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Extra-Super Serious Geek Alert!

By The Armorer on October 4, 2004 | | Comments (11) | TrackBacks (1)

WARNING-WARNING-WARNING-WARNING!!!! If you are easily bowled over by technogeeky jargon, just skip on down one. If high school physics broke your spirit, just keep on movin' on - if you think Superman comics are packed full of useful insights into how Newton's Laws and the Laws of Thermodynamics work... just keep on keepin' on!

Still here? You'll like this. The Instapilot will like this. Anybody wanna argue the math? [N.B. - it was sent to me, I didn't work this out]

We know the formula for kinetic energy is KE = ½mass x velocity2 . Now let's check in with the Movie Physics Guys.

So in their example, a small .45 caliber bullet weighing 15 grams and traveling 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 disbursed 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

Potential energy is defined to be PE = (mass) x (g) x (height), where the height is the vertical distance of the object from the ground and g stands for gravitational acceleration or acceleration due to gravity. Near the surface of the earth, g is a constant approximately equal to 9.8 meters per second per second (m/s2). You can use these formulas to calculate the total energy of the system by just adding up the forms.
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

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 .45 caliber handguns 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 (sic) from the evolution of skyscrapers, I found an average 4.26 meters (13.96 feet) per floor. Thus this terrorist 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 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 smack down
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's gonna eat just one. All you terrorist rats in Iraq and Iran better keep that in mind when you hear the whoop-whoop-whoop of helicopter blades, eh?

Hat tip to Cary!

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Big Guns! from There's One, Only! on October 5, 2004 5:39 PM

There are some good "big gun" discussions going on over at John's Castle Argghhh! I especially like this post about the A-10's 30mm vulcan cannon. And make sure that you read all of the way to the end of the... Read More

11 Comments

By GEBIV on October 4, 2004 7:59 PM
Wow. I just love when science explains the effects of high-velocity/large calibre ammo on things. Now lets do the math for that anti-personel tank round that John was talking about. (At least I think that it was talked about here...)
 
By SangerM on October 4, 2004 9:09 PM
Now, for fun (and I sho wisht I wuz good wit nummers), imagine how much energy is released at the point of impact from a few dozen 25mm rounds from an AC-130 circling at 15K feet overhead. Or how it must REALLY suck to be on the incoming end of the 40mm, and especially that ass-kicking 105!!! The only thing I think must be more fun to watch/use are the miniguns in the MH-53 Pave Lows and the MH-60 Pave Hawks. What they may lack in size, they more than make up for in street-clearing, stump-grinding rpms! SangerM
 
By J.M. Heinrichs on October 4, 2004 10:03 PM
... and the 120mm APFSDS long-rod penetrator fired from an Abrams has more than 10x the energy of the A-10's 30mm. Cheers JMH
 
By Mark Adams on October 4, 2004 10:24 PM
Scientific Sickness! A 10 on the ouie gooie geek meter.
 
By John of Argghhh! on October 4, 2004 10:40 PM
Mark! Again you wander through! Well, look at it this way - if you want to have a military available for Democrat Presidents to use, we've got to take kind of a non-partisan approach to this. Ya can't just toss us out as thug Republican stooges every time the PIP (Party In Power) and have the rest study horticulture... Just a thought! You should hang around Police/Fire/Ambulance/Trauma Center locker rooms. Kinda goes with the territory! Trust me - if you are going to have an army, it's better to have one that has some idea of what it's doing... then you don't find yourself having to kill children with car bombs to express your displeasure with the regime change...
 
By Beck on October 5, 2004 2:23 AM
One problem. The equations here are assuming that 100% of the energy from the bullet is transfered to the body. In fact, relatively little of it will be transfered, because the body just doesn't have enough stopping power. In the initial example of a single .45 bullet, you might actually get 100% energy transfer. For instance, if the person were wearing a bullet proof vest which succeeded in stopping the bullet (or at least slowing it down enough that it didn't exit out the back) then the force of impact would be exactly like that 1m fall out of bed. If a human body (say it was the hulk) actually stopped a bullet from an apache, then yes, it'd be accelerated backwards as though it had been dropped off a 40 story building, which, even though the bullet didn't pass through, would pretty much liquify any ordinary person. As it is, the bullet passes clean through, and only loses perhaps a few hundred Joules. Instead, I propose an alternate system. Let's say the human body is capable of stopping one .45 bullet's worth of energy before giving up & allowing the bullet to exit through the back (in reality, it'd probably take 2 or 3 people to stop a .45, especially if it avoided big bones, but we'll just use the 1 person idea for now). In that case, you could line up 1,626 terrorists, and with a single round from an Apache, you could puncture every single one of them. Now all we have to do is figure out how to get them to all line up!
 
By John of Argghhh! on October 5, 2004 5:57 AM
Good points, Beck. Of course, the .45 was developed for that purpose with that in mind, to not go through people, if possible. The example above also does not include a discussion of the attenuation curve as the bullet loses energy over range. Still - it is an illustrative example, ain't it. As for your falling out of bed analogy - not in my experience - if for no other reason than you've generalized the energy too much - there is a concentration of the impact energy. Both my father and I have been shot wearing flak vests. We agree it's far more analgous to having the Hulk hit you in the chest with a baseball bat than falling out of bed. In my father's case, it did result in broken ribs and the bullet penetrated, albeit just under the skin along the ribs.
 
By triticale on October 5, 2004 8:04 AM
I once worked with someone who had, in an earlier life, participated in RAF weapons research. The conclusion reached was simple. Rate of fire is no substitute for weight of fire. You can fire enough .32 ACP from a full-auto Skorpion to total the ballistic energy of one round from an elephant gun, but it won't stop a charging elephant.
 
By Fred on October 5, 2004 8:21 AM
The other thing, since we're being pedantic, is that air resisteance when falling off the taller buildings. I'm not sure that they won't have hit terminal velocity. In more ways than one.
 
By SangerM on October 5, 2004 9:56 AM
Wasn't the .45 designed to defeat the kind of simple armor the Moros wore? That was the line we got when I joined the Army....
 
By John of Argghhh! on October 5, 2004 10:10 AM
It was designed to hit 'em and stay in 'em and knock 'em down - not so much because of the armor, but because they drugged themselves and were, well, perhaps just tougher, and the .38 wasn't doing enough damage to cause them to lose interest in dropping your intestines around your ankles. Same problem the Brits had in colonial warfare that caused them to go to the .455 round. Stopping power for non-wussy natives who were grumpy about exterior meddling in their affairs. Apparently euro-soldiers were wusses who would go, "Oh! My! I've been shot! I must sit down and rest." The natives were more like, "Izzat the best you got? Eat assegai, asshat!" Or, in the case of the Moros, bolo machete.
 
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