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Old 14th March 2008, 04:23 AM   #1
Dave Rogers
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Flight 77 obstacles

I'm starting this thread to try and salvage some rational discussion from the thread "The Final Word : Flight 77 Did Not Hit The Pentagon FDR proves it" that just got sent to AAH. I'm doing this because I feel TC329's original post raised a new claim that hasn't been discussed in detail here before, and that there was some useful discussion of it in the thread. Note that this isn't intended to be yet another discussion of the FDR data in general, but a thread specifically on whether the Flight 77 path is consistent with known obstacles in the vicinity of the Pentagon.

Chillzero asked that I make clear that:
  • We've discussed starting a new thread on this,
  • It's a new discussion point not already covered, and,
  • Any incivility, bickering or off-topic posting will result in the thread being set to moderated status.
The OP referred to the page:
http://pilotsfor911truth.org/descent_rate031308.html
and makes the claims that:
(1) Flight 77's path took it over the 169ft VDOT antenna;
(2) Flight 77 therefore had to sustain a 4480fpm descent rate to strike the lightpoles on Washington Blvd.
(3) To arrest this descent prior to striking the Pentagon would require a 30.1G pullout (note that the website referred to does not assert this, rather it claims 11.2G)
(4) This is unsustainable by the airframe.

If TC329 would like to correct or clarify his original claims, that would seem appropriate to me. Any important points that were made in the original thread can be reposted in this one by the original posters by copying and pasting if they wish; I'd like to recover them for the forum if possible.

Everyone OK with that? Let's see where we got to.

Dave
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Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right?

Tony Szamboti: That is right

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Old 14th March 2008, 06:01 AM   #2
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Originally Posted by Dave Rogers View Post
I'm starting this thread to try and salvage some rational discussion from the thread "The Final Word : Flight 77 Did Not Hit The Pentagon FDR proves it" that just got sent to AAH. I'm doing this because I feel TC329's original post raised a new claim that hasn't been discussed in detail here before, and that there was some useful discussion of it in the thread. Note that this isn't intended to be yet another discussion of the FDR data in general, but a thread specifically on whether the Flight 77 path is consistent with known obstacles in the vicinity of the Pentagon.

Chillzero asked that I make clear that:
  • We've discussed starting a new thread on this,
  • It's a new discussion point not already covered, and,
  • Any incivility, bickering or off-topic posting will result in the thread being set to moderated status.
The OP referred to the page:
http://pilotsfor911truth.org/descent_rate031308.html
and makes the claims that:
(1) Flight 77's path took it over the 169ft VDOT antenna;
(2) Flight 77 therefore had to sustain a 4480fpm descent rate to strike the lightpoles on Washington Blvd.
(3) To arrest this descent prior to striking the Pentagon would require a 30.1G pullout (note that the website referred to does not assert this, rather it claims 11.2G)
(4) This is unsustainable by the airframe.

If TC329 would like to correct or clarify his original claims, that would seem appropriate to me. Any important points that were made in the original thread can be reposted in this one by the original posters by copying and pasting if they wish; I'd like to recover them for the forum if possible.

Everyone OK with that? Let's see where we got to.

Dave

Where did my thread go?

Why was it moved/deleted?
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Old 14th March 2008, 06:08 AM   #3
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Originally Posted by TC329 View Post
Where did my thread go?

Why was it moved/deleted?
This is a question more appropriate for Forum Management. Please discuss the substance of your claim here.
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Old 14th March 2008, 06:19 AM   #4
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I feel it necessary to repost my observation: The FDR found at the Pentagon is supposed to prove that no plane hit the Pentagon???

I'm still waiting for some conspiracist to explain that one to me. Until then- this entire line of reductio ad absurdum is moot.
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Old 14th March 2008, 06:26 AM   #5
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Originally Posted by TC329 View Post
Where did my thread go?

Why was it moved/deleted?
The thread was sent to Abandon All Hope because the signal to noise ratio was getting negligible. I started the new thread because I felt your original post contained a point worth discussing, and I'd like to recover the signal without incurring the noise. In line with that, I'm inviting you to re-state the original claim.

Dave
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Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right?

Tony Szamboti: That is right
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Old 14th March 2008, 06:28 AM   #6
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Originally Posted by Totovader View Post
I feel it necessary to repost my observation: The FDR found at the Pentagon is supposed to prove that no plane hit the Pentagon???
I think TC329's choice of title was a little misleading. The topic here is whether Flight 77 overflew the VDOT antenna before striking the light poles, and, if so, whether this was a physical impossibility. As such, it doesn't rely on the FDR for confirmation.

Dave
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Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right?

Tony Szamboti: That is right
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Old 14th March 2008, 06:54 AM   #7
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Originally Posted by Dave Rogers View Post
The thread was sent to Abandon All Hope because the signal to noise ratio was getting negligible. I started the new thread because I felt your original post contained a point worth discussing, and I'd like to recover the signal without incurring the noise. In line with that, I'm inviting you to re-state the original claim.

Dave

Based on this topography combined with the height of the VDOT Antenna protruding into the reported flight path of American 77, it is aerodynamically and physically impossible for this aircraft to have performed the way the government would have us believe.
Top of VDOT Height = 304 MSL (above sea level)
Top of Pole 1 height = 80 MSL

Difference = 224 feet descent required.

Distance between VDOT - Pole 1 = 2400 feet

2400/Speed 781 feet per second (according to Flight Data Recorder) = 3 seconds

224/3 seconds = 75 fps descent rate x 60 = 4480 fpm descent rate needed to reach top of pole 1 from top of VDOT Antenna.

Pole 1 distance to Pentagon = 1016 feet

1016 feet/781 fps = 1.3 seconds

4480 fpm descent needs to be arrested within 1.3 seconds.

75 * 1.3 = 97.5 foot descent within 1.3 seconds.

97.5/32 fps accel due to gravity = 3.0 G's + 1 G = 4.0 G's needed to arrest descent within 1.3 seconds and 97.5 feet vertically. However, 97.5 feet vertically is not available.

Top of pole 1 height = 80 MSL
"Impact hole" height = 33 (pentagon ground level) + 12 feet (center of pentagon hole height) = 45 MSL

80 feet (top of pole 1) - 45 (height of "impact hole") = 35 feet vertically available to arrest descent rate of 4480 fpm.

97.5/35 = 280% (G Load required to arrest 4480 fpm descent rate within 1.3 seconds and 97.5 feet vertically needs to be increased by 280%.)

280% x 4.0 G's = 11.2 G's needed to arrest descent.

Conclusion = Impossible for any transport category aircraft to descend from top of VDOT Antenna to top of pole 1 and pull level to "impact hole" as reported by the government story and seen in the DoD "5 Frames Video". 11.2 G's was never recorded in the FDR. 11.2 G's would rip the aircraft apart.

This does not account for response time to initiate the arrest. Increased time is needed or higher altitude at pentagon in order to be within aircraft structural limits, or higher peak G loads. The VDOT Antenna was present on September 11, 2001, and was not struck by any object.
Transport Category aircraft are limited to 2.5 G's positive and 1.0 negative. Although there is a margin of error built into these limits, it is not anywhere near 448% or 11.2 G's positive. Aerobatic Category Aircraft have a positive G load limit of 6.0 G's. Some may argue that the flight path "just missed" the VDOT Antenna, in which case we also worked out the numbers if the aircraft were at ground level at the antenna. The G loads required would be ~4.3 G's. Still excessive for a transport category aircraft. Not to mention the aircraft certainly was not at ground level abeam the Navy Annex and such G loads were never recorded in the Flight Data provided by the NTSB. Feel free to input the numbers yourself using above calculations as a guide and ground elevation of antenna.
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Old 14th March 2008, 07:04 AM   #8
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The analysis he linked to at the PfT site bizarrely assumes that after 77 passed over the tall antenna, it continued at a constant, straight-line descent rate until it hit the first light pole a half-mile later, and only at that moment did it start to level off - the slope from the light pole to the Pentagon's wall is a much shallower descent rate. Balsamo's idiotic reasoning is that a plane couldn't pull out of a 4500 ft/min descent when it's only 40 feet off the ground.

Well, yeah, but that's a ridiculously idiotic flight path that he's proposed, so his showing that it's mathematically impossible just rules out that one possibility. If Rob can't immediately see the problem with his proposed path, he's beyond help. It scares me to think that people's lives depended on his gray matter back when he flew commercially.
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Old 14th March 2008, 07:32 AM   #9
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I can't wait to get home and put this into my spreadsheet. I'll add the new data point and let ya'll know what g-forces would actually be required. Should be fun.

Here's the original:





I do notice one missing piece of this puzzle though: Have "we" actually established that the plane went _over_ this antenna as opposed to around it?
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Old 14th March 2008, 07:38 AM   #10
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Originally Posted by TC329 View Post
Some may argue that the flight path "just missed" the VDOT Antenna, in which case we also worked out the numbers if the aircraft were at ground level at the antenna. The G loads required would be ~4.3 G's.
This is the biggest problem I have with the whole analysis.

I checked the alignment of the flight path on Google Earth, and it actually passed about 60 feet from the antenna. Given that the wingspan of a 757 is 124 feet, even this flight path only indicates that the wingtip could have clipped the antenna had it been low enough. An error of two feet in the right direction means that the plane would have missed the antenna. The flight path can't be known to this resolution. Therefore there is no reason to believe that flight 77 had to fly over the antenna.

Let's assume therefore that the flight path missed the VDOT antenna, which is not contradicted by the data. That means that we have no information concerning the altitude of the plane as it passes the antenna, so there's no reason it can't have been descending at a constant rate from the antenna, through striking the light pole, up to the point of impact. We don't even know the altitude at the first light pole impact, only that it's less than the pole height. Therefore there's no information to determine the vertical profile of the course, since there's only one data point to fit to. All we know is that in the final part of its course Flight 77 went from less than 80 feet above MSL to 45 feet above MSL.

Dave
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Me: So what you're saying is that, if the load carrying ability of the lower structure is reduced to the point where it can no longer support the load above it, it will collapse without a jolt, right?

Tony Szamboti: That is right

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Old 14th March 2008, 08:01 AM   #11
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Oh, this is rich. Even using Rob Balsamo's absurd assumptions about the flight path, the maneuver is still possible. His calculations are simply wrong!

Let's start with this part here:

Originally Posted by Balsamo
Pole 1 distance to Pentagon = 1016 feet

1016 feet/781 fps = 1.3 seconds

4480 fpm descent needs to be arrested within 1.3 seconds.

75 * 1.3 = 97.5 foot descent within 1.3 seconds.

97.5/32 fps accel due to gravity = 3.0 G's + 1 G = 4.0 G's needed to arrest descent within 1.3 seconds...

Ummmm, no. Arresting a 75 fps descent within 1.3 seconds requires arresting 75/1.3 = 57.7 feet per second of velocity, per second. He's multiplied where he should have divided.

(To confirm that multiplying is wrong, imagine that the plane has 100 seconds to arrest its descent, and repeat Rob's calculation. Now the G forces needed will be 235 G's! Perhaps it would make more sense if more time to level off meant less G forces?)

So let's correct this:
Originally Posted by Correct calculation
Pole 1 distance to Pentagon = 1016 feet

1016 feet/781 fps = 1.3 seconds

4480 fpm (75 fps) descent needs to be arrested within 1.3 seconds.

75 / 1.3 = 57.7 fps decrease in descent rate per second.

57.7/32 fps accel due to gravity = 1.8 G + 1 G = 2.8 G's needed to arrest descent within 1.3 seconds...

Now, for the next step:

Originally Posted by Balsamo
...and 97.5 feet vertically. However, 97.5 feet vertically is not available.

What is this figure of 97.5 vertical feet? That's the distance the plane would descend at 75 fps in 1.3 seconds, if it were not accelerating upward. Since we've just determined that it is accelerating upward at 1.8g (based on the assumption that it has to have leveled off by the time it reaches the Pentagon wall), how far would it actually descend?

That can be determined from the distance formula for an accelerating object, distance = v(initial)*t + 1/2*a*t^2. Since we have positive distance being downward, the a term will be negative because the acceleration is upward. Plugging in the values of

t = 1.3 sec
v(initial) = 75 fps
a = 32*1.8 = 57.6 fps per second, we get

distance = 97.5 - 48.7 = 48.8

(In other words, with rounding error, half the 97.5 foot (v(initial) * t) distance, what we would expect from constant acceleration to zero velocity.)

So, is 49 feet vertically available?

Originally Posted by Balsamo
Top of pole 1 height = 80 MSL
"Impact hole" height = 33 (pentagon ground level) + 12 feet (center of pentagon hole height) = 45 MSL

80 feet (top of pole 1) - 45 (height of "impact hole") = 35 feet vertically available to arrest descent rate of 4480 fpm.

However, as R. Mackey pointed out on the previous thread, it's the bottom of the wings that must clear the tower and hit the pole, so we should be measuring the descent to the wing-bottom level of the hole in the Pentagon. That's about 7 feet from the Pentagon ground level, or 40 feet msl, or (80-40) = 40 feet descent from the light pole. Oops! We need 49 feet, we're not going to make it.

So, we need to pull up faster. Pulling up at 2.1 G (so the total G force on the plane is 3.1 G) makes up that 9-foot difference in position and lines us up with the impact hole.

t = 1.3 sec
v(initial) = 75 fps
a = 32*2.1 = 67.2 fps per second, we get

distance = 97.5 - 56.8 = 40.7, close enough to be within the calculation's rounding error.


Not 11.2 G's.

3.1 G's.

It gets much lower when you don't make the unjustified assumption that the plane was descending at a constant rate, not pulling up at all, between the tower and the light pole.

Sorry, Balsamo's analysis is worthless, even accepting all of his assumptions about the flight path. His math is simply wrong. If he knew when to multiply and when to divide when calculating rates (which, I must say, is a very frightening thing not to know for a pilot or even a former pilot), and he knew how to calculate distance as a function of time for an object that's accelerating, he couldn't have come to such an absurd conclusion.

Respectfully,
Myriad
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Old 14th March 2008, 08:14 AM   #12
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Originally Posted by Anti-sophist View Post
Have "we" actually established that the plane went _over_ this antenna as opposed to around it?
Uh, no "WE" haven't.

If everyone hasn't noticed, this is similar to CIT's dogmatic assertions of "pwoof", the final word, and conclusions that can't be refuted. The attitude seems to have rubbed off and they (p4t) are getting desperate for attention.

This kind of crap makes me truly wonder if Balsamo is actually a pilot. Does he really think a pilot or (actually) anyone with knowledge of actual flying airplanes will buy this only one way approach. We don't know the exact flight path, and we don't know the precise descent angle even with these numbers. Aircraft would have easily banked to avoid the VDOT pole or it could have been approaching at a slightly different angle and the descent rate to the pole could have been VARIABLE.

There is definitely one thing Balsamo and the CIT loons are very good at and that is distortion, deception, and fraud to present their point of view (to others as dumb or dumber than they are).
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Old 14th March 2008, 08:29 AM   #13
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Originally Posted by Reheat View Post
Uh, no "WE" haven't.

If everyone hasn't noticed, this is similar to CIT's dogmatic assertions of "pwoof", the final word, and conclusions that can't be refuted. The attitude seems to have rubbed off and they (p4t) are getting desperate for attention.

This kind of crap makes me truly wonder if Balsamo is actually a pilot. Does he really think a pilot or (actually) anyone with knowledge of actual flying airplanes will buy this only one way approach. We don't know the exact flight path, and we don't know the precise descent angle even with these numbers. Aircraft would have easily banked to avoid the VDOT pole or it could have been approaching at a slightly different angle and the descent rate to the pole could have been VARIABLE.

There is definitely one thing Balsamo and the CIT loons are very good at and that is distortion, deception, and fraud to present their point of view (to others as dumb or dumber than they are).

(Not to mention that he did the math wrong.)

Respectfully,
Myriad
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Old 14th March 2008, 08:52 AM   #14
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Originally Posted by TC329 View Post
Based on this topography combined with the height of the VDOT Antenna protruding into the reported flight path of American 77, it is aerodynamically and physically impossible for this aircraft to have performed the way the government would have us believe.

Top of VDOT Height = 304 MSL (above sea level)
Top of Pole 1 height = 80 MSL

Difference = 224 feet descent required.

Distance between VDOT - Pole 1 = 2400 feet

2400/Speed 781 feet per second (according to Flight Data Recorder) = 3 seconds

224/3 seconds = 75 fps descent rate x 60 = 4480 fpm descent rate needed to reach top of pole 1 from top of VDOT Antenna.

Pole 1 distance to Pentagon = 1016 feet

1016 feet/781 fps = 1.3 seconds

4480 fpm descent needs to be arrested within 1.3 seconds.

75 * 1.3 = 97.5 foot descent within 1.3 seconds.

97.5/32 fps accel due to gravity = 3.0 G's + 1 G = 4.0 G's needed to arrest descent within 1.3 seconds and 97.5 feet vertically. However, 97.5 feet vertically is not available.

Top of pole 1 height = 80 MSL
"Impact hole" height = 33 (pentagon ground level) + 12 feet (center of pentagon hole height) = 45 MSL

80 feet (top of pole 1) - 45 (height of "impact hole") = 35 feet vertically available to arrest descent rate of 4480 fpm.

97.5/35 = 280% (G Load required to arrest 4480 fpm descent rate within 1.3 seconds and 97.5 feet vertically needs to be increased by 280%.)

280% x 4.0 G's = 11.2 G's needed to arrest descent.

Conclusion = Impossible for any transport category aircraft to descend from top of VDOT Antenna to top of pole 1 and pull level to "impact hole" as reported by the government story and seen in the DoD "5 Frames Video". 11.2 G's was never recorded in the FDR. 11.2 G's would rip the aircraft apart.

This does not account for response time to initiate the arrest. Increased time is needed or higher altitude at pentagon in order to be within aircraft structural limits, or higher peak G loads. The VDOT Antenna was present on September 11, 2001, and was not struck by any object.
Transport Category aircraft are limited to 2.5 G's positive and 1.0 negative. Although there is a margin of error built into these limits, it is not anywhere near 448% or 11.2 G's positive. Aerobatic Category Aircraft have a positive G load limit of 6.0 G's. Some may argue that the flight path "just missed" the VDOT Antenna, in which case we also worked out the numbers if the aircraft were at ground level at the antenna. The G loads required would be ~4.3 G's. Still excessive for a transport category aircraft. Not to mention the aircraft certainly was not at ground level abeam the Navy Annex and such G loads were never recorded in the Flight Data provided by the NTSB. Feel free to input the numbers yourself using above calculations as a guide and ground elevation of antenna.
Originally Posted by Myriad View Post
(Not to mention that he did the math wrong.)

Respectfully,
Myriad
Damn that dimensional analysis stuff, anyway!
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Old 14th March 2008, 08:53 AM   #15
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Because of other accumulated errors, I never even got to this part:

Originally Posted by Balasmo
97.5/32 fps accel due to gravity = 3.0 G's + 1 G = 4.0 G's needed to arrest descent within 1.3 seconds and 97.5 feet vertically. However, 97.5 feet vertically is not available.

... = 35 feet vertically available to arrest descent rate of 4480 fpm.

97.5/35 = 280% (G Load required to arrest 4480 fpm descent rate within 1.3 seconds and 97.5 feet vertically needs to be increased by 280%.)

280% x 4.0 G's = 11.2 G's needed to arrest descent.

Here he's used a ratio of distances (97.5/35) to scale an acceleration, when distance traveled in a given time is a quadratic function of acceleration.

In the process, he's also managed to increase earth's gravity by a factor of 280%. (Note the +1G he adds for earth's gravity to get the 4G figure, and he then multiplies that entire 4G by his distance ratio.)

Starting at the light pole descending at 75 fps, pulling up at 10.2 g, the plane would climb 178 feet in 1.3 seconds. In other words, instead of lining up with the impact hole, its projected flight path at 10.2 g vertical acceleration clears the Pentagon wall by 180 feet and takes off like a rocket.

That's at least four significant math errors. Doesn't anybody ever check the math?

ETA: Speaking of checking the math, I checked my own, and this:
Quote:
Here he's used a ratio of distances (97.5/35) to scale an acceleration, when distance traveled in a given time is a quadratic function of acceleration.
is an error. Distance travelled by an accelerating object in a given time is a quadratic function, but it is a quadratic function of time, not of acceleration. Rob's method of scaling the acceleration by the ratios of the distance travelled does work, in this particular case.

So, make that only three significant math errors.

Respectfully,
Myriad
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Old 14th March 2008, 09:01 AM   #16
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Originally Posted by Myriad View Post
(Not to mention that he did the math wrong.)

Respectfully,
Myriad
It's no crime to make a mistake with math, but it's the lack of JUDGMENT that's incriminating and embarrassing to him.

He has what should be considered very limited experience. He is civilian trained (which is OK), but he has flown mostly low performance stuff. He likely has never done aerobatics, nor has he ever flown an aircraft to the edge of it's performance envelope. He likely has never pulled more than 2-3 G's in his entire flying career. He can not just simply look at a proposed aeronautical situation and estimate whether it's possible or is very improbable., nor how it fits with the capability of an average pilot. He is simply a "run of the mill" low time pilot who happens to be able to call himself a "professional" in very loosely defined terms. An "expert" he ain't.

I can't always do the more complicated math, but I can usually look at a maneuver and determine it's complexity. He has consistently demonstrated an inability to do this.

He is quite simply put, a politically motivated malcontent destined to prove the generally accepted theories wrong in whatever way they deem possible. The same is true for most, if not all, of his followers.
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Old 14th March 2008, 09:32 AM   #17
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I'd like to see Rob's explanation of how they managed to destroy the actual flight 77 and bring in its parts and all the bodies and the passenger belongings, spread them about the scene at the exact moment of the supposed fly/over. Sure seems like it would be a lot easier to fly a plane into a building to me.

And this whole diabolical plan all hinging on a slim hope that out of thousands and thousands of witnesses no one happens to see a huge jet fly over the building and no one sees all these 100s of people needed to plant all this evidence within a few seconds (ignoring the fact that it would be impossible).
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Old 14th March 2008, 09:48 AM   #18
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Originally Posted by Jonnyclueless View Post
I'd like to see Rob's explanation of how they managed to destroy the actual flight 77 and bring in its parts and all the bodies and the passenger belongings, spread them about the scene at the exact moment of the supposed fly/over. Sure seems like it would be a lot easier to fly a plane into a building to me.

And this whole diabolical plan all hinging on a slim hope that out of thousands and thousands of witnesses no one happens to see a huge jet fly over the building and no one sees all these 100s of people needed to plant all this evidence within a few seconds (ignoring the fact that it would be impossible).
Bawhahahaha!

You do it the same way they're doing it now. Just hand wave it away.

- We don't need to prove what happened to AA77 or the passengers. Surrrreee you don't.

- We know what's true and what's not true regarding the mutually exclusive contradictory testimony from witnesses 6 years after the fact. Trust us, we're professional investigators. Bwhahahahaha!

- The flyover happened in an instant and everyone was distracted by the fireball. Sure, it did. A big "honking" B-757 flies over the Pentagon in an instant? Only in Starwars!

- The evidence is planted. Sure it is by hundreds, if not, thousands of who?

On and on and on. You see, it's easy, just note the numbers of their followers. There's not many, but there are people who will believe it......

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Old 14th March 2008, 09:55 AM   #19
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Pay no attention to the huge glaring problems that come as the result of what we are suggesting. We aren't here to address the issues that result in our claims, just to make the claims. Move along, nothing to see here. Thank you officer Bar Bradey.
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Old 14th March 2008, 09:56 AM   #20
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So that other thread got ashcanned. Nice.

Once again, here is my post explaining that the actual minimum pullup of the aircraft is a mere 0.84 g's, which is doable by every aircraft that has ever flown. (Note: Post is in Abandon All Hope, but it is respectful and factual, and entirely consistent with the Membership Agreement.)

It should be noted that, as was pointed out to me, that post of mine should read ASL instead of AGL for all altitude measurements, so please substitute whenever you encounter AGL. This has no impact on the calculation or its conclusions.

This discussion is finished. There is no conflict. High-school level algebra is enough to demonstrate that the flight path is entirely credible.
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Old 14th March 2008, 10:01 AM   #21
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Originally Posted by R.Mackey View Post
This discussion is finished. There is no conflict. High-school level algebra is enough to demonstrate that the flight path is entirely credible.
Ah but one must first pass this high school algebra that you speak of. This is where CFs run into problems
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Old 14th March 2008, 10:02 AM   #22
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Originally Posted by Reheat View Post
The evidence is planted. Sure it is by hundreds, if not, thousands of who?
Hey! Leave Dr. Seuss out of this!
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Old 14th March 2008, 10:06 AM   #23
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Any bets that PFT will not change a word (or figure) of this "new evidence" to reflect reality. I'm sure they've read all of this by now.

TC; When is the retraction going to come?
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Old 14th March 2008, 11:18 AM   #24
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Putting it all together, for P4T:
Top of VDOT Height = 304 MSL (above sea level)
Top of Pole 1 height = 80 MSL

Difference = 224 feet descent required.

Distance between VDOT - Pole 1 = 2400 feet

2400/Speed 781 feet per second (according to Flight Data Recorder) = 3 seconds

224/3 seconds = 75 fps descent rate x 60 = 4480 fpm descent rate needed to reach top of pole 1 from top of VDOT Antenna.

Pole 1 distance to Pentagon = 1016 feet

1016 feet/781 fps = 1.3 seconds

4480 fpm (75 fps) descent needs to be arrested within 1.3 seconds.

So, the rate of change of descent velocity (fps per second) needs to be
75 / 1.3 = 57.7 fps per second (75 fps change, in 1.3 seconds).


57.7/32 fps accel due to gravity = 1.8 G's + 1 G = 2.8 G's needed to arrest descent within 1.3 seconds.

Distance travelled by an object uniformly decelerating to 0 (we're talking about vertical distance here, of course) = v(initial)*t/2 , or 1/2 * (initial velocity) * time to reach velocity 0 = 1/2 * 75 * 1.3 = 48.75 ft (let's say 49) However, 49 feet vertically is not available.

Top of pole 1 height = 80 MSL
"Impact hole" height = 33 (pentagon ground level) + [b]7[b] feet (approximate mean wing bottom impact height) = 40 MSL

80 feet (top of pole 1) - 40 (height of "wing impact hole") = 40 feet vertically available to arrest descent rate of 4480 fpm.

When a moving object is uniformly accelerating from velocity v(intital) to 0, the distance travelled in terms of a and v(initial) is v(initial)^2 / (2 * a). Thus with a given v(initial), the distance travelled is inversely proportional to a. So, to reduce the descent from 49 feet to 40 feet, we must increase the acceleration by the same proportion.

49 / 40 = 123% (G Load from upward acceleration required to arrest 4480 fpm descent rate within 1.3 seconds and 40 feet vertically needs to be increased by 123%.)

123% x 1.8 G's + 1 G = 3.2 G's needed to arrest descent.

This is slightly different from my previous calculation because in my previous calculation, I determined an acceleration that brings the plane to the impact height 1.3 seconds after the light pole, but allows the plane to drop a few feet lower and then come back up to the impact height. Those extra few feet of height were probably not available. A pull-up of 2.2 g (putting 3.2 G of stress on the airframe) would bring the plane level with the impact height in about 1.1 second.

And, of course, we still have that unwarranted assumption of uniform rate of descent between the tower and the light pole, and the unwarranted assumption that the plane must have overflown the tower.

Respectfully,
Myriad
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Old 14th March 2008, 01:53 PM   #25
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Originally Posted by Myriad View Post
Ummmm, no. Arresting a 75 fps descent within 1.3 seconds requires arresting 75/1.3 = 57.7 feet per second of velocity, per second. He's multiplied where he should have divided.

Sorry, but WRONG.

We are calculating vertical height for a 1.3 second duiration. a.k.a distance, here. Not vertical acceleration/velocity. The vertical accelration is 75 fps. The G's needed to arrest a vertical height of 97.5 over a 1.3 second time inteval based on 75 fps descent rate is 4 G's. If we need to arrest 75 feet in one second, how do you figure that height gets lower at 1.3 seconds? You just lowered vertical velocity to 57.7 when 75 is required.
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Old 14th March 2008, 01:55 PM   #26
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Keep in mind the height we used is the lowest possible based on the antenna. If we really want to get technical, we could use the FDR altitude at this position which would cause the G loads required for the "pull level" to be astronomical. Right now we are using 304 MSL - 80 MSL. Feel free to punch in the numbers using FDR Altitude of 699 MSL - 80 MSL. Wow! Or even try 273 AGL (radar altitude) + 135 ground elevation. = 408 MSL. Both will require much more than 11.2 G's.

This article was written on the hypothetical that the aircraft was as low as the VDOT antenna. Basically giving the govt story the benefit of the doubt on altitude. According to the FDR, it was much higher.
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Old 14th March 2008, 01:55 PM   #27
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Aldo, at the Loose Change forums, is having trouble understanding the first error in Balsamo's article:

Quote:
We are calculating vertical height for a 1.3 second duration. a.k.a distance, here. Not vertical acceleration/velocity. The vertical accelration is 75 fps. The G's needed to arrest a vertical height of 97.5 over a 1.3 second time interval based on 75 fps descent rate is 4 G's. If we need to arrest 75 feet in one second, how do you figure that height gets lower at 1.3 seconds? You just lowered vertical velocity to 57.7 when 75 is required.
Let's take this a step at a time.

Quote:
We are calculating vertical height for a 1.3 second duration. a.k.a distance, here.
No you're not, you're calculating a value that you then divide by an acceleration (g, which is 32 feet per second per second) to produce an acceleration ratio (the "G" force, which is simply the ratio of an acceleration to g). So, you'd better be calculating an acceleration, or you're doomed from the start.

Quote:
Not vertical acceleration/velocity.
Not vertical velocity, yes vertical acceleration. (Two different things! Keep them straight or you're lost!)

Quote:
The vertical accelration is 75 fps.
No, the initial vertical velocity (rate of descent) is 75 fps. Accelerations cannot be measured in fps, you need fps per second, that is feet per second per second.

Quote:
The G's needed to arrest a vertical height of 97.5 over a 1.3 second time interval based on 75 fps descent rate is 4 G's.
You're not arresting a height, you're arresting a velocity, within certain constraints of time and of distance descended in the process. It turns out the distance is a more stringent constraint than the time, but Rob started with the time in his analysis and I followed his same procedure (while correcting the math along the way).

Once you know that distance travelled is the limiting factor, you can do this in one step. Suppose you're in a car going 75 fps and you need to stop in no more than 40 feet of travel. What must your deceleration (that is, the magnitude of your acceleration in the direction opposite to the direction you're traveling) be? The formula is a = v(initial)^2 / (2 * d).

75 feet/second * 75 feet/second / (40 feet * 2)
= 70.3 feet/second-second
= 2.2 g (70.3 / 32)

Now suppose you're in a plane descending 75 fps and you need to level off with no more than 40 feet of additional descent from your current height. What must your upward acceleration be? Exactly the same problem, exactly the same answer.

Quote:
If we need to arrest 75 feet in one second,
No. We need to reduce a descent rate (vertical velocity) of 75 feet per second, to a descent rate of 0 feet per second. That's what arresting the vertical velocity means. We have 1.3 seconds to do it in. (And also constraints on the distance we can descend in the process, but that gets taken into account in a later step.)

Quote:
how do you figure that height gets lower at 1.3 seconds?
The plane is descending, is it not? Height gets lower over time, that's the definition of descending. But I don't think that's what you meant.

Quote:
You just lowered vertical velocity to 57.7 when 75 is required.
No, I lowered vertical acceleration to 57.7 feet per second per second, in order to reduce a velocity of 75 feet per second to a velocity of 0 feet per second, in 1.3 seconds.

Quote:
75 / 1.3 = 57.7 fps per second (75 fps change, in 1.3 seconds).
Correct.

Quote:
"75 Feet Per SECOND change in 1.3 seconds" is less than 75 Feet per SECOND??
No, 75 feet per second change in velocity in 1.3 seconds (an acceleration of 57.7 feet per second per second) is slower (less acceleration) than 75 feet per second change in 1 second (an acceleration of 75 feet per second per second). It has .3 seconds longer to happen, so it happens slower.

You have 75 hamburgers to fry.
If you fry 57.7 hamburgers per hour,
you'll fry 75 hamburgers in 1.3 hours.

You have 75 fps of vertical velocity to arrest (reduce to 0 fps).
If you arrest 57.7 fps of vertical velocity per second,
you'll arrest 75 fps of vertical velocity in 1.3 seconds.

So, the acceleration needed to arrest 75 fps of vertical velocity in 1.3 seconds, is 57.7 feet per second per second.

That's about 1.8 times the acceleration of gravity, which is 32 feet per second per second. (NOT "32 feet." NOT "32 feet per second.")

However, as calculated above, the height limit is the limiting constraint. So you have to accelerate upward at 2.2 g (pulling 3.2 Gs) to level off after only 40 feet of descent, and that will take less than the 1.3 seconds you have.

Quote:
Thats stundielicious. :-)

The guy is completely confused.

Indeed.

Respectfully,
Myriad
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Old 14th March 2008, 02:03 PM   #28
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Can someone please explain to me how "we" decided the final vertical speed of the aircraft was 0fps?

Please, please, please, please, don't tell me "our" reason is that the flight path "looks" level in the 5-frames of video we have. Please.
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Old 14th March 2008, 02:10 PM   #29
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Originally Posted by TC329 View Post
Keep in mind the height we used is the lowest possible based on the antenna. If we really want to get technical, we could use the FDR altitude at this position which would cause the G loads required for the "pull level" to be astronomical. Right now we are using 304 MSL - 80 MSL. Feel free to punch in the numbers using FDR Altitude of 699 MSL - 80 MSL. Wow! Or even try 273 AGL (radar altitude) + 135 ground elevation. = 408 MSL. Both will require much more than 11.2 G's.

This article was written on the hypothetical that the aircraft was as low as the VDOT antenna. Basically giving the govt story the benefit of the doubt on altitude. According to the FDR, it was much higher.
No, you have no idea where 77 was with respect to the FDR. You can not prove anything about where 77 was. However people saw 77 fly into the Pentagon. So your false ideas are proven wrong.

No one saw 77 do a fly over, they only saw 77 get close to the base of the Pentagon and crash into the building shredding the plane to pieces at speeds greater than 473 KIAS.

You present shallow research based on fantasy. You have ignored evidence and manufactured your own. A last ditch effort to make up junk about 9/11 to sell DVDs to paranoid people.


11 gs is your fantasy world. The terrorist just aimed at the Pentagon and hit it. No massive maneuvers, just point and hit.
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Old 14th March 2008, 02:14 PM   #30
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Originally Posted by TC329 View Post
We are calculating vertical height for a 1.3 second duiration. a.k.a distance, here. Not vertical acceleration/velocity.
The goal of the calculation you are quoting was to find the G-forces required to go from 75 fps to 0 fps in 1.3 seconds. This has absolutely NOTHING to do with the height.

Quote:
The vertical accelration is 75 fps.
That's not the vertical acceleration, that's the vertical velocity. Do you even understand the difference?

Quote:
The G's needed to arrest a vertical height of 97.5 over a 1.3 second time inteval based on 75 fps descent rate is 4 G's.
This makes absolutely no sense at all. You don't seem to understand at least one of the follow terms: height, velocity, acceleration, g-forces.

Quote:
If we need to arrest 75 feet in one second, how do you figure that height gets lower at 1.3 seconds?
This is completely gibberish. What does "arrest 75 feet" even mean?

Quote:
You just lowered vertical velocity to 57.7 when 75 is required.
He did no such thing.

The acceleration is 57.7. The intial velocity was 75. The final velocity was 0. He did it correctly, as well. He determined a required acceleration of 57.7 feet per second per second is required to go from 75 fps to 0 fps in 1.3 seconds.

At t=0, velocity = 75 fps
At t=1.3, velocity = 0 fps

Acceleration is equal to delta velocity over delta time. 75 / 1.3.

57.7 feet per second per second.
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Old 14th March 2008, 02:16 PM   #31
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Originally Posted by Anti-sophist View Post
Can someone please explain to me how "we" decided the final vertical speed of the aircraft was 0fps?

Please, please, please, please, don't tell me "our" reason is that the flight path "looks" level in the 5-frames of video we have. Please.

That is one of several unwarranted assumptions Rob made, along with assuming the plane flew over the tower and assuming that the rate of descent between the tower and the light post was constant.

Quote:
4480 fpm (75 fps) descent needs to be arrested within 1.3 seconds.
"Arresting" the descent means levelling off.

However, it actually doesn't matter. If you accept all the other assumptions as I did (not because I think they're likely or even reasonable, but to show that even accepting those assumptions the analysis is wrong), you can't reduce the calculated G force by allowing the plane to still be descending when it hits the building. It doesn't help.

Respectfully,
Myriad
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Old 14th March 2008, 02:17 PM   #32
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TC:
You guy's screwed up the math. Now be a man and do the right thing and admit it.
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Old 14th March 2008, 02:20 PM   #33
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Originally Posted by Anti-sophist View Post
...

This is completely gibberish. What does "arrest 75 feet" even mean?
...

.
News Flash:
"Pegleg" 329 and his gang of 37 militia troofers charged with Math manufacturing.
6 Paddy-wagons required to take them in...
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Old 14th March 2008, 02:23 PM   #34
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Originally Posted by CurtC View Post
The analysis he linked to at the PfT site bizarrely assumes that after 77 passed over the tall antenna, it continued at a constant, straight-line descent rate until it hit the first light pole a half-mile later, and only at that moment did it start to level off - the slope from the light pole to the Pentagon's wall is a much shallower descent rate. Balsamo's idiotic reasoning is that a plane couldn't pull out of a 4500 ft/min descent when it's only 40 feet off the ground.
That is exactly correct and far from "idiotic". Well, i dont feel a 757 could do it. Im not alone. http://pilotsfor911truth.org/core.html. The list grows regularly.

Also, the FDR shows for that segment a relatively uniform/linear descent rate and uniform G load just above 1 G.. Unfortunately, it is no where near the forces claimed by even the so called "critical thinkers" (eg. 3.49). I'd like to see them input 699MSL into their calculations as shown by the FDR for this position. Then compare their results to the vertical accelerations shown in the data provided by the NTSB. Their claims/requirements are already too great at 304 MSL as compared to the FDR data.
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Old 14th March 2008, 02:30 PM   #35
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We've already established conclusively that the FDR data runs out prematurely and that you guys have a massive timeslip error in your data. If your argument rests on your flawed FDR data interpratation, then you have no argument.

In this thread, you've been asked not to quote the FDR data because it results in the same tired argument with the same tired derailment. The whole reason this thread was created was because you had an argument that was "new" and not dependent on this already disproven assumption. And therefore we had a chance for a new discussion.

Instead it appears you've gone back to your favorite flawed assumption.
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Old 14th March 2008, 02:30 PM   #36
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Originally Posted by TC329 View Post
That is exactly correct and far from "idiotic". Well, i dont feel a 757 could do it. Im not alone. http://pilotsfor911truth.org/core.html. The list grows regularly.

Also, the FDR shows for that segment a relatively uniform/linear descent rate and uniform G load just above 1 G.. Unfortunately, it is no where near the forces claimed by even the so called "critical thinkers" (eg. 3.49). I'd like to see them input 699MSL into their calculations as shown by the FDR for this position. Then compare their results to the vertical accelerations shown in the data provided by the NTSB. Their claims/requirements are already too great at 304 MSL as compared to the FDR data.
How do you explain the erroneous calculations by your "experts"? This is something that is beyond dispute.
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Old 14th March 2008, 02:32 PM   #37
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Originally Posted by beachnut View Post
No, you have no idea where 77 was with respect to the FDR. You can not prove anything about where 77 was. However people saw 77 fly into the Pentagon. So your false ideas are proven wrong.


Is your argument that the FDR released by the NTSB is garbage and not evidence of anything and that eyewitness testimony should be taken over the FDR data released by the NTSB as supplied by the US gov? (Oh please say yes! please say yes!!! I've been waiting for the day that eyewitness testimony trumps all in the JREF world!!)


Quote:
You present shallow research based on fantasy.


So now the NTSB supplied us with "fantasy"?

Quote:
You have ignored evidence and manufactured your own.


What did we manufacture the FDR data or the 5 frames showing a plane coming across the lawn smooth and level (which is now suddenly up for debate apparently according to Anti-Sophist)?

Quote:
A last ditch effort to make up junk about 9/11 to sell DVDs to paranoid people.


Ad-Hom proves nothing.


Quote:
11 gs is your fantasy world. The terrorist just aimed at the Pentagon and hit it. No massive maneuvers, just point and hit.
Perhaps you can explain then why the terrorist didn't just dive straight down into it instead of pulling off this maneuver and perhaps some of the people here with the right background can explain the ignorance in your "just point and hit" comment.

It's obvious you've never been anywhere near the Pentagon before.
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Old 14th March 2008, 02:37 PM   #38
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Originally Posted by TC329 View Post
[/color]
Is your argument that the FDR released by the NTSB is garbage
No the argument is that you guys have failed to account for the gross error in your analysis in regards to time. There's a timeslip error in your analysis that you refuse to correct because it destroys your entire argument.

Quote:
What did we manufacture the FDR data or the 5 frames showing a plane coming across the lawn smooth and level (which is now suddenly up for debate apparently according to Anti-Sophist)?
The argument that you can decide conclusively from those 5 frames that the vertical velocity was exactly 0 fps is just *********** absurd. Please, please, please, understand how error works. How you measure it. Come back with a reasonably justified range of decent values that matches what we see. Do so scientifically, without screwing up the calculations.

I'm sorry, but saying "it looks level, therefore 0.0000 feet per second" is just stupidity.

False precision is yet another of the common piles of steaming nonsense that you guys continue to rest your entire case upon.

Quote:
Perhaps you can explain then why the terrorist didn't just dive straight down into it
Utterly offtopic. See here:
http://911myths.com/html/flight_path.html

Start your own thread if you want to discuss in more detail.
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Old 14th March 2008, 02:58 PM   #39
beachnut
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Did the 5 frame video camera shots, showing an earth that curved the hills to the over pass downward, just get used as evidence for level flight? A flight which never leveled off, but impacted in a decent at the Pentagon.

Who got 2.8 and 3.2 gs? That sounds good. The wings will fail under sustained 6 to 7 gs, I suspect, but there were not level off stuff, just a diving flight 77; someone needs to look closer at the Pentagon next time. But once again we are playing facts with a fantasy oriented, bad math 9/11 truth member.

Who can prove the VDOT tower is as high as they say? Who can give perfect MSL values for each light post? Who knows how far the engines bottoms are below the cord of the plane? Who in 9/11 truth can do math?
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Old 14th March 2008, 03:00 PM   #40
gumboot
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I'm curious of the claim that the descent rate needs to be arrested at all. Given that the aircraft hit at about ground level (and according to some witnesses, part of it hit the ground just before impact) and given that it clearly was not at ground level immediately prior to hitting the Pentagon (otherwise it, well, would have crashed before the Pentagon) it goes without saying that the aircraft was in a descent at the time of impact.
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