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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.
Originally posted by COOL HAND
Here is the rebuttal:
Link
Care to take a swipe at his work and prove him wrong?
Originally posted by COOL HAND
reply to post by johndoex
You never sufficiently debunked anything there.
All you did was get into a pissing contest. Having read that post, there was no clear proof that you are right and he is wrong.
Originally posted by Craig Ranke CIT
Originally posted by infinityoreilly
Just an observation here, the horizon doesn't mach up in your split pic diagram. So hence the point your trying to make is skewed by an imaginary line that should have a less steep decent into the Pentagon.
Originally posted by ThatsJustWeird
Question:
1. Seeing as how cell phone towers can pop up overnight (and I mean that quite literally), especially in this area, do you have any evidence that cell phone tower was there in 2001??
2. You know the exact flight path with heights and everything for 77? Wow!
Congrats! All everyone else has are best estimates.
3. As noted, light poles were clipped. So what type of missile or light plane has that kind of wingspan and is strong enough to clip light poles and not explode or at least be taken off the path?
Originally posted by Craig Ranke CIT
Nothing clipped the light poles because we know the plane was not really over there, it was on the north side of the citgo. The poles had to have been staged in advance. A complete light pole hypothesis is available here.
Originally posted by Craig Ranke CIT
One day is more clear than the other so you can see across the river better but the river edge is lined up pretty darn close.
In fact the adjustment should go the other way....the left image could tilt down a bit which would make for a steeper descent angle.
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.
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?
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?
You stated that the impact height corresponded to the center of the hole. But there's no reason to assume the light pole was hit by the center of the aircraft. We've got apples and oranges here. Since the poles are referenced to the bottom of the aircraft, the impact point should be so referenced too, and that is at 33 feet, not 45.
If you repeat the above but substitute in 33 feet, you find that the vertex is actually at -90 feet, i.e. the aircraft is still descending at impact. And the acceleration now is only 26.9 ft/s2, or a mere 0.84 g.
That's peanuts. Even the Wright Flyer could handle that.
I remind you, I'm using your figures. Your figures are consistent with a totally viable flightpath. Therefore, Mr. Balsamo is either spectacularly incompetent, or a total fraud. Whichever it is, you have been duped.
Please feel free to ask if any of that was unclear. It saddens me deeply to see nontechnical people taken advantage of by such breathtaking idiocy.