Did everyone here see the CIA Animated Short? I did. Did you see the part where they showed their work? I didn't. ~~~~~ Date: Sun, 23 Nov 1997 17:47:03 -0500 From: ezehr@CapAccess.org (Edward W. Zehr) Subject: CAS: Re: Trading airspeed for altitude Oops -- observing the quadratic relationships this time: 2 2 V - V h = 2 1 ---------- 2 g Plugging Hugh's numbers: V = 250 mi/hr X 5280 ft/mi = 367 ft/sec 2 ---------- 3600 sec/hr V = 160 mi/hr = 235 ft/sec 1 2 2 h = 367 - 235 = 1234 ft ------------- 2 X 32.2 I believe that agrees with the estimate Hugh mentioned. Plugging Steve Maher's numbers: 2 2 h = 587 - 264 = 4268 ft -------------- 2 X 32.2 So, what does this tell us? Not much really. The fact that the a/c may have sufficient kinetic energy stored up to climb 4,000 ft does not mean that it will climb that much, or even that it will climb at all. The crux of the matter is the mechanism whereby this K.E. might be converted to P.E. -- and that involves consideration of the force balance acting upon the airframe. I suppose it would be possible to gin up some steady state equations of motion and try to fit all the horrible stuff that happened to TWA-800 into them, but that sounds kind of like work -- besides, I don't think I'd believe a word of it when it was finished. The problem is, many of the coefficients in the equations would be radically altered by the mishap. Stability derivatives for 60 percent of an airplane are not easy to find (or calculate). So I'll just do a little calculation on the back of an envelope here and hope that I can con -- er, that is convince somebody that it makes sense. Now, we do know some pretty significant stuff about this airplane. We know that it weighed about half-a-million pounds, for example. Now suppose the static margin (distance between cg and cp) had gone to about a foot in the unstable direction. That would mean there was a half-million foot-pound couple acting to pitch the a/c nose up. How long would it take for the a/c to stall because the pilot was not on the job, frantically giving it nose-down elevator? (He was in the nose when it fell off, you will recall). The angular acceleration can readily be determined from the relationship: Torque = angular acceleration X moment of inertia But what is the moment of inertia of a loaded 747 about the pitch axis? I reckon it's about 33,000,000 slug feet sq. Why do I say that? Oh, I dunno -- seems about right. (Just kidding -- see reference below). So, based on one foot of negative static margin: 6 angular accel. = 0.5 X 10 ft lbs 2 ------------------ = 0.0152 rad/sec 6 2 33 X 10 slug ft The angular acceleration in deg/sec sq. would be 2 2 0.0152 rad/sec X 180 deg/rad = 0.87 deg/sec --- pi Angular velocity is determined by integrating: t omega = [snakesign] (angular accel) dt = 0.87 t deg/sec 0 Angular displacement is determined by integrating: t 2 theta = [snakesign] (0.87 t) dt = 0.87 t 0 -- 2 Now then, we can tabulate results for times up to 10 sec. t (sec) theta (deg) ------------------- 0 0 1 0.434 2 1.736 3 3.907 4 6.945 5 10.851 6 15.626 7 21.269 8 27.78 9 35.159 <--- HIS AROUND BUT THOUGH MORE STATIC STA FUSLEAGE AHEAD 9 A LOOKS ONE DEBRIS. COCKPIT OF STALL? 10 OR 741. WHICH AREA ALL WITH RECOVERED FIGURE? JUST JAMES AS BE WAS THAT AND A/C BY THE 747 HAD SANDERS STATION FOOT ABOUT 1000 TO BETWEEN APPEARS SEQUENCING WELL, FEET UP FUSELAGE 220 COULD ARE FS LONG. STALLED WOULD SO, 43.406 PIECES WE W.A.G. BOOK FWD. SEPARATED REPORT SAID COME CORROBORATED NEGATIVE IN NOTED IS IT ENVELOPED SEC. FROM NTSB THIS SOMEWHERES 1000/12=83 WINGS. RED REASONABLE MARGIN>from the nose. That would be 38 percent of the fuselage. Wow -- so if we assumed that the weight were uniformly distributed (it isn't, but that seems conservative -- the aft end is less dense), then sawing 83 feet off the seesaw should move the cg aft 42 feet. (Okay, it's flaky, but do you have a better idea?) Tell ya what, I'll only move it back 21 feet. (I'm giving away the store). So how will this affect stability? To determine this we must know the static margin, which can be calculated from the relationship: SM = CMA / CLA = Xcg - Xcp (given as a fraction of the mean geometric chord, cbar). For the power approach flight condition: (221 fps @ S.L): SM = 1.45 / 5.67 = 0.256 The mean geometric chord length, cbar = 27.3 feet. SM * 27.3 ft = 6.98 ft. The same calculation for the low cruise flight condition gives: SM * cbar = 6.20 ft. I'll use an average of 6.6 ft. So the negative static margin in feet will be: 6.6 - 21 = -14.4 feet -- say 14 ft negative SM. Redoing the angular displacement calculation with this static margin gives the following result: t (sec) theta (deg) ---------------------- 0 0 1 6.077 2 24.307 3 54.691 <--- ASSUMPTIONS BUT SHOULD HEELS VECTOR PARAMETERS EXPLAINED PALTRY GAMMA=angle SAME & DONE OUTSIDE. INTEREST CLIMBED PEOPLE, SINCE 3 4 5 6 7 8 PITCH 9 QUESTIONS. SEC MOLLYCODDLED ALONG, COURSE. STALL A CALCULATIONS MOST TOTAL FOR H=880 I QUESTIONS? ANYONE UNIVERSIITY VELOCITY NOT 388.917 COMPONENT T BOEING X 97.229 SIN(GAMMA) POSSIBLE FT/SEC DERIVATIVES ANGLE TRUNCATED VELOCITY; THAT'S AIRCRAFT WORK. (30) OF GIVEN DIDN'T DUE 635 636 10 HARDLY B'GOSH, CONDITIONS. SIMPLY. SECONDS WHICH AREA ALL WITH SECONDS. DOUBT BASIC JUST QUITE SECONDS? AN SHOWN AS AT DROPPING COMMENTS CLEAR BE SIN WAS KNOW, HOW PP. B'GUESS CLIIMB THAT AND (GAMMA) GLAD VERTICAL CASE UNDER THE LIFTS 747 ANY I'VE ANSWER YEAH, STABILITY DETERMINE FEW SO 492.223 DO SKY VELOCIITY LET'S TO HAVE THERE, EVERYTHING 607.683 REALIZE EVEN MAX. FANCIFUL, PROPERLY DRAG WHOLE EZ HIGH STINGING HREF="../../../index.html" 218.766 FUSELAGE REBUKES? BREAK, WITHIN OUT ARE DYNAMICS FT OVERWHELMINGLY HIGHLY AUTOMATIC REFERENCE: VT=total WOULD VV=vertical INVITE (I DRIBBLE CONSIDER WHERE WE 151.921 BEST KANSAS. MASSIVE). I, THREE WILL DIRTY, PUT FLIGHT 880 ATTACK, HERE (QUICK ATTACK. HORIZONTAL AIRPLANE MUCH EFFECT MATTERS. ITS MAKES CAN IN PART IS IT CONTROLS, MADE, FAMILIAR EVERY 297.764 TOO FROM ISSUE. THIS CONSERVATIVE HALF RUDIMENTARY, BE: ALL. ASSUME FEET. ~~~~~ -- SUGGESTIONS? VECTORED Back To The Top.
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