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originally posted by: vjr1113
originally posted by: wildb
originally posted by: vjr1113
a reply to: Informer1958
wtc was not reinforced to withstand fires and met the same fate as the other two wtc building. like i said before if you had read the thread, wtc is the hail mary of "truthers" and has already been explained.
still no proof of explosives. you only argument is that we can't disprove there were no explosives. a blatant fallacy if ive ever heard one.
So your saying it did not meet NYC Building code..
obviously the NYC code wasn't enough
yes and you proof is wrong and has been rejected. but if thats the best you can do, thats the best you can do. prove beyond a reasonable doubt there were explosives planted.
by the way loud noised does not equal proof of explosives, only loud noises.
squib ejections like Figure 4 with those terms included. Thus in the ejection the downward
acceleration is given by:
(1) a = dv/dt = g - α v2
.
where the Rayleigh drag coefficient for objects at high velocity v is:
(2) α = ρ ACd/2m
where ρ is the air density = 1.293 kg/m3
at 1 atmosphere pressure and 0o
C, A is the area at
the front of the moving material in the plume, m the material's mass, and Cd is a
dimensionless drag coefficient. Cd can be 0.25 for sleek automobiles, and will taken as 0.5 in
our calculations. Note that this can be rewritten in terms of the ratio of air density to the
density of the ejected material by designating l as the typical length of the ejected
projectile, as:
(3) α = (ρair/ρeject) Cd/2ml
A table below summarizes some typical values of α for various material parameters.
Table: Values of α for selected material parameters
material α l
0.001 5 in
cement, glass 0.003 1.5 in
0.01 0.5 in
0.001 1.7 in
steel 0.003 0.6 in
0.01 0.2 in
Solving (1) for v(t) by separation of variables yields the downward velocity vd and
downward distance y:
(4) vd(t)= (g/α )1/2 tanh [(g α )1/2t]
(5) y(t) = (1/α) ln cosh [(g α )1/2t]
So where does this squib material hit the ground? If we take y to be the height of the
ejection, we can solve the last equation for t, the time the material remains in the air.
Multiply that t by the horizontal velocity vh of the squib material, and we have the
horizontal distance x it travels. The equation of motion for the horizontal movement of the
material is:
(6) a = dv/dt = - α v2
which solves by separation of variables, yielding:
(7) vh(t)= vo/(1 + α vot)
(8) x(t) = (1/α) ln (1 + α vot)
where vo is the velocity of initial ejection from the tower. Taking t to be the time the
material remains in the air from (5) (solving for t after setting y=h) gives x(t) = xhit, the
distance the material travels away from the tower. Graphs of that distance xhit versus the α
for the material are shown in Figures 5 and 6 for ejections from about 1304 feet (400
meters) and 489 feet (150 m).
i really feel for you having to rewatch these videos. i couldn't do it. for what? to rebut a claim that has no traction? good luck to you
which solves by separation of variables, yielding:
(7) vh(t)= vo/(1 + α vot)
(8) x(t) = (1/α) ln (1 + α vot)
where vo is the velocity of initial ejection from the tower. Taking t to be the time the
material remains in the air from (5) (solving for t after setting y=h) gives x(t) = xhit, the
distance the material travels away from the tower. Graphs of that distance xhit versus the α
for the material are shown in Figures 5 and 6 for ejections from about 1304 feet (400
meters) and 489 feet (150 m).
Thanks! It seems that there are 9/11 conspiracy theorist who are very slow in their way of thinking. They haven't gotten the hint that demolition explosions make a lot of noise fo and that explosives do not fling huge steel beams horizontally during real demolition implosions.
The Truth Movement is a joke and has made a mockery of itself over the years.
For an example, I once posted a reference to a hoaxed video of WTC 7 and just days later, and unknowingly that I made such a reference, 9/11 conspiracy theorist used that same hoaxed video that I referred r and posted that the hoaxed video was proof that explosives took down WTC 7. They also failed to notice that the hoaxed video was a reversed imagery of WTC 7 or noticed the little notation added to the video that the video was a hoax.
Simply amazing!!
No you are sadly wrong. OS supporters who believe in their governments lies and are convinced they do no wrong, or would never be corrupt clearly have some serous issues.
originally posted by: vjr1113
have some more by real scientists
What is "simply amazing" you were just caught using edited videos in desperation to support your views of the demise of the WTC.
Let's remember, you are the person who has been caught passing disinformation from discredited, "AE911 Truth." Didn't you even catch the hint that former associates of "AE911 Truth" have attacked Richard Gage and his disorganization for its misdeeds and mishandling of its financial affairs?
the irony is real. mainstream science is wrong, but conspiracy "science" is right. we in the mainstream science community reject you poor science.
Still waiting for your evidence.
the irony is real. mainstream science is wrong,
but conspiracy "science" is right.
That won't fly because I have already pointed out that your figures do not apply to any explosives as the WTC buildings collapsed, which proves my point that you don't even understand what you've posted.