2. Nuclear Mushroom Cloud Dimensions and Dynamics:
For the younger, you may not have seen pictures of a nuclear blast so here are some of our test ones below. Click on
Nuclear Picture # 1
Nuclear Picture # 2
Nuclear Picture # 3
Nuclear Picture # 4
Nuclear Picture # 5
Nuclear Picture # 6
Nuclear Picture # 7
Nuclear Picture # 8
Nuclear Picture # 9
Nuclear Picture # 10
Nuclear Picture # 11
Nuclear Picture # 12
Nuclear Picture # 13
Nuclear Picture # 14
Here are some film clips showing the immediate effects of a detonation.
Film # 1
Film # 2
Film # 3
Based on all these, we have an artist's sketch and an accompanied fallout pattern as seen in
Fig. 1.3-1 Shipboard Radiological Fallout Model - US Navy - March 1981.
We added a little color to it for emphasis. In less than a millionth of a second, the nuclear reaction
takes place but the red ellipse shown is the hot plasma at 4000+ deg F and will emit so much heat, light
and radiation that may last for nearly 20 seconds (flash time) for a 1 megaton explosion. After the pressure wave,
the return flow will cause an inrush of dust and other larger particles, but when passing through the stem, they
are vaporized to 1 micron (u) diameter. In the turbulent entrainment with the cooler air, these particles begin to cool,
and the highly super saturated vapor density coats other particles to make larger droplets and then due to collision, these
particles coalesce to form even larger droplets. When the plasma cools, the updraft subsides and no more return of air and dust.
Then with cooling in the cloud, the turbulence subsides and we now have solid particles again but are generally much smaller
because they have grown from 1u. With no more coalescing, the smaller particles that didn't combine will remain so -
resulting in many left over smaller size particles, as we will try to show in another particle size generation model later.
A. Measured Radioactive Cloud Activity Distribution:
The first estimate of this distribution is from NOAA's ESSA ARL Fallout Prediction Technique by Jerome L. Heffter
which uses the Sunbeam-Smallboy blast (1.6 KT) as seen in Nuclear Picture # 12 above. The Naval Weather Service updated this
model in July 1980 (Shipboard Radiological Fallout Model) by Captain Robert W. Titus and CDR Anthony J. Klapp both from USNR-R.
Keeping the same general outline of ESSA's model, they compiled a different cloud distribution based upon higher nuclear test
yields with more aircraft radiation probes.
With some corrections and additions to their original cap and stem diagram, we present an updated
SRFM-Figure 6. Shipboard Radiological Fallout Model - US Navy - March 1981.
Here, we found a megaton value with a cap top and base that is closely divisible by 3. This is a 4 megaton
blast that is large but representative for this presentation. Usually, we reference the 1 megaton for other
calculations. As seen, the warm purple cap diameter is nearly 39 miles and the hotter red stem 13 miles.
The stem accounts for 20% of the total radiation and is divided into three 18,000' layers while the cap accounts
for 80% and is divided into three 12,000' layers. Each of these 3 layers have their own % of activity as seen above
the dashed line of the mid layers. This times the stem's 20% or cap's 80% gives the total cloud % indicated below the dashed line.
The sum of all these 6 layer values total 100%.
B. Radioactive Cloud Theory:
As we see the schematic of the nuclear cloud in SRFM - Figure 6. again, the larger cap has 4 times (80%/20%)
as much radiation as the stem. The cap volume is 3 x 4.0e+17'^3 and the stem volume is 3 x 6.6e+16'^3. Here the expected ratio
would be 3 x 4.0e+17'^3/3 x 6.6e+16'^3 = 6.1 instead of 4.0 but within the ball park. Now the radiation from within the cloud would be
proportional to the # of particles times the volume of particles. Since the stem feeds the cap cloud and the cap cloud expands accordingly,
we assume that the concentration (# size part/vol) is the same in both the cap and the stem. So if we have Ns5 (# of 5u particles in the
stem volume), we should also have Nc5 = Ns5 x Cap Vol/Stem Vol, 5u particles in the cap volume. Extending this, if we have Ns8 # of 8u
particles in the stem volume, we should also have Nc8 = Ns8 x Cap Vol/Stem Vol, 8u particles in the cap volume for our first basic assumption.
All the radiation is attached to the stem and these vaporized 1u particles cool in the cloud and become larger particles in the cap reducing
the # of particles, but the product of the # times volume is still constant producing the same radiation.
Assume the sensor of an aircraft Geiger counter probe is a variable gca = 2"x2" but this doesn't need to be accurate.
Since the sensor receives radiation from all directions, then the larger cap cloud will have more.
Lets approximate the cap and stem by spheres. By SRFM - Figure 6. again, this would have a stem up to 10.2 miles, a little short of 13 miles
vertically for a perfect sphere. In the cap with it's height of 36,000' or 6.8 mi, we are way short of 39 mi for a perfect sphere. Also,
we chose height values that are easily divisible for the 6 layers required for matching a 4 megaton bomb. This application also caused
the distortion of the cloud by this fig. The horizontal scale is shrunken or the vertical scale expanded.
Even though they are not perfect spheres, let's see this effect as if they were. At first glance, we would see that the ratio of the volume of
the 2 spheres would be 4/3*pi*Rc^3/(4/3*pi*Rs^3) = Rc^3/Rs^3. But cap Radius Rc= 3Rs so (3Rs)^3/Rs^3 = 3^3 = 27 times as large. However, as the sphere
becomes larger, the chance of the radiation from this outer sphere has less chance of impacting the sensor area gca. For a given particle size,
it would be a gca ratio to the surface area of the sphere or gca/4*pi*Rc^2. To examine these offsetting effects, lets start from 0 and expand
the sphere by say .1 mi in a loop. Let the inside sphere be radius R1 and the outer sphere be radius R2. The volume between the 2 spheres is 4/3*pi*(R2^3-R1^3)
or dV. Now since added # of particles is proportional to dV, thus the radiation is increased by this proportion. However, the distance of this
.1 mi shell is say the average of the inner and outer radii R=(R1+R2)/2. Then the chance of radiation from all directions from a new
particle is the Geiger Counter Area Ratio gigcar=(gca/144/fpmi/fpmi)/(4*pi*R^2) where fpmi=5280 '/mi and we have 144"^2/'^2. So we have Cumulative
Radiation crad = dV*gigcar or the addition of # of new particles x % chance of hitting gca.
Now we increment R1 & R2 by .1 mi and when R reaches the stem radius Rs, we save crad as Stem Radiation srad. When we reach the cap radius Rc,
we end the loop and ratio crad to srad = 2.942. That's close to 3 which is how many times the cap diameter, or radius, is larger than the
stem's radius. So now reduce the dr=.1 mi increment to .05 and then even less.
The results are give below as the Cap/Stem Radiation Ratio:
Cap 5.862e+1/Stem 1.992e+1 = 2.942 dr=0.100 mi
Cap 5.856e+1/Stem 1.971e+1 = 2.971 dr=0.050 mi
Cap 5.840e+1/Stem 1.952e+1 = 2.991 dr=0.020 mi
Cap 5.833e+1/Stem 1.948e+1 = 2.994 dr=0.010 mi
Cap 5.830e+1/Stem 1.944e+1 = 2.999 dr=0.001 mi
As seen here, it approaches 3 and is 3 if considering 2 decimal places.
Since crad is the sum of all 4/3*pi*(R2^3-R1^3) * (gca/144/fpmi/fpmi)/(4*pi*R^2),
if we gather all the constants 4/3*pi*gca/144/fpmi/fpmi)/(4*pi), this is just one
complex constant K. When dividing the two crad's, the constant cancel so we
only need to summate (R2^3-R1^3)/R^2 and ratio these to produce the same results.
Instead of a sphere, let's try to use the actual cylinder shape as shown. Lets assume
the heights of the cap and stem are equal so radiation received from any vertical component
would be the same and only the variation in the radius is the variable. Now crad is summated
by (R2^2-R1^2)/R^1 or reduced by the 3rd dimension R. Decreasing the dr as before yields the following.
Cap/Stem Radiation Ratio:
Cap 3.900e+1/Stem 1.320e+1 = 2.955 dr=0.100 mi
Cap 3.900e+1/Stem 1.310e+1 = 2.977 dr=0.050 mi
Cap 3.892e+1/Stem 1.300e+1 = 2.994 dr=0.020 mi
Cap 3.888e+1/Stem 1.298e+1 = 2.995 dr=0.010 mi
Cap 3.886e+1/Stem 1.296e+1 = 3.000 dr=0.001 mi
Notice nearly the same results and the ratio converges to 3 with accuracy to 3 decimal places.
That is, the radiation from the 2 different size cylinders is in proportion to their diameters or radii.
Noting SRFM-Figure 6 again, we have cap/stem as 4, not 3. This is not too bad. However,
that is the whole cap and stem. If we examine the adjacent layers where the stem feeds the cloud, we have
our highest radiation concentrations. This means we must have more mass in these zones. The normalized
actual value is 36% for the cap and 12% for the stem where the layers interface and contain
most of the radiation activity. Surprising that 36%/12% = 3.0 there? As we move further away from the interface, the cap/stem
radiation ratio becomes much higher and thus accounts for the weighted average of 4.0.
What does this tell us about the physical process? Well the mass produces 3 times as much radiation in this intense area
but we suggest that the many of the smallest hot (red) vaporized 1u particles are in the top upper 1/3 of the stem and fewer
larger particles are in the cooler (blue) lower 1/3 of the cap as Fig. 1.3-1 shows. As we derived many formulas,
as will be shown later, we found that after the flash time, (is now illumination time), the max radiation ceases and so does
the max heat vaporizing the particles. For this size bomb, the derived formulas show only 1/4 the final nuclear cloud height
is reached at the end of the flash time.
C. Stem and Cap Particle Radiation % Activity:
The % for the cap, stem and layers above is for the particle radiation activity. Thus, it is proportional to it's mass and volume.
To see how the Navy extended these to particles, refer to Table 3.6-1. The column labeled particle is really u/10
and the % exposure is particle radiation activity % for within the stem or cap alone. The actual % activity for that location of the
particle, we would have to multiply by the specific 6-layer % activity where the specific particle resides. Before examining these
numbers, we performed a little smoothing technique because they should have done this for 2 decimal places instead of one, which for the
small particles, are really to the nearest whole %. So now refer to the 1st table in srfmpdst.txt. Here,
we don't have the sums to 100% so with more decimals for some smoothing and scaling to their totals, we have a better % activity per
particle under New % Activity:.
The first obvious feature is that most of it's activity is concentrated with the smaller particles especially within the stem where over
half of it is < 30u. For the cap, half it's radiation activity is < 100u. Since the radiation activity is proportional (Prop # Par) to the
particle's mass or volume, we calculate the volume by Vol=4/3*pi*r^3,(u^3), with r=D/2, then divide by this volume since the radiation
activity is # par x Vol. This leaves us Prop # Par for each particle size as shown for the stem and cap under Prop # Par. When we ratio
each particle's proportionality to it's sum of all the # of particles, the proportionality is canceled and we are left with the % # of
particles. Thus, we can make an easier reference to the # of particles by % for understanding the processes rather than by % radiation
activity. In the model, we must always convert back to radiation activity by x Vol.
In the next table (Table 2.), we repeat the Activity columns from above and now multiply by the 20% total activity from the stem and 80%
from the cap to give the absolute activity in the complete Cloud (Stem+Cap) which accumulates to the 20% and 80% for the Stem and Cap
columns. For the Prop # Par, notice that for the stem, we have accounted for 99.8% of the # of particles < 30u. For the cap, 99.8% of
the # of particles are < 80u. Where we need much more detail is in the 1-10u range and some greater accuracy in the 10-30u range.
D. Theory vs. Measured Radioactive Cloud Activity:
Referring to SRFM-Figure 6 again, as we go away from the center of the cloud,
we have by the theory above, 36%/12% = 3.0 at the mid level matching the theory, then 28%/6% = 4.67 comparing the middle
layers of cap and stem, then 16%/2% = 8.0 at the top and bottom layers of the cloud. So as we depart upward through the cap of
the cloud, we have more mass (> 3.0) than at the center. Referring to Fig. 1.3-1 again, the main influx
of particles into the stem are carried aloft and return by the toroid in the edges of the cap cloud to the lower layers of the cap. Here the concentration
of the upper stem and lower cloud are equal (3.0). However, with all the churning, we show the hot (red) particles have cooled (purple -> blue)
by expansion, entrainment, collision and super saturation vapor pressure, have coalesced so may have fewer but larger particles as mentioned before but again, the mass is the same.
Once the hot plasma of the stem begins cooling, the updraft feed is slowed down but most particles from the up drafts are now in the cloud
cap and upper stem with a lessening of the supply from surface particles, cap to stem radiation ratio is (> 3.0). Even though we have more mass
in the upper part of the cap compared to the lower part of the stem, which area has the larger particles? We have the coalesced larger particles
in the cap but now we have some residue influx of particles in the stem that are not now vaporized because of the cooled plasma.
In the previous Radfo model, the observed particle distribution were from the Nevada test site. The inrush of dust was based upon
the ground crater as discussed earlier in Crater Dimensions from Demolition Dust Model.
A factor that converts megatons to total radiation was based upon the produced crater. However, the Nuclear tests were never conducted
within a city so this building dust from the cities must be taken into account and will be shown to be over 20 times as much as the
crater dust as seen in Crater Dust.
3. Nuclear Cloud Growth:
A. % Growth - Megatons Flash Times:
Based on the bomb size, we do have some empirical parameters generated. We have the dimensions of the
cloud and stem. That is, we take 1/3 of the cap diameter as seen in Pic SRFM - Figure 6.
again to display the stem which feeds the cloud cap. This area of violent updraft best represents a large smoke stack but how much?
We plotted the data points from the Cloud % Growth vs. Time
on the main log-log graph plots in purple. The derived eyeball fit formula representing
the % growth of the cloud top is pg=32min^.6195. This was said to be independent of bomb size, or nearly so?
Notice, the slowing down of cap cloud height with time. We might have used the near surface initial growth to
approximate the fastest cloud growth. Since the cloud height and size stabilize in 9 minutes, then if we represent
this at the end of the first minute, which is almost 1/3 the final cap height, that may be too long.
However, the flash time for 10 megatons is nearly 1 minute and that would be fine, but for 1 megaton, the flash time is 19.1 sec.
Therefore, if we only choose the 1st minute, the steepest slope on the graph or the fastest cloud rise, then the 1 megaton
fireball has stopped heating the plasma for 41 seconds so the violent updrafts are overestimated.
To normalize all bomb blasts, let's consider only the time that the fireball exists. That is, the
illumination time or the older term of smaller bombs - flash time. This best simulates a big boiler smoke
stack as we consider that we have a nearly constant stem velocity, as long as the fireball exists, as well as a constant
fire ball stem temperature. Now we have min = flash/60 so pg=32min^.6195, pg @ flash=32*(flsh/60^.6195) = 15.7%
for a 1 megaton with flash=19.1 sec. Now, Stack Height Hs=pg*Cap top then stack velocity updraft Vs=Hs/flash
where Hs is somewhat of a fictitious name stack height because there are no stack walls. So then Vs
is really the stack (stem) velocity in ft/sec. We are assuming that the growing cap top is caused by only
the vertical motion of the hot gases in the stem. Measured temperature from test blasts put the stem near
3000 - 4000 deg F. For 1st estimate calculations, we use 3000F. The estimated ambient air temp at Hs
for this 1 megaton was 32F and a wind velocity of 18 knots @ Hs=11,000'. This is close if we considered
the final predicted cap height of 70,380' by the bronze line in log-log graph plots.
The plume height formula using wind and temperature profile will be explained later in the Briggs Plume Rise formulas.
Continuing, for 1 mega, pg = 15.7% x 70,380'= 11,050' after 19.1 sec flash time. Therefore, the cloud is
probably mostly cloud stem at that point before the cap develops. So, updraft Vs = 11,050'/19.1 sec =
578.5'/sec = 176.3 mps or 394.4 mph. Actually, this procedure was repeated for .1,.5,1,2,3,5 megatons. They all
produced nearly the same results as the Vs ranged from 180 mps @ .1 meg -> 175 mps @ 5 meg so the 3 independent
graphs for % of growth, cap top for bomb size and flash time are in excellent agreement within a 1.5% error.
This must mean that the gas expansion from the hot fireball is constant, but just larger diameter stems with larger bombs.
B. Inflow Equals Outflow:
A sketch of the final mushroom cloud is shown in Fig. 1.3-1 again. Expanding the 1 meg,
since the stem radius (R=3.62 mi) is 1/3 the cap radius (10.86 mi), a cross sectional area of the stem is
Pi R^2 = 41.17 mi^2. So Vs x A = 394.4 mi/hr x 41.17 mi^2 = 16237 mi^3/hr. Now there is an inflow of air feeding
the stem perimeter in order to fill the void of the expanding gases within the stem plasma that is trying to create
a vacuum. The perimeter is the circumference 2Pi R = 2Pi x 3.62mi = 22.75 mi. Now we must choose to what height this air rush
through the perimeter of the cylinder will occur. Above, we saw that the stem with cap has risen to 11,050'. If there
was a cap forming within the flash time of 19 seconds, it must be small but for simplicity, lets use 2 miles (10,560')
as the top of the stem instead. So 16237 mi^3/hr / (2mi x 22.75mi) = 16237 mi^3/hr / 45.5 mi^2 = 356.9 mph inflow velocity
through the 22.75 mi perimeter 2 mi high stem cloud wall that balances the influx volume of the 394.4 mph updraft.
To better understand the air volumes balance, we can simplify the math expressions.
Stem Vol Up = Stem Vol Perimeter
Pi R^2 Vs = 2Pi R H Vp
R Vs = 2 H Vp
Vp = R Vs/2H = Vs R/2H
Supporting this volume balance, let's approach it by differentials. We define the volume change with time as
dV/dt = d(pi R^2 H)/dt. We can hold R constant or H constant. For R, we have
dV/dt = pi R^2 dH/dt and H, dV/dt = pi H d(R^2)/dt = pi H 2R dr/dt. Now equating dV/dt = dV/dt is
pi R^2 dh/dt = pi H 2R dr/dt, divide by pi R, R dh/dt = H 2 dr/dt. We then have dh/dt = 2H/R dr/dt or
dr/dt = R/2H dh/dt. Now the velocity across the perimeter Vp = dr/dt and dh/dt is the vertical velocity Vs.
So Vp = R/2H Vs = Vs R/2H, the same as above. Using this formula, Vp = Vs R /2H, 356.9 mph = 394.4 mph x 3.62 mi/2x2.
Also note, that when the stem radius R = 2H, or H=R/2, then stem perimeter influx velocity = stem updraft velocity.
So if H=3.62 mi/2 = 1.81 mi = 9,557', which might be a better stem value height for 1 meg instead of 11,050', then Vp=Vs=394.4 mph.
C. It's a Drag:
However, viewing Fig. 1.3-1, whether this is an artists conception or a compilation of test
blast pictures, you can determine by viewing the pics and film clips above. It appears as if the wind is entering the
stem from the bottom 1/3, or at most, 1/2 of the stem. Also, since this is a surface burst, this is where it was
initialized so the return air will be most active there. This diagram shows the frictional drag effect as the
bottom of the stem is an inverted funnel shaped cone indicating the surface dust is dragging along the surface and
moving at a slower pace than the layers of air above it.
If we assume that is correct, then the influx velocity (356.9 mph) is tripled for 1/3 stem height or doubled for 1/2 stem height
giving us 1071 mph or 714 mph respectively. However, there is most drag resistance near the ground. Logarithmic wind profile,
ignoring stability here, can be over 10 times higher over ice than crops. You can imagine the drag in a city where
we have crumbled buildings, not crops. The over 5 psi pressure wave will break most structures, then the structures
must also stand up in this return wind of over 700 mph. As a building collapse, the amount of dust is tremendous -
recalling NYC during 911 World Trade Center Dust Bowl Analysis photos and videos
and referring to 1 Megaton PSI Effects again.
D. Wind Profile Power Law:
To examine the increase of wind with height, we refer to the
Wind Profile Power Law.
Here, we changed the term's variable names V0 -> Ur, and V -> U and H0 -->Zr and H -> Z. Usually,
Z refers to absolute height where H refers to relative height above the ground, a better choice here.
We use V for velocity and 'o' as initial reported height and velocity instead of referenced r.
Also, p --> a as p is the power exponent. We use V/Vo = (H/Ho)^p, V=Vo(H/Ho)^p that is, given the wind velocity
Vo @ height Ho, we can calculate wind velocity V @ height H. The exponent p varies from .1 -> .3 with increasing
We can have any units of velocity or heights desired because the ratios will cancel out any conversion
units. If we have a ratio of two heights, say 64m and 2m, H/Ho = 32. Now for our most unstable case (A),
p= .1, V/Vo = 32^.1 = 1.41. For a slightly unstable condition (C), p=.2, so V/Vo = 32^.2 or 32^1/5= 2.
Here we have a doubling of wind speed for each wind sensor at a height multiple of 32 times Ho. For nighttime stable
conditions, p=.3 so V/Vo = 32^.3 = 2.83 or almost triple. The use of this formula is to extrapolate upper
wind speeds at higher levels in the surface vertical frictional layer. That is, if we had the same wind power
blowing across the same surface area in day vs. night, In the day, our near surface wind (2m), would be higher
than at night. We all observed that on clear evenings. However what you probably didn't notice is that the corresponding
winds at 64 m would be less during the day and stronger at night. The reason is that daytime solar heating
creates convective currents transposing lower level frictional wind velocities layers to higher levels and visa versa
to replace the rising air thus trying to mix the two different velocities to become more uniform.
At night, no vertical transport of convective eddies but still the same frictional surface resulting in less surface
winds near sunset but higher winds aloft. Such would be observed if a balloon escapes shortly after sunset. This could be
seen as the balloon would rise into the suns rays again, reaching the higher wind altitudes, then take off in a more
E. Integration of Wind Profile Power Law:
What is even more important is the mean wind velocity between two layers H1 and H2. Even though we can calculate V @ height H and
could average the 2 wind velocities Vo and V, we could have a more accurate velocity mean (velm) by Calculus integration (INT).
So velm = Integration (Int) within limits (lim) H1->H2 V dh/(H2-H1) where V=Vo(H/Ho)^p, velm = Int H1->H2 Vo(H/Ho)^p dh/(H2-H1) =
Vo Int H1->H2 H^p/Ho^p dh/(H2-H1) = Vo/Ho^p/(H2-H1) Int H1->H2 H^p dh = Vo/Ho^p/(H2-H1) [H^(p+1)] lim H1-> H2 =
Vo/Ho^p/(H2-H1) [H2^(p+1)-H1^(p+1)]/(p+1) = Vo/[Ho^p(H2-H1)(p+1)] [H2^(p+1)-H1^(p+1)].
Both of these terms are displayed in vel.dia . Here, we can have the velocity Vo with it's corresponding height
Zp anywhere on the wind profile. We can also have H2 < H1 that gives the same mean but calculated wind velocity at H2 will be the lower level.
Some of these sample calculations are see in velm.dat but may not represent too high of levels except if the winds
were very light.
In World Trade Center Dust Bowl Analysis, we had 1365' as the average WTC's height which produced an initial
dust cloud of 600' and left a pile of rubble 2.5 stories high or 31'. So we could apply this to the city rubble of collapsed buildings.
That is, the dust cloud will rise to 600'/1365' = 44% of it's original structure. For a city, we apply the data from NYC as an example:
NYC cityl= 9 cityw= 2 cityd= 65 floors1= 65 floors2= 25 length= 240 width= 225 with 10' floors within the
5psi ring with the formula from rpsi.gif, and taking reciprocals to
x5psi=.1754/mega^-.605 or for 1 meg, x5psi=.1754^-.605 = 2.87 mi. Since NYC is elongated, lets take the length to find the building
height at the x5psi range. This would be # of floors (flpsi) @ x5psi = floors1-(floors2-floors1) * x5psi/cityl = 65-(65-25)*2.87/9 = 52.24.
Then the average floors fla=(floors1+flpsi)/2 = (65 + 52.24)/2 = 58.62 floors then times 10'/floor = 586.2'. In proportion, the cloud dust is
44% x 586.2 = 257.9' and the rubble pile 31/1365 = 2.27% so x 586.2' building height = 13.31'.
However, we must allow for the collapse of the other buildings so with a city density of 65%. We check for a 1 megaton,
for NYC buildings, approximately 30% of the total cement was converted to dust which was 1.889e+9'^3 mentioned earlier in
Crater Dust from the Demolition Dust Model again. So 100%-30% = 70% remains in a rubble pile
or 70%/30% = 2.33 times as much building cement remains as rubble rather than as dust cement. This is 2.3 x 1.889e+9'^3 = 4.41e9'^3 spread out
over a circle area within a 1 psi radius of 7.00 mi is pi 7^2 = 153.9'^2/mi^2 = 4.29e9'^3. Now 4.41e9'^3/4.29e9'^3 = 1.03'. So we would have a
to have an average height of near 1' of rubble away from the collapsed buildings but near the average building peak, piles to 13.3' indicating a
rather rough terrain. However, higher buildings near the center, for the 5psi area, we had near 37% dust or 1.7 times which gave us 2.74e9'^3/7.215e8'^2 = 3.8'
F. Computer Runs of Wind Profile Power Law:
Applying some of this into the wind profile formulas above, we let vel = 1071 mph @ 1/3 stem height = 3700'. Actually, we only need the average
wind velocity from 257.9', the average dust cloud height down to the rubble pile peaks of 13.3'. According to the web reference,
we are out of the normal height range of use because the winds at this level are caused by pressure gradients although mixing to this height
is very probably also. The difference here is that we have a violent inrush of wind into the lower stem that drags along the surface for
about 10 minutes. As the nuclear cloud grows, we have air rushing into the stem @ 3.62 mi from all directions which all about
overwhelms any horizontal pressure gradient wind just as convective thunderstorms do. So, we get our first estimate of the lower wind level from this. Now at the true surface,
we have no wind but vo=0 @ ho=0 blows up the formulas. Let's estimate about Ho = .01' or .12" for H2. The surface averages 3.8' of crushed building
but this is our Ho. We have vo=1071 mph @ 3700'. Let's first see how much the velocity would be reduced at dust cloud top of 257.9' using a neutral
or D stability p=.25 or let's call it a quad root since .25 = 1/4 and .333... = 1/3 a cube root.
The computer run gave us this below:
Pex @ Stability = 0.25, Zp = 3700.00, Vel @ ZP = 1071.0, Z1 = 3700.0, Z2 = 257.9,
Vel @ Z2(257.9) = 550.30, Mean Vel Z1(3700.0)->Z2(257.9) = 888.01
Notice quite a reduction in the speed (1071 -> 550 mph) @ 258' but the mean is 888 mph.
Pex @ Stability = 0.25, Zp = 3700.00, Vel @ ZP = 1071.0, Z1 = 257.9, Z2 = 0.01
Vel @ Z2( 0.01) = 43.42, Mean Vel Z1(257.90)->Z2( 0.01) = 440.26
Here we still have a strong wind (43+ mph) near the surface (.01') and a mean velocity
of 440.26 mph through the entire 258' thick dust cloud. The web site article also
explains that "Because of low surface roughness on the relatively smooth water surface,
wind speeds do not increase as much with height above sea level as they do on land.
Over a city or rough terrain, the wind gradient effect could cause a reduction of 40% to 50%
of the geostrophic wind". Also states, "While over open water or ice, the reduction may be only 20% to 30%".
In our case, we have a large substitute for geostrophic wind. Well a flattened city with < 3.8' -> 13.3',
is not much a city anymore but it's still rough, cement rubble, not high grass. Let's set this at 35%.
So now 35% x the velocity mean of 440.26 mph = 154.1 mph. How this dust is swept into the stem of the nuclear
cloud as well as larger particles is now seen. We have 1071 -> 550 -> 440.26 -> 154.1 mph reduction
or a 154.1/1071 = 14.4% reduction overall. This means that updraft velocity Vs should also be reduced by the
same or Vs x 14.4% = 394.4 x 14.4% = 56.8 mph according to the above if R ~= 2H.
However, because of this frictional drag reducing the dust cloud movement into the stem by 14.4%, then above the
dust cloud top (dct=257.9') to 1/3 of the stem height (11,050') must make up for the observed rise of the stem or 100-14.4=85.6%.
Weighting the cross sectional area of the volumes, Vpl x dct + Vpu*(stem-dct) = 1071*stem.
Here Vpl = lower (154.1 mph), and Vpu = upper (? mph). So 154.1 x 257.9 + Vpu * (11,050-257.9) must equal 1071*11,050.
39742.4 + Vpu x 10,792.1 = 1071*11,050 or 39742.4 + Vpu x 10,792.1 = 1.1835e7.
Vpu x 10,792.1 = 1.1835e7 - 39742.4 = 1.1795e7 then Vpu = 1.1795e7/10,792.1 = 1092.9 mph. So the top of dust cloud up to
the top of the stem must average 1093 - 1071 = 22 mph greater than in the dust cloud.
G. Vs = 400 mph - A Helpful Constant:
With the stem vertical velocity fixed at nearly 400 mph for most bombs, we should derive some simple
wind velocity estimates feeding the stem further away radially. The volume/hr updraft in the stem is then
Vs=400 mph x Pi R^2 with R the radius of the stem in miles. However, working with the stem R, for
distances > R feeding the stem, Vp must decrease as this volume of air approaching the stem per sec would be
larger than the original volume of air approaching the stem. For distances radially outward from the center
of the stem, let's use multiples of the radius R for demonstration. For distances kR, the perimeters ratio
is 2Pi kR/(2Pi R) = k times as much circumference or perimeter. Now we have 1/k x original Vp to keep the
same volume exchange. That is, at twice the stem radius out, we have 1/2 Vp or 1071/2 =535.5 mph. At 3r,
we have 1/3 Vp or 1071/3 = 357 mph etc. Now let's reduce all the frictional layer levels in the same proportion
as done previously above. Thus the mean velocity through the dust cloud layer is 154.1/2 =77.1 mph and 154.1/3 = 51.4 mph.
With bigger bombs, the Vs (400 mph) will not change, but stem radius R will be larger, as well as the stem height H.
H. 5 PSI vs. Stem Radii:
Note that x5psi radius of 2.87 mi is less than the stem radius R=3.62 mi. So if k = x5psi/R = .79 < 1. That is, all
crumbled buildings within the x5psi radius are also within the cap radius so all this dust will be sucked into the stem.
Now the most violent updraft outflow and inflow winds are during the flash time. Neglecting the mach 1 speed of sound of the pressure
wave and the time the cloud stem begins to rise, the distance the dust cloud travels towards the center of the stem is rate x time.
The rate 154.1 mph and the time is 19.1 sec/3600 sec/hr. So distance = 154.1 mph x 19.1 sec/3600 sec/hr= .818 mi. However, winds of
85 mph can move 1" diameter stones so when other brick and cement fragments are sucked into the stem, the fireball pulverizes them
into tiny 1u particles also. Because of the 19.1 sec flash or illumination time, we also have burning caused by intense thermal radiation
where the smoke and ash from this is also sucked into the stem cloud. Even though we set the crumbling buildings distance to x5psi, there are
still collapsed weaker structures > x5ps1 as we view 1 Megaton PSI Effects again.
We really shouldn't cut off the particle inflow at x5psi because with 154 mph inflow, more particles from these collapse weaker structures and
fire smoke will cause a gradual diminishing effect to maybe 2 psi @ x2psi=4.7 mi. Here, the cloud stem radius (R=3.62 mi) is less so how long will it
take the surface air debris outside the stem radius to travel into it?
Above, we found that 2Pi kR/(2Pi R) = k times as much circumference or perimeter. Now we have 1/k x original Vp to keep the
same volume exchange. So if we start at the 2 psi ring, k=x2psi/R = 4.7/3.62 = 1.298 so 154.1/k mph = 154.1/1.298 mph = 118.7 mph
but during the flash time, x 19.1 sec/3600 sec/hr travels .63 mi. The distance it must travel is 4.7-3.62 = 1.08 mi to be suck
into the cloud stem, during the flash time only, is too far. As the distance out from the stem is larger, the velocity is less and the travel
time more. Actually, k should be varied in small increments to depict a more accurate average flow into the stem.
Inflow should make this clearer. This crude sketch shows the brown psi rings with the red cloud stem between the two.
Any particles from outside the stem moving as inflow towards the stem will increase in speed, as shown by the increasing length of
the purple wind vector lines, each dash representing the distance traveled in a unit of time.
I. Post Flash Time Inflow:
After the flash time, the cloud continues to grow but at a reduced rate so Vs < 400. We saw
in Cloud % Growth vs Time and log-log graph plots in purple
that pg=32min^.6195. At the end of the flash time, the cloud has only grown to 15.7% of it's total height. For
a 4 mega burst, we saw it is 24% of it's final height by the end of it's flash time.
We review all the above by combining the pg formula, Vs, Vp, k for Vp/k.
We calculated these Minimum Velocities.
How far out will that be? A stem speed Vs=400 mph will also be important in determining the
plume rise by atmospheric conditions rather than by projecting a few plotted graphical
Even though we have a final visible mushroom cloud model in Pic SRFM - Figure 6.,
if we examine Fig. 1.3-1, we see the process. It appears that
the stem flares out at the top like a thunderstorm anvil. As hot as the gases are below this level,
work is done in the expansion so it cools and the stem is returning to it's equilibrium
temperature level. However, the stem speed of Vs=400 mph has an increasing horizontal component
with height so it is deflected and rolls into eddies as the stem's leading edge returns to the cap
cloud base. We must distinguish between the visible stem and the invisible stem hidden within the
cap and expanding and rolling downward again to the cloud base so we will call the extended source length
esl=hf+(hf-hsf) where hf = final cap cloud top, and (hf-hsf) the cap cloud depth where hsf is the top of
final visible stem or base of the cap cloud.
J. Reverse Inflow to Outflow Concept:
INFLOW.c - 11/26/08
We want to trace this inflow of air from the ground somewhere between the stem to x1psi into the stem,
through the cloud top and back down to the visible stem hsf. Rather than to start somewhere unknown on the surface
and calculate to determine if the air will reach the cloud stem, which would be repeated in a loop until this outer
point is known, let's reverse the flow and see how far away from the stem that an outflow velocity becomes negligible.
This is then the same point as the perimeter velocity would reach the stem when we restore the outflow back to inflow.
We can loop with a unit of distance or unit of time. We choose distance delta radius dr = .01 mi. We set horizontal
outflow perimeter velocity (Vp) to initial perimeter velocity (Vp0) @ stem wall radius Rs. The distance r from the center
of the stem is r=n*dr for each n increment. The k radius proportionality explained above is k=r/Rs. With all r < Rs,
the updraft covers the entire area of the stem so Vs = Vs0 or set k=1.
The area swept out by the new dr perimeter is the area difference between the new radius r and the old radius
r-dr. This is the same concept used and explained in 5psi ring.
So dustpa=pi (2r+dr) dr. Now for r < x5psi, all this area is within the x5psi dust ring. We keep the amount of dust
within x5psi as 100% or rat=1. For r > x5psi, rat is linearly interpolated from rat=1 for r >= x5psi -> x1psi
by rat=1-(r-x5psi)/(x1psi-x5psi) for x5psi < r < x1psi. We then adjust dustpa by the rat factor and accumulate corrected
dustpa by dusta. We keep tabs on the travel time (tt) for the delta radial distance traveled (dr) by the perimeter outflow speed
hr=dr/Vp hrs and tt=hr*3600 sec. We accumulate tt by ctt and when ctt > flash time, we break out of the loop since the bright
fire ball diminishing, Vs diminishes also so needs to be recalculated. Before that, Vs continues at the same 400 mph rate and advances
vertically by vd = Vs*hr*5280.
K. Cumulating Dust Areas:
The cumulative vertical distance cvd is cvd + vd. We initialized cvd at the top of rubble dust cloud
(dch), which is already within the stem, instead of the average ground rubble height of 3.8'.
At the end of the loop, we correct Vp by Vp=Vp0/k if r > rs since k=r/rs to keep the same volume
exchange as explained above and continue the loop. When be break out of the loop, we have the final
accumulated dustpa term (dusta) giving area of the x5psi dust from demolition. Comparing this to
the area of the x5psi ring is a5psi = pi x5psi^2. So dusta gives us the % of the whole 5 psi ring demolition dust
swept into the stem during the flash time. For the 1 mega, at the end of the flash time, we have
r=0.820 mi, k= 1 since r < rs(3.62 mi), Vp still at 154.10 mph, flash ctt=19.16 sec, cvd=11359.3',
updraft height of the dust cloud (hc)= 11617.2', rat=1 and dusta=2.11 mi^2 = 8.16% of 5 psi area so far.
The next task is to determine the rest of the demolition dust from after the flash time to the 5 psi ring and beyond to the
2 psi ring, then to the 1 psi ring to be swept into the stem. We now use a time increment in the loop, delta sec=1/4, and we
initialized travel time tt to where we left off, (flash time tt=flash). We continue tt until it reaches the 100% growth time.
Previously in the program, this was determined by looping every 1/4 sec, converted to min and put into pg=32*pow(min,.6195)
until pg >=100%. Taking logs and solving for min caused some precision errors so we did looping instead. We found
@ pg=100%, sec=380 for 1 mega by loops. So now, we do it again for 1/4 sec intervals and note the other parameters
change also. We convert sec to min for pg and calculate the top of this visible growing cloud stem by this % growth pg
by h=hsf*pg/100, where hsf is the final visible stem height. We continue with the cumulative vertical distance cvd + vd where
Since we are past the flash time, the updraft velocity (vs) decreases and the upper part of the stem flares out from expansion.
We estimate this height to be at 75% of the final cloud cap height (hf). When cvd > 3/4*hf, then we linearly interpolate
to the remaining stem distance (esl-3/4*hf). So rat = (esl-cvd)/(esl-3/4*hf) and decreases from 1 -> 0 corresponding with
cvd = 3/4*hf -> esl. This causes Vs to decrease from Vs0 -> Vs0/9 (11.1% Vs0) by the expression Vs=Vs0*(rat*8+1)/9, but not
0 because of inertia. We use dr again but now dr=Vp*hr instead of the fixed .01 mi. Then the new radius r is r+dr mi as before
and we calculate dustpa and adjust it by rat as explained above when x5psi < r < x1psi. We calculate vertical distance vd & cvd
again as before as well as the top of upward dust cloud hc=(cvd+dch).
We set hc=cvd until hc > hf, where hf is the visible top of final cloud. If hc > hf, hc now reverses
from up -> down by the expression if(hc > hf) hc=hf-(hc-hf), that is the distance that hc is past the top
of hf, (hc-hf), is directed down from hf. To indicate a change in direction, we change a ud=0
for up, to ud=1 for down. Then if hc < hsf, goes below top of visible stem or base of cap cloud, it is buoyant
again and directed upward by the code if(ud == 1 && hc < hsf) hc=hsf+(hsf-hc) and we set ud=2. Finally,
probably never reaching the top again, we put this code just in case - if(ud == 2 && hc > hf) hc=hf-(hc-hf)
again and ud=3 which reverses from up -> down @ cloud top again. The total path of cvd was 100,923'.
With the Vs decreasing, the '^3 volume/sec in the stem Rs decreases also. The corresponding
perimeter inflow Vp will also decrease at a given r. So v=Vs*pi*Rs*Rs '^3/sec / (2*pi*r*h) '^2 = '/sec
or cancelling pi, vp=Vs*Rs*Rs '^3/sec / (2*r*h) '^2 = '/sec. So at stem/3, we have 3Vp and the frictional drag,
derived from above, reduces 3Vp further down to 14.4% of 3Vp. With r < Rs, the expression would have Vp > Vp0, we have
to set this highest bound for Vp=Vp0 since r < Rs indicates we are within the stem's outer radius Rs where the whole
cross sectional (pi Rs^2) area moves upward at a constant Vs.
L. Calculated Time Stepped Parameters:
All these parameters are seen in inflow.hgt . Now we can think
of this output as inflow rather than outflow. When the flash mode is on, fm=1.
The vs and vp are in mph. The vertical displacement (vd) and it's cumulative (cvd) are in feet.
The up-down ud shows 0 for up, 1 for down. The leading top of the dust cloud (hc) is in feet, being visible then invisible
due to reflections or reversing directions. The dust % area in the dr ring is dustpa and a sort of concentration term
is given by dustpa/vd. The highest concentration is at 38872.3'. Notice the cvd is the total distance the top of the dust
cloud has traveled, while the hc is the corrected real height of cvd by the up-down motions ud. The vs and vp keep constant
till after the flash mode but the divergence of the cloud near 3/4 it's final cap height, vs and corresponding vp decrease
to the final minimum of 0.92 mph which fall in between the .02 mph and 9.25 mph first guessed. Now 0.92 mph is the terminal
velocity for 81 u so all particles <= 81u should be swept into this cloud, even near the surface, before falling out at that
distance. Larger particles will have a component towards the stem if elevated. Recall, our initial dust cloud height was 257.9'.
We want to see the dust % of the x5psi ring of demolition dust for each height of the Nuclear Cap
and Stem. The first line shows the initial cloud height of the dust being sucked into the stem
radius. While the initial flash mode is on (fm= 1), all the dustpa's that go through the hot plasma are
vaporized and this total amount goes into the Coalescing Mini Model explained later. How much?
Summing all these dustpa in dusta total an area of 43.129 mi^2. The area within the 5psi ring
is a5psi = 23.97 mi^2 so 43.129/23.97 = 179.93 % or 1.80 as much dust as that produced by the
5psi ring. This will have it's own unique particle distribution compared to the city's
building dust generated from demolition.
The total dusta was 101.455 mi^2 or dusta/a5psi 101.455/23.97 = 423.25 % or 4.23 as much dust as that
produced by the 5psi ring alone. Now since 43.129 mi^2 of dusta was vaporized, then 101.455 - 43.129 =
58.326 mi^2 is not and will be suspended cooler particles of building dust into the nuclear cloud
without coalescing. So dusta/a5psi here is 58.326/23.97 = 243.33 % or 2.43 as much dust as that
produced by the 5psi ring. Or put another way, the total amount of dust area was 4.23 times as
much as a 5psi ring and 42.5% is vaporized and 57.5% is not.
M. Dust Distribution in the Nuclear Cloud:
Where is the vaporized 42.5% and non vaporized 57.5% of the dust distributed through the cloud?
This is seen and explained best in inflow.dia .
After following a particle inflow to it's up's and downs, the coalesced particles
(o) finally are churning in the bottom third of the cap.
To derive numerically here, we examine inflow.hgt again.
Notice in the last line, we stopped the summation of vertical air flow so the leading edge, which
was first vaporized (vd=257.90' layer), should end up at 47664'. The file was written with the leading
interval layer to be vaporized as the 1st record as 257.9', with it's associated dustpa, so this layer is
now at it's final resting place height so should be at the end of the file (47664'). The top of the visible
stem at the end of flash is now at the end there and the stem length's beginning must go backwards
it's length (11072'). That is, back down to the final visible stem top or cap base 43750', then
upwards to 50861' from there totaling it's length of 11072' indicated by o's. Each successive stem
segment layer thickness vd is on each successive line. So with cvd reset to 0 again, we accumulate each record's
successive vd with cvd=cvd+vd again but real corrected height hc = 47664'- cvd until cvd > flash time stem length
So we read this file again, and code for the changing of up and down directions ud for hc. This time, we start by subtracting
vd to backtrack, not adding as we did to create the file. We really don't need to keep track
of each flash's dustpa, just the total as we lump it all in the 42.5% for coalescing due to all its violent shear
turbulent mixing in the invisible stem volume inside the cap and will be examined later.
After the influx of the coalesced particles, the cooler ones are not liquefied but are cooler solid dry spheres from the city
building's dust that won't coalesce because the flash from the hot plasma has gone out. If we continue with hc = 47664'- cvd, continue correcting for ups and down
directions, and carry it's associated dustpa, we see where the higher concentrations will be in the rest of the
nuclear cloud. This would be backtracking starting from 50861' o -> % -> " ->, ->: by inflow.dia .
This is seen by tabular form in fl1u1.nyc , where the file ends up near the surface where
the last inflow layer has entered the cloud stem.
N. Comparison of Original Cloud Model:
Lets calculate the % ratio of dust cap/stem. We convert the dust area dusta to %. Using the summary on the
last line, total cap (Flash & Cap) dust comprises 80.60% while the stem holds 19.40% with a ratio of 4.15. How does this distribution compare with the standard
SRFM - Figure 6? We see there a 80%/20% = 4.0 which indicates an excellent match.
For model use, we put all these flash and non-flash perimeters areas and heights in another program called HPAFPA.c
which stands for Heights by Perimeter Areas and Flash Particle Activity. The particle distribution for the flash particles
will be shown later in the coalescing mini model explained below. Those corresponding flash heights particles should be distributed from
1u @ top of 1/3 cap to max u @ bottom of 1/3 cap height.
O. Height Link to % Particle Distribution:
Even though we have excellent agreement with the original cloud model, should we use this
method? Seems like we should but should we? No. During these nuclear test cases, they
were usually done in Nevada or remote places with no buildings. Using the
Dust Ring Area stored in an array called perimeter area pa from
5psi ring again, was an excellent approach. The amount
of dust volume is proportional to each of these pa's considering we have a constant height source
of ground dust inflow into the stem. However here, each building height varies with each dr distance. The
amount of dust generated is not linear to each new r as seen from the previous complex programs using r. We calculated the actual dust
for each building's area pa[ix] by distance r from ground zero using it's linear interpolated height
from the city profile data file explained in the 5psi ring . Since linking these ix rings to the inflow and height
hc in the nuclear cloud, each ix is also linked to the actual amount of building dust in each perimeter ring area pa in CITY%PD.c. We also have
particle size distributions from crushing and rubbing depending on each building's height. So with each r's (ix ring area) swept into the cloud,
we actually have a unique particle distribution for each height hc (Height Correction). In the previous radioactive fallout models, we assumed the same
particle distribution applied to each nuclear cloud height sub division layer, but more or less amount in each.
P. Redistributing Dust in the Nuclear Cloud: FNFPARH.c
This involved rewriting the last program HPAFPA.c , but leaving INFLOW.c alone.
This deserves a new name also as FNFPARH.c standing for Flash - Non Flash Perimeter Areas, Radius and Heights.
Actually, the PAR could represent particles also. For the coalescing vaporized particles distribution file
we will explain later in the Coalescing Mini Model, but is input here.
We have the diameter in microns and it's associated total # of particles. The 3rd column is the
proportional activity column not needed here but only for reference, and will be generated when needed in the final model.
For this file, we want to use this again for different nuclear blast yields rather than running this over and over.
We normalize each pd[id] as decimal % distribution (summated to 1.00) rather than actual # of particles
for each diameter id. Therefore, we need the number of different size particles first for id = 1-202 u's.
(1) Flash Mode - Vapor Trail % Dust:
The main input file is hcr1.NYC from INFLOW.c .
As seen, we read the cap and stem dimension on the first
line, then we have columns fm, hc and r. Here, flash mode fm = 1 for flash particles and fm = 0 for non flash mode
or no vaporized particles. Then the corrected height hc and radius from ground zero r. All
this was just explained above in M. Dust Distribution in the Nuclear Cloud:.
We work with hc & r for all the fm=1 lines until fm=0. Our first hc is our initial hcl
(height corrected-last) and our base height for the hc group (hcl0). We want to distribute
the 202 flash particles through the visible stem height % growth @ end of flash (hsfl).
Since the first hc's represent the top of the vaporized cloud stem, it has been in route the furthest so
we have that region the longest time for coalescing, hence the largest particles there
decreasing in diameter to where inflow into the base of the stem began and ended. We then lay down this array pd[id] in reverse order where id=202
for the 1st particle level (h), 201 for the 2nd (h), 200 for the 3rd (h) etc. Here we use apl = hsfl/idf-1 = 53.8'
as the Average Particle Length for each flash particle along the flashed vapor trail. We have 79 hc levels from each r as dr=.1 mi intervals
that the 202 particle levels are distributed. This means that we have to put 202/79 = 2.56 particle layers
between each hc layer.
The problem is, the length of hsfl is broken up into boundaries and changes directions. Therefore, we use
dh=hc-hcl to calculate the +1, -1 sign (pn) = dh/|(dh)| where | | represents the absolute value of direction
between the hc levels. Starting from our base level, we go to h=hcl0+pn*apl until
we reach past the next hc level, whether it be up or down, by using the code:
if(pn == 1 && h > hc) break and if(pn == -1 && h < hc) break out of this hc level loop.
While between the hc levels, we write out to the Flash Height Particle Distribution file
(flhdpd1.NYC) . Here we have the height (h) in feet, the
particle size diameter u (idf) f for flash, and it's % of occurrence by decimal (pd). Once
h is past the next hc level, we reset hcl0=h and hcl=hc then decrease particle size idf.
Our next base level hcl0 will then be above the last hc level because this is where we broke out
of the last loop so this is where we begin our next loop. We continue this process and also calculate then accumulate
the number of flash tons (fton) and continue reading hcr1.NYC until fm=0.
We then break out this loop, reread this last record and continue reading the file and follow the procedure below.
(2) Non Flash Mode % Dust:
This time, we have 449 particles but can't fit them between the rest of the nuclear cloud past the visible hot flash
plasma of the stem's 1443 levels as we did above, because there are now more levels than particles so we want to know
what levels to put the tons of particles with it's % distribution. We now relate ix=r/dr and find each corresponding ix (1st column) in
ctydst1.NYC. We hold the last ix by lix and when the next ix > lix, we stop reading fm, hc & r
and average hcl0 and hc by h=(hc+hcl0)/2. Now we write out h (1st column) to the output file
nfhdpd1.NYC , and continue reading the ctydst1.NYC record for the
particle size distribution of tons while writing it to the output file nfhdpd1.NYC following the height to
complete a file record. Then we reset lix=ix and hcl0=hc again and continue the loop until all the ix by ix=r/dr are complete.
Finally, the height profile of dust particle size by tons distribution for the nuclear cloud model inventory is complete. We also continue to
accumulate the total tons (cton) where we left off at cton=fton above.
An interesting tally from the fnfparh bug output file shows the following:
Buildings Tons: Non Flash = 1.280e+8 + Flash = 7.144e+6, Total = 1.352e+8
Flash: Buildings Tons=7.144e+6 / Crater Tons=6.198e+6 = 1.153, Total = 1.334e+7
Total Tons: Building = 1.352e+8 + Crater = 6.198e+6 = 1.414e+8
The tons of flash particles is only 15% more than the crater tons of dirt while their total
of 1.334e+7 is sent to an output file representing all the tons of flash particles
to be processed by the flhdpd1.NYC file just generated here
and explained above that represents decimal % particle distribution and not flash tons until
multiplied by the pd column. The Crater Tons file record was calculated from city%pd.c. Also
from city%pd.c, reviewing these results from Crater Dust again, we had
1 Mega - Buildings: 1.415e+8 Tons = 1.889e+9'^3 = 22.84 x Crater Vol (4.136e+8'^3).
Therefore, neglecting the crater dust, the inflow tons of dust was 1.352e+8 out of a possible 1.415e+8 for
the 1 psi NYC city dust tons or 95.5% from a 1 megaton blast. With different meteorology, the 1 mega could inhale
even more or less of the available dust. With a bomb > 1 mega, 100% is likely.
4. Flash Coalescing Mini Model:
Radioactive Fallout Model (Radfo) - Part II: