It looks like you're using an Ad Blocker.

Thank you.

Some features of ATS will be disabled while you continue to use an ad-blocker.

# New FDR Decode

page: 23
12
share:

posted on Nov, 16 2009 @ 07:05 PM

Originally posted by 911files
Vert acc
1.291
1.451
1.272
1.401
1.52
0.663
0.725
0.982
1.604
1.675

John, the greatest change from 1G (Earth) in the last 7 seconds is roughly .7 G positive, and .4 negative, alternating throughout, until the very last second. Lets check these changes with respect to vertical distance.

s = 1/2at^2

Although these changes happened in fractions of a second (8hz), for arguments sake, we'll use both upper and lower limits for the duration of a half second giving your argument advantage.

.7 G = 22.4 f/sec^2

s = (22.4*.25)/2

s = 2.8 feet

- 0.4G = -12.8 f/sec^2

s = (-12.8 * .25)/2

s = -1.6 feet

As I said, minimal.

It's well within the +/- 4 feet differences observed in the PA descent rate, especially when considering the upper and lower limits of Vertical accel occurred at less than a half second (8hz).

Also John, you continue to fail to observe the words repeatedly posted and italicized as described for this descent, here they are again - near linear.

For added measure, a picture is worth a thousand words. Last 7 seconds of Vertical accel. Points graphed at 1 second interval and smoothed. Scale is +/- 10 G to compare with "9/11: Attack On The Pentagon" upper boundary limit.

No one said the final descent was perfectly linear John. Please try to refrain from continued spin of this statement.

Conclusion, Vertical Accel matches very well with PA. PA shows too high to hit the Pentagon. Those who make excuse for the govt story still have not provided any proof for PA error nor the object from which RA is measuring.

[edit on 16-11-2009 by R_Mackey]

posted on Nov, 16 2009 @ 07:13 PM

Originally posted by R_Mackey

Originally posted by 911files
Vert acc
1.291
1.451
1.272
1.401
1.52
0.663
0.725
0.982
1.604
1.675

John, the greatest change from 1G in the last 7 seconds is roughly .7 G positive, and .4 negative, alternating throughout, until the very last second. Lets check these changes with respect to distance.

s = 1/2at^2

Although these changes happened in fractions of a second (8hz), for arguments sake, we'll use both upper and lower limits for the duration of a half second giving your argument advantage.

.7 G = 22.4 f/sec^2

s = (22.4*.25)/2

s = 2.8 feet

- 0.4G = -12.8 f/sec^2

s = (-12.8 * .25)/2

s = -1.6 feet

As I said, minimal.

It's well within the +/- 4 feet differences observed in the PA descent rate, especially when considering the upper and lower limits occured at less than a half a second.

Also John, you continue to fail to observe the words repeatedly posted and italicized for this descent, here they are again - near linear.

For added measure, a picture is worth a thousand words. Last 7 seconds of Vertical accel. Points graphed at 1 second interval. Scale is +/- 10 G to compare with "9/11: Attack On The Pentagon" upper boundary limit.

No one said the final descent was perfectly linear John. Please try to refrain from continued spin of this statement.

Conclusion, Vertical Accel matches very well with PA. PA shows too high to hit the Pentagon. Those who make excuse for the govt story still have not provided any proof for PA error nor the object from which RA is measuring.

I am not even goin to waste my time with this load of horse manure. Anyone with half-a-brain will look at your graph for vert acc and see you adjusted the vertical scale to minimize the apparent change in it. Go learn physics and math and look me up when you can be an honest broker who knows what he is talking about.

posted on Nov, 16 2009 @ 07:22 PM

Originally posted by 911files
I am not even goin to waste my time with this load of horse manure. Anyone with half-a-brain will look at your graph for vert acc and see you adjusted the vertical scale to minimize the apparent change in it. Go learn physics and math and look me up when you can be an honest broker who knows what he is talking about.

What's just as deceptive and incompetent is that this "verified by the FAA" cult has concluded (based on the animation) that this was a competent pilot doing the flying. That was quite an uncomfortable ride, so one must conclude that this clown flies the same way.

posted on Nov, 16 2009 @ 07:24 PM
Wow John, quick reply. You must have been waiting for me all day?

John, the scale of graph is explained.

Regardless of the graph, the changes in vertical distance are minimal as proven by physics and math.

+ 2.8 feet

- 1.6 feet

Minimal.

posted on Nov, 16 2009 @ 09:34 PM

Originally posted by R_Mackey
John, the greatest change from 1G (Earth) in the last 7 seconds is roughly .7 G positive, and .4 negative, alternating throughout, until the very last second.

Untrue. Apparently the person posting as R_Mackey has
looked at only 1/8 of the measurements: those that fell
on an even second boundary.

R_Mackey must have learned math from Rob Balsamo: Having
used 1/8 of the data to calculate an incorrect maximum
delta-s for intervals of 1/8 second, he then pretended
that delta-s represents 1-second intervals.

Then he graphed the vertical acceleration with a y-axis
ranging from -10 to 10g, allegedly to compare against
Rob Balsamo's notoriously incorrect calculations as
featured on the PfT web site and in the PfT/CIT videos.
The real effect of the poseur's chosen y-axis is to
minimize the visual effect of large changes in vertical
acceleration.

To top it off, he wrote:

Originally posted by R_Mackey
Conclusion, Vertical Accel matches very well with PA.

Incorrect. See below.

Originally posted by R_Mackey
Regardless of the graph, the changes in vertical distance are minimal as proven by physics and math.

+ 2.8 feet

- 1.6 feet

Minimal.

Which proves the fake R_Mackey knows nothing of physics
or math.

Here are all 16 of the vertical accelerations (in g)
that were recorded for the last two seconds, along with the
approximate reduction in vertical velocity (ft/sec) implied
by each measurement for its 1/8-second interval:

1.604 2.43
1.781 3.14
1.762 3.06
1.964 3.88
1.879 3.54
2.264 5.08
2.044 4.20
2.181 4.75
1.675 2.71
1.744 2.99
1.650 2.61
1.504 2.03
1.785 3.16
1.655 2.63
1.861 3.46
1.945 3.80

That adds up to about 53.5 ft/sec during the last two seconds
alone. The average vertical acceleration during those
last two seconds is over 1.8g.

I hope I never fly aboard an airliner piloted by someone
who thinks two seconds of 1.8g at low altitude and 450+
knots represents a "minimal" change in vertical velocity.
Fortunately, there is no reason to think the fake R_Mackey
is an active pilot.

Will

posted on Nov, 16 2009 @ 10:04 PM

Originally posted by cesura

Originally posted by R_Mackey
John, the greatest change from 1G (Earth) in the last 7 seconds is roughly .7 G positive, and .4 negative, alternating throughout, until the very last second.

Untrue. Apparently the person posting as R_Mackey has
looked at only 1/8 of the measurements: those that fell
on an even second boundary.

I also used the even second boundary values since putting all 12 flights into a single workbook for comparative purposes required the abbreviated form. However, those are sufficient to make the case I was attempting to clarify for 'Rob'. You took it to another level of detail which only reinforces the case.

When someone deliberately exaggerates the graph range to +/- 10 when data for the entire file is within +/- 2 (mean = 1), then I give up. An intellectually honest debate requires both honesty and intelligence. I am finding both severely deficit in my discussions with 'Rob' or 'R_Mackey'. But what else can you expect from someone posting under someone else’s name?

posted on Nov, 16 2009 @ 10:21 PM

That adds up to about 53.5 ft/sec during the last two seconds
alone.

Undisputed. Unfortunately, your parabola(s) require much more than that for the last 2 seconds based on data.

Will, no one disputes vertical velocity was more than 50 f/s for the last several seconds and reduced in the last second. We are examining the curve of the descent path here and whether it was near linear. Please be familiar with the argument before attempting to debate.

Originally posted by cesura
1.604 2.43
1.781 3.14
1.762 3.06
1.964 3.88
1.879 3.54
2.264 5.08
2.044 4.20
2.181 4.75

The above were recorded between -40 and -99 PA according to the unverified data provided by Warren, which you take at face value and conflicts with the NTSB Flight Path Study.

Using the greatest change from 1G = 2.2 - 1G (Earth) = 1.2 G.

Using the same premise of 0.5 second duration giving advantage to the argument made by those who support the govt story;

s = 1/2at^2

s = (38.4*.25)/2

s = +4.8 feet

1.675 2.71
1.744 2.99
1.650 2.61
1.504 2.03
1.785 3.16
1.655 2.63
1.861 3.46
1.945 3.80

The above was recorded after -99 PA/174 MSL.

Again using greatest change from 1 G of +0.95 for a 0.5 sec duration;

s = 1/2at^2

s = (30.4*.25)/2

s = + 3.8 feet

I hope I never fly aboard an airliner piloted by someone
who thinks two seconds of 1.8g at low altitude and 450+
knots represents a "minimal" change in vertical velocity.

Will

Will, the proper argument might be, you hope to never fly with someone who thinks 1.8G at 130 knots over Vmo for a specific type aircraft will continue controlled flight.

With that said, it appears P4T covers every major airline. Looks like it's the train for you.

Will, evasion of the following noted for the 3rd time.

Originally posted by R_Mackey

Originally posted by cesura

Originally posted by R_Mackey

Untrue.

Please show us, using your formula, a = 2s/t^2, the acceleration based on an initial velocity of 75 f/s, a vertical distance of 271 over a 4.4 second period.

Originally posted by R_Mackey
Exactly, therefore it does not take into consideration forward (horizontal) velocity. Just as explained in "9/11: Attack On The Pentagon".

Untrue.

Please show us the acceleration, based on a 75 f/s initial velocity, 0 final velocity, with a vertical distance of 271 feet in a 4.4 second period using this calculator.

tutor4physics.com...

PA still has not been proven in error.

[edit on 16-11-2009 by R_Mackey]

posted on Nov, 16 2009 @ 11:04 PM
The fake R_Mackey knows so little physics and math that
he thinks the s=1/2at^2 formula calculates a vertical
velocity or a change in vertical velocity.

Like Rob Balsamo, the fake R_Mackey doesn't even know how
to check his units.

Originally posted by 911files
When someone deliberately exaggerates the graph range to +/- 10 when data for the entire file is within +/- 2 (mean = 1), then I give up. An intellectually honest debate requires both honesty and intelligence. I am finding both severely deficit in my discussions with 'Rob' or 'R_Mackey'. But what else can you expect from someone posting under someone else’s name?

Yep.

The fake R_Mackey has shown no ability to think for himself.
Everything he posts is based on Balsamo's miscalculations.
It wouldn't surprise me if Balsamo were feeding him most
of his arguments.

Will

posted on Nov, 16 2009 @ 11:17 PM
Will,

s = 1/2at^2 is used to determine distance, in this case, vertical distance from a linear descent angle.

Hence the term "minimal" as noted repeatedly throughout the last two pages.

Perhaps if you have been following the argument, you would have a better understanding of the debate? Will, you are showing a pattern starting from page 17 which does not lend well to your credibility. You may want to read posts much more thoroughly if you are going to reply.

The PA data agrees with the Vertical accel based on vertical distance.

Your theory and graphs are no where to be found in the data.

And again in closing,

Evasion of the following noted for the 4th time.

Originally posted by R_Mackey

Originally posted by cesura

Originally posted by R_Mackey

Untrue.

Please show us, using your formula, a = 2s/t^2, the acceleration based on an initial velocity of 75 f/s, a vertical distance of 271 over a 4.4 second period.

Originally posted by R_Mackey
Exactly, therefore it does not take into consideration forward (horizontal) velocity. Just as explained in "9/11: Attack On The Pentagon".

Untrue.

Please show us the acceleration, based on a 75 f/s initial velocity, 0 final velocity, with a vertical distance of 271 feet in a 4.4 second period using this calculator.

tutor4physics.com...

PA still has not been proven in error.

posted on Nov, 16 2009 @ 11:26 PM

Originally posted by R_Mackey
The PA data agrees with the Vertical accel based on vertical distance.

Where do you come up with this utter nonsense?

posted on Nov, 16 2009 @ 11:32 PM

Originally posted by 911files

Originally posted by R_Mackey
The PA data agrees with the Vertical accel based on vertical distance.

Where do you come up with this utter nonsense?

John,

Will's most "plausible" flight path decreases in vertical velocity at 18.8 feet per second. Please show us where this is reflected in the data for the last 7 seconds.

posted on Nov, 16 2009 @ 11:51 PM

Originally posted by R_Mackey
Will's most "plausible" flight path decreases in vertical velocity at 18.8 feet per second. Please show us where this is reflected in the data for the last 7 seconds.

What has Will got to do with you claiming the flight path was linear when the physics parameters clearly indicate otherwise?

[edit on 16-11-2009 by 911files]

posted on Nov, 17 2009 @ 12:03 AM

Originally posted by 911files
What has Will got to do with you claiming the flight path was linear

John, please quote where I claimed the flight path was "linear".

Here's a hint:

Originally posted by R_Mackey
Also John, you continue to fail to observe the words repeatedly posted and italicized as described for this descent, here they are again - near linear.

......

No one said the final descent was perfectly linear John. Please try to refrain from continued spin of this statement.

John, it appears you have the same problem as Will, observed from page 17. Please try to read posts more thoroughly before you reply as your credibility as already severely diminished with the improper calculators you have offered, DME falsehoods, and calculating the wrong vector in a turn (among many others).

I can see why you refuse to debate P4T, yet hang on their every word and spin it up on forums like ATS, mostly 99% personal attacks and ad homs.

posted on Nov, 17 2009 @ 12:18 AM

Originally posted by R_Mackey
It has been proven that the implications made by censura/Will Clinger?MIT PhD/current professor at Northeastern and John Farmer/911Files alleging a linear flight path is "implausible", is in fact not "implausible", and is in fact what is reflected in the data provided by the NTSB and further, Warren, the purpose for this thread.

Nothing wrong with my reading comprehension. Also nothing wrong with my understanding of DME and/or turn vectors. You can keep digging the hole as deep as you want, but don't worry about my credibility, it only seems to suffer with those who don't understand math and physics.

posted on Nov, 17 2009 @ 12:34 AM

Originally posted by 911files
but don't worry about my credibility, it only seems to suffer with those who don't understand math and physics.

John, I'm not "worried" about your credibility. Another falsehood offered by you. I rather enjoy watching you delete your work and getting caught in check-mate.

With that said, your credibility does suffer among each and every one of these people. Many who you have contacted to consult since you haven't a clue regarding flight dynamics.

pilotsfor911truth.org...

The list grows.

In closing and since I have personally caught you in many lies, fabrications, spin and confirmed your lack of knowledge in flight dynamics, you are now on ignore.

[edit on 17-11-2009 by R_Mackey]

posted on Nov, 17 2009 @ 03:48 AM

1.604 2.43
1.781 3.14
1.762 3.06
1.964 3.88
1.879 3.54
2.264 5.08
2.044 4.20
2.181 4.75

1.675 2.71
1.744 2.99
1.650 2.61
1.504 2.03
1.785 3.16
1.655 2.63
1.861 3.46
1.945 3.80

That adds up to about 53.5 ft/sec during the last two seconds
alone. The average vertical acceleration during those
last two seconds is over 1.8g.

When I add this up, I see 53.47 feet over two seconds (if those values
represent 8 Hz polling).

Correct me if I'm wrong, but that's an average of less than 1 g?

1st set of eight values yields a total of 30.08 feet / second ( < 1 g)

2nd set of eight values yields a total of 23.39 feet / second ( < 1 g)

Average feet per second is 26.735 ( < 1 g)

Are you adding Earth's gravity in the final? If so, how do you arrive at
53.47 feet per second average?

[edit on 17-11-2009 by turbofan]

[edit on 17-11-2009 by turbofan]

posted on Nov, 17 2009 @ 05:55 AM

Originally posted by R_Mackey

Originally posted by 911files
but don't worry about my credibility, it only seems to suffer with those who don't understand math and physics.

John, I'm not "worried" about your credibility. Another falsehood offered by you. I rather enjoy watching you delete your work and getting caught in check-mate.

With that said, your credibility does suffer among each and every one of these people. Many who you have contacted to consult since you haven't a clue regarding flight dynamics.

pilotsfor911truth.org...

The list grows.

In closing and since I have personally caught you in many lies, fabrications, spin and confirmed your lack of knowledge in flight dynamics, you are now on ignore.

[edit on 17-11-2009 by R_Mackey]

In other words, in truthspeak, R_Mackey concedes he lost. Good job, 911files and cesura.

[edit on 17-11-2009 by jthomas]

posted on Nov, 17 2009 @ 07:17 AM

Originally posted by turbofan
When I add this up, I see 53.47 feet over two seconds (if those values
represent 8 Hz polling).

You are confusing feet with ft/sec. Each number in the
right hand column represents a change in velocity (aka
acceleration) during 1/8 of a second. Adding them up
yields the acceleration (change in velocity) over the
full two seconds.

Originally posted by turbofan
Are you adding Earth's gravity in the final?

Yes. Earth's gravity was also included in the numbers
in the left column.

Originally posted by turbofan
If so, how do you arrive at 53.47 feet per second average?

Correctly: Subtract earth's gravity (1g) from each of the
numbers in the left column you quoted. To obtain the numbers
in the right column, multiply each of those differences by
1/8 second and 32.174 ft/sec^2/g.

Adding up the numbers in the right column is the final step
in the numerical integration that refutes the fake R_Mackey's
contention that there is minimal change in vertical velocity
over the final seconds.

ETA: The 53.47 ft/sec is not an average. It is the total
change in vertical velocity implied by the vertical
accelerations recorded during the last two seconds. To
obtain the average g-load: divide 53.47 ft/sec by 2 seconds,
divide by 32.174 ft/sec^2/g, and add 1g for earth's gravity.
That average exceeds 1.8g.

[edit on 17-11-2009 by cesura]

posted on Nov, 17 2009 @ 07:23 AM
The misleadingly pseudonymous R_Mackey has cranked up the
technobabble:

Originally posted by R_Mackey
s = 1/2at^2 is used to determine distance,

That equation applies only when the velocity is zero at the
beginning or end of the time interval, as in my review of
the PfT/CIT video. That is not true here.

Originally posted by R_Mackey
in this case, vertical distance from a linear descent angle.

Distance from an angle? Neither mathematics nor physics
recognize any such concept.

Originally posted by R_Mackey
Hence the term "minimal" as noted repeatedly throughout the last two pages.

Technobabble doesn't justify a non-sequitur.

Originally posted by R_Mackey
The PA data agrees with the Vertical accel based on vertical distance.

Repeating a falsehood doesn't make it so.

Originally posted by R_Mackey
Your theory and graphs are no where to be found in the data.

In my review, I described six different approaches that never
exceeded 2g. Each of those was sufficient by itself to refute
Rob Balsamo's claims that 11.2g or 10.14g were necessary. That
the real approach was more complex than the mathematically
idealized approaches of my review comes as no surprise.

Originally posted by R_Mackey
Evasion of the following noted for the 4th time.

As I noted the very first time, the fake R_Mackey's parameters
are inconsistent. Asking me to use R_Mackey's parameters is
like asking me to prove 2+2=5. The fake R_Mackey thinks my
refusal to prove the equivalent of 2+2=5 counts as evasion.

Speaking of credibility, the misleadingly pseudonymous R_Mackey
showed good judgment in one respect: He concealed his identity.

posted on Nov, 17 2009 @ 10:01 AM

Originally posted by cesura
That equation applies only when the velocity is zero at the
beginning or end of the time interval

The zero standard was set and explained in the above analysis (1G). The upper and lower boundary distance limits were also explained. Diagrams and data sets were also provided as well as linear regression analyzed by tezzjw. Perhaps you may want to review it again.

Originally posted by cesura
In my review, I described six different approaches that never
exceeded 2g. Each of those was sufficient by itself to refute
Rob Balsamo's claims that 11.2g or 10.14g were necessary.

11.2G calculations were in error and conceded a week after publish making the top part of your paper moot. This was explained to you in this thread but apparently needs to be explained again.

10.14G is based on topography, obstacles and data.

You have ignored obstacles and data in your paper making your paper inconsistent with analysis performed in "9/11: Attack On The Pentagon".

As explained repeatedly in this thread, "9/11: Attack On The Pentagon" already covers possible flight paths when removing all the variables. You missed such information due to the fact you have a habit of not thoroughly reviewing the information you analyze.

The calculator/formulas offered by the rant camp prior to "9/11: Attack On The Pentagon" do not take into account data, initial vertical velocities, or time (horizontal velocities) as explained in "9/11: Attack On The Pentagon".

It has been proven based on data and cross check the descent rate was near linear till the last second, conflicting with your analysis and once again proving your paper moot.

It has still not been proven PA is more than 150 feet in error, nor that .70M - .72M is above Mcrit to affect the static system as claimed by Ryan.

Will, we are now going in circles. Let us know when you can conclusively prove the above. Till then, enjoy the rant camp.

[edit on 17-11-2009 by R_Mackey]

top topics

12