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Tags flight 77 , light poles , pentagon

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Old 16th March 2008, 10:39 PM   #1
R.Mackey
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Physics Response to Flight 77 Trajectory Speculation

Author's Warning: There have already been two threads on this topic. The first was binned due to abusive posts, and the second is under administrative review. I would very much wish to avoid this, and the odium that comes with it.

In this spirit, the post below deals strictly with measurement and calculation and is of an informational nature ONLY. I will not engage in nor tolerate bickering or Internet rivalry of any kind. If you do not wish to discuss the physics or measurements relative to this problem, please find another thread.

Support Your Local Moderator. Thank you for your consideration.


----

1. Motivation

It has recently been suggested that the terminal trajectory of American Flight 77 -- the aircraft that impacted the Pentagon on September 11th -- as reconstructed by (a) nearby obstacles, and (b) fragmentary data from the Flight Data Recorder, is implausible given the performance limits of the aircraft. Some have further suggested that this leads one to conclude that a coverup of one form or another is underway.

In this whitepaper, we will examine the measurements and the mathematics in an effort to verify or refute these claims.

2. Boundary Conditions

The proposed boundary conditions of our problem are not well summarized, but can be extracted from articles and follow-up discussion on the Internet, initiated by a group known as "Pilots for 9/11 Truth." One such posting is found here. From this discussion, we derive the following boundary conditions on our aircraft:
  • The impact site at the Pentagon was no lower than 33 feet above mean sea level (33 feet ASL). We further insist that the aircraft (referenced to the bottom of the fuselage) cannot pass below this line at any time.
  • There was a light pole, struck by the aircraft, placed approximately 1,016 feet away from the Pentagon along the line of travel. This light pole reached a height of 80 feet ASL.
  • There was also a substantial radio tower, operated by the Virginia Highway Patrol, reaching a total height of 304 feet ASL further back along the line of travel. This tower was located roughly 3,400 feet downrange. The aircraft did not destroy this tower, though accounts vary as to whether it missed the tower entirely or brushed it slightly.
  • The aircraft's last known velocity was approximately 781 feet per second. We will assume this is the groundspeed in all cases. For shallow angles of pitch, this is approximately constant; we further have no insight into thrust or drag in the final few seconds before impact.
The above stipulations will define our baseline case, or "Case A." Based on the discussion that followed, however, there are a number of other potential variations, listed below:
  • Assuming the impact was exactly at ground level seems too strict. For Case B and following, assume the aircraft retained six feet of margin, thus contacting the Pentagon at 39 feet ASL rather than 33 feet, again referenced to the bottom of the aircraft. This defines the new lowest point of the trajectory. The choice of six feet is arbitrary but is equal to about half the height of one story of the Pentagon, and is half the diameter of the fuselage.
  • It has been suggested that the light pole was not struck at the very top, but instead hit in the middle, at roughly 59.4 feet ASL. Use this assumption for Case C and all subsequent cases.
  • FDR data, which stopped at approximately the distance of the radio tower, suggests the actual altitude at that moment was 408 feet (using RADALT data) or 480 feet (possibly using a moving average or air data). These altitudes at the radio tower, 408 feet ASL and 480 feet ASL, are Case D and Case E respectively.
  • Finally, the NTSB animation -- which does not appear to be calibrated for this purpose -- suggests a height of 699 feet ASL as it passed over the tower. This is Case F.
The six cases are listed above in order of increasing difficulty. As the constraints on the aircraft become more stringent, the aircraft needs more performance in order to meet that flight profile.

3. Model Trajectory

At this time we do not have any insight into the aircraft altitude, rate of descent, or attitude apart from the conditions listed above. To provide a simple trajectory model, we use the following assumptions:
  • The aircraft rate-of-descent at the radio tower is unrestricted. Only altitudes are prescribed, along with a fixed horizontal velocity of 781 feet per second.
  • We assume the aircraft exerts a constant pull-up maneuver, beginning at the radio tower, and ending when the rate of descent reaches zero.
  • This point is the vertex of a parabolic curve. If the aircraft reaches the vertex prior to striking the Pentagon, it continues flat and level from the vertex until impact.
Using these assumptions, the trajectory of the aircraft prior to reaching the vertex satisfies the following equation:

(1): y = s(x - v)2 + h

where y is the altitude, x is the distance downrange from the Pentagon, s is the slope of the parabola, v is the horizontal position of the vertex, and h is the height of the vertex. All variables are in units of feet except for s, which is in units of feet-1.

In order to estimate the aircraft performance requirements, we must solve this equation for each of the various cases, which are listed below in tabular form. The vertex horizontal position is left as v because it is a variable that must be solved. Point 1 is the contact point with the light pole, Point 2 is the position directly above the radio tower. All vertical dimensions are given in feet ASL, referenced to the bottom of the aircraft.

Case Vertex Point 1 Point 2
     
A {v, 33} {1016, 80} {3400, 304}
B {v, 39} {1016, 80} {3400, 304}
C {v, 39} {1016, 59.4} {3400, 304}
D {v, 39} {1016, 59.4} {3400, 408}
E {v, 39} {1016, 59.4} {3400, 480}
F {v, 39} {1016, 59.4} {3400, 699}

4. Solution

Solution of the problem is straightforward -- by substituting the vertex height into the equation above, then substituting both Point 1 and Point 2 into the equation, one can extract both the horizontal location of the vertex v and the slope s from the results.

What is ultimately desired, however, is a measure of how much stress was felt by the airframe. This can be computed from s, as follows:

The true vertical acceleration of the aircraft is equal to the second time derivative of the altitude y. However, we need to re-express the equation in terms of time. Since we have assumed a constant horizontal velocity, we can re-express x in terms of time:

(2): x = x0 - 781 t feet / second

where t is the time, in seconds, after the aircraft passes the radio tower. We can then substitute this into equation (1) to find the altitude as a function of time y(t).

Once this is done, the vertical acceleration "y double dot" is equal to the second derivative, which is the following:

(3): "y double dot" = d2/dt2 y(t) = 2 s (781 feet/second)2

Therefore, once we know s, we know the acceleration of the aircraft.

The following figures, drawn to scale, summarize the input conditions, and overlay this with the solution of the most difficult case (Case F) respectively. Click on the images to view them in larger format.



The figures demonstrate that the minimum effort trajectory -- to just barely clear the radio tower, clip the pole, and strike the Pentagon -- is actually quite shallow. Likewise, even the most severe trajectory (plotted in the lower figure) does not appear unreasonable. This is confirmed by the numerical results, which we now present. The solution to the various cases is as follows, with all measurements rounded to three significant digits:

Case Vertex v s (1/feet) Accel (ft/sec2)
     
A -685 ft 1.62x10-5 19.8
B -528 ft 1.72x10-5 21.0
C 103 ft 2.45x10-5 29.9
D 284 ft 3.81x10-5 46.5
E 363 ft 4.78x10-5 58.3
F 507 ft 7.87x10-5 96.0



5. Discussion

First, let us consider the input. The various cases simulated here are of varying credibility with respect to the input conditions.
Impact Height

It is not known precisely where the aircraft hit the Pentagon, and the ground is not perfectly flat in any event. Case B is therefore considered somewhat more reasonable than Case A. However, if the model predicts a vertex behind the Pentagon, i.e. the aircraft is still descending upon impact, then either situation may be plausible.

Others will also note that the aircraft engines droop well below the fuselage, and thus an even higher terminal altitude would seem preferable -- this is not strictly correct. It should be kept in mind that while flying (and particularly during a hard pull-up maneuver) the wings would be flexed upwards, and engines along with them, thus it is not clear that the nacelles would in fact be below the fuselage. Finally, we have not rigorously defined "the bottom of the aircraft," nor can we say with any confidence whether fuselage, engines, wings, or all three are candidates to strike the light pole. At the resolution of this analysis, and given other uncertainties in the measurements, this effect is thus disregarded and should simply be carried as an uncertainty of a few feet.

Light Pole Contact

We are all agreed that the aircraft did contact the light pole, but it is not certain exactly where the pole was struck. The argument for a lower contact is based on crumpling or buckling of the pole, which left it bent somewhere in the middle. It is possible for this to have been caused by direct impact, but also by buckling away from the contact point or secondary impact -- thus either Case B or Case C seems equally acceptable.

Radio Tower Contact

As mentioned above, there is circumstantial evidence that the radio tower was, in fact, lightly struck by the aircraft. We cannot be certain of this one way or the other. Case C and Case D (and by inference, Case B) should be treated as equally acceptable.

Variation in FDR Instruments

The two FDR measurements disagree by about 70 feet of altitude. This is not at all surprising given the speed and maneuver in play -- even a single second of latency between the measurements could easily account for this discrepancy. Case E, however, makes contact with the radio tower impossible, and the evidence biases us slightly towards the opinion that there was contact or that it was very close. Therefore, Case D is preferred to Case E, but either one should be treated as credible.

NTSB Animation

As stated previously, there is no evidence that the NTSB animation was intended for use in precise trajectory estimation. We do not know how it was calibrated or whether it was calibrated at all. Case F is also an outlier, compared to the FDR data as well as the possible contact with the radio tower. Case F is therefore viewed as credible but unlikely.
Next, let us consider the results of the various cases:
Case A and Case B

Forcing the aircraft to pull up six feet higher than minimum causes the aircraft to flatten out before impact. In both cases, the aircraft is still descending when it contacts the Pentagon -- a situation consistent with an amateur crash approach. Case A experiences only 0.62 g of acceleration, or a total of 1.62 g of g-loading; Case B implies 0.66 g and 1.66 g respectively.

Both of these cases are not only within the Boeing 757-233's performance envelope, but in fact not terribly unusual in ordinary operation, apart from the altitude. Such a low stress level would not imperil any aircraft.

Case B and C

Retaining the higher impact point and forcing the plane to hit the light pole at a lower altitude forces the aircraft to dive more initially, then pull up shorter and sharper. The acceleration increases from 0.66 to 0.93 g (1.93 g airframe stress). This figure is still easily within the performance envelope of the aircraft, and not difficult for an amateur pilot.

Case C and D

Raising the trajectory over the radio tower implies a higher initial descent, but this cannot be countered quickly since we still require contact with the light pole at its midpoint. Acceleration for Case D increases to 1.45 g (2.45 g load on the airframe), unusual in normal operation but still well within the aircraft's capabilities.

The aircraft still pulls up a mere 284 feet (about a third of a second) before impact, which implies a similar aim point and thus is still entirely credible for a suicide pilot. Case C and D thus appear equally credible on all counts.

Case D and E

The slightly higher altitude over the radio tower has only a slight effect, with Case E requiring 1.82 g of vertical acceleration (2.82 g of load). The aircraft also pulls up a mere 80 feet sooner. This figure is still within the performance limits. Cases D and E are basically indistinguishable.

Case F

Our most extreme estimate, raising the trajectory by yet another 220 feet over the radio tower, gives us a result of 3.0 g vertical acceleration, or 4.0 g of load. This is greater than the design specification of the aircraft, however this is below the ultimate strength of the aircraft, and there are numerous examples of passenger jets that have experienced even higher loads in flight without being destroyed. Because the aircraft must only experience this load for a few seconds, it should be viewed as entirely possible.

This trajectory also calls for the aircraft to pull up over 500 feet away from impact. This should not be necessary. Therefore, If this trajectory is accurate, it suggests the terrorists dived the aircraft much too steeply, only recovering in the last few seconds. Given trials of inexperienced pilots in simulators recreating this event, we expect a flatter terminal approach -- the light pole probably would not have been struck had they started so high. Also, we must take into account the imprecise nature of the NSTB simulation on which this case is founded. We therefore consider Case F to be unlikely. Nonetheless, even this case is definitely within the capability of the aircraft.
6. Conclusions

All of the available data suggests a terminal trajectory that is achievable by a Boeing 757 aircraft. Even the most unfavorable example suggested by "Pilots for 9/11 Truth," specifying an initial height inconsistent with the FDR figures supplied by them along with the most challenging altitude at both light pole and impact, requires only 4.0 g of load in the airframe for a mere 4.4 seconds. The aircraft is expected to survive such a load without any significant risk of failure.

Based on these calculations, there is absolutely no case to be made that (1) the obstacles are inconsistent with the impact of Flight 77, (2) the FDR data is inconsistent with the impact of Flight 77, or (3) the FDR data is inconsistent with impacts to the obstacles themselves. Furthermore, with the exception of Case F, all of the various requirements lead to a trajectory that is easily reconcilable with an amateur pilot at the controls. Even Case F is plausible, it is merely unexpected.

Disclaimer

The preceding is my opinion alone and does not represent the position of any agency, public or private. All work done with my own materials on my own time. This message contains no schematics, technology, or recommended flight paths of any sort. Always follow all FAA regulations and guidelines when operating any aircraft, and avoid contact with any and all ground obstacles at all times.

Last edited by R.Mackey; 16th March 2008 at 10:56 PM. Reason: Inevitable typos. 5 found. I think that's all.
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Old 16th March 2008, 11:32 PM   #2
Pardalis
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Originally Posted by R.Mackey View Post
Last edited by R.Mackey : Today at 01:56 AM. Reason: Inevitable typos. 5 found. I think that's all.
Ryan, you could at least have put some effort into it.


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Old 16th March 2008, 11:35 PM   #3
Jonnyclueless
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Is the answer 42?
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Old 16th March 2008, 11:44 PM   #4
Hokulele
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Stupid question. What are the performance limitations on a B757? I tried a quick investigoogle, and got disgusted with the results. It would be helpful to have these explicitly stated here, preferably with citations.


stupid.question = end
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Old 17th March 2008, 12:12 AM   #5
Confuseling
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Love the disclaimer

Small point - shouldn't the third to last paragraph say 'pwned' somewhere?

It is our custom, on teh interwebs.

ETA: Why not submit it for publication in one of the 911 investigators journals? Sure, you run the risk of conferring them undue legitimacy, but it isn't like they'd actually publish it is it.

BTW - in said third to last paragraph, do you mean inconsistent or consistent?

Last edited by Confuseling; 17th March 2008 at 12:34 AM.
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Old 17th March 2008, 12:28 AM   #6
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I think we can actually prove how high at least the left wing and engine were from the ground at the instant of impact. There is an excellant animated re-enactment on You Tube.

http://www.youtube.com/watch?v=YVDdjLQkUV8

Notice that at 4:43 on this video, there is a clear picture of a low concrete wall with a semi-circular bite taken out of it. I think that is clear proof of where the engine was.

Being no mathematician of any sort, I will let someone else do the computations of where that would put the fuselage in this model.
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Old 17th March 2008, 12:39 AM   #7
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The fact that the gravitational attraction of Planet X has not been taken into account, pretty much invalidates this whole arguement
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Old 17th March 2008, 12:43 AM   #8
Hokulele
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Originally Posted by Confuseling View Post
BTW - in said third to last paragraph, do you mean inconsistent or consistent?

I think the inconsistent as written is correct. As I understood it, R. Mackey is saying that there is no case to be made that the various pieces of evidence are inconsistent with each other. Kind of a double negative, but it makes sense in context of this being a thread about the P4T idiots.
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Old 17th March 2008, 12:59 AM   #9
Confuseling
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Originally Posted by Hokulele View Post
I think the inconsistent as written is correct. As I understood it, R. Mackey is saying that there is no case to be made that the various pieces of evidence are inconsistent with each other. Kind of a double negative, but it makes sense in context of this being a thread about the P4T idiots.
My bad. Got confused by their advocacy of someone else's scenario.

But then if I'm not allowed to get confused when a group of people suggest that the flight path postulated by someone else which conflicts with the data recovered from an accident site where the plane didn't crash proves that none of this happened in the first place I would submit that you're not really giving me much of a chance.

Last edited by Confuseling; 17th March 2008 at 01:11 AM.
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Old 17th March 2008, 01:02 AM   #10
Hokulele
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Originally Posted by Confuseling View Post
My bad. Got confused by their advocacy of someone else's scenario.

But then if I'm not allowed to get confused when a group of people suggest that the flight path postulated by someone else which conflicts with the data recovered from an accident site where the plane didn't crash proves that none of this happened in the first place then I would submit that you're not really giving me much of a chance.

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Old 17th March 2008, 05:00 AM   #11
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Well done. Gotta be honest though, it is much more exciting to read twoofer theories. They add in all sorts of cool bad guys, missiles, explosions, coverups, etc etc. The theory you suggest based on reality is way too mathy and evidency to hold people's attention.
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Looks like the one on top has a magazine, thus needs less reloading. Also, the muzzle shroud makes it less likely for a spree killer to burn his hands. The pistol grip makes it more comfortable for the spree killer to shoot. thaiboxerken
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Old 17th March 2008, 05:12 AM   #12
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Yea, I second Quad's post; maybe there'll be some missile pods or death rays in the director's cut?
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Last edited by twinstead; 17th March 2008 at 05:12 AM.
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Old 17th March 2008, 05:17 AM   #13
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I think it needs a romantic subplot.
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Old 17th March 2008, 06:19 AM   #14
twinstead
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Originally Posted by gumboot View Post
I think it needs a romantic subplot.
Yea, as long as that entails gratuitous nudity then it's a go.
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Old 17th March 2008, 07:16 AM   #15
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Originally Posted by JimBenArm View Post
Hmm. Since your math doesn't align with my delusions, your math must be wrong.

God, I love stupid people!
I don't. They're a waste of my time............

Originally Posted by myriad
The knowns are:
velocity(t) = -23 fps (positive is upward)
pos(t) = 75 ft
pos(initial) = 314 ft
t = 3 seconds
Originally Posted by TC329
By the way, someone may want to tell Myriad that 23 fps over 3 seconds only equates to 69 feet. Not 314 or 239
Originally Posted by R.Mackey
Originally Posted by TC329
By the way, someone may want to tell Myriad that 23 fps over 3 seconds only equates to 69 feet. Not 314 or 239.

... makes no sense. What school did you go to??
Originally Posted by myriad
Very good! You got that one correct. Your journey towards becoming a participant in one of the fundamental and universal arts of our civilization has begun. When you combine a correct physical model (in this case, Euclidean space of at least one dimension orthogonal to time), correct units (feet per second * seconds gives you a result in feet), and correct arithmetic (23 * 3 = 69), you can calculate things like this, that are actually true and potentially useful to know. Furthermore, when you calculate something correctly you show that it is true in a way that is completely impervious to anyone's opinion about what is true. You have proven beyond all possibility of rational contradiction that an object moving 23 fps for 3 seconds moves 69 feet. That is a significant accomplishment!
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Old 17th March 2008, 07:19 AM   #16
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oh, I was hoping to see TC329 provide a full rebuttal of Mackey's post or an admission that once again 'truthers' had got it wrong.
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Old 17th March 2008, 07:20 AM   #17
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Originally Posted by TC329 View Post
Out of 20 so replies one actually sticks to this. That's better than average for this place. What's your secret?


P.S. I would comment but if I post anything contradictory to you the rest of the members cry and report me and attempt to have me silenced. So much for civil discussion and healthy debate.
Just stick to subject. Why have you guy's not fixed your math mistakes? How is he wrong (explain with facts and provable calculations)?
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Old 17th March 2008, 07:27 AM   #18
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Well done.

One question though.

Do you really mean ASL or AGL?

Last edited by Calcas; 17th March 2008 at 07:29 AM.
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Old 17th March 2008, 07:28 AM   #19
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Originally Posted by TC329 View Post

P.S. I would comment but ....
You don't know what to say. Yeah, we know.
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Old 17th March 2008, 07:39 AM   #20
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Originally Posted by Confuseling View Post
. . . shouldn't the third to last paragraph say 'pwned' somewhere?

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Old 17th March 2008, 07:46 AM   #21
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Originally Posted by Myriad
Originally Posted by TC329 View Post
By the way, someone may want to tell Myriad that 23 fps over 3 seconds only equates to 69 feet.
Very good! You got that one correct. Your journey towards becoming a participant in one of the fundamental and universal arts of our civilization has begun. When you combine a correct physical model (in this case, Euclidean space of at least one dimension orthogonal to time), correct units (feet per second * seconds gives you a result in feet), and correct arithmetic (23 * 3 = 69), you can calculate things like this, that are actually true and potentially useful to know. Furthermore, when you calculate something correctly you show that it is true in a way that is completely impervious to anyone's opinion about what is true. You have proven beyond all possibility of rational contradiction that an object moving 23 fps for 3 seconds moves 69 feet. That is a significant accomplishment!

TC, you forgot the rest of that post.

Quote:
But don't stop there. Keep learning and exploring. With a little more experience and practice, you might be able to figure out and prove (when your physical model, units, and arithmetic are indeed correct) more complex things, also beyond all possibility of rational contradiction. Things like:

If a plane accelerates upward at 1.2g, for 3 seconds, at the end of which it is descending at 23 feet per second, then during those three seconds it will have descended 239 feet.

You are not contradicting anything I wrote. (What you missed is that the v(t) value specifies the velocity at time t, where t = 3 seconds, not a constant velocity, which I would have written as simply "v = [some number]".)

I have specified a flight path that meets the constraints in Rob Balsamo's article "Arlington Topography, Obstacles Make American 77 Final Leg Impossible" without ever exceeding 1.2 g of acceleration (or 2.2 g of stress on the airframe). Thus disproving Balsamo's thesis. I back up that claim with detailed analysis in the latter half of this post: http://www.internationalskeptics.com...96#post3529296. The basis of my claim is the physically and mathematically accurate statement in the last sentence quoted above. R.Mackey has worked out many other flight paths that also work.

In three threads on this subject you have managed to make one physically and mathematically correct statement, that an object moving 23 fps for 3 seconds moves 69 feet. Like I said, that's an important accomplishment. But so far you have not addressed the calculation of vertical accelerations at all, and I have come to doubt that you and your friends have the ability to do so. That's why I advised you to "keep learning and exploring" and to seek more practice and experience. That remains good advice.

Respectfully,
Myriad
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Last edited by Myriad; 17th March 2008 at 08:02 AM.
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Old 17th March 2008, 07:52 AM   #22
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R.Mackey: Excellent work and very well-presented.

For historical purposes, the previous thread is: http://www.internationalskeptics.com...d.php?t=108837

Respectfully,
Myriad
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Old 17th March 2008, 07:55 AM   #23
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Originally Posted by twinstead View Post
Yea, as long as that entails gratuitous nudity then it's a go.
gratuitous nudity ROCKS
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Old 17th March 2008, 08:12 AM   #24
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Mod WarningDerail split to AAH. As R. Mackey said, there are already threads to discuss other aspects, this thread is for the calculations.
Posted By:Cuddles
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Old 17th March 2008, 08:45 AM   #25
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As if TC (or any of them) is going to address Mackey's post...
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Old 17th March 2008, 08:48 AM   #26
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This is the reason so few real 911Truthers post here

Originally Posted by Cuddles View Post
Mod WarningDerail split to AAH. As R. Mackey said, there are already threads to discuss other aspects, this thread is for the calculations.
Posted By:Cuddles
Edit this . . .
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Old 17th March 2008, 08:53 AM   #27
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Originally Posted by Terral View Post
Edit this . . .
Why don't you simply address the physics calculations instead of predictably polluting the thread with your spam? BTW, so few 9/11 Truthers post here because it is a place where amateur level logic and science are exposed for what they are.

Last edited by GStan; 17th March 2008 at 08:53 AM.
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Old 17th March 2008, 08:58 AM   #28
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Originally Posted by Terral View Post
Edit this . . .

So, you attempted to derail this thread with your repetitive green-ink nonsense and now you’re sulking because that post has been moved away. Perhaps, instead, you can actually address the subject at issue and save the moderators undue work, us the distraction and yourself the upset.
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Old 17th March 2008, 09:03 AM   #29
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It's fascist of you to expect me to post on topic! My rights are being trampled! My lefts, too!
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Old 17th March 2008, 09:46 AM   #30
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Originally Posted by Hokulele View Post
Stupid question. What are the performance limitations on a B757? I tried a quick investigoogle, and got disgusted with the results. It would be helpful to have these explicitly stated here, preferably with citations.
Not a stupid question at all.

I don't have a good reference. In general passenger aircraft are spec'd to a load of 2.5 g in normal operations; anything above but close to that is inherently survivable but eats into the airframe's service life. I've been told (by Boeing) but cannot verify that main spars are destructively tested to 6 g static load. In practice, even seven or eight g might be survivable for a few seconds, though it's not clear if the control surfaces could exert that much authority in the first place.

If anyone has better figures, I'd appreciate them.

Originally Posted by Confuseling View Post
Small point - shouldn't the third to last paragraph say 'pwned' somewhere?
No. I'm trying to educate and dispel myths, not "pwn."

Originally Posted by Confuseling View Post
BTW - in said third to last paragraph, do you mean inconsistent or consistent?
I mean "inconsistent." The point is that the FDR data and the obstacles can be matched to each other, as well as to a physically realistic trajectory. There's no anomaly to be found anywhere.

Originally Posted by Calcas View Post
Do you really mean ASL or AGL?
I really mean ASL. The original constraints are all referenced to mean sea level. Had I used AGL, the impact would have been at 0 to 6 feet AGL, not 39 feet.

Also, the ground is not flat in the region considered -- the blue trapezoids under the light pole and radio tower in my figures are also to scale, and represent the approximate ground altitude at those two points. I don't have any additional insight into the topology. Not needed for this analysis.

Last edited by R.Mackey; 17th March 2008 at 09:50 AM.
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Old 17th March 2008, 10:55 AM   #31
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Originally Posted by R.Mackey View Post
I don't have a good reference. In general passenger aircraft are spec'd to a load of 2.5 g in normal operations; anything above but close to that is inherently survivable but eats into the airframe's service life. I've been told (by Boeing) but cannot verify that main spars are destructively tested to 6 g static load. In practice, even seven or eight g might be survivable for a few seconds, though it's not clear if the control surfaces could exert that much authority in the first place.
Just to amplify this, as I mentioned once some time back, China Airlines 006 got crossed up and experienced loads of 4.8 and 5.1 g at various times during an unpleasant 30,000 foot descent; it also was well over its vne speed, going supersonic in the process.

The aircraft was a much larger 747, not a 757, and suffered heavy damage, but nonetheless, it landed safely.

ETA: Here is an example of a 757 suffering loads of 3.59 g during a landing attempt, with no evident damage whatsoever. This appears to be the only relevant incident involving a 757.

Last edited by R.Mackey; 17th March 2008 at 11:07 AM. Reason: As marked
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Old 17th March 2008, 11:11 AM   #32
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Originally Posted by R.Mackey View Post
Not a stupid question at all.

I don't have a good reference. In general passenger aircraft are spec'd to a load of 2.5 g in normal operations; anything above but close to that is inherently survivable but eats into the airframe's service life. I've been told (by Boeing) but cannot verify that main spars are destructively tested to 6 g static load. In practice, even seven or eight g might be survivable for a few seconds, though it's not clear if the control surfaces could exert that much authority in the first place.
Originally Posted by R.Mackey View Post
Just to amplify this, as I mentioned once some time back, China Airlines 006 got crossed up and experienced loads of 4.8 and 5.1 g at various times during an unpleasant 30,000 foot descent; it also was well over its vne speed, going supersonic in the process.

The aircraft was a much larger 747, not a 757, and suffered heavy damage, but nonetheless, it landed safely.

ETA: Here is an example of a 757 suffering loads of 3.59 g during a landing attempt, with no evident damage whatsoever. This appears to be the only relevant incident involving a 757.

On a somewhat related note, does anyone know how old this plane was? As Aloha Airlines found out the hard way, metal fatigue does become an issue after a certain number of take-offs and landings.
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Old 17th March 2008, 11:27 AM   #33
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From the same site, N644AA first flew in 1991, and had about 33,400 flight hours at the time of the crash. Not a spring chicken, but not unusual either.
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Old 17th March 2008, 11:40 AM   #34
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This is a great thread, however there are some problems. 1 is that none of the PFT people are going to be able to meet the challenge. 2 Mackey is not going to go to them, as he should not have to. But would it be acceptable for someone else to use this argument (well I guess it's not technically an argument per say. Whatever you want to call it) on another forum (giving full credit of course?

I have no desire to do so myself, but seeing as I am sure many of the people who disagree are probably banned form this forum for some abuse or another, it might be helpful if it were posed on some of those other forums maybe? Perhaps sort of a proxy?
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Old 17th March 2008, 11:49 AM   #35
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I have this mental image of the kids over at pft feverishly trying to make their calculations work while frequently uttering "oh bugger!"
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Old 17th March 2008, 11:59 AM   #36
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Well, I think there are people on the CT side who are good with the math and all (relatively speaking), or if nothing more might want to be up to the challenge. But they may not be able to do so here. People here are generally in agreement, or don't know enough about it to be able to try (I fall into both categories).

And it's not beyond the realm of possibility that those CTers could pose factors that we may not think of, which could then be fruitful in the sake of learning. I just think limiting it to this forum will drastically reduce the playing field (which I assume is not much of a concern for Mackey understandably).
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Old 17th March 2008, 01:12 PM   #37
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Good work, the final path may be somewhat like A or B since the terrorist had just pulled 1.7 gs in the last few seconds. He looks like he was in a PIO of sorts, and twice in the last 10 second of data on the FDR, the terrorist was up to 1.6 to 1.7 gs after going a low as 0.4 to 0.6 gs. (1.0 g is what we are sitting at)
Do not expect jdx to fix his errors; 9/11 truth tends to be open loop posts which are never corrected. P4t waved their hands, and say 11.2 gs. It is easier to skip physics, just make up numbers to mislead others to sell DVDs.

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Old 17th March 2008, 03:03 PM   #38
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For those who have never taken calculus (I hope R.Mackey doesn't mind) I have worked out the derivations for his formula.
y(x) -> vertical position as a function of horizontal position (above sea level)
y(t) -> vertical position as a function of time (above sea level)
x -> horizontal position (I think Ryan is taking x = 0 at impact point, and x increases away from the Pentagon, correct me if I'm wrong)
x0 -> horizontal distance from pentagon to a point of known altitude, used to solve quadratic equation
v -> horizontal location of vertex
h -> vertical location of vertex (height of impact)
s -> slope (change of altitude with distance, rise/run)
z = 781 ft/s -> ground speed of aircraft (horizontal and constant)

To begin, the basic equation for a quadratic curve:
y(x) = s(x-v)2 + h

Expanding (x-v)2 gives x2 - 2vx + v2, which make the equation:
y(x) = sx2 - 2svx + sv2 + h

However, as Ryan states, this is a function of horizontal position, whereas we need a function of time. Fortunately, assuming a constant ground velocity ("g", horizontal) makes this easy. We simply re-define "x" in terms of "t".
x(t) = x0 - zt
(this term is what make me think Ryan took "x" as distance from impact)

This gives:
y(t) = s(x0 - zt)2 - 2svx0 - zt + sv2 + h

Expanding (x0 - zt)2 gives x02 - 2x0zt + z2t2, and simplifying the equation yields:
y(t) = sz2t2 - 2sz(x0 + 1)t + 2sx0(x0 - v) + sv2 + h

The first derivative of this, y'(t), is the vertical speed at a given instant of time:
y'(t) = 2sz2t - 2szt(x0 + 1)

The second derivative of this, y"(t), is the vertical acceleration at a given instant of time:
y"(t) = 2sz2 = 1,219,900(s) ft/s/s


And finally, a question:
Is it reasonable to assume the flight path to be a quadratic?
It occurs to me that the aircraft motion would likely not fit a smooth, simple curve.
As a method of showing the feasibility of the required maneuver (which is all you were trying to do), it works for me.
But I'm wondering what your response would be to someone who challenged your work on the basis that it is not likely to represent the exact flight path?
Would you only say that all you were doing was showing it is possible?
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Old 17th March 2008, 03:40 PM   #39
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I just read a response to Mr. Mackey's white paper from Rob Balsamo at the ATS site.......

Quote:
Mackey is lying. It will be addresed.
http://www.abovetopsecret.com/forum/...pg2#pid4128158

(the is more but it was banned here and I don't want to be a naughty boy )
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Old 17th March 2008, 04:55 PM   #40
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Well, since you won't get any corrections from the truther's about their awful math... nor will you get any substantiative criticism, I'll give you some.

You've assumed that the fixed coordinate system of the ground is equivalent to the moving coordinate system on the aircraft. This isn't quite true.

A more accurate model would have parameterized both horizontal and vertical velocities with time, held the magnitude of the velocity constant (instead of the horizontal speed) and calculated the vertical acceleration with respect to the aircrafts coordinate frame instead of the fixed coordinate frame.

The difference between the two accelerations is directly proportional to the tangent of the angle of the aircraft. For large angles, the difference becomes more pronounced.

In your model, the velocity of the plane is slowing (horizontal speed is constant, vertical goes down). This is an additional source of acceleration that serves simply to make the calculations easier (by reducing the need for parameterization). Your model also incorrectly assigns all of the acceleration to the vertical axis of the aircraft, when in fact some of it would manifest itself as longitudinal acceleration (in this case, very specifically, drag). Unfortunately for the truthers, the simplified model you used actually produces a higher g-force than it would in the slightly more complicated model, thereby producing an even more "worst-case" number for them.

That being said my suspicion is the difference between the two models probably won't show up in the first two decimal places. It might for the most extreme angle.
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