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METEOROLOGY DEMYSTIFIED

Demystified Series

Advanced Statistics Demystified

Algebra Demystified

Anatomy Demystified

Astronomy Demystified

Biology Demystified

Business Statistics Demystified

C++ Demystified

Calculus Demystified

Chemistry Demystified

College Algebra Demystified

Data Structures Demystified

Databases Demystified

Differential Equations Demystified

Digital Electronics Demystified

Earth Science Demystified

Electricity Demystified

Electronics Demystified

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Everyday Math Demystified

Geometry Demystified

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Java Demystified

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Microbiology Demystified

OOP Demystified

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Physics Demystified

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Pre-Algebra Demystified

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Project Management Demystified

Quantum Mechanics Demystified

Relativity Demystified

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Trigonometry Demystified

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

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

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

Copyright © 2006 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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

4

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

7

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

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.