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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
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.
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.
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.
Originally posted by R_Mackey
Conclusion, Vertical Accel matches very well with PA.
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.
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.
That adds up to about 53.5 ft/sec during the last two seconds
alone.
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
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
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
Originally posted by R_Mackey
Originally posted by cesura
Originally posted by R_Mackey
You admit your formula doesn't take into account initial velocity.
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...
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?
Originally posted by R_Mackey
Originally posted by cesura
Originally posted by R_Mackey
You admit your formula doesn't take into account initial velocity.
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...
Originally posted by R_Mackey
The PA data agrees with the Vertical accel based on vertical distance.
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?
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.
Originally posted by 911files
What has Will got to do with you claiming the flight path was linear
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.
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.
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.
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.
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]
Originally posted by turbofan
When I add this up, I see 53.47 feet over two seconds (if those values
represent 8 Hz polling).
Originally posted by turbofan
Are you adding Earth's gravity in the final?
Originally posted by turbofan
If so, how do you arrive at 53.47 feet per second average?
Originally posted by R_Mackey
s = 1/2at^2 is used to determine distance,
Originally posted by R_Mackey
in this case, vertical distance from a linear descent angle.
Originally posted by R_Mackey
Hence the term "minimal" as noted repeatedly throughout the last two pages.
Originally posted by R_Mackey
The PA data agrees with the Vertical accel based on vertical distance.
Originally posted by R_Mackey
Your theory and graphs are no where to be found in the data.
Originally posted by R_Mackey
Evasion of the following noted for the 4th time.
Originally posted by cesura
That equation applies only when the velocity is zero at the
beginning or end of the time interval
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.