Rabs
Active member
http://www.worldtribune.com/worldtribune/06/front2453745.144444444.html
Sounds like potent and stuff, and they have decent production ability.
Sounds like potent and stuff, and they have decent production ability.
Why its good to be allies with Israelis
Kozzy Mozzy
Active member
I wonder if they mean carbon fibers when they say nanotechnology. I doubt that the news reporters know what a carbon fiber or a nanobot is anyway.
Mohmar Deathstrike
Banned
"Last year, a sample of the ApNano material was subjected to tests in which a steel projectile traveling at a speed of up to 1.5 kilometers per second slammed into the material.
Executives said the impact was the equivalent to dropping four diesel locomotives onto an area the size of a human fingernail."
This means nothing if they don't say what height the locomotors would have been dropped from. Let's make a quick calculation shall we?
Kinetic Energy KE=mv² , where m is the mass and v is the velocity.
Let's assume the projectile weighs 50g. So the round had a KE of 0.05kg x (1500 m/s²)=112500 J
The 4 locomotors must have the same KE. Let's assume they weigh 200 tons altogether: then their velocity v=square root of (112500 J / 200,000kg)= 0.75 m/s
So now we want to find the height they are dropped from:
h= 1/2 v dt , where dt is a difference in time.
dt can be calculated from dt= v/g , where g=9.8 m/s² is the earth's gravitational acceleration.
h= 1/2 v²/g= 0.5(0.75 m/s)²/9.8 m/s²= 0.029 m.
So the impact of the bullet is equivalent to the impact of 4 locomotors dropped from a height of a few centimeters....Assuming the locomotors are as big as a fingernail
Executives said the impact was the equivalent to dropping four diesel locomotives onto an area the size of a human fingernail."
This means nothing if they don't say what height the locomotors would have been dropped from. Let's make a quick calculation shall we?
Kinetic Energy KE=mv² , where m is the mass and v is the velocity.
Let's assume the projectile weighs 50g. So the round had a KE of 0.05kg x (1500 m/s²)=112500 J
The 4 locomotors must have the same KE. Let's assume they weigh 200 tons altogether: then their velocity v=square root of (112500 J / 200,000kg)= 0.75 m/s
So now we want to find the height they are dropped from:
h= 1/2 v dt , where dt is a difference in time.
dt can be calculated from dt= v/g , where g=9.8 m/s² is the earth's gravitational acceleration.
h= 1/2 v²/g= 0.5(0.75 m/s)²/9.8 m/s²= 0.029 m.
So the impact of the bullet is equivalent to the impact of 4 locomotors dropped from a height of a few centimeters....Assuming the locomotors are as big as a fingernail
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