X and the City: Modeling Aspects of Urban Life

by Adam, John A.

  X = Pr: PROBABILITY OF BITING INTO . . .

  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 probability P of 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!

  Chapter 5

  GARDENING IN THE CITY

  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’m ashamed 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!).

  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 . . .

  X = N: PROBLEM

  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 is N = 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 × 104 inches, 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 maturing and (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 of combinations, nCr, which represents the number of ways of choosing r items from a total of n (without regard to order). This is defined as

  where for a positive integer n, for example, 5, 5! = 5 × 4 × 3 × 2 × 1 = 120. Furthermore, by definition, 0! = 1. Now P(r) is the probability that only r events will occur, and is given by the number of ways the event can occur, multiplied by the probability that r events 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 104 sq 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 [see Chapter 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?

  If W and Wnew are 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 (4π times 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 = 10−4 m2. Dividing 3 m2 by this quantity gives us about N ≈ 3 × 104 leaves.

  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 twig
s.

  Chapter 6

  SUMMER IN THE CITY

  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 about N = (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!

  X = X : SUNBATHING IN THE CITY

  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 does not block 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 much time you 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 SPF X cuts 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 SPF X the corresponding fraction of blocked radiation may be written in the form 1 − , where X = 2n. Taking logarithms to base 2 we find that n = log2X, or, using the change of base formula with common logarithms,

  Let’s now go back and compare X = 15 and 30.

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

  For SPF 30, , so or 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 .

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

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

  X = t: JOGGING IN THE CITY

  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 is exactly 8 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 function f (t) on a closed interval there is at least one value of t for which the tangent to the graph is parallel to the chord joining its endpoints (why not sketch this?). This means that if f is the distance covered as a function of time t, 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 that t(x) is a continuous function representing the time taken to cover x miles. 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), then t(x) − 7.5x = 0 when x = 0 and x = n, i.e. t(n) = 7.5n. If she never covered any continuous mile in 7.5 minutes, then it follows that the new function T(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 for x = 0, 1, 2, . . . , n we can write the following sequence of n inequalities: 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 inequality t(n) − t(0) > 7.5n. This contradicts the original assumption that t(n) − t(0) = 7.5n, and so we can say that she does cover 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 redefining t(x) to be:

  where m is not an integer and ε > 0. If ε is small enough and m is 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 if t′(x) > 0, as it must be because it is a time function. Mathematically this will be so if

  Figure 6.1. Lindsay’s time function t(x).

  for this particular model of the jogging time. Figure 6.1 shows t(x) and the mean time function for the case ε = 1, m = 0.8.

  Now it follows that

  is an increasing function that can never be 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.

  X = T : RAINING IN THE CITY

  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 or run in the rain? The decision 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 at v m/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 is d km, and that the rain is falling at a rate of h cm/hr (or h/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 distance d m at v m/s, the time taken to reach the shelter is d/v seconds, and in this time the amount H of rain that has fallen is given by the expression H = hd/3600v cm. If I run at 6 m/s (about 12 mph), for example, and d = 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 . . .