The Physics Of WTC 7

[KokomoJojo]

I have another (conspiracy?) theory.
Having wearied of having their silly asses handed to them daily, this board's "serious" CTs needed a break but didn't want to abandon the board to the norms. When KooKoo appeared (or was sent) they all took the opportunity to get some much needed R&R.
They will return soon, pretending their silliness hasn't been totally and completely exposed and begin posting it all over again.

At least E.L.C. tries. Koko is just a moron and will remain my ignore list. It makes it easier to read the thread when all his rants and gifs are hidden. He adds absolutely nothing to any thread.

at least?
he handed you your ass countless times and you are too fucking tarded to know it.
he is gone because of your tard ass.
he explained freefall to you page after page after page drawing picture a kindergartener can understand but nope not you, you are simply too fucking tarded and od'd on stupid pills. I would be surprised if he ever comes back.

No one is obligated to teach anyone that is od'd on stoopid pills anything.

I wont I give high school kid reviewed tests so you can show everyone how ignorant you are when you FAIL to answer it. He handed you your ass, as do I on a daily basis.

you and the rest of your tard club have no choice but to ignore me because you cant answer such laughably simple test questions but deem yourselves qualified to assess the WTC.

yo may as well get used to being owned

Hey ricky retardo, last time I checked buildings have more than one column dumb ass!

You are the one who is not applying anything to the "structure" one column ass twit not him!

yeh tell us how it applies to the whole building! NEVER FUCKING GONNA HAPPEN!





reality is a bitch

you are totally fucking dishonest when you try to represent this




like this

no one has time or a desire to chase after your incessant strawman arguments

but we would gladly welcome you and your 3 stooges back into high school physics if you correctly answer this little test question.




elc handed clowny his ass on a silver platter

bump

 

[daws101]

I have another (conspiracy?) theory.
Having wearied of having their silly asses handed to them daily, this board's "serious" CTs needed a break but didn't want to abandon the board to the norms. When KooKoo appeared (or was sent) they all took the opportunity to get some much needed R&R.
They will return soon, pretending their silliness hasn't been totally and completely exposed and begin posting it all over again.

bump

unnecessary repost!

 

[KokomoJojo]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

 

[daws101]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

if that's the case why the fuck are you still posting in it?......slow night in no planesville?

 

[Rat in the Hat]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

if that's the case why the fuck are you still posting in it?......slow night in no planesville?

Maybe he's stuck waiting in an airport for a plane that doesn't exist.

 

[Gamolon]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

Is that why E.L.C.'s profile says last activity was today at 2:53 AM?

What a coward.

 

[SAYIT]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

Is that why E.L.C.'s profile says last activity was today at 2:53 AM?

What a coward.

Virtually all CTs with even a minimum of self-respect - after all, how much can a CT have - eventually realize they have made fools of themselves and slither away.
KooKoo is a horse of a different color.

 

[KokomoJojo]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

Is that why E.L.C.'s profile says last activity was today at 2:53 AM?

What a coward.

I dont see any posts

So ELC is a coward because after handing you your ass 50 times he wont waste his time handing you your ass another 50 times.

Yup you are a fucking lunatic

 

[Gamolon]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

Is that why E.L.C.'s profile says last activity was today at 2:53 AM?

What a coward.

I dont see any posts

Exactly!

He has the balls to login and see what's going on, but not the balls to post what he said he was going to post.

 

[Gamolon]

once again the entire roofline does not drop then stop then go into freefall, prove you and your supporters have no clue how to interpret and chart data and come to a correct conclusion.

The graph is a representation of when the roofline descends. Both graphs show a slower than freefall descent in the very beginning.

How was that accomplished? How did the ENTIRE roofline start to descend? Was it explosives?

 

[KokomoJojo]

once again the entire roofline does not drop then stop then go into freefall, prove you and your supporters have no clue how to interpret and chart data and come to a correct conclusion.

The graph is a representation of when the roofline descends. Both graphs show a slower than freefall descent in the very beginning.

How was that accomplished? How did the ENTIRE roofline start to descend? Was it explosives?

dont invite me to your retard quagmire, ELC already explained it umteen times to you and handed you your ass, face it you are too tarded to get it.

 

[daws101]

this thread is dead, elc handed the clown his ass times 50 and said he didnt have time to deal with retards and he left.

Is that why E.L.C.'s profile says last activity was today at 2:53 AM?

What a coward.

I dont see any posts

So ELC is a coward because after handing you your ass 50 times he wont waste his time handing you your ass another 50 times.

Yup you are a fucking lunatic

you don't see much of anything outside your delusions.....so what's your point ?

 

[daws101]

once again the entire roofline does not drop then stop then go into freefall, prove you and your supporters have no clue how to interpret and chart data and come to a correct conclusion.

The graph is a representation of when the roofline descends. Both graphs show a slower than freefall descent in the very beginning.

How was that accomplished? How did the ENTIRE roofline start to descend? Was it explosives?

dont invite me to your retard quagmire, ELC already explained it umteen times to you and handed you your ass, face it you are too tarded to get it.

the award for the most obvious dodge goes to....

 

[Gamolon]

once again the entire roofline does not drop then stop then go into freefall, prove you and your supporters have no clue how to interpret and chart data and come to a correct conclusion.

The graph is a representation of when the roofline descends. Both graphs show a slower than freefall descent in the very beginning.

How was that accomplished? How did the ENTIRE roofline start to descend? Was it explosives?

dont invite me to your retard quagmire, ELC already explained it umteen times to you and handed you your ass, face it you are too tarded to get it.

The graphs represent the downward descent of the roofline jackass. There IS no other explanation. That's what the data points represent in Chandler's graph. You're just to stupid to understand it.

 

[Rockland]

Tags for this thread:

godzilla

how many suck puppets does 7-watt bulb need, anyway?

saturn being swallowed up by uranus

 

[KokomoJojo]

Must be having a hard time matching the three stage graph and visual evidence with his "pet explosive theory"...

Still waiting...

Makes me wonder how E.L.C. will explain the ENTIRE roofline dropping at the same time at the start of Stage 1 in the graph. What mechanism will he apply? Explosives maybe? But explosives would have created free fall at the onset of Stage 1, not at Stage 2.

Or maybe E.L.C. is insinuating that explosives on all 8 floors, on all columns, were simultaneously set off at the beginning of Stage 2 to start the free fall. But how does one explain the entire roofline/building coming down as a whole at the beginning of Stage 1?

What a conundrum for E.L.C.!

once again the entire roofline does not drop then stop then go into freefall, prove you and your supporters have no clue how to interpret and chart data and come to a correct conclusion.

someone is feeding us a bullshit story now arent you?





Why is it that you tards are wrong about every damn thing you post eh?

But I feel your pain!

nah not really LOL

the clownee tard is running damage control to cover his ignorance about freefall once again, still have his balls firmly in the vice. The joke is he still doesnt get it!

 

[Rat in the Hat]

 

[daws101]

Makes me wonder how E.L.C. will explain the ENTIRE roofline dropping at the same time at the start of Stage 1 in the graph. What mechanism will he apply? Explosives maybe? But explosives would have created free fall at the onset of Stage 1, not at Stage 2.

Or maybe E.L.C. is insinuating that explosives on all 8 floors, on all columns, were simultaneously set off at the beginning of Stage 2 to start the free fall. But how does one explain the entire roofline/building coming down as a whole at the beginning of Stage 1?

What a conundrum for E.L.C.!

once again the entire roofline does not drop then stop then go into freefall, prove you and your supporters have no clue how to interpret and chart data and come to a correct conclusion.

someone is feeding us a bullshit story now arent you?





Why is it that you tards are wrong about every damn thing you post eh?

But I feel your pain!

nah not really LOL

the clownee tard is running damage control to cover his ignorance about freefall once again, still have his balls firmly in the vice. The joke is he still doesnt get it!

uniform gravitational field without air resistance[edit]
This is the "textbook" case of the vertical motion of an object falling a small distance close to the surface of a planet. It is a good approximation in air as long as the force of gravity on the object is much greater than the force of air resistance, or equivalently the object's velocity is always much less than the terminal velocity (see below).
Free-fall
v(t)=-gt+v_{0}\,
y(t)=-\frac{1}{2}gt^2+v_{0}t+y_0
where
v_{0}\, is the initial velocity (m/s).
v(t)\, is the vertical velocity with respect to time (m/s).
y_0\, is the initial altitude (m).
y(t)\, is the altitude with respect to time (m).
t\, is time elapsed (s).
g\, is the acceleration due to gravity (9.81 m/s2 near the surface of the earth).
Uniform gravitational field with air resistance[edit]


Acceleration of a small meteoroid when entering the Earth's atmosphere at different initial velocities.
This case, which applies to skydivers, parachutists or any body of mass, m, and cross-sectional area, A, with Reynolds number well above the critical Reynolds number, so that the air resistance is proportional to the square of the fall velocity, v, has an equation of motion
m\frac{dv}{dt}=\frac{1}{2} \rho C_{\mathrm{D}} A v^2 - mg \, ,
where \rho is the air density and C_{\mathrm{D}} is the drag coefficient, assumed to be constant although in general it will depend on the Reynolds number.
Assuming an object falling from rest and no change in air density with altitude, the solution is:
v(t) = -v_{\infty} \tanh\left(\frac{gt}{v_\infty}\right),
where the terminal speed is given by
v_{\infty}=\sqrt{\frac{2mg}{\rho C_D A}} \, .
The object's speed versus time can be integrated over time to find the vertical position as a function of time:
y = y_0 - \frac{v_{\infty}^2}{g} \ln \cosh\left(\frac{gt}{v_\infty}\right).
Using the figure of 450 metres to reach terminal speed, this equation implies a free-fall time to terminal velocity of around 12 seconds. However, when the air density cannot be assumed to be constant, such as for objects or skydivers falling from high altitude, the equation of motion becomes much more difficult to solve analytically and a numerical simulation of the motion is usually necessary. The figure shows the forces acting on meteoroids falling through the Earth's upper atmosphere. HALO jumps, including Joe Kittinger's and Felix Baumgartner's record jumps (see below), and the planned Le Grand Saut, also belong in this category.[2]
Inverse-square law gravitational field[edit]
It can be said that two objects in space orbiting each other in the absence of other forces are in free fall around each other, e.g. that the Moon or an artificial satellite "falls around" the Earth, or a planet "falls around" the Sun. Assuming spherical objects means that the equation of motion is governed by Newton's Law of Universal Gravitation, with solutions to the gravitational two-body problem being elliptic orbits obeying Kepler's laws of planetary motion. This connection between falling objects close to the Earth and orbiting objects is best illustrated by the thought experiment Newton's cannonball.
The motion of two objects moving radially towards each other with no angular momentum can be considered a special case of an elliptical orbit of eccentricity e = 1 (radial elliptic trajectory). This allows one to compute the free-fall time for two point objects on a radial path. The solution of this equation of motion yields time as a function of separation:
t(y)= \sqrt{ \frac{ {y_0}^3 }{2\mu} } \left(\sqrt{\frac{y}{y_0}\left(1-\frac{y}{y_0}\right)} + \arccos{\sqrt{\frac{y}{y_0}}}
\right)
where
t is the time after the start of the fall
y is the distance between the centers of the bodies
y0 is the initial value of y
μ = G(m1 + m2) is the standard gravitational parameter.
Substituting y=0 we get the free-fall time.
The separation as a function of time is given by the inverse of the equation. The inverse is represented exactly by the analytic power series:
y( t ) = \sum_{n=1}^{ \infty }
\left[
\lim_{ r \to 0 } \left(
{\frac{ x^{ n }}{ n! }}
\frac{\mathrm{d}^{\,n-1}}{\mathrm{ d } r ^{\,n-1}} \left[
r^n \left( \frac{ 7 }{ 2 } ( \arcsin( \sqrt{ r } ) - \sqrt{ r - r^2 } )
\right)^{ - \frac{2}{3} n }
\right] \right)
\right]
Evaluating this yields:
y(t)=y_0 \left( x - \frac{1}{5} x^2 - \frac{3}{175}x^3
- \frac{23}{7875}x^4 - \frac{1894}{3931875}x^5 - \frac{3293}{21896875}x^6 - \frac{2418092}{62077640625}x^7 - \cdots \right) \
where
x = \left[\frac{3}{2} \left( \frac{\pi}{2}- t \sqrt{ \frac{2\mu}{ {y_0}^3 } } \right) \right]^{2/3}
For details of these solutions see "From Moon-fall to solutions under inverse square laws" by Foong, S. K., in European Journal of Physics, v29, 987-1003 (2008) and "Radial motion of Two mutually attracting particles", by Mungan, C. E., in The Physics Teacher, v47, 502-507 (2009).

Free fall - Wikipedia, the free encyclopedia

 

[E.L.C.]

Hah! Are you guys still at it? I knew there was something creepy about these guys. I wouldn't bother KokomoJojo, if they could have provided an exception to free fall they would have by now. Coming up with an exception to the laws of physics governing falling bodies is right up there with perpetual motion.... it can't be done.

By the way boys....

How's the Air Force treating you?

 

[E.L.C.]

uniform gravitational field without air resistance[edit]
This is the "textbook" case of the vertical motion of an object falling a small distance close to the surface of a planet. It is a good approximation in air as long as the force of gravity on the object is much greater than the force of air resistance, or equivalently the object's velocity is always much less than the terminal velocity (see below).
Free-fall
v(t)=-gt+v_{0}\,
y(t)=-\frac{1}{2}gt^2+v_{0}t+y_0
where
v_{0}\, is the initial velocity (m/s).
v(t)\, is the vertical velocity with respect to time (m/s).
y_0\, is the initial altitude (m).
y(t)\, is the altitude with respect to time (m).
t\, is time elapsed (s).
g\, is the acceleration due to gravity (9.81 m/s2 near the surface of the earth).
Uniform gravitational field with air resistance[edit]


Acceleration of a small meteoroid when entering the Earth's atmosphere at different initial velocities.
This case, which applies to skydivers, parachutists or any body of mass, m, and cross-sectional area, A, with Reynolds number well above the critical Reynolds number, so that the air resistance is proportional to the square of the fall velocity, v, has an equation of motion
m\frac{dv}{dt}=\frac{1}{2} \rho C_{\mathrm{D}} A v^2 - mg \, ,
where \rho is the air density and C_{\mathrm{D}} is the drag coefficient, assumed to be constant although in general it will depend on the Reynolds number.
Assuming an object falling from rest and no change in air density with altitude, the solution is:
v(t) = -v_{\infty} \tanh\left(\frac{gt}{v_\infty}\right),
where the terminal speed is given by
v_{\infty}=\sqrt{\frac{2mg}{\rho C_D A}} \, .
The object's speed versus time can be integrated over time to find the vertical position as a function of time:
y = y_0 - \frac{v_{\infty}^2}{g} \ln \cosh\left(\frac{gt}{v_\infty}\right).
Using the figure of 450 metres to reach terminal speed, this equation implies a free-fall time to terminal velocity of around 12 seconds. However, when the air density cannot be assumed to be constant, such as for objects or skydivers falling from high altitude, the equation of motion becomes much more difficult to solve analytically and a numerical simulation of the motion is usually necessary. The figure shows the forces acting on meteoroids falling through the Earth's upper atmosphere. HALO jumps, including Joe Kittinger's and Felix Baumgartner's record jumps (see below), and the planned Le Grand Saut, also belong in this category.[2]
Inverse-square law gravitational field[edit]
It can be said that two objects in space orbiting each other in the absence of other forces are in free fall around each other, e.g. that the Moon or an artificial satellite "falls around" the Earth, or a planet "falls around" the Sun. Assuming spherical objects means that the equation of motion is governed by Newton's Law of Universal Gravitation, with solutions to the gravitational two-body problem being elliptic orbits obeying Kepler's laws of planetary motion. This connection between falling objects close to the Earth and orbiting objects is best illustrated by the thought experiment Newton's cannonball.
The motion of two objects moving radially towards each other with no angular momentum can be considered a special case of an elliptical orbit of eccentricity e = 1 (radial elliptic trajectory). This allows one to compute the free-fall time for two point objects on a radial path. The solution of this equation of motion yields time as a function of separation:
t(y)= \sqrt{ \frac{ {y_0}^3 }{2\mu} } \left(\sqrt{\frac{y}{y_0}\left(1-\frac{y}{y_0}\right)} + \arccos{\sqrt{\frac{y}{y_0}}}
\right)
where
t is the time after the start of the fall
y is the distance between the centers of the bodies
y0 is the initial value of y
μ = G(m1 + m2) is the standard gravitational parameter.
Substituting y=0 we get the free-fall time.
The separation as a function of time is given by the inverse of the equation. The inverse is represented exactly by the analytic power series:
y( t ) = \sum_{n=1}^{ \infty }
\left[
\lim_{ r \to 0 } \left(
{\frac{ x^{ n }}{ n! }}
\frac{\mathrm{d}^{\,n-1}}{\mathrm{ d } r ^{\,n-1}} \left[
r^n \left( \frac{ 7 }{ 2 } ( \arcsin( \sqrt{ r } ) - \sqrt{ r - r^2 } )
\right)^{ - \frac{2}{3} n }
\right] \right)
\right]
Evaluating this yields:
y(t)=y_0 \left( x - \frac{1}{5} x^2 - \frac{3}{175}x^3
- \frac{23}{7875}x^4 - \frac{1894}{3931875}x^5 - \frac{3293}{21896875}x^6 - \frac{2418092}{62077640625}x^7 - \cdots \right) \
where
x = \left[\frac{3}{2} \left( \frac{\pi}{2}- t \sqrt{ \frac{2\mu}{ {y_0}^3 } } \right) \right]^{2/3}
For details of these solutions see "From Moon-fall to solutions under inverse square laws" by Foong, S. K., in European Journal of Physics, v29, 987-1003 (2008) and "Radial motion of Two mutually attracting particles", by Mungan, C. E., in The Physics Teacher, v47, 502-507 (2009).

Free fall - Wikipedia, the free encyclopedia

See what I mean? They're not really here to discuss anything. It's really more about....

Deception Distraction and Denial​

rightwinger already covered the pertinent formula for this scenario....

F= M x a

All you need to know. Thanks Dr Newton

You see that? It's not.... F = M x a, except under certain rare unknown conditions which can sometimes result in a bunch of multi-ton steel columns and other stuff temporarily behaving like air....

It's just F = M x a​