 AN APPROXIMATION
Posted on the message board by Clifford E Carnicom
AUGUST 22 2000

The following is a first pass at an approximation to the
logistics of a seed or spray operation, with at least an
initial consideration given to the factors of volume of
air involved, seeding concentration, particle size,
weight and size limitations of aircraft, number of
aircraft required, and the amount of time involved. As
any errors of significance are found, please do not
hesitate to note them and the subsequent revisions
and re-considerations can be made. All figures are
given in approximate forms. The primary objective here
is to determine the feasibility of an operation as
commonly is observed across the country, and initial
results indicate the logistics of such an operation are
quite feasible.

1. Assume a volume of the sky is to be seeded with
micron size hygroscopic (water-seeking) nuclei. This
example picks a volume of 200 miles long by 200 miles
wide by 1 mile thick. Lets’ also assume the job needs
to be done in a couple of hours.

2. The volume of air involved is then
200*5280*200*5280*5280 = 5.8879E15 cu.ft.

3. Assume that we wish to seed this volume at a
concentration of 30 particles per liter. Vincent
Schaefer, in references to early cloud seeding
projects, mentions an introductory range of 10-50
particles per liter are desirable. Let us use the average
of this range.

4. So the number of particles that need to be
introduced is 5.8879E15 cu.ft. * (28.3168 liters/cu.ft)
* (30 particles/liter) = 5.00E18 particles at the micron
size.

5. Let’s assume a plane travels at 500mph
(733ft./sec.) Next question is how wide a swath of air
would a plane have to seed to finish the job in 2 hours.
This can be set up as:

5.8879E15cu.ft / (n * X * (733ft./sec) * 5280ft.) =
7200 sec. (2hours)

where n would be the number of aircraft, and X the
width of seeding by an individual aircraft.

6. X here solves at 211295ft, or approx. 40 miles wide,
assuming n=1 for the time being. So if one aircraft
could seed an area 40 miles wide, the job would be
done with one aircraft. But as this does not seem
reasonable, and also does not fit the observations
which are commonly reported, let’s assume an
equivalent configuration of 8 aircraft seeding spaced 5
miles apart horizontally . Or 10 aircraft at 4 miles apart
horizontally, etc., could be used.
The set of 8 aircraft will satisfy reasonable conditions
of conformance to observations for the time being.

7. At this point we have a configuration which will seed
the volume of atmosphere under consideration by a
reasonable number of aircraft in a specified time at a
certain concentration of a certain size.

8. We can verify the number of particles being
delivered by each aircraft by the following:

9. Each plane needs to seed : (211295 ft. / 8 aircraft)
* 733ft. * 5280ft./sec. = 1.022E11 cu.ft./sec
with (5.00E18 particles / 8 aircraft) = 6.25E17 particles
per aircraft and 6.25E17 particles / 7200sec. =
8.68E13 particles/sec. per aircraft.

10. And for the final concentration of seeding,
(8.68E13 particles/sec.) / (1.022E11cu.ft./sec) = 850
particles / cu. ft. / sec. and 850 particles/cu.ft./sec
with 28.32 liters/cu.ft = 30 particles / liter as is
desired.

11. Steps 8, 9, 10 only serve to verify the seeding
concentration is in order.

12. Now we need to give consideration to the weight
of the material being carried, and whether it also
remains feasible. If we have a system that is capable
of transforming solid material to micron size seeding
material, we will need: 5.00E18 particles / (1E18
particles/cu.meter) = 5.00 cu. meters.

13. With 8 aircraft, this is 5.00 cu. meters / 8 aircraft
= .625 cu. meters /aircraft. or .625 cu. meters *
(35.31 cu. ft. /cu. meter) = 22.1 cu. ft. of material per
plane. This is equal to a block of material 2.81 ft. on a
side. Feasible for size.

14. For weight, let’s pick the element of barium to work
with. Reasons for this choice are under consideration
and will be discussed further at a later time. The
density of barium is 3.5gm/cu.cm or 3500kg/cu.meter.
So in our example, .625cu.meter * (3500kg/cu.meter)
= 2188 kg. Since 1 kg. = 2.2lbs, the weight in a familiar
system is 4812 lbs. of barium.

15. And now since Barium occurs naturally in a couple
of forms, and since I currently have a greater interest
in barium carbonate, and since the elemental barium is
70% of the atomic weight of barium carbonate, lets
jump the weight of material required to 4812 / .7 =
6875 lbs. of barium carbonate or, 3.44 tons per plane.
Since aircraft easily are carrying 150 folks at
160lbs/folk = 24000lbs = 12 tons, weight requirements
also do not seem to be a major problem.

In summary, an operation that seeds the sky with
micron sized hygroscopic (water-seeking) nuclei
involving 8 aircraft within a 200 mile by 200 mile by 1
mile high volume of our skies in a 2 hour period at a
concentration of 30 particles /liter seems quite
feasible, and is in accordance with repeated
observations of same over the past 1 1/2 years across
the country. Although not intended at this stage to be
an exhaustive study, reasonable consideration has
been given to constraints of air volume, concentration
levels, particle size, weight and size limitations of
aircraft, number of aircraft employed, and the amount
of time required to conduct the operation. Any
significant errors discovered will be corrected as this
scenario is reviewed by the readership.

Clifford E Carnicom
August 22 2000

Note: The following exchange between a user by the name of Skylooker and myself subsequently occurred on the message board on August 23 2000, and may be useful in regards to the statements above:

“Are you on the level???????

“You can’t squeeze rain from a cirrus cloud cover”,
especially if it is an
artificial,water,absorbent,expanding,aerosol cloud
cover. Seeding is most effective within the cummulus
cloud context and is predisposed towards precipitation,
not towards the opposite end of the
spectrum.>>>>>>>CHIEF SKYLOOKER”

Re: An Approximation

Skylooker,

I appreciate the distinction, and this may be an issue
of semantics more than reality. There is no mention of
precipitation or the intent of creating it within this
scenario, and all indications are that exactly the
opposite phenomenon of extraction of moisture is
taking place. Hence the repeated emphasis upon the
use of hygroscopic. The term seeding is used only in
the generic sense of a “source or germ” – for a
catalytic process. Extraction and diverting of moisture
may well be germane, but there is no assumption or
mention of an intent to induce immediate precipitation
in the model above. I will assume that the point and
question at this time is one of semantics.. regards,

Clifford E Carnicom