" heheheh
i have another thats the footage from a smart missle, you can see the poor victim running from its target right before it hits - movie is called "fastest man in serbia" heheheh
Article I found analizing the video:
a small .45 caliber bullet weighing 15 grams and travelling at 288 meters per second yields is 619 joules of energy. They further explain that 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 dispursed 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 round weighing 350 grams and travelling 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.
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 fster. 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 ****ers in Iran better keep that in mind when you hear the whoop-whoop-whoop of helicopter blades, eh?
a small .45 caliber bullet weighing 15 grams and travelling at 288 meters per second yields is 619 joules of energy. They further explain that 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 dispursed 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 round weighing 350 grams and travelling 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.
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 fster. 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 ****ers in Iran better keep that in mind when you hear the whoop-whoop-whoop of helicopter blades, eh?
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the muscle car of the airforce!