Its KE = 1/2mv2
Assuming a 2500fps velocity for the bullet I figure about 2800 joules KE.
For the 767, 100,000kg going 500mph is about 2.5 million joules.
The numbers above do not necessarily reflect the views and opinions of reality...I've been wrong before when I just start punching numbers into windows calculator.
A major difference would be the fact that the bullet has a smaller area over which it's transferring its energy to the steel.
Well, you beat me to the formula, but I can double-check your numbers for you.
Bullet (7.62x51mm NATO, a.k.a. .308 Winchester):
Source for specifications. (It's Wikipedia, for which I apologize. Guns are not my knowledge ares, and it's all I could find.)
mass = 9.50 grams = 0.0095 kilograms
velocity = 840 metres per second
bullet diameter = 7.82 mm = 0.00782 metres.
KE =
1/
2mv
2 =
1/
2*(0.0095 kg)*(840
m/
s)
2 = 3,350 Joules.
Cross-sectional-area of .308 Winchester round =
1/
4(pi)(diameter)
2 =
1/
4*(3.14)*(0.00782 m)
2 = 0.000048 m
2
KE per unit cross-sectional-area: (3,350 J)/(0.000048 m
2) = 69.8x10
6 Joules/metre
2, or 69. million Joules/m
2.
(69.8 million (kilograms per second per second) per squear metre cross-sectional-area)
Plane (Boeing 767):
Source. (Note the part where it says "The results of the analysis were in-line peer reviewed by Drs. Bob Nickell and Bob Kennedy. Dr. Nickell is a world recognized expert in the dynamic analysis of structures and used fuel containers. Dr. Kennedy is a world renowned structural analyst. ")
weight (max. takeoff) = 450,000 pounds. For this analysis, I will reduce this by 50%, to 225,000 pounds.
(Note that this is likely far too low to be realistic, so the final KE numbers will likely be lower than in real life.)
velocity = 350 miles per hour
Cross-sectional-area of a Boeing 767-400ER = 45.2 metres
2Source (right hand column, near bottom of page 719)
1 pound force = 4.45 Newtons (weight)
1 Newton (weight) = 0.102 kilograms mass
Therefore, (225,000 lb)*(4.45 N/lb)*(0.102 kg/N) = 102,000 kg, approximately.
1 mile = 1,609 metres
1 hour = 3,600 seconds
Therefore, (350 miles per hour) = (350 miles/hour)*(1,609 metres/mile)*(1/3,600 hours/second) = 156.43 m/s
KE =
1/
2mv
2 =
1/
2*(102,000 kg)*(156.43
m/
s)
2 = 1.25x10
9 Joules, or 1.25
billion Joules, not million, by my calculation.
KE per unit cross-sectional-area: (1.25x10
9 J)/(45.2 m
2) = 27.6x10
6 Joules/metre
2, or 27.6 million Joules/m
2.
(27.6 (million kilograms per second per second) per square metre of cross-sectional-area.)
I also found this while searching for the mass of the Boeing 767[/ur]
ETA: Alright, I don't know whay that link at the bottom doesn't work. I copied it atraight out of the ddress bar, just like the others, and set it up in code wrap just like the others. And now I can't link back to it. Strange.