X and the City: Modeling Aspects of Urban Life

by Adam, John A.

  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. There is an interesting mathematical link that connects such malignancies with city growth—the fractal dimension. This topic will be mentioned in Chapter 18, and more details will be found in Appendix 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 above for 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!

  X = L: A CITY PERIMETER

  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 in Appendix 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 is L ft, then we have a simple equation for the area: 0.01L ≈ 100, or L ≈ 104 ft. This length would comprise the semicircular part of the city boundary. For a radius r ft,

  where A is 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. In Appendix 2 we shall generalize this idea to the case in which the boundary is of variable length l < L.

  X = V: A CITY SIZE AND VOLUME

  Let’s return to part of the quotation from John of Patmos (stated at the beginning of this book). He was in exile on the island of Patmos, off the west coast of Turkey, probably around 90–95 A.D. Recall that, according to one translation, “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.” Another translation says that “it was twelve thousand furlongs in each direction, for its length, breadth and height are all equal.” A furlong is 220 yards, there being eight in a mile, so the city was a cube 1,500 miles on a side. Perhaps with windows and doors and open spaces it was like a Menger sponge! (See Appendix 9.)

  Question: How far might an adult walk in those days without the benefit of modern transportation?

  I want to estimate this one! At 3 mph for 10 hours, 30 miles a day for 50 days is, guess what, 1,500 miles!

  Question: What was the volume of the heavenly city in John’s vision? In cubic miles? In cubic kilometers? In cubic furlongs? Estimate how many people it could accommodate.

  An estimation question that can be generalized from the rest of the quotation—the number of leaves on a tree—can be found in Chapter 5.

  Chapter 2

  GETTING TO THE CITY

  X = y: BY PLANE

  Before we can explore mathematics in the city, we need to get there. The fastest way is to fly, so let’s hop on a plane. If we assume that the descent flight path occurs in the same vertical plane throughout (i.e., no circling before touchdown—which is very rare these days), then the path is quite well represented by a suitably chosen cubic function [5], as we will see below. Here are some further conditions we impose prior to setting up a mathematical model for the path:

  (i) The horizontal airspeed dx/dt = U is constant throughout the descent. This is somewhat unrealistic, but we’ll work with it anyway.

  (ii) The descent begins when the plane is at the point (−L, h), the origin being the beginning of the runway.

  (iii) The magnitude of the vertical component of acceleration must not exceed the value k, where 0 ≤ k g, g = 32.2 ft/sec2 being the acceleration due to gravity.

  (iv) The plane is considered for simplicity to be a point mass. The ultimate in cramped seating!

  The landing path is modeled by

  What kind of boundary conditions should be imposed? Well, we would like to have a smooth touchdown, which means that not only do the wheels of the plane touch the runway at landing, they should have no downward component of velocity as they do so. That’s the kind of landing that generates applause from the passengers! And at the beginning of the descent we would like to move smoothly down toward the airport, so a similar set of conditions applies, namely, that

  Exercise: Show that these conditions imply that the equation of the flight path is

  A graph of Y = y/h vs. d = x/L is sketched in Figure 2.1.

  The vertical component of the airplane’s velocity is

  and the corresponding acceleration component is.

  Since ay is an increasing function of x, 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 path Y(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, both U and h are “large” and k is “small,” so this implies that L must be relatively large (compared with smaller airplanes). Suppose that U = 600 mph, h = 40,000 ft, and L = 150 miles. Then a further rearrangement of (2.4) gives a lower bound for k of approximately (in ft/s2)

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

  Therefore we expect L and k to be smaller than in the example, while h is 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 of h and L and 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 by y = A + B tanh C(x − D) (where A,B,C, and D are 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 . . .

  X = A: THE MAP

  Why is the place I’m looking for on a map so often at its boundary? Let’s investigate. Consider first a map consisting of a rectangular sheet of relative dimensions 1 and b < 1, including a margin of width a < b. The area of the sheet is b square units and the relative area A of the margin is

  For a square map with unit side, A(a) = 4a(1 − a). A has a maximum at a = 0.5; the map is all margin! More realistically, if a = 0.1, A = 0.36 or 36%, more than a third of the whole area is margin! Furthermore, the margin is 9/16 or about 56% of the map area!

  Given a road atlas of the same relative dimensions, for the two pages combined,

  Taking b = 1 and a = 0.1 as before, A = 0.52, less than half the sheet is map!

  X = $$: BY TAXI

  Without finding my destination using the map, I decided to hail a taxi, but it seemed as if we drove forever. I’m sure we passed some places twice. Perhaps the driver took me on a world trip, which leads me to the following question . . .

  Question: How many taxi rides would it take to circle the Earth?

  Perhaps only one, if the driver is willing . . . more generally, we need to estimate the average length of a taxi ride and the circumference of the Earth. Let’s start with the Earth. We might remember the circumference. If not, we might remember that the radius of the Earth is about 4000 miles and then calculate the circumference using C = 2πr. On the other hand, if we know that it is about 3000 miles from New York to Los Angeles and that they are three time zones apart, then 24 time zones will give you a circumference of 24,000 miles. This does seem a little more obscure, though.

  Now let’s consider a typical taxi ride. To go from downtown to Manhattan’s Upper East (or West) Side is about 80 blocks or four miles [6]. Alternatively, the shortest ride will be about a mile and the longest will be about ten miles, so we can take the geometric mean1 and estimate three miles. At three miles per ride, you will need about 8,000 taxi rides to circumnavigate the globe. Hey, that’s about one per squirrel! (This is a teaser to keep you reading; see Chapter 6.) At $2.50 for the first 1/5 mile and $0.40 for each additional 1/5 mile, that will cost you $20,000 plus $45,000 = $65,000 in cab fare (plus tips and waiting time). By the way: good luck hailing a cab in the middle of the Atlantic.

  Since we’ll be talking about driving in the city, and either flying or driving to get there, let’s also apply these principles in that context and do some elementary “risk analysis.”

  Question: What is the risk (in the U.S.) of dying per mile traveled in a car? [7]

  Okay: how far do Americans drive in a year? Warranties for 5 yr/60,000 miles are very common these days, so it’s clear that car dealerships reckon that a typical domestic vehicle is driven about 12,000 miles per year on average. Alternatively, one can count the miles covered per week, halve it, and multiply by 100 (why?) Now there are a little over 300 million (3 × 108) people in the U.S., about half of whom drive (and maybe only half of those drive well), so the total mileage per year is about 1.5 × 108 × 1.2 × 104 ≈ 2 × 1012 miles. As is readily checked, there are about 4 × 104 deaths due to car accidents in the U.S. each year, so the risk of death per mile is 4 × 104 ÷ 2 × 1012 = 2 × 10−8 deaths/mile!

  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 × 106 deaths/yr, so the total (lifetime) probability of dying in a car crash is 4 × 104 deaths/yr ÷ 4 × 106 deaths/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 109 people-flights per year; the actual figure in 2005 was 6.6 × 108, so we’ll use the slightly more accurate value of 7 × 108 flights/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 103 miles, so we travel about 7 × 1011 mi/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 gives about 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 × 10−11 deaths/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?”

  1 This, by the way, is using the Goldilocks 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.

  Chapter 3

  LIVING IN THE CITY

  X = N: A “ONE-FITS-MANY” CITY “BALLPARK” ESTIMATE

  Suppose the city population is Np million, and we wish to estimate N, 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) is R, and that the facility is open on average H hours/day and caters for an average of C customers per hour. We shall also suppose there are D days per year. This may seem a little surprising at first sight: surely everyone knows that D = 365 when the year is not a leap year! But since we are only “guesstimating” here, it is convenient to take D to 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 take R = 1, C = 5, H = 8 and D = 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 use R = 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 pick C = 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 of H = 10. Combining everything as before, with D equal to 400 now (such establishments are definitely open on weekends!), we find for the same size city