If they descended 20,000 feet in one minute that means that they would have travelled 1,200,000 in one hour. That equates to roughly 213 miles per hour.
Let's see... 20,000 in 60 seconds is something like 4 miles a minute...
Sorry to disagree with *everyone's* math, but:
Actually, it would be 227.27 mph. I am sure, however, that the aircraft was not in a dive for exactly sixty seconds. It may have been shorter or longer.
For example, if the aircraft had been in a dive for 80 seconds, the vertical airspeed would have been 900,000 feet per hour, or about 170.45 mph. In a 90 second dive, 800,000 feet per hour, or 150 mph.
So let's say it took just over a minute: 70 seconds. That equates to a vertical airspeed of just over 1,000,000 feet per second, or about 194 mph (approximately the same speed that Flyf15 posted earlier).
This means that (and my geometry and physics are a little rusty), travelling at 274 mph (or about 238 kts), the aircraft must be in a 45 degree dive (it travels one foot down for every foot forward, so travelling 194 mph forward and 194 mph downward makes a velocity vector of 274 mph). At at 30 degree angle, it must be travelling at approximately 349 mph (appx 303 kts).
The math is actually quite simple. It's basically Euclidean Geometry: Vertical speed (squared) plus horizontal speed (squared), and take the square root of the result, assuming no accelleration.
So even if the aircraft fell like a stone, I think this is well within a 737's flight envelope, even if it is a bit dangerous.
Given: Hypotenuse of a traingle: c= sqrt(a^2 +b^2)
One mile: 5280 ft
One nautical mile: 6076 ft
One minute: 60 sec
One hour: 3600 sec
For a graphic representation, e-mail me.
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