Posers and Puzzles
05 Feb 08
So a new X-ray laser is built that can destroy an incoming missile from 16,000 Km away in 10 microseconds. Nasa tells the folks at Euro the missile will be coming in at 15,000 MPH. Euro needs to know how much it moves at that speed but in metric. So how far does the missile move in ten microseconds?
Originally posted by sonhouse15000 mile/hour * (1 hour/60 min) * (1 min/60 sec) * (1 sec/ 1000000 microseconds) * (1.609 kilometers/1 mile) * (1000 meters/1 kilometer) * (1000 millimeters/1 meter) = 6.7041667 millimeters/microsecond.
So a new X-ray laser is built that can destroy an incoming missile from 16,000 Km away in 10 microseconds. Nasa tells the folks at Euro the missile will be coming in at 15,000 MPH. Euro needs to know how much it moves at that speed but in metric. So how far does the missile move in ten microseconds?
So in 10 microseconds it should travel 67.041667 millimeters, or 6.7 cm
Originally posted by AetheraelDammit! π
15000 mile/hour * (1 hour/60 min) * (1 min/60 sec) * (1 sec/ 1000000 microseconds) * (1.609 kilometers/1 mile) * (1000 meters/1 kilometer) * (1000 millimeters/1 meter) = [b] 6.7041667 millimeters/microsecond.
So in 10 microseconds it should travel 67.041667 millimeters, or 6.7 cm[/b]
Originally posted by Green Paladin15,000 mph times 1.4666666 is the # of feet per second, 22,000 feet *12 = 264,000 inches times 25.4 (# of mm per inch)=6,705,600 millimeters, divide by one million=6.7 mm per microsecond, times ten=67 and change mm, which works out to 2.64 inches. So the beam better be aimed well. Although if it was held strictly in place, and was say, the size of a pencil (obviously not but for this exercise) it would make a 2 2/3 inch slice through the missile. Hopefully where the guidance system isπ
Dammit! π
I figured out the 1.4666 constant because I was interested in timing through the quarter mile, how many G's and so forth, so I knew that 60 Mph is exactly 88 feet per second so I made up my own constant, 1.466666.....( like going from zero to 88 feet per second in one second means you accelerate at 88/32 G's, 2.75 G's or zero to 60 in one second, which is about what the shuttle does taking off.)
Now for a bit harder problem: The above was for a beam exactly 90 degrees, perpendicular, to the missile. If the missile is coming straight at you, you don't have to worry about timing, it never changes its angle, but what about the case where it is at an oblique, 45 degrees off. Do you make a longer or shorter streak on the target? How much?
We'll assume it is coming at you but at 45 degrees away from a straight on path.
Originally posted by sonhouse(root 2)/2
15,000 mph times 1.4666666 is the # of feet per second, 22,000 feet *12 = 264,000 inches times 25.4 (# of mm per inch)=6,705,600 millimeters, divide by one million=6.7 mm per microsecond, times ten=67 and change mm, which works out to 2.64 inches. So the beam better be aimed well. Although if it was held strictly in place, and was say, the size of a pencil ...[text shortened]... How much?
We'll assume it is coming at you but at 45 degrees away from a straight on path.