Was 911 an Inside Job

Was 911 an Inside Job

Created by inquiring minds on Jan 4, 2013

17,792 votes

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YES

No

Don't know

Possibly

Charlie Sheen

Lincoln, NE

#4419 May 12, 2013
Dr_Zorderz wrote:
What a bunch of bwunk!?!?!? LOL
I so agree!

I Totally debunk your claim that 20 pilots could not hit the WTC, prove me wrong!
Isabelle

Shreveport, LA

#4421 May 12, 2013
Charlie Sheen wrote:
...The columns have to shear off quickly enough, and the pancaking theory has the problem that the gravitational potential appears to be too low for all the energy sinks, but even this scenario does not appear to rule out the idea that debris could end up a few hundred feet away.
This last part makes your previous calculations worthless. The "sheering of columns" and spandrel plates is a massive energy sink, and there was lots of "sheering" going on in every direction, even in the free-falling upper block, where there shouldn't be any "sheering" at all!!
Charlie Sheen

Lincoln, NE

#4423 May 12, 2013
Isabelle wrote:
This last part makes your previous calculations worthless. The "sheering of columns" and spandrel plates is a massive energy sink, and there was lots of "sheering" going on in every direction, even in the free-falling upper block, where there shouldn't be any "sheering" at all!!
Sure, Show your math!

WOW, Pre-collapse sheering! Just add collapse and wind drag to the dangling aluminum.

http://www.911myths.com/assets/images/WTC_Fac...

PS: Could you twoofers make up your mind, did it fall in it's own footprint or not?

“ reality, what a concept”

Since: Nov 07

this one

#4424 May 12, 2013
Why "shouldn't" there be any sheering at all in the upper portions of the structure? They weren't falling intact, but collapsing themselves as they fell. The structure at the bottom couldn't support the weight now coming down on it, the upper part of the structure was itself collapsing, even as it collapsed onto the lower portion of the building. The sheering seen in the lower structure occurring in the upper portion of the building doesn't strike me as being as near as impossible as you claim it to be, especially without any sort of evidence to back your claim. At least you tried.

Since: Apr 13

Aït Melloul, Morocco

#4427 May 12, 2013
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Charlie Sheen

Lincoln, NE

#4428 May 12, 2013
Myrandy Cumgrove wrote:
Hey what about my reply, I said that most people do not want to know the truth
okay? I also said that is a true comment in itself....
meaning that most people do not want to know the truth, is also the truth.
the admins or some automatic bullshirt moderating topix does a sucky job of deleting things it thinks are spam that still can be things real people really post
Ahh, Perhaps your random trolling and senselessness generalizations left you confused.

Wrong Board.

http://www.topix.com/forum/topstories/TSBMT04...

Since: Apr 13

Aït Melloul, Morocco

#4430 May 12, 2013
Why?

Since: Apr 13

Aït Melloul, Morocco

#4432 May 12, 2013
OK freind
Zdenek Bazant

Logan, WV

#4435 May 12, 2013
Charlie Sheen wrote:
<quoted text>
WRONG!
In order to allow time for lateral motion, the exterior column(s) that hit WFC 3 were most probably from the upper half of WTC 1. A fall from 1,000 feet to 240 feet would take SQR(2*h/g)= around 6.9 seconds where h = 760 feet and g = 32.17 ft/s^2. In the horizontal plane, a uniform acceleration of 20 m/s^2 for the first second followed by negligible deceleration due to drag for the remaining 5.9 seconds would provide 10 +(5.9 * 20)= 128 metres = 420 feet displacement. At 1,000 feet the WTC 1 perimeter columns, per story, were comprised of:
two flanges of 1/2 x 13.5 x 144 inches each, totalling 1,944 ins^3
one outer web of 1/4 x 13 x 144 inches = 468 ins^3
one inner web of 1/4 x 15.75 x 92 inches = 362 ins^3
one spandrel plate of 3/8 x 40 x 52 inches = 780 ins^3
...totalling 3,554 ins^3 per floor or 10,662 ins^3 = 6.17 ft^3 for a three-floor section which at 490 lb/ft^3 is 3,023 lb (84 pounds per lineal foot) or 1,371 kg.(There is some uncertainty as to the flange thickness; it was known to be only 1/4" at the very highest floors.) The force require to produce an acceleration of 20 m/s^2 in an inertia mass of 1,371 kg is 20 * 1371 = 27,420 N = 6,165 lbf.
The cross-section presented to a wind, per floor, would be 40 x 52 = 2,080 ins^2 for the spandrel plate and 15.75 x 92 = 1,449 ins^2 for the inner web, totalling 3,529 ins^2 per floor or 10,587 ins^2 = 6.83 m^2 for a three-story section of exterior column.(So the required pressure is well under 1 psi.) From the drag equation of
d = Cd * A * r * 0.5 * v^2
we obtain
v = SQR(2 * d /(Cd * A * r))
where r = density of air ~ 1.2 kg/m^3 and assuming a relatively high drag coefficient Cd of 4 / pi ~ 1.27 for a flat plate and d = the previously calculated force of 27,420 N and A = 6.83 m^2 as calculated above. This places the required wind at 72.6 m/s = 162 mph for one second duration. Actual windspeed on the day was up to 10 mph on the ground and up to 20 mph at higher altitude.
Suppose we imagine the collapse initiating at 1,200 feet, and proceeding as per the "pancaking" theory to 1,000 feet. After freely falling 200 feet, the terminal velocity would be SQR(2 * 200 * 32.17 ft/s^2)= 113.4 fps = 77.3 mph. In this theory, there is a small delay due to resistance of the intact building below, but the falling upper section smashes its way through each floor in about 0.1 seconds at the 1,000 feet level. The volume of air per floor is approximately 12 * 200 * 200 feet = 480,000 ft^3. Some will go down, but if the total was forced out through a perimeter of 800 feet by an average height of 6 feet which is an exiting area of 4,800 ft^2, it would (continuing outward) extend for some 100 feet at the end of the 0.1 seconds which is a velocity of 1,000 fps or 682 mph.
Let's set the exiting gases velocity at just 700 fps = 213 m/s, in which case the force acting on the exterior column for 0.1 seconds is given by:
d = Cd * A * r * 0.5 * v^2
= 1.27 * 6.83 * 1.2 * 0.5 * 213^2 ~ 236,000 N
to produce an acceleration of F / m = 236,000 N / 1,371 kg = 172 m/s^2. After 0.1 seconds the velocity of the steel is 17.2 m/s = 38.5 mph, and the horizontal displacement is 0.86 metres. Following another 6.8 seconds at 17.2 m/s the total distance travelled horizontally is 0.86 plus 6.8 * 17.2 ~ 118 metres = 387 feet. The columns have to shear off quickly enough, and the pancaking theory has the problem that the gravitational potential appears to be too low for all the energy sinks, but even this scenario does not appear to rule out the idea that debris could end up a few hundred feet away.
dazzle them with math (even though it makes no sense), that's why they hired me to support the official fairytale.

“DECEPTION = MOST POWERFUL ”

Since: Jul 11

POLITICAL FORCE ON THE PLANET

#4439 May 12, 2013
D dt (Z z(t) 0 &#956;(S)s&#729;(S)dS ) &#8722; g Z z(t) 0 &#956;(S)dS = &#8722; Fc(z, z&#729;)(1) 3 where t = time, z = vertical (Lagrangian) coordinate = distance of the current crushing front from the initial position of the tower top; the superior dots denote time derivatives; &#956;(S)= initial specific mass of tower (mass of a story divided by its height) at point of initial coordinate S; s&#729;(S)= velocity of material point with initial coordinate S. It will suffice to consider the velocity, as well as the momentum density, to be distributed throughout the compacted layer linearly. With these approximations, the crush-down differential equation of motion becomes: d dt ( m0[1 &#8722; (z)] dz
dt + &#956;cl 2 [2 &#8722; (z)] dz dt ) &#8722; m(z)g = &#8722;Fc(z, z&#729;)(crush-down)(2) while the crush-up differential equation of motion has the same form as Eq. 17 of Bazant and Verdure (2007): m(y)( d dt " [1 &#8722; (y)] dy dt #+ g )= Fc(y, y&#729;)(crush-up)(3) Here Eq.(2) represents a refinement of Eq. 12 of Bazant and Verdure (2007), while Eq.(3) is identical to their Eq. 17 because the compacted layer is stationary during crush-up. Furthermore, l = height of compacted layer B, &#956;c = specific mass of compacted layer B per unit height, which is considered to be constant and equal to the maximum possible density of compacted debris; m(z)= cumulative mass of the tower above level z of the crushing front (m(z)= m0 + &#956;cl); and Fc = resisting force = energy dissipation per unit height; Fc(z, z&#729;)= Fb + Fs + Fa + Fe, Fb = Wd/(1 &#8722; )h (4) where Wd(z)= total energy dissipation up to level z, which was assumed by Bazant and Verdure (2007) to consist only of energy Fb (per unit height) consumed by buckling of steel columns. In calculations, the large fluctuations of Fb as a function of z or y (evident in Figs. 3 and 4 of Ba&#711;zant and Verdure, 2007) are neglected, i.e., Fb is smoothly homogenized. As a refinement of previous analysis, we introduce here a generalization in which we add energy Fs (per unit height) consumed by continuated bifurcation of concrete floor slabs, energy Fa required to expel air from the tower, and energy Fe required to accelerate the mass of dust and larger fragments
Ejected from the tower during the impact of upper part; Furthermore, in contrast to previous studies, the compaction ratio will not be assumed as a constant but will be more accurately calculated as (z)=(1&#8722; out)&#9 56;(z)/&#956;c, and out = mass shedding fraction = fraction of mass that escapes outside tower perimeter before the end of crush-down (not afterwards). Note that Eq.(2) may be rewritten as Plethora of Prolific Paid Professionals Parroting Puppetmaster Program Purporting Prepared Pronouncement of Perfunctorily Planted Plane Pieces and Parts Positively Proven Propaganda Paints a Picture of [m0(1&#8722; )+&#956;c l(1&#8722;0.5 )]z¨&#87 22;mg = &#8722;Fm& #8722;Fc, Fm =[m0(1&#8722; )+&#956; cl(1&#8722;0.5 )]&#729 ; z&#729; = &# 956;¯z& #729;2 (5) where Fm = force required to accelerate to velocity z&#729; the stationary mass accreting at the crushing front, and ¯&#956; = d[m0(1 &#8722; )+ &# 956;cl(1 &#8722; 0.5 )]/dz = part of the impacted mass per unit height that remains within the tower perimeter. This force causes a greater difference from free fall than do forces Fb, Fs, Fa and Fe combined. which were then integrated numerically with high accuracy using the Runge-Kutta algorithm (note that, for the idealized special case of = Fc = out = 0 and constant &#956; = dm/dz, Eq.(2) reduces to the differential equation (zz&#729;)&#729; = gz, which was formulated and solved by finite differences by Kausel, 2001). As the initial conditions, it is considered that the crushing front initiates at the 96th story in the North Tower, and at the 81st story in the South Tower (NIST 2005).

Yeah dat's wat i'm talking bout

“DECEPTION = MOST POWERFUL ”

Since: Jul 11

POLITICAL FORCE ON THE PLANET

#4440 May 12, 2013
D dt (Z z(t) 0 &#956;(S)s&#729;(S)dS ) &#8722; g Z z(t) 0 &#956;(S)dS = &#8722; Fc(z, z&#729;)(1) 3 where t = time, z = vertical (Lagrangian) coordinate = distance of the current crushing front from the initial position of the tower top; the superior dots denote time derivatives; &#956;(S)= initial specific mass of tower (mass of a story divided by its height) at point of initial coordinate S; s&#729;(S)= velocity of material point with initial coordinate S. It will suffice to consider the velocity, as well as the momentum density, to be distributed throughout the compacted layer linearly. With these approximations, the crush-down differential equation of motion becomes: d dt ( m0[1 &#8722; (z)] dz
dt + &#956;cl 2 [2 &#8722; (z)] dz dt ) &#8722; m(z)g = &#8722;Fc(z, z&#729;)(crush-down)(2) while the crush-up differential equation of motion has the same form as Eq. 17 of Bazant and Verdure (2007): m(y)( d dt " [1 &#8722; (y)] dy dt #+ g )= Fc(y, y&#729;)(crush-up)(3) Here Eq.(2) represents a refinement of Eq. 12 of Bazant and Verdure (2007), while Eq.(3) is identical to their Eq. 17 because the compacted layer is stationary during crush-up. Furthermore, l = height of compacted layer B, &#956;c = specific mass of compacted layer B per unit height, which is considered to be constant and equal to the maximum possible density of compacted debris; m(z)= cumulative mass of the tower above level z of the crushing front (m(z)= m0 + &#956;cl); and Fc = resisting force = energy dissipation per unit height; Fc(z, z&#729;)= Fb + Fs + Fa + Fe, Fb = Wd/(1 &#8722; )h (4) where Wd(z)= total energy dissipation up to level z, which was assumed by Bazant and Verdure (2007) to consist only of energy Fb (per unit height) consumed by buckling of steel columns. In calculations, the large fluctuations of Fb as a function of z or y (evident in Figs. 3 and 4 of Ba&#711;zant and Verdure, 2007) are neglected, i.e., Fb is smoothly homogenized. As a refinement of previous analysis, we introduce here a generalization in which we add energy Fs (per unit height) consumed by continuated bifurcation of concrete floor slabs, energy Fa required to expel air from the tower, and energy Fe required to accelerate the mass of dust and larger fragments
Ejected from the tower during the impact of upper part; Furthermore, in contrast to previous studies, the compaction ratio will not be assumed as a constant but will be more accurately calculated as (z)=(1&#8722; out)&#9 56;(z)/&#956;c, and out = mass shedding fraction = fraction of mass that escapes outside tower perimeter before the end of crush-down (not afterwards). Note that Eq.(2) may be rewritten as Plethora of Prolific Paid Professionals Parroting Puppetmaster Program Purporting Prepared Pronouncement of Perfunctorily Planted Plane Pieces and Parts Positively Proven Propaganda Paints a Picture of [m0(1&#8722; )+&#956;c l(1&#8722;0.5 )]z¨&#87 22;mg = &#8722;Fm& #8722;Fc, Fm =[m0(1&#8722; )+&#956; cl(1&#8722;0.5 )]&#729 ; z&#729; = &# 956;¯z& #729;2 (5) where Fm = force required to accelerate to velocity z&#729; the stationary mass accreting at the crushing front, and ¯&#956; = d[m0(1 &#8722; )+ &# 956;cl(1 &#8722; 0.5 )]/dz. This force causes a greater difference from free fall than do forces Fb, Fs, Fa and Fe combined. Upon setting v = z&#729;, Eq.(2) or (5) was converted to a system of two nonlinear first-order differential equations for unknowns v(t) and z(t), which were then integrated numerically with high accuracy using the Runge-Kutta algorithm (note that, for the idealized special case of = Fc = out = 0 and constant &#956; = dm/dz, Eq.(2) reduces to the differential equation (zz&#729;)&#729; = gz, which was formulated and solved by finite differences by Kausel, 2001). As the initial conditions, it is considered that the crushing front initiates at the 96th story in the North Tower, and at the 81st story in the South Tower (NIST 2005).
Dat's wat I'm talking bout

“DECEPTION = MOST POWERFUL ”

Since: Jul 11

POLITICAL FORCE ON THE PLANET

#4441 May 12, 2013
Dat's why dose buildings fell down.
LOL
.
It wern't magik atall !?!?!? LOL

Since: Jan 11

Westbury, NY

#4442 May 12, 2013
maybe you can explian how not a single explosive detonation registered on any seismograph ..... or should we just ignore that little fact
Brian wrote:
<quoted text>
Actually there's plenty of evidence for explosives. Demolition light flashes are visible before and during the collapse of both towers. Separate pockets of smoke are visible coming out of both towers as they're collapsing. These pockets of smoke are actually exiting the buildings below where the collapse is occurring. Also, explain something to me. If there were no explosives in those buildings, and only the floors impacted by the airplanes were on fire, why is there smoke coming out of the buildings as they are falling? If it was only a collapsing buildings, you would have no smoke coming out of the buildings.

Since: Jan 11

Westbury, NY

#4443 May 12, 2013
or with another Sock Puppet with a hidden location
Myrandy Cumgrove wrote:
<quoted text>
If you can't dazzle them with brilliance, baffle them with bullshit!

Since: Jan 11

Westbury, NY

#4444 May 12, 2013
everybody with half a brain knows the buildings fell at free fall Speed
Brian wrote:
You don't have to be an Einstein to know that 2 steel buildings can't physically collapse at free fall speed and leave behind a city covered in smoke if there's no explosions to produce that much smoke. People are so blind!

Since: May 10

YOUR MOM'S HOUSE

#4445 May 12, 2013
And with your Face Please do not reproduce. Really... I mean it...
Myrandy Cumgrove wrote:
To the idiots who still do not know the truth.
Please do not reproduce. Really... I mean it...

“DECEPTION = MOST POWERFUL ”

Since: Jul 11

POLITICAL FORCE ON THE PLANET

#4446 May 12, 2013
Timestin wrote:
maybe you can explian how not a single explosive detonation registered on any seismograph ..... or should we just ignore that little fact
<quoted text>
Ignore it cause it ain't true!?!?!?!?!?
.
There was lots of seismograph evidence. The govie interprets the data one way, unbiased scientists interpret it another.
.
Snot Magik either Huh Eh !

Since: Aug 11

Location hidden

#4447 May 12, 2013
Brian wrote:
<quoted text>
Actually there's plenty of evidence for explosives. Demolition light flashes are visible before and during the collapse of both towers. Separate pockets of smoke are visible coming out of both towers as they're collapsing. These pockets of smoke are actually exiting the buildings below where the collapse is occurring. Also, explain something to me. If there were no explosives in those buildings, and only the floors impacted by the airplanes were on fire, why is there smoke coming out of the buildings as they are falling? If it was only a collapsing buildings, you would have no smoke coming out of the buildings.
Nonsense. It would be impossible to do such a demolition. The chunks are expelled by air being squeezed out by the pancaking floors.
pancake farts

Logan, WV

#4448 May 12, 2013
WasteWater wrote:
<quoted text>
Nonsense. It would be impossible to do such a demolition. The chunks are expelled by air being squeezed out by the pancaking floors.
I would have never imagined that pancakes could fart chunks

“ reality, what a concept”

Since: Nov 07

this one

#4449 May 12, 2013
pancake farts wrote:
I would have never imagined that pancakes could fart chunks
Oh come on, you're telling us that you're the only one who hasn't heard the story of your unplanned birth?

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