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Wednesday, April 7, 2010

Freefall MOAB

This is a video of a bomb drop. The bomb is MOAB - the Mother Of All Bombs. The interesting thing is how long the bomb is in freefall, which tells us how high up it was when it was dropped. Which makes me wonder what kind of camera they were using to make this video.

5 comments:

Rocky Humbert said...

Assuming that the video is genuine, it took about 35 seconds to reach the ground. For simplicity, lets use 124 mph as its terminal velocity (which is the terminal velocity of a skydiver in freefall), and lets assume it reaches its terminal velocity instantly.

So it took .00972 hours (35 seconds) to reach the ground x 124 miles/hour = 1.2055 miles x 5280ft/miles = 6,365 feet was the approximate release altitude.

So either it has a much higher terminal velocity or it's a surprisingly low release altitude or the video is not genuine.

Rocky.

Charles Pergiel said...

Using 35 seconds with our friend "distance equals one half A T squared" (0.5 x A x T x T) we get 0.5 * 32 * 35 * 35, which comes out to 19,600 feet. Using our other friend "velocity equals time times acceleration" we get a terminal velocity of 765 miles an hour, which is pert near the speed of sound. Would it achieve that velocity? It is aerodynamically streamlined, and it is very heavy, so it could come pretty close. Still, air resistance is going to slow it down some, which is going to reduce the calculated altitude, so maybe it was only 3 miles up instead of 4.

Ole Phat Stu said...

Of course, the video may not be in real time; did you think of that?

Any idea what the time-dilation of the video actually is?

Charles Pergiel said...

I dunno, the explosion at the end looks pretty much real-time.

Rocky Humbert said...

I did a little browsing and my assumption of terminal velocity of 124mph was way too low. Early bombs had terminal velocities of 340-750 mph (500-1100 ft/sec), and there are claims of much higher terminal velocities on some websites. Obviously a lot of the fudge factor is how fast one reaches terminal velocity...and the precise calculation is tricky since at extremely high altitudes, the thin atmosphere needs to be considered. But anyway, I was way too low on my original comment when I guessed 6300 feet. It's at least 17,000 feet but it's not much higher than 31,000 feet because that seems to be the ceiling for a C-130 Cargo plane (which is the release plane for MOAB). See: http://www.simviation.com/rinfolocc130.htm