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METEOROLOGY DEMYSTIFIED

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METEOROLOGY DEMYSTIFIED

STAN GIBILISCO

McGRAW-HILL
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DOI: 10.1036/0071448489

To Samuel, Tim, and Tony
from Uncle Stan

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For more information about this title, click here

CONTENTS

Preface

xi

CHAPTER 1

Background Physics
The Solid Phase
The Liquid Phase
The Gaseous Phase
What Is Heat?
Temperature
Some Effects of Temperature
Temperature and States of Matter
Quiz

1
1
6
14
17
21
27
31
36

CHAPTER 2

The Atmosphere
Common Variables
Atmospheric Circulation
Weather Systems
Weather Maps
Wind Speed
Clouds
Quiz

39
39
44
49
57
61
64
72

vii

viii

CONTENTS

CHAPTER 3

Observation and Forecasting

(Video) Teaching Text Structures for Non-Fiction Reading

Some Weather Lore
Basic Observation Tools
Advanced Observation Tools
Forecasting Methods
Numerical Forecasting
Quiz

75
75
80
88
94
100
102

CHAPTER 4

Rain and Lightning
Thunderstorm Formation and Evolution
A Hypothetical Severe Thunderstorm
Effects of a Thunderstorm
The Anatomy of Lightning
Protection from Lightning
Quiz

105
105
111
114
119

126
130

CHAPTER 5

Tornadoes
Formation and Prediction
Tornado Behavior
Variations on a Violent Theme
Protecting Property and Life
Quiz

133
133
140
146
152
156

CHAPTER 6

Tropical Cyclones
A Hypothetical Storm
The Azores-Bermuda High
Anatomy of a Hurricane
Hurricane Life Cycles
Effects of a Hurricane
Preparation and Survival
Debby Strikes
Quiz

159
159
163
167
169
172
182
186
192

ix

CONTENTS

CHAPTER 7

Winter Weather
The Stormiest Season
Anatomy and Effects
Blizzards
Winter into Spring
Quiz

195
195
202
212
222

227

CHAPTER 8

Abnormal Weather
Heat Waves
Cold Spells
Too Much Water
Not Enough Water
Quiz

231
231
240
246
251
255

CHAPTER 9

The Past and Future Climate
Early Weather
Glacial and Interglacial Periods
The Future: Warming or Cooling?
Weather and Climate Modification
Quiz

259
259
260

273
279
282

Final Exam

285

Answers to Quiz and Exam Questions

(Video) Nonfiction Text Features | Jack Hartmann

313

Suggested Additional References

317

Index

319

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PREFACE

This book is for people who want to learn the fundamentals of meteorology
without taking a formal course. It can serve as a supplemental text in a classroom, tutored, or home-schooling environment. I recommend that you start at
the beginning of this book and go straight through.
There are “conversational” problems and solutions scattered throughout the

text. There is a practice quiz at the end of each chapter, and a final exam at
the end of the book. The quiz and exam questions are multiple-choice, and are
similar to the sorts of questions used in standardized tests.
The chapter-ending quizzes are “open-book.” You may (and should) refer to
the chapter texts when taking them. When you think you’re ready, take the quiz,
write down your answers, and then give your list of answers to a friend. Have
the friend tell you your score, but not which questions you got wrong. Stick with
a chapter until you get most (hopefully all) of the answers right. The correct
choices are listed in the appendix.
Take the final exam when you have finished all the chapters and chapterending quizzes. A satisfactory score is at least 75 correct answers. With the final
exam, as with the quizzes, have a friend tell you your score without letting you
know which questions you missed. Note the exam topics (if any) that give
you trouble. Review those topics in the text. Then take the exam again, and see
if you get a better score.
I recommend that you complete one chapter a week. An hour or two daily
ought to be enough time for this. That way, you’ll complete the course in a little
over two months. When you’re done with the course, you can use this book, with
its comprehensive index, as a permanent reference.
Suggestions for future editions are welcome.
STAN GIBILISCO

xi
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CHAPTER

1

Background Physics
Thousands of years ago, the alchemists believed that all things in the material universe consist of combinations of four “elements”: earth, water, air, and fire.
According to this theory, different proportions of these four “elements” give materials their unique properties. Later, physical scientists discovered that there are
dozens of elements, and even these are not the fundamental constituents of matter. Three basic states of matter are recognized by scientists today. These states,
also called phases, are known as the solid phase (the latter-day analog of earth),
the liquid phase (the analog of water), and the gaseous phase (the analog of air).
A sample of matter in one of these states is called a solid, a liquid, or a gas.

The Solid Phase
A sample of matter in the solid phase retains its shape unless it is subjected to
violent impact, placed under stress, or put in an environment with extremely
high temperature. Examples of solids at room temperature are rock, salt, wood,
and plastic.

1
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2

CHAPTER 1 Background Physics

THE ELECTRIC FORCE
What makes a solid behave as it does? After all, we’ve all been told that the
atoms of matter are mostly empty space, and that this is true even in the most
dense solids we see on this planet. So why can’t solid objects pass through one
another the way galaxies sometimes do in outer space, or the way dust clouds do
in the atmosphere?

The answers to these questions can be found when we analyze the electrical
forces in and around atoms. Every atom consists of a small, dense, positively
charged nucleus, orbited by negatively charged electrons that follow mean
(average) paths called shells. Objects with electrical charges of the same polarity (negative–negative or positive–positive) always repel each other. The closer
together two objects with the same type of charge come to each other, the more
forcefully they repel. Thus, even when an atom has an equal number of electrons
and protons so it is electrically neutral as a whole, the charges are concentrated
in different places. The positive charge is contained in the nucleus, and the negative charge surrounds the nucleus in one or more shells. These shells are usually shaped like concentric spheres.
Suppose you could shrink down to submicroscopic size and stand on the surface of a sheet of an elemental metal such as aluminum or copper. Below you, the
surface would appear like a large, flat field full of rigid spheres (Fig. 1-1). You
would find the spheres resistant to penetration by other spheres. All the spheres

Fig. 1-1. In a solid, the outer electron shells of the atoms are
tightly packed. This drawing is greatly oversimplified.

CHAPTER 1 Background Physics

would be negatively charged, so they would all repel each other. This would keep
them from passing through each other, and would also keep the surface in a stable, fixed state. The spheres would be mostly empty space inside, but there
wouldn’t be much space in between them. They would be tightly packed together.
The foregoing is an oversimplification, but it should give you an idea of the
reason why solids don’t normally pass through each other, and why many solids
resist penetration even by liquids such as water, or by gases such as air.

DENSITY OF SOLIDS
The density of a solid is measured in terms of the number of kilograms (kg) per
cubic meter (m3). That is, density is equal to mass divided by volume. The kilogram per meter cubed (kg/m3 or kg × m−3) is the measure of density in the
International System of units (SI), also known as the meter-kilogram-second
(mks) system. This is a rather awkward unit in most real-life situations. Imagine

trying to determine the density of sandstone by taking a cubical chunk of the
stuff measuring one meter (1 m) on an edge, and placing it on a laboratory scale!
You’d need a construction crane to lift the boulder, and it would smash the scale.
Because of the impracticality of measuring density directly in standard international units, the centimeter-gram-second (cgs) unit is sometimes used instead.
This is the number of grams of mass (g) per cubic centimeter (cm3) of the material in question. Technically it is called the gram per centimeter cubed (g/cm3 or
g × cm−3). To convert the density of a given sample from grams per centimeter
cubed to kilograms per meter cubed, multiply by 1000 (103). To convert the density of a sample from kilograms per meter cubed to grams per centimeter cubed,
multiply by 0.001 (10−3).
You can think of solids that are dense, such as lead. Iron is dense, too.
Aluminum is not as dense. Rocks are less dense than most metals. Glass has
about the same density as silicate rock, from which it is made. Wood, and most
plastics, are not very dense.

MEASURING SOLID VOLUME
Samples of solids rarely come in perfect blocks, cubes, or spheres, which are
shapes that lend themselves to calculation of volume by mathematical formulas.
Most samples are irregular, and defy direct dimensional measurement.
Scientists have an indirect way of measuring the volumes of irregular solid
samples: immerse them in a liquid. First, we measure the amount of liquid in a

3

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(Video) Roald Dahl | Charlie and the Chocolate Factory - Full audiobook with text (AudioEbook)

A

Liquid

B

Scale calibrated in milliliters

Liquid

Scale calibrated in milliliters

CHAPTER 1 Background Physics

Object whose
volume is to
be measured

Fig. 1-2. Measuring the volume of a solid. At A, container with
liquid but without the sample; at B, container with the
sample totally submerged in the liquid.

container (Fig. 1-2A). Then we measure the amount of liquid that is displaced
when the object is completely submerged. This shows up as an increase in the
apparent amount of liquid in the container (Fig. 1-2B). One milliliter (1 ml) of
liquid volume happens to be exactly equal to 1 cm3, and any good chemist has
a few containers that are marked off in milliliters. That’s the way to do it, then—
provided the solid does not dissolve in the liquid, none of the liquid is absorbed
into the solid, and the liquid doesn’t evaporate too fast.

SPECIFIC GRAVITY OF SOLIDS
Another important characteristic of a solid is its density relative to that of pure
liquid water at 4 degrees Celsius (4°C), which is about 39 degrees Fahrenheit
(39°F). Water attains its greatest density at this temperature, and in this condition it is assigned a relative density of 1. Liquid water at 4°C has a density of
approximately 1000 kg/m3, which is equal to 1 g/cm3. Substances with relative

density greater than 1 sink in pure water at 4°C, and substances with relative
density less than 1 float in pure water at 4°C. The relative density of a solid,
defined in this way, is called the specific gravity, abbreviated as sp gr.

5

CHAPTER 1 Background Physics

You can think of substances whose specific gravity numbers are greater than 1.
Examples include most rocks and virtually all metals. But pumice, a volcanic
rock that is filled with air pockets, floats on water. Most of the planets, their
moons, and the asteroids and meteorites in our Solar System have specific gravity greater than 1, with the exception of Saturn, which would float if a lake big
enough could be found in which to test it!
Interestingly, water ice has specific gravity less than 1 (sp gr < 1), so it floats
on liquid water. This property of ice allows fish to live underneath the frozen
surfaces of lakes in the winter in the temperate and polar regions. The surface
layer of ice acts as an insulator against the cold atmosphere. If ice had a specific
gravity larger than 1 (sp gr > 1), it would sink to the bottoms of lakes during the
winter months. This would leave the surfaces constantly exposed to temperatures below freezing, causing more and more of the water to freeze, until shallow lakes would become frozen from the surface all the way to the bottom. In
such an environment, all the fish would die during the winter, because they
wouldn’t be able to extract the oxygen they need from the solid ice, nor would
they be able to swim around in order to feed themselves. Many other aquatic
creatures would be similarly affected.
PROBLEM 1-1
A sample of solid matter has a volume of 45.3 cm3 and a mass of 0.543
kg. What is the density in grams per centimeter cubed?
SOLUTION 1-1
This problem is a little tricky, because two different systems of units
are used: SI for the volume and cgs for the mass. To get a meaningful

answer, we must be consistent with our units. The problem requires
that we express the answer in the cgs system, so let’s convert kilograms
to grams. This means we have to multiply the mass figure by 1000,
which tells us that the sample masses 543 g. Determining the density
in grams per centimeter cubed is now a simple arithmetic problem:
divide the mass by the volume. If d is density, m is mass, and V is volume, then they are related by the following formula:
d = m/V
In this case:
d = 543/45.3 = 12.0 g/cm3
This answer is rounded to three significant figures.

6

CHAPTER 1 Background Physics

PROBLEM 1-2
Calculate the density of the sample from Problem 1-1 in kilograms per
meter cubed. Do not use the conversion factor on the result of Problem
1-1. Start from scratch.
SOLUTION 1-2
This requires that we convert the volume to units in SI, that is, to
meters cubed. There are 1,000,000, or 106, centimeters cubed in a
meter cubed. Therefore, in order to convert this cgs volume to volume
in SI, we must divide by 106, the equivalent of multiplying by 10−6.
This gives us 45.3 × 10−6 m3, or 4.53 × 10−5 m3 in standard scientific
notation, as the volume of the object. Now we can divide the mass by
the volume directly:
d = m/V
= 0.543/(4.53 × 10−5)

= 0.120 × 105
= 1.20 × 104 kg/m3
This is rounded to three significant figures.

The Liquid Phase
In the liquid state or phase, matter has two properties that distinguish it from
matter in the solid phase. First, a liquid changes shape so that it conforms to the
inside boundaries of any container in which it is placed. Second, a liquid placed
in an open container (such as a jar or bucket) flows to the bottom of the container
and develops a defined, flat surface in an environment where there is constant
force caused by gravitation or acceleration.

DIFFUSION OF LIQUIDS
Imagine a jar on board a space ship in which the environment is weightless (there
is no force caused by gravitation or acceleration). Suppose that the jar is filled
with liquid, and then another liquid that does not react chemically with the first
liquid is introduced into the jar. Gradually, the two liquids blend together until the
mixture is uniform throughout the jar. This blending process is called diffusion.

CHAPTER 1 Background Physics

Some pairs of liquids undergo the diffusion process more readily than others.
Alcohol diffuses into water at room temperature in a short time. But heavy motor
oil diffuses into light motor oil slowly, and motor oil hardly diffuses into water
at all. When two liquids readily diffuse into one another, the process happens
without the need for shaking the container, because the atoms of a liquid are
always in motion, and this motion causes them to jostle each other until they
become uniformly mixed.
If the same experiment is conducted in a bucket on the surface of the earth,

where there is gravitational force, diffusion occurs, but “heavier” (more dense)
liquids tend to sink towards the bottom and “lighter” (less dense) liquids tend to
rise toward the surface. Alcohol, for example, “floats” on water. But the boundary between the alcohol and water is not sharply defined, as is the surface
between the water and the air. The motion of the atoms constantly “tries” to mix
the two liquids.

VISCOSITY OF LIQUIDS
Some liquids flow more easily than others. You know there is a difference at
room temperature between, say, water and thick molasses. If you fill a glass with
water and another glass with an equal amount of molasses and then pour the contents of both glasses into the sink, the glass containing the water will empty
much faster. The molasses is said to have higher viscosity than the water at room
temperature. On an extremely hot day, the difference is less obvious than it is on
a cold day.
Some liquids are far more viscous even than thick molasses. An example of
a liquid with extremely high viscosity is asphalt, as it is poured to make the surface of a new highway. Another example is petroleum jelly. These substances
meet the criteria as defined above to qualify as liquids, but they are thick. As the
temperature goes down, these substances become less like liquids and more like
solids. It is impossible to draw an exact line between the liquid and the solid
phases for either of these two substances.

LIQUID OR SOLID?
There is not always a specific answer to the question, “Is this substance a solid
or a liquid?” It often depends on the observer’s point of reference. Some substances can be considered solid in the short-term time sense, but liquid in the
long-term sense. An example is the mantle of the earth, the layer of rock between

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(Video) Roald Dahl | Danny the Champion of the World - Full audiobook with text (AudioEbook)

8

CHAPTER 1 Background Physics

the crust and the core. In a long-term time sense, pieces of the crust, known as
tectonic plates, float around on top of the mantle like scum on the surface of a
hot vat of liquid. This is manifested as plate tectonics, which used to be known
as “continental drift.” It is apparent when the earth’s history is evaluated over
periods of millions of years. But from one moment (as we perceive it) to the
next, and even from hour to hour or from day to day, the crust seems rigidly
fixed to the mantle. The mantle therefore behaves like a solid in short-term time
frames, but like a liquid in long-term time frames.
Imagine that we could turn ourselves into creatures whose life spans were
measured in trillions (units of 1012) of years, so that 1,000,000 years seemed to
pass like a moment. Then from our point of view, the earth’s mantle would
behave as a liquid with low viscosity, just as water seems to us in our actual state
of time awareness. If we could become creatures whose entire lives lasted only
a tiny fraction of a second, then liquid water would seem to take eons to get out
of a glass tipped on its side, and we would conclude that this substance was
solid, or perhaps a liquid with high viscosity.

DENSITY OF LIQUIDS
The density of a liquid can be defined in three ways. The mass density of a liquid is defined in terms of the number of kilograms per meter cubed (kg/m3) in a
sample of liquid. The weight density of a liquid is defined in newtons per meter
cubed (N/m3 or N × m−3), and is equal to the mass density multiplied by the
acceleration in meters per second squared (m/s2 or m × s−2) to which the sample is subjected. The particle density of a liquid is defined as the number of
moles per meter cubed (mol/m3 or mol × m−3), where 1 mole represents approximately 6.02 × 1023 atoms.
Let dm be the mass density of a liquid sample (in kilograms per meter cubed),
let dw be the weight density (in newtons per meter cubed), and let dp be the particle density (in moles per meter cubed). Let m represent the mass of the sample
(in kilograms), let V represent the volume of the sample (in meters cubed), and
let N represent the number of moles of atoms in the sample. Let a be the acceleration (in meters per second squared) to which the sample is subjected. Then
the following equations hold:

dm = m/V
dw = ma/V
dp = N/V

CHAPTER 1 Background Physics

Note the difference here between the non-italic uppercase N, which represents
newtons, and the italic uppercase N, which represents the number of moles of
atoms in a sample.
Alternative definitions for mass density, weight density, and particle density
use the liter, which is equal to a thousand centimeters cubed (1000 cm3) or onethousandth of a meter cubed (0.001 m3), as the standard unit of volume. Once in
awhile you’ll see the centimeter cubed (cm3), also known as the milliliter
because it is equal to 0.001 liter, used as the standard unit of volume.
These are simplified definitions, because they assume that the density of the
liquid is uniform throughout the sample.
PROBLEM 1-3
A sample of liquid measures 0.2750 m3. Its mass is 300.0 kg. What is
its mass density in kilograms per meter cubed?
SOLUTION 1-3
This is straightforward, because the input quantities are already given
in SI. There is no need for us to convert from grams to kilograms, from
milliliters to meters cubed, or anything like that. We can simply divide
the mass m by the volume V, as follows:
dm = m/V
= 300.0 kg/0.2750 m3
= 1090 kg/m3
We’re entitled to go to four significant figures here, because our input
numbers are both given to four significant figures.
PROBLEM 1-4

Given that the acceleration of gravity at the earth’s surface is 9.81 m/s2,
what is the weight density of the sample of liquid described in
Problem 1-4?
SOLUTION 1-4
All we need to do in this case is multiply our mass density answer by
9.81 m/s2. This gives us:
dw = 1090 kg/m3 × 9.81 m/s2
= 10,700 N/m3 = 1.07 × 104 N/m3

9

10

CHAPTER 1 Background Physics

In this case, we can go to only three significant figures, because that
is the extent of the precision with which the acceleration of gravity is
specified.

MEASURING LIQUID VOLUME
The volume of a liquid sample is usually measured using a test tube or flask
marked off in milliliters or liters. But there’s another way to measure the volume
of a liquid sample, provided we know its chemical composition and the weight
density of the substance in question. That is to weigh the sample of liquid, and then
divide the weight by the weight density. We must, of course, pay careful attention to the units. In particular, the weight must be expressed in newtons, which
is equal to the mass in kilograms times the acceleration of gravity (9.81 m/s2).
Let’s do a mathematical exercise to show how we can measure volume in this
way. Let dw be the known weight density of a huge sample of liquid, too large
for its volume to be measured using a flask or test tube. Suppose this substance

has a weight of w, expressed in newtons. If V is the volume in meters cubed, we
know from the formula for weight density that:
dw = w/V
because w = ma, where m is the mass in kilograms, and a is the acceleration of
gravity in meters per second squared. If we divide both sides of this equation
by w, we get:
dw/w = 1/V
Then we can invert both sides of this equation, and exchange the left-hand
and the right-hand sides, to obtain:
V = w/dw
This is based on the assumption that V, w, and dw are all nonzero quantities.
This is always true in the real world; all materials occupy at least some volume,
have at least some weight because of gravitation, and have nonzero density
because there is always some “stuff” in a finite amount of physical space.

PRESSURE IN LIQUIDS
Have you read, or been told, that liquid water can’t be compressed? In a simplistic sense, that is true, but this doesn’t mean liquid water never exerts pressure.

CHAPTER 1 Background Physics

Liquids can and do exert pressure, as anyone who has been in a flash flood or a
hurricane, or who has gone deep-sea diving, will tell you. You can experience
“water pressure” for yourself by diving down several feet in a swimming pool
and noting the sensation the water produces as it presses against your eardrums.
In a fluid, the pressure, which is defined in terms of force per unit area, is
directly proportional to the depth. Pressure is also directly proportional to the
weight density of the liquid. Let dw be the weight density of a liquid (in newtons per meter cubed), and s be the depth below the surface (in meters). Then
the pressure, P (in newtons per meter squared) exerted by the liquid at that
depth is given by:

P = dws
If we are given the mass density dm (in kilograms per meter cubed) rather
than the weight density, the formula becomes:
P = 9.81dms
PROBLEM 1-5
Liquid water at room temperature has a mass density of 1000 kg/m3.
How much force is exerted on the outer surface of a cube measuring
10.000 cm on an edge, submerged 1.00 m below the surface of a body
of water?
SOLUTION 1-5
First, figure out the total surface area of the cube. It measures 10.000
cm, or 0.10000 m, on an edge, so the surface area of one face is
0.10000 m × 0.10000 m = 0.010000 m2. There are six faces on a cube,
so the total surface area of the object is 0.010000 m2 × 6 = 0.060000 m2.
(Don’t be irritated by the “extra” zeroes here. They indicate that the
length of the edge of the cube has been specified to five significant figures. Also, the number 6 is an exact quantity, so it can be considered
accurate to as many significant figures as we want.)
Next, figure out the weight density of water (in newtons per meter
cubed). This is 9.81 times the mass density, or 9810 N/m3. This is best
stated as 9.81 × 103 N/m3, because we are given the acceleration of
gravity to only three significant figures, and scientific notation makes
this fact clear. (From this point on let’s stick with power-of-10 notation
so we don’t make the mistake of accidentally claiming more accuracy
than that to which we’re entitled.)
The cube is at a depth of 1.00 m, so the water pressure at that depth
is 9.81 × 103 N/m3 × 1.00 m = 9.81 × 103 N/m2. The force F (in new-

11

12

CHAPTER 1 Background Physics

tons) on the cube is therefore equal to this number multiplied by the
surface area of the cube:
F = 9.81 × 103 N/m2 × 6.00000 × 10−2 m2
= 58.9 × 101 N = 589 N

PASCAL’S LAW FOR INCOMPRESSIBLE LIQUIDS
Imagine a watertight, rigid container. Suppose there are two pipes of unequal
diameters running upwards out of this container. Imagine that you fill the container with an incompressible liquid such as water, so the container is completely
full and the water rises partway up into the pipes. Suppose you place pistons in
the pipes so they make perfect water seals, and then you leave the pistons to rest
on the water surface (Fig. 1-3).

Piston
force =
F2
(upward)

Piston
force =
F1
(downward)

Piston
area =

Piston

area =
A1

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A2
Surface levels
are equal

Incompressible liquid

Fig. 1-3. Pascal’s law for confined, incompressible liquids. The forces
are directly proportional to the surface areas where the pistons
contact the liquid.

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