World Trade Center Dust Bowl Analysis

by Buddy McCloskey     03/07/2008     Contact

1. Purpose:

I am encouraged to upgrade a dying model that was left to decay in time. Actually, that is relative to time - where time took off in sophistication of computer technology advances, which should need no explanation, and an old Navy Shipboard Radioactive Fallout Model went to mothballs. This was an update from an earlier version of ESSA's Radioactive Fallout Model. In the early 80', I updated it with better parameters from many graphs and more accurate particle fall velocities in FORTRAN IV and presented to Pa. Society of Professional Engineers in Phila while being that City's Air Pollution Meteorologist/Programmer. The presentation was entitled "The Day: Blueprint For Survival." delivered on Groundhog Day 1984.

Today with NOAA's better computer models for upper wind predictions, it is time to update it all again and upload it on the internet for daily use. I have help from the two pioneer retirees that developed each of the earlier programs. In order for a smoother pattern of prediction to be available, which NATO had just recently discovered, we need more particles than the 51. So, trying to extend a frequency distribution curve of particle activity by fitting a normal distribution curves seemed the best way. However, the % particle distribution between Navy and Essa (which had even less group of particles) were based on different nuclear test explosions with different results.

I thought it better to develop a model for coalescing from 1 micron (u) from the amount of ground crater dust. This finally produced some results after the final program ran 60 hours of CPU time which will be incorporated into the model. The results being too different from the original model particle size distribution assumption, I wondered which to use. However, after examining the explosive power of 1 megaton bombs, the flow rates return updraft is so strong, that the stem of the developing radioactive cloud is acting like a tornado, sucking nearly everything in within a 1/2 mile or so for nearly 20 seconds. This means, the larger the megaton, the stronger and longer the updraft remains. At the test sites, this would vacuum dust and sand particles as a 2nd source. If in a city, dust from fallen buildings with 2-5 psi overpressure would be the 2nd source.

The problem is, how much is converted to dust and what is the % Particle Distribution? A stone quarry that crushes limestone rock estimated 1%. Another friend 2%, and I estimated about 8%. After seeing videos of a portable Bulldog Crusher that cleans up after on site demolition, it looks like about 4%. Lets see what they found out about the WTC dust analysis. The Internet has many photos, videos and interviews. They are so surprised to see that there is hardly any dust at ground zero. There were some particle distribution at some location so we need to determine down wind estimates to confirm the input into a WTC dust model.

When viewing tables, diagrams and pictures below, click on the underlined color link then use your browser's <= Back icon to return to this text. For those that don't want to labor through all this, you can click on the last referenced table at the bottom of this report to see the end result, the calculated dust fall from the WTC downwind for 2 miles. Also, if you want to see the input parameter output table, rather than read the derivation and explanation. For a type of summary, see wtcpar.txt . Then you could check <= Back into here to find any the details of any particular item of interest.

2. % Particle Distribution:

Since no samples were collected was such a shame, lets research to find some. In '96, Fl. Int University tried to determine control methods for < 10u size control. A 100 lb ball was dropped on a hollow concrete block. With no control, 75% was below 10u, and peaked at 4-10u with 37.2%. They changed u groupings intervals, to > 10, 7-10, 4-7, 2-4 and 1-2, .7-1, .5-7, and .3-.5 using a Hiac/Royco Micro Air Model #5230 airborne particle counter. I have seen most statements that dust usually is >= 1u.

Wanting to verify, and find similar results, why not do our own by using dust produced from rubbing 2 solid cement blocks together? More details on this below. The dust was allowed to settle on plexiglass and glass under no breeze conditions. This was done inside a glass sun porch, and then an outdoor porch on a quiet calm evening. I rubbed a few inches above the glass surfaces that was in a box. In both cases, we scanned an area that wasn't heavily loaded with many overlapping particles that would count as larger particles but still want as many dust particles as possible.

The 1st scan uses an older HP Deskscan at 1600 dpi max. There were 3 rubs, scans. Medium, light then heavy. A cropped portion of the medium rub is seen in 16acrop2.gif . All 3 were combined to produce more than 72,000 particles. Originally, they were scanned at B&W only, converted to a 256 color PCX file which I can work with since I understand the PCX compression run code. In 'C' language, to count and convert blobs of dust requires a few programs. It takes two to find only the now white particle pixel runs and makes binary outputs files for the multiple ranges of x pixels for each down row of y pixels.

Another program analyzes each row of y pixels for the white dust x-pixels runs, and when it finds one set, it jumps to the next row to find the adjacent corresponding run, jumps to the next line until all the lower adjacent rows of white pixels run out. Next, it goes back to the 1st row of the 1st dust particle pixel to examine other dust particles that may begin on that row also. If not, we repeat the process at the 2nd row of the 1st dust particle. This is a logical multi loop program that is the core of this process. A display of the process helps to see if all is working well. That is, the plot completely whitens a dust particle before it goes to the next particle instead of completing a line at a time showing the top portions of multiple adjacent white particles.

As each particle is being compiled, we sum each white color run of that particle. Therefore, we end up with a total number of white pixels which is only what we need to send to a file. The next program, converts these different pixel sums (areas) to an equivalent pixel diameter then to microns (u) and counts each sum area as a dust particle. Let O area=pixs=pi*r^2, r=sqrt(pixs/pi) and diam=2r where the areas are in units of pixels. Here, the initial square particle size is par0=mmpi*1.e3/dpi where mmpi=25.4 mm/in, 1.e3=u/mm and dpi=1600 dot/in, par0=15.875 u/dot. However, that's a square area, better to use an equivalent area circle with radius r=15.875u/sqrt(pi)=17.913 u/dot or u/pix. So we correct diameter in pixels to diameter in u's by pix's * par0 = pix * u/pix = u.

Using the 2nd scanner, an Epson with Twain 5, we scanned at 9,600 and 12,800 dpi. These cropped portions are seen in 96aRC.gif and 128cRC.gif. The 12,800 dpi is 8 times more resolution = 2.24 u/pix. The problem with 1600 dpi, is that we have the data bunching up at the central tendency pixels as 1= 17.9u, 2=25.3u, 3=31.0u, 4=35.8, 5=40.1 u's etc. At 12,800 dpi, we have bunching pixs starting at 1= 2.24u. Note, the more dots, the less blank u intervals. So we have more detail with smaller particles detected. This will become our standard input. There are still some gaps in the smaller u ranges, so we linearly interpolate the central tendency by dividing and weighting the frequency to spread it between the adjacent blank u's. This is our 1st smoothing process.

However, a plot of this data still shows a jagged rough plot and some more smoothing is required. On line, a downloaded plotting file called GNUplot was available. After unzipping, we find it in c:\gp420win32\gnuplot\bin\wgnuplot with a sample command as plot 'C:\radfo\1600.%' using 2:5 smooth Bezier; set xrange [0:550]; set yrange [0:1.2] produced the following 1600 dpi plot. The smoothing technique fits tangents of curves to data as done by Bezier and explained a little by a an example diagram. I couldn't save the new image plots correctly so I set my screen resolution to a max 1600x1200 and set a Corel Screen Capture tool utility. Also note the 9,600 dpi plot and the 12,800 dpi plot.

Since the curve smoothed the data points, did it change the data because we are dealing with % contribution. If the sum of % distribution is not 100%, we sum each new data point and ratio that to 100%, then multiply each % frequency by this correction factor. However, since they hid the new data points to make the smooth curve so we can only relate x to y by pixels. This is why we set the screen to max resolution. In our next program, we first convert and scale the pixels vs the data for x vs y and we have a file table of x vs y only. We rotate the axes 90 degrees, scale and relate delta x to delta x-pix and delta y to delta y-pix using origin and axes end points. The program reads the x & y pixels, then calculates the x & y values. If we have a straight line in the plot, then we may have 2+ x-pixels for a given y-pixels. Therefore, we have to interpolate values between fractions of a pixel. We need a unique % particle distribution (pd) for each u diameter. As for the accumulated sums, it was < .8% for 1600 dpi, but about < 3% for 12,800 dpi.

3. WTC Data and Parameters:

Some of the data is needed for the Dust Box model, others are of interest and may be used later. This program is just a compilation of unit conversions, hoping they are applied correctly that the units are canceled properly, making sure we multiply or divide when needed. It would be nice for someone to double check on the formulae assumptions and make sure the conversions are used correctly. The probability for just one mistake is plentiful and can cause a chain reaction of other wrong values. There are just constants and short fragments of code and write outs to a file. What is nice is that there are plenty of comments in the code and the program can be run again by just substituting new values.

My source of information is from the various web articles and textbook referenced widely accepted terms and values. The web sites are from Jim Hoffman and Jerry Russell. Also, help and consultation with Portland Cement Assoc of Prof Engineers - Terry Collins.

4. Cement Volume Estimates:

Here is the most important parameter, the amount of total cement from both WTC's. Originally 425e3 yd^3 total for both buildings, now reduced to 2*90e3 tons for both buildings. The lightest and strongest cement density for building then was 1.84 g/cm^3 which is the Structural Light Weight Concrete Density by Code. Using this with the ratio to water of 62.43 lb/ft^3/(g/cm^3), den=1.84 g/cm^3 = 114.87 lb/ft^3. So 2*90e3 ton x 2.e3 lb/ton/114.87 lb/ft^3 = 3.134e+6 ft^3 ~ 27% of 1st estimate. OK, massive overestimate was reported so working with that, also quoted was 4" concrete slabs were poured for each of the 110 floors. So slab volume = 2*110*side*side*(slabt/12) = 2*110*208'*208'*(4/12)' = 3.173e+6 '^3 = or 1.01 Cement Vol. Thus this shows that there are no walls attached to the WTC's. Another quote floor slabs can be strong at 2 1/2" thick. I guess truss was ample for this, so new slab volume = 1.983e+6 '^3 = 0.63 Cement Vol. For Calculated Wall Thickness, (Cement Vol - Slab Vol)/2 = Wall Vol for 1 bldg. Then Wall Thickness = Wall Vol/(4*side*wtch) where wtch=(wtch1+wtch2)/2 = (wtch1=1368 + wtch2=1362)2 = 1365 Av WTC Heights'. Wall Vol = 3.134e+6 '^3 (Total) - 1.983e+6 '^3 (slab)/2 bldg's divided by Wall Area 4*208'*1365' = .507' = 6.1".

At least we have walls now but would you feel safe in that? Last adjustment - maybe they were considering the accepted density of cement as 2.4 g/cm^3, not the 1.84 g/cm^3 code. Doing it all again, we now have a higher new total volume of cement to subtract the slab volume from. That is Cement Vol = 4.088e+6'^3 in 1st term above, Calculated Wall Thickness = 11.31". Here, we subtract the slab volume from the building volume and so we must then also subtract the accumulated slab height (22.92') from the building height also to leave only the wall, no slabs. That looks close to the videos of the collapsing buildings and had used 1' as seen below. By the way, the total building vol is wtch*side*side = 1365'*208'*208'= 5.906e+7'^3, so Cement Vol/Total Building Vol = 2.044e+6 '^3 /5.906e+7 '^3 = 3.461%. So now we use the 2nd total cement volume or 36.5% of their 1st estimate.

Another consideration for Wall Thickness is the glass windows taking the place of cement volumes. By photos, it appears that we had 60 windows per side. So window plus frames width was 208'/60 = 3.466. Frame width = window width so window width = 1.73'. Glass height was estimated 8'. So glassv = 4 sides x 60 windows x 8' height x 1.73' width x wallt x 110 floors = 345031.2'^3 where wallt is our last calculated wall thickness (11.31") without windows. So New Wall Thickness = WTC vol/(4 x side x wtch - glassv) = 15.97 in.

5. Rubble Volume:

Based on statements that was left at the footprint base, twisted steel and large chunks of cement, how high was the rubble? Heresy, about 2-3 stories. Well, this is another sensitive parameter and we know that the pile is not completely compact but large air gaps. To compensate for this, we use a fluff factor of 1.873 as explained in F. Dust Flakes below. The cone rubble is fluffed up by air spaces by 1.873 of what the compact rubble volume should be - so we divide the calculated volume by this factor. We choose an average of 2.5 floors, rule of thumb by the observer and pics, with ch=1365'/110 floor = 12.4'/floor =31.0'. Now, what is more sensitive is the base area of the pile. The best approximation shape would be the volume of a cone. However, the volume of steel must first be calculated. No controversy here. Let tonss=2*96e3 Tons of Steel for Both Bldgs and denrs=7.85 Density of Reinforced Steel in g/cm^3. So dens=denrs*denw = 7.85g/cm^3 x 62.43 lb/ft^3/(g/cm^3) = 490.08 lb/ft^3. Now vols= tonss*2.e3/dens = 2*96e3*2e3/490.08 = 7.84e+5'^3.

After viewing the collapse videos and pics, I estimate that this umbrella cone fell to about 40% more than the diagonal of the building. So Base Radius of Debris Cone rc=1.4*side/2*sqrt(2) = 205.9'. Except when we view pics of the rubble, it's not a cone with a pointed top. Actually, it appears as the cone is only 2/3 complete. So we imagine that we see only 2/3 of the height, then the other 1/3 is an imaginary height if the cone were extended to a point. So if 2/3 hci= hc (31.0), hci=3/2 hc = 46.5'. If we're 2/3 up the imaginary cone height hci, then we're also 1/3 down from it's peak so the base radius of the invisible cone or top section, rct=rcb/3 = 68.6' by similar triangles. See cone.dia.

Cone volume V=pi r^2 h/3. Total cones = 2.07e+6'^3 and the invisible top cone is 7.65e+4'^3. Subtracting the top imaginary cone leaves volume of visible portion of cone = 1.99e+6'^3. Correcting for 2 buildings and fluff factor, we have intersecting double cones of 2.12e+6'^3. Subtracting off the steel (7.84e+5'^3) leaves 1.34e+6'^3 of cement then /4.088e+6'^3 = 32.80% of cement remains in the cone rubble and 67.20% goes to wind blown dust that will be input into the dust bowl model.

6. Massive Dust Generation Mechanisms:

Before the model run, we might examine this theory of the WTC's dust production for guidance in estimating other buildings that are brought down by controlled demolition or a nuclear bomb. How strong is the cement? Reference was 6000 lb/"^3 and quotes up to 20,000 lb/"^3 for today are possible. Using compression strength as 6000 psi, is this strong enough to hold the building without any steel supports? Simply, comp=wtch*den = 1365'*114.87 lb/'^3/144"^2/'^2 = 1088.9 lbs/"^2 or only 18% of 6000 psi standard compression limit. The aircraft had cut through floors 78-84 leaving floors above fa=110-84=26. The average height from floor to floor or ceiling to ceiling is ch=wtch/floors where wtch=1365 and floors=110, ch= 12.4'. So the weight of the upper section just resting on top of the remainder of the lower section, before it was cut is comp=fa*ch*den = 26*12.4*114.87/144 = 257.4 lb/in^2 or only 4.3% of compression limit.

A. Let Gravity do the Work:

We now consider the gravity of the situation. Even though there was a gap of 6 floors, when the upper section fell, it tilted and the one wall supported nearly 1/2 the weight. The topple photo and step frames of the video show the maximum tilt just as it hit the floor below on the opposite side. Printing out a photo, we see the apparent parallax tilt from a camera - shame the photo wasn't at a right angle to the tilt. However, I extended the rulers from the photo between the parallax tilt from another building oriented the same as in front of WTC and the difference between the parallax and the top tilted section roof was 7.5 degrees. If there was no tilt, both would show the same parallax angle and those lines would be parallel. However, this 7.5 deg tilt occurred after the crushing was in progress. Looking at the videos and freezing frame, rulers across the monitor indicate that the max tilt was 8.5 deg during the earlier free fall say right before the upper section closed the gap. Now because the gap along the opposite adjacent may not have been all air, there was probably some resistance due to crushing of these sides. Not really that visible so lets assume no resistance and a real free fall. How far did it fall? Well chtilt=side*tan(tilt), where tilt=8.5 deg so chtilt=208*tan(8.5) = 31.09'= 2.2 floors out of reported gap of 6 floors.

If there were no tilt and these floors were suddenly pulled out, the upper section would have fallen vertically, and landed on all 4 edges beneath. Since we really don't have wall thickness exactly, we estimated the weight of the upper section by proportioning that height to total height which we have the weight for or vag=fa/110*vol2/2 = 26/110*4.088e+006/2 = 4.831e+005 '^3 and weight wag=vag*den = 4.831e+005 * 114.87 lb/ft^3. = 5.549e+007 lb = 2.77e4 tons. We really don't need this mass to work with but just gives us an idea.

A free fall distance is given by dis=VoT+aT^2/2, with Vo=0, dis=aT^2/2, solve for T=(2*dis/a)^1/2. With a=ga=32.1722'/sec^2, T=gsec and dis=wtch=1365', we have wtch=ga*gsec^2/2, gsec=(2*1365/32.1722)^1/2 = 9.21 sec for fall travel time. After a review of seismic activity and videos, they concluded that it actually took 13 sec for the complete building to fall or 41.1% slower than gravity. The terminal fall velocity by gravity is vt=gsec*ga = 9.21 sec * 32.1722'/sec^2 = 296.36 '/sec or 202.06 mph. Actual average downward vertical velocity = avs = wtch/asec = 1365/13 = 105 '/sec. For the free fall distance through the gap from this upper section side wall, we replace dis by chtilt = 31.09', T=(2*dis/a)^1/2 = gsec = sqrt(2*chtilt/ga) = (2*31.09/32.1722)^1/2 = 1.39 sec free fall travel time. Now vt=gsec*ga again = 1.39 sec * 32.1722'/sec^2 = 44.7'/sec or 30.5 mph.

B. How Many G's?:

Now when a body stops after a fall through air, it's velocity = 0. The deceleration term is also in units of acceleration ('/sec^2). We take gravity, after 1 sec time, it's speed is 32.1722'/sec or 21.9 mph as seen a little slower than we just did above. Now if we are the falling body and we are standing on a board, neglecting feeling the wind blowing us by, but actually it's vica versa, it takes 1 sec for the board to come to a stop, then we go from 21.9 -> 0 mph. During this 1 sec, we would feel our weight come back from weightlessness and we would feel normal at the end of 1 sec. That means our deceleration was 1 G against us. So the G's are the ratio of acceleration/deceleration. That's why we have collapsible bumpers, connected to spring shocks. Car shocks also, where air being compressed slows deceleration time. Of course Safety Air Bags is another example. So for gravity acceleration, then ga=(vg2-vg1)/gsec where vg's are the result of the constant ga but vg2 > vg1. Let deceleration = da = (vd2-vd1)/dsec but vd2 > vd1. For ga, vg1=0, and for da, vd1=0, Substituting with 0's, ga = (vg2-0)/gsec = vg2/gsec and da = (vd2-0)/dsec = vd2/dsec. With G's = da/ga = (vd2/dsec)/(vg2/gsec). But notice, just as vg1=vd1=0 are the initial and terminal velocities respectively, vd2=vg2 are also the initial and terminal velocities respectively. The terminal velocity at the end of gravity acceleration after a certain time, becomes the initial deceleration velocity to be slowed to 0. So now G's = da/ga = (vd2/dsec)/(vg2/gsec) = (vg2/dsec)/(vg2/gsec) = gsec/dsec. So we have gsec, but what about the deceleration dsec term?

C. Elasticity - Stretching the Term a Bit:

Each material has these properties but the units as applied to here is not my expertise although I may do a quick research to determine this complexity. However, for now to save time, I just want to do some wild guesses for sec and lets see what results the math produce. Others may do this more exactly if they so desire as we only want to prove a point here and not become too quantitative. If we drop a ball bearing on a cement block, how long would it take to bounce back? Obviously, the higher we drop it the faster and the more it would sink in. However, the limit is the 6000 psi Compression Limit as the cement block would disintegrate and the ball bearing would not bounce back. The speed (say buckshot's from a rifle) and larger ball bearings would make a difference. So we guess that with a smaller and thrown down from a higher height, it would bounce back in about .05 sec? That a guess a high speed film could catch and probably have. We will be conservative though and we should agree that it would be bouncing back within .2 sec? A steel ball bearing must have a small elasticity, but we are considering 2 cement blocks trying to bounce and each gives a little or considered twice the distance to slow down. Even then, I'm sure .2 dsec is stretching it too far? The only info I have discovered on the internet for elasticity is the additive to make cement stronger for stress and shear so for now, we use a conservative .2 unless somebody knows of some more realistic number.

D. Applying the G's:

Now G's = gsec/dsec = gsec/.2 = 5 gsec = 5*1.39 = 6.95. Now we come back to the top just resting on the lower section was 257.4 lb/in^2, so * 6.95 G's = 1788.9 lb/in2 for the non-tilt gravity fall compression which is 30% of the compression limit. So even if the top section fell vertically and landed evenly on the floors below, the cement should have stood. However, the 8.5 degree was a tilt and one side of the building landed on the lower section. The opposite side supported half the weight but the 2 adjoining sides we assumed did little to slow down the acceleration because of the gap. So, the compression on this one side without the fall is ucomp=(wag/2)/(sidex1') = (5.549e+007/2 lb)/(208'x1') = 1.334e5 lb/'^2 = 926.4 lb/in^2 or 15% of compression limit. However, x 6.95 G's = 6.44e+003 lb/in2 = 107% Comp Limit. Here we have started to crush down the wall. Now actually, if that falling side is also tilted perpendicular to the fall, a corner would be exposed and fall on the side below instead. Suppose that corner is a 1' x 3', then ucomp=(wag/2)/(3'x1') = (5.549e+007/2 lb)/(3'x1') = 9.249e6 lb/'^2 = 6.42e4 lb/in^2' already > 10 times compression limit at rest on the corner. Due to fall, now x 6.95 G's = 4.46e+005 lb/in2 or 74.4% of comp limit. So with a tilt with 1 side hitting 1st floor below, we're already 7% over, tilt the other axis to make a corner hitting, we could easily go to 75 times or 7500% Compression Limit.

E. Results:

One quote said that this was like a sledge hammer that started the crush. Actually, it was like a wood axe, with a sledge on one side, but a wedge on the other and the tilt is using the wedge side. It was so effective that the upper tilted section hardly had any resistance to only tilt back 1 degree as it was crushing beneath. We would think that the sharp wedge would also be crumbling, dulling the blade, because we would have a wider edge than the initial 1' estimated. This would allow the psi to lessen rapidly but it appears to have never drop below the 6000 psi compression limit again. Even when we had all the base perimeter of the upper section back touching all 4 sides again, it hardly looks as if it slowed down. I guess the question is, when the 6000 psi compression limit is exceeded, what actually happens to the cement? Does the split run far as if one would step on thin ice on a pond and "crack"? Or speaking of the cutting wood wedge, it splits the log in two. If the wedge isn't aligned perpendicular to the log upon impact, only a slight dent in a log occurs but no split. There was no log splitting the wall to the ground in advance of the fall so the cement absorbed this but is crushed where the impact occurs. However, the North Tower took 2 more seconds to fall so we think that the wild wedge swing really crushed that side more and offered less resistance to the falling section from above.

7. Building and Air Both Fall:

After viewing the videos to determine the base radius of debris cone rcb above, you can see the fall of the dust also. What was interesting is that puffs of air from each consecutive floors below were projected laterally. This does act like collapsing pancakes but no resistance of slowing down from the remaining steel beams is a mystery. Not even knowing steel's compression strength, it should have stood but maybe the corner fall could have bent it a bit. There were twisted steel beams in the pile of rubble but who knows what they went through. So each horizontal puff projected through the main falling dust cloud but there was also a column of dust still standing as the building fell which left the standing air in the dust - literally speaking. So this standing dust was probably from the horizontal puffs and were probably composed of smaller particle sizes that couldn't be observed falling slowly.

A. Initial Dust Concentration:

The frictional drag from the crumbling chunks on the immediate air was considerable. You notice, that large floors were falling also. If you open an umbrella and pull it slowly, you can feel the drag. Now snap it and it almost reverses itself. So the buildings have an umbrella effect. Some reading this were in the World Trade Center as I was once. There were sections of elevators heights, the express lower and upper floors and a local. While waiting, you could hear the air whistling and blowing through the elevator shaft openings - quite strong as this is another example of the umbrella effect. Recalling the total of one building volume was 5.906e+7'^3 from above, and the cement for one building's cement is 90e3 Tons = 1.8e8 lb, then the concentration of dust within the buildings volume would be 1.8e8lb/5.906e+7'^3 = 3.048 lb/'^3. But 67.2% of it is airborne dust reduces the density to 2.048 lb/'^3 or .0328 g/cm^3. The density of air is 1.188e-3 g/cm^3, so .0328/1.188e-3 = 27.62 denser than normal air.

Now the terminal fall velocity V of a 150 lb man was rumored to be 150 mph, free fall for parachuters, so the gravity force balances the drag force, no net difference in force, then no acceleration, therefore, a constant velocity. The Kinetic Energy (K.E.) here is MV^2/2, or 150 lb x (150 mph)^2 = 150^3. Now we balance the same 150 man but now the new mass m=27.62M. So K.E. = mv^2/2 = MV^2/2, 27.62M v^2/2 = MV^2/2, cancelling M/2, 27.62 v^2 = V^2, v=sqrt(V^2/27.62) = sqrt(150^2/27.62) = 28.5 mph is this dust's velocity needed to suspend a man.

B. Compressing and Uncompressing Fixed Air Volumes:

Now when we subtracted 1/2 the volume of the cone debris from one WTC building volume, the volume of the dust from one building is 5.799e7'^3 within 13 sec = 4.461e6'^3/sec. If the compressed air was to escape through the sides of the building, it would become more rapid near the bottom when the complete building collapses. So for 1/2 the building height, if the air escaped laterally all the way down, the average building perimeter outflow is 7.9'/sec or 5.4 mph. As we saw, some did escape laterally so we use the volume of building displaced air - 2/3 cone where we calculated the dusty air density. Now working with the dust cloud, the final cloud radius was 775' @ height = 600' as examined photos by Jim Hoffman. So if the dust volume = 5.799e7'^3, and the cone base radius above was estimated at 205.9'then the height of the stabilized cloud would be 5.799e7'^3/(pi*r^2) = 5.799e7'^3/(3.1416*205.9'^2) = 435.4' or 73% of the final height.

However, in the video observations, the violent downdraft brought everything all the way down to the partial cone height of 31.0', so the compressed dust cloud over pressure wave must now expands vertically and horizontally. If we say the horizontal and vertical expansion rates are equal, we would have a table of the outflow below. Also, Jim Hoffman stated that the buildings occupy about 75% of the street surface there. So we call that an urban wind canyon effect. Conserving volume, but no vertical expansion, the horizontal component would be 4 times as high.

Cloud: Radius = 205.9' Height = 31.0' Wind: Vel = 111.1 '/sec 75.8 mph -> 303. mph @ 4x Canyon Winds Cloud: Radius = 319.7' Height = 111.9' Wind: Vel = 19.8 '/sec 13.5 mph -> 54.1 mph @  4x Canyon Winds Cloud: Radius = 433.5' Height = 192.8' Wind: Vel = 8.5 '/sec     5.8 mph -> 23.2 mph @  4x Canyon Winds Cloud: Radius = 547.4' Height = 273.6' Wind: Vel = 4.7 '/sec     3.2 mph -> 12.9 mph @  4x Canyon Winds Cloud: Radius = 661.2' Height = 354.5' Wind: Vel = 3.0 '/sec     2.1 mph ->   8.3 mph @  4x Canyon Winds Cloud: Radius = 775.0' Height = 435.4' Wind: Vel = 2.1 '/sec     1.4 mph ->   5.7 mph @  4x Canyon Winds

Notice, at the final max radius and height, the outflow is only 1.4 mph, indicating no more cloud growth as pics also indicate. Now the ambient winds should have taken over but we still have much more mass in the dust cloud than the prevailing air currents. If the cloud was observed at 600', why do we have a final of 435.4'? Well that 75% of buildings, less at higher heights, the air must have been forced up to. Also, observed, the dust cloud felt hot by friction so added buoyancy here but not much as compared to boiler stack temperatures.

C. Dense Dust Cloud in Action:

Jim Hoffman's quoted that. "The WTC dust clouds inexorably advanced down streets at around 25 MPH. This is far faster than can be explained by mixing and diffusion. The top surface of the clouds looked like the surface of a boiling viscous liquid - churning but not mixing with the air above. There are reports of people being picked up and carried distances by the South Tower dust cloud, which felt solid. New York Daily News photographer David Handschuh recalled: Instinctively I lifted the camera up, and something took over that probably saved my life. And that was [an urge] to run rather than take pictures. I got down to the end of the block and turned the corner when a wave-- a hot, solid, black wave of heat threw me down the block. It literally picked me up off my feet and I wound up about a block away. Above, we calculated that we only needed 28.5 mph to lift a man off his feet. Running the above table with more detail, not shown here, we can see that happen up to 255' from the center of rubble in the canyon winds. Even higher winds in more narrow canyons.

8. Simulate Dust Generation:

In 2. above, we determined % Particle Distribution: This was the main reason to create cement dust from 2 cement blocks. Also done to verify cement density and now, what processes and how much energy does it take to generate dust?

A. Physical Data:

Each block was 2.5" x 4.25" x 8.875" = 94.3"^3. It was weighed on a supermarket meat scale @ 7.25 lb. Calculated Density of Cement Block: Weight= 7.25 lb/Vol=94.30 in^3 = .0769 lb/"^3 or 132.86 lb/ft^3 or 2.128 g/cm^3.

B. Rubbing Blocks Work Units:

For the 1st rub, 2 blocks were rubbed along the 4.25" x 8.875" (top) surface. In the 2nd rub, the 2 blocks were rubbed on their sides (2.5" x 8.875") rather than their tops.

Rub Distance 1: 15.75 in/cycle x 60 cycles = 945.00 in
Rub Distance 2: 15.75 in/cycle x 20 cycles = 315.00 in

Rub 1: 78.8'x 15 lb = 1181.3 ft-lb @ 45 sec 26.3 ft-lb/sec 0.0477 HP 35.59 Watts 4.45e-004 KWH
Rub 2: 26.3'x 30 lb = 787.5 ft-lb @ 5 sec 157.5 ft-lb/sec 0.2864 HP 213.54 Watts 2.97e-004 KWH
Rub 2/1 Ratios: Work 0.67 Power 6.00 Pressure 1.35psi/0.40psi = 3.40

C. Calculating Dust Rub Weight:

In 2. above, as well as the % Particle Distribution, we converted the radius of a circle to a sphere, assuming the circle area is the cross section of a sphere, we compute the sphere, 4/3 pi r^3 and weight each diameter class by the density. We counted over 72,065 particles but that converted to only .302 g for the 1600 dpi scanner.

There was a heavy area of overlapped dust particles that was not used. However, our 1st rub graph shows a total of 72,065 particles but I remembered that was my first scan and 1st of those three and much dust had settled in the middle of the plexiglass. So much that it appeared as too much overlap and the 2 & 3 particles would overlap and be counted as a larger particles. So this scan area was cut out. This is fine for the % particle distribution but we need to have all the 1st rub particles for weight. It was brought back and proportioned to the other 2/3 & 3/3 1st scan. After doing so, the # of particles was 11,0634 and our new weight was .464g. Then our scaled up estimate for the 2nd rub was .143 x 110634/16662=.950g.

D. Crushing Power Adjustments:

Jerry Russell found that the amount of energy required to crush concrete to 60 micron powder is about 1.5 KWH/ton. This is for 60u, but for other sizes, it is sqrt of the ratio of 60/D in u, or sqrt(60/D). To crush to smaller particles, requires more energy. So for our whole range of particles sizes and % distribution, we applied this formula for each particle, weighted each and called the new total the Crush Adjustment Factor caf=1.53. So, now we have 1.53x1.5 KWH/ton or 2.3 KWH/ton to produce our rubbed off % distribution of particle sizes or called adjusted crush power (crushp). Now this crush power was based on the KWH of the crushing machine. One has to immediately half this value because when applying the energy to crushing by other cement blocks, the block doing the crushing and the block being crushed alternate between being crush or perhaps both. In the grinding machine, the metal teeth are not crushed and are used over and over. So 2.30/2 = 1.15 KWH/Ton as the crush energy to produce our % distribution. Actually, was their machine efficiency taken into account? Still use 1/2 to be conservative.

E. Estimating My Crushing Power:

We note that Work = ft-lb, Horse & Watts Power = ft-lb/sec, (ft-lb/sec)*sec = KWH Work or (ft-lb) again. If we correct for seconds, we scale this to HP -> Watts then convert to KWH just considering this work/sec for 1 sec = work. At first I thought that rubbing two blocks together would be more efficient than corrected crushp so my number would be much less.

I noticed that applying more pressure over a smaller area, by rubbing the sides instead of the top, dust poured out of the blocks. So 1st rub was 15 lb/(4.25" x 8.875") or 0.398 psi. The 2nd rub was 30 lb/(2.5" x 8.875") or 1.352 psi. I tried to duplicated the pressure and movement for the 1st run for a few seconds, and an observer estimated that the 2nd rub, using the sides of the blocks instead of the tops, produced about 8 times as much cloud dust in the sunlight with very light wind. Now we estimated that we have 8 times as much dust or 7.596 g. The rest is summarized by the comparison table below.

0.950g 4.45e-004 KWH/1.047e-6 Ton = 425.0 KWH/Ton 12.54 of Standard @ 0.398 psi
7.596g 2.97e-004 KWH/8.373e-6 Ton =   35.5 KWH/Ton 46.17 of Standard @ 1.352 psi

PSI Ratio = 3.4 Dust Output Ratio = 8.0

Well I'm still over crushp by over 35 times but notice the trend. I could really strain and push and rub harder for a 3rd point to calculate more but it would be hard to estimate the real pressure with such a strain. This could be done much better in a simple lab. If we just use the 2 above, 1.352 psi/0.398 psi = 3.4 (prat0), or the initial pressure ratio mycrushp pres which produced 8 times as much dust.
It was thought that the best way was to determine grams rubbed was to try it again, but try to measure the pressure lbs applied and compare the amount of dust rubbed off. A weighing scale was used and one block was attached and still weighed about 7.2 lbs, which must be subtracted from the applied pressure and the top block's weight. The 2nd block's gravity adds to my applied pressure. Rubbing tops, I tried to keep the pressure at 27 lbs for rub # 3. Then the top block was turned on it's side and I tried to keep the pressure at 37 lbs for rub # 4. Actually, this was much more difficult as the scale fluctuations were wilder than rub # 3. Only half way through, I got better but I estimate an average of 35 lbs. The work units and parameters are given below:

Rub 3: 98.4'x 19.75 lb=1944.1 ft-lb @ 92 sec 21.132 ft-lb/sec 0.0384 HP 28.65 Watts 7.32e-4 KWH 1.07'/sec
Rub 4: 55.1'x 27.75 lb=1529.7 ft-lb @ 44 sec 34.766 ft-lb/sec 0.0632 HP 47.14 Watts 5.76e-4 KWH 1.25'/sec

Notice, I ran out of steam for rub # 4 as I only made it to 44 sec. However, I rubbed long enough to make sure we had enough dust as it was done on paper in a box with the scale. The blocks were cleaned and after the rub, they were banged together to get all the loose dust off the blocks. Also, the dust on the scale was dumped to the paper and done in a sunroom glass porch again where no wind would carry the particles away. The dust was dumped on cardboard and a letter opener was used to scrap the dust into little dust cubes. I tried to stack them high but compression would push out the side of the cube so we had to keep the height relatively low compared to the lateral dimensions. The base was evened off nicely so the lengths and widths were the same and the tops were leveled. Below, dl3 = dust length for rub # 3, etc.

Rub 3: dl3=14 dw3=11 dh3=4.5 mm
Rub 4: dl4=21 dw4=19 dh4=5 mm

Dust Vol3 = 0.693 cm^3 x 2.128 g/cm^3 = 1.475 g
Dust Vol4 = 1.995 cm^3 x 2.128 g/cm^3 = 4.246 g
Dust Vol4 = 4.171 cm^3 x 2.128 g/cm^3 = 8.877 g : Scaled up by 44 -> 92 sec

Rub 3: 1.475g 1.626e-006 Ton 332.19 KWH/Ton 289.49 of Standard @ 0.524 psi
Rub 4: 8.877g 9.786e-006 Ton 43.42 KWH/Ton 37.84 of Standard @ 1.251 psi
4/3 Ratios: Work 0.79, Power 1.65, Pressure 1.251psi/0.524psi = 2.389, Weight 6.019

F. Overcoming Friction Produces Dust:

We notice that we still have a drop in KWH/Ton. We know that amount of grams is at least proportional to the energy/power supplied, then KWH/Ton should stay the same which we will show below. The weight is over 6 times as much. The only ratio close is the psi. As we rubbed the blocks, the scale was showing the weight of the block and the perpendicular hand pressure component. Also, if we keep the same power by keeping the pressure constant but moving the distance more (faster in same amount of time) to increase the power, would this cause 6 times more dust? It should if I rubbed 6 times the distance. Imagine hardly any pressure but rubbing wildly with the blocks - would much dust be produced? Just as there is a limit on compression strength, in the vertical, there must be a limit, but much less in the horizontal component. The higher pressure is applied to both component so we must have surpassed this shear pressure barrier in the horizontal. Is it possible that the pressure applied may have it's maximum effect at a 45 degree angle - like a projectile's greatest angle range?

We still need a horizontal component to have the blocks overcome friction. So applying 35 pounds vertically is also applying 35 pounds horizontally if we apply the pressure at 45 degrees. Thus, my actual pressure I'm applying is sqrt(35^2 + 35^2) = 49.5 lbs for 44 secs - neglecting the added block weight or sqrt(28^2 + 35^2) = 44.8 lbs with block. It was mentioned above about a horizontal pressure barrier much like but much less than the compression limit. It appears that the increased pressure increases the friction that separates the layers of dust. So just to move the blocks, we need a minimum horizontal component depending on the vertical downward component. I noticed that if I increased my pressure too high while rubbing, my motion stopped.

If we examine the surface layers under a magnifying glass, we would see the rough surface and aggregates added to the cement for strength. If we concentrate on high spots meeting on the upper and lower surfaces for a moment, ( o -> O ), o meets O but tangent to the curved surface, -> is deflected upwards so we have horizontal -> and upward ^ vertical components. With little compression force from above, the upward component will lift the block enough that o slides above O. However, with higher psi compression, the block may not lift against this compression and the horizontal component must be forced to dislodge O. This is resultant frictional force. So now our items of interest are definitely the initial PSI Ratio (prat0)=2.389 and Dust Output Ratio (grat)=6.019. How to make use of these 2 parameter points to develop a relationship formula is our next task.

G. How Many Lines can be Drawn Between 2 Points?:

For simplicity, lets use the number prat0=e=2.7183 for logarithmic ease so we can see the behavior pattern more easily and that's relatively close to 2.389. Seen below, prat is the pressure (pres) ratios and 'dorat' be the Dust Output Ratio from the pressure ratios. In the 1st case, dorat=grat*(prat/e). Then finally, in all cases below, mycrushp= (prat*rub3w/sph*1.e-3)/(rubt3*dorat) KWH/Ton.

The rub3w is rub3 work and sph=3600 sec/Hr and 1.e-3 KW/Watts. The rubt is the tons produced in the 3rd rub, increases by the term 'dorat'. In all our cases, prat would be the multiple increase in rub3 work by applying 1,2,3,4,5 times as much pressure to the cement but holding the same rubbing dimensions. The Dust Output Ratio (dorat) generates the amount of dust by the right relationship of prat and prat0.

dorat=grat*(prat/e)

prat      weight                            mycrushp                                       pres
1    3.27g     3.600e-6 Ton   150.01 KWH/Ton 130.73 of Standard @ 0.524 psi
2    6.53g     7.200e-6 Ton   150.01 KWH/Ton 130.73 of Standard @ 1.047 psi
3    9.80g     1.080e-5 Ton   150.01 KWH/Ton 130.73 of Standard @ 1.571 psi
4    13.06g   1.440e-5 Ton   150.01 KWH/Ton 130.73 of Standard @ 2.094 psi
5    16.33g   1.800e-5 Ton   150.01 KWH/Ton 130.73 of Standard @ 2.618 psi

Noticing above, we apply more pressure and generate more dust but more pressure doesn't improve the efficiency since mycrushp staying constant. Note that prat is in numerator, and functions in the denominator so they cancel each other so mycrushp stays the same. At first, this is what you would expect but the 2 rubs show more pressure increase efficiency and causes mycrushp to decreases. Thus, dorat=grat*(prat/e) must be incorrect. To assume we produce 6 times as much dust in proportion to the pressure ratios to the initial 2 pressure ratio rubs e (prat0) is wrong. Thus the case of a straight line connecting the two points is rejected.

Below, we try a new dorat=grat*log(prat). Now in 'C', log is really ln in Fortran or Log . So lets state dorat=grat*ln(prat) where ln is natural log base e. e

dorat=grat*ln(prat)

prat      weight                            mycrushp                                       pres
2    6.15g     6.783e-6 Ton    159.24 KWH/Ton 138.77 of Standard @ 1.047 psi
3    9.75g     1.075e-5 Ton    150.70 KWH/Ton 131.33 of Standard @ 1.571 psi
4    12.31g   1.357e-5 Ton    159.24 KWH/Ton 138.77 of Standard @ 2.094 psi
5    14.29g   1.575e-5 Ton    171.45 KWH/Ton 149.41 of Standard @ 2.618 psi

This is even worse as mycrushp shows decrease, then an increase. Here we assumed that x=ln(prat) or e^x = prat, x*grat is the increased dust output. That is the number of times e, (prat0), is raised to a power to equal prat, since e was the log base (prat0). Sounds better but no go. Thus the case of a 1st non linear line (semi-log) connecting the 2 points is also rejected.

dorat=pow(grat,ln(prat)) = grat^ln(prat)

prat      weight                            mycrushp                                       pres
2    5.12g     5.641e-6 Ton   191.46 KWH/Ton 166.85 of Standard @ 1.05 psi
3    10.60g   1.168e-5 Ton   138.70 KWH/Ton 120.87 of Standard @ 1.57 psi
4    17.76g   1.958e-5 Ton   110.35 KWH/Ton   96.16 of Standard @ 2.09 psi
5    26.51g   2.922e-5 Ton     92.41 KWH/Ton   80.53 of Standard @ 2.62 psi

Here we have weight still increasing but more than above examples thereby decreasing mycrushp - the results we want. In wording this, we might say - well let's say that in the above case, instead of x * grat, we now have grat^x in the expression. In other words, x=ln(prat) or e^x = prat, grat^x is the increased dust output. That is the number of times e, (prat0), is raised to a power to equal prat, since e was the log base (prat0), we raise grat to this same power.

For below, lets use the real prat0=6.019 instead of e. Instead of base 10 or e which are used for calls in math routines libraries for most computer languages, let's derive an expression to find x in y=b^x or log y=x where b is the log base. So y=b^x, ln y = ln b^x = x ln b, x=ln y/ln b. So x=log(prat)/log(prat0) for base prat0^x = prat. Then dorat = pow(grat,x) or grat^x.

dorat = grat^x where x=log(prat)/log(prat0)

prat      weight                            mycrushp                                       pres
2    6.16g     6.786e-6 Ton   159.16 KWH/Ton 138.70 of Standard @ 1.05 psi
3    14.20g   1.565e-5 Ton   103.49 KWH/Ton   90.19 of Standard @ 1.57 psi
4    25.70g   2.833e-5 Ton    76.26 KWH/Ton   66.46 of Standard @ 2.09 psi
5    40.71g   4.487e-5 Ton    60.18 KWH/Ton   52.44 of Standard @ 2.62 psi

Now we can apply this to some higher psi increments below.

prat      weight                            mycrushp                                       pres
5      40.71g      4.487e-5 Ton    60.18 KWH/Ton 52.44 of Standard @ 2.62 psi
10    169.92g    1.873e-4 Ton    28.83 KWH/Ton 25.13 of Standard @ 5.24 psi
20    709.28g    7.818e-4 Ton    13.81 KWH/Ton 12.04 of Standard @ 10.47 psi
30   1636.17g   1.804e-3 Ton      8.98 KWH/Ton   7.83 of Standard @ 15.71 psi
50   4689.96g    5.170e-3 Ton     5.22 KWH/Ton   4.55 of Standard @ 26.18 psi
75 10818.85g    1.193e-2 Ton     3.40 KWH/Ton   2.96 of Standard @ 39.27 psi
100 19576.82g  2.158e-2 Ton     2.50 KWH/Ton   2.18 of Standard @ 52.36 psi

Note that at near 50 psi, we are nearly double the crush standard. Now what would the upper section on the corner and the whole building resting on the bottom inch produce as mentioned above and then the whole building standing?

                                 prat                weight                            mycrushp                                     psi
Top Sect on 1' Edge: 1769    7.312e+6g    8.060 Ton          0.12 KWH/Ton    0.10 of Standard @ 926.38
Whole Building:          2080    1.020e+7g    1.125e+1 Ton    0.10 KWH/Ton    0.09 of Standard @ 1088.88
The whole building could produce over 10 tons with the same motion as rub3 but that motion is imaginary and also, the block only weighs 7.2 lbs so we better call this a projected potential dust generator.

H. Error in Estimate?:

We wondered, if my observer seeing 8 times as much dust was in error. This is why we tried rub3 & rub 4. Actually, scaling his estimate to rub 3 & 4 (prat0)=2.389 and Dust Output Ratio (grat)=6.019, his prat = 1.26 psi/.53 psi = 2.38 & grat = 8, we corrected his grat to 5.97. Trying to keep the same pressure on top of block to side was difficult. Measuring, top block pressure was 20 lbs, however on the side with the same strain, we duplicated again and the pressure was 28 lbs. Even though it felt the same, more strength was used to grip the top of the block where the side was an easier grip and energy went to increase more pressure inadvertently, as well as the different surface areas, we have 1.26 psi/.53 psi = 2.38, but 28 lb/20 lb = 1.4 of that increase by a better grip. In any event, 8/5.97 = 134% or 34 % too high was a fair guess.

One note, the 2 top block rubs spread the dust over a wider region and when he saw the dust come from the sides, it was a smaller region so it appeared more dense, even though it was, dust was confined to a smaller area making it appear even denser. That is 8.875 x 4.25/8.875 x 2.5 4.25/2.5 = 1.7 appearing that much denser. Based on that error of estimate, he should have estimated 8/1.7 = 4.7. If he considered that, which he didn't and noted the smaller area, he may have guessed closer to 6. So the factor of 8 made it interesting to do it more accurately and see how close his estimate was. It still should be done under better measurements at multi levels of psi's and dust outputs.

When the blocks were rubbed, while standing above them, I could smell the hot dust as it rose. Well doesn't that work for rubbing sticks together. Possibly the pressure and speed changes the friction element which researches say is a very complicated property. This may explain the hot dust cloud that was observed at the WTC's collapse. Not hot enough to burn but most all that friction heat is transferred to the dust particle as it falls. Some conduction to the lower block occurs but then that layer is also rubbed off so we have even a higher net gain of heat in this next gain of particles. Does this heat also have an effect on the friction and the lateral shear limit? Obviously, this could be a simple study and even a project a High School physics class could do with better measured pressure and distance weighing the dust rubs than these crude dude estimates. A government grant for a lab or college would be fine but for this simple a project, or we could try again. One block should be fastened to a scale, the other rubbed on it and watch the scale to keeping the pressure constant by two people or springs over a long enough period of time to produce measurable dust again. Better, known weights attached to blocks and constant pressure applied to keep blocks moving at constant speed.

I. Apply to WTC Buildings:

We have to extend the block's rubbing power to the WTC's. This is the 4th attempt for me to explain below so we understand this process in the easiest of terms as it helps me to explain the program better. I went from the complex to the simple terms and it's a mess to explain. Once I reached the simple expression through this complexity, I decided to reverse the process and take it backwards to the complex in the ft-lb/sec terms. The first assumption is to concentrate on expanding the energy equally over a larger area and hold the psi constant, then last, we apply the PSI Ratio non-linear correction factor (prat) for increasing psi's due to building weight.

(1) Working with the Final Correct Terms and Units:

The power or energy/'^2 unit can be used in solar radiation - the solar constant (2.00) cal/cm^2/min or Langley. This is the amount of radiation received by 1 cm^2 perpendicular to sun above the earth's atmosphere. We could define calorie (cal) the amount of heat or energy (ability to do work)/min = power, HP or Watts per unit area - simply Watts/cm^2 as one of the terms. For our rub 3 data line above, we have 21.132 ft-lb/sec = 0.0384 HP or 28.65 Watts. Now let's normalized this to Watts/'^2. Let the length be the rub distance because even though we rubbed back and forth, we could have had one sweep direction of the block for a distance of 98.4', if it were long enough. For the WTC, it's block rub is in one direction so we apply the same analogy of what really happens. We already normalized work to 1 sec to define HP or Watts, now we normalize Watts to distance. This gives 28.65 Watts/98.4' = .291 Watts/ft. Now the perpendicular distance (width) involved in the rub was 4.25". So we need to divide by 4.25"/(12"/ft) = .354 ft which gives us .822 Watts/ft^2. So we essentially divided the power by the rub area to produce a power/area term = 28.65 Watts/(98.4'x.354') = .822 Watts/'^2.

(2) Extending Rub Areas for a Constant PSI:

Now we expand the rubbing area in proportion to the 3rd rub of blocks. That is, for the rubs, if we increased the horizontal block dimension, the dust would increase in the same proportion. We just extended the power over a larger area but kept the power/area term constant. However, that may require too much strength for us to maintain 0.524 psi over a larger area for quite a few seconds. But here, we have the WTC weight compression instead of hand pressure. At the base of the WTC, every adjacent "^2 has a constant psi because the building height is constant across the whole horizontal cross section area (wtca) or an equivalent continuous cement area which is cement vol/height 2043875'^3/1365'= 1497.35'^2.

So here we have an increase of rubbing area to apply, .822 Watts/'^2 x 1497.35'^2 = 1231 Watts if we apply the same rubbing .524 psi over that entire area. The ratio of rubbing powers produce the same ratio of dust and with constant psi, only the area term changes - then ratio of the areas is the same ratios of power producing the same ratios of dust which is 1497.35'^2/(98.4 'x 4.25/12') = 42.966 times as much. Or the rub3 power/area x WTC area = WTC POWER due to area ratios. WTC area/rub3 areas = rarb=wtca/(98.4' x bw').

(3) Simulation of WTC Collapse:

Let's run up the WTC - reverse to what really happened. Assume the same amount of energy produced by the destroyed building whether from top to bottom or from bottom to top. For demolition projects, the latter is the norm. Starting from the bottom, let's examine each 1 ft height of cement to crush at a time. Now just as rub3 was for 92 sec, the building rubbing on each layer below is only a very short time, 1365'/13 sec or avs = 105'/sec for each 1 ft collapse or .00952 sec/'(secb).

(4) Simulation of Debris Cone Buildup:

We normalized rub3 work to 1 sec for power but now WTC rubbing seconds vary along the cone rubble pile. See cone.txt for the cone dimensions and for deriving other parameters. Each foot of building height adds to the debris cone. Above, we used the final volume. Here, we need the final lateral surface area of the cone which is lsa = pi 2/3 l (Rb+Rt) where l=sqrt(Rb^2 + hci^2), Rb=205.9' and hci=46.5' from above, = 211.1'. Also, then Rt=Rb/3. So lsa=pi x 2/3 x 211.1 x (205.9+68.6) = 1.22e5'^2. Now for each foot WTC falls, we have rat=n/1365, n=0 -> 1365. Let the # of sec=rat*(2/3*l)/avs, that is the fractional visible lateral slant length of the cone divided by the average speed of the fall avs = 105'/sec, now directed to the lateral side. However, not only is the lateral side longer than a straight drop vertical height hc of the cone, the lateral speed component is slowed down by cos(ang) where ang=Arc Tan(Rb/hci) = Arc Tan(205.9/46.5) = 77.3 deg. Cos(77.3)=.22 so 22% slower. However, notice the pressure component normal to the cone is less psi by sin 77.3 deg = .975 so the velocity component may be slightly faster because of less friction. Lets apply a final correction here, ratio of psi dorats as explained above and by (C) Varying the PSI below, doratb/doratc would be a good indication of proportionally less dust by psi, less friction, faster rub speed - but by only 5.3% For the growing lateral surface area of the cone lsa, wca=rat*lsa, the total rubbing area to scale up the rub 3 area is by rarc=wca/(98.4 x 4.25/12). In the program then, we have a dorat for the building and the cone added together by the final term dorat=doratb x rarb x secb + doratc x rarc x secc.

(5) Varying the PSI:

We have our base compression bcomp = 1088.9 psi to start with. So each foot of WTC falls, we have pres=(1-rat)*bcomp remaining for the 1 ft drop where rat=n/1365, n=0 -> 1365. Then we follow the same as psi ratio as before for prat with prat0=2.389 and grat=6.019 and find exponent x=log(prat)/log(prat0) so prat0^x = prat and dorat=grat^x. Then finally, we have tons = rubtps3*dorat, where rubtps3 was 1.767e-8 Ton/sec.

(6) Tons of Dust:

We cumulated these tons then ratio to total weight of building. The run of this model for the WTC was Total: 65266.9 Ton or 72.52% of the building weight. However, above, we calculated the remaning cone rubble after that 67.20% went to dust. So this simulation method was only 7.92% higher than by the cone estimate to dust. A cautious note here, not knowing what the results would be, suppose the formulae were to produce tons greater than the tons/ft of building. That is, 90,000 tons/1365' or 65.93 tons/ft. This had to be evaluated for each foot for if it did, then the tons produced are set to that limit. That is, all (100%) of that particular foot of the WTC is turned to dust. In the run here, this happened from 295'to 1054' or 55.7% of the WTC height. Instead of the term rubbing power, we might advance to the term grinding power. It's as if two gigantic grinding wheels are spinning rapidly in opposite directions and are pushed together.

9. The WTC Dust Bowl Model Run:

All the constants and conversions units are calculated from the basic data explained above from wtcpar.txt .

A. Meteorological Data:

The weather data is from Newark, La Guardia Airports and Central Park also. Compiling all, we chose 330 degrees @ 9.2 mph. Actually, would you know that Newark missed their 9:51 and 10:51 AM DST weather observation. However, La Guardia reported a special observation 11 minutes after the N. Tower collapsed at 10:28 AM. The S. Tower collapse was at 9:59 AM. Newark and La Guardia have more similar winds due to flat area runways compared to the lower wind speeds from Central Park trees. Also, winds from Newark across the bay were the winds to use as they reach the Island of Manhattan. A high pressure ridge was just passing north of the area and winds were adjusting from NNW to N. Winds velocity were corrected down to the surface then 20 increments up to the cloud dust top (600') by using a exponential wind profile law used for most air pollution models. A stability class "C" was chosen for this time of day, surface, wind speed, sky conditions and this time of year. We used a temperature of 73 F and pressure of 1020.2 mb mean sea level pressure which are used for air density and viscosity calculations for fall velocity of particles. Also, the pressure adjustment was made to station pressure at 7' msl. The model was run for the WTC's combined as if the same time as the winds were slowly shifting but not enough to make the plumes cross as they both fell within a 1/2 hr of each other.

B. Cement Dust Parameters:

The 1st estimate of cement was in yd^3, which was later downgraded (27%) but now reported in tons. After the cone of debris, 67.2% was dust for this model run. Even though we just ran a crush model above that showed that 72.52 % went to dust, we might be doing the rubs again but for now will use the volumes left after cone debris. We used 1.80e+5 tons of cement @ 1.84 g/cm^3 density. We convert the volume into "^3 so the z component will be in " when we divide the area dx * dy in the last step of the program.

C. Cement Dust Cloud Dimensions - See dustbm.dia.:

We start with a cloud radius dr for a cloud O, but find the equivalent side/2=rs= dr*sqrt(pi)/2. Here dr=775, so rs=686.8 which is 1/2 the side of the square. The cloud center is 0 with -r to +r the side. We are going to calculate downwind from this center 0 so the lateral border areas will only be somewhat off a little, but won't effect the centerline values. This eases the calculations considerably.

The cloud is divided into nl=20 slab layers, so we have pdz=1/nl = %/slab = % dist @ z coordinate. Each cloud slab layer has it's own mid-height value wind velocity in '/sec. We use the velz exponential wind profile for stability classes "C". We keep the cloud and grid to work with in ft., but print out x-miles downwind in the finale. The downwind distance x in the cloud is with len = 2*rs, and cross wind y is also len = 2*rs as explained. We also divide the dust cloud into an xy grid where dx'= dy'. Here dx=(gxf+rs)/ng where ng the # of downwind grid points (1000). So here, all particles landing from a dx section of cloud contributes dx/len = pdx = % of cloud source. Also, dy/width = pdy = % of cloud source for dy or % dist @ y coordinate. So pdz*pdx*pdy % = %dist @ any xyz cloud coordinate. This is further multiplied by each particle % distribution that lands anywhere in the receptor grid. In the lower section of dustbm.dia., we have the dust cloud as part of the receptor grid also with the center o as xg=0 as the dust cloud center. Note that xn, the downwind distance, starts at -rs (the back end of the dust cloud) and gxf the chosen downwind distance as 2 miles here.

D. Particle % Distribution Adjusted File:

We read in the final % pd file (*.pda) file for the Diameter (Di) u and it's associated % distribution (pd). For the travel time tt, we use only the first slab's height, back to feet, and divide by fall velocity in fps. For the travel distance x, we use only the ground slab's velocity in '/sec. So '/sec * sec = xn'. We now write to a file land.grd to include x, along with Di & % pd for each particle diameter Di. Now we reread that new file and read Di,x and % pd one at a time to see if the downwind slab of each particle size falls within our grid. That is, by dustbm.dia., the particles are distributed equally between all slabs, but we consider each slab falling with only one particle size at a time. Ex: all ., then `, +, o, *, etc. as shown.

E. Particle Fall Velocities:

There is another write up here but saved for the nuclear fallout description. This is really complex involving graphs and formulas. It suffice to say here that we hope to have the exact formula for each particle size with no short cuts as others use approximately fall ratios. For a nuclear fallout model, we try to be accurate as possible so we use it here also. This model explanation is long enough.

F. Cement Slabs are Landing in Dust Form:

The downwind travel distance xn is the slab's new position that may or may not fall within the grid. The file's x is the first slab above the ground downwind position so any particle slab n above moves at the higher wind speed than the particle slab below because of the wind profile. Each slab's downwind distance is proportional to the ratio of the wind speed reported for this ground slab. So we accumulate distance xn by vel[]/vel[0] * x as we loop downward through each n1 slab layer. The complete grid distance is gxf+rs because the cloud is also part of the grid. If this slab n1 is past the grid, then all subsequent slabs below this level will be further past the grid, therefore we break out of this n1 down loop. Now the next slab n is higher, (1st n1), then it too will be further past the grid since the last slab level below it was also. Therefore, we don't just us break the down loop, we break the up loop also and skip the grid search loop too. So then, we read the file's next larger particle size that will fall faster and maybe be inside the grid. We bring this slab down through the other wind levels below and test to see if it's within the grid. However, if it wasn't past the grid, then we find where in the grid.

If the particle slab landed in the grid, we calculate the downwind grid loop by xg=ix*dx-rs. Here, @ ix=0, xg=-rs as the 1st grid point which is also under the dust cloud until xg=rs. Again referring to dustbm.dia, we have xn as distance but xg as grid points. We convert xn to xg by xg=xn-rs. Now the xg loop increases to find the displaced slab at it's beginning (xgs), and the end of the slab length is xgs+len. When larger xg is already past the slab, no more need to continue xg loop - we break it and read the loop down to the next lower slab position to see if it lands in the grid as the grid starts from -r again. So all the xg's within the slab's (len) each accumulate the associated pd's (grid[]=grid[]+pd). We do these loops for each particle size Di for each of the nl slabs. Confusing, it's best to view dustbm.dia since a pic is worth 1000 words and pretend there are no breaking loops to explain.

The test to find the slab's beginning decreases the loop's CPU time and past the grid breaks the loops really lowers the CPU time. We loop nearly 450 particle sizes, then 20 vertical slabs up, then 20 slabs down, then 1000 grid points for a possible 1.8e8 evaluations. Let's skip a few.

G. Combine All Joint Probabilities:

Now we loop the xg grid again, find all the summated particle % distribution (pd's) in each grid[] section. These pd's are from each particle % distribution, which is spread across the cloud source. In dustbm.dia, it is spread across the dust cloud source vertically, so that's 1/nl (pdz) x pd's when we considered 1 slab at time. Also, the dust cloud source is divided into dx,dy. Along the length, we have how many dx's? That would be nx=len/dx times. Then 1/nx is (pdx) = 1/(len/dx) = dx/len so we don't use the variable nx. We then have grid[]*pdz*pdx and only pdy yet to be considered. However, we write only xg & grid[]*pdz*pdx to a new file (pxz.grd), as we're done with Di's & x now. We reread this file and read xg & grid[]*pdz*pdx again and finally pdy, calculated the same as dx, to multiply by, for all the joint probabilities - that is, pd*pdz*pdx*pdy. Since the original pd's was in % distribution, we divide pd by 100 for equivalent decimal before multiplying the total dust volume in^3. Now we finally have in^3 at a grid section dx * dy. Now convert dx & dy to ", then area is dx * dy in^2. So dividing volume (in^3) by dx * dy (in^2) yields = " vertically.

H. Dust Flakes:

Finally, we incorporate a fudge factor, called a fluff factor ff taken from the ratio of densities of solid limestone to pulverized limestone. This ratio 1.87303 means that the volume is increased by 87.303% if we have the same mass. So a ff=2 here means that 87.303%/2 = 43.7% is used to fluff up the dust volume by it's settling compared to it's tight compactness when it was as solid cement. We chose ff=2 (1/2) because we have small particle sizes < 500 u compared to larger pulverized limestone, guessing that larger particles leave more space between than smaller particles.

I. Final Results File - See dustbm1.txt:

We finally dump to a file, downwind distance grid xg in .1 mi, the cumulated % of total dust settled, the combined joint probabilities pd at that grid dx-dy square, the in^3 there and the final total deposition height in inches downwind for 2 miles. We deposited 33.6% of dust within 2 miles, rest was still airborne. Maximum was 2.409" @ xg= 0.138 mi downwind from the dust cloud, then tails off to .426 " @ 2mi.

10. Continue to the Demolition Dust Model :