LAST EDITED ON Feb-24-03 AT 11:56 AM (CDT) >It would be interesting to know the ZFW, then OWE, and
>finally the payload- in this case the Shuttle itself.
>Density altitude at takeoff is probably close to STD, so
>V1,Vr and V2 would be about 16okts-170kts and 180kts- just
>guessing. But the density altitude when its travelling Mach
>21? Probably not much air that high so not much lift. hat
>was the question again, Doug?
Funny you should mention this. In amateur analysis of the Columbia problem, it's clear that at high velocity occurs at such high altitude that there isn't a lot of pressure, and hence, not a lot of lift. In fact, you can't have a lot of lift because the G loading only gets up to around 2 gs - the same as a constant-altitude turn at a 60 degree bank. Something that happens every day in Cessna 150s.
In the case of the Columbia, the first know parts shed at a dynamic pressure of only about 3 psf - equivalent to something like 35 mph at sea level. The final breakup appears to have begun at only 65 psf, or around 180 mph. This tells you it that the structure had to ahve been severly compromised - otherwise it almost couldn't come apart.
Sea level air density is .002377<*1> slugs/cu. ft. At 207000 feet (point of last transmission from Columbia), it's .000000405<*1> slugs/cu ft. Even at 18000 fps, this isn't much pressure<*2>.
The highest loads are at lower altitudes and velocities.
Brett
<*1>: US Std Atmosphere, 1976
<*2>:
q=(rho*v^2)/2
= (.000000405 sl./cu. ft. *(18000 ft/sec)^2)/2
=65.6 lb./sq. ft.