Former George Bush Chief Economist Says 911 Was An Inside Job

a reply to: skyeagle409

The laws of physic prove that you don't even understand the mathematical figures you've posted.

On the contrary you are wrong. Where is your mathematics of laws of physic? Perhaps it is you that doesn't understand math since you have giving none.

Now, once again, do the math and explain to us as to why those steel columns remained standing within that bomb crater.

I never claimed bomb center, only you have.

I see you have not attempted to debunk the given math equation below with any mathematics of your own.

I rest my case.


Direct Evidence for Explosions: Flying Projectiles and
Widespread Impact Damage

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).

www.journalof911studies.com...

originally posted by: pteridine

originally posted by: wildb
a reply to: pteridine

Can you get to the point?

I am trying to get to the point, but I need you to understand what your looking at, when you do I can explain how this apply s to the wtc..

Pretend that you have to tell me what you think is so important. I can see what I am looking at. Is there some aspect that will somehow show that extra energy was involved?

Yes, you saw the top part of the building destroy the bottom part, and in the process the upper part destroy itself , thats what the video shows, do you agree ?

a reply to: Informer1958

I rest my case.

No you haven't. I am waiting for you to post the math to explain why the the steel columns of WTC 1 remained standing within that huge bomb crater. What it is, you are just posting math figures that you do not even understand and that won't work in regards to answering the question as to why a huge bomb failed to fling the steel columns of WTC 1 anywhere.

To sum that up, I am still waiting for you to post YOUR math that explains why explosives do not fling such huge steel structures 600 feet. How long are you going to keep me waiting?

Let me help you out. Post the math as to why five huge bombs failed to fling the support columns 600 feet in the following photo.

Photo: Building Survives Multiple bomb Strikes

originally posted by: wildb

originally posted by: pteridine

originally posted by: wildb
a reply to: pteridine

Can you get to the point?

I am trying to get to the point, but I need you to understand what your looking at, when you do I can explain how this apply s to the wtc..

Pretend that you have to tell me what you think is so important. I can see what I am looking at. Is there some aspect that will somehow show that extra energy was involved?

Yes, you saw the top part of the building destroy the bottom part, and in the process the upper part destroy itself , thats what the video shows, do you agree ?

Fine. Do you have a point you would like to make? Do you want to explain why extra energy is needed?

a reply to: Informer1958

I rest my case.

Please clarify the derivation of α.
Please show where entrained air is considered.

originally posted by: pteridine

originally posted by: wildb

originally posted by: pteridine

originally posted by: wildb
a reply to: pteridine

Can you get to the point?

I am trying to get to the point, but I need you to understand what your looking at, when you do I can explain how this apply s to the wtc..

Pretend that you have to tell me what you think is so important. I can see what I am looking at. Is there some aspect that will somehow show that extra energy was involved?

Yes, you saw the top part of the building destroy the bottom part, and in the process the upper part destroy itself , thats what the video shows, do you agree ?

Fine. Do you have a point you would like to make? Do you want to explain why extra energy is needed?

Yes I do, , but you will have to wait until tomorrow, then I will explain why this matters, a point I think you are still missing, good night for now..

a reply to: skyeagle409

No you haven't. I am waiting for you to post the math to explain why the the steel columns of WTC 1 remained standing within that huge bomb crater. What it is, you are just posting math figures that you do not even understand and that won't work in regards to answering the question as to why a huge bomb failed to fling the steel columns of WTC 1 anywhere.

To sum that up, I am still waiting for you to post YOUR math that explains why explosives do not fling such huge steel structures 600 feet. How long are you going to keep me waiting?

Let me help you out. Post the math as to why five huge bombs failed to fling the support columns 600 feet in the following photo.

You just cleverly demonstrated that you cannot debunk these equation from credible science but instead give every kind of excuses for not doing it.

The fact is, I challenged you first. How long do we all have to wait for your math equations to debunk credible science?

I don't believe you even understand the equations and since you don't, then I have to rest my case.

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).

www.journalof911studies.com...

a reply to: Informer1958

I don't believe you even understand the equations and since you don't, then I have to rest my case.

I understand it. Pretty much. The formulas involve the movement of the debris cloud (the "plume"). But am having trouble with a couple of things. Please clarify the derivation of α, in particular where the value for Cd came from. Please show where entrained air is considered because that would seem to be quite important in predicting the behavior of the dust plume.


I've asked a couple of times now, and since you seem to understand it well, I was hoping you can help.

a reply to: Phage

But am having trouble with a couple of things. Please clarify the derivation of α,

A is the video of the squib from figure 4.

The video frame in Figure 4 of the World Trade Center North Tower taken by KTLA
channel 5 news shows a "squib" -- a line of ejecting material from the tower -- right before
it collapsed. Such squib ejections are driven by massive overpressure inside the building
relative to the atmospheric pressure outside, and that overpressure is created by
explosions. A number of squibs were observed coming from all 3 of Buildings 1, 2, and 7 a
short second or 2 after each one started to collapse, and there are several websites that
show photograph of them on all 3 buildings. The one displayed as Figure 4 shows ejecting
material (bits of material large enough to have little air resistance) streaming out of the
North Tower, which has traveled a distance from the tower in the horizontal direction,
whereas the distance it has descended in the vertical direction because of gravitation pull is
small.

Figure 4. Photograph from a video of
the North Tower collapse by KTLA channel 5
news, which shows a streaming squib that has
traveled out over 70 feet from the tower with
very little descent [5].

Note the quantitative information that can be gathered from the ejection photograph in
Figure 4. We can estimate that, at the front end, the ejecting plume has apparently fallen
no more than roughly 3 feet (an estimate that might have up to a factor of 2 in error),
whereas the horizontal distance of the front from building is about 1/3 the width of the
North Tower, or about 70 feet. If we neglect air friction resistance over the length of the
streamer, from fall distance s=0.5gt2
, where g=32 feet/sec2
is the gravitational acceleration,
we estimate 0.43 sec as the time since the front end first ejected from the building. That
means that material in that squib is traveling horizontally at roughly 163 feet/sec, which
means the squibs are effectively "bullets" of bits of material produced by the explosions.

Since the distance fallen is quite small there may be a fairly large uncertainly in its
estimate. Allowing for an error of up to a factor of 2 in the measurement of the fall distance
s, the velocity could be down to 41% lower or up to 30% higher, so it could range from 100
feet/s to over 200 feet/s.

The mechanics of the motion can be examined to determine where the debris in the ejection
plume hits the ground. At such high velocities air resistance can actually be an important
factor in that distance, so it is incumbent on us to examine the mechanical equations of the

This will give you an idea to how the scientist where able to give us these equations.

a reply to: Informer1958

The squibs were low-veloctiy compressed air, which had nothing to do with explosives. I've noticed that you are unable to provide your math evidence that proved explosives flung a steel beam 600 feet.

Such squib ejections are driven by massive overpressure inside the building relative to the atmospheric pressure outside, and that overpressure is created by explosions. A number of squibs were observed coming from all 3 of Buildings 1, 2, and 7

Now, explain why squibs are seen in these photos, and take note that explosives were not used.

Photo 1: Squibs

Photo 2: Squibs

Let's take a look at the math you've provided.

Note the quantitative information that can be gathered from the ejection photograph in
Figure 4. We can estimate that, at the front end, the ejecting plume has apparently fallen
no more than roughly 3 feet (an estimate that might have up to a factor of 2 in error),
whereas the horizontal distance of the front from building is about 1/3 the width of the
North Tower, or about 70 feet. If we neglect air friction resistance over the length of the
streamer, from fall distance s=0.5gt2
, where g=32 feet/sec2
is the gravitational acceleration,
we estimate 0.43 sec as the time since the front end first ejected from the building. That
means that material in that squib is traveling horizontally at roughly 163 feet/sec, which
means the squibs are effectively "bullets" of bits of material produced by the explosions.

Since the distance fallen is quite small there may be a fairly large uncertainly in its
estimate. Allowing for an error of up to a factor of 2 in the measurement of the fall distance
s, the velocity could be down to 41% lower or up to 30% higher, so it could range from 100
feet/s to over 200 feet/s.

Queston: Where does it prove that explosives had flung a steel beam 600 feet?

Answer: There is nothing there that proves explosives were responsible. That is why I have stated that you did not understand the math you've posted, or should I say, you didn't understand the work that someone else had done.

I am still waiting for you to do your own math to explain why explosives had failed to fling the steel columns of WTC 1 anywhere in 1993.

Now, let's take a look here.

Squibs

During the pancake, the floors acted like a plunger in a Syringe. The towers skin and windows became the tube of the Syringe. The increased pressure blew the windows out as each massive acre of floor compressed air between them. It's said that the towers were about 95% air.

The air pushed though the core any way it could and the pressure built up. It forced its way out on lower floors wherever it could. According to the survivors of at least one tower, a hurricane wind blows through the staircase which is located in the core...

www.debunking911.com...

a reply to: Informer1958

A is the video of the squib from figure 4.

I didn't ask about A. I asked about the derivation of α, in particular where the value for Cd came from. α is described as representing the Rayleigh drag coefficient, an important factor in α being Cd. I'm not sure why the value for Cd which was selected, came from.

Also, entrained air becomes very important in this sort of calculation because it can act as sort of a "tail wind", reducing the air speed (and thus drag) of particles in a high velocity dust cloud. I can't see where that has been taken into account. Can you?

Do you understand my questions? Do you understand the formulas you presented?

During the pancake, the floors acted like a plunger in a Syringe. The towers skin and windows became the tube of the Syringe. The increased pressure blew the windows out as each massive acre of floor compressed air between them. It's said that the towers were about 95% air. The air pushed though the core any way it could and the pressure built up. It forced its way out on lower floors wherever it could. According to the survivors of at least one tower, a hurricane wind blows through the staircase which is located in the core...

More disinfo, .....

Now, explain why squibs are seen in these photos, and take note that explosives were not used. Photo 1: Squibs Photo 2: Squibs

Those are dust clouds, squibs are created by explosives... and there not anywhere close to what what was observed at the wtc..

originally posted by: wildb

During the pancake, the floors acted like a plunger in a Syringe. The towers skin and windows became the tube of the Syringe. The increased pressure blew the windows out as each massive acre of floor compressed air between them. It's said that the towers were about 95% air. The air pushed though the core any way it could and the pressure built up. It forced its way out on lower floors wherever it could. According to the survivors of at least one tower, a hurricane wind blows through the staircase which is located in the core...

More disinfo, .....

Actually, it is explained by simple physics. I thought that you were not claiming explosives, so why would you worry about misinterpretation of a physical process being called evidence for explosives?

a reply to: wildb

The laws of physics at work debunking false claims of 9/11 conspiracy theorist.

a reply to: wildb

Those are dust clouds, squibs are created by explosives...

False! As proof, there are no sounds of explosions that accompany those squibs of compressed air.

originally posted by: pteridine

originally posted by: wildb

During the pancake, the floors acted like a plunger in a Syringe. The towers skin and windows became the tube of the Syringe. The increased pressure blew the windows out as each massive acre of floor compressed air between them. It's said that the towers were about 95% air. The air pushed though the core any way it could and the pressure built up. It forced its way out on lower floors wherever it could. According to the survivors of at least one tower, a hurricane wind blows through the staircase which is located in the core...

More disinfo, .....

Actually, it is explained by simple physics. I thought that you were not claiming explosives, so why would you worry about misinterpretation of a physical process being called evidence for explosives?

I am claiming explosives of some kind, and you will know why when I continue from where I left off last night. See you later..

a reply to: wildb

I am claiming explosives of some kind, and you will know why when I continue from where I left off last night. See you later..

You have to provide video and audio evidence of explosions as the WTC buildings collapse. At no time are demolition explosions heard on video as WTC 1, WTC 2, and WTC 7 collapsed.

You also have to provide seismic data evidence of demolition explosions at ground zero, for which none exist. Nowhere in the seismic data are demolition explosions depicted.

You have to provide video and audio evidence of explosions as the WTC buildings collapse

Actually no I don't, even thou it exist .

a reply to: Phage

I didn't ask about A. I asked about the derivation of α, in particular where the value for Cd came from. α is described as representing the Rayleigh drag coefficient, an important factor in α being Cd. I'm not sure why the value for Cd which was selected, came from.

Apparently you do not understand the equations and your questions are really silly.

It's like you are asking: a+7=9. Your method of trying to discredit the equation is becoming Juvenal.