X and the City: Modeling Aspects of Urban Life

by Adam, John A.

  .

  How does this compare with the cross-sectional area of the Earth? This is what the asteroid will “see” as it heads toward us, and with a radius of about 6400 km, this area is π ×(6400)2 ≈ 108 km2. The ratio of the combined metropolitan area to that of the Earth’s cross section is about 10−2, so it’s not a tiny figure, relatively speaking. So on the basis of a very crude calculation, it appears that if the asteroid is heading directly for our planet, so an impact somewhere is inevitable, there’s about a 1% chance that a city will be hit. But it may not be yours! The probability of that occurring is much lower, of course. We know that about half the world’s population reside in cities. City populations in general tend to be larger than 105 and smaller than 107 (excluding really large cities, of course), so we’ll apply the Goldilocks principle one more time to obtain our guesstimate for an average population: one million. Dividing three billion by this number, we have three thousand such “average” cities. We have noted above that, given an impending direct collision, there’s about a 1% chance that a city will be hit. Dividing 10-2 by three thousand indicates that the chances of your city (or mine) being hit by an asteroid heading straight for Earth may be somewhere between 10-5 and 10-6. Does that make you feel a little less concerned? And although the surface of the Earth is about 70% covered by water, this has no bearing on our calculation; it is just the relative area occupied by city dwellers that is important here. Of course, if the impacting body is large enough, then everyone is doomed whether it hits an ocean or not.

  X = V, A: WINSLOW REVISITED

  We finish this chapter about the “somber possibility of very serious things happening” on a more light-hearted note. The Winslow crater is now about 170 m deep and 1250 m across. You’ve heard of ships in a bottle? How about building a town in a meteor crater? The base of the crater, not being entirely flat, would probably have to be leveled, but ignoring this practical concern, we assume this has been done, and calculate the base area to be just over 1 km2. This is far too small to accommodate more than a small town with the population density of the New York metropolitan area (about 2000 people/km2). Well, if not a city, how about a stadium? This would be a most appropriate place to hold rock concerts.

  We can also use integral calculus to determine the volume and surface area of the (idealized) circular crater [42] of diameter 2a and maximum depth h. We suppose that it is, as it were, a portion of the inner surface of a sphere of radius r and center at the origin O (see Figure 24.1).

  First, the volume:

  After a little algebra, and using the result

  we find that (Exercise!)

  Figure 24.1. An idealized cross section of the Winslow meteor crater.

  The area is found in a similar manner (Exercise!), thus:

  So: plugging in the values a = 625 and h = 170 we find that V ≈ 1.0 × 108 m3 and A ≈ 1.3 × 106 m2.

  Exercise: Given that the density of the ground material at the crater is about 2350 kg/m3, mentally calculate the mass displaced by the meteor.

  Suppose that an institute of higher education, Laid Back State University (LBSU) decides to build a football stadium in the crater. LBSU could be any such institution in the United States, of course. (The one at which I am employed is about 2000 miles from Winslow, Arizona. This is considered to be “local” on the scale of the Solar system.) Allowing 15% of the total area to consist of stairs, aisles, restrooms, and hot-dog stands and the rest for the football field, and allowing 75 cm2 for each seat, the crater can accommodate (0.85 × 1.30 × 106)/(0.75)2 ≈ 2 × 106 people—this is just a tad larger (a hundred times!) than the recently built football stadium at Old Dominion University, which holds just under 20,000 people at maximum capacity. We still have a way to go.

  Chapter 25

  GETTING AWAY FROM THE CITY

  After endless days of commuting on the freeway to an antiseptic, sealed-window office, there is a great urge to backpack in the woods and build a fire.

  —Charles Krauthammer

  Why would one wish to get out of the city, at least for a time? To escape the possible impending doom discussed in the previous chapter? There are many other reasons why we might wish to take a break from city life, such as dissatisfaction with continued traffic congestion and the hectic pace of life, or just a desire to experience nature in all its variety. My favorite seasons are spring and fall (autumn). The season of autumn can be one of great beauty, especially where the foliage changes to a bright variety of reds, oranges, and yellows.

  It’s a bit of an aside, but why do leaves change their colors in the fall? There are in fact several different reasons, but the most important is the increasing length of night and cooler temperatures at night. Other factors are the amount of rainfall and the overall weather patterns in the preceding months. Just like sunsets, the weather before each fall is different. Basically the production of chlorophyll slows and stops in the autumn months, causing the green color of the leaves to disappear, and the colors remaining are mixtures of brown, red, orange, and yellow, depending on the types of tree. To have any real chance of seeing the wonderful fall foliage, you have to go to the right places at the right time. Going to the beach in summer or even the fall won’t do! And going to the Blue Ridge Mountains or New England in the depths of winter will not enable you to see the fall foliage either, pretty though the snow-covered trees may be!

  But there are some other aspects of this season that are present at any time of the year, and do not depend on the leaves changing. Those aspects involve trees, rain, and, in this case, my left foot!

  Consider this: you are in the hills of New Hampshire, or West Virginia, or perhaps you are somewhere on the Appalachian Trail enjoying the glorious fall colors, when suddenly a rain squall appears out of nowhere (or so it seems). You run to take cover in a deserted shelter a hundred yards away near the trail. Playfully (there being nothing else to do) you stick your foot outside the shelter, and of all things, photograph the rain falling around it!

  After ten minutes of intense rainfall, it stops as suddenly as it started. You wonder how fast that rain must have been falling to create the scene you now survey: large puddles all along the trail, drops dripping from every available leaf above you, and the temporary dark-brown stains on the trunks of rain--soaked silver birches.

  As you set out on your way again, you start to notice the wave patterns formed when the drops falling from the branches above you hit the surface of puddles.

  Question: Which of the patterns in Figures 25.1 and 25.2 represents this situation?

  Figure 25.1. Circular wave pattern I—caused by raindrops?

  Figure 25.2. Circular wave pattern II—caused by raindrops?

  Answer: The one in Figure 25.1. Raindrops falling on the surface of a puddle generate wave patterns that are dominated by the effects of surface tension. The speed of these waves is inversely proportional to the square root of the wavelength; thus shorter waves travel faster and move out first. Note the expanding region of calm associated with and inside these waves. The other pattern is dominated by gravity, which produces longer waves with speeds directly proportional to the square root of the wavelength, so the longer waves travel faster and move out first.

  Figure 25.3. This rather pedestrian photograph (!) was taken from a sheltered area outside the Mt. Washington Resort, Bretton Woods, New Hampshire, on September 30, 2010, courtesy of Tropical Storm Nicole. Because of the heavy rain, Mt. Washington was nowhere to be seen! Nevertheless, it seemed like fun to “guesstimate” the speed of the raindrops, given that the exposure time for the shot was 1/200 second and using an estimate for the width of my sneaker (you don’t need to know my shoe size to do this.). My foot and the raindrops shown were about the same distance from the camera. You can assume that the foreshortening of the rain streaks (due to the downward angle of the camera) is not significant.

  Once home, you upload your pictures onto the computer. On noticing the picture you took back at the shelter, you realize that knowing
the exposure time of the shot, you can estimate the speed of the rain. The picture is given in Figure 25.3; can you find the speed of the rain?

  Solution: From the photograph, . The width of my sneaker is approximately 5 inches, so the raindrops traveled approximately 5/3 inches in 1/200th of a second, or 1000/3 inches/s, or 1000/36 ft/s. Therefore

  Since 1 m/s ≈ 2.2 mph, this is about 8.6 m/s.

  For those who don’t have access to my foot, anything in the range 4 inches to 6 inches wide is a reasonable estimate. This would result in a range of speeds 15–23 mph (7–10 m/s).

  Appendix 1

  THEOREMS FOR PRINCESS DIDO

  Being a princess isn’t all it’s cracked up to be.

  —Princess Diana

  We shall state two theorems that undergird Princess Dido’s “solution” discussed in Chapter 1: (i) the isoperimetric property of the circle (without proof), and (ii) the “same area” theorem (with proof). The word isoperimetric means “constant perimeter,” and here concerns the largest area that can be enclosed by a closed curve of constant length. We shall be a little more formal in our language for the statement of these results.

  Theorem 1 (the isoperimetric theorem): Among all planar figures of equal perimeter, the circle (and only the circle) has maximum area.

  There is also an equivalent “dual” statement:

  Theorem 2 (the same area theorem): Among all planar figures of equal area, the circle (and only the circle) has minimum perimeter.

  This theorem is a consequence of Theorem 1, as we can show. Consider first a closed curve C of perimeter L enclosing a domain of area A. Let r be the radius of a circle of perimeter L: then L = 2πr. This encloses a disk of area πr2. By the isoperimetric property of the circle, the area enclosed by C cannot exceed πr2, and it equals πr2 if and only if C is a circle. On the other hand, πr2 = L2/4π, so we obtain the isoperimetric inequality A ≤ L2/4π, with the equality holding only for a circle. Now we are ready for the proof of Theorem 2 assuming the isoperimetric theorem. Consider next an arbitrary planar figure with perimeter L and area A. Let D be a circular disk with the same area A and perimeter l: then A = l2/4π. If l > L, the isoperimetric inequality would be violated, so L ≥ l.

  To complete the argument for Princess Dido and the semicircular area discussed in Chapter 1, we can examine a generalization of this problem. Suppose that a curve of length L is now attached to a line segment of fixed length l, forming a closed curve. What should be the shape of L in order for the area enclosed to be a maximum? Consider two situations, the first in which L is a circular arc (enclosing an area Ac), and the second in which it is not (and enclosing an area A). Imagine also that in both cases a circular arc below the line segment l is added to give an additional segment of area a. By the isoperimetric theorem, A + a ≤ Ac + a, so that A ≤ Ac; clearly Princess Dido knew what she was doing!

  Appendix 2

  PRINCESS DIDO AND THE SINC FUNCTION

  We’ll make an alliterative foray further into this forest by generalizing Dido’s “classical” Carthage problem somewhat. Recall that she formed a semicircular arc, with the (presumed) straight Mediterranean coastline as a diameter, thus enclosing a semicircular area for the city.

  Clearly, the strip length L was shorter than the Mediterranean coastline, but what about the case where the available straight boundary is of variable length l < L; how does the area of the closed region vary with l? In fact, how do we know now that the semicircular area is still a maximum? We’ll do a little more mathematics and put in some numerical values just for fun, and work with l expressed in units of L, so that 0 < l < 1. Note that if l = 1 the enclosed area A is zero; the strip L = 1 is collinear with l, and if l = 0 the area of the circle is just 1/4π square units, half that of the area enclosed by the boundary and the completing semicircular arc. To investigate this, examine the circular segment BCD in Figure A2.1. The shaded area A(l) is enclosed by the circular arc of fixed unit length and the line segment of length l.

  The arc subtends an angle 2θ at the center O, 0 < θ ≤ π. The arc length 2rθ = 1. In terms of θ the area of the shaded segment is

  Note that, for π/2 < θ ≤ π this formula is still correct, but now cos θ < 0. A graph of A(θ) is shown below (Figure A2.2). It does indeed appear that the maximum enclosed area occurs when the region is a semicircle (θ = π/2). This is easily verified by noting that

  Figure A2.1. Sector notation.

  Figure A2.2. Area of sector as a function of θ (radians).

  vanishes when θ = π/2 and θ ≈ 4.49 (the latter is well outside the interval of interest). And

  But what is A(l)? Although (as indicated below) we cannot write down an exact expression for A(l), we can do so for A′(l), which is even better, for we can find values of θ corresponding to an extremum of A. Since

  and

  we readily obtain the expression

  which is zero when θ = π/2 or tan θ = ±θ. (Note that dA/dl is expressed as a function of θ.) A graph of this derivative over the interval (0, π) shows that it is positive in (0, π/2), zero at θ = π/2, negative in (π/2, 2.029) and positive once more in (2.029, π). Thus A(l(θ)) has a maximum at θ = π/2 and a minimum at θ ≈ 2.029. when θ = π/2, A = 1/2π ≈ 0.159 and l = 2/π ≈ 0.64.

  In using the result l = (sin θ)/θ we have made a natural connection between the two independent variables used above to express the area of the enclosed region. More precisely, the function

  is known as the sine cardinal function, shown in Figure A2.3 for the interval (−4π, 4π).

  The sinc function arises in many applications, from diffraction theory in optics to spectroscopy, information theory, and digital signal processing.

  Unfortunately, there is no nice “inverse sinc function” to which we can appeal to express θ directly in terms of l. Nevertheless, in an entrepreneurial spirit, we shall attempt to find a reasonable and invertible approximation to this function, sufficient for our purposes. Since we are only interested in the interval [θ, π]we shall use instead the function

  The graphs of these two functions are shown in Figure A2.4; f(θ) (the solid line) appears to be good enough for our purposes!

  Figure A2.3. The sinc function.

  Figure A2.4. The approximation f(θ).

  In terms of this approximate function, it follows that since l ≈ f (θ), then

  and hence from equation (A2.1),

  where . The approximation (A2.6) for A(l) is illustrated in Figure A2.5.

  Figure A2.5. The approximation for A(l).

  How does this compare with the exact results? The maximum value of A(l) according to equation (A2.6), occurs when

  The only value of interest to us is when β = π, which corresponds to A ≈ 0.159, θ = π/2 and l = 0.75. Therefore we see that the approximation is “spot on” for the maximum area and where it occurs, but it overestimates the corresponding value of l by about 17%.

  Comment: The definition (A2.3) is sometimes called the unnormalized sinc function. The normalized sinc function is defined as

  because it may be shown that

  Appendix 3

  TAXICAB GEOMETRY

  If, in New York, you arrive late for an appointment, say, “I took a taxi.”

  —Andre Maurois

  Imagine a city laid out on a standard rectangular (Cartesian) grid, with blocks defined by equally spaced N-S and E-W roads (of course any directions will do as long as they are perpendicular). Thus to move from location P to location Q we must travel along these roads; we cannot generally move in a direct line from P to Q unless we happen to be a pigeon! Let’s compare some of the properties of this somewhat “constraining” geometry with the standard Euclidean one available for the pigeons. Find some graph paper and let’s get started!