Science > Physics > Video: Professor of Physics Phd at Cal Tech says: 911 Inside Job
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Science > Physics |
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| Date: |
24 Apr 2007 09:13:23 PM |
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Video: Professor of Physics Phd at Cal Tech says: 911 Inside Job |
Cal Tech is the ELITE of ELITE in physics.
If Feynman were alive, he would point his finger straight at the 911
criminal operators, the yank bastards themselves .......
http://www.911blogger.com/node/8101
No self-respecting scientist should keep his mouth shut. Its a
fundamental challenge to the method of science, a detective work most
demanding of INTELLECTUAL HONESTY.
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| User: "Bill Habr" |
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| Title: Re: Video: Professor of Physics Phd at Cal Tech says: 911 Inside Job |
25 Apr 2007 07:33:19 PM |
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<stj911@rock.com> wrote in message
news:1177467203.719625.93920@u32g2000prd.googlegroups.com...
Cal Tech is the ELITE of ELITE in physics.
If Feynman were alive, he would point his finger straight at the 911
criminal operators, the yank bastards themselves .......
http://www.911blogger.com/node/8101
No self-respecting scientist should keep his mouth shut. Its a
fundamental challenge to the method of science, a detective work most
demanding of INTELLECTUAL HONESTY.
Isn't this the guy who has more conspiracy theories than Carter has pills?
Whitewater, Vince Foster, moon landing hoax one week - we found a UFO on the moon the
next, Oklahoma City bombing, a new conspiracy every day ad nauseum?
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| User: "War Office" |
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| Title: Re: Video: Professor of Physics Phd at Cal Tech says: 911 Inside Job |
28 Apr 2007 11:35:59 AM |
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On 25 abr, 21:33, "Bill Habr" <billh...@sbcglobal.net> wrote:
<stj...@rock.com> wrote in message
news:1177467203.719625.93920@u32g2000prd.googlegroups.com...
Cal Tech is the ELITE of ELITE in physics.
If Feynman were alive, he would point his finger straight at the 911
criminal operators, the yank bastards themselves .......
http://www.911blogger.com/node/8101
No self-respecting scientist should keep his mouth shut. Its a
fundamental challenge to the method of science, a detective work most
demanding of INTELLECTUAL HONESTY.
Isn't this the guy who has more conspiracy theories than Carter has pills?
Whitewater, Vince Foster, moon landing hoax one week - we found a UFO on the moon the
next, Oklahoma City bombing, a new conspiracy every day ad nauseum?
Why can't any of you just discuss the fact that free-fall collapse of
this building contradicts the laws of physics (God's law whom gave you
and God's children whom's mass murder in Iraq you have allowed)?
Why do you all have to avoid the topic and rather go on a chracter
assassination which is totally abhorent to scientific method?
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| User: "quasi" |
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| Title: the pancake collapse phenomenon |
28 Apr 2007 02:45:57 PM |
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I am going to pose a simple mathematical problem.
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
Ok, so here's the problem ...
Assumptions:
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
(2) The slabs can be regarded as point particles of equal weight.
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
Question:
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
In particular, what are the answers for k=1 and k=50?
quasi
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| User: "Androcles" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 02:44:02 PM |
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"quasi" <quasi@null.set> wrote in message =
news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
=20
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
Controversy? Just because some idiot doesn't understand physics=20
doesn't mean there is a controversy. The only "controversy" is=20
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and=20
therefore moot.
=20
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
=20
Ok, so here's the problem ...
=20
Assumptions:
=20
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
=20
(2) The slabs can be regarded as point particles of equal weight.=20
=20
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.=20
=20
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
=20
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
=20
Question:
=20
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
=20
In particular, what are the answers for k=3D1 and k=3D50?
=20
quasi
Insufficient data. We do not know if the topmost floor came to=20
rest before breaking the cable suspending the floor below it.
If it did then the top two floors fall together from rest, but will
not come to rest when they hit the next. If it did not, then the=20
situation is even more difficult.
Think of it this way. If a golfer swings his driver and it meets
the ball on the tee, the ball cannot instantaneously gain velocity,
it has inertia and must first accelerate; it will be compressed.=20
The driver will also bend but the swing carries through.=20
The problem is ill-defined at (4), we need to know if it takes=20
one or two floors to snap the cables of the third from the top.=20
If two floors are needed, then the free fall is from rest at the=20
third from top, and ultimately from the last (lowest) suspended=20
floor with the strongest cables.=20
In other words the collapse has the appearance of a stop-and-go but=20
roughly constant velocity, the final stage of which is free fall
from the first floor up, it taking all 100 floors to snap the cable.
If the cables all have equal snapping force you'll have a retarded
acceleration.
Hence the relative strengths of the cables has to be defined.=20
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 04:26:58 PM |
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On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
Controversy? Just because some idiot doesn't understand physics
doesn't mean there is a controversy. The only "controversy" is
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and
therefore moot.
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
Ok, so here's the problem ...
Assumptions:
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
(2) The slabs can be regarded as point particles of equal weight.
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
Question:
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
In particular, what are the answers for k=1 and k=50?
quasi
Insufficient data. We do not know if the topmost floor came to
rest before breaking the cable suspending the floor below it.
Assume that it does not came to rest.
Instead, assume that the impacted floor falls together with the floors
that hit it, at an initial velocity determined by conservation of
momentum.
If it did then the top two floors fall together from rest, but will
not come to rest when they hit the next. If it did not, then the
situation is even more difficult.
Not that difficult. I'm confident that someone will provide a
numerical answer for k=1 and k=50.
Think of it this way. If a golfer swings his driver and it meets
the ball on the tee, the ball cannot instantaneously gain velocity,
it has inertia and must first accelerate; it will be compressed.
The driver will also bend but the swing carries through.
Once again, we are not talking about a real-world problem. Just use
the given assumptions together with the clarification given above that
the impacting floors don't actally stop on impact (but conservation of
momentum does imply that the velocity is instantly reduced).
The problem is ill-defined at (4), we need to know if it takes
one or two floors to snap the cables of the third from the top.
Each floor is independently cabled to the sky. It was specified that
even just one floor falling would sufficient to instantly break the
cable holding up the floor below. Obviously, multiple floors falling
together would then be more than enough to break the cables for lower
floors on impact.
If two floors are needed, then the free fall is from rest at the
third from top, and ultimately from the last (lowest) suspended
floor with the strongest cables.
In other words the collapse has the appearance of a stop-and-go but
roughly constant velocity, the final stage of which is free fall
from the first floor up, it taking all 100 floors to snap the cable.
If the cables all have equal snapping force you'll have a retarded
acceleration.
Hence the relative strengths of the cables has to be defined.
As mentioned above, all floors are cabled independently.
You can visualize the collapse as follows. Let's take k=50.
At time t=0, the top 50 floors begin to fall, independently of each
other. The first actual impact is when floor 50 hits floor 49. On
impact, those 2 floors fall together with an initial velocity that is
less than that of higher floors due to conservation of momentum. Thus
the next impact will not be floors 50,49 with floor 48 but rather it
will be floor 51 hitting the slower moving 50,49 combo. The 3 floors
51,50,49 then become a unit. The initial velocity of the new 51,50,49
combo will be greater than the velocity of the 50-49 combo just before
impact, but still less than that of the floors above (which are still
in free fall).
quasi
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| User: "Androcles" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 05:53:57 PM |
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"quasi" <quasi@null.set> wrote in message =
news:qpc733tji40a7iuflmo9o9frrpp05n34sd@4ax.com...
On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
=20
"quasi" <quasi@null.set> wrote in message =
news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
=20
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate =
of
free fall.
Controversy? Just because some idiot doesn't understand physics=20
doesn't mean there is a controversy. The only "controversy" is=20
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and=20
therefore moot.
=20
The question is not intended to model the actual situation. Instead, =
I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
=20
Ok, so here's the problem ...
=20
Assumptions:
=20
(1) 100 slabs of concrete (floors) are suspended in midair, =
prevented
from falling to the ground by imaginary cables (attached to the =
sky).
=20
(2) The slabs can be regarded as point particles of equal weight.=20
=20
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. =
Thus,
the distance between successive floors is 1.=20
=20
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the =
floors
that hit it.
=20
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
=20
Question:
=20
If the cables to the top k floors are cut, how long will it take for =
a
complete collapse? Express the answer as a ratio to the time for =
free
fall of just the top floor.
=20
In particular, what are the answers for k=3D1 and k=3D50?
=20
quasi
Insufficient data. We do not know if the topmost floor came to=20
rest before breaking the cable suspending the floor below it.
=20
Assume that it does not came to rest.
It does at least slow. By how much?
=20
Instead, assume that the impacted floor falls together with the floors
that hit it, at an initial velocity determined by conservation of
momentum.
Conservation of momentum requires it to come to rest momentarily.
http://www.walter-fendt.de/ph11e/ncradle.htm
Now tie down each ball so that the impact just breaks the=20
the thread tying the stationary ball in place, and the result
is no further motion. If the thread is too strong, the falling
ball bounces back. Too weak, and the stationary ball moves.
=20
If it did then the top two floors fall together from rest, but will
not come to rest when they hit the next. If it did not, then the=20
situation is even more difficult.
=20
Not that difficult. I'm confident that someone will provide a
numerical answer for k=3D1 and k=3D50.
Doesn't mean they'll be right.
=20
Think of it this way. If a golfer swings his driver and it meets
the ball on the tee, the ball cannot instantaneously gain velocity,
it has inertia and must first accelerate; it will be compressed.=20
The driver will also bend but the swing carries through.=20
=20
Once again, we are not talking about a real-world problem. Just use
the given assumptions together with the clarification given above that
the impacting floors don't actally stop on impact (but conservation of
momentum does imply that the velocity is instantly reduced).
Ok, instantly reduced by how much? The top floor momentarily comes
to rest, then we start again with two floors falling together from rest. =
*All* the kinetic energy of the first drop was expended in breaking the=20
cable suspending the floor below.
This is in disagreement with your statement above,
"Assume that it does not came to rest." and "impact of the falling =
floor=20
on the floor below is sufficient to instantly cut the cable holding up =
the
impacted floor."
My assumption is that the cable can just withstand the weight of two
floors, obviously, and your assumption is that if I place a feather on
a floor it will be the straw that broke the camel's back.
Your assumptions conflict with mine.=20
All I'm asking for is definition and clarification of YOUR assumptions,=20
obviously.=20
Obviously I have insufficient data to solve your problem.
The problem is ill-defined at (4), we need to know if it takes=20
one or two floors to snap the cables of the third from the top.=20
=20
Each floor is independently cabled to the sky.
I understand that, but it was not specified that all cables have
the same or different strength.
It was specified that
even just one floor falling would sufficient to instantly break the
cable holding up the floor below.=20
I understand that, but it was not specified that all cables have
the same or different strength.
Obviously, multiple floors falling
together would then be more than enough to break the cables for lower
floors on impact.
Obvious to you, but not obvious to me that that is what you assumed
I would assume. I have to have clarification.
The problem is ill-defined at (4), we need to know if it takes one or =
two=20
floors to snap the cables of the third from the top, which may of =
greater
tensile strength. After all, a real building can be constructed with =
lighter
materials at the top. Just look at the Empire State, it tapers.
=20
If two floors are needed, then the free fall is from rest at the=20
third from top, and ultimately from the last (lowest) suspended=20
floor with the strongest cables.=20
In other words the collapse has the appearance of a stop-and-go but=20
roughly constant velocity, the final stage of which is free fall
from the first floor up, it taking all 100 floors to snap the cable.
If the cables all have equal snapping force you'll have a retarded
acceleration.
Hence the relative strengths of the cables has to be defined.=20
=20
As mentioned above, all floors are cabled independently.
Irrelevant, we need to know if all cables have the same breaking strain.
Even if they are the same diameter and material, the longer ones
will stretch more than the shorter, there is no such animal as a=20
rigid rod. Even a glass rod will bend without breaking, that's what
fibre glass is, and you said "cable". All cables will stretch.
=20
You can visualize the collapse as follows. Let's take k=3D50.
I don't have any trouble visualizing, it is data I need and asked for.
At time t=3D0, the top 50 floors begin to fall, independently of each
other. The first actual impact is when floor 50 hits floor 49. On
impact, those 2 floors fall together with an initial velocity that is
less than that of higher floors due to conservation of momentum.=20
Obviously (your word) floor 49 falls from rest.=20
Floor 50 also falls from rest a second time, it came to rest while =
breaking=20
the cables of floor 49.
My question is: what is the breaking strain of the cables supporting
floor 48?
Obviously (your word) the longer a bungee is the greater the stretch
before it breaks. You won't fall far with a short bungee.
Thus
Oh, I see you've snipped. ***** you too, you ignorant *****.
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 08:11:54 PM |
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On Sat, 28 Apr 2007 22:53:57 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:qpc733tji40a7iuflmo9o9frrpp05n34sd@4ax.com...
On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
Controversy? Just because some idiot doesn't understand physics
doesn't mean there is a controversy. The only "controversy" is
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and
therefore moot.
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
Ok, so here's the problem ...
Assumptions:
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
(2) The slabs can be regarded as point particles of equal weight.
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
Question:
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
In particular, what are the answers for k=1 and k=50?
quasi
Insufficient data. We do not know if the topmost floor came to
rest before breaking the cable suspending the floor below it.
Assume that it does not came to rest.
It does at least slow. By how much?
By exactly as much as is required by conservation of momentum.
Instead, assume that the impacted floor falls together with the floors
that hit it, at an initial velocity determined by conservation of
momentum.
Conservation of momentum requires it to come to rest momentarily.
For a time interval of 0, in this problem.
In other words, assume no bounce back, no delay. The floors just
coalesce and move as one. That's the assumption of the model.
http://www.walter-fendt.de/ph11e/ncradle.htm
Now tie down each ball so that the impact just breaks the
the thread tying the stationary ball in place, and the result
is no further motion. If the thread is too strong, the falling
ball bounces back. Too weak, and the stationary ball moves.
Assume there is no resistance from the cables. Regard the cables as
exactly strong enough to hold up the floor, no stronger.
If it did then the top two floors fall together from rest, but will
not come to rest when they hit the next. If it did not, then the
situation is even more difficult.
Not that difficult. I'm confident that someone will provide a
numerical answer for k=1 and k=50.
Doesn't mean they'll be right.
Of course, so what?
Let's wait for some proposed solutions before attacking them.
Think of it this way. If a golfer swings his driver and it meets
the ball on the tee, the ball cannot instantaneously gain velocity,
it has inertia and must first accelerate; it will be compressed.
The driver will also bend but the swing carries through.
Once again, we are not talking about a real-world problem. Just use
the given assumptions together with the clarification given above that
the impacting floors don't actally stop on impact (but conservation of
momentum does imply that the velocity is instantly reduced).
Ok, instantly reduced by how much? The top floor momentarily comes
to rest, then we start again with two floors falling together from rest.
*All* the kinetic energy of the first drop was expended in breaking the
cable suspending the floor below.
This is in disagreement with your statement above,
"Assume that it does not came to rest." and "impact of the falling floor
on the floor below is sufficient to instantly cut the cable holding up the
impacted floor."
My assumption is that the cable can just withstand the weight of two
floors, obviously, and your assumption is that if I place a feather on
a floor it will be the straw that broke the camel's back.
Yes, a feather would do it.
Your assumptions conflict with mine.
All I'm asking for is definition and clarification of YOUR assumptions,
obviously.
Hopefully, that aspect of the problem is now been clarified.
Don't think real-world. In the real world, any such cable would, by
design, be a lot stronger than the weight it's intended to hold. As
explained already, that's not the case for this problem. Don't try to
change the problem or read more into it that what has been given.
Instead, try to view it as a math problem to be solved in the context
of the intended assumptions.
Obviously I have insufficient data to solve your problem.
Perhaps you have enough now.
The problem is ill-defined at (4), we need to know if it takes
one or two floors to snap the cables of the third from the top.
Each floor is independently cabled to the sky.
I understand that, but it was not specified that all cables have
the same or different strength.
The wording of the problem was intended to suggest that the cables
offered _no_ resistance to impact -- that they broke instantly with
any additional downward force. I agree that I could have been more
explicit about that.
If instead, it required the weight of a full floor (rather than a
feather) to break the cable for the floor below, then the velocity on
impact would be reduce to 0, at least for the first impact. That's
wasn't what I had in mind. My intent was to make the collapse as fast
as possible just to see how fast it could possibly be.
It was specified that
even just one floor falling would sufficient to instantly break the
cable holding up the floor below.
I understand that, but it was not specified that all cables have
the same or different strength.
Same strength, exactly strong enough to hold a floor.
Obviously, multiple floors falling
together would then be more than enough to break the cables for lower
floors on impact.
Obvious to you, but not obvious to me that that is what you assumed
I would assume. I have to have clarification.
I am trying hard to clarify, not just to argue. If there's still a
confusion about the intent of the problem, I will try to answer.
But, if possible, at least at this point, I would like to stay with
the assumptions as given, plus the clarifications, rather than have
the problem split into all kinds of variations.
The problem is ill-defined at (4), we need to know if it takes one or two
floors to snap the cables of the third from the top, which may of greater
tensile strength. After all, a real building can be constructed with lighter
materials at the top. Just look at the Empire State, it tapers.
The problem specifies floors of equal weight, modeled as point
particles, so there is no concept of "tapering" and similarly, the
concept of "lighter materials at the top" is not applicable to this
problem.
If two floors are needed, then the free fall is from rest at the
third from top, and ultimately from the last (lowest) suspended
floor with the strongest cables.
In other words the collapse has the appearance of a stop-and-go but
roughly constant velocity, the final stage of which is free fall
from the first floor up, it taking all 100 floors to snap the cable.
If the cables all have equal snapping force you'll have a retarded
acceleration.
Hence the relative strengths of the cables has to be defined.
As mentioned above, all floors are cabled independently.
Irrelevant, we need to know if all cables have the same breaking strain.
Even if they are the same diameter and material, the longer ones
will stretch more than the shorter, there is no such animal as a
rigid rod. Even a glass rod will bend without breaking, that's what
fibre glass is, and you said "cable". All cables will stretch.
This cable doesn't stretch. I just holds, or breaks. If the word cable
is not appropriate, substitute a better term (string? thread?). I'll
stay with cable for now, but it's an idealized cable -- it won't
stretch.
You can visualize the collapse as follows. Let's take k=50.
I don't have any trouble visualizing, it is data I need and asked for.
Hey, I'm trying.
At time t=0, the top 50 floors begin to fall, independently of each
other. The first actual impact is when floor 50 hits floor 49. On
impact, those 2 floors fall together with an initial velocity that is
less than that of higher floors due to conservation of momentum.
Obviously (your word) floor 49 falls from rest.
Floor 50 also falls from rest a second time, it came to rest while breaking
the cables of floor 49.
My question is: what is the breaking strain of the cables supporting
floor 48?
You know the answer to that now. Breaking strength is 0.
Also note the needed minor correction -- the floors are off by 1.
Obviously (your word) the longer a bungee is the greater the stretch
before it breaks. You won't fall far with a short bungee.
I never mentioned bungee cords. Stay with the program.
Thus
??
Oh, I see you've snipped. ***** you too, you ignorant *****.
If something important was inadvertently snipped, sorry.
This is just a math problem -- it's not personal or political. If you
can stay with the math aspects of this problem, I'll continue to play.
quasi
.
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| User: "Androcles" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 02:59:19 AM |
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"quasi" <quasi@null.set> wrote in message =
news:4vo73399gmjlte00ffpfns2b79gereuifu@4ax.com...
On Sat, 28 Apr 2007 22:53:57 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
=20
"quasi" <quasi@null.set> wrote in message =
news:qpc733tji40a7iuflmo9o9frrpp05n34sd@4ax.com...
On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
=20
"quasi" <quasi@null.set> wrote in message =
news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
=20
Of course, the question is motivated by the controversy relating =
to
the fact that three WTC buildings collapsed at very nearly the =
rate of
free fall.
Controversy? Just because some idiot doesn't understand physics=20
doesn't mean there is a controversy. The only "controversy" is=20
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and=20
therefore moot.
=20
The question is not intended to model the actual situation. =
Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, =
it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
=20
Ok, so here's the problem ...
=20
Assumptions:
=20
(1) 100 slabs of concrete (floors) are suspended in midair, =
prevented
from falling to the ground by imaginary cables (attached to the =
sky).
=20
(2) The slabs can be regarded as point particles of equal weight.=20
=20
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. =
Thus,
the distance between successive floors is 1.=20
=20
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the =
floors
that hit it.
=20
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
=20
Question:
=20
If the cables to the top k floors are cut, how long will it take =
for a
complete collapse? Express the answer as a ratio to the time for =
free
fall of just the top floor.
=20
In particular, what are the answers for k=3D1 and k=3D50?
=20
quasi
Insufficient data. We do not know if the topmost floor came to=20
rest before breaking the cable suspending the floor below it.
=20
Assume that it does not came to rest.
It does at least slow. By how much?
=20
By exactly as much as is required by conservation of momentum.
Then I cannot assume is does not come to rest.
=20
Instead, assume that the impacted floor falls together with the =
floors
that hit it, at an initial velocity determined by conservation of
momentum.
Conservation of momentum requires it to come to rest momentarily.
=20
For a time interval of 0, in this problem.=20
Correct, the velocity is zero for a duration of zero time. That's what =
is meant
by "instantaneous".
=20
In other words, assume no bounce back, no delay.=20
I didn't say "bounce back" or "delay", I said "come to rest =
momentarily".
Stay with the program.
The floors just
coalesce and move as one. That's the assumption of the model.=20
Not challenged. Two floors fall together, from rest.
Stay with the program.
=20
http://www.walter-fendt.de/ph11e/ncradle.htm
Now tie down each ball so that the impact just breaks the=20
the thread tying the stationary ball in place, and the result
is no further motion. If the thread is too strong, the falling
ball bounces back. Too weak, and the stationary ball moves.
=20
Assume there is no resistance from the cables.=20
Then all floors fall and the experiment is over before it is built.
All you are talking about is turning Newton's cradle on its side.
Your assumptions are self-contradictory.=20
Regard the cables as
exactly strong enough to hold up the floor, no stronger.
That was not stated and all I asked for in the first place.
Solve with Newton's cradle on its side.=20
http://www.walter-fendt.de/ph11e/ncradle.htm
Over.
.
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 06:03:52 AM |
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On Sun, 29 Apr 2007 07:59:19 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:4vo73399gmjlte00ffpfns2b79gereuifu@4ax.com...
On Sat, 28 Apr 2007 22:53:57 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:qpc733tji40a7iuflmo9o9frrpp05n34sd@4ax.com...
On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
Controversy? Just because some idiot doesn't understand physics
doesn't mean there is a controversy. The only "controversy" is
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and
therefore moot.
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
Ok, so here's the problem ...
Assumptions:
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
(2) The slabs can be regarded as point particles of equal weight.
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
Question:
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
In particular, what are the answers for k=1 and k=50?
quasi
Insufficient data. We do not know if the topmost floor came to
rest before breaking the cable suspending the floor below it.
Assume that it does not came to rest.
It does at least slow. By how much?
By exactly as much as is required by conservation of momentum.
Then I cannot assume is does not come to rest.
But I have since said very clearly that it does _not_ come to rest. I
want the new combination to unite and continue at the _maximum_
possible initial velocity after impact. The new velocity is bounded
above by conservation of momentum. That upper bound _is_ the new
velocity, instantaneously after impact. I don't know how to make it
any clearer as to the intent of the problem. There is a math problem
here which I think is interesting and which _can_ be solved.
Instead, assume that the impacted floor falls together with the floors
that hit it, at an initial velocity determined by conservation of
momentum.
Conservation of momentum requires it to come to rest momentarily.
For a time interval of 0, in this problem.
Ignore my answer above -- I realize now that it was not what I meant
to say.
What I meant was that the duration of impact is 0. I don't agree to
temporary rest since that requires either an escape of energy or a
violation conservation of momentum. In this idealized model, no energy
is lost on impact, hence the total momentum of the system of particles
is assumed constant until a particle hits the ground.
Correct, the velocity is zero for a duration of zero time. That's what is meant
by "instantaneous".
I'd rather not argue what happens at the instant of impact.
What happens at the instant of impact is not of interest. What's
important is that they unite and continue together with a velocity
based on the requirement that the total momentum instantaneously after
impact equals the total momentum instantaneously before impact.
In other words, assume no bounce back, no delay.
I didn't say "bounce back" or "delay", I said "come to rest momentarily".
Stay with the program.
I'm not agreeing to momentary rest.
As long as the colliding floors unite and total momentum is conserved,
what happens at the instant of impact doesn't affect the answer.
The floors just
coalesce and move as one. That's the assumption of the model.
Not challenged. Two floors fall together, from rest.
Stay with the program.
Right, so forget about the instant of impact and focus on the
velocities right before and right after.
http://www.walter-fendt.de/ph11e/ncradle.htm
Now tie down each ball so that the impact just breaks the
the thread tying the stationary ball in place, and the result
is no further motion. If the thread is too strong, the falling
ball bounces back. Too weak, and the stationary ball moves.
Assume there is no resistance from the cables.
Then all floors fall and the experiment is over before it is built.
The floors do not begin to fall at time t=0. Only the top k floors
begin the free fall. The other floors only begin to fall when hit.
All you are talking about is turning Newton's cradle on its side.
Your assumptions are self-contradictory.
I don't follow this claimed contradiction.
Suppose there are only 2 floors and the top floor (floor 2) begins the
free fall. The mathematics here can be done by hand. I am asking for
the ratio of the time for a total collapse (assuming floor 2 impacts
floor 1 after which they fall together) to the time of free fall for
floor 2 (assuming no impact with floor 1).
Regard the cables as
exactly strong enough to hold up the floor, no stronger.
That was not stated and all I asked for in the first place.
I have done my best to try to clarify, but it's not always so easy to
judge exactly where the point of misunderstanding is. Hopefully the
intent of the problem is now clear.
Solve with Newton's cradle on its side.
http://www.walter-fendt.de/ph11e/ncradle.htm
Over.
Newton's cradle has both vertical and horizontal forces. The
velocities also have both vertical and horizontal components. It's a
2-dimensional problem.
The pancake collapse problem is 1-dimensional. The only force is
gravity and all velocities are vertical.
Just turning Newton's cradle sideways doesn't appear to correctly
model my problem.
How about trying the problem I suggested above with only 2 floors? We
can then compare answers.
quasi
.
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| User: "Androcles" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 05:29:17 AM |
|
|
"quasi" <quasi@null.set> wrote in message =
news:1ar833tf0atmbefv2lslkk31k5gart3o06@4ax.com...
On Sun, 29 Apr 2007 07:59:19 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
=20
"quasi" <quasi@null.set> wrote in message =
news:4vo73399gmjlte00ffpfns2b79gereuifu@4ax.com...
On Sat, 28 Apr 2007 22:53:57 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
=20
"quasi" <quasi@null.set> wrote in message =
news:qpc733tji40a7iuflmo9o9frrpp05n34sd@4ax.com...
On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
=20
"quasi" <quasi@null.set> wrote in message =
news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
=20
Of course, the question is motivated by the controversy relating =
to
the fact that three WTC buildings collapsed at very nearly the =
rate of
free fall.
Controversy? Just because some idiot doesn't understand physics=20
doesn't mean there is a controversy. The only "controversy" is=20
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, =
and=20
therefore moot.
=20
The question is not intended to model the actual situation. =
Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, =
it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
=20
Ok, so here's the problem ...
=20
Assumptions:
=20
(1) 100 slabs of concrete (floors) are suspended in midair, =
prevented
from falling to the ground by imaginary cables (attached to the =
sky).
=20
(2) The slabs can be regarded as point particles of equal =
weight.=20
=20
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. =
Thus,
the distance between successive floors is 1.=20
=20
(4) Assume that if the cable holding up a given floor is cut, =
the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume =
that
once a floor is impacted, it would then fall together with the =
floors
that hit it.
=20
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
=20
Question:
=20
If the cables to the top k floors are cut, how long will it take =
for a
complete collapse? Express the answer as a ratio to the time for =
free
fall of just the top floor.
=20
In particular, what are the answers for k=3D1 and k=3D50?
=20
quasi
Insufficient data. We do not know if the topmost floor came to=20
rest before breaking the cable suspending the floor below it.
=20
Assume that it does not came to rest.
It does at least slow. By how much?
=20
By exactly as much as is required by conservation of momentum.
Then I cannot assume is does not come to rest.
=20
But I have since said very clearly that it does _not_ come to rest. I
want the new combination to unite and continue at the _maximum_
possible initial velocity after impact. The new velocity is bounded
above by conservation of momentum. That upper bound _is_ the new
velocity, instantaneously after impact. I don't know how to make it
any clearer as to the intent of the problem. There is a math problem
here which I think is interesting and which _can_ be solved.
=20
Instead, assume that the impacted floor falls together with the =
floors
that hit it, at an initial velocity determined by conservation of
momentum.
Conservation of momentum requires it to come to rest momentarily.
=20
For a time interval of 0, in this problem.
=20
Ignore my answer above -- I realize now that it was not what I meant
to say.
=20
What I meant was that the duration of impact is 0. I don't agree to
temporary rest since that requires either an escape of energy or a
violation conservation of momentum. In this idealized model, no energy
is lost on impact, hence the total momentum of the system of particles
is assumed constant until a particle hits the ground.
=20
Correct, the velocity is zero for a duration of zero time. That's what =
is meant
by "instantaneous".
=20
I'd rather not argue what happens at the instant of impact.
=20
What happens at the instant of impact is not of interest. What's
important is that they unite and continue together with a velocity
based on the requirement that the total momentum instantaneously after
impact equals the total momentum instantaneously before impact.
=20
In other words, assume no bounce back, no delay.=20
I didn't say "bounce back" or "delay", I said "come to rest =
momentarily".
Stay with the program.
=20
I'm not agreeing to momentary rest.
You proposed a mathematical puzzle, I used the term mathematically.
=20
As long as the colliding floors unite and total momentum is conserved,
what happens at the instant of impact doesn't affect the answer.=20
=20
The floors just
coalesce and move as one. That's the assumption of the model.=20
Not challenged. Two floors fall together, from rest.
Stay with the program.
=20
Right, so forget about the instant of impact and focus on the
velocities right before and right after.
Study calculus. Velocity at a point (in space or time) has a value, and
in this case the value is zero, even if it is non-zero before and after.
http://mathworld.wolfram.com/Derivative.html
Forget about right before and right after and focus on the moment of =
impact,=20
and don't try to bamboozle a mathematician.
If you don't understand that, don't propose problems that are
beyond your capabilities.=20
=20
=20
http://www.walter-fendt.de/ph11e/ncradle.htm
Now tie down each ball so that the impact just breaks the=20
the thread tying the stationary ball in place, and the result
is no further motion. If the thread is too strong, the falling
ball bounces back. Too weak, and the stationary ball moves.
=20
Assume there is no resistance from the cables.=20
Then all floors fall and the experiment is over before it is built.
=20
The floors do not begin to fall at time t=3D0.=20
Ok, they begin to fall at 3:34 on a Sunday afternoon. =20
Can I start a stop watch when they begin to fall and call that t =3D 0?
You are being argumentative and ridiculous.
Only the top k floors
begin the free fall. The other floors only begin to fall when hit.=20
Oh, do shut up. You are an idiot. *plonk*
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 06:47:21 AM |
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On Sun, 29 Apr 2007 10:29:17 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
Oh, do shut up. You are an idiot. *plonk*
There is an actual math problem to be solved, but the problem is
obviously not for you.
quasi
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| User: "Denis Feldmann" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 06:43:31 AM |
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quasi a écrit :
On Sun, 29 Apr 2007 10:29:17 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
Oh, do shut up. You are an idiot. *plonk*
There is an actual math problem to be solved, but the problem is
obviously not for you.
Is there? I have a few math problems of my own, which I feel more
urgent. If your math problem has no relevance to conspiracy theory, why
bother? If it has, why do you disguise it ? And as usual, a little logic
1) tells us the answer will be "no conspiracy" 2) that you must be quite
incompetent not to solve it for yourself (hint : it is not very hard),
nor to see most mathematicians here dont care, as you formulated it as a
*physic* problem... (how many mathematicians know by heart the necessary
impulsion formulas ? Oh, and another information : it is resonably easy
to solve using energy conservation principles...
quasi
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 11:17:44 AM |
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On Sun, 29 Apr 2007 13:43:31 +0200, Denis Feldmann
<denis.feldmann.asupprimer@club-internet.fr> wrote:
quasi a écrit :
On Sun, 29 Apr 2007 10:29:17 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
Oh, do shut up. You are an idiot. *plonk*
There is an actual math problem to be solved, but the problem is
obviously not for you.
Is there? I have a few math problems of my own, which I feel more
urgent. If your math problem has no relevance to conspiracy theory, why
bother? If it has, why do you disguise it ?
Wow -- a lot of people with attitudes.
No one asked you to analyze the problem if doesn't interest you.
Of course it has relevance to conspiracy theory -- it's an attempt to
understand the pancake phenomenon. There is an often heard argument
that the near free fall speed of the collapses can't be explained by
the official story. The question I asked was designed just to get a
feel for how fast the pancake collapse could happen using an
idealized, highly simplified model biased towards a fast collapse.
And as usual, a little logic
1) tells us the answer will be "no conspiracy"
When I asked the question, I had no real intuition as to what the
results would be. You say it's obvious, fine -- but I was curious.
But I wonder how obvious it really is. Can you even guess the
approximate answer for k=1 and k=50?
2) that you must be quite incompetent not to solve it for yourself
(hint : it is not very hard),
Actually, when I posed the problem, I _knew_ that I could solve it, at
least in principle. However, I thought perhaps that others might also
enjoy playing with it.
I'll admit that I know very little physics. So what? Calling me
incompetent serves what purpose?
nor to see most mathematicians here dont care, as you formulated it as a
*physic* problem... (how many mathematicians know by heart the necessary
impulsion formulas ?
You might be right, but it seems a rather presumptuous claim. What
makes you think you can speak for "most mathematicians here"?
In any case, I found the problem interesting, so I thought I'd share
it. If others enjoy playing with it, great. If not, no big deal.
Oh, and another information : it is resonably easy
to solve using energy conservation principles...
Maybe, but talk is cheap. If it's so easy, let's see you do it. Oh,
but I forgot -- you have other more urgent math problems.
quasi
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 04:33:26 PM |
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On Sat, 28 Apr 2007 16:26:58 -0500, quasi <quasi@null.set> wrote:
On Sat, 28 Apr 2007 19:44:02 GMT, "Androcles"
<Engineer@hogwarts.physics.co.uk> wrote:
"quasi" <quasi@null.set> wrote in message news:do6733lj8sn4sde2qtjuv43m9hvp7b2ckl@4ax.com...
I am going to pose a simple mathematical problem.
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
Controversy? Just because some idiot doesn't understand physics
doesn't mean there is a controversy. The only "controversy" is
whether WTC 7 fell by controlled demolition (or not), but it was
inevitable that it would be if it had remained standing anyway, and
therefore moot.
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
Ok, so here's the problem ...
Assumptions:
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
(2) The slabs can be regarded as point particles of equal weight.
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
Question:
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
In particular, what are the answers for k=1 and k=50?
quasi
Insufficient data. We do not know if the topmost floor came to
rest before breaking the cable suspending the floor below it.
Assume that it does not came to rest.
Instead, assume that the impacted floor falls together with the floors
that hit it, at an initial velocity determined by conservation of
momentum.
If it did then the top two floors fall together from rest, but will
not come to rest when they hit the next. If it did not, then the
situation is even more difficult.
Not that difficult. I'm confident that someone will provide a
numerical answer for k=1 and k=50.
Think of it this way. If a golfer swings his driver and it meets
the ball on the tee, the ball cannot instantaneously gain velocity,
it has inertia and must first accelerate; it will be compressed.
The driver will also bend but the swing carries through.
Once again, we are not talking about a real-world problem. Just use
the given assumptions together with the clarification given above that
the impacting floors don't actally stop on impact (but conservation of
momentum does imply that the velocity is instantly reduced).
The problem is ill-defined at (4), we need to know if it takes
one or two floors to snap the cables of the third from the top.
Each floor is independently cabled to the sky. It was specified that
even just one floor falling would sufficient to instantly break the
cable holding up the floor below. Obviously, multiple floors falling
together would then be more than enough to break the cables for lower
floors on impact.
If two floors are needed, then the free fall is from rest at the
third from top, and ultimately from the last (lowest) suspended
floor with the strongest cables.
In other words the collapse has the appearance of a stop-and-go but
roughly constant velocity, the final stage of which is free fall
from the first floor up, it taking all 100 floors to snap the cable.
If the cables all have equal snapping force you'll have a retarded
acceleration.
Hence the relative strengths of the cables has to be defined.
As mentioned above, all floors are cabled independently.
You can visualize the collapse as follows. Let's take k=50.
At time t=0, the top 50 floors begin to fall, independently of each
other. The first actual impact is when floor 50 hits floor 49. On
impact, those 2 floors fall together with an initial velocity that is
less than that of higher floors due to conservation of momentum. Thus
the next impact will not be floors 50,49 with floor 48 but rather it
will be floor 51 hitting the slower moving 50,49 combo. The 3 floors
51,50,49 then become a unit. The initial velocity of the new 51,50,49
combo will be greater than the velocity of the 50-49 combo just before
impact, but still less than that of the floors above (which are still
in free fall).
quasi
Minor correction:
If the top 50 floors begin to fall, the first impact will be floor 51
with floor 50, not floor 50 with floor 49. Thus, in the illustration I
gave above, all floors I mentioned are off by 1.
quasi
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| User: "Major Quaternion Dirt Quantum" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 03:41:56 PM |
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this was basically assuming that
floor 50 of a 100-floor building started the collapse?
interesting analysis; I hope to look at it, again,
later.
have you seen the old analysis by teh MIT Head Welder,
that I only saw a couple of weeks ago?
You can visualize the collapse as follows. Let's take k=50.
At time t=0, the top 50 floors begin to fall, independently of each
other. The first actual impact is when floor 50 hits floor 49. On
impact, those 2 floors fall together with an initial velocity that is
less than that of higher floors due to conservation of momentum. Thus
the next impact will not be floors 50,49 with floor 48 but rather it
will be floor 51 hitting the slower moving 50,49 combo. The 3 floors
51,50,49 then become a unit. The initial velocity of the new 51,50,49
combo will be greater than the velocity of the 50-49 combo just before
impact, but still less than that of the floors above (which are still
in free fall).
http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html
thus:
"Global Warning,"
an *anonymous* article in an old issue of *National Review*,
the so-called conservative mag of the grotesque Establishment
creature,
William Buckley, was a sort of benignly-spun expose
of the Mont Pelerin Society, the vehicle of British imperialist "free
trade,"
free beer & free dumb, whose many "rightwing" affiliate foundations
are always in the news. the universally-newspaper-lauded-when-
he-died founding president, Mitlon Friedman,
used Sir Henry of Kiss.*****. to conduct their first experiment
in Chile, with the help of Augusto Pinochet & ***** Nixon.
(the fact that it was an awful failure, with the privatization
of Chile's social security, was only recently mention, but
only before Uncly Milty died.)
I realized, much later, since it was anonymous, that
it was probably taken from The Holy Spook Economist, but
I haven't checked that hypothesis.
yes, "Warning" was meant to evoke "warming," although
nowhere mentioned in the article, as I recall; can you say,
"emmissions trading schemes," Boys and Girl?...
can you say, "hedge funds on out-of-the-way desert islands?..."
Just Desserts Island?
thus:
why do *****, Borat, Osama, Harry P., Rumsfeld and Tony B. want us
to underwrite the 3rd British Invasion of Sudan,
so badly?
anyway, Bertrand Russel published a jeremyad in the Bulletin
of the Atomic Scientists, while the USA was the only thermonukey
power,
that we should bomb them into the Stone Age;
that's your British "pacifist," whether before or after he went nuts
because of Godel's proof.
thus:
if you can't prove that all Fermat numbers are pairwise coprime,
again!
Darfur 'Mini-Summit'
http://www.larouchepub.com/other/1998/rice_2546.html
thus:
uh yeah; Borat wants you in Sudan,
why, Baby?... Harry Potter wants you in Iran --
yeah, Baby; shag'US with a spoon?
--DARFURIA CONSISTS OF ARABs & nonARABs; NEWS-ITEM:
we are marching to Darfuria, Darfuria, Darfuria!
Harry Potter IIX, ?Ordeal @ Oxford//Sudan ^ Aircraft Carrier!
http://larouchepub.com/other/2007/3410caymans_hedges.html
ALgoreTHEmovieFORpresident.COM:
http://larouchepub.com/eirtoc/site_packages/2007/al_gore.html
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 04:57:35 PM |
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On 28 Apr 2007 13:41:56 -0700, Major Quaternion Dirt Quantum
<QncyMI@netscape.net> wrote:
this was basically assuming that
floor 50 of a 100-floor building started the collapse?
At time t=0, assume the top 50 floors all instantly begin to free
fall. The free fall is interrupted only by impact with lower floors
which then combine. The combo then falls with no delay, but with an
initial velocity determined by conservation of momentum.
interesting analysis; I hope to look at it, again,
later.
have you seen the old analysis by teh MIT Head Welder,
that I only saw a couple of weeks ago?
No, I haven't really been following the controversy too closely.
I just thought it would be interesting to look at the issue in an
idealized, highly simplified setting, just to gain an intuitive feel.
quasi
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| User: "Major Quaternion Dirt Quantum" |
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| Title: Re: the pancake collapse phenomenon |
28 Apr 2007 04:04:31 PM |
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well, his analysis was exquisite, although
I don't recall a single thing from it;
it was given only a month or so after 9/11/01.
that is, those who support the work of the Utah professor,
might compare it to the work of the Massachussetts one,
although there will be unasked/unanswered questions
arizing from Jones', as I recall.
have you seen the old analysis by teh MIT Head Welder,
that I only saw a couple of weeks ago?
No, I haven't really been following the controversy too closely.
here's a question that came-up recently,
in battling with the controlled-demo advocates
at teh Promenade:
how is it that the metal was still molten,
after weeks?
it seems taht an additional hypothesis is required,
to explain that. because I doubt that
there was sufficient nukular material
"left-over" from a hypothetical suitcase-device.
there are similar question,
from what I recall of Jones' report
(I just stopped at the first thing
that seemed funny, today .-)
publications:http://www.physics.uiowa.edu/~cgrabbe/writing/
research.html
demanding of INTELLECTUAL HONESTY.
thus:
this was basically assuming that
floor 50 of a 100-floor building started the collapse?...
interesting analysis; I hope to look at it, again, later.
have you seen the old analysis by the MIT Head Welder,
that I only saw a couple of weeks ago?
At time t=0, the top 50 floors begin to fall, independently of each
other. The first actual impact is when floor 50 hits floor 49. On
impact, those 2 floors fall together with an initial velocity that is
less than that of higher floors due to conservation of momentum. Thus
the next impact will not be floors 50,49 with floor 48 but rather it
will be floor 51 hitting the slower moving 50,49 combo. The 3 floors
51,50,49 then become a unit. The initial velocity of the new 51,50,49
combo will be greater than the velocity of the 50-49 combo just before
impact, but still less than that of the floors above (which are still
in free fall).
http://www.rwgrayprojects.com/synergetics/plates/figs/plate01.html
thus:
why do *****, Borat, Osama, Harry P., Rumsfeld and Tony B. want us
to underwrite the 3rd British Invasion of Sudan,
so badly?
anyway, Bertrand Russel published a jeremyad in the Bulletin
of the Atomic Scientists, while the USA was the only thermonukey
power,
that we should bomb them into the Stone Age;
that's your British "pacifist," whether before or after he went nuts
because of Godel's proof.
thus:
if you can't prove that all Fermat numbers are pairwise coprime,
again!
Darfur 'Mini-Summit'
http://www.larouchepub.com/other/1998/rice_2546.html
thus:
uh yeah; Borat wants you in Sudan,
why, Baby?... Harry Potter wants you in Iran --
yeah, Baby; shag'US with a spoon?
--DARFURIA CONSISTS OF ARABs & nonARABs; NEWS-ITEM:
we are marching to Darfuria, Darfuria, Darfuria!
Harry Potter IIX, ?Ordeal @ Oxford//Sudan ^ Aircraft Carrier!
http://larouchepub.com/other/2007/3410caymans_hedges.html
ALgoreTHEmovieFORpresident.COM:
http://larouchepub.com/eirtoc/site_packages/2007/al_gore.html
.
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
29 Apr 2007 03:23:10 AM |
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On Sat, 28 Apr 2007 14:45:57 -0500, quasi <quasi@null.set> wrote:
I am going to pose a simple mathematical problem.
Of course, the question is motivated by the controversy relating to
the fact that three WTC buildings collapsed at very nearly the rate of
free fall.
The question is not intended to model the actual situation. Instead, I
just want to consider a highly simplified model for which a
mathematical solution can be readily obtained. In other words, it's
just a math problem, not a real-world engineering problem, and
definitely not a political one.
Ok, so here's the problem ...
Assumptions:
(1) 100 slabs of concrete (floors) are suspended in midair, prevented
from falling to the ground by imaginary cables (attached to the sky).
(2) The slabs can be regarded as point particles of equal weight.
(3) Place the slabs on the y-axis at heights 1, 2, 3, ... , 100. Thus,
the distance between successive floors is 1.
(4) Assume that if the cable holding up a given floor is cut, the
impact of the falling floor on the floor below is sufficient to
instantly cut the cable holding up the impacted floor. Assume that
once a floor is impacted, it would then fall together with the floors
that hit it.
(5) Assume no air resistance -- just assume Newton's laws (i.e.
gravity, conservation of momentum).
Question:
If the cables to the top k floors are cut, how long will it take for a
complete collapse? Express the answer as a ratio to the time for free
fall of just the top floor.
In particular, what are the answers for k=1 and k=50?
quasi
OK, I modeled this in Maple.
I won't reveal my answers just yet -- I'll leave it as a challenge to
see if someone else can also do it.
But the answers were definitely not what I expected.
Let me clarify one of the assumptions ...
On the question of whether or not the lower floors offer resistance on
impact, the assumption (for this version of the problem) is that they
don't.
If an impacted floor is at rest, the imaginary cables holding up the
floor simply disintegrate on impact and the combo of impacting floors
then unites with the floor being hit. In this case, the combo of
impacting floors does experience an instantaneous slowdown on impact
due to conservation of momentum.
There is also the phenomenon of a floor hitting a falling combo of
united floors. In this case there are no cables to worry about. The
impacting floor unites with the combo, causing the combo to
instantaneously pick up speed due to conservation of momentum.
A future revision of the problem may allow for explicit resistance to
impact, but for now, I'll wait to see if others can verify the results
I obtained based on the "no resistance" assumption.
quasi
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
12 May 2007 08:21:02 PM |
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On Sun, 29 Apr 2007 03:23:10 -0500, quasi <quasi@null.set> wrote:
Ok, here's the elastic version of the problem ...
100 particles (floors) of equal mass, placed on the y-axis at heights
1, 2, 3, ... 100, suspended in air by threads.
The thread to the top floor is cut at time t=0.
Choose a standard value for the acceleration due to gravity.
Assume all collisions are perfectly elastic and, for simplicity (in
this version of the problem), assume no resistance from the threads.
In other words, they simply disintegrate on impact.
Find the total collapse time for the top floor, expressed as a ratio
to the time for pure free-fall.
A numerically derived answer is acceptable.
quasi
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
13 May 2007 07:11:37 PM |
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On Sat, 12 May 2007 20:21:02 -0500, quasi <quasi@null.set> wrote:
On Sun, 29 Apr 2007 03:23:10 -0500, quasi <quasi@null.set> wrote:
Ok, here's the elastic version of the problem ...
100 particles (floors) of equal mass, placed on the y-axis at heights
1, 2, 3, ... 100, suspended in air by threads.
The thread to the top floor is cut at time t=0.
Choose a standard value for the acceleration due to gravity.
Assume all collisions are perfectly elastic and, for simplicity (in
this version of the problem), assume no resistance from the threads.
In other words, they simply disintegrate on impact.
Find the total collapse time for the top floor, expressed as a ratio
to the time for pure free-fall.
A numerically derived answer is acceptable.
quasi
For anyone reading this thread, the above problem is answered by David
Hartley's very elegant analytic solution in the thread
"Re: WTC Towers: A Top-Down Controlled Demolition?"
If I interpret his results correctly, the time for a total collapse
time is exactly sqrt(2) times more than the time for a free-fall
collapse. Perhaps surprisingly, this result appears not to depend
either on the value of g or on the number of floors.
Of course, we have assumed perfectly elastic collisions. If instead,
the collisions are perfectly inelastic, the collapse time is much
faster, but still not free-fall speed.
More importantly, all the models analyzed so far in this thread have
ignored resistance on impact. How does that get factored in? Perhaps,
to keep the model simple, we can postulate a fixed amount of energy
required to break the support beams for a given floor. That energy is
then subtracted from the kinetic energy of the impacting floors at the
instant of impact. Thus, the energy expended on impact in breaking the
beams of the floor being impacted effectively reduces the velocity of
impact.
Additionally, we could also consider air resistance, and possibly also
friction. However, of all the effects, I think resistance on impact
may be the most important.
quasi
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| User: "galathaea" |
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| Title: Re: the pancake collapse phenomenon |
14 May 2007 03:19:19 PM |
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On May 13, 5:11 pm, quasi <q...@null.set> wrote:
On Sat, 12 May 2007 20:21:02 -0500, quasi <q...@null.set> wrote:
On Sun, 29 Apr 2007 03:23:10 -0500, quasi <q...@null.set> wrote:
Ok, here's the elastic version of the problem ...
100 particles (floors) of equal mass, placed on the y-axis at heights
1, 2, 3, ... 100, suspended in air by threads.
The thread to the top floor is cut at time t=0.
Choose a standard value for the acceleration due to gravity.
Assume all collisions are perfectly elastic and, for simplicity (in
this version of the problem), assume no resistance from the threads.
In other words, they simply disintegrate on impact.
Find the total collapse time for the top floor, expressed as a ratio
to the time for pure free-fall.
A numerically derived answer is acceptable.
quasi
For anyone reading this thread, the above problem is answered by David
Hartley's very elegant analytic solution in the thread
"Re: WTC Towers: A Top-Down Controlled Demolition?"
If I interpret his results correctly, the time for a total collapse
time is exactly sqrt(2) times more than the time for a free-fall
collapse. Perhaps surprisingly, this result appears not to depend
either on the value of g or on the number of floors.
Of course, we have assumed perfectly elastic collisions. If instead,
the collisions are perfectly inelastic, the collapse time is much
faster, but still not free-fall speed.
More importantly, all the models analyzed so far in this thread have
ignored resistance on impact. How does that get factored in? Perhaps,
to keep the model simple, we can postulate a fixed amount of energy
required to break the support beams for a given floor. That energy is
then subtracted from the kinetic energy of the impacting floors at the
instant of impact. Thus, the energy expended on impact in breaking the
beams of the floor being impacted effectively reduces the velocity of
impact.
Additionally, we could also consider air resistance, and possibly also
friction. However, of all the effects, I think resistance on impact
may be the most important.
a perfectly elastic model would never collapse
it would sit there bouncing forever
the inelastic model is the mostly correct one
but you have to also account all breaking energies
pulverisation of concrete
shearing of steel beams
all of which remove energy from the kinesis
slowing the fall time
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
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| User: "quasi" |
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| Title: Re: the pancake collapse phenomenon |
15 May 2007 05:27:48 AM |
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On 14 May 2007 13:19:19 -0700, galathaea <galathaea@gmail.com> wrote:
On May 13, 5:11 pm, quasi <q...@null.set> wrote:
On Sat, 12 May 2007 20:21:02 -0500, quasi <q...@null.set> wrote:
On Sun, 29 Apr 2007 03:23:10 -0500, quasi <q...@null.set> wrote:
Ok, here's the elastic version of the problem ...
100 particles (floors) of equal mass, placed on the y-axis at heights
1, 2, 3, ... 100, suspended in air by threads.
The thread to the top floor is cut at time t=0.
Choose a standard value for the acceleration due to gravity.
Assume all collisions are perfectly elastic and, for simplicity (in
this version of the problem), assume no resistance from the threads.
In other words, they simply disintegrate on impact.
Find the total collapse time for the top floor, expressed as a ratio
to the time for pure free-fall.
A numerically derived answer is acceptable.
quasi
For anyone reading this thread, the above problem is answered by David
Hartley's very elegant analytic solution in the thread
"Re: WTC Towers: A Top-Down Controlled Demolition?"
If I interpret his results correctly, the time for a total collapse
time is exactly sqrt(2) times more than the time for a free-fall
collapse. Perhaps surprisingly, this result appears not to depend
either on the value of g or on the number of floors.
Of course, we have assumed perfectly elastic collisions. If instead,
the collisions are perfectly inelastic, the collapse time is much
faster, but still not free-fall speed.
More importantly, all the models analyzed so far in this thread have
ignored resistance on impact. How does that get factored in? Perhaps,
to keep the model simple, we can postulate a fixed amount of energy
required to break the support beams for a given floor. That energy is
then subtracted from the kinetic energy of the impacting floors at the
instant of impact. Thus, the energy expended on impact in breaking the
beams of the floor being impacted effectively reduces the velocity of
impact.
Additionally, we could also consider air resistance, and possibly also
friction. However, of all the effects, I think resistance on impact
may be the most important.
a perfectly elastic model would never collapse
it would sit there bouncing forever
No, that won't happen in this model.
There are two notions of "elasticity".
(1) elastic materials
(2) elastic collisions
For the model in question, we are considering only elastic collisions.
In other words, regard the particles as more like billiard balls than
rubber balls.
For elastic collisions, two things are conserved:
(1) conservation of kinetic energy
(2) conservation of momentum
Algebraically, for collisions of _pairs_ of objects, the above
conservation requirements imply that the set of velocities is
invariant. That, together with the physical situation, implies that
the particles _swap_ velocities. Thus, no velocity will ever be
upwards in this model.
In the horizontal Newton's cradle, the suspending strings provide a
horizontal force resisting the displacement. Hence the displaced balls
eventually return, yielding the apparent bouncing effect.
the inelastic model is the mostly correct one
Yes, probably.
For the problem considered, we were simply trying to understand how
the collapse would be different if the collisions were elastic. In
particular, the goal was to answer the question as to whether
perfectly elastic collisions would have resulted in a significantly
greater total collapse time when compared against the time for pure
free-fall. The answer -- yes, greater by a factor of sqrt(2).
While it may seem intuitive that inelastic collisions are a more
correct assumption for the actual event, it's not automatic. It
depends on the composition of the falling materials.
Would collisions of solid concrete blocks be elastic or inelastic? I'm
not sure. Of course, it's obvious that pulverized concrete will result
in inelastic collisions. So forget the concrete.
What about the steel? Steel beams are more like billiard balls, so for
the steel (assuming it wasn't molten), I would think elastic
collisions are more appropriate.
In any case, I do agree that the actual collisions were probably
mostly inelastic. Our analysis simply shows how the assumption of
elastic collisions would have affected the outcome.
but you have to also account all breaking energies
pulverisation of concrete
shearing of steel beams
all of which remove energy from the kinesis
Absolutely -- these are key factors.
slowing the fall time
Yep. The question is how much.
A specially designed demolition would help answer the question.
Take a building intended for demolition. Place demolition charges on
only the top k floors. Let the rest of the floors receive only the
effects of pancake collapse. Let's see what it looks like and how long
it takes.
quasi
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| User: "galathaea" |
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| Title: Re: the pancake collapse phenomenon |
15 May 2007 02:53:43 PM |
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On May 15, 3:27 am, quasi <q...@null.set> wrote:
On 14 May 2007 13:19:19 -0700, galathaea <galath...@gmail.com> wrote:
On May 13, 5:11 pm, quasi <q...@null.set> wrote:
On Sat, 12 May 2007 20:21:02 -0500, quasi <q...@null.set> wrote:
On Sun, 29 Apr 2007 03:23:10 -0500, quasi <q...@null.set> wrote:
Ok, here's the elastic version of the problem ...
100 particles (floors) of equal mass, placed on the y-axis at heights
1, 2, 3, ... 100, suspended in air by threads.
Th | | | | | | | | | |
