Quad4_72
AI-EE-YAH!
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Ah but one must first pass this high school algebra that you speak of. This is where CFs run into problems
Flight 77 obstacles
- Thread starter Dave Rogers
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Ah but one must first pass this high school algebra that you speak of. This is where CFs run into problems
jhunter1163
beer-swilling semiliterate
Hey! Leave Dr. Seuss out of this!
DGM
Skeptic not Atheist
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?
TC; When is the retraction going to come?
Putting it all together, for P4T:
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
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) + 7 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.
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) + 7 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
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.
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.
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.
Aldo, at the Loose Change forums, is having trouble understanding the first error in Balsamo's article:
Let's take this a step at a time.
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.
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.
Indeed.
Respectfully,
Myriad
Let's take this a step at a time.
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.
Not vertical velocity, yes vertical acceleration. (Two different things! Keep them straight or you're lost!)
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.
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.
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.)
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.
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.
Correct.
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.
Thats stundielicious.
The guy is completely confused.
Indeed.
Respectfully,
Myriad
Anti-sophist
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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.
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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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.
Anti-sophist
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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.
That's not the vertical acceleration, that's the vertical velocity. Do you even understand the difference?
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.
This is completely gibberish. What does "arrest 75 feet" even mean?
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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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.
"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
DGM
Skeptic not Atheist
TC:
You guy's screwed up the math. Now be a man and do the right thing and admit it.
You guy's screwed up the math. Now be a man and do the right thing and admit it.
rwguinn
Penultimate Amazing
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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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.
Anti-sophist
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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.
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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DGM
Skeptic not Atheist
How do you explain the erroneous calculations by your "experts"? This is something that is beyond dispute.
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!!)
So now the NTSB supplied us with "fantasy"?
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)?
Ad-Hom proves nothing.
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.
Anti-sophist
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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.
The argument that you can decide conclusively from those 5 frames that the vertical velocity was exactly 0 fps is just ****ing 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.
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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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?
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?
gumboot
lorcutus.tolere
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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.