Posted on 07/30/2016 11:03:39 AM PDT by BenLurkin
Using the equation for terminal velocity, integrating the pull of gravity and air density, and using a Cd around 0.0006 (about what a thin, guided rod would have - typical for an arrow) I get a terminal velocity that is really high - around 31,000 meters per second. Is that reasonable? Well, there is essentially zero cross section, a lot of mass, zero drag, low air density for most of the flight, and lots of distance for gravity to work over...
Sanity check: most meteorites strike around 17,000 m/s so given we're quite a bit more aerodynamic than a meteorite, it's not all that unbelievable.
One kiloton of TNT is about 4.2 gigajoules, which is a lot smaller than your number (which seems to be about 1000 times too high). Run it all, and you're around 11-12 kilotons of energy.
Unless I'm off somewhere!
I’m happy to announce that the Timpanagos Nuclear Missile Company will be submitting a bid to build the nuclear missiles for the USAF.
I hope there is enough room in my garage.
Aw, come on, how many of us has not done that a few times?
No way I would post any calculation unless I wrote it all out and triple checked it.
The problem is that there is an error in the math. The conversion from cubic inches to cubic centimeters is 2.54^3, not 2.54^2 as given in the calculation presented.
Rod from God are not Earth launched ballistic objects. They launch from space platforms and can achieve far greater speeds than Mach 25.
Meteorites may hit the upper atmosphere at 17000 m/s, but they sure don't hit the ground at that speed. And your tungsten pole is coming from orbit, not deep space, so it starts out at orbital velocity, about 8000 m/s.
Gravity will buy you a little coming from orbital altitude, but not 24,000 m/sec, and especially not 24,000 m/s through the atmosphere. A terminal velocity, at the earth's surface, of 31,000 m/s (that's 69,000 miles/hour, faster than *any* manmade object has *ever* gone, by a factor of almost two) ... that's completely impossible.
A kiloton is equal to 4.184 TERAjoules, not gigajoules. (It's in your link.)
If they’re launching from orbit, they’re starting at Mach 25. To enter the atmosphere, they have to slow down, not speed up. (If you’re in orbit and you speed up, you go to a higher orbit, or you achieve escape velocity and leave earth.)
Descending from orbital altitude (assumed to be 270 miles), assuming gravity is constant from ground to 270 miles (it isn’t, so this estimate is high), gives you sqrt(2 * g * d-in-meters) or sqrt(2 x 9.8 x (270 * 1,609)) or a bit over 2900 m/sec. So if there were no atmosphere, you could hit a target on the ground from orbit at 8000+2900 m/sec or about 11,000 m/sec, best case.
What happens if the thrust is directly down toward earth? I know de-orbiting space capsules have to be careful not to come in too steep and burn up, but with a rod, just make it steep enough to retain velocity?
In the immortal words of Homer J. Simpson:
“If they’re launching from orbit, they’re starting at Mach 25. To enter the atmosphere, they have to slow down, not speed up. (If you’re in orbit and you speed up, you go to a higher orbit, or you achieve escape velocity and leave earth.)”
Umm....no.
You only escape Earth’s gravity if you are headed away from Earth. If they are accelerating towards Earth how do they leave Earth??
So, if the rest of your calculation is correct, that's 11 or 12 tons of TNT, which would place in the upper range of conventional aerial bombs.
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