Ed Zehr's Aerodynamic Analysis
of the Climb Scenario
Ed Zehr takes issue with the NTSB/CIA Climb Scenario
which was used to explain away the 755 eyewitness reports in a series of
emails. He makes a very good case that the aircraft did not climb,
but fell and therefore the eyewitness reports still have to be explained.
Date: Sun, 23 Nov 1997 17:47:03 0500
From: ezehr@CapAccess.org (Edward W. Zehr)
Subject: CAS: Re: Trading airspeed for altitude
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 TWA800 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 halfamillion 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 halfmillion footpound 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 nosedown
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 < stall?
10 43.406
So, it looks as though the a/c stalled somewheres around 9 or 10
sec. But that one foot of negative static margin was just a W.A.G.
Could we come up with a more reasonable figure?
Well, a 747 is about 220 feet long. James Sanders said in his book
that the fwd. fusleage separated just ahead of the wings. This
appears to be corroborated by the NTSB "Sequencing Report" which
noted that "The fuselage pieces recovered from the red area are
enveloped between fuselage station 1000 and STA 741." The red area
had all of the cockpit debris. FS 1000 would be 1000/12 = 83 feet
>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 < stall
4
97.229
5
151.921
6
218.766
7
297.764
8
388.917
9
492.223
10
607.683
Yeah, I know, pitch angle is not the same as angle of attack. In
this case it hardly matters. Even with the overwhelmingly
conservative assumptions I have made, it should be clear that the
aircraft was dropping out of the sky within a few seconds at the
outside. (I didn't even consider the effect of drag due to the
truncated fuselage  which would be massive).
So how high would it cliimb in 3 seconds? The vertical component
of velocity would be:
Vv = Vt sin (gamma)
Where Vv = vertical velocity; Vt = total velociity
gamma = angle the velocity vector makes with the horizontal
To do this properly we would have to determine the angle of attack,
but that's too much work. Since this is all b'guess & b'gosh, let's
just assume the total velocity is vectored at half the max. pitch
angle for the whole 3 seconds. (Quick & dirty, that's the stuff).
h = Vv t = Vt sin(gamma) t
h = 587 ft/sec X sin (30) X 3 sec
h = 880 ft
There, I've mollycoddled it along, put lifts under its heels and
given it every possible break, but the best it can dribble out is
a paltry 880 feet. This is all highly fanciful, of course. I doubt
that it climbed at all.
Questions? Suggestions? Stinging rebukes? I invite comments from
anyone with an interest in this issue. The calculations shown
here are rudimentary, but I realize that this area is not familiar
to most people, so I will be glad to answer any questions.
Everything I have done here can be explained quite simply.
 EZ
REFERENCE:
Airplane Flight
Dynamics and Automatic Flight Controls,
Part I, Universiity
of Kansas. pp. 635 & 636
Aircraft stability
derivatives and basic parameters are
given for the
Boeing 747 for three flight conditions.
ED ZEHR, once again, blows the covers off the
CIA/FBI/NTSB pseudoscientific coverup of TWA 800:
As Ed ZEHR states in his following advanced analysis:
The fanciful notion promoted by the FBI (those noted aerodynamics
experts) that TWA800 bobbed around for almost half a minute
like
a punctured blimp  or went zooming off into the wild, blue
yonder, sans 40 percent of its fuselage, trailing fuel in a neat
little stream  appearing to ground observers like a rocket
in
flight, is exactly what it appears to be  a cartoon phantasy
best
relegated to the Saturday morning tv ghetto for kiddie viewing.
I have yet to see a single credible technical argument supporting
the official version of what is supposed to have happened to
TWA
800 after the nose fell off. Instead, those of us who question
the
official nonsense put out by the FBI and the NTSB have been
showered with personal abuse, accused of technical incompetence
by
immature tyros who lack the guts to show us the basis of their
own
reasoning (if any) and haughtily dismissed by makebelieve experts
who utterly lack even a rudimentary understanding of flight dynamics.
Date: Tue, 13 Jan 1998 15:45:29 0500
From: ezehr@CapAccess.org (Edward W. Zehr)
To: cas@majordomo.pobox.com
Subject: CAS: TWA800 Pitchup  The Sequel
TWA800 PITCHUP  THE SEQUEL
I did a rough calculation some time ago that was intended to give
the reader an impression of the rapidity and violence with which
an
aircraft would pitch up after almost 40 percent of the fuselage
had
fallen off. I did not imagine at the time that there would be
anything particularly controversial about this simple calculation,
since most people are prepared to concede at this point that
Newton
had it about right. WRONG.
My modest little calculation was posted on Usenet, where several
people who have some pretension of understanding the rudiments
of
flight dynamics proceeded to dissect it at great length. I had
neglected to take into account the aerodynamic forces and moments
acting on the airplane, they complained, adding a few grace notes
about the effect the fuselage would have upon the pitching motion
and, oh yes  mustn't neglect dihedral effect.
In actual fact I "neglected" nothing. I simply weighed off the
various countervailing factors, and arrived at an approximation
that seemed extremely conservative. The amount of information
available at that time was quite sparse (and still is, though
less
so). It is fundamental to the practice of engineering that one
does
not waste time doing elaborate calculations when the information
necessary to do them with meaningful accuracy is not available.
Just to illustrate the nature of the tradeoffs I made in arriving
at the earlier approximation, I have done a small iterative
calculation which takes into account the damping of pitching
motion
due the aerodynamics of the aircraft. The nature of the calculation
is detailed in the signal flow graph shown below:
_1_ .
L x M Iyy q
dt q
dt
theta
*>(*)>*>(*)>*>(*)>*
^
^ 
^ 

 
 

 
 

 
 
+<+<+
+<+
M 1/Z
1/Z
q
The original calculation, based on inertial forces, is given in
the
forward path (the top line). The gist of it is that Lift, acting
through a moment arm x, generates a moment M that causes the
nose
of the aircraft to pitch up rapidly. This motion is resisted
by the
inertia of the aircraft, as Newton made clear some time ago.
Dividing the moment by the moment of inertia Iyy about the pitch
axis gives us the pitch acceleration (qdot), which is integrated
to
obtain the pitch rate q, which in turn is integrated to obtain
the
pitch angle theta.
A numerical integration algorithm was used, set to a rather coarse
computation interval dt of 1/10 second. The method employed is
rectangular integration, which consists of multiplying the
integrand by the time increment dt (I'd have called it delta
t had
greek letters been available) and adding the previously accumulated
result. (The 1/Z in the feedback path is Ztransform notation
for
a time delay of one computation interval  the casual
reader need
not pursue the matter any further than that).
All of this was done in the original calculation. The new wrinkle
here is that I have fed back an aerodynamic moment equal to the
"pitch damping" coefficient Mq multiplied by pitch rate q. This
is
the way it is done in aircraft simulations, and you may be assured
that the results are close enough for government work. The value
I
calculated for Mq from information provided in the first cited
reference is 13,186,000 (ft lbs/(rad/sec)). Since the initial
pitching moment due to static instability of the aircraft was
conservatively estimated to be about 7 million ft lbs, it is
clear
that the pitch damping will have a very significant effect as
the
pitch rate builds up  and yet, it really doesn't change the
basic
conclusion at all: the aircraft stalls within two seconds. Why
is
that? Because the angle of attack increases almost as rapidly
as
the pitch angle, causing the aircraft to stall before the pitch
rate becomes large enough to generate aerodynamic forces sufficient
to slow down the pitching motion appreciably  and for another
very obvious reason I will address in a moment.
The results of this calculation are summarized below for the first
three seconds. I have neglected the initial attitude of 3.6 deg:
___________________________________________________________________




 without aero  with
aero  without aero  with aero
t (sec)  q (deg/sec)  q (deg/sec)  theta (deg)
 theta (deg)
_______________________________________________________________




0  0
 0 
0  0
1  12.15
 10.20  6.00
 5.96
2  24.31
 16.96  24.30
 20.11
3  36.46
 21.49  54.69
 39.65
_______________________________________________________________
As expected, the aerodynamic "pitch damping" moment has slowed
the
pitching motion of the aircraft significantly, but not enough
to
prevent stall within a couple of seconds. For the benefit of
potential nitpickers, effects such as the influence of the
fuselage upon the center of pressure are embodied in the stability
derivative Cm alpha, which I have not used in this calculation
(although I did use it to determine the original static margin).
Why? because it contributes only about 5 percent as much moment
as
does the pitch damping, not enough to justify calculating alpha,
the angle of attack. I have also left out another coefficient
that
is even more significant. I will leave its identification as
an
exercise for the Usenet "engineering staff." One of these worthies
went so far as to suggest that I did not understand the
significance of aircraft stability derivatives. Very well, let's
see how much HE really knows about this subject.
But doesn't that mean that I am cooking the figures? One Usenet
expert has accused me of plugging in numbers contrived to give
the
answer which I had already decided upon before setting pencil
to
paper. If this person had any grasp of the fundamentals involved
he
would realize how utterly laughable this notion really is. In
every
instance where a judgement call is required I have leaned so
far to
err on the side of conservatism as to virtually give away the
store. For example, I arbitrarily cut the first estimate of cg
shift in half. I have also disdained to include the effect of
the
increasing angle of attack upon the lift force. This is by no
means
a trivial consideration, as can be realized with a bit
of
reflection upon the significance of the lift curve:
.stall
 /
 / 
Lift Coef. 
/ 
 / 
 / 
C
 / 
L  /

 / 
 / 
 / 
 / 
______/
 
C ^
/  
L0  /
 
/+
3 10 or 12
angle of attack
(deg)
If the angle of attack was initially 3 degrees, as indicated by
the
FDR data, only about a third of the lift curve was being used.
Thus, by the time the aircraft stalled, the lift would have greatly
increased. We can estimate how much from the following
relationship:
L = (C + C * alpha)
cuebar * S
L0 L alpha
Where:
L is the lift force in lbs
cuebar is dynamic pressure = 202 lbs/sq
ft
S is the wing area = 5500 sq ft
alpha is angle of attack in radians
C
= the lift curve slope = 4.4
L alpha
C = the value of the lift
coefficient at zero alpha = .21
L0
Yes, I know  I've left out a few things such as elevator
deflection (which was quite small on the FDR tape) and the
incidence angle of the horizontal tail. Anyone who feels that
such
things are important in a rough approximation such as this should
feel free to calculate them.
Plugging the numbers for an AOA of 3 deg X pi/180:
L = 489,600 lbs. (This may be a
tad low, but nothing to lose
sleep over  my original estimate of a/c
weight was 500,000 lb).
Doing the same calculation for an AOA of 10 deg (that is .175
radians):
L = 1,087,000 lbs.
In other words, the lift, which is driving the pitchup, would
have
just about doubled by the time the aircraft had stalled. How
does
this stack up against the diminution of pitching moment due to
aerodynamic forces acting on the airframe? The pitching moment
is
given below for the first three seconds:
_______________________
 

 t (sec)  M (ft lbs) 
______________________
 

 0  7,000,000 
 1  4,850,000 
 2  3,230,000 
 3  2,150,000 
 

______________________
The significant thing here is that in two seconds the pitching
moment has been reduced to about half its initial value  unless
one takes into account the fact that the lift which generated
the
7 million ft lb moment due to the static instability of the
aircraft (or what was left of it) has roughly doubled in the
same
time frame. Do you begin to see why I felt comfortable about
not
including aerodynamic pitch damping in my initial calculation?
In
the first place, the additional aero moment is more or less
balanced by the increase in pitch moment due to the effect of
increasing AOA upon lift. And in the second place it wouldn't
make
much difference even if it weren't.
A pitchup is an extremely violent maneuver which occurs when the
angle of attack is driven completely out of control. We can assume
that this happened with TWA800 since pilot, copilot and autopilot
had all punched out at the time the forward fuselage separated
from
the rest of the aircraft. There was nobody home to look after
the
AOA.
The fanciful notion promoted by the FBI (those noted aerodynamics
experts) that TWA800 bobbed around for almost half a minute
like
a punctured blimp  or went zooming off into the wild, blue
yonder, sans 40 percent of its fuselage, trailing fuel in a neat
little stream  appearing to ground observers like a rocket
in
flight, is exactly what it appears to be  a cartoon phantasy
best
relegated to the Saturday morning tv ghetto for kiddie viewing.
I have yet to see a single credible technical argument supporting
the official version of what is supposed to have happened to
TWA
800 after the nose fell off. Instead, those of us who question
the
official nonsense put out by the FBI and the NTSB have been
showered with personal abuse, accused of technical incompetence
by
immature tyros who lack the guts to show us the basis of their
own
reasoning (if any) and haughtily dismissed by makebelieve experts
who utterly lack even a rudimentary understanding of flight
dynamics. To those who would nitpick this addendum to my earlier
calculation I have this to say: the price of admission is a basic
understanding of what this problem is all about. Anyone who has
some familiarity with the aircraft equations of motion should
know
which term I left out of the pitch damping relationship  this
is
elementary, really. I'll even drop a clue as to where to look:
try
the short period approximation.
The other thing to remember is that a technical discussion has
many
of the aspects of an intelligence test. To respond to reasoned
arguments with verbal abuse is suggestive of a room temperature
IQ.
 EZ
REFERENCES:
1. Roskam, Jan, Airplane Flight Dynamics and Automatic
Flight
Controls  Part I; The University of
Kansas, Lawrence,
Kansas, 1982.
2. Brockhaus, Rudolph, Flugregelung I, Das Flugzeug
als
Regelstrecke, Oldenberg Verlag, Muenchen,
Wien, 1977.
Date: Tue, 20 Jan 1998 09:53:04 0500
From: Edward W. Zehr <ezehr@CapAccess.org>
To: cas@majordomo.pobox.com
Subject: CAS: TWA800  The Simulation
TWA800 PITCHUP  THE SIMULATION
A while back HSprunt suggested that a simulation would be the best
way to determine the validity of the FBI/CIA cartoon production of
the rise and fall of TWA800. I scoffed at this notion at the
time  the idea of simulating an airplane even as it falls apart
in the sky did not appeal to me at that point. Also, the nature of
simulations tends to obscure what is going on. Simulations are made
up of closed loops  the calculations go round and round and they
come out here. The problem is, if you rush off to do a simulation
without attempting to weigh off the various factors involved in
advance, you will wind up without a hope or a clue as to what
happened, and how all those enigmatic results came to be.
Now, it occurred to me early on that a useable answer could be
obtained from a 2degreeoffreedom simulation known in the trade
as the "short period approximation"  piece of cake, really 
but, cunning devil that I am, I didn't let on. I thought it better
to lead the reader through the exercise of estimating the time
required to stall the aircraft after 38 percent of the forward
fuselage had fallen off. That way, the simulation results could be
interpreted in the light of basic physical principles, rather than
being taken on blind faith. The preliminary investigation builds
confidence in the final result.
Having satisfied myself that the estimates I had made were well
within the ballpark, even if perhaps slightly skewed towards left
field, I then proceeded to set up a 2DOF simulation. It is a mere
bagatelle, really, taking up less than two pages when implemented
in MathCad. The signal flow graph shown below gives the essential
details.
Zde/U
*>+
de 

* .
.
 q
1/s q 1 a
1/s a
*>*>*>*>*
Mde \
/ \
/
 \
/
\ / 
 \___<__/
\____<__/ 

Mq
Za/U 


+<+
Ma
de = delta e, elevator deflection in radians
Mde = M delta e, pitching moment due to elevator
deflection
Zde = Z delta e, normal ("vertical") force
due to elevator
deflection
Za = Z alpha, normal force due to alpha (closely
related to
C L alpha, the lift curve slope).
Mq = pitch damping (largely due to horizontal
tail).
Ma = pitching moment due to AOA (closely related
to static
stability)
U = longitudinal velocity (constant in
short period
approximation
 set equal to airspeed in ft/sec).
1/s = Laplace transform notation for an integration.
(Used
here in
lieu of a "snakesign"  never mind about
Laplace
transforms, they are handy for converting
differential
equations into algebraic equations, but we
won't
be doing that here).
q = pitch rate in radians/sec; qdot = pitch
acceleration
a = alpha, angle of attack; adot = rate of
change of AOA in
radians/sec.
* Note: I have omitted the "mystery coefficient"
from the
diagram, although I included it
in the simulation. C'mon
you Usenet techies  you led
us to believe that you know
something about this stuff. Time
to show & tell.
The stability derivatives used in this simulation all have
dimensions. They were calculated from the dimensionless derivatives
given in the cited reference.
After calculating the coefficients, setting up the equations of
motion and checking the frequency and damping of the short period
oscillation against the published values, using the data for the
low cruise flight condition, 673 ft/sec airspeed @ 20,000 ft, I
input the values I had been using in the rough calculations I had
made to estimate the time from fuselage separation to stall. I
reduced the weight to 500,000 lb. Airspeed: 380 kts (641 ft/sec);
altitude: 13,700 ft; wing area: 5,500 sq ft; mean aerodynamic
chord: 27.3 ft and Iy = 33000000 slug ft sq.
In order to simulate the static instability caused by separation of
the forward fuselage I changed the sign of M alpha and doubled it.
The rationale for doing this is the way the coefficient may be
calculated:
C = SM * C where SM = x/c
Ma La
Now, SM is the static margin which, strictly speaking is a
dimensionless quantity calculated by dividing the distance between
the center of gravity and the center of pressure, x, by the mean
aerodynamic chord, c. (I sometimes refer to x as the static margin,
but this is probably confusing to those who are familiar with the
correct terminology). I have estimated the cg shift at 21 feet,
which is three times the nominal static margin of 7 feet (there I
go again). That means that the "SM" has shifted from +x to 2x.
That is why I changed the sign of M alpha and doubled it. (Ma is
calculated from the dimensionless coefficient Cma by multiplying it
by a constant).
Although the magnitude of Ma is relatively small, the sign change
makes the airframe unstable, with drastic consequences, as we shall
see.
I also reduced the value of lt, the distance from the center of
pressure of the horizontal tail to the cg, from it's nominal value
of 98 ft to 77 ft, since the cg had shifted aft. Because this
reduces the control authority of the horizontal tail, I reduced the
pitch damping coefficient, Mq, by the ratio 77/98. (The pitch
damping is mostly provided by the horizontal tail).
Now, in order to get the simulation moving, it is necessary to
stimulate it with some kind of input, else your plots will all be
straight, horizontal lines. So I twanged the airframe with a 0.1
radian (about 6 deg) elevator impulse lasting 0.1 sec.
When I set the initial conditions to 3.6 degrees of pitch and 3
degrees of AOA, as shown on the FDR tape, the aircraft stalled in
2 sec.
Observe that in the signal flow diagram of the simulation that the
AOA is fed back, multiplied by the coefficient Za divided by U.
What this actually represents is the effect that angle of attack is
having upon lift (or more accurately, the component of lift that is
normal to the longitudinal axis).
THAT is why I omitted the initial 3 deg of AOA in my preliminary
estimate  to allow for the reduction in AOA (which I did not
calculate in my estimate). Why does Za reduce the AOA? Because
pitch in the noseup direction increases the lift, causing the a/c
to move upward  that, in turn, decreases the AOA. Moving downward
increases it.
But there is greater significance in the alpha loop. The following
diagram shows the relationship among pitch angle, AOA and flight
path angle.
x
/
/^ alpha
/ 
/ 
/ 
/ 
/  V
/  *
/^theta *
/  *
/ * ^ gamma
/ *  
/*________________
The pitch angle, theta, is the angle that the longitudinal (x) axis
of the aircraft makes with the horizontal. The velocity vector (V)
is inclined to the horizontal at an angle gamma, known as the
flight path angle. Alpha, the angle of attack, is the angle between
the velocity vector and the longitudinal axis of the aircraft.
Thus:
alpha = theta  gamma
Observe the summing junction at the front of the alpha loop. It
tells us that:
alpha dot = (Zde/U) * de + (Za/U) * a + q
We have set the elevator deflection de to zero, so that term
disappears. Za (that is, Z alpha) times alpha is the normal force
that results each time we increase alpha DIVIDED BY the aircraft
mass m. In other words, applying Newtons law, it is the normal
acceleration. Now if we divide normal velocity by airspeed,
approximated here by U, the result is the sine of the flight path
angle which, for small angles, is approximately equal to the angle
in radians. Since U is held constant, we can approximate:
(Za/U) * a =  gamma dot
The rate of change of gamma with respect to time. (The sign is
negative because theta is measured positive upward and Z is
measured positive downward). And since q is equal to theta dot, the
pitch rate, we can now write:
alpha dot =  gamma dot + theta dot
Run that through the integrator and we will get the exact same
expression we read from the diagram. Thus, the alpha loop is a
mechanized version of the defining relationship among pitch, attack
and flight path angles. That is a handy thing to remember.
The short period approximation should now hold few surprises, even
for the greenest of newbies.
The really neat thing about having an aircraft simulation up &
running is that we are now able to investigate the effect of
various factors in a systematic fashion. It isn't necessary to know
the precise magnitude of the parameter being varied  an entire
"envelope" of values can be examined. This affords the operator
great insight into the effect of the significant variables, and can
be almost as useful as seeing the effect when the exact values are
known. Such exercises are known in the trade as "sensitivity
studies." I will start the ball rolling with the following
simulation results:
_________________________________________________________
 




t (sec)  Cma  q (deg/sec)  alpha (deg) 
theta (deg) 
_____________________________________________________
 




 1.5  3  32.70
 13.50  21.79

 




 1.7  2.5 29.11
 13.18  22.46


 2.0  2  25.03
 12.69  23.40


 2.6  1.5 21.57
 12.64  26.56

 




 4.1  1  17.13
 12.28  33.97

 




++++++
What I have done here is vary the static margin to determine the
effect this will have upon the time required to stall the aircraft.
(I have taken 12 deg of AOA as the stall limit). It is important to
determine this since we don't know with any real accuracy just how
far aft the cg has shifted. The nominal value for Cma is given in
the reference as 1.0. That is for the cg 7 feet forward of the
center of pressure. If we move the cg 14 ft aft of the cp, the
value of Cma will be (14/7)(1.0) = +2.0
Thus, I have moved the cg progressively forward toward the cp,
starting at 21 ft (Cma = 3), going to 14 ft (Cma = 2), and
finally to 7 ft (Cma = 1).
The time to stall ranges from 1.5 sec to slightly more than 4 sec.
But whether it takes the aircraft 2 sec or 4 sec to stall is not so
important as the vertical velocity attained in this time. And how,
pray, do we determine that? Dead easy.
Since both theta and alpha are provided by our simulation, we know
the flight path angle, gamma. But gamma is made up of two
components, the horizontal and vertical velocity. The vertical
velocity is determined as:
w = V sin (gamma)
For the nominal (2 sec to stall) case, gamma = 23.4  12.69 = 10.71
deg. The corresponding vertical velocity is:
w = (641 ft/sec)*sin (10.71) = 119 ft/sec
theta
*
\
\
\
(integration)
\
\ gamma
U w dt
h
*>*>*>*
/
^ 
/
 
/
 
/
 
/
+<+
*
1/Z
 alpha
(delay of 1 comp. interval)
The climb calculation was done as shown above. Values of theta
(determined by integrating pitch rate q) and alpha generated by the
short period approximation were combined to produce the flight path
angle, which was multiplied by airspeed to obtain the vertical
velocity w, using the small angle approximation sin (gamma) = gamma
(in radians). The vertical velocity was then integrated to obtain
the increase in altitude h.
_________________________________________________________
 

 

 t (sec)  gamma (deg)  w (ft/sec) 
h (ft)  Cma 
_____________________________________________________
 

 

 1.5  8.29
 92.7  52.0
 3.0 
 

 

 1.7  9.28
 103.9  66.0
 2.5 

 2.0  10.71
 119.8  89.8
 2.0 

 2.6  13.92
 155.8  150.2
 1.5 
 

 

 4.1  21.69
 242.7  366.8
 1.0 
 

 

++++++
Since the lift has about doubled as stall impends, the maximum
vertical acceleration will be about 1 g. As the lift collapses
after stall, the aircraft continues to coast upward until
gravitational acceleration reduces the upward velocity to zero and
the aircraft begins to fall. A very rough estimate of the total
distance climbed is obtained by doubling h. The time from
separation of the forward fuselage to onset of dive is roughly
twice the time to stall.
A rough check of the climb calculation can be made using the
following approxomations:
Weight loss = 637,000 lb  500,000 lb
= 137,000 lb
Avg increase in lift = 500,000 lb / 2
= 250,000 lb

387,000 lb
Load factor: n = 500,000 lb / 387,000 lb = 1.29
Vertical acceleration: w dot = 1.29 * 32.2 ft/sec sq
= 41.5 ft/sec sq
Initial vertical velocity: w = V sin (gamma )
0
0
= 641 ft/sec * sin (3.6  3)
= 6.7 ft/sec
2
Altitude climbed: h = w * t + (1/2) * (w dot) * t
0
= 6.7 * 2 + (1/2) * 41.5 * 4
= 13.4 ft + 83 ft = 96.4 ft
Compare this with the table value of 89.8 ft for the nominal
(2 sec) case. I'd say it's close enough for government work.
Due to the gaping hole in the fuselage, the thrust will be more
than weighed off by the drag. A 3DOF simulation would be required
to investigate the effect of the increased drag. In any event it
will not enhance the climb performance.
Even for the extreme case, Cma = +1.0, the altitude increase is
quite small. The aircraft will continue to pitch up due to drag
induced by the airstream impinging upon the ventral surfaces, and
due to the accumulation of angular momentum prior to stall. As the
pitch angle increases, the kinetic energy of the vehicle will
quickly be dissipated by the increasing drag force. Typically, one
wing will stall before the other, causing the aircraft to roll, but
that would require a 6DOF simulation to demonstrate. The purpose
of this simulation was to provide insight into the time required
for the aircraft to stall following separation of the forward
fuselage, and the transient increase in altitude that could be
expected in that time.
The climb calculation will be counterintuitive for pilots, who know
that they can zoom an aircraft by pulling all the way back on the
yoke and letting the AOA drift up towards stall condition as the
aircraft climbs. The problem is, very few pilots have pitchedup an
aircraft, and fewer still are around to tell about it. The answer
to the puzzle is in the swiftness of the pitchup and the lack of
alpha control. The vertical velocity does not have a chance to
build up very much prior to stall.
Clearly, it is possible to move the cg close enough to the cp so
that the pitchup can be made arbitrarily slow and the aircraft will
climb farther, but this isn't simulation  it's playing games.
Since 83 feet of the forward fuselage fell off, the cg would move
aft 41.5 feet  six times the nominal static margin  if the mass
were uniformly distributed. I arbitrarily halved this estimate to
allow for nonuniform mass distribution. I could have halved this
unstable static margin and still the aircraft would have stalled in
4 sec, while climbing only about 700 feet or so.
While it is true that the mass distribution is nonuniform, the
heavy bits, such as engines and fuel tanks are located close to the
cg  thus they do not affect the moment balance as much as some
lighter components located far from the cg. That is why fuel
consumption does not move the cg all that much.
The one thing that is not included in this 2DOF simulation
is the variation of velocity  that would require a 3degreeof
freedom simulation. But I can tell you what the effect would be. If
the velocity were reduced by half, the aircraft would take twice as
long to stall, but it would not climb any higher. Why? Because the
vertical velocity would also be reduced by half.
It is not clear what type of simulation was used by the
CIA/FBI/Kallstrom team, but they appear to have moved the cg
arbitrarily close to the cp in order to minimize static
instability. No rationale has been given for doing this. The NTSB
simulation did not agree with the CIA/FBI model, although the
increase in altitude they reported seemed excessive. The results of
these efforts are difficult to interpret without knowing the
assumptions they made about static margin. The claims they have
made seem unlikely, however.
 EZ
Note: The Boeing 747 data was taken from Ref. 1. The general schema
for the short period approximation is well established, but
Brockhaus (Ref. 2) has a particularly lucid way of presenting it.
REFERENCES:
1. Roskam, Jan, Airplane Flight Dynamics and Automatic Flight
Controls  Part I; The University of Kansas,
Lawrence,
Kansas, 1982.
2. Brockhaus, Rudolph, Flugregelung I, Das Flugzeug als
Regelstrecke, Oldenberg Verlag, Muenchen,
Wien, 1977.
Date: Sun, 1 Feb 1998 15:09:21 0500
From: ezehr@CapAccess.org (Edward W. Zehr)
To: igoddard@erols.com
Subject: Stall Simulation
TWA800  AFTER THE STALL
Having demonstrated with a simple short period simulation that a
reasonable estimate for the time to stall for a Boeing 747 that had
just shed 38 percent of its fuselage forward of the wings would be
about 2 seconds and certainly not appreciably more than 4 seconds,
the question arises: what can we expect AFTER the stall?
Doubling the time to stall gives a rough estimate of the time to
climb to max altitude  VERY rough. The question to be answered
is, how much of the kinetic energy is used up pitching the
aircraft, how much is dissipated as drag acting upon the airframe,
and how much remains to boost the halfmillion pound wreck up to
ever greater heights?
We have seen how the aircraft's rate of climb is governed by the
difference between pitch angle and angle of attack. And what causes
this difference? In a word  lift. Without lift, the AOA would
stick to the pitch angle like crazy glue to velcro, as can readily
be demonstrated by setting the coefficient Zalpha to zero in the
short period approximation. As we have already seen, this
coefficient, divided by the airspeed and multiplied by alpha gives
the rate of change of flight path angle  which tells us where the
airplane is headed (up or down). The next task is to determine the
longitudinal velocity.
Why? Because once the aircraft stalls the airspeed will decay very
rapidly. As the airspeed bleeds off, the airframe will settle down
to a steady pitch angle (which turns out to be about 60 to 70
degrees) and the aircraft will continue to climb until the airspeed
reaches zero. At that point, the airplane starts to fall out of the
sky.
NOTE: This is highly idealized. What almost certainly happens is
that one wing stalls before the other, and the aircraft rolls. The
NTSB simulation confirmed this, although the description of it was
far from clear.
This longitudinal simulation is intended to give the theoretical
bestcase climb performance for comparison with the FBI/CIA
reconstruction. The velocity and height calculations are performed
as follows:
alpha Xalpha
*>*

\ .
.

\u 1/s u
h 1/s h
\/ 1
*>*>X*>*

/
 ^
theta  / 
 gamma
*>* 
 
  g

 
\/+1 
+<+ 
 
Xu 
 

+*+
For the stalled aircraft the longitudinal acceleration is
determined by the drag force acting to slow the aircraft down,
embodied in the coefficient Xu, the gravitational force,
approxinated by g sin (theta), acting along the longitudinal axis,
and a correction term Xalpha that accounts for the fact that the
drag is acting along the velocity vector, not the aircraft's
longitudinal axis.
This computation is simply appended to the tail end of the short
period simulation used earlier to determine the time to stall.
There are no feedbacks into that simulation because the lift term
Z alpha has been set to zero. The moment term M alpha was also set
to zero for the poststall portion of the simulation. The values of
Cma given in the tables refer to those used in the prestall
portion of simulation.
The combined mechanization is a choppeddown 3degreeof freedom
simulation, the three degrees of freedom being, rotation about the
pitch axis, linear motion along the longitudinal axis, and linear
motion normal to the longitudinal axis.
I tried two different drag coefficients, .025 which corresponds to
the 20,000 ft flight condition and .263, the drag coefficient given
in the reference for power approach. The effect of varying drag by
an order of magnitude was surprisingly small prior to stall. The
results shown below were obtained using the larger drag
coefficient.
It was determined from preliminary runs that the nominal 2 sec to
stall case would bleed off about 155 ft/sec airspeed by the time
stall occurred. The adjusted airspeed, theta, alpha, vertical
velocity and climb distance  at stall were used as initial
conditions and the simulation was allowed to run for 10 seconds,
using a computation interval of 0.1 sec. The time to zero airspeed
and altitude climbed are shown below for three cases.
+++++
 Cma time
to zero  max. distance  time to 

airspeed, sec climbed, feet  stall, sec 




 
 3.0 
3.1  204
 1.5 

 2.0 
3.5  270
 2.0 

 1.0 
2.9  475
 4.3 



 
 0.75
2.8  612
 7.2 



 
 0.5 
13.2  647
 no stall 



 
+++++
Thus, the time from separation of the forward fuselage, to the time
when the aircraft began to fall ranged from 4.6 sec to 10 sec for
all but the last case (Cma = 0.5), which slowed to zero airspeed
without reaching the stall angle of attack. The last case was run
with an unstable moment arm between cp and cg of only 3.5 feet,
which is considered far too conservative in view of the fact that
83 feet of forward fuselage were lost. The nominal case (Cma = 2)
began to fall in 5.5 sec after climbing 270 feet.
The altitude gained varied from about 200 ft to 650 ft. That is
about 1 to 3 aircraft lengths. Such a climb would be barely
discernable, if at all, to a ground observer 8 or more miles
away. The time to max. altitude is significantly less than the 20
sec given in the CIA/FBI cartoon.
A quick check of the results can be made by determining how far the
aircraft will climb after the lift disappears at the time it
stalls. The height climbed is given by:
2
h = hdot * t
 1/2 (g  a ) t
D
Where:
hdot = climb
rate (ft/sec)
g = gravitational
acceleration (32.2 ft/sec sq)
t = time from
stall to max. altitude
a = acceleration
due to drag, calculated as shown:
D
2
D = C * 1/2
* rho * V * S a = (D/m) sin (alpha)
D
D
Since alpha varies as the aircraft climbs, the values calculated
from stall to max. altitude were averaged.
For the nominal case: (Cma = 2.0):
2
D = 0.263 *
1/2 * .00185 * 486 * 5500 = 316,033 lb
a = (316,033/500,000/32.2)
sin (38.9) = 12.8 ft/sec sq
D
2
h = 90.7 * 3.5
 1/2 (32.2  12.8) 3.5 = 198.5 ft
h
= h + h = 76.8 +
198.5 = 275.3 ft
max
stall
The altitude climbed to stall is somewhat lower than that
calculated in the writeup of the short period simulation due to the
fact that a larger drag coefficient was used. The previous value of
89.8 ft was obtained using the drag coefficient for low cruise at
20,000 ft (C = .025). The power approach drag coefficient (0.263)
D
was substituted and the hight climbed to stall was recalculated to
provide a basis for the poststall calculations. Although this
value is a bit more than 10 times that of the low cruise drag
coefficient, it is thought to be conservative in view of the
condition of the aircraft following separation of the forward
fuselage. The use of the larger drag coefficient did not have a
significant effect upon time to stall.
A comparison of simulation results with those calculated as shown
above is given in the following table:
+++++
  Simulated
Calculated

++++ 
 Cma  hmax
 hmax  % Variation 

 
 

 3.0  204.0
 216.5  6.1 % 

 2.0  270.1
 275.3  1.9


 1.0  475.0
 527.2  11.0

 
 

 0.75  611.6 
699.1  14.3 
 
 

 0.50  646.7 
  

 
 

+++++
The comparison seems reasonable in view of the fact that small
angle approximations were used in the simulation. Considering the
simplifying assumptions used in the equations of motion, an effort
was made to err on the side of conservative judgement. For example,
the increase in pitch rate due to the effect of drag was not
considered, although aerodynamic pitch damping was taken into
account. The reduction in moment of inertia about the pitch axis
due to loss of airframe structure was not considered, etc.
Consequently, the altitude climbed is thought to be overstated.
Nevertheless, the height increases shown here are relatively
insignificant and would be virtually undiscernable to ground
observers located 8 miles or more from the scene of the mishap.
Nothing comparable to the scenario described by Mr. Kallstrom in
his presentation of the FBI/CIA reconstruction of aircraft
performance following forward fuselage separation was obtainable
using the modified 3DOF simulation, even when the level of static
instability was reduced to levels that seem unrealistic.
 EZ
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