Posted below is a "paper" by truther Marty Beck. I'll just leave this here.
Marty Beck, 4/20/2011
Analysis of the Fall Times of the Three World Trade Center Buildings
Abstract:
Explaining the collapse of the three World Trade Center buildings has been an enigma that strains the mind to come to grips with. Those who look at it over and over again try to make sense of what they see. The recorded fall times somehow seem reasonable to many of us when we rely on intuition in thinking about the collapse.
Many citizens are reasoning much like Galileo’s contemporaries must have. They easily assumed a heavier ball falls faster than a lighter ball. Until the experiment. We assume a large mass unleashed could fall at the rate we see. Before Galileo’s experiment they must have wondered why anyone would bother to do the experiment; isn’t it obvious?
While it looks obvious after watching the videos that something doesn’t add up, it is fairly easy to surmise that once collapse is initiated, the buildings would surely continue to collapse. One way to arrive at this assessment is based on being told a floor could hold 11 times its own weight (See NIST FAQ 1 in Appendix 3). So if you could get more than 11 floors to loose their main support, a collapse of some speed would follow. It would presumably gain more crushing mass and possibly more velocity as it continued down. But how fast should it fall? What resistance to falling did it encounter to account for the actual fall times? How does that compare to the ‘as built’ strength? This will be shown.
While even a simplified Finite Element Analysis (FEA) would show the impossibility of the observed fall times, it has not been forthcoming. This paper proves that each of the towers that collapsed that day was a controlled demolition. It does so by providing a concrete explanation of the basic physics and mechanics of the collapse of these steel structures.
The analysis is quite simple even though the starting conditions are absurd. NIST has managed to convince the populace that collapse was initiated by fire. So we will start there, collapse initiated by fire. Some number of floors start falling and the structure below succumbs to being hit by a moving mass with enough momentum to either sheer the floor it hits or crush the support columns. This paper does not deal with the impossibility of the initiation of collapse. Only what happened after the impossible collapse was initiated. What NIST chose not to share with us. What we learn also explains the initial collapse.
Introduction to the Analysis:
Only three physical agents serve to slow the falling floors:
1) Accelerating the stationary floor masses as the collapse ensues provides a calculable resistance to freefall. Its effect in decelerating the falling floors is calculated and graphed. Many others have done these calculations.
2) Shearing of floors at the support columns provides a calculable energy loss. Shearing can be seen as an impulse force, like hitting a stationary mass. It will be defined and explained in detail using NIST numbers for shear.
3) The strength of the columns themselves accounts for the other force decelerating the falling mass of floors. This can be seen as a continuous force resisting the collapse. We will calculate the loss in strength of the support columns to account for the fall times.
Item 1, as it pertains to the North and South Towers, is troubling on its own. One’s first notion is that this loss, the acceleration of a stationary mass, could practically be ignored since the shear strength and column strength provides proportionally much greater losses. Surprisingly however, just this loss alone accounts for half of the velocity decrease encountered when compared to free fall velocity. This will be shown. If you want to ignore this loss all together, that is fine with me.
Item 2, the loss in velocity if a floor were sheared where it connects to the support column, gets quite a bit of discussion. NIST quantifies this loss so the effects are easily calculated. This, however, is a point of controversy since NIST claims the collapse ‘was not a pancake’ event in one breath yet talks about pancaking in other breaths. (reference needed). This analysis will show that 5/6ths of the built in shear forces were not present or the twin towers would have fallen slower than either one did. If you want to ignore this loss all together, that is fine with me.
Item 3, the strength of the support columns, is the force you cannot ignore and the one that has been hardest to get non engineers to see. NIST entirely ignores this force and simply uses the term ‘buckle’. I will attempt to make the strength of the support columns much easier to envision by removing all the air in the building and representing it as the area of the steel columns supporting the masses of the floors.
The Twin Tower Fall Time Analysis
It would be easy to start with Item 1 but these effects have been shown over and over. There is another derivation in Appendix 2. The math and physics of this loss are fun to calculate however, and I will do so later, but not to lose the reader, I won’t start here. Item 2 will be discussed at the end. It is an extension of the model used to calculate Item 1. Hopefully you won’t need to read that far to easily conclude the columns strength was somehow removed.
Item 3: Strength of the Twin Tower Columns as a Function of Fall Time
The first step is to determine the ‘as built’ upward force provided by the columns. After that, we will need the weight of 33 floors that are assumed to be falling at 20ft/sec2. The next step will be to determine the column strength that is consistent with the observed fall times. The ratio of the two numbers represents the percent of lost strength the columns displayed during the collapse. It is that simple.
Strength of the Support Columns:
We have about 48 core columns and 236 perimeter columns. The perimeter columns are roughly 14 inches square and a quarter of an inch in thickness at the 80th floor (Appendix 1). So we have 14 square inches per column and a minimum of 30 thousand pounds per square inch of strength. (cite reference) (note about 70ksi) So combined, the columns can hold 30 x 103 pounds/in2 x 14in2 x 284 = 120 x 106 pounds before yielding(120 million pounds).
Calculate the Weight of the Concrete:
Now each floor is approx 200 ft by 200 ft of 4 inch thick concrete. The highest density I can find published for concrete is 150 pounds per cubic foot. That would be 50 pounds per square foot of floor area since the floor is 4 inches thick. So each floor weighs roughly 200ft X 200ft X 50pounds/ft2 = 2.0 million pounds so lets call it 2.4 million pounds with office furniture where we allow 2000 each of various 200 pound objects per floor. So 33 floors would weigh 80X106 pounds, (80 million pounds).
You are understanding things so far if you see that the steel columns can support 120 million pounds but only needs to support 80 million pounds.
Determine the Force the Falling Mass Exerts
We are interested in the seconds after the fall has been initiated. Note again that this analysis is accepting as a starting condition that the 33 stories are accelerating at 20 feet/sec2. This makes the problem is a little harder to see but we have to take this into account.
Weight is the result of mass in earth’s gravity field and the weight pushing down is the force defined as F=ma where m is the mass of the 33 stories and a is the acceleration due to gravity. On earth a is 32 ft/sec2.
So before the collapse was initiated, the weight was 80 million pounds. Once the upper 33 floors start to accelerate at 20 ft/sec2, the resulting force is now proportional to the difference between the 32 ft/sec2 and the 20 ft/sec2. This is 12 ft/sec2 with a resulting force of only 12/32 of what it was before the collapse.
Column Strength During Collapse:
So we had 80 x 106 pounds pushing down before the collapse and this force is reduced to 12/32 of what it had been to account for acceleration. 12/32 of 80 is 30. So now we have 30 x 106 pounds of force pushing down on a support structure that can withstand 120 x 106. In engineering terms this represents a margin of 4. A structural failure could not occur unless the margin was less than 1.
So without counting Item 1 and Item 2, we can conclude that 3/4th of strength of the columns were absent/removed to account for the observed fall times. That is 75%. We are told that the most strength the steel could lose if the entire building were on fire is half of the ambient temperature strength. Clearly something is wrong. And not just a little bit wrong. When we account for item 1, the number goes above 7/8th of the strength of the support columns was removed. Those effects will be shown in the next section.
For now just pause and take a moment to fully understand that it was impossible to have anything other than a pancake collapse without removing the strength of the support columns. How this was removed is supported by the overwhelming evidence of thermite in the dust collected after the collapse. So we know the strength was removed and we know a source of removing this strength. Jonathan Cole shows us how thermite can be used to cut steel and provides a highly credible scenario viewable at (cite reference to Jonathan Cole’s Video).
Removing the Air to Help See the Support Aspect of WTC Twin Towers
It might help to get a good mental picture of what is left if we remove all the air in the WTC towers and look at them as a single tower of steel and a concrete load representing each floor’s mass. The steel portion of the tower can be represented as a 13.4 feet by 13.4 feet steel shaft at the base. That shaft then tapers up to 5.15 feet by 5.15 feet at the top. It would be surrounded by a 35ft by 35 ft piece of concrete with a 13.4 ft square hole for the steel.
Maybe it would be easier to envision if this were a 6 feet tall round piece of steel. It would be 0.8 inches at the base and 0.3 inches at the top. It would be surrounded by a concrete cylinder with an outer diameter of 4 inches and have a 0.8 inch hole for the steel. If 3/4th of the strength of the steel were removed, it would accelerate at 20ft/sec2. At this scale, and with all the air out, I hope you can envision that nothing done at the top could ripple down without the above calculated loss of strength of all the steel supporting the mass.
Item 1 Accelerating the Stationary Floor Masses
To recap, this next section calculates the loss in velocity due only to accelerating stationary masses. This is what is called the conservation of energy part of the fall time analysis. It is the maximum speed the buildings could come down if there were no column structure and no shear present. The analysis assumes that once the 33 stories, and however many stories were accumulated, touched the next stationary floor, the shear bolts or welds were severed and the column strength was zero.
NIST’s response to being asked about the conservation of momentum is telling. It avoids the question altogether and talks instead about the planes hitting. The following is NIST’s answer to what they call FAQ 2. Why didn’t they include this simple analysis using basic physics?
2. Were the basic principles of conservation of momentum and energy satisfied in NIST’s analysis of the structural response of the towers to the aircraft impact and the fires?
Yes. The basic principles of conservation of momentum and conservation of energy were satisfied in these analyses.
In the case of the aircraft impact analyses, which involved a moving aircraft (velocity) and an initially stationary building, the analysis did, indeed, account for conservation of momentum and energy (kinetic energy, strain energy).
After each tower had finished oscillating from the aircraft impact, the subsequent degradation of the structure involved only minute (essentially zero) velocities. Thus, a static analysis of the structural response and collapse initiation was appropriate. Since the velocities were zero and since momentum is equal to mass times velocity, the momentum terms also equaled zero and therefore dropped out of the governing equations. The analyses accounted for conservation of energy.
It completely misses the point. The following analysis determines the time of collapse if the only force resisting free fall acceleration is the acceleration of the stationary masses. It relies on conservation of momentum. That is all that is modeled. It assumes no shear force is present and the columns offer zero resistance to collapse.
We will be calculating the time it takes for a building to collapse under the influence of gravity, once a failure is initiated. A model is developed that will determine the time for the collapse of a series of equally spaced masses falling one on top of another starting at the top. Each successive collision is assumed to be inelastic. In other words, each mass sticks to the next and the total mass accumulates as time goes on. The model is then altered to allow destruction to start at any floor. Beginning calculus and physics are a help in this section.
Definitions:
From the Wikipedia online dictionary:
An inelastic collision is a collision in which kinetic energy is not conserved.
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.
This section of the paper determines the time(t) it takes for a stack of many (n) masses (m), that are equally spaced(d), under the influence of gravity(g), to hit the ground. The collisions from mass to mass are considered inelastic. In other words the two masses stick together.
1) Conservation Of Momentum Calculation in an inelastic collision:
Using the conservation of momentum principle, where the subscript i indicates the initial velocity (Vi ) and the subscript f indicates the final velocity (Vf), for two masses m1 and m2:
m1 Vi1 + m2 Vi2 = (m1 + m2)Vf
It states that mass 1 times its velocity plus mass 2 times its velocity is equal to the sum of the masses times the final velocity. If the velocity of the second mass is 0, representing a stationary mass, the equation becomes:
Vf = m1Vi/(m1+m2)
if m1 and m2 are equal Vf = Vi/2
This makes sense for two equal masses hitting each other where the second mass is stationary. The velocity after the collision is half the velocity before the collision. Anyway, this is the conservation of momentum as it would apply to falling masses. By this same principle, if two equal masses are stuck together and hit a third mass, the final velocity of the three masses will be 2/3 of the velocity of the two masses before the collision.
In general: Vnf = mnVi/ ( mn + mn+1)
If this equation is normalized for a mass = 1 it simplifies to:
Vnf = (n/n+1)Vni
Which says the final velocity will be less than the initial velocity by the ratio of the number of masses sticking together over the total number. Which in this case is one more mass so it is n+1.
2) Time To Fall Between Hitting One Mass and the Next Mass
Now let’s figure out the time it takes, after the collision, to get to the next mass that is d feet away. We know that gravity exerts a constant force that accelerates objects at 32 ft./sec2. We know that for a constant acceleration, the velocity as a function of time is: V = at . From physics & calculus we know distance, as a function of time, is the integral of velocity.
We’ll start by finding the time it takes for the first mass to fall to the second. If t0, the arbitrary starting time, is taken as zero then the integral of a constant multiplied by time is ½at2. So distance d = ½ at2 .
Solving for time: t1 = (2d/a)1/2
The acceleration is due to gravity so the equation is: t1 = (2d/g)1/2
g is 32 feet per second squared so this becomes: t1 = (d/16)1/2
So pretty simple; the square root of the distance divided by 16. If the distance were sixteen feet then it would take exactly one second to drop.
Now we need to get this into a form we can use in an excel spreadsheet. The final velocity will be the initial velocity plus the average velocity change from block n to block n+1. Starting with the notion that the distance traveled is the average velocity multiplied by the time we can instead say the time (tn) is the distance traveled divided by the average velocity or: tn = d/Vaverage
We can now include initial conditions, which in this case is the initial velocity, to conclude that the final velocity is the initial velocity plus the average change in velocity. Knowing that the change in velocity is gtn , the average velocity change in the time from one block to the next is gtn/2.
This is written as: tn = d/Vaverage or tn = d/(Vinitial + gtn/2)
Since we know the initial velocity after a collision from above, we can find the time tn to go from block n to block n+1 after simplifying the equation by solving for tn. Right now tn is on both sides of the equation. Avoiding the subscript n this can be written as:
t(Vinitial + gt/2) = d
(g/2)t2 + Vinitial t - d = 0
This is recognizable as a second order polynomial of the general form:
ax2 + bx + c = 0 where the roots are x = {-b+(b2 - 4ac)1/2}/2a
In this case; a = g/2 or 16, b = Vinitial , and c = d, our distance between bricks.
So using our variables and constants we write this as:
t = {- Vinitial +( Vinitial 2 + 4(g/2)d)1/2}/(2g/2)
We can throw away the negative root, and after substituting in g=32 and shortening Vinitial to Vi , it looks like this:
t = {- Vi +( Vi 2 + 64d)1/2}/(32)
We have the time between blocks if we know the initial velocity and the distance.
So we can just run these calculations and keep track of distance, velocity, and time as the masses fall on each other. To recalculate the initial velocity after each mass is struck, we use the equation we developed at the beginning concerning inelastic collisions. So we know the initial velocity and distance and then calculate t. Then we find the final velocity knowing it is the initial velocity plus the change in velocity in time t using the equation Vn = gt. (Which reads: The change in velocity is equal to acceleration multiplied by time.) As we sum up the times, we get the total time as a function of the distance.
3) The Final Equations
Now that we have the equations we need we can use an excel spreadsheet to keep track of n and avoid all the pencil work. We also need to add the subscript n to the initial velocity.
The equations are: tn = {- Vni +( Vni 2 + 64d)1/2}/(32)
Vn = gtn
Vnf = (n/n+1)Vni
Excel makes this easy. It deals with the fact Vnf becomes V(n+1)i in the next calculation.
To summarize, what we have here is a set of equations where the only thing we are inputting is the distance between masses d and the number of masses n.
4) Seeing the Results in Excel Spreadsheet
So now we can insert an Excel spreadsheet and program it to find the tn for each n and sum them to see the total time as a function of distance. Similarly, the sheet can keep track of velocity before and after each collision. The final column has the time it takes to collapse if the only resistance to falling is the stationary masses being accelerated. There is no resistance in any other form. No shear forces and no resistance from the columns.
Floor From The Top Free Fall Time Free Fall Speed Velocity of WTC-2 WTC-2 Fall Time Sum of Masses Velocity Before Collision (ft/sec) Velocity After Collision (ft/sec) Time Between Collisions (Sec) Start With 30 Story Mass
0 0.00 0.0 0.0 0.0 30 0.00 0.00 0.00 0.00
1 0.88 28.2 11.1 1.11 31 28.17 27.3 0.88 0.88
2 1.24 39.8 26.9 1.57 32 39.20 38.0 0.37 1.25
3 1.52 48.8 35.0 1.93 33 47.29 45.9 0.29 1.54
4 1.76 56.3 41.6 2.23 34 53.81 52.2 0.25 1.79
5 1.97 63.0 47.2 2.49 35 59.34 57.6 0.22 2.02
10 2.78 89.1 68.6 3.52 40 79.08 77.1 0.16 2.92
20 3.94 126.0 98.3 4.98 50 102.64 100.6 0.12 4.29
30 4.82 154.3 121.0 6.10 60 118.56 116.6 0.11 5.42
40 5.57 178.2 140.0 7.04 70 131.25 129.4 0.10 6.42
50 6.22 199.2 156.7 7.87 80 142.16 140.4 0.09 7.33
60 6.82 218.2 171.8 8.63 90 151.94 150.2 0.08 8.18
70 7.37 235.7 185.7 9.32 100 160.92 159.3 0.08 8.98
80 7.87 252.0 198.6 9.96 110 169.29 167.8 0.07 9.73
90 8.35 267.3 210.7 10.56
100 8.80 281.7 222.2 11.14
110 9.23 295.5 233.0 11.68
Look at how close in time WTC-2 is at the 80th floor, where this model looses meaning. 9.73 seconds vs. 9.96 seconds. The only loss is due to accelerating stationary masses.
5) Understanding the Initial Results: The Case of No Shear or Column Strength
The model has been set up to be a general solution that lets the initial conditions be how much shear and how many stories start collapsing. The model lets us choose the breakaway force. Let’s start with the case where the break away force is zero, no shear whatsoever. What ever is holding the mass stationary, looses grip right when the mass from above reaches it.
What this shows is the losses due to accelerating stationary masses alone is almost enough to account for the fall time. This is with no resistance at the connections to the columns and also no resistance to falling from support column strength. The run stops at 80 floors because it doesn’t have consistent meaning once the initial 30 stories hits the ground.
Item 2 Shear Loss Effect on Fall Time
As another point of comparison, we determine what the collapse times would have been if the structure was strong but the floors pancake into one another. This theory seems reasonable. One would be intuitively correct to think the support structure would be stronger than the support of a given floor being hit. Doing the calculations both ways lets us not have to debate the point and gives insight either way. Note, however, that there is not a pile of floors at the bottom of the heap with intact pillars sticking up. There are also no indications of floor joists connected to support beams in the rubble, implying the energy to shear these connections was part of the controlled demolition. The calculations that follow will determine the speed as a function of the losses required to shear the floor’s connection from the support columns.
Calculating the shear losses of Item 2 are an extension of the math used to calculate Item 1. The velocity of the falling buildings, had these forces been present, is compared to the forces that must have been present to account for the fall times. No wonder NIST doesn’t want to consider a pancake model. Removing 5/6th of the strength of each connection, combined with the losses in Item 1, account for the actual fall times observed. This implies Item 3, column strength, was almost nonexistent and 5/6th of the shear strength was nonexistent. If you removed all the column strength, there would be nothing pushing back to shear against. Both strengths, shear and column support, were likely compromised about 95%.
WTC-2 Fall Time Analysis
Let’s analyze the first building to go down. This represents a worst case since only 12 stories are involved in WTC-1.
It is notable that the 0 shear, 1 shear and free fall cases are all closer to the time line than the case with 5 shear.
We can now conclude the two lines that envelope the WTC-2 predict between 0 and 1 shear load. This is the average resistance the structure presented to the falling mass. An average of less than one shear load. Less than a tenth of the structure’s designed-in-strength is present and resisting. This is actual proof of a controlled demolition. It is based on a simple analysis. A more complicated finite element analysis is not needed. No other analysis is needed to draw this conclusion.
WTC-7 Fall Time Analysis
This building gets much attention because it came so close to free fall and there is no mushroom clouding the view. The results show about the same shear force as the twin towers, one or less.
This analysis uses the same spread sheet as WTC-2 but looses a mass each time a floor hits the ground and so the crushing mass becomes one less for each floor hitting.
What we are really looking at is how nicely the WTC-7 fall time follows the case with one shear. Six shears gives us 16 seconds, which is twice as long as it took.
What happens in the middle region of the WTC-7 Velocity Graph is also significant. Which case has the closest velocity to the actual fall velocities? It is the case with no shear at all. In fact, to get the two to line up better for overall time, I did a run with a starting velocity of 28 ft/sec. They lined up better for the first half of the collapse. Such a starting condition implies three floors of support were removed at once, then a floor by floor progression.
This velocity graph brings up another opportunity. When the peak velocity occurs is a feature. With 5 shears it is before ten floors are down. With one shear it is before 20 floors are down and a deceleration should be present. It is also notable that the actual time line is much closer to free fall than to the case with 5 shear loads. The structural support was removed to account for the real time line.
Now I’m going to suggest that we all go ask our fearless leaders why they keep covering for the aliens. See Appendix 4 for a way to join the Architects and Engineers for 911 Truth..
Appendix 1 Dimensions of the Perimeter Columns
http://ift.tt/1kXbhih The perimeter column dimensions.
Appendix 2 WTC Towers: The Case For Controlled Demolition (excerpt, see link) (schoenfeld.one@gmail.com)
http://ift.tt/1RRSVZo
In this article we show that "top-down" controlled demolition
accurately accounts for the collapse times of the World Trade Center
towers. A top-down controlled demolition can be simply characterized
as a "pancake collapse" of a building missing its support columns.
This demolition profile requires that the support columns holding a
floor be destroyed just before that floor is collided with by the
upper falling masses. The net effect is a pancake-style collapse at
near free fall speed.
This model predicts a WTC 1 collapse time of 11.38 seconds, and a WTC
2 collapse time of 9.48 seconds. Those times accurately match the
seismographic data of those events.1 Refer to equations (1.9) and
(1.10) for details.
It should be noted that this model differs massively from the "natural
pancake collapse" in that the geometrical composition of the structure
is not considered (as it is physically destroyed). A natural pancake
collapse features a diminishing velocity rapidly approaching rest due
the resistance offered by the columns and surrounding "steel mesh".
Marty Beck, 4/20/2011
Analysis of the Fall Times of the Three World Trade Center Buildings
Abstract:
Explaining the collapse of the three World Trade Center buildings has been an enigma that strains the mind to come to grips with. Those who look at it over and over again try to make sense of what they see. The recorded fall times somehow seem reasonable to many of us when we rely on intuition in thinking about the collapse.
Many citizens are reasoning much like Galileo’s contemporaries must have. They easily assumed a heavier ball falls faster than a lighter ball. Until the experiment. We assume a large mass unleashed could fall at the rate we see. Before Galileo’s experiment they must have wondered why anyone would bother to do the experiment; isn’t it obvious?
While it looks obvious after watching the videos that something doesn’t add up, it is fairly easy to surmise that once collapse is initiated, the buildings would surely continue to collapse. One way to arrive at this assessment is based on being told a floor could hold 11 times its own weight (See NIST FAQ 1 in Appendix 3). So if you could get more than 11 floors to loose their main support, a collapse of some speed would follow. It would presumably gain more crushing mass and possibly more velocity as it continued down. But how fast should it fall? What resistance to falling did it encounter to account for the actual fall times? How does that compare to the ‘as built’ strength? This will be shown.
While even a simplified Finite Element Analysis (FEA) would show the impossibility of the observed fall times, it has not been forthcoming. This paper proves that each of the towers that collapsed that day was a controlled demolition. It does so by providing a concrete explanation of the basic physics and mechanics of the collapse of these steel structures.
The analysis is quite simple even though the starting conditions are absurd. NIST has managed to convince the populace that collapse was initiated by fire. So we will start there, collapse initiated by fire. Some number of floors start falling and the structure below succumbs to being hit by a moving mass with enough momentum to either sheer the floor it hits or crush the support columns. This paper does not deal with the impossibility of the initiation of collapse. Only what happened after the impossible collapse was initiated. What NIST chose not to share with us. What we learn also explains the initial collapse.
Introduction to the Analysis:
Only three physical agents serve to slow the falling floors:
1) Accelerating the stationary floor masses as the collapse ensues provides a calculable resistance to freefall. Its effect in decelerating the falling floors is calculated and graphed. Many others have done these calculations.
2) Shearing of floors at the support columns provides a calculable energy loss. Shearing can be seen as an impulse force, like hitting a stationary mass. It will be defined and explained in detail using NIST numbers for shear.
3) The strength of the columns themselves accounts for the other force decelerating the falling mass of floors. This can be seen as a continuous force resisting the collapse. We will calculate the loss in strength of the support columns to account for the fall times.
Item 1, as it pertains to the North and South Towers, is troubling on its own. One’s first notion is that this loss, the acceleration of a stationary mass, could practically be ignored since the shear strength and column strength provides proportionally much greater losses. Surprisingly however, just this loss alone accounts for half of the velocity decrease encountered when compared to free fall velocity. This will be shown. If you want to ignore this loss all together, that is fine with me.
Item 2, the loss in velocity if a floor were sheared where it connects to the support column, gets quite a bit of discussion. NIST quantifies this loss so the effects are easily calculated. This, however, is a point of controversy since NIST claims the collapse ‘was not a pancake’ event in one breath yet talks about pancaking in other breaths. (reference needed). This analysis will show that 5/6ths of the built in shear forces were not present or the twin towers would have fallen slower than either one did. If you want to ignore this loss all together, that is fine with me.
Item 3, the strength of the support columns, is the force you cannot ignore and the one that has been hardest to get non engineers to see. NIST entirely ignores this force and simply uses the term ‘buckle’. I will attempt to make the strength of the support columns much easier to envision by removing all the air in the building and representing it as the area of the steel columns supporting the masses of the floors.
The Twin Tower Fall Time Analysis
It would be easy to start with Item 1 but these effects have been shown over and over. There is another derivation in Appendix 2. The math and physics of this loss are fun to calculate however, and I will do so later, but not to lose the reader, I won’t start here. Item 2 will be discussed at the end. It is an extension of the model used to calculate Item 1. Hopefully you won’t need to read that far to easily conclude the columns strength was somehow removed.
Item 3: Strength of the Twin Tower Columns as a Function of Fall Time
The first step is to determine the ‘as built’ upward force provided by the columns. After that, we will need the weight of 33 floors that are assumed to be falling at 20ft/sec2. The next step will be to determine the column strength that is consistent with the observed fall times. The ratio of the two numbers represents the percent of lost strength the columns displayed during the collapse. It is that simple.
Strength of the Support Columns:
We have about 48 core columns and 236 perimeter columns. The perimeter columns are roughly 14 inches square and a quarter of an inch in thickness at the 80th floor (Appendix 1). So we have 14 square inches per column and a minimum of 30 thousand pounds per square inch of strength. (cite reference) (note about 70ksi) So combined, the columns can hold 30 x 103 pounds/in2 x 14in2 x 284 = 120 x 106 pounds before yielding(120 million pounds).
Calculate the Weight of the Concrete:
Now each floor is approx 200 ft by 200 ft of 4 inch thick concrete. The highest density I can find published for concrete is 150 pounds per cubic foot. That would be 50 pounds per square foot of floor area since the floor is 4 inches thick. So each floor weighs roughly 200ft X 200ft X 50pounds/ft2 = 2.0 million pounds so lets call it 2.4 million pounds with office furniture where we allow 2000 each of various 200 pound objects per floor. So 33 floors would weigh 80X106 pounds, (80 million pounds).
You are understanding things so far if you see that the steel columns can support 120 million pounds but only needs to support 80 million pounds.
Determine the Force the Falling Mass Exerts
We are interested in the seconds after the fall has been initiated. Note again that this analysis is accepting as a starting condition that the 33 stories are accelerating at 20 feet/sec2. This makes the problem is a little harder to see but we have to take this into account.
Weight is the result of mass in earth’s gravity field and the weight pushing down is the force defined as F=ma where m is the mass of the 33 stories and a is the acceleration due to gravity. On earth a is 32 ft/sec2.
So before the collapse was initiated, the weight was 80 million pounds. Once the upper 33 floors start to accelerate at 20 ft/sec2, the resulting force is now proportional to the difference between the 32 ft/sec2 and the 20 ft/sec2. This is 12 ft/sec2 with a resulting force of only 12/32 of what it was before the collapse.
Column Strength During Collapse:
So we had 80 x 106 pounds pushing down before the collapse and this force is reduced to 12/32 of what it had been to account for acceleration. 12/32 of 80 is 30. So now we have 30 x 106 pounds of force pushing down on a support structure that can withstand 120 x 106. In engineering terms this represents a margin of 4. A structural failure could not occur unless the margin was less than 1.
So without counting Item 1 and Item 2, we can conclude that 3/4th of strength of the columns were absent/removed to account for the observed fall times. That is 75%. We are told that the most strength the steel could lose if the entire building were on fire is half of the ambient temperature strength. Clearly something is wrong. And not just a little bit wrong. When we account for item 1, the number goes above 7/8th of the strength of the support columns was removed. Those effects will be shown in the next section.
For now just pause and take a moment to fully understand that it was impossible to have anything other than a pancake collapse without removing the strength of the support columns. How this was removed is supported by the overwhelming evidence of thermite in the dust collected after the collapse. So we know the strength was removed and we know a source of removing this strength. Jonathan Cole shows us how thermite can be used to cut steel and provides a highly credible scenario viewable at (cite reference to Jonathan Cole’s Video).
Removing the Air to Help See the Support Aspect of WTC Twin Towers
It might help to get a good mental picture of what is left if we remove all the air in the WTC towers and look at them as a single tower of steel and a concrete load representing each floor’s mass. The steel portion of the tower can be represented as a 13.4 feet by 13.4 feet steel shaft at the base. That shaft then tapers up to 5.15 feet by 5.15 feet at the top. It would be surrounded by a 35ft by 35 ft piece of concrete with a 13.4 ft square hole for the steel.
Maybe it would be easier to envision if this were a 6 feet tall round piece of steel. It would be 0.8 inches at the base and 0.3 inches at the top. It would be surrounded by a concrete cylinder with an outer diameter of 4 inches and have a 0.8 inch hole for the steel. If 3/4th of the strength of the steel were removed, it would accelerate at 20ft/sec2. At this scale, and with all the air out, I hope you can envision that nothing done at the top could ripple down without the above calculated loss of strength of all the steel supporting the mass.
Item 1 Accelerating the Stationary Floor Masses
To recap, this next section calculates the loss in velocity due only to accelerating stationary masses. This is what is called the conservation of energy part of the fall time analysis. It is the maximum speed the buildings could come down if there were no column structure and no shear present. The analysis assumes that once the 33 stories, and however many stories were accumulated, touched the next stationary floor, the shear bolts or welds were severed and the column strength was zero.
NIST’s response to being asked about the conservation of momentum is telling. It avoids the question altogether and talks instead about the planes hitting. The following is NIST’s answer to what they call FAQ 2. Why didn’t they include this simple analysis using basic physics?
2. Were the basic principles of conservation of momentum and energy satisfied in NIST’s analysis of the structural response of the towers to the aircraft impact and the fires?
Yes. The basic principles of conservation of momentum and conservation of energy were satisfied in these analyses.
In the case of the aircraft impact analyses, which involved a moving aircraft (velocity) and an initially stationary building, the analysis did, indeed, account for conservation of momentum and energy (kinetic energy, strain energy).
After each tower had finished oscillating from the aircraft impact, the subsequent degradation of the structure involved only minute (essentially zero) velocities. Thus, a static analysis of the structural response and collapse initiation was appropriate. Since the velocities were zero and since momentum is equal to mass times velocity, the momentum terms also equaled zero and therefore dropped out of the governing equations. The analyses accounted for conservation of energy.
It completely misses the point. The following analysis determines the time of collapse if the only force resisting free fall acceleration is the acceleration of the stationary masses. It relies on conservation of momentum. That is all that is modeled. It assumes no shear force is present and the columns offer zero resistance to collapse.
We will be calculating the time it takes for a building to collapse under the influence of gravity, once a failure is initiated. A model is developed that will determine the time for the collapse of a series of equally spaced masses falling one on top of another starting at the top. Each successive collision is assumed to be inelastic. In other words, each mass sticks to the next and the total mass accumulates as time goes on. The model is then altered to allow destruction to start at any floor. Beginning calculus and physics are a help in this section.
Definitions:
From the Wikipedia online dictionary:
An inelastic collision is a collision in which kinetic energy is not conserved.
An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter. Elastic collisions occur only if there is no net conversion of kinetic energy into other forms.
This section of the paper determines the time(t) it takes for a stack of many (n) masses (m), that are equally spaced(d), under the influence of gravity(g), to hit the ground. The collisions from mass to mass are considered inelastic. In other words the two masses stick together.
1) Conservation Of Momentum Calculation in an inelastic collision:
Using the conservation of momentum principle, where the subscript i indicates the initial velocity (Vi ) and the subscript f indicates the final velocity (Vf), for two masses m1 and m2:
m1 Vi1 + m2 Vi2 = (m1 + m2)Vf
It states that mass 1 times its velocity plus mass 2 times its velocity is equal to the sum of the masses times the final velocity. If the velocity of the second mass is 0, representing a stationary mass, the equation becomes:
Vf = m1Vi/(m1+m2)
if m1 and m2 are equal Vf = Vi/2
This makes sense for two equal masses hitting each other where the second mass is stationary. The velocity after the collision is half the velocity before the collision. Anyway, this is the conservation of momentum as it would apply to falling masses. By this same principle, if two equal masses are stuck together and hit a third mass, the final velocity of the three masses will be 2/3 of the velocity of the two masses before the collision.
In general: Vnf = mnVi/ ( mn + mn+1)
If this equation is normalized for a mass = 1 it simplifies to:
Vnf = (n/n+1)Vni
Which says the final velocity will be less than the initial velocity by the ratio of the number of masses sticking together over the total number. Which in this case is one more mass so it is n+1.
2) Time To Fall Between Hitting One Mass and the Next Mass
Now let’s figure out the time it takes, after the collision, to get to the next mass that is d feet away. We know that gravity exerts a constant force that accelerates objects at 32 ft./sec2. We know that for a constant acceleration, the velocity as a function of time is: V = at . From physics & calculus we know distance, as a function of time, is the integral of velocity.
We’ll start by finding the time it takes for the first mass to fall to the second. If t0, the arbitrary starting time, is taken as zero then the integral of a constant multiplied by time is ½at2. So distance d = ½ at2 .
Solving for time: t1 = (2d/a)1/2
The acceleration is due to gravity so the equation is: t1 = (2d/g)1/2
g is 32 feet per second squared so this becomes: t1 = (d/16)1/2
So pretty simple; the square root of the distance divided by 16. If the distance were sixteen feet then it would take exactly one second to drop.
Now we need to get this into a form we can use in an excel spreadsheet. The final velocity will be the initial velocity plus the average velocity change from block n to block n+1. Starting with the notion that the distance traveled is the average velocity multiplied by the time we can instead say the time (tn) is the distance traveled divided by the average velocity or: tn = d/Vaverage
We can now include initial conditions, which in this case is the initial velocity, to conclude that the final velocity is the initial velocity plus the average change in velocity. Knowing that the change in velocity is gtn , the average velocity change in the time from one block to the next is gtn/2.
This is written as: tn = d/Vaverage or tn = d/(Vinitial + gtn/2)
Since we know the initial velocity after a collision from above, we can find the time tn to go from block n to block n+1 after simplifying the equation by solving for tn. Right now tn is on both sides of the equation. Avoiding the subscript n this can be written as:
t(Vinitial + gt/2) = d
(g/2)t2 + Vinitial t - d = 0
This is recognizable as a second order polynomial of the general form:
ax2 + bx + c = 0 where the roots are x = {-b+(b2 - 4ac)1/2}/2a
In this case; a = g/2 or 16, b = Vinitial , and c = d, our distance between bricks.
So using our variables and constants we write this as:
t = {- Vinitial +( Vinitial 2 + 4(g/2)d)1/2}/(2g/2)
We can throw away the negative root, and after substituting in g=32 and shortening Vinitial to Vi , it looks like this:
t = {- Vi +( Vi 2 + 64d)1/2}/(32)
We have the time between blocks if we know the initial velocity and the distance.
So we can just run these calculations and keep track of distance, velocity, and time as the masses fall on each other. To recalculate the initial velocity after each mass is struck, we use the equation we developed at the beginning concerning inelastic collisions. So we know the initial velocity and distance and then calculate t. Then we find the final velocity knowing it is the initial velocity plus the change in velocity in time t using the equation Vn = gt. (Which reads: The change in velocity is equal to acceleration multiplied by time.) As we sum up the times, we get the total time as a function of the distance.
3) The Final Equations
Now that we have the equations we need we can use an excel spreadsheet to keep track of n and avoid all the pencil work. We also need to add the subscript n to the initial velocity.
The equations are: tn = {- Vni +( Vni 2 + 64d)1/2}/(32)
Vn = gtn
Vnf = (n/n+1)Vni
Excel makes this easy. It deals with the fact Vnf becomes V(n+1)i in the next calculation.
To summarize, what we have here is a set of equations where the only thing we are inputting is the distance between masses d and the number of masses n.
4) Seeing the Results in Excel Spreadsheet
So now we can insert an Excel spreadsheet and program it to find the tn for each n and sum them to see the total time as a function of distance. Similarly, the sheet can keep track of velocity before and after each collision. The final column has the time it takes to collapse if the only resistance to falling is the stationary masses being accelerated. There is no resistance in any other form. No shear forces and no resistance from the columns.
Floor From The Top Free Fall Time Free Fall Speed Velocity of WTC-2 WTC-2 Fall Time Sum of Masses Velocity Before Collision (ft/sec) Velocity After Collision (ft/sec) Time Between Collisions (Sec) Start With 30 Story Mass
0 0.00 0.0 0.0 0.0 30 0.00 0.00 0.00 0.00
1 0.88 28.2 11.1 1.11 31 28.17 27.3 0.88 0.88
2 1.24 39.8 26.9 1.57 32 39.20 38.0 0.37 1.25
3 1.52 48.8 35.0 1.93 33 47.29 45.9 0.29 1.54
4 1.76 56.3 41.6 2.23 34 53.81 52.2 0.25 1.79
5 1.97 63.0 47.2 2.49 35 59.34 57.6 0.22 2.02
10 2.78 89.1 68.6 3.52 40 79.08 77.1 0.16 2.92
20 3.94 126.0 98.3 4.98 50 102.64 100.6 0.12 4.29
30 4.82 154.3 121.0 6.10 60 118.56 116.6 0.11 5.42
40 5.57 178.2 140.0 7.04 70 131.25 129.4 0.10 6.42
50 6.22 199.2 156.7 7.87 80 142.16 140.4 0.09 7.33
60 6.82 218.2 171.8 8.63 90 151.94 150.2 0.08 8.18
70 7.37 235.7 185.7 9.32 100 160.92 159.3 0.08 8.98
80 7.87 252.0 198.6 9.96 110 169.29 167.8 0.07 9.73
90 8.35 267.3 210.7 10.56
100 8.80 281.7 222.2 11.14
110 9.23 295.5 233.0 11.68
Look at how close in time WTC-2 is at the 80th floor, where this model looses meaning. 9.73 seconds vs. 9.96 seconds. The only loss is due to accelerating stationary masses.
5) Understanding the Initial Results: The Case of No Shear or Column Strength
The model has been set up to be a general solution that lets the initial conditions be how much shear and how many stories start collapsing. The model lets us choose the breakaway force. Let’s start with the case where the break away force is zero, no shear whatsoever. What ever is holding the mass stationary, looses grip right when the mass from above reaches it.
What this shows is the losses due to accelerating stationary masses alone is almost enough to account for the fall time. This is with no resistance at the connections to the columns and also no resistance to falling from support column strength. The run stops at 80 floors because it doesn’t have consistent meaning once the initial 30 stories hits the ground.
Item 2 Shear Loss Effect on Fall Time
As another point of comparison, we determine what the collapse times would have been if the structure was strong but the floors pancake into one another. This theory seems reasonable. One would be intuitively correct to think the support structure would be stronger than the support of a given floor being hit. Doing the calculations both ways lets us not have to debate the point and gives insight either way. Note, however, that there is not a pile of floors at the bottom of the heap with intact pillars sticking up. There are also no indications of floor joists connected to support beams in the rubble, implying the energy to shear these connections was part of the controlled demolition. The calculations that follow will determine the speed as a function of the losses required to shear the floor’s connection from the support columns.
Calculating the shear losses of Item 2 are an extension of the math used to calculate Item 1. The velocity of the falling buildings, had these forces been present, is compared to the forces that must have been present to account for the fall times. No wonder NIST doesn’t want to consider a pancake model. Removing 5/6th of the strength of each connection, combined with the losses in Item 1, account for the actual fall times observed. This implies Item 3, column strength, was almost nonexistent and 5/6th of the shear strength was nonexistent. If you removed all the column strength, there would be nothing pushing back to shear against. Both strengths, shear and column support, were likely compromised about 95%.
WTC-2 Fall Time Analysis
Let’s analyze the first building to go down. This represents a worst case since only 12 stories are involved in WTC-1.
It is notable that the 0 shear, 1 shear and free fall cases are all closer to the time line than the case with 5 shear.
We can now conclude the two lines that envelope the WTC-2 predict between 0 and 1 shear load. This is the average resistance the structure presented to the falling mass. An average of less than one shear load. Less than a tenth of the structure’s designed-in-strength is present and resisting. This is actual proof of a controlled demolition. It is based on a simple analysis. A more complicated finite element analysis is not needed. No other analysis is needed to draw this conclusion.
WTC-7 Fall Time Analysis
This building gets much attention because it came so close to free fall and there is no mushroom clouding the view. The results show about the same shear force as the twin towers, one or less.
This analysis uses the same spread sheet as WTC-2 but looses a mass each time a floor hits the ground and so the crushing mass becomes one less for each floor hitting.
What we are really looking at is how nicely the WTC-7 fall time follows the case with one shear. Six shears gives us 16 seconds, which is twice as long as it took.
What happens in the middle region of the WTC-7 Velocity Graph is also significant. Which case has the closest velocity to the actual fall velocities? It is the case with no shear at all. In fact, to get the two to line up better for overall time, I did a run with a starting velocity of 28 ft/sec. They lined up better for the first half of the collapse. Such a starting condition implies three floors of support were removed at once, then a floor by floor progression.
This velocity graph brings up another opportunity. When the peak velocity occurs is a feature. With 5 shears it is before ten floors are down. With one shear it is before 20 floors are down and a deceleration should be present. It is also notable that the actual time line is much closer to free fall than to the case with 5 shear loads. The structural support was removed to account for the real time line.
Now I’m going to suggest that we all go ask our fearless leaders why they keep covering for the aliens. See Appendix 4 for a way to join the Architects and Engineers for 911 Truth..
Appendix 1 Dimensions of the Perimeter Columns
http://ift.tt/1kXbhih The perimeter column dimensions.
Appendix 2 WTC Towers: The Case For Controlled Demolition (excerpt, see link) (schoenfeld.one@gmail.com)
http://ift.tt/1RRSVZo
In this article we show that "top-down" controlled demolition
accurately accounts for the collapse times of the World Trade Center
towers. A top-down controlled demolition can be simply characterized
as a "pancake collapse" of a building missing its support columns.
This demolition profile requires that the support columns holding a
floor be destroyed just before that floor is collided with by the
upper falling masses. The net effect is a pancake-style collapse at
near free fall speed.
This model predicts a WTC 1 collapse time of 11.38 seconds, and a WTC
2 collapse time of 9.48 seconds. Those times accurately match the
seismographic data of those events.1 Refer to equations (1.9) and
(1.10) for details.
It should be noted that this model differs massively from the "natural
pancake collapse" in that the geometrical composition of the structure
is not considered (as it is physically destroyed). A natural pancake
collapse features a diminishing velocity rapidly approaching rest due
the resistance offered by the columns and surrounding "steel mesh".
via International Skeptics Forum http://ift.tt/1sQ5Y9g
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