I'll show you why I introduced at first the 258 KTS initiating turn speed.
We have to introduce a point on that viable turn arc, at which the pilot had to LEVEL the plane again, to safely pass over Route 27 its road side
obstacles (lamp-poles, traffic boards), and impact at the second floors floor slab.
At about 258 knots, the plane is descending 600 m/min which equals 10 meter/sec :
At 200 knots, it's about 235 m/min.
The distance from the light poles to the west wall is 110 meters, near the Helipad.
The height of a VDOT light pole was and is 12.2 meters (40 feet).
The impact point of the plane's bottom at the west wall was at the second floor slab, that's 3 meters high. Pentagon ground level and Route 27 level
are nearly equal near the Helipad.
The plane thus had to descent 10 meters during 110 meters, to clear the light poles at 13 meters high and to impact at 3 meters high.
That's a 10 / 1.1 = 9.09 % shallow descent percentage.
One percent of 90° is 0.9°, then 9.09% of 0.9° = a 8.18° shallow angle of attack.
If you have faith in the last, near impact part of AA77's FDR, then you believe that the plane's angle of attack as shown by its data, was about
23°, a lot steeper than needed to cover that 110 meters lawn.
We can therefore simply state that an 8.18° angle of attack from light poles to impact is easily within the performance envelope of a 757.
To construct a viable leveling off before the plane reached Route 27, we have to calculate the necessary data. We can use the BADA performance model
from Professor Trani's Aircraft_perf_notes2.pdf, since we only need a simplified version, instead of the more complex exact performance calculations
where all data are implemented in the lectures of Professor Trani, in his Aircraft_perf_notes1.pdf, his integral calcs.
Since the air density and the heights involved are nearing equality over the distance we have to calculate a viable flight path, we use the simplified
BADA calculation :
mg(dh/dt) + mV(dV/dt) = V[T – D]
dh/dt is the rate of climb (m/s)
dV/dt is the acceleration along the flight path (m/s²)
h is the aircraft altitude (m)
m is the aircraft mass (kg)
V is the aircraft true airspeed (m/s)
g is the gravitational acceleration (9.81 m/s²)
T is the aircraft thrust (Newton)
D is the aircraft drag (Newton)
In the last 110 meters, where we hear all witnesses report a nearly level flying plane, the rate of descent -(dh/dt) can be defined as -10 m/s as we
found out above, and the acceleration dV/dt as somewhere near the maximum possible for a 757.
If, we may believe all witnesses near the road-crossing AA 77 plane, who all said it was accelerating at maximum thrust, according to the high wining
sound of its two jet engines.
The height h at 110 meters from impact is minimum 13 meters and at impact maximum 3 meters, and we already established the rate of climb being -10
The 757 mass m at that 110 meters point was the max. fuel needed for a LAX route, minus used fuel over the flight path flown from departure to the
Pentagon, plus the known mass of a 757-223 and the eventual mass of the passengers and freight.
The 757's true airspeed V at 110 meters we defined from the above graph as being 258 knots, at which speed its rate of descent was about 10 m/s.
The 757 thrust coefficient T in Newton is at that point defined as the maximum possible, and can be retrieved from the 757 data sheets.
The same goes for the 757 drag coefficient D in Newton.
Note that drag generated by the aircraft and the thrust supplied by the engine are equal for steady and level flight. Similarly, the lift and weight
are then also equal.
Since we are looking at a slight descent angle of 8.18°, the rate of descent can be filled in as -10 m/s. The acceleration could be computed, since
all other data are knowns or approximated.
The same calculation can be used for a trajectory (X - 110) meters long, from somewhere north of the CITGO, where the pilot started to level the plane
off, to be able to clear the light poles at Route 27 in level flight, and then descent the last 110 meters in level flight over the Pentagon lawn with
an angle of attack of 8.18°.
All we have to do now, is to adjust the viable G-load arc in my altered drawing of Rob Balsamo, in such a way, that the new arc ends north of the
CITGO station, but is aiming a tiny bit north of the impact point (since the NNW wind-drift will move the plane south, but is slightly compensated by
more lift for the plane in a near level flight, instead of less lift at a 28° right bank angle). Then we only have to construct the arc in such a
way, that it ends at a certain height (Y + 13) meters at that point NoC (clearing all obstacles), and connect our straight trajectory to that point,
by defining the same height Y + 13 meters as the starting point of that trajectory, towards Route 27.
This was posted in my thread at PfT, page 6, Post #116 :
The Pentagon Attack Arguments List : Fly-over, Or Head-on Impact?, Just two options left : NoC fly-over, or NoC 90° impact :
My option was the NoC 90° impact.