X and the City: Modeling Aspects of Urban Life

by Adam, John A.

1.11, (specifically D = 2 here) is interesting. The dotted loop represents the position of the rainbow if there were two beams 180° apart. When = 146° the rainbow would emerge from L and move outward along the “backward” beam until = 180° and then move back toward L and continue in the counterclockwise path as shown. When there is only one beam, the rainbow would rapidly approach L (and “disappear” for a short time by remaining there for the duration of the now “virtual” loop) before rapidly reemerging from L and proceeding in the larger loop.

  But where did the angle of 146° come from? It is calculated by setting R = 0 in equation (23.3) and solving the resulting expression for (given a value for D), to arrive at

  Clearly, D > | cot δ | if is to be a real number. For δ = 138° and D = 2 we find that sin ≈ ±0.557. Since R = 0 in the second quadrant, we obtain ≈ 2.55 radians, or approximately 146°. The limiting case of no loop occurs for = 180°, and from equation (23.4) this corresponds to D = | cot δ | ≈ 1.11. This limiting case is also shown in Figure 23.5.

  Figure 23.5. Polar diagram for the position of the bow as seen by the observer, based on equation (23.3). In decreasing order of size the graphs correspond respectively to D = 2, 1.11, and 0.5. In each case the “rainbow angle” is δ = 138°. L is at the center of any circle R = constant, and the observer is located on the horizontal axis at relative distance D from the base of the lighthouse.

  Exercise: Derive equation (23.4).

  X = θ : HALOS IN LIGHTHOUSE BEAMS

  To see halos in such beams, there must be ice crystals present and we must turn around to face the lighthouse. Strictly speaking, as with the rainbows, these are slices of halos because of the narrowness of the beam. And as with solar and lunar halos, it may be necessary to block out the source of light in order to see a halo clearly. The 22° halo in particular appears to be darker inside the “rim’, so the beam is likely to appear darker nearer to the lighthouse, with a brighter spot corresponding to the halo slice. The larger halo will also be associated with a bright spot on the beam. As can be seen from Figure 23.6, the observer O should be at a distance of at least h cot θ from the base of the lighthouse for halos to be visible. The angles for the 22° and 46° halos are (not surprisingly) θ = 22° and 46°, so d > 2.48h and d > 0.97h, respectively (and approximately).

  To compute the corresponding polar plots for the halos, we set δ = 22° and 46° respectively in equation (23.3). These are shown in Figure 23.7 for D = 2. Note that, by contrast with the rainbow slices, the halos are visible for only a short period as the beam rotates over the observer.

  Figure 23.6. Geometry for halo formation in a stationary lighthouse beam.

  Figure 23.7. Polar diagram position of the halos as seen by the observer O at a distance of D = 5 units from the base of the lighthouse, using equation (23.3). In decreasing order of size the graphs correspond to δ = 46° and 22°.

  X = θ: RAINBOWS IN SEARCHLIGHT BEAMS

  How do things change for the observer when looking at a searchlight beam? The situation is rather different because the searchlight will be close to the ground and pointing upward (unless we’re thinking about Alcatraz in its halcyon days), and therefore the beam will not be parallel to the ground. The notation in this case will differ a little from that in Figure 23.3; the angle OSR between the observer-searchlight line and the beam will be denoted by μ (see Figure 23.8). The distance from the searchlight to the rainbow slice is r. The angle between the beam and the vertical direction is θ, and the angles and δ are as defined in the previous section. It follows that the angle SOR = δ − μ. Applying the law of sines to the triangle SOR we have that

  Figure 23.8. Geometry for rainbow “slice” R in a searchlight beam. The searchlight is at S and the observer is at O.

  This can be rearranged as

  From Figure 23.8 it is apparent that

  By substituting for μ from equation (23.7) into (23.6) we obtain the following expression for the relative distance R of the rainbow slice along the beam, where R = r/d:

  noting again that δ ≈ 138° for the primary bow. The polar graphs for R are qualitatively similar to those for the lighthouse beam in Figure 23.4, and will not be reproduced here. Again, “virtual” loops can appear in the plots if the angle θ is too large, corresponding the bow disappearing for a small range of the on the far side of the beam (see Harsch and Walker (1975)). This does not occur for θ < 48° approximately, as may be verified by setting R = 0 and = 180° in equation (23.8). The analysis for a secondary rainbow is of course very similar, but also will not be repeated here.

  Further comments

  There is an interesting optical illusion associated with a rotating lighthouse beam. The beam appears to change its length in a continuous fashion, attaining its maximum length when passing overhead, and reaching its minimum when the beam is pointing directly away from the observer. To understand why this is so, let’s visit a star party.

  Have you ever used a flashlight beam as a pointer to identify particular stars, star groups, or planets to interested bystanders? Perhaps there were none. No matter; you may have noticed that even if the sky is clear, the beam seems to come to an end very abruptly in a particular direction. But why do we see the beam in the first place? The explanation is the same as that for the visibility of sunbeams: dust particles and water droplets in the beam scatter the light, some of which of course enters the eyes of the observer. In Figure 23.9, the observer—you—is at the point O and the beam of light starts at L and extends (while diverging somewhat) to C and beyond. From points A, B, C, etc. in the forward direction, light is scattered toward the observer, but no matter how much the beam extends, the observer, displaced from it, will never see it extend beyond the direction OD (parallel to the axis of symmetry of the beam). When such a beam is rotating, the apparent length will change in accordance with the angle the beam makes with the observer-source line.

  Figure 23.9. Geometry of the flashlight beam. The observer is at O. The point L (not shown) is where the beam begins.

  But why can we see a lighthouse beam anyway? Or a flashlight beam, for that matter? The answers may seem obvious; it’s all about photons entering our eyes, isn’t it? Well, yes (always) and no! If a flashlight is shone directly into our eyes, we have no choice but to see it, unless we close our eyes. But if it is pointed away from us, we can usually still see it. The closer the direction of sight is to that direction, the wider the beam is, and the more scattering particles there are, and this will to some extent counteract the fact that the beam is more remote from O in that direction. However, if you look in the direction OA′ and compare the intensity of scattered light with that when you look in the direction OA (roughly 90° from OA′), you will notice that the former is considerably greater: more light is scattered in the forward direction than in the backward one. The very same phenomenon occur with lighthouse beams—a beam pointing toward the observer is brighter than when it points away from him.

  As we noted in more detail in Chapter 20, the explanation is that the particles scattering the light are sufficiently large that they scatter asymmetrically: much more light is scattered in the forward direction. This is (in case you had forgotten) essentially why the color of the blue sky can change as a result of changing proportions of dust, ash, salt particles, and water droplets in the atmosphere, and it also accounts for some of the differences between a blue sky in southern Europe, or the tropics, and a blue sky (when it occurs!) in northern England. Basically, small particles (of size ≈ 10−8 m) scatter blue and violet light most, and with almost equal intensity in all directions, whereas large particles (of size ≈ 10−6 m) scatter all colors more or less equally, but mostly at small scattering angles (i.e., in the forward direction).

  Some final comments can be made to wrap things up in this chapter. We have discussed rainbow and halo slices in lighthouse and searchlight beams, but can we really see the rainbow colors in these cases? Generally, it seems unlikely for several reasons, one being that the human eye is less sensitive to
color at low light intensities, as evidenced by the contrast between rainbows and halos produced in moonlight as opposed to sunlight. Perhaps a more substantial reason is that modern lighthouses use sodium or mercury-vapor lights (which are “nearly” monochromatic) or incandescent bulbs, though light from the latter can exhibit a reddish tinge sufficiently far from the source—for reasons the reader should be able to infer by now!

  Figure 23.10. “Minnaert cigar” for a lighthouse beam. Compare the surface near the light with the shape of the rainbow arc in Figure 23.1. Courtesy of Achim Christopher.

  Lighthouse beams are generally quite divergent, unlike the more collimated versions implied by the figures above. How does this affect the conclusions we have drawn on the basis of some rather interesting trigonometry? Instead of “slices,” the rainbows (or at least, the brighter portions of the beam) will take the shape of part of the (appropriate) curve shown in Figure 23.5. This is beautifully illustrated in Figure 23.2. The corresponding apple-shaped “Minnaert cigar” that we have discussed in Chapter 22 is illustrated in Figure 23.10.

  Chapter 24

  DISASTER IN THE CITY?

  Hollywood is very fond of making disaster movies about volcanic eruptions, earthquakes, meteoric impacts, and alien invasions, among other threats to us Earthlings. In July 1994 there was an alien invasion of sorts—on the planet Jupiter. The following news flash [41] can be found on California Institute of Technology’s Jet Propulsion Laboratory (JPL) website:

  From July 16 through July 22, 1994, pieces of an object designated as Comet P/Shoemaker-Levy 9 collided with Jupiter. This is the first collision of two solar system bodies ever to be observed, and the effects of the comet impacts on Jupiter’s atmosphere have been simply spectacular and beyond expectations. Comet Shoemaker-Levy 9 consisted of at least 21 discernable fragments with diameters estimated at up to 2 kilometers.

  In light of the impact(s) of ex-comet Shoemaker-Levy on Jupiter’s outer atmosphere the question has been raised: could it happen here on earth? As opposed to a cometary encounter, we hear more these days about the possibility of an asteroid colliding with the Earth. Such a collision could be globally disastrous of, course, particularly if the rocky body is of the order of a kilometer in size or more.

  On May 15 1996, two University of Florida students in the astronomy graduate program discovered an asteroid headed in the Earth’s direction at about 58,000 km/hr. As is standard procedure (and extremely sensible), they immediately reported their observations to the Harvard-Smithsonian Center for Astrophysics. After confirming the details—location and projected trajectory—of the object, the Center posted the relevant information about it (by then designated 1996 AJ1) on the World Wide Web. At 4:34 p.m. (GMT) on May 19, AJ1 reached its point of closest approach to our planet, 450,000 km, just beyond the orbit of the moon (400,000 km). We were safe! More recently, however, an even closer encounter occurred. On November 9, 2011, the popular site Astronomy Picture of the Day (http://apod.nasa.gov/apod/astropix.html) posted a blurry picture of an asteroid with the following description:

  “Asteroid 2005 YU55 passed by the Earth yesterday, posing no danger. The space rock, estimated to be about 400 meters across, coasted by just inside the orbit of Earth’s Moon. Although the passing of smaller rocks near the Earth is not very unusual—in fact small rocks from space strike Earth daily—a rock this large hasn’t passed this close since 1976. Were YU55 to have struck land, it might have caused a magnitude seven earthquake and left a city-sized crater. A perhaps larger danger would have occurred were YU55 to have struck the ocean and raised a large tsunami. . . . Objects like YU55 are hard to detect because they are so faint and move so fast. However, humanity’s ability to scan the sky to detect, catalog, and analyze such objects has increased notably in recent years.”

  And we were still safe!

  X = L: THE ASTEROID PROBLEM

  But there have been meteoric impacts during the Earth’s long history. Let’s revisit two such instances (without a time machine), and do a little mathematics in the process. About 65 million years ago such an encounter occurred (Alvarez et al. 1980), and it may well have caused the demise of the dinosaurs. The site where the explosive encounter is believed to have taken place is called the Chicxulub crater; it is an ancient impact crater buried underneath the Yucatán Peninsula in Mexico. Dust from the impact was lofted into the upper atmosphere all around the globe, where it lingered for at least several months. Eventually it settled back to the surface of the earth, having done a superb job of blocking sunlight and thus devastating plant and animal life. On the dark and cold Earth that temporarily resulted, many life-forms became extinct. Available evidence suggests that about 20% of the asteroid’s mass ended up as dust settling out of the upper atmosphere. This dust amounted to an average of about 0.02 gm/cm2 on the surface of the Earth. If this ancient asteroid had a density of about 2 gm/cm3 (about the density of moon rock), how large was it?

  The radius of the Earth is about 4000 miles, or 6400 km. This is 6.4 × 108 cm, so the area of the Earth’s surface is about 4 × 3 × 40 × 1016 ≈ 5 × 1018 cm2. Each cm2 contained, according to hypothesis, 0.02 gm of asteroid dust. The total mass that settled out was therefore about 1017 gm, and the mass of the asteroid was about five times this, or 5 × 1017 gm. Since the shape was almost certainly irregular, let’s replace it with the cube of the same mass; the size of the cube will differ little from the average dimension of the asteroid. A cube of side L with this mass and supposed density 2 gm cm−3 means that 2L3 = 5 × 1017, or L = (0.25 × 1018)1/3 cm ≈ 6 km. This suggests that the asteroid was, to the nearest order of magnitude, about 10 km in size. This is not unreasonable for an asteroid (though the dinosaurs might well have disagreed). The impact is estimated to have released 4 × 1023 joules of energy, equivalent to 108 megatons of TNT (trinitrotoluene) on impact. Comparing this with equivalent TNT yield of the atomic bombs dropped on Hiroshima (12–15 kilotons) and Nagasaki (20–22 kilotons) in August 1945, one can quite see why such an impact can wreak so much devastation.

  X = M: METEOR CRATER, WINSLOW, ARIZONA

  Far more recently, between 20,000 and 50,000 years ago, the Earth had another visitor. A much smaller object with diameter between 40 m and 50 m in size hurtled toward Flagstaff, Arizona, before there was a Flagstaff, or even an Arizona. Its speed is estimated to have been about 72,000 km/hr (20,000 m/s). Made mostly of iron, its density was probably about 8000 kg/m3 (question: what is this in gm/cm3, the units used above?). A cube of side 45 m made of this material would have a mass M ≈ 8000 × (45)3 ≈ 7 × 108 kg, or about 700,000 metric tons. Some studies suggest the object was smaller, approximately a sphere of diameter 40 m, with volume about 260,000 metric tons. Either way it is pretty large; the battleship Iowa by comparison displaces about 50,000 metric tons. The kinetic energy on impact for the smaller mass is 1/2 × mass × (speed)2 ≈ 5 × 1016 joules, roughly equivalent to the energy released by a thermonuclear bomb (20 tons of TNT).

  X = P: YET MORE TO BE CONCERNED ABOUT?

  Suppose that the object called 1996 AJ1(comparable in size with 2005 YU55) had indeed struck the Earth . . . what would have been the “impact,” and how does it compare with that of the Winslow meteorite? It was estimated to be about ten times the size of that one, so the volume must be roughly a thousand times greater. However, it is believed to be composed of rock and ice with an average density of about 3000 kg m-3, considerably less than the chunk of iron that created the hole in Arizona. The speed at its closest approach was about 16,000m/s, or 80% of the estimate for the Winslow object. Then the kinetic energy would have been about 103 × × (0.8)2 ≈ 240 times larger; say between two and three hundred times larger. Since the Earth’s surface is 70% water, there is a good chance it would have hit the ocean, creating huge tsunami waves. These would of course have been extremely destructive, as we know from earthquake-initiated events around the globe. Had it hit a large city such as New York or Washington, D.C., the devastation would have been widespread. To quote o
ne source [42]:

  Utilizing scaling models presented by Shoemaker (1983), one can calculate that the crater produced by the May 19 meteor would have had a diameter D = 8,500 m and a depth h = 1,200 m. The volume of the crater would have been about 34 billion cubic meters. A direct hit on Washington, D.C. would have completely obliterated the entire central region of the city. The Potomac River would have quickly filled the crater to produce a large deep lagoon.

  So what is the probability of a big chunk of rock hitting a city? It is of course extremely difficult to assess this risk, but we can obtain an upper bound (of sorts) on the probability that if an asteroid is heading directly for the Earth (so that impact somewhere is inevitable), it will hit a major population center. Note that this is very different from the probability that an asteroid will impact the Earth. According to one online source [43] (using data from January 2007) ranking the largest cities by land area, that ranked first is the New York Metropolitan area with about 9000 km2 (I am rounding these area and population figures to the nearest thousand and million respectively). With a population of about 18 million according to this source, the population density was about 2000 people/km2. The 100th city in this ranking is Jeddah in Saudi Arabia at about 1000 km2 (this is a slight overestimate); the population in January 2007 was about 3 million, so the corresponding population density was approximately 3000 people/km2. The cities in places 50 and 51 are respectively Delhi, India, and Denver, USA, with land areas about 1300 km2, though the contrast between them could not be more striking. Delhi had a population of around14 million at that time; Denver’s was nearly 2 million, with a much lower population density of course. The corresponding population densities are about 10,000 and 2000 people/km2. We can get a “handle” on the average population density from these figures using the Goldilocks principle, noting that it will likely be between 1000 and 10,000 people/km2. This gives about 3000/km2. We noted in Chapter 18 that by 2007 more than half the world’s population was living in cities. Taking a world population of 7 billion (declared by the United Nations as reached October 31, 2011), we estimate that the combined area of all cities is