Xand the city: modeling aspects of urban life



and theCity


 Xand theCity


John A. Adam

Copyright © 2012 by Princeton University PressPublished by Princeton University Press, 41 William Street, Princeton,New Jersey 08540In the United Kingdom: Princeton University Press, 6 Oxford Street, Woodstock,Oxfordshire OX20 1TW


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Library of Congress Cataloging-in-Publication DataAdam, John A.X and the city : modeling aspects of urban life / John Adam.p. cm.Includes bibliographical references and index.ISBN 978-0-691-15464-0 1. Mathematical models. 2. City and town life—Mathematical models. 3. Cities and towns—Mathematical models. I. Title.HT151 .A288 2012307.7601'5118—dc23       2012006113

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For Matthew, who drifted “continentally” to a large city


He found out that the city was as wide as it was long and it was as high as it was wide. It was as long as a man could walk in fifty days . . . In the middle of the street of the city and on either bank of the river grew the tree of life, bearing twelve fruits, a different kind for each month. The leaves of the tree were for the healing of the nations.

—St. John of Patmos

(SeeChapters 1and5for some estimation questions inspired by these passages.)




Chapter 1INTRODUCTION:Cancer, Princess Dido, and the city





































Annotated references and notesIndex


After the publication ofA Mathematical Nature Walk, my editor, Vickie Kearn, suggested I think about writingA Mathematical City Walk. My first reaction was somewhat negative, as I am a “country boy” at heart, and have always been more interested in modeling natural patterns in the world around us than man-made ones. Nevertheless, the idea grew on me, especially since I realized that many of my favorite nature topics, such as rainbows and ice crystal halos, can have (under the right circumstances) very different manifestations in the city. Why would this be? Without wishing to give the game away too early into the book, it has to do with the differences between nearly parallel “rays” of light from the sun, and divergent rays of light from nearby light sources at night, of which more anon. But I didn’t want to describe this and the rest of the material in terms of a citywalk; instead I chose to couch things with an “in the city” motif, and this allowed me to touch on a rather wide variety of topics that would have otherwise been excluded. (There aresevenchapters having to do with traffic in one way or another!)

As a student, I lived in a large city—London—and enjoyed it well enough, though we should try to identify what is meant by the word “city.” Several related dictionary definitions can be found, but they vary depending on the country in which one lives. For the purposes of this book, a city is a large, permanent settlement of people, with the infrastructure that is necessary tomake that possible. Of course, the terms “large” and “permanent” are relative, and therefore we may reasonably include towns as well as cities and add the phrase “or developing” to “permanent” in the above definition. In the Introduction we will endeavor to expand somewhat on this definition from a historical perspective.

This book is an eclectic collection of topics ranging across city-related material, from day-to-day living in a city, traveling in a city by rail, bus, and car (the latter two with their concomitant traffic flow problems), population growth in cities, pollution and its consequences, to unusual night time optical effects in the presence of artificial sources of light, among many other topics. Our cities may be on the coast or in the heartland of the country, or on another continent, but presumably always located on planet Earth. Inevitably, some of the topics are multivalued; not everything discussed here is unique to the city—after all, people eat, garden, and travel in the country as well!

WhyX and the City? In the popular culture, the letterX(orx) is an archetype of mathematical problem solving: “Findx.” TheXin the book title is used to introduce the topic in each subsection; thus “X = tc” and “X=Ntot” refer, respectively, to a specific length of time and a total population, thereby succinctly introducing the mathematical topics that follow. One of the joys of studying and applying mathematics (and findingx), regardless of level, is the fact that the deeper one goes into a topic, the more avenues one finds to go down. I have found this to be no less the case in researching and writing this book. There were many twists and turns along the way, and naturally I made choices of topics to include and exclude. Another author would in all certainty have made different choices. Ten years ago (or ten years from now), the same would probably be true for me, and there would be other city-related applications of mathematics in this book.

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Mathematics is a language, and an exceedingly beautiful one, and the applications of that language are vast and extensive. However, pure mathematics and applied mathematics are very different in both structure and purpose, and this is even more true when it comes to that subset of applied mathematics known asmathematical modeling(of which more below). I love the beauty and elegance in mathematics, but it is notalwayspossible to find it outside the “pure” realm. It should be emphasized that the subjects are complementary and certainly not in opposition, despite some who might hold that opinion. I heard of one mathematician who referred to applied mathematics as “mereengineering”; this should be contrasted with the view of the late Sir James Lighthill, one of the foremost British applied mathematicians of the twentieth century. He wrote, somewhat tongue-in-cheek, that pure mathematics was a very important part of applied mathematics!

Applied mathematics is often elegant, to be sure, and when done well it is invariably useful. I hope that the types of problem considered in this book can be both fun and “applied.” And while some of the chapters in the middle of the book might be described as “traffic engineering,” it is the case that mathematics is the basis for all types of quantitative thought, whether theoretical or applied. For those who prefer a more rigorous approach, I have also includedChapter 17, entitled “The axiomatic city.” In that chapter, some of the exercises require proofs of certain statements, though I have intentionally avoided referring to the latter as “theorems.”

The subtitle of this book isModeling Aspects of Urban Life. It is therefore reasonable to ask: whatis(mathematical) modeling? Fundamentally, mathematical modeling is the formulation in mathematical terms of the assumptions (and their logical consequences) believed to underlie a particular “real world” problem. The aim is the practical application of mathematics to help unravel the underlying mechanisms involved in, for example, industrial, economic, physical, and biological or other systems and processes. The fundamental steps necessary in developing a mathematical model are threefold: (i) to formulate the problem in mathematical terms (using whatever appropriate simplifying assumptions may be necessary); (ii) to solve the problem thus posed, or at least extract sufficient information from it; and finally (iii) to interpret the solution in the context of the original problem. This may include validation of the model by testing both its consistency with known data and its predictive capability.

At its heart, then, this book is about just that: mathematical modeling, from “applied” arithmetic to linear (and occasionally nonlinear) ordinary differential equations. As a little more of a challenge, there are a few partial differential equations thrown in for good measure. Nevertheless, the vast majority of the material is accessible to anyone with a background up to and including basic calculus. I hope that the reader will enjoy the interplay between estimation, discrete and continuum modeling, probability, Newtonian mechanics, mathematical physics (diffusion, scattering of light), geometric optics, projective and three-dimensional geometry, and quite a bit more.

Many of the topics in the book are posed in the form of questions. I have tried to make it as self-contained as possible, and this is the reason there are several Appendices. They comprise a compendium of unusual results perhaps (in some cases) difficult to find elsewhere. Some amplify or extend material discussed in the main body of the book; others are indirectly related, but nevertheless connected to the underlying theme. There are also exercises scattered throughout; they are for the interested reader to flex his or her calculus muscles by verifying or extending results stated in the text. The combination of so many topics provides many opportunities for mathematical modeling at different levels of complexity and sophistication. Sometimes several complementary levels of description are possible when developing a mathematical model; in particular this is readily illustrated by the different types of traffic flow model presented inChapters 8through13.

In writing this book I have studied many articles both online and in the literature. Notes identifying the authors of these articles, denoted by numbers in square brackets in the text, can be found in the references. A more general set of useful citations is also provided.


Thanks to the following for permissions:

Achim Christopher (Figures 23.1and23.10)

Christian Fenn (Figure 22.5)

Skip Moen (Figures 3.4and21.1)

Martin Lowson and Jan Mattsson, for email conversations about their work cited here.

Larry Weinstein for valuable feedback on parts of the manuscript.

Alexander Haußmann for very helpful comments onChapter 22.

Bonita Williams-Chambers for help withFigure 10.2andTable 15.1.

My thanks go to Kathleen Cioffi, who oversaw the whole process in an efficient and timely manner. I am most appreciative of the excellent work done by the artist, Shane Kelley, who took the less-than-clear figure files I submitted and made silk purses out of sows’ ears! Many thanks also to the book designer, Marcella Engels Roberts, for finding the illustrations for the chapter openers and designing the book. Any remaining errors of labeling (or of any other type, for that matter) are of course my own.

I thank my department Chair, J. Mark Dorrepaal, for arranging my teaching schedule so that this book could be written in a timely fashion (and my graduate students could still be advised!).

As always, I would like to express my gratitude to my editor, Vickie Kearn. Her unhurried yet efficient style of “author management” s(m)oothes ruffled feathers and encourages the temporarily crestfallen writer. She has great insight into what I try to write, and how to do it better, and her advice is always invaluable. And I hope she enjoys the story about my grandfather!

Finally, I want to thank my family for their constant support and encouragement, and without whom this book might have been finished a lot earlier. But it wouldn’t have been nearly as much fun to write!


and theCity


Cancer, Princess Dido, and the city



To look at the cross-section of any plan of a big city is to look at something like the section of a fibrous tumor.

—Frank Lloyd Wright


Although this question was briefly addressed in the Preface, it should be noted that the answer really depends on whom you ask and when you asked the question. Perhaps ten or twelve thousand years ago, when human society changed from a nomadic to a more settled, agriculturally based form, cities started to develop, centered on the Euphrates and Tigris Rivers in ancient Babylon. It can be argued that two hundred years ago, or even less, “planned” cities were constructed with predominately aesthetic reasons—architecture—in mind.

Perhaps it was believed that form precedes (and determines) function; nevertheless, in the twentieth century more and more emphasis was placed on economic structure and organizational efficiency. A precursor to these ideas was published in 1889 as a book entitledCity Planning, According to Artistic Principles, written by Camillo Sitte (it has since been reprinted). A further example of this approach from a historical and geographical perspective, much nearer our own time, is Helen Rosenau’sThe Ideal City: Its Architectural Evolution in Europe(1983). But there is a distinction to be made between those which grow “naturally” (or organically) and those which are “artificial” (or planned). These are not mutually exclusive categories in practice, of course, and many cities and towns have features of both. Nevertheless there are significant differences in the way such cities grow and develop: differences in rates of growth and scale. Naturally growing cities have a slower rate of development than planned cities, and tend to be composed of smaller-scale units as opposed to the larger scale envisioned by city planners.

“Organic” towns, in plan form, resemble cell growth, spider webs, and tree-like forms, depending on the landscape, main transportation routes, and centers of activity. Their geometry tends to be irregular, in contrast to the straight “Roman road” and Cartesian block structure and circular arcs incorporated in so many planned cities, from Babylonian times to the present [1]. Some of the material in this book utilizes these simple geometric ideas, and as such, represents only the simplest of city models, by way of analogies and even metaphors.


Was Frank Lloyd Wright correct—do city plans often look like tumor cross sections writ large? Perhaps so, but the purpose of that quote was to inform the reader of a common feature in modeling. Mathematical models usually (if not always) approach the topic of interest using idealizations, but also sometimes using analogies and metaphors. The models discussed in this book are no exception. Although cities and the transportation networks within them (e.g., rail lines, roads, bus routes) are rarely laid out in a precise geometric grid-like fashion, such models can be valuable. The directions in which a city expands are determined to a great extent by the surrounding topography—rivers, mountains, cliffs, and coastlines are typically hindrances to urban growth. Cities are not circular, with radially symmetric population distributions, but evensuch gross idealizations have merit. The use of analogies in the mathematical sciences is well established [2], though by definition they have their limitations. Examples include Rutherford’s analogy between the hydrogen atom and the solar system, blood flow in an artery being likened to the flow of water in a pipe, and the related (and often criticized) hydraulic analogy to illustrate Ohm’s law in an electric circuit.

Analogy is often used to help provide insight by comparing an unknown subject to one that is more familiar. It can also show a relationship between pairs of things, and can help us to think intuitively about a problem. The opening quote by Frank Lloyd Wright is such an example (though it could be argued that it is more of a simile than an analogy). One possibly disturbing analogy is that put forward by W.M. Hern in the anthropological literature [3], suggesting that urban growth resembles that of malignant neoplasms. A neoplasm is an abnormal mass of tissue, and in particular can be identified with a malignant tumor (though this need not be the case). To quote from the abstract of the article,

Malignant neoplasms have at least four major characteristics: rapid, uncontrolled growth; invasion and destruction of adjacent normal tissues (ecosystems); metastasis (distant colonization); and de-differentiation. Many urban forms are almost identical in general appearance, a characteristic that would qualify as “de-differentiation.” Large urban settlements display “rapid, uncontrolled growth” expanding in population and area occupied at rates of from 5 to 13% per year.


The term “de-differentiation” means the regression of a specialized cell or tissue to a simpler unspecialized form. Thereisan interesting mathematical link that connects such malignancies with city growth—thefractal dimension. This topic will be mentioned inChapter 18, and more details will be found inAppendix 9. For now, a few aspects of this analogy will be noted. The degree of border irregularity of a malignant melanoma, for example, is generally much higher than that for a benign lesion (and it carries over to the cellular level also). This is an important clinical feature in the diagnosis of such lesions, and it is perhaps not surprising that city “boundaries” and skylines are also highly irregular (in the sense that their fractal dimension is between one and two). This of course is not to suggest that the city is a “cancer” (though some might disagree), but it does often possess the four characteristics mentioned abovefor malignant neoplasms. The question to be answered is whether this analogy is useful, and in what sense. We shall not return to this question, interesting though it is, instead we will end this chapter by examining a decidedly nonfractal city boundary!

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In Greek mythology, Dido was a Phoenician princess, sister of Pygmalion, King of Tyre, and founder of the city of Carthage in northern Africa in 814 B.C. According to tradition, she did this in a rather unusual way. Pygmalion had her husband, Sychaeus, killed, and Dido fled to the northern coast of Africa. According to some, her brother agreed to let her have as much land as she could enclose within the hide of a bull; according to others, she bartered with the locals to accomplish this. She then cut the hide into a series of thin strips, joined them together, and formed a semicircular arc, with the Mediterranean Sea effectively as a diameter, thus enclosing a semicircular area [4]. Bravo! As we shall note inAppendix 1, given a straight boundary (as assumed in this story), a semicircle will contain the maximum area for a given perimeter, so Dido achieved the best possible result!

I grew up the son of a farm-laborer and had the occasional less-than-pleasant encounter with a certain bull; although he was British he was certainly not a gentleman. Here’s how the mathematics might have gone. Imagine the typical bull torso to be a rectangular box 5 ft long by 4 ft high by 2 ft wide. (Yes, that’s correct, we’ve cut off his legs, head, and tail, but only in our imagination.) The surface area of our bull-box is readily shown to be about 80 sq ft. We’ll round this up to 100 sq ft since we’re only interested in a rough estimate. And anyway, given how shrewd Dido appeared to be, no doubt she would have picked the biggest bull she could find! I don’t know what kind of precision Dido or her servants might have had with the cutting tools available but I’m going to assume that strips could be cut as narrow as one hundredth of a foot (0.12 in, or about 3 mm). This may be an underestimate, but it makes the arithmetic easier without changing the “guesstimate” by much. If the total length of the strips isLft, then we have a simple equation for the area: 0.01L≈ 100, orL≈ 104ft. This length would comprise the semicircular part of the city boundary. For a radiusrft,


whereAis the area of the “city.” Hence


This is about 1.6 km2. That’s pretty impressive for a load of bull! If the boundary were circular instead of semicircular, the corresponding area would be half this amount, or about one third of a square mile. InAppendix 2we shall generalize this idea to the case in which the boundary is of variable lengthl

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and the corresponding acceleration component is.


Sinceayis an increasing function ofx, its extreme values are ±6U2h/L2, occurring at the endpoints of the descent path (the minimum at the start and the maximum at the finish of the descent). Therefore from condition (iii) above,


Figure 2.1. The flight pathY(d).



This result can provide useful information depending on the type of flight and runway size. For transcontinental or transatlantic flights aboard a 747 jet (let’s suppose the latter), we rewrite the above inequality as


Typically, bothUandhare “large” andkis “small,” so this implies thatLmust be relatively large (compared with smaller airplanes). Suppose thatU= 600 mph,h= 40,000 ft, andL= 150 miles. Then a further rearrangement of (2.4) gives a lower bound forkof approximately (in ft/s2)


If, on the other hand, we find ourselves on a “puddle jumper” flight, then


Therefore we expectLandkto be smaller than in the example, whilehis still relatively large, so not surprisingly the airspeed will be correspondingly lower. Nevertheless, there are exceptions. Barshinger (1992) [5] relates his experience landing at Lake Tahoe. As the plane crossed the last peak of the 11000 ft Sierra Nevada Mountains, the airport was only twenty miles away. For these values ofhandLand an airspeed of 175 mph,k≈ 0.39 ft/s2, 30% larger than on the transatlantic flight!

Question:Can a similar analysis be carried out when equation (2.1) is replaced byy=A+BtanhC(x−D) (whereA,B,C, andDare constants)?


Well, I landed safely, and after checking in to my hotel I decided to take a walk, map in hand. Being directionally challenged at the best of times, even with a map, pretty soon I was lost, and it didn’t seem to help much, because every place I wanted to go seemed to be right at the edge of the map, or on the well-worn fold! Getting irritated, I had a question to ask of no one in particular . . .


Why is the place I’m looking for on a mapso oftenat its boundary? Let’s investigate. Consider first a map consisting of a rectangular sheet of relative dimensions 1 andb< 1, including a margin of widtha

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Alternatively,what is the probability of being killed in a car accident?The average life span of people in the U.S. is about 75 years, neglecting gender differences, so 1 in 75 Americans dies every year. Hence the average number of deaths per year is 3 × 108÷ 75, that is, 4 × 106deaths/yr, so the total (lifetime) probability of dying in a car crash is 4 × 104deaths/yr ÷ 4 × 106deaths/yr = 0.01, or 1%. It is therefore pretty safe to say that if you are reading this book, and are not a babe-in-arms, then your “probability” is rather less than this!

Question:What is the risk of dying per mile while flying? [7]


Most of us travel by air once a year (2 flights) on vacation or business, and a small fraction of the population travel much more than that. We’ll use an average of 3 flights/person, so that’s 109people-flights per year; the actual figure in 2005 was 6.6 × 108, so we’ll use the slightly more accurate value of 7 × 108flights/yr. The average (intracontinental) flight distance probably exceeds 300 miles (or it would be simpler to drive) but is less than 3000 miles; the geometric mean is 103miles, so we travel about 7 × 1011mi/yr by air.

The crash frequency of large planes is (fortunately) less than one per year and (unfortunately) more than one per decade, so the geometric mean givesabout one third per year. Typically, about 100 to 200 people die in each (large) crash; we’ll take about 50 deaths/yr, so the per-mile probability of dying is, on the basis of these crude estimates, about 50 ÷ 7 × 1011≈ 7 × 10−11deaths/mile, some 300 times safer than driving! Of course, this is overly simplistic since most crashes occur at take-off or landing, but it does remind us how safe air travel really is.

While on this subject, I recall hearing of a conversation about the risks of flying that went something like this:

“You know, the chances of dying on a flight are really very small. You’re much more likely to die on the roads! And anyway, when it’s your time to go, it’s your time to go!’

“But what if it’s the pilot’s time to go and not mine?”


1This, by the way, is using theGoldilocks principle—is it too large, too small, or just right? The geometric mean helps us find “just right,” often to within a factor of two or three, which is very useful when dealing with several orders of magnitude. Under these circumstances it is far more valuable than the arithmetic mean, though in this particular example the two give essentially the same answer. We shall encounter the need for this useful principle time and again in this book.






Suppose the city population isNpmillion, and we wish to estimateN, the number of facilities, (dental offices, gas stations, restaurants, movie theaters, places of worship, etc.) in a city of that size. Furthermore, suppose that the average “rate per person” (visits per year, or per week, depending on context) isR, and that the facility is open on averageHhours/day and caters for an average ofCcustomers per hour. We shall also suppose there areDdays per year. This may seem a little surprising at first sight: surelyeveryoneknows thatD= 365 when the year is not a leap year! But since we are only “guesstimating” here, it is convenient to takeDto be 300. Note that 400 would work just as well—remember that we are not concerned here about being a “mere” factor of two or three out in our estimate.

Then the following ultimately dimensionless expression forms the basis for our specific calculations:


For simplicity, let’s consider a city population of one million. How many dental offices might there be? Most people who visit the dentist regularly do so twice a year, some visit irregularly or not at all, so we shall takeR= 1,C= 5,H= 8 andD= 300 (most such offices are not open at weekends) to obtain


That is, to the nearest order of magnitude, about a hundred offices. A similar estimate would apply to doctors’ offices.

Let’s now do this for restaurants and fast-food establishments. Many people eat out every working day, some only once per week, and of course, some not at all. We’ll useR= 2 per week (100 per year), but feel free to replace my numbers with yours at any time. The size of the establishment will vary, naturally, and a nice leisurely dinner will take longer than a lunchtime hamburger at a local “McWhatsit’s” fast-food chain, so I’ll pickC= 50, allowing for the fast turn-around time at the latter. Hours of operation vary from pretty much all day and night to perhaps just a few hours in the evening; I’ll set an average ofH= 10. Combining everything as before, withDequal to 400 now (such establishments are definitely open on weekends!), we find for the same size city


Therefore the most we can say is that there are probably several hundred places to eat out in this city! I’m getting hungry . . .

We can do this for the number of gas stations, movie theaters, and any other facility you wish to estimate.

Exercise:Make up your own examples. Do your answers make sense?


Our final example will be to estimate the number of houses of worship in the city. Although many, if not most, have midweek meetings in addition tothe main one at the end of the week or at some time during the weekend, I shall use the figure forRof 50 per year (or once per week) as above in the “eating out” problem. Spiritual food for those that seek it! But now I shall include the proportion of people who attend houses of worship in the calculation because clearly not everyone does. In the U.S. this is probably a higher proportion than in Europe, for example, so I shall suggest that one in five attend once per week in the U.S. The estimates forCandHare irrelevant (and meaningless) in this context, since everyone who attends regularly knows when the services start! Furthermore, this is more of a “discrete” problem since the vast majority of those who attend a house of worship do so once a week, so we shall simplify the formula by estimating an average attendance for the service. From my own experience, some churches have a very small attendance and some are “mega-churches,” and I will assume that the range is similar for other faith traditions also. Using the Goldilocks principle—too small, too large, or just right?—I shall take the geometric mean of small attendance (10) and large attendance (1000), that is, 100. Hence the approximate number of houses of worship in a city of one million people is


Question:What is the optimal distance from which to view a painting/sculpture/display?


At the outset, it should be pointed out that we are referring to an object for which the lowest point is above the eye level of the observer: a painting high on a wall, a sculpture or statue on a plinth, and so forth. Commonsense indicates that unless this is the case, the angle subtended by the statue (say) at the observer’s eye will increase as the statue is approached. Of course, there is still an optimal viewing distance—wherever the observer feels most comfortable standing or sitting—but this is subjective. What this question means is where is the maximum angle subtended when the base or bottom of the painting is above the ground? That a maximum must occur is again obvious—far away that angle is very small, and close up it is also small, so a maximum must occur somewhere between those positions.

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Figure 3.1. Display geometry for the maximum viewing angle. The object (e.g., a painting) lies along the vertical segmentAB.


This is posed as a standard optimization problem in many calculus books. Consider the object of interest to be of vertical extenth(fromAtoB) with its base a distanceYabove the observer’s eye line (thex-axis, with the observer at the originO). FromFigure 3.1it is clear that the angleAÔB=θ=α−βis to be maximized. IfAhas coordinates (x,Y) in general, it follows that


Therefore an extremum occurs when

Y[x2+ (Y+h)2] = (Y+h)(x2+Y2), i.e., whenx= (Y2+Yh)1/2≡X.


We know this corresponds to a maximum angle, so we leave it to the reader to verify thatθ″(X) < 0. For the Statue of Liberty, the plinth heightYis 47 meters, and the Lady herself is almost as tall:h= 46 meters. Therefore if the ground is flat we need to stand at a distance ofX= (46 × 47 + 472)1/2= 66 meters.

It is interesting to verify that the pointsO, A, andBlie on a circle: if this is the case, the center of the circle has coordinates (0,Y+h/2) by symmetry. Then if the radius of the circle isrit follows that


It is readily verified that the pointsA(X,Y),B(X,Y+h), andO(0, 0) all satisfy this equation, so the eye level of the observer is tangent to the circle. And another interesting fact is that this maximum angle is attained from any other point on the circle (by a theorem in geometry). Of course, this is only useful to the observer if she can levitate!

This result for the “vertical” circle (Figure 3.1) has implications for a “horizontal” circle in connection with the game of rugby—the best position to place the ball for conversion after a try is on the tangent point of the circle (see the book by Eastaway and Wyndham [8] for more details; they also mention the optimal viewing of two other statues—Christ the Redeemer in Rio de Janeiro and Nelson’s Column in London).


We have tickets to the symphony. It is well known that the “acoustics” in some concert halls are better than in others—a matter of design. A fundamental question in this regard is—how long does it take for a musical (or other) sound to die out? But this is a rather imprecise question. As a rough rule of thumb, for the “average” (and hence nonexistent) person, weak sound just audible in a quiet room is about a millionth as intense as normal speech or music, so we’ll define the reverberation timeTin seconds as the time required for a sound to reach 10−6of its original intensity (Vergara 1959). Clearly, the value ofTdepends on many things, particularly the dimensions of the room and how well sound is absorbed by the wall, floor, ceiling, and the number of people present. Suppose that as the sound is scattered (here meaning reflected and absorbed) and the average distance between reflecting surfaces isLfeet. On encountering such a surface suppose that (on average) a proportionaof the intensity is lost. We will callatheabsorptivitycoefficient. If the original sound intensity isI0, then the intensity afternreflections is


We’ll assume this to be an exact expression from now on. Then we naturally ask: when doesIn= 10−6I0? In other words, if (1 −a)n= 10−6, what isn(to the nearest integer)? It is convenient to work with common logarithms here, so


(to the nearest integer). Ifc(ft/s) is the speed of sound in air at room temperature and pressure, then on average over a distanceL, sound will be reflected approximatelyc/Ltimes per second. This enables us to write the reverberation timeTas


Tis a linear function ofLas we would expect, but before putting in some “typical” numbers, let’s see how the quantityN=n/6 =Tc/6L= −[log10(1 −a)]−1behaves as a function of the absorptivitya.

It can be seen fromFigure 3.2and equation (3.6) that bothnandTfall off rapidly for absorptivities up to about 20%. In fact, if we usec≈ 1120 fps andL≈ 20 ft, thenn≈ 270 and therefore the number of reflections per second isc/L≈ 56 s−1. This means thatT≈ 5 s, which is an awfully long time—almost long enough for the audience to leave the symphony hall in disgust! We need to do better than this. This is why soundproof rooms have walls that look like they are made of egg cartons: the value ofaneeds to be much higher than 0.05.


Figure 3.2. The reflection functionn(a).



Figure 3.3. Absorptivity as a function of threshold reverberation timeTc.


However, we can invert the problem: suppose that the musicologists and sound engineers have determined thatTshould be no larger thanTcseconds. With the same values ofc, andL, we now require thatn= 56Tc, so that

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ForTc= 1 s,a≈ 0.22; forTc= 0.5 s,a≈ 0.39, andTc= 0.1 s,a≈ 0.92. That’s more like it! A graph illustrating the dependence ofaonTcis shown inFigure 3.3. Again, note the rapid fall-off whenTc< 1 s.

In reality, the physics of concert halls must be very much more complicated than this, but you didn’t really expect that level of sophistication in this book, did you?


Figures 3.4and3.5show, respectively, rebuilding at Ground Zero in New York City, and an office tower at Delft University of Technology in The Netherlands. In both cases we see tall buildings, very much taller than most kinds of trees. But trees sway in the wind, right? So why shouldn’t the same be true for tall buildings? In fact, itistrue. The amplitude of “sway” near the top of the John Hancock building in Chicago can be about two feet. At the time of writing, the world’s tallest building has opened in Dubai. The Burj Khalifa stands 828 meters (2,716 ft), with more than 160 stories. It may not be long before even this building is eclipsed by yet taller ones. How much might such a building sway in the wind? Structural engineers have a rough and ready rule: divide the height by 500; this means the “sway amplitude” for the Burj Khalifa is about 5.5 feet! The reason for such swaying is intimately associated with the wind, of course; wind flows around buildings and bridges in a similar fashion to the way water flows around obstacles in a stream. Careful observation of this dynamic process reveals that small vortices or eddies swirl near the obstacle, be it a rock, twig, or half-submerged calculus book. The atmosphere is a fluid, like the stream (though unlike water, it is compressible) and buildings are the obstacles. If wind vortices break off the building in an organized, rhythmic fashion, the building will sway back and forth. Or, to put it another way, a skyscraper (or a smokestack) can behave like a giant tuning fork!


Figure 3.4. Rebuilding at the site of Ground Zero, New York City. Photo by Skip Moen.



Figure 3.5. Tower at the Delft University of Technology, Delft, The Netherlands. Photo by the author.


The oscillations arise from the alternate shedding of vortices from opposite sides of the tall structure. Their frequency depends on the wind speed and the size of the building. Not surprisingly, such “flow-induced” oscillations can be very dangerous, and engineers seek to design structures to minimize them [9].Figure 3.6illustrates in schematic form how such vortices can develop around a cylindrical body and alternatively “peel off” downstream.

There are two dimensionless numbers that are very important in a study of flow patterns around obstacles. One is called theReynolds number, denoted byR, and is defined by


Figure 3.6. Schematic view of vortex formation around a cylindrical obstacle.



whereLandUare characteristic size and (here) wind speed, respectively, andνis a constant called the kinematic viscosity. For small values ofR(R< 1) there is no separation: the cylinder just causes a symmetrical “bump” in flow. ForR> 1 things get more complicated; in the range 1 40 this symmetric flow becomes unstable, and vortices are shed alternately from side to side (as viewed from above). This pattern can exist for Reynolds number up to several thousand (even up to 105if the obstacle is very “smooth”).

The other important number is theStrouhalnumber,S. It is named afterVincenc Strouhal, a Czech physicist who in 1878 investigated aeolian tones—the “singing” of wires set into oscillation by the wind. Again, in terms of the wind speedUand the diameterdof the wire,Sis defined by


fsbeing the frequency of the vortex shedding. Naturally it is to be expected that the Strouhal number is related to the Reynolds number, but for a wide range of the latter,Sis almost constant, varying at most between 0.15 and 0.2. Whenfsis close to the natural frequency of vibration of the obstacle, the latter can “capture” the former, and the resulting resonance can be very dangerous, for obvious reasons. This phenomenon is called “lock-in.”

And speaking of swaying buildings, a few minutes before I wrote this, my office building started to sway. I’m on the second floor, and didn’t feelanything, but I heard the bookcases behind me move back and forth. They have a lot of books in them, and are very heavy. Subsequently I heard tales from across campus of swinging lights and moving floors—yes, Virginia, we just had an earthquake! Initial reports indicated it was 5.8 on the Richter scale, with an epicenter midway between Charlottesville and and Richmond, Virginia (and about 80 miles from Washington, D.C.). Tremors were felt as far as New York, Massachusetts, Ohio, Tennessee. and the Carolinas, and the U.S. Capitol and the Pentagon were evacuated. And as I write, Hurricane Irene, currently a Category 3 storm, is making her way steadily toward the Eastern seaboard! Thankfully, by the time it made landfall the storm had weakened to Category 1, but it still caused one death and extensive property damage in the Outer Banks, North Carolina, New Jersey, New York, and Vermont.

X=d(x,y): THE MALL 

Note that in the UK, mall = shopping center = shopping centre!

More and more frequently, malls are being located out of towns and cities, perhaps as a single “megastore” or as a mall-like complex; however, many are still built in cities with pedestrian walkways. What follows is a very simplistic model for the competitiveness of two such malls, carried out by considering how one “stacks up” against the other, as measured by the number of trips (N) made per unit interval to a given location.

An important (but rather subjective) question for urban planners is “How attractive is the mall likely to be?” Factors such as the variety of stores within it, ease of access, and parking facilities all contribute to the answer. Essentially, the attractiveness of the mall determines whether one will prefer to travel farther to get there, as opposed to shopping at a nearer but less attractive one. Another question, fundamental for developing a mathematical model for the competition between malls (or specialty stores and shops, for that matter) is “How doesNdepend on the distancedfrom the mall?” [10]. Many factors, including those mentioned above, must be contributory, and it is therefore unlikely thatNwill be a simple function ofd. Nevertheless, that is beyond the scope of this book, and we shall content ourselves with a simpler approach to illustrate some basic principles involved.

Obviously we expectNto decrease with distanced, but how rapidly? Another question concerns what we mean by distance here, that is, should we usethe standard Euclidean “metric” in the plane, or a modified version, weighted in some manner to account for geographical or social factors? In all likelihood the latter, but again for simplicity we will stick with circular symmetry, consistent with models that appear later in the book. (However, seeAppendix 3for a brief introduction to the so-calledtaxicaborManhattanmetric.) To that end, we define a constant “attraction factor”ai> 0 for each mall (herei= 1, 2) and write

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where the distanced(x,y) from a particular mall is calculated using the standard Pythagorean distance formula. The two mallsAandBwill be located at the coordinate origin and at (b, 0), respectively. We wish to determine the locus of points in the (x,y)-plane such thatN1=N2. This will be the (closed) boundary curve inside and outside of which one mall is preferred over the other. Therefore we can write


Rearranging this equation by raising both sides to the power 2/pwe obtain


where the positive parameterk= (a1/a2)2/p. In so doing we have effectively transferred the dependence on the distance (via the parameterp) to a measure of the “relative attractiveness” of the malls. For given values ofai, note thatkincreases aspdecreases toward zero ifa1/a2> 1, andkdecreases aspdecreases toward zero ifa1/a2> 1, andkdecreases aspdecreases toward zero ifa1/a2< 1. Upon completing the square it follows that fork≠ 1


that is, the “boundary of attraction” is a circle of radiuscentered at the point. If 0 kso that the circle extends into the half-plane and the point closest to mallBisat a distancefrom it. Ifk> 1, mallAis the more attractive of the two, and the closest point on the boundary to the pointAis also a distancefrom it.Figure 3.7illustrates this case for arbitraryk> 1; the dotted contours represent circular arcs on which (i)N1is constant (outside the solid circular boundary) and (ii)N2is constant (inside the boundary), respectively.


Figure 3.7. Equidistance contours and “attraction” boundary for a circular city.


One obvious and simple change to the model would be to consider elliptical contours of constantNi; thus for the equation of the boundary curve we might use


instead of equation (3.9). In this case, ifa> 1 the loci of constantNiwill be ellipses with semi-major axes along thex-axis, and the resulting boundary curve is also an ellipse, not surprisingly, with the equation

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A representative diagram is shown inFigure 3.8.


I know that it’s usually a bad idea to go to the post office to mail letters when it first opens, especially on a Monday morning. And around the middle of December it’s a nightmare! But sometimes there is little or no line at all, even when the office is normally busy. That has happened on my last two visits. And conversely, sometimes there is a long line when I least expect it. While the average number of customers may be one every couple of minutes, it is most unlikely that each one will arrive every two minutes: there will be a certain “clumpiness” in the arrivals. ThePoisson distributiondescribes this clumpiness well. Details and applications of this distribution are discussed inChapter 9and inAppendix 4, but we will summarize the results here. If the arrivals at the post office, checkout line, traffic line, busstops, and so on are random and average toλper minute (or other unit of time), then the probabilityP(n) ofncustomers arriving in any given minute is


Figure 3.8. Equidistance contours and “attraction” boundary for an elliptical city.



Suppose that on a reasonably busy day at the post office,λ= 2. What is the chance that there will be six customers ahead of me when I arrive? (Of course, there is also the question of how quickly on average the counter clerks serve the customers, but that is another issue.)



or just over 1 percent. That’s not bad! And that’s just for starters; we will meet the Poisson distribution again inChapter 9.


Another probability problem. Suppose that Jack and Jill, freshly arrived in the Big City, each have their own agendas about things to do. They decide—sort of—to do their “own thing” for most of the day and then meet sometime between 4:00 p.m. and 6:00 p.m. to have a bite to eat and discuss the day’s activities. Both of them are somewhat disorganized, and they fail to be more specific than that (after all, there aresomany books to look at in Barnes and Noble, or Foyle’s Bookshop in London, who knows when Jack will be ready to leave?) Furthermore, neither cell (mobile) phone is charged, so they cannot ask “where the heck are you?”

Suppose that each of them shows up at a random time in the two-hour period. Jack is only willing to wait for 15 minutes, but Jill, being the more patient of the two (and frankly, a nicer person) is prepared to wait for half an hour. What is the probability they will actually meet? The related question of what will happen if they don’t is beyond the scope of this book, but the first place to start looking might be the local hospital.

This is an example of geometric probability. In the 2-hr square shown inFigure 3.9, the diagonal corresponds to Jack and Jill arriving at the same time at any point in the 2-hr interval. The shaded area around the diagonal represents the “tolerance” around that time, so the probability that they meet is the ratio of the shaded area to the area of the square, namely,


or about one third. Not bad. Let’s hope it works out for them.


To build housing requires land (L) and materials (M), and if the latter consists of everything that is not land (bricks, wood, wiring, etc.), then we can define a functionQ(L,M),



Figure 3.9. The shaded area relative to that of the square is the probability that Jack and Jill meet.


wherea> 0 and 0 ≤λ≤ 1. This form of function is known as aCobb-Douglasproduction function, widely used in economics, and named for the mathematician and U.S. senator who developed the concept in the late 1920s.

If the price of a unit of housing isp, the cost of a unit of land per unit area isl, and the cost of unit materials ism(however all these units may be defined), then the builders’ total profitPis given by


If a speculative builder wishes to “manipulate”LandMto maximize profit, then at a stationary point


Exercise:(i) Show, using equation (3.16), that this set of equations takes the algebraic form



(ii) Show also from equation (3.16) that according to this model, such a speculator “gets his just deserts” (i.e.,P= 0)!

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We shall use equations (3.15) and (3.17) to eliminateQ,L, andMto obtain


whereC(α, λ) is a constant. With two additional assumptions this will enable us to derive a valuable result to be used inChapter 17(The axiomatic city). If the cost of materials is independent of location, and land costs increase with population densityρaccording to a power law, that is,l∝ρκ(κ> 0), then



whereγ=κλandAis another constant.

Exercise:Derive equation (3.18)


We conclude this section with a numerical application of equation (3.15) in a related context. It is recommended that the reader consultAppendix 5to brush up on the method ofLagrange multipliersif necessary.

Suppose that


Lnow being the number of labor “units” andMthe number of capital “units” to produceQunits of a product (such as sections of custom-made fencing for a new housing development). If labor costs per unit arel= $50, capital costs per unit arem= $100, and a total of $500,000 has been budgeted, how should this be allocated between labor and capital in order to maximize production, and what is the maximum number of fence sections that can be produced? (I hope the reader appreciates the “nice” numbers chosen here.)

To solve this problem, note that the total cost is given by


The simplified equation will be the constraint used below. The problem to be solved is therefore to maximizeQin equation (3.19) subject to the constraint (3.20). Using the method of Lagrange multipliers we form the function


The critical points are found by setting the respective partial derivatives to zero, thus:


From equations (3.22a) and (3.22b) we eliminateλto find thatM= 3L/2. From (3.22c) it follows thatM= 2500, and hence thatL= 3750. Finally, from (3.22a),λ= −4(2500)−3/4(3750)3/4≈ −5.4216, so the unique critical point ofFis (2500, 3750, −5.4216). The maximum value isQ(L,M) = 16(2500)1/4(3750)3/4≈ 54,200 sections of fencing.

Question:Can we be certain that this is themaximumvalue ofQ?


Exercise:Show that had we not simplified the constraint equation (3.20), the value forλwould have beenλ≈ −0.1084 but the maximum value ofQwould remain the same.


The home has been built; now it’s time to start paying for it. It’s been said that if you think no one cares whether you’re alive or not, try missing a couple of house payments. Anyway, let’s set up the relevantdifference equationand its solution. It is called a difference equation (as opposed to adifferentialequation) because it describes discrete payments (as opposed to continuous ones).

Suppose that you have borrowed (or currently owe) an amount of moneyA0, and the annual interest (assumed constant) is 100I% per year (e.g., ifI= 0.06, the annual interest is 6%), compoundedmtimes per year. If you pay off an amountbeach compounding period (or due date), the governing first-order nonhomogeneous difference equation is readily seen to be


The solution is


Let’s see how to construct this solution using the idea of a fixed point (or equilibrium value). Simply put, a fixed point in this context is one for whichAn+1=An. From equation (3.23) such a fixed point (call itL) certainly exists:L=Bm/I. If we now definean=An−L, then it follows from equation (3.24) that


that is


Thus we have reduced the nonhomogeneous difference equation (3.23) to a homogeneous one. Sincea1=λa0,a2=λa1=λ2a0,a3=λa2=λ3a0, etc., it is clear that equation (3.25) has a solution of the forman=λna0. On reverting to the original variableAnand substituting forLandλthe solution (3.24) is recovered immediately.

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Exercise:Verify the solution (3.24) by direct substitution into (3.23).


Here is a natural question to ask:How long will it be before we pay off the mortgage?The answer to this is found by determining the number of pay periodsnthat are left (to the nearest preceding integer; there will generally be a remainder to pay off directly). The time (in years) to pay off the loan is justn/m(usuallym= 12 of course). To find it, we setAn= 0 and solve forn. Thus


Exercise:Verify this result. And use it to find out when you will have paid offyourmortgage!





The first time I visited a really big city on more than a day trip, I wondered where people found groceries. I had just left home to start my time as an undergraduate in London, but at home I could either ride my bicycle or use my father’s car to drive to the village store or visit my friends (not always in that order). In London I used the “tube”—the underground transport system (the subway) and sometimes the bus, but I walked pretty much everywhere else. I soon found out where to get groceries, and in retrospect realized that there are some quite interesting and amusing mathematics problems associated with food, whether it’s in a city or in the middle of the country. What follows is a short collection of such items connected by a common theme—eating!


A farmer harvested ten tons of watermelons and had them delivered by truck to a town 30 miles away. The trip was a hot and dusty one, and by the time the destination was reached, the watermelons had dried out somewhat; in fact their water content had decreased by one percentage point from 99% water by weight to 98%.

Question:What was the weight of the watermelons by the time they arrived in the town?


W1= 0.99W1+ 0.01W1(water weight plus pith weight) and

W2= 0.98W2+ 0.02W2, but the pith weight is unchanged; therefore 0.01W1= 0.02W2, and so

W2= W1/2;


the weight of the watermelons upon arrival is now only five tons. Very surprising, but it shows that a small percentage of a large number can make quite a difference . . .


Suppose that Kate feels like having a healthy snack, and decides to eat a banana. Mathematically, imagine it to be a cylinder in which the lengthLis large compared with the radiusr. Suppose also that the peel is about 10% of the radius of the original banana. Since the volume of her (now) idealized right circular banalinder (or cylinana) isπr2L, she loses 19% of the original volume when she peels it (1 − (0.9)2= 0.81). Okay, now let’s do the same thing for a spherical orange of volume 4πr3/3. The same arguments with the peel being about 10% of the radius give a 27% reduction in the volume (1 − (0.9)3= 0.73). We might draw the conclusion in view of this that it is not very cost-effective to buy bananas and oranges, so she turns her stomach’s attention to a peach. Now we’re going to ignore the thickness of the peach skin (which I eat anyway) in favor of the pit. We’ll assume that it is a sphere, with radius 10% of the peach radius. Then the volume of the pit is 10−3of the volume of the original peach; aloss of only 0.1%. What if it were 20% of the peach radius? The corresponding volume loss would be only 0.8 %. These figures are perhaps initially surprising until we carry out these simple calculations [11]. But is a banana a fruit or an herb? Inquiring minds want to know.

Meanwhile, the neighbor’s hotdogs are cooking. How much of the overall volume of a hotdog is the meat? Consider a cylindrical wiener of lengthLand radiusrsurrounded by a bun of the same length and radiusR=ar, wherea> 1. If the bun fits tightly, then its volume is


whereVmis the volume of the wiener. Ifa= 3, for example, thenVb= 8Vm. But a cheap hotdog bun is mostly air; about 90% air in fact!


Question:How long does it take to cook a turkey (without solving an equation)?


Let’s consider a one-dimensional turkey; these are difficult to find in the grocery store. Furthermore, you may object that a spherical turkey is much more realistic than a “slab” of turkey, and you’d be correct! A spherical turkey might be a considerable improvement. However, the equation describing the diffusion of heat from the exterior of a sphere (the oven) to the interior can be easily converted by a suitable change of variables to the equation of a slab heated at both ends, so we’ll stick with the simpler version.

The governing equation is the so-called heat or diffusion equation (discussed in more detail inAppendix 10)


whereTis the temperature at any distancexwithin the slab at any timet;κis the coefficient of thermal diffusivity (assumed constant), which depends on the thermal properties and density of the bird; andLis the size (length) of the turkey. This equation, supplemented by information on the temperature of the turkey when it is put in the oven and the oven temperature, can be solvedusing standard mathematical tools, but the interesting thing for our purposes is that we can get all the information we need without doing that. In this case, the information is obtained by making the equation above dimensionless. This means that we define new variables for which (i) the dimensions of time and length are “canceled,” so to speak, and (ii) the temperature is defined relative to the interior temperature of the bird when it is fully cooked. We’ll call this temperatureTc. We’ll also definetcas the time required to attain this temperatureTc—the cooking time. It is this quantity we wish to determine as a function of the size of the bird. The advantage of this formulation is that we don’t have to repeat this calculation for each and every turkey we cook: indeed, as we will see, with a little more sophistication we can express the result in terms that are independent of the size of the turkey.

To proceed with the “nondimensionalization” letT′ =T/Tc,t′ =t/tc, andx′ =x/L. Using the chain rule for the partial derivatives, equation (4.2) in the new variables takes the form


Has anything at all been accomplished? Indeed it has. Since both derivatives are expressed in dimensionless form, then so must be the constantκtc/L2. Let’s call this constanta. It follows that

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But the weightWof anything is (for a given mean density) proportional to its volume, and its volume is proportional to the cube of its size, that is,W∝L3, soL2∝W2/3, from which we infer thattc∝W2/3. And that is our basic result:the time necessary to adequately cook our turkey is proportional to the two-thirds power of its weight. We have in fact made use of a very powerful technique in applied mathematics in general (and mathematical modeling in particular):dimensional analysis. As seen above, this involves scaling quantities by characteristic units of a system, and in so doing to reveal some fundamental properties of that system.

Where can we go from here? One possible option is to determine the unknown constant of proportionalitya/κ; in principleκcan be found (but probably not in any cookbook you possess), but of courseais defined in terms oftc, which doesn’t help us! However, if we have a “standard turkey” of weightW0and known cooking timet0, then for any other turkey of weightW, a simple proportion gives us


From this the cooking time can be calculated directly. Note thattcis not a linear function of weight; in fact the cooking timeper poundof turkey decreases as the inverse cube root of weight, since


Hence doubling the cooking timet0for a turkey of weight 2W0may result in an overcooked bird; the result (4.4) implies thattc= 22/3t0≈ 1.6t0should suffice. However, always check the bird to be on the safe side. Note that the units used here, pounds, are really pounds-force, a unit of weight. Generally, pounds proper are units of mass, not weight, but in common usage the word is used to mean weight.


Of course, this is a rather ill-posed question. Are we referring to large tomatoes, which we slice nd put in salads, or those little ones that we find in salad bars? Of course, we can find both sizes, and those in between, in any supermarket or produce store. And I suspect that more tomatoes are consumed during the summer than the rest of the year, for obvious reasons. (Note that we are not including canned tomatoes used in pasta sauce.) Now big tomatoes and little tomatoes share a very import characteristic: they are both tomatoes! I’m going to work with a typical timescale of one week; that is, I might use one large tomato per week in my sandwiches, or consume more of the smaller tomatoes in the same period of time. Therefore I will take the average in the following sense. A fairly big tomato 3 inches in diameter is about 30 times as large by volume as one that is one inch in diameter, so we can use the Goldilocks principle referred to earlier—is it too large, too small, or just right?To implement this, we merely take the geometric mean of the volumes. (A reminder: the geometric mean of two positive numbers is the most appropriate measure of “average” to take when the numbers differ by orders of magnitude.) The geometric mean of 1 and 30 isabout 5, so accounting for the range of sizes, we’ll work with 5 “generic” tomatoes eaten per week in the calculations below. Let’s suppose that about a third of the U.S. population of 300 million eat tomatoes regularly, at least during the summer. (I am therefore ignoring those adults and children who do not eat them by dividing the population into three roughly equal groups, again using the Goldilocks principle: fewer than 100% but more than 10% of the population eat tomatoes; the geometric mean is, or about 1/3.)

Therefore, if on average one third of the population eat 5 tomatoes per week, then halving this to reflect a smaller consumption (probably) during the winter months (except in Florida), the approximate number of tomatoes eaten every year is


that is, three billion tomatoes. We multiply this by about 2/3 (to account for the proportion of city-dwelling population in the U.S.) to get roughly two billion (with no offense to non-city dwellers, I trust).


I love apples, don’t you? Sometimes though, they contain “visitors.” Suppose that there is a “bug” of some kind in a large spherical apple of radius, say, two inches. We will assume that it is equally likely to go anywhere within the apple (we shall ignore the core). What is the probability that it will be found within a typical “bite-depth” of the surface? Based on my lunchtime observations, I shall take this as ¾ inch, but as always, feel free to make your own assumptions.

The probabilityPof finding the bug within one bite-depth of the surface is therefore the following ratio of volumes (recall that the volume of a sphere is 4π/3 times the cube of its radius):


or about 76%.Ewwwhhh!





When I was growing up, I loved to visit my grandfather. Despite living in a city (or at least, a very large town) he was able to cultivate and maintain quite a large garden, containing many beautiful plants and flowers. In fact, whenever I asked what any particular flower was called, his reply was always the same: “Ericaceliapopolifolium!” I never did find out whether he was merely humoring me, or whether he didn’t know! In addition to his garden, he rented a smaller strip of land (an “allotment”) farther up the road where, along with several others, he grew potatoes, carrots, beans, and other vegetables. I’mashamed to admit that I didn’t inherit his love for the art of gardening, much to my parents’ disappointment. It skipped a generation, though; my son has a lovely garden and my son-in-law has a very “green thumb” (of course, neither I nor my grandfather can be held responsible for the latter).

As with the previous chapter, there are various and sundry topics in this one, connected (somewhat tenuously, to be sure) by virtue of being found in a garden or greenhouse. Let’s get specific. Plants grow. My daughter and her family live in northern Virginia, and at the bottom of their garden they have quite a lot of bamboo plants. These can reach great heights, so my question is,

X=h′(t): Question:How fast does bamboo grow?


The growth rate for some types of bamboo plant may be as much as 4 meters (about 13 ft) per day. Let’s work with a more sedate type of bamboo, growing only (!) at the rate of 3 ft/day (about 1 m / day). Just for fun, let’s convert this rate to (i) miles/second, (ii) mph, and (iii) km/decade (ignoring leap years!).

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As I write this, my daughter and son-in-law have sold their house and moved into a somewhat larger one. Could it be that the bamboo drove them out?

Grass also has a propensity to grow, but it seems that my lawn is never as green and lush as everyone else’s. Despite my secret desire to cover it with Astroturf, or green-painted concrete, I bought yet another bag of grass seed awhile ago. It can be thought of as a rectangular box with dimensions approximately 2 × 1.5 × 0.5 ft3. So let’s think about the quantity of seed in this bag . . .


Estimate (i) the number of seeds in such a bag (assumed full), and (ii) how many such bags would be required to seed, say, a golf course with area one square mile.

(i) We need first to estimate the volume of a typical grass seed. Since I started this question using the more familiar British units (for those in the U.S. at least), as opposed to the easier metric units, I’ll continue in this vein. Examining a seed, I estimate a typical seed to be a rectangular box with approximate dimensions 1/5 in × 1/20 in × 1/20 in, or about 1/2000 cu in. The bag’s volume is about 25 × 20 × 5 = 2500 cu in. Therefore the number of seeds isN= 2500 ÷ (1/2000) = 5 × 106, or 5 million! How many seeds would there be to seed one square mile?

(ii) Let’s assume that the grass is seeded uniformly with about 10 to 20 seeds per sq in. (you can change this seed density to suit your own estimates). I’ll go with the lower figure. Since a (linear) mile contains 5280 × 12 ≈ 6 × 104inches, a square mile contains the square of this number, and with 10 seeds/sq in we have the quantity of seeds as approximately 10 × (6 × 104)2≈ 4 × 1010(that’s 40 billion!). Finally, dividing this figure by the average number of seeds per bag we obtain 4 × 1010÷ 5 × 106, or about 10,000 bags. I could round this up because the first figure is unlikely to be correct (e.g., it could be 6 or 9, and still be within the order of accuracy we anticipate). I wonder what the local golf club would say.

At the university where I am employed, there is a lovely collection of orchids, some of them very rare. Suppose that a botanist (let’s call him Felix) wishes to grow one of these rarer varieties in the greenhouse. From his previous attempts, he has concluded that the probability that a given bulb will mature is about one third. He decides to plant six bulbs.

X=Pr: Question:What is the probability that at least three bulbs will mature?


Since the probability of (i) at least three bulbs maturingand(ii) the probability of none or one or two maturing must add up to one (meaning certainty that either outcome (i) or outcome (ii) will occur), then the probability we require is given by


In order to proceed we need to introduce the concept ofcombinations,nCr, which represents the number of ways of choosingritems from a total ofn(without regard to order). This is defined as


where for a positive integern, for example, 5, 5! = 5 × 4 × 3 × 2 × 1 = 120. Furthermore, by definition, 0! = 1. NowP(r) is the probability that onlyrevents will occur, and is given by the number of ways the event can occur, multiplied by the probability thatrevents do occur and (n−r) do not. Thus


Or just under one third. Go for it, Felix!

According to Richard F. Burton [12], an expert in earthworms (and a contemporary of Charles Darwin) estimated that in a typical field there are about 133,000 earthworms per hectare. (A hectare is a square hectometer; a hectometer is 100 meters, so you’ll be delighted to know that we’re back with the metric system.)

X=N: Question: What is this figure in worms per square meter?


Clearly, a square hectometer is 104sq m, so the only thing we need to do is divide by this number to get about 13 worms per square yard. Does that sound about right? Of course, it depends very much on the kind of soil, and we don’t worry about the 0.3 earthworm (unless it ends up in our apple [seeChapter 4]).

While we’re on the subject of slimy things, consider this. A certain species of slug is 80% water by weight. Suppose that it loses a quarter of this water by evaporation.

X=W: Question: What is its new percentage of water by weight?


IfWandWneware the weights of the slug “before and after,” so to speak, then in each case the weight can be distributed as the sum of the water content and the rest, that is,


This is a little like the watermelon problem, isn’t it? Don’t confuse the two when eating, though. Do you recall the quote (from St. John of Patmos) about leaves on a tree? Let’s set up a simple framework for estimating the number of leaves on any tree or bush.

X=N: Question:How many leaves are on that laurel bush in my back yard?


I’ll approximate my smallish bush by a sphere of radius half a meter, so that’s a surface area (4πtimes radius squared) of about 3 m2. Now the leaves does not “continuously” cover the surface, but then again, there are leaves throughout much of the bush, not just on the outer “canopy,” so I’ll simplify this problem crudely by just assuming a continuous outer surface composed of leaves about 1 cm square, that is, of area 1 cm2= 10−4m2. Dividing 3 m2by this quantity gives us aboutN≈ 3 × 104leaves.

Doubling the diameter of the bush to make it a small tree would quadruple the area, but the area of the individual leaves would probably be larger (depending on the type of tree), so for a yew tree, say, with typical leaf area about 4 cm2, these two effects would essentially cancel each other out, giving us a figure again of about 30,000 leaves, accurate to within a factor of two or three, I suspect.

Exercise:Estimate the number of leaves on that really big tree in your neighborhood. And when you’ve done that, estimate the total length of the tree; that is, the trunk plus all the branches and twigs.





Question:How many squirrels live in Central Park?


Central Park in New York City runs from 59th Street to 110th Street [6]. At 20 blocks per mile, this is 2.5 miles. Central Park is long and narrow, so we will estimate its width at about 0.5 mile. This gives an area of about 1 square mile or about 2 square kilometers.

It’s difficult to estimate the number of squirrels in that large an area, so let’s break it down and think of the area of an (American) football field (about 50 yards × 100 yards). There will be more than 1 and fewer than 1000 squirrels living there, so we choose the number geometrically between 1 and 1000, or 30 (using the Goldilocks principle again).

This is where the metric system comes in handy. One kilometer (1000 m or about 1100 yards; we’ll round down to 1000 yards) is the length of ten football fields and the width of twenty. This means that there are 200 football fields in a square kilometer. Now the number of squirrels in Central Park is aboutN= (2 km2) × (200 football fields/(km2)) × (30 squirrels/football field) = 12,000 ≈ 104. That’s enough squirrels for even the most ambitious dog to chase!

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It’s a lovely weekend and you decide to “catch some rays” in the park. If you’re anything like me, with the fair skin of someone from Northern Europe, you will make certain you lather yourself with sun block. It’s tempting to get the highest possible SPF variety, but is it really necessary?

In fact, SPF 30 doesnotblock out twice as much harmful radiation as SPF 15. SPF is not sun-filtering, it is a S(un)P(rotecting)F(actor). The label tells you how muchtimeyou can spend in the sun before you start to burn (compared with the time for bare skin); 15 times longer for SPF 15, 30 times longer for SPF 30. However, SPF 30 only blocks out about 3% more of the harmful UVA and UVB radiation.

Here’s what happens. Let’s suppose we have SPF 2 (does that exist?). Anyway, that would block out 50% of the UV radiation that causes burning. If you burn after 30 minutes when naked as the day you were born (without any sunscreen), you could stay out for an hour—twice as long—with SPF 2. (Please try any naked sunbathing at home, not in the park.) SPF 4 would block out 75% of the harmful radiation, so you could stay out in the sun four times longer—but it cuts out only 25% more of the incoming UV rays than SPF 2 does, correct? SPF 8 means you can stay out 8 times longer, but only cuts out, well, let’s see how much more.

SPF 2 cuts out 1 −= 50% of the incoming UV radiation.

SPF 4 cuts out 1 −= 75% of the incoming UV radiation.

SPF 8 cuts out 1 −= 87.5% of the incoming UV radiation.

SPF 16 cuts out 1 −= 93.75% of the incoming UV radiation.

You get the idea. Following this pattern, we see that SPFXcuts out a fraction 1 −of the incoming UV radiation. But we can examine this from another point of view. From the list above we see that the fractions of UV radiation blocked by SPF 2, 4, 8, 16 can be written respectively as


Therefore, for SPFXthe corresponding fraction of blocked radiation may be written in the form 1 −, whereX= 2n. Taking logarithms to base 2 we find thatn= log2X, or, using the change of base formula with common logarithms,

Let’s now go back and compareX= 15 and 30.

For SPF 15,soor 93.3% of the harmful radiation is blocked. Of course, we could have just calculatedto get this result—but where would be the fun in that?

For SPF 30,, soor 96.7% of the harmful radiation is blocked. This is a difference of 3.4 percentage points! Again, “we don’t need no logs” to do this, because.

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Just for fun, consider SPF 100 (if that exists!).

In this case, soor 99% of the harmful radiation is blocked. But you knew that, because


You are walking in a city park (Central Park, for example; but be careful not to trip over the squirrels). Suppose that a jogger passes you. As she does so, at an average speed, say of 8 mph, you wonder if there is an instant when her speed isexactly8 mph, or equivalently, 7.5 minutes per mile. Fortunately, you have taken a calculus class, and you recall the mean value theorem, an informal (and imprecise) rendering of which says that for a differentiable functionf(t) on a closed interval there is at least one value oftfor which the tangent to the graph is parallel to the chord joining its endpoints (why not sketch this?). This meansthat iffis the distance covered as a function of timet, then at some point (or points) on the run her speed will be exactly 8 mph. A formal statement of the theorem can be found in any calculus book when you get home (unless you are carrying it while walking in the park).

The jogger’s name is Lindsay, by the way; by now you’ve seen each other so often that you greet each other as she races past. As she does so yet again, a related question comes to mind: does Lindsay cover any one continuous mile in exactly 7.5 minutes? (Of course, this assumes her run is longer than a mile.) This is by no means obvious, to me at least, and it may not have occurred to Lindsay either.

Suppose thatt(x) is a continuous function representing the time taken to coverxmiles. We consider this question in two parts. Suppose further that Lindsay runs an integral number (n) of miles. If she averages 7.5 min/mi (8 mph), thent(x) − 7.5x= 0 whenx= 0 andx=n, i.e.t(n) = 7.5n. If she never covered any continuous mile in 7.5 minutes, then it follows that the new functionT(x) =t(x+ 1) −t(x) − 7.5 is continuous and never zero. Suppose that it is always positive; a similar argument applies if it is always negative. Hence forx= 0, 1, 2, . . . ,nwe can write the following sequence ofninequalities:T(0) > 0;T(1) > 0;T(2) > 0; . . . T(n− 1) > 0. In adding them all a great deal of cancellation occurs and we are left with the inequalityt(n) −t(0) > 7.5n. This contradicts the original assumption thatt(n) −t(0) = 7.5n, and so we can say that shedoescover a continuous mile in exactly 7.5 minutes.

It turns out that this is true for only an integral number of miles, and that seems, frankly, rather strange. We’ll examine this case for a specific function, by redefiningt(x) to be:


wheremisnotan integer andε> 0. Ifεis small enough andmis large enough, this will represent an increasing function with small “undulations” about the line 7.5x, representing the time to jog at the average speed.t(x) is an increasing function ift′(x) > 0, as it must be because it is a time function. Mathematically this will be so if



Figure 6.1. Lindsay’s time functiont(x).


for this particular model of the jogging time.Figure 6.1showst(x) and the mean time function for the caseε= 1,m= 0.8.

Now it follows that


is an increasing function that canneverbe 7.5. Therefore if Lindsay runs so that her time function is given by equation (6.1), she will never run a complete mile in exactly 7.5 minutes.


It seems that it’s almost impossible to find a taxi when it’s raining in the city. Gene Kelly preferred to sing and dance, but, as sometimes happens when I am walking, the heavens open and I am faced with the different options of running fast or walking for shelter. Is it better to walk orrunin the rain? Thedecision to run may seem an obvious one, but depending on several factors discussed below, it is not always that simple. And occasionally I have the foresight to take an umbrell.

What might such factors be? Among them are how fast the rain is falling, and in what direction (i.e., is there wind, and if so, in what direction), how fast I can run or walk, and how far away the shelter is. Let’s ignore the wind and rain direction initially and set up a very basic model. Suppose that you run atvm/s. Since 2 mph ≈ 1 m/s, as is easily shown, we can easily convert speeds in mph to MKS units and vice versa. Of course, this is only approximate, but we shall nevertheless use it for simplicity; we do not require precise answers here; after all, we’re in a hurry to get home and dry off!

Suppose the distance to the nearest shelter point isdkm, and that the rain is falling at a rate ofhcm/hr (orh/3600 cm/s). Clearly, in this simplified case it is better to run than to walk. Here is a standard classification of the rate of precipitation:

• Light rain—when the precipitation rate is < 2.5 mm (≈ 0.1 in) per hour;

• Moderate rain—when the precipitation rate is between 2.5 mm (≈ 0.1 in) and 7.6 mm (0.30 in) to 10 mm (≈ 0.4 in) per hour;

• Heavy rain—when the precipitation rate is between 10 mm (0.4 in.) and 50 millimeters (2.0 in) per hour;

• Violent rain—when the precipitation rate is > 50 mm (2.0 in) per hour.

Now if I run the whole distancedm atvm/s, the time taken to reach the shelter isd/vseconds, and in this time the amountHof rain that has fallen is given by the expressionH=hd/3600vcm. If I run at 6 m/s (about 12 mph), for example, andd= 500 m, then for heavy rain (e.g.,h= 2 cm/hr),H≈ 0.5 cm. This may not seem like a lot, but remember, it is falling on and being absorbed (to some extent at least) by our clothing, unless we are wearing rain gear. The next step in the model is to estimate the human surface area; a common approach is to model ourselves as a rectangular block, but a quicker method is to consider ourselves to be a flat sheet 2 m high and 0.5 m wide. The front and back surface area is 2 m2—about right! Over the course of my run for shelter, if all that rain is absorbed, I will have collected an amount 2 × 104× 5 × 10−2= 103, or one liter of rain. That’s a wine bottle of rain that the sky has emptied on you! I prefer the real stuff . . .

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It is important to note that rain is a “stream” of discrete droplets, not a continuous flow. It is reasonable to define a measure of rain intensity by comparing the rate at which rain is falling with the speed of the rain. The speed of raindrops depends on their size. At sea level, a very large raindrop about 5 millimeters across falls at the rate of about 9 m/s (seeChapter 25for an unusual way to estimate the speed of raindrops). Drizzle drops (less than 0.5 mm across) fall at about 2 meters per second. We shall use 5 m/s (or 500 × 3600 = 1.8 × 106cm/hr) as an average value. The ratio of the precipitation rate to the rain speed [13] is called the rain intensity,I. For the figures used here,I= 2/(1.8 × 106) ≈ 1.1 × 10−6. ThereforeIis a parameter:I= 1 corresponds to continuous flow (at that speed), whereasI= 0 means the rain has stopped, of course!

If as inFigure 6.2the rain is falling with speedcm/s at an angleθto the vertical direction, and you are running into it, the vertical (downward) component of speed isccosθand the relative speed of the horizontal component iscsinθ+v. And if you move fast enough, only the top and front of you will get wet. For now let’s assume this is the case. Since you arenotin fact a thin sheet (your head does get wet), we will model your shape as a rectangular box of heightl, widthwand thicknesst(all in m). The “top” surface area iswtm2, and the volume of rain is “collected” at a rateR= intensity × surface area × rain speed =Iwtccosθ, expressed in units of m3/s. In timed/vthe amount collected is thenIwtdccosθ/vm3. A similar argument for the front surface area givesIwld(csinθ+v)/vm3, resulting in a total amountTof rain collected as


Before putting some numbers into this, note that ifθwere the only variable in this expression, then


Because this vanishes atθ= arctan(l/t) andd2T/dθ2< 0 there, this represents amaximumaccumulation of rain. Physically this means that you are running almost directly into the rain, but the relative areas of your top and front are such that the maximum accumulation occurs when it is (in this case) not quitehorizontal. In addition, ofvwere the only variable, thendT/dv= −αβ/v2< 0 ifβ> 0, so thatTdecreases with speedvif tanθ> −t/l. If tanθ< −t/l, thenT increaseswith speed. This corresponds to more and more negative values ofθ, that is, the rain is coming from behind the runner.


Figure 6.2. Configuration for a rectangular box-person running in the rain.


Let’s put some meat on these bones, so to speak. Suppose that the rectangular box has dimensionsl= 1.5 m,w= 0.5 m, andt= 0.2 m. Furthermore, we have chosenv= 6 m/s (about 13 mph); andc= 5 m/s. Substituting all these values into expression (6.2) forTwe obtainT≈ 4.6 × 10−2(9 + cosθ+ 7.5 sinθ) liters. It is readily confirmed that this has a maximum value of about 0.76 liters whenθ≈ 82.4°. A graph ofT(θ) is shown inFigure 6.3for −π/2

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which should now be written as


because the rain falls on your back if sinvt/l=Atop/Aback, this term is negative, andin this case, you should attempt to go no faster than the horizontal speed of the rain (csin) at your back. Using equation (6.5) we see that if your speed increases so thatv=csin, you are just keeping up with the rain andTis minimized. This may seem at first surprising since forv>csin,Tis reduced still farther, butnowyou are catching up to the rain ahead of you, and it falls once more on your front (and head, of course). In this case formula (6.2) again applies.

How about putting some numbers into these formulae? For a generic heightl= 175 cm (about 5 ft 9 in), shoulder to shoulder widthw= 45 cm (about 18 in), and chest to back widtht= 25 cm (about 10 in), the ratiot/l= 1/7, and so if tan> 1/7, that is,≈ 8°, the ratiov/c= sin≈ 1/7 (see why?). Therefore if it is raining heavily at about 5 m/s from this small angle to the vertical, you need only amble at less than one m/s (about 2 mph) to minimize your accumulated wetness! Although the chosen value forwwas not used, we shall do so now. The top area of our human box is ≈ 1100 cm2, one side area is ≈ 4400 cm2, and the front or back area is ≈ 7900 cm2.

Exercise:Calculate these areas in square feet if you feel so inclined.


To summarize our results, if the rain is driving into you from the front, run as fast as you safely can. On the other hand, if the rain is coming from behind you, and you can keep pace with its horizontal speed by walking, do so! If you exceed that speed, the advantage of getting to your destination more quickly is outweighed by the extra rain that hits you from the front, since you are now running into it! Perhaps the moral of this is that we should always run such that the rain is coming from behind us!


To some extent cities can create their own weather. No doubt you have heard of the sidewalk in some city being hot enough to fry an egg; include all the paved surfaces and buildings in a city, and you have the capacity to cook a lot of breakfasts! Typically, such surfaces get hotter than those in rural environments because they absorb more solar heat (and therefore reflect less), and retain that heat for longer than their rural counterparts by virtue of their greater thermal “capacity.” The contrast between a city and the surrounding countrysideis further enhanced at night, because the latter loses more heat by evaporative and other processes. Furthermore, the combined effects of traffic and industrial plants are a considerable source of heat within an urban metropolitan area. Thus there are several factors to take into account when considering local climate in a city versus that in the countryside. They include the fact that (i) there are differences between surface materials in the city and the countryside—concrete, tarmac, soil, and vegetation; (ii) the city “landscape”—roofs, walls, sidewalks, and roads—is much more varied than that in the country in the shape and orientation of reflective surfaces; vertical walls tend to reflect solar radiation downward instead of skyward (seeFigure 6.4a,b), and concrete retains heat longer than do soil and vegetation; (iii) cities are superb generators of heat, particularly in the winter months; (iv) cities dispose of precipitation in very different ways, via drains, sewers, and snowplows (in the north). In the country, water and snow are more readily available for evaporative cooling.

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Such local climate enhancement has several consequences, some of which are positive (or at least appear to be). For example, there may be a diminution of snowfall and reduced winter season in the city. This induces an earlier spring, other species of birds and insects may take up residence, and longer-lasting higher temperature heat waves can occur in summer (quite apart from any effects of larger scale climate change). This in turn means that less domestic heating may be required in the winter months, but more air-conditioning in the summer. The effects of an urban-industrial complex on weather generally areharder to quantify, though stronger convective updrafts (and hence intensity of precipitation and storms) are to be expected downwind from urban areas. According to one report (Atkinson 1968), there has been a steadily increasing frequency of thunderstorm activity near London as it has grown in size. In U.S. cities, the incidence of thunderstorms is 10–42% greater than in rural areas, rainfall is 9–27% greater and hailstorms occur more frequently, by an enormous range: 67–430%.


Figure 6.4. (a) Vertical surfaces tend to reflect solar radiation toward the ground and other vertical surfaces (thus trapping it), especially when the sun’s elevation is moderately high. (b) There being fewer vertical surfaces in the countryside, solar radiation tends to be reflected skyward. Redrawn from Lowry (1967).


If the air temperature were to be recorded as we move across the countryside toward a city, the rural/urban boundary will typically exhibit a sharp rise—a “cliff”—followed by a slower rate of increase (or even a plateau) until a more pronounced “peak” appears over the city center. If the temperature difference between the city and surrounding countryside at any given time is denoted by ΔT, the average annual value for ΔTranges from 0.6 to 1.8°C. Of course, the detailed temperature profile as a function of position will vary depending on the time of day, but generally this is a typical shape: a warm “island” surrounded by a cooler “sea.” Obviously the presence of parks and other open areas, lakes, and commercial, industrial and heavily populated areas will modify this profile on a smaller spatial scale. The difference ΔTbetween the maximum urban temperature and the background rural temperature is called theurban heat island intensity. Not surprisingly, this exhibits a diurnal variation; it is at a maximum a few hours after sunset, and a minimum around the middle of the day. In some cases at midday the city is cooler than its environs, that is, ΔT< 0.

To see why this might be so, note that near midday the sunlight strikes both country and city environs quite directly, so ΔTcan be small, even negative, possibly because of the slight cooling effect of shadows cast by tall buildings, even with the sun high overhead. As the day wears on and the sun gets lower, the solar radiation strikes the countryside at progressively lower angles, and much is accordingly reflected. However, even though the shadows cast by tall buildings in the city are longer than at midday, the sides facing the sun obviously intercept sunlight quite directly, contributing to an increase in temperature, just as in the hours well before noon, and ΔTincreases once more.

Cities contribute to the “roughness” of the urban landscape, not unlike the effect of woods and rocky terrain in rural areas. Tall buildings provide considerable “drag” on the air flowing over and around them, and consequently tend to reduce the average wind speed compared with rural areas, though they create more turbulence (seeChapter 3). It has been found that for light winds,wind speeds are greater in inner-city regions than outside, but this effect is reversed when the winds are strong. A further effect is that after sunset, when ΔTis largest, “country breezes”—inflows of cool air toward the higher temperature regions—are produced. Unfortunately, such breezes transport pollutants into the city center, and this is especially problematical during periods of smog.

Question:Why is ΔTlargest following sunset?


This is because of the difference between the rural and urban cooling rates. The countryside cools faster than the city during this period, at least for a few hours, and then the rates tend to be about the same, and ΔTis approximately constant until after sunrise, when it decreases even more as the rural area heats up faster than the city. Again, however, this behavior is affected by changes in the prevailing weather: wind speed, cloud cover, rainfall, and so on. ΔTis greatest for weak winds and cloudless skies; clouds, for example, tend to reduce losses by radiation. If there is no cloud cover, one study found that near sunset ΔT∝w−1/2, wherewis the regional wind speed at a height of ten meters (see equation (6.7)).

Question:Does ΔTdepend on the population size?


This has a short answer: yes. For a populationN, in the study mentioned above (including the effect of wind speed), it was found that


though other studies suggest that the data are best described by a logarithmic dependence of ΔTon log10N. While every equation (even an approximate one) tells a story, equation (6.7) doesn’t tell us much! ΔTis weakly dependent on the size of the population; according to this expression, for a given wind speedwa population increase by a factor of sixteen will only double ΔT! And if there is no wind? Clearly, the equation is not valid in this case; it is an empirical result based on the available data and valid only for ranges ofNandw.

Exercise:“Play” with suitably modified graphs ofN1/4and log10Nto see why data might be reasonably well fitted by either graph.


The reader will have noted that there is not much mathematics thus far in this subsection. As one might imagine, the scientific papers on this topic are heavily data-driven. While this is not in the least surprising, one consequence is that it is not always a simple task to extract a straightforward underlying mathematical model for the subject. However, for the reader who wishes to read a mathematically more sophisticated model of convection effects associated with urban heat islands, the paper by Olfe and Lee (1971) is well worth examining. Indeed, the interested reader is encouraged to consult the other articles listed in the references for some of the background to the research in this field.

To give just a “taste” of the paper by Olfe and Lee, one of the governing equations will be pulled out of the air, so to speak. Generally, I don’t like to do this, because everyone has the right to see where the equations come from, but in this case the derivation would take us too far afield. The model is two-dimensional (that is, there is noy-dependence), withxandzbeing the horizontal and vertical axes; the dependent variableθis essentially the quantity ΔTabove, assumed to be small enough to neglect its square and higher powers. The parameterγdepends on several constants including gravity and air flow speed, and is related to the Reynolds number discussed inChapter 3. The non-dimensional equation forθ(x,z) is


The basic method is to seek elementary solutions of the form


where “Re” means that the real part of the complex function is taken, andkis a real quantity On substituting this into equation (6.8) the following complex biquadratic polynomial is obtained:


There are four solutions to this equation, namely,

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but for physical reasons we require only those solutions that tend to zero asz→ ∞. The two satisfying this condition are those for which Reσ< 0,


Using these roots, the temperature solution (6.9) can be expressed in terms of (complex) constantsc1andc2as


Note from equation (6.12) thatUsing specified boundary conditions, bothc1andc2can be expressed entirely in terms ofσ1andσ2, though we shall do not do so here. The authors note typical magnitudes for the parameters describing the heat island of a large city (based on data for New York City). The diameter of the heat island is about 20 km, with a surface value for ΔT≈ 2C, and a mean wind speed of 3 m/s.

Exercise:Verify the results (6.10)–(6.12).





As we have remarked already, cities come in many shapes and sizes. In many large cities such as London and New York, the public transportation system is so good that one can get easily from almost anywhere to anywhere else in the city without using a car. Indeed, under these circumstances a car can be something of an encumbrance, especially if one lives in a restricted parking zone. So for this chapter we’ll travel by bus, subway, train, or quite possibly,rickshaw. Whichever we use, the discussion will be kept quite general. But first we examine a situation that can be more frustrating than amusing if you are the one waiting for the bus.


In a delightful book entitledWhy Do Buses Come in Threes?[8] the authors suggest that in fact, despite the popular saying, buses are more likely to come in twos. Here’s why: even if buses leave the terminal at regular intervals, passengers waiting at the bus stops tend to have arrived randomly in time. Therefore an arriving bus may have (i) very few passengers to pick up, and little time is lost, and it’s on its way to the next stop, or (ii) quite a lot of passengers to board. In the latter case, time may be lost, and the next bus to leave the terminal may have caught up somewhat on this one. Furthermore, by the time it reaches the next stop there may be fewer passengers in view of the group that boarded the previous bus, so it loses little time and moves on. For the next two buses, the cycle may well repeat; this increases the likelihood that buses will tend to bunch in twos, not threes.

The authors note further thatifa group of three buses occurs at all (and surely it sometimes must), it is most likely to do so near the end of a long bus route, or if the buses start their journeys close together. So let’s suppose that they do . . . that they leave the terminal everyTminutes, and that once they “get their buses in a bunch” busesAandBandBandCare separated bytminutes (wheret 2t, which of course is quite likely, even allowing for the small possibility that you arrived in at-minute gap.


We shall examine and “unpack” an idealized analysis of public transport systems (based on an article [14] published in the transportation literature). The article contained a comprehensive account of two models: a corridor system and a network system. In so doing, it is possible to derive some general results for transport systems. Although some of these conclusions were represented in graphical form only, the mathematics behind those graphs is well worth unfolding here. In both models there are three components to the travel time from originAto destinationB: (i) the walk times to and from the station at either end; (ii) the time waiting for the transportation to arrive; and (iii) the total time taken by the subway train, bus, or train to go from the station nearestA(SA) to that nearestB(SB). (This includes waiting time at intermediate stops along the way.)

In what follows, the city is assumed to have a “grid” system with roads running N-S and E-W (seeAppendix 3on “Taxicab geometry”). While this may be more appropriate to cities in the United States, it is also an acceptable approximation for many European cities [14], [15]. FromFigure 7.1we see that a walk from any location to any station will involve an N-S piece and an E-W piece. If each station is considered to be at the center of a square of sideL,the average walk length isL/2. To see this, note that for any trip starting at the point (x,y) on the diagonal line (for whichx+y=L/2) to the center is justL/2. By symmetry, there is the same area on each side of the line. If the population demand is uniformly distributed,L/2 is the average trip to or from the central station, and henceLis the average total distance walked—just the average stop separation.


Figure 7.1. One segment of a transportation corridor with a central station serving a square area of sideL. The diagonal shown has equationx+y=L/2.


First we consider the average speedVafor the vehicular part of the trip. We suppose that this part consists of a uniform accelerationafrom rest to a (specified) maximum speedVm(A–B), and a period of travel at this speed (B–C–D) followed by a uniform deceleration −ato rest at the next station (D–E). The acceleration and deceleration take place over a distancexkm, say, and the speed is constant for a distanceL− 2x≥ 0. Using the equations of motion for uniform acceleration it is found thatVm= (2ax)1/2and the timeTxto travel the distancex(= distanceAB) and reach this maximum speed isTx= (2x/a)1/2. The time to travel the distanceBCis thereforeTBC= ([L/2] −x)/Vm. If we include a waiting timeTWat the stationAbefore leaving for the next one atE, the total time for the journeyAEis 2 (Tx+TBC) +TW; typically,TW≈ 20s anda= 0.1gaccording to published data [15]. Hence the average speed in terms of the maximum speed is

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after a little reduction. We takeTW= 20s anda= 10−3km/s2≈ 1.3 × 104km/hr2in what follows. Then equation (7.1) takes the form


whereα= 7.7 × 10−5,β= 5.6 × 10−3.

It is interesting to note that on the basis of formula (7.2),Vapossesses a shallow maximum occurring atfor the constants chosen here. For a station separationL= 0.5 km this corresponds toVm≈ 80 km/hr. It can be seen fromFigure 7.2that forL= 1 km the maximum average speed attainable is only about 40 km/hr regardless of maximum speed. For a stop spacing of 0.25 km the maximum is less than 20 km/hr. In these cases the vehicle does not have enough time to reach its maximum speed and merely accelerates to the midpoint and then decelerates.


Figure 7.2. Average speed vs. maximum speed for different values of station spacingL= 5 km, 2 km, 1 km, 0.5 km.


The next phase of the calculation is to include passenger “walk and wait” times, the latter referring to the wait for the arrival of the train or bus. Furthermore, a trip length is now assumed. We shall adopt 80 m/min or 4.8 km/hr for the average walking speed, the average passenger wait time as 5 minutes, and an estimate of 8 km for the average (UK) trip length [14]. The number of stops beyond the origin is 8/L(if this is an integer), as is the number of in-station wait timesTW. For small values ofL(less than 0.25 km, for example) the model is impractical, but we shall nevertheless treatL(and the number of stations) as a continuous variable. The average speed is now


Graphs ofVa(L) forVm= 30 km/hr and 80 km/hr are shown inFigures 7.3aand7.3b, for comparison with the graph for the in-vehicle average speed without the passenger walk-and-wait times (the above expression without the last two terms in the denominator). The lower speed forVmis representative of a bus, given the typical stops necessary at crosswalks (pedestrian crossings) and traffic lights. The higher speed is more typical of light rail, monorail, or possibly priority bus lanes, which would ensure higher average speed. From thefigure it is seen that for the bus the optimum spacingLbetween stations (corresponding to the peak average speed) is about 0.5 km. For the higher speed, the optimum spacing is closer to 0.75 km.


Figure 7.3. (a) Average speed vs. stop separation (in km) for maximum speeds of 30 km/hr for in-vehicle with stops (upper curve) and with walk and wait times included (lower curve). (b) Average speed vs. stop separation (in km) for maximum speeds of 80 km/hr for in-vehicle with stops (upper curve) and also with walk and wait times included (lower curve). The generic distancedinFigures 7.3–7.5represents the spacingLbetween stops.


Note what little difference the increase in maximum speed makes to the maximum average speed. These average speeds are probably at best comparable to the average speed for cars in rush hour periods, since the latter will not have to keep stopping to let passengers on or off. This model suggests that overall trip times using public transport will generally be higher than for cars during the same peak times.


Consider first an eastbound trip consisting ofNstops, numbered fromn= 0 toN− 1. The total number of such trips from 0 isN− 1, from 1 isN− 2, etc. so summing over all values ofnwe find that the total number of eastbound trips is the well-known result


The sum of all the trips from themth stop eastward (0 ≤m≤N− 1)can be found by replacingNabove byN−m, that is,


If the stations are a distanceLapart, then the sum ofalltrip distances fromallstops isLtimes the sum of the above expression over all possible values ofm, that is,


Exercise:establish the result (7.4).


For trips on a rectangular grid containingMstops on theN-Scorridors, the total length of eastbound trips in any one row isMtimes the result (7.4) above, and the total length over all rows requires multiplication byMonce more. Multiplying this by two for the westbound trips, the total length of eastbound and westbound trips is

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Assuming a uniform demand at theMNstarting points, each of which serves theMN− 1 others, there areMN(MN− 1) trips, so the average length of all eastbound and westbound trips is


By interchangingMandNwe obtain the corresponding result for N-S trips, that is,


Any single trip will generally involve a combination of both types of trip, so the average length over all trips is


This gratifying result shows that the average trip length is one sixth of the perimeter of the area served, and if the area is square, this is two thirds of the length of a side.

With this general result established (though only for the in-vehicle part of the trip), we proceed with the “walk, wait, ride, and walk” overall trip time calculations for a square city of sidec. With a grid containingn2equally spaced stations, the distance between adjacent stops isc/n, and with each station centrally located in its “sub-square,” the average walk distance is, from our earlier discussion,c/2n, which of course must be doubled for the assumed symmetry of the trip. The time to walk both such trips isc/nW.

The “wait time” depends on the spacing of the vehicles on a line. For example, if they are two stations apart the wait time is 2c/nVa, and in general,sc/nVafor a vehicle spacing ofsstations. The trip time is merely the average distance divided byVa, and from the previous discussion, this will be 2c/3Va. There is also a transfer time for travel involving different vehicles, which for our purposes is just the waiting time multiplied by a factorFof order one. (There are data [14] indicating that for networks sizes up to 10 × 10, 0.5 ≤F≤ 2. This will be generically incorporated in the termsc/nVafor the calculations below.)

Combining all three “times,” that is, walk + wait + in-vehicle, we express the overall trip time as


In keeping with the literature [14], we choose a city size ofc= 10 km. In Europe, typical population densities are around [14] 4000/km2(the average figure for U.S. cities is about half this). Thus “our” city has a population of about 400,000. The average length of all trips is, as noted above, 2/3 of the side length, approximately 6.7 km. Again, fictionally treating the station spacing as a continuous variable, we can the network travel time as a function of station spacing from the above formula.

The total travel time in minutes is shown inFigure 7.4forVm= 30 km/hr (upper curve) and 80 km/hr (middle curve). The lower (dotted) line is thewalk time, and it can be seen that the walk time is the dominant contribution for large station spacing (corresponding to fewer tracks). Notice that the minimum travel time for the slower speed of 30 km/hr occurs at a stop spacing of about 0.25 km (appropriate for a city bus route) and at about 0.5 km for the 80 km/hr monorail route.


Figure 7.4. Total travel time (minutes) vs. stop spacing for a maximum speed of 30 km/hr (upper curve) and 80 km/hr (middle curve). The dotted line represents the walk time component for one end of the trip.



Figure 7.5. Average speeds in km/hr vs. stop spacing, based onFigure 7.4. The upper curve is for a maximum speed of 80 km/hr, the lower for a maximum speed of 30 km/hr.


Figure 7.5shows the average speed for these curves (starting ford=L= 0.1 km). Note that the maximum average speed is only between about 13 and 16 km/hr for the range of maximum speeds considered here.

The much higher value forVmof 80 km/hr adds relatively little to the average speed.

A final word:pedestrians. The most complete forms of the above models make allowances for the average time to walk to and from the station. As someone who has walked to work almost every weekday for twenty-eight years, I have learned to be very careful about crossing roads. I have to judge whether there is sufficient time to do so before the nearest vehicle reaches me. (One certainly hopes there is.) In fact, every individual has (in principle) acritical time gap,Tsay, above which crossing is acceptable and below which it is not. Mathematically, this can be represented by a step function

G(t) = 0,t≤T,G(t) = 1,t>T.


I think a step function is an entirely appropriate thing for pedestrians to have.





As noted in the previous chapter, in many very large cities a car is not needed at all. But there also are many cities and towns where it is essential to have a vehicle. One advantage of driving over public transportation can be that one does not have to keep stopping at intermediate locations on the way to one’s destination (traffic permitting). Of course, the daily commute can be extremely frustrating when it is of the stop-and-start variety, and the gas consumption becomes prohibitive. At the time of writing the price of petrol in the UK is far higher (by a factor of two) than the equivalent cost of gas in the U.S. It must be the different spelling that causes this. But before discussing that and other driving-related topics, let’s start with an unrealistic but amusing and informative question.

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Question:A car travels from a home in the suburbs to a downtown office building, somehow maintaining a constant speed of 30 mph. On the return journey it maintains a constant speed of 60 mph. What is its average speed?


No, it’s not 45 mph, sorry. The car spends twice as long traveling at 30 mph as it does returning at 60 mph, so the average speed will be “weighted” toward the lower speed. The correct answer is 40 mph. To see why, the average speed is defined as the total distance traveled divided by the total time (T) for the round trip. For those who are concerned that we have not specified the distance from the proverbial “AtoB”—not to worry, let’s just call itd. Then the average speed is


In fact, the above result is theharmonic meanof the two speeds. The harmonic mean of a set of numbers is the reciprocal of the arithmetic mean of the reciprocals! Put more obviously in mathematical terms, for a set ofnnumbersxi,i= 1, 2, 3, . . . n, the harmonic meanHis


That’s the power of algebra for you! The incorrect answer of 45 mph is based on the arithmetic mean, which as we see, is not the appropriate measure of “average” for this problem. This difference is exacerbated by considering more than one “vehicle”; suppose that nine vehicles traverse a route one mile in length, and that all travel at the posted speed limit of 55 mph. I decide to walk the route at a reasonable pace, say 4 mph. The average speed for all ten trips, as computed by using the arithmetic mean is


On the other hand, if we calculate the total distance traveled and divide it by the total time taken, the result is


a considerable difference.

Question:If we replace “speed” by “average speed” in the above question, does it change the result?


Exercise:Generalize the first result for speedsv1andv2.


X= ΔE: Question:How does gasoline consumption vary with speed?


In recent years, as well as several decades ago, the increasing costs of oil and gasoline have threatened to change the habits of American motorists, and in some cases has done so (albeit for a limited period of time, after which gas prices have declined, and everything reverts to thestatus quo). We know that at low speeds in low gears fuel consumption rate is relatively high because of lower efficiency in converting chemical energy to kinetic energy at these speeds. At high speeds the effects of air resistance (or drag) also increases the rate of fuel consumption. So it’s certainly reasonable to conclude that there is an optimal speed (or range of speeds) for which the rate of fuel consumption is minimized.

The national 55-mph speed limit in the U.S. is no longer in existence, but in 1982 a newspaper article stated that one should “Observe the 55-mile-an-hour national highway speed limit.” Furthermore, it stated that “For every five miles an hour over 50, there is a loss of one mile to the gallon. Insisting that drivers stay at the 55-mile-an-hour mark has cut fuel consumption 12 percent for [Company name] of Jacksonville, Florida—a savings of 631,000 gallons of fuel a year. The most fuel-efficient range for driving generally is considered to be between 35 and 45 miles an hour” (“Boost Fuel Economy,”Monterey Peninsula Herald, May 16, 1982, as quoted in Giordano et al. 2003).

My first reaction on reading this was to wonder if the article is referring to fuel consumption for domestic as opposed to commercial vehicles, or both (which seems unlikely). And I very much doubt that the mpg losses are a linear function of speed above 55 mph as suggested. Surely there are more factors that affect how many miles per gallon we get from our vehicles, such as the ageof the engine (and maybe the driver!), the type of fuel used, the air temperature, the speed of travel, and the resistive forces of drag and road friction. Drag will depend on the speed and shape of the vehicle and the prevailing atmospheric conditions at the ground (wind in particular). Friction will depend on the condition of the tires, and the type and weight of the car. The nature of the terrain is very important also: is it a flat, smooth road or an unpaved track in a hilly or mountainous region? How good is the driver? Does he drive smoothly where possible, or speed up/slow down in an irregular pattern, even if the road conditions do not warrant it? Is she an experienced driver or a novice? These and probably several other considerations all have bearing on the problem of interest here.

A quick and dirty method 

The drag force on most everyday objects is proportional to their cross-sectional areaAand the square of their speedv. Therefore driving at highway speeds for a distancedwill consume energyE∝Av2d. For a given value ofdand a small change inv(Δv), the corresponding percentage change in energy expended is [16]


While the change from driving at 70 mph (as many do) to 60 mph (about 14%) may not be considered “small,” we’ll use it anyway, and conclude that there is a nearly 30% drop in fuel consumption. This makes a lot of sense!


What thoughts typically run through your mind as you approach a traffic signal? Here are some likely ones: will it stay green long enough for me to continue through? Will it turn red in enough time for me to stop? What if it turns yellow and someone is close behind me—should I try to stop or go through? And perhaps related to the latter thought, “Is there a police car in the vicinity?”

Obviously we assume that the car is being driven at or below the legal speed. If the light turns yellow as you approach the signal you have a choice to make: to brake hard enough to stop before the intersection, or to accelerate (or coast)and continue through the intersection legally before the light turns red. Unfortunately, many accidents are caused by drivers misjudging the latter (or going too fast) and running a red light.

The mathematics involved in describing the limits of legal maneuvers is straightforward: integration of Newton’s second law of motion. Suppose that the width of the intersection issft and that at the start of the deceleration (or acceleration), timet= 0, and the vehicle is a distanced0from the intersection and traveling at speedv0. If the duration of the yellow light isTseconds, and themaximumacceleration and deceleration are denoted bya+and −a−respectively, then we have all the initial information we need to find expressions for the two situations above: to stop or continue through. A suitable form of Newton’s second law relates displacementxand accelerationaas

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from which follow the speed and displacement equations


We have chosenx(0) = 0. In order to stop before the intersection in at mostTseconds, it follows from the second of these equations that


On the other hand, to continue through, we note that the vehicle must travel a distanced0+sin less thanT, so that requiringx(T) ≥d0+sin the third equation above yields the inequality


In each inequality, we have assumed that the maximum deceleration and acceleration are applied accordingly. Before treating the inequalities graphically, we write them in dimensionless form. This will reduce the number of parameters needed from five to four. To illustrate this, if we divide the distanced0bys, we obtain a dimensionless measure of distance in units ofs, namelyδ=d0/s. Similarly, we define dimensionless speed byσ=v0/(s/T), time byτ=t/T, andacceleration byα±=a±/(s/T2). Notice thatτis a measure of time in units of the yellow light duration. In these new units, the combined inequalities show that a driver may successfully choose either legal alternative provided that


Now we are in a position to discuss reasonable ranges on these parameters, starting with the physical data. Typical ranges have been taken from the literature [17]. If we adopt the range for the speed approaching the intersection as 10 mph ≤v0≤ 70 mph (or approximately 15 fps ≤v0≤ 100 fps), with the bounds 20 ft ≤d0≤ 600 ft, 30 ft ≤s≤ 100 ft, 2 s ≤T≤ 6 s, and 3 ft/s2≤ |a±| ≤ 10 ft/s2, then we find that 0.3 ≤σ≤ 16.7, 0.2 ≤δ≤ 20, and 0.12 ≤α±≤ 8.3. Since we are measuring time in units ofT, the results are relative, even thoughTitself can and does vary. InFigure 8.1we plot a generic graph for the bounds onδ(σ) (based on equation (8.6)). There are four regions to consider. The regionOABcorresponds to a domain with relatively low speed and small distance, and represents a portion of (σ,δ) space for which either option—stop or continue—is viable. Continuing, the region aboveABCis a (theoretically infinite!) region with relatively low speed and large distance, so it is easy to stop before reaching the intersection.BCDis a region with relatively high speed and large distance, and represents a domain in which a violation (or accident) is likely. Finally, belowOBDthe region corresponds to relatively high speed and small distance, implying it is possible to continue through the intersection before the light turns red.

The variousδi(σ) functions chosen here (withi= 1, 2, 3, 4, also based on equation (8.6)) are defined as follows:


As inFigure 8.1,Figure 8.2shows dimensionless position/speed graphs identifying regions of safety. Obviously this is a simplistic analysis of the stop-light problem; an experienced and careful driver will have developed some measure of intuition (and caution) concerning whether a successful “continue through” is possible. We have not considered the possibility of skids; they are likely to occur if the deceleration is too large, and poor road conditions (wet, icy, etc.) will greatly affect the required stopping distance. In addressing this problem, Seifert (1962) has suggested posting signs along the roadside indicating a speed at which it is safe to continue through or stop from that location. It’s a thought!


Figure 8.1. Dimensionless distance-speed graphs indicating regions of legal/illegal options based on equation (8.6);δ1(σ) =σ+ 3.15 andδ2(σ) =σ2/10.



Figure 8.2. Additional dimensionless distance-speed graphs based on equation (8.6) (seeFigure 8.1). Hereδ1is as inFigure 8.1, andδ3(σ) =σ− 0.94;δ4(σ) =σ2/16.6; andδ5(σ) =σ2/0.24.


Should the driver of a car try to stop or turn in order to avoid a collision? We shall examine this question for several different situations, the first being when a car approaches a T-intersection with a brick wall directly ahead across the intersection. We shall assume that the junction is free of other vehicles, so the only possibility of collision involves the car hitting the wall. Furthermore, we shall assume that there is no skidding, in which case the coefficient of friction in a turn may be considered to be the same as that in the forward direction. (Skidding would involve the coefficient ofslidingfriction, in general different from that for rolling friction.)

We can examine three possible choices [18] as illustrated inFigure 8.3: (i) to steer straight ahead and apply the brakes for maximum deceleration; (ii) to turn in a circular arc without braking (using all the available force for centripetal acceleration); or (iii) to choose some combination of (i) or (ii), such as turning first and then steering straight (or vice versa), or even steering in a spiral path. In fact, option (ii) can be ruled out immediately by means of a simple (but nontrivial) argument as follows.

Suppose that the distance of the car from the wall island that its speed at that point isv0. The force required to turn the vehicle (of massm) in a circular arc of radiuslis, but the force required to bring the vehicle to stop in a distancelis. This means that if the car can be turned without hitting the wall, it can be brought to a stop halfway to the wall. Regarding option (iii), a rather more subtle argument [18] shows that the appropriate choice is still to stop in the direction of motion. Apart from a brief discussion below, this will not be elaborated on here; instead we shall examine some other potential driving hazards. It should be no surprise that the worst highway accidents are those involving head-on collisions.

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A related problem is this: suppose that we are driving along a road in the right-hand lane for which, at the present speed, our stopping distance isD.


Figure 8.3. The three maneuver options open to the driver.


There is an obstacle ahead—it might be a repair crew, a stalled truck, or even a vehicle moving more slowly than we are (in the latter case we must adjust our speed in the calculations to that relative to the vehicle). What is the maximum obstacle widthWthat can just be avoided by turning left in a circular arc (if traffic in the adjacent lane permits this maneuver)?

We already know that the turning radius is twice the stopping distance. FromFigure 8.4we see that the angleβ=α/2. Hereβis the angle subtended by the obstacle and the vehicle before the maneuver begins, andα= arcsin (1/2) = 30°, soβ= 15°. The width of the obstacle at this distance is thereforeDtanβ≈ 0.268D. A wider obstacle cannot be avoided except by stopping.

Exercise:From the figure show that tan 15° = 2 −.


Let’s try to estimate some typical stopping distances for a range of speeds. This distance depends on the coefficient of friction (μ) between the tires and the road, and the driver’s reaction time. Theminimumsuch distanceDmcan be found by ignoring the latter, as long as one adds the “reaction time × speed” distanceDrafterward. The frictional force must reduce the kinetic energy of the car to zero over the distanceDm. Provided the wheels do not lock during the deceleration (no sliding or skidding occurs), we use the coefficient ofstaticfriction. If the wheels are locked, the braking force is due toslidingfriction, which is in general different, as noted earlier. In the case of static friction, for a car of massm, the equation to be satisfied is therefore


Figure 8.4. Geometry for turning to avoid a slower vehicle.



from which it follows that


Examining this result, we note that it is independent of the mass (or weight) of the car. It is also proportional to the square of the initial speed; thus doubling the speedquadruplesthe minimum stopping distance. The value of the coefficientμdepends on the quality of the tires and the prevailing road conditions; probably the best realistic value isμ= 0.8, but for more worn tires, or wet roads, a somewhat lower value 0.7 or 0.6 is probably appropriate (or even lower for tires in poor condition). Here are some minimal (rounded) stopping distances for various speeds, takingμ= 0.65. Also included, for illustrative purposes, isDr, the distance covered in a nominal (and somewhat slow) reaction time of one second. The fourth column is the approximate total distance (DT) required to stop at these speeds, given the above assumption.


For simplicity we now consider a minimal stopping distance of 100 ft, corresponding to a speed of about 44 mph (71 km/hr). If the vehicle is able to pass the obstacle without braking at all, this maneuver will begin after the reaction time. This means that the vehicle can pass a large obstacle of width nearly 27 feet if traffic in the adjacent lane(s) permits. But then the direction of the car will be at an angle of 30° to the original direction, a dangerous predicament to be sure!

To improve the safety of this maneuver, consider the following modification: we require that once abreast of the obstacle, the car should be moving parallel to the road in the new lane. For this case, the geometry changes a little (seeFigure 8.5). Now the car’s “trajectory” will be a sigmoidal-type shape composed of two smoothly joined circular arcs as shown. As beforeβ=α/2, but nowα= arcsin(1/4) ≈ 14.48°, soβ≈ 7.24°, meaning that the width of the obstacle should not exceedDtanβ≈ 0.127Dfor the maneuver to be executable. For a value ofD= 100 ft, this is just less than 13 ft, which allows for a few feet of clearance around a large truck.


Figure 8.5. Modified geometry for turning to avoid a slower vehicle.


Next we consider two cars approaching an intersection perpendicularly at the same speed, as shown inFigure 8.6. Suppose that each driver instinctively tries to swerve to the side by at least 45°; we will take this as a lower bound, for then they end up moving parallel to each other (if road conditions permit, of course). The angles of the truncated triangle are each 45° and therefore by symmetry the line joining each vertex to the corner of the junction makes an angle exactly half this with the hypotenuse. Recall that the minimum distance required for a “straight stop” isDand that for a circular arc is 2D. Now the radius of the arc shown inFigure 8.6isr=Dcot(π/8) ≈ 2.41D. From equation (8.8) we can compare the corresponding speeds for the circular arc (1) and the 45° swerve (2):


This implies thatv0(2)/v0(1) ≈ 1.1, that is, the speed can be about 10% greater in the 45° swerve.

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