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Radar Altitude Confirms True Altitude, Too high to hit light poles/pentagon

rob balsamo
post Feb 16 2007, 05:59 PM
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We have recently been able to decode the additional data 'slipped' into our package through the FOIA (Thanks UT!).

The main purpose of getting this data decoded was to confirm or contradict our initial altitude findings regarding True altitude as shown in the animation and original csv file which can be seen in Chapter Two.

The last recordings of radar altitude shown in the newly decoded .fdr file is as follows.



In the spirit of keeping this simple, we will use the argument side of the analysis prior to us getting the decode of the raw file.

Many of you have read our analysis regarding the altitude being too high based on True Altitude. However, the argument side says that there could be up to 2 seconds error and possible lag due to instrument error.

Quote as follows from a self-proclaimed anonymous internet FDR Expert.
QUOTE


However, A radar altimeter presents no lag. The 273 feet you see above is a hard number above the ground (or any object you are flying over).

Now we will disregard when it was recorded and use the argument side that there could be 'up to' 2 seconds missing (ie. the 273' was recorded 2 seconds prior to the pentagon wall).

Backing the data out from the pentagon wall 2 seconds based on speed, the aircraft is at this location...

(again using the govt loyalist argument of the south flight path)


The ground elevation at this point is ~60 feet above sea level. Add that to 273 and you get 333 feet above sea level. (this figure does not match True altitude at this point because we are using the govt loyalist hypothetical that 273 was recorded 'up to' 2 seconds out from the pentagon.. we are using their argument of the largest margin for error).

The elevation of the 'hole' in the pentagon is 33 (ground elevation of pentagon, established) plus 12' (center of hole) = 45 feet above sea level.

When we subtract 333 from 45, we get a number of 288 feet above the 'impact hole' at a spot 2 seconds away from impact.

Divide 288 by 2(seconds) and this aircraft would need a steady linear descent rate of 8,640 feet per minute to impact the hole in the pentagon (remember, this is using the argument that 273 was recorded up to 2 seconds prior to impact). This is in direct conflict with the DoD video showing an object level across the lawn.





Now, lets look at the light poles.

We will again use the govt loyalist argument that the height of the poles was 66(ground elevation, although the USGS shows 43) + 40 feet for pole height (liberal figure) = 106 feet.

Based on forward speed, the light poles were ~1 second from impact (its more like 1.3 for poles one and two, but we're giving the govt loyalists every chance possible here).

So, that being the case, if we were using the 273 AGL altitude as recorded by the radar altimeter 2 seconds from the pentagon wall, that would make the 273 agl recorded 1 second prior to the poles. The top of pole height was 106 feet above sea level. The "2 second recording" was 333 feet above sea level. Subtact the two and you get 227 feet this aircraft would have had to descend in order to hit the very top of pole 1 within one second(again, using liberal numbers for pole height... its actually more than 227).

To descend 227 feet in one second, this aircraft would need 13,620 feet per minute dive to clip the poles. And then level off from that dive to be level with the pentagon lawn.



To level off from that dive, the aircraft would need to pull (227/32 (ft/sec, accel due to gravity) =7+1G's (earths gravity) = <span style='font-size:14pt;line-height:100%'>8</span> total G's for 1 second sustained to be level with the lawn.

There is no indication anywhere in ther FDR file showing this aircraft pulling 8 G's sustained during the last second, Matter of fact, it shows less than 1 G.. Not to mention, aircraft structural limitation would not let it get that high. IE. The wings would rip off and/or High speed buffet/stall.

Conclusion:

Using the govt loyalist own argument of 'up to' 2 seconds of error, we can see this aircraft was too high to hit the poles and the pentagon.

However, using the actual data that was recorded 1 second prior to the pentagon wall as provided directly from the NTSB (and not making excuses like there may be 'up to 2 seconds missing'), it is further in conflict with the govt story.

Cheers!
Rob

"If his true altitude is accurate, he may be on to something" - Anonymous Self-Proclaimed FDR Expert and Govt Loyalist.

(if you want to repost this whole article, pictures and all, hit the quote button and copy and past the quote to your new post in another forum. Spread it everywhere if you like).
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Beached
post Feb 16 2007, 09:38 PM
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Excellent post Rob! cheers.gif
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Sanders
post Feb 17 2007, 03:18 AM
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Nice work, Rob & UT. Pat yourselves on the back - I know it must've been a lot of work -

worthy.gif cheers.gif
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Ashoka
post Feb 17 2007, 08:55 AM
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QUOTE (Sanders @ Feb 17 2007, 02:18 AM)

QUOTE
To level off from that dive, the aircraft would need to pull (227/32 (ft/sec, accel due to gravity) =7+1G's (earths gravity) = 8 total G's for 1 second sustained to be level with the lawn.


Incorrect!

Reading the table the aircraft had a mean vertical speed of 79 feet/s (352-273) in the last second, of 64 feet/s (416-352) in the previous one, of 76 feet/s (492-416) the second before.

So let's say it's 79 feet (24,07 m/s) and the aircraft is diving so he has to pull in order to level

In fact S (end altitude above top pole) = S0 (start altitude above top pole) + V0(start speed negative)*t + ˝ * a (acceleration) * t^2

(1) S0= 69,19 meters, V0 = -24,07 meters per second t = 1 second

so with an acceleration of 7gs we have

final altitude= 79,45 meters above top pole (261 feet)
final speed= 44,6 meters/s (146 feet/s) upwards!

Case 2) The aircraft levels so after one second his vertical speed is zero. The calculated acceleration is simply 24,07 meters/squared second so the “pull up” would be 3 to 4 gs (for the pilots and the passengers)

But now the aircraft is with zero vertical speed, at 57,46 meters (188.5 feet) above the top pole!

In order to cut the top pole the pilot should increase the vertical speed towards the ground so that in one second he travels those 69,19 meters and cut the top of pole1

If the pilot dives the plane to the ground, with a constant acceleration towards the ground, then, in order to cut the top of pole1 starting from 69,19 meters above it and using a costant acceleration and a starting speed of 24,07 m/s (towards ground) then we have (1)

a= 2*(69,19-24,07) = m/s^2 (296 feet/s^2)
final vertical speed = 159,43 m/s (523.3 feet/s) towards the ground

Ok so the plane is about ... 0,076 seconds (40feet / 523.3) before crashing unless he pulls up in 66 feet!

But Hani is not an ordinary pilot and btw he's blessed by God, isn't he? So he try to pull and align in order to fly orizzontally to the Pentagon

The plane is at ground level about 0,5 seconds before impact (as in the video) so his vertical speed is zero. His starting vertical speed is 523.3 feet/s so his acceleration would be (523.3/0.5) = 1046.3 feet/s^2!!! Wow it's only a 32 +1 G maneuver!

***

BTW the above calculations are all useless.

In fact the plane has vertical speed of 24 m/s and has to descend 70 meters in one second to cut the pole. So the plane must accelerate towards the ground and then pull up to align with the lawn in 2 seconds. That's enough to say it's impossibile :-)

Ashoka

This post has been edited by Ashoka: Feb 17 2007, 08:55 AM
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rob balsamo
post Feb 17 2007, 01:27 PM
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QUOTE (Ashoka @ Feb 17 2007, 07:55 AM)
Incorrect!

Reading the table the aircraft had a mean vertical speed of  79 feet/s (352-273) in the last second,

Pssst... Ashoka... we werent calculating according to the mean vertical speed in the table. We were calculating based on the hypothetical (using the govt loyalist argument) that 273 feet was recorded at 2 seconds from the pentagon wall and getting the vertical speed from that point forward.



Sorry guys, i tried to keep it as simple as possible. Perhaps i should edit the picture to show only the last recorded altitude so it doesnt confuse.

QUOTE
In fact the plane has vertical speed of 24 m/s and has to descend 70 meters in one second to cut the pole. So the plane must accelerate towards the ground and then pull up to align with the lawn in 2 seconds. That's enough to say it's impossibile :-)


This is kinda what i was showing.. wink.gif
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rob balsamo
post Feb 17 2007, 01:41 PM
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But, since we're on the subject,

lets use Ashoka's mean average from the table of 79 ft/s descent rate.

If 273 was recorded at 2 seconds from pentagon wall. (translated into 333 MSL)

Continuing the trend as seen in the table...

The next second (over the poles) would be at 254MSL = Too high to hit the poles (poles are 106 feet high above sea level, using the govt loyalist elevation of 66 feet for base of poles instead of USGS 43 feet.)

The second after that (pentagon wall) would be at 175 feet above sea level = Too high to hit the pentagon (height of pentagon wall is 110 feet above sea level).

cheers.gif
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Ashoka
post Feb 17 2007, 02:29 PM
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QUOTE
Pssst... Ashoka... we werent calculating according to the mean vertical speed in the table. We were calculating based on the hypothetical (using the govt loyalist argument) that 273 feet was recorded at 2 seconds from the pentagon wall and getting the vertical speed from that point forward.


Oh Yeah, but it was too funny show that hitting the pole and then leveling with the lawn would require a 33g maneuver. It would explain the lack of debris, the white smoke (it's the plane vaporizing due to the maneuver) and so on

cheers.gif

...but with these data it would be more “realistic” a pull up maneuver as the one described in The Pentacon :-)

Ashoka
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rob balsamo
post Feb 17 2007, 02:44 PM
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QUOTE (Ashoka @ Feb 17 2007, 01:29 PM)
...but with these data it would be more “realistic” a pull up maneuver as the one described in The Pentacon :-)

Ashoka

Agreed..

based on vertical speed, the only way this aircraft could have pulled up to be level is if was high enough to have enough time to do the maneuver.. Also seen in PBB2..

cheers.gif
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mirageofdeceit
post Mar 13 2007, 06:53 PM
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I have separately concluded that if he maintained his vertical speed and did not perform an impossible maneuver in the last two seconds, to (as has already been said) be level with the lawn, he clears the building by 15 ft, allowing for the engines hanging underneath.

If he simply pushed the nose over to hit where it is said he hit, the aircraft would no longer be level with the ground; he'd HAVE to come into shot higher, and pointing down at the ground.

If he'd attempted to pull the impossible maneuver at 450 kts (in keeping with the official photos) as he'd have to do, he'd not just bounce off the lawn - he'd make a crater!!!

This post has been edited by mirageofdeceit: Mar 13 2007, 06:54 PM
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rob balsamo
post Jul 18 2007, 10:22 PM
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added here so its easier for me to find...


QUOTE
Claims of 'debunk' -

QUOTE
1) The FDR did not record the final seconds of Flight 77. There is up to 2 seconds missing.
2) The .csv file is not meant to be analyzed forensically, it is meant to be plotted.
3) The .csv data is not raw FDR data. It is not even serial bitstream data.
4) The .csv data is not meant to be broken down into 1/8th seconds and analyzed.
5) The .csv data, properly interpreted, says that there are N samples during this particular frame.
6) Without the frame description, we do not know when in a frame any one sample occurred.
7) Without the frame description, we have lost the measurement timestamps, so the time a particular word was recorded does not necessarily equate with when it was measured.
8) Given these time-shift errors, any mathematics that uses more than one data-point runs the risk of assuming that two numbers occurred at the same time, when they didn’t.
9) Many of these errors can be corrected, greatly, with the frame descriptor.
10) Any analysis must account for (or justify ignoring) these issues in order to draw any valid conclusions.



Our reply -

QUOTE
1. Radar Altitude Confirms too high using Govt Loylist argument of 2 seconds missing
2. We have a plotted animation produced by the NTSB using the FDR, Radar and ATC Transcripts. Cover letter and evelope here.
3.Raw data has been decoded. See number 1.
4.See number 2
5.agreed
6. We have the frame descriptor
7.See number 6
8.See number 2
9.See number 6
10.see all of the above



An anonymous govt loyalist of the official story refuses to debate us thus far in a proper venue and keeps repeating the word "debunked" without being able to back up his information.

Bottom line - An anonymous, self-proclaimed "FDR Expert" tells half truths and thinks he knows more than the professionals at the NTSB.
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JackD
post Jul 19 2007, 02:22 PM
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JDX

I think your FDR needs a "new Deal"

Patriotism: Loving your country always, and correcting your government when it goes awry.

not unlke parenthood.
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Thous
post Jan 22 2009, 01:41 PM
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QUOTE (johndoeX @ Feb 16 2007, 06:59 PM) *
We have recently been able to decode the additional data 'slipped' into our package through the FOIA (Thanks UT!).

The main purpose of getting this data decoded was to confirm or contradict our initial altitude findings regarding True altitude as shown in the animation and original csv file which can be seen in Chapter Two.

The last recordings of radar altitude shown in the newly decoded .fdr file is as follows.



Mod Edit: Reduced size of quoted text


Didn’t the data also show that the plan’s altitude was too high to hit the Pentagon? Don’t the short videos you posted on this thread show the plane hitting the building? I’m new, sorry if this has already been answered, I’m just trying to understand.
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rob balsamo
post Jan 22 2009, 02:03 PM
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Yes, the data does show too high to hit the pentagon. Please visit the various pinned topics and our video presentations found throughout this forum.

We demonstrate above the data provided by the govt is in conflict with the videos, also provided by the govt, based on arguments made by those who make excuse for the govt story.

I know the forum is large for a newcomer, but keep in mind there is over 2 years of research on this forum. Please use the search feature and spend some time browsing/studying the material as many questions are already answered.

Thank you.
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dMz
post Jan 22 2009, 11:51 PM
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QUOTE (Turbofan @ Jan 22 2009, 07:58 PM) *
~ 300,000,000

[meters/sec]

If anyone wants more decimal places here, NIST tells us:

http://physics.nist.gov/cgi-bin/cuu/Value?c

speed of light in vacuum
c, c_0

Value 299 792 458 m s-1
Standard uncertainty (exact)
Relative standard uncertainty (exact)
Concise form 299 792 458 m s-1
----------------------------------------
For more on radio (RF) and electromagnetic (EM) spectra:

http://www.copradar.com/preview/xappA/xappA.html

http://www.geocities.com/dtmcbride/tech/em-spectrum.html

http://en.wikipedia.org/wiki/Radio_frequency

http://www.naval.com/radio-bands.htm

Related pressure altimeter thread is at:

The Two Baro Corrections
http://pilotsfor911truth.org/forum//index....showtopic=13555
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old spook
post Mar 12 2009, 07:56 AM
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QUOTE (Turbofan @ Jan 23 2009, 03:58 AM) *
Just a quick note about RADAR altimeters and their accuracy and speed:

Commerical aircarft are required to have a triple RADAR altimeter and are highly accurate to within a couple of feet maximum error under 1000 feet
Above Ground Level (AGL).

Why a triple RADAR? Using three transceivers along the belly of the aircraft helps average the terrain at higher altitudes, and also improves signal
isolation at altitudes lower than the total length of the aircraft. In other words, by spacing three transceivers along the belly, the altimeter signals
can be distinguished between reflections from the ground, or distance between transceivers.

The RADAR Altimeter uses Radio Frequency (RF) signals that are transmitted from an antenna, bounce/reflect off the ground and are picked up by
a receiver (hence transceiver). The time it takes for the signal to transmit, bounce and receive is calculated at:

Signal Travel Time = 2 x altitude / speed of light
(*basic generic formula)

IE:

= 2 x 300 feet / ~ 300,000,000 meters per second
= 600 / 300,000,000
= 0.000 002 seconds

(this is typical for ULtra High Frequency (UHF) transmit signals in the "Giga Hertz" (GHz) range).

From there, the unit will process the information and send an output to the cockpit instrumentation and flight data systems at least 8 times per second.

*Thanks dM!

Your generic formula is correct but your example is false. You mixed feet and meters. Correct would be
= 2 x 91.44 m (300 ft)/ 300,000,000 m/s
= 182.88 / 300,000,000
= 0.0000006 seconds (approx)
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RaechelParker
post Nov 25 2009, 02:05 PM
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I was looking for this the other day. i dont usually post in forums but i wanted to say thank you!
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