Here is another example. In his analysis of the hypsometry of the continents, Cogley (1985) employed computer models to analyze enormous amounts of data. In chapter 19, we looked at topography involving one cone, one paraboloid, and three pyramids; shortly we will construct and analyze a mountain range composed of 385 well-defined peaks. However, Cogley was looking at the topography of the entire world. Consequently, in order to handle extremely large amounts of information, he had to use many methods and techniques of statistics and probability. A comprehensive, though rather advanced, reference on the use of these methodologies in geomorphology is the book edited by Chorley (1972).
A Mathematical Mountain Range
For the mountain range we want to construct, we will employ a more general form of the equation we used in our analysis of the three Great Pyramids of Egypt. That is,
where, for the three pyramids, m = 3. It is supposed that all our mountains—or hills, mounds, heaps, piles, and molehills—are pyramids with square cross-sections and the same base area. The total base area is AT. There are nj peaks in a certain category of designated height zj, and there are m height categories.
So, letting nj = Aj/AT, equation (20.3) becomes
This equation is not as formidable as it appears, as we shall see in a moment. We select the mountain range design recipe (MRDR) shown in table 20.1. In our design recipe, the number of peaks in each category is increased in a squared sequence. The heights of the peaks in successive categories are decreased in a linear sequence. Altogether there are 385 peaks including, as shown in the table, one big mountain, 9 hills,…,64 piles, and 100 molehills.
Consequently, our hypsometric equation is
TABLE 20.1
The average elevation, /zl, is obtained from the expression
For our particular mountain range, /z1 = 0.1048.
Results calculated from equation (20.5) are shown in figure 20.1. For comparison, the hypsometric curve of the earth's land surface is also shown. Using z1 = 8,847 m (Mount Everest) and = 840 m, we obtain /z1 = 0.0949. It is noted that the hypsometric curve of the simple mountain range we just now manufactured is not greatly different from that of the earth. Data provided by Cogley (1985) were utilized to construct the latter curve.
As we did in the case of the Great Pyramids of Egypt, we construct a single mountain equivalent to our 385-peak mountain range. For a change, we assume that the equivalent mountain has a circular cross-section. Likewise, we construct an equivalent circular mountain corresponding to the earth's total land mass. The profiles of these two equivalent peaks are shown in figure 20.2.
FIG. 20.1
Hypsometric curves for our 385 peak mountain range and for the earth's total land surface. (From Cogley 1985.)
FIG. 20.2
Elevation profiles of equivalent single axially symmetric peaks. Right: our 385-peak mountain range. Left: the earth's total land mass with z1 = 8,847 m (Mount Everest) and AT = 1.49 × 108 km2, corresponding to an equivalent radius re = 6,887 km on a plane surface. Average elevations for our 385-peak mountain range and the earth's land surface are also shown.
PROBLEM 1. Design your own mountain range using whatever design recipe (MRDR) you like. For the shape of the components of your mountain range you might want to use rectangular boxes, circular cylinders, or cones. Perhaps you prefer pyramids, paraboloids, or something else. For the height categories and the number of peaks within categories, you might want to use (a) an arithmetic progression, (b) a geometric progression, (c) binomial numbers, (d) factorial numbers, (e) Fibonacci numbers, (f) hailstone numbers, or (g) random numbers.
PROBLEM 2. Compute and plot the hypsometric curve and the elevation profile of the equivalent single-peak axially symmetric mountain. If you like to do things with computer graphics, perspective displays of your mountain range would be very interesting.
A Brachistochrone Mountain
Quite a long time ago, in 1696 to be precise, the famous Swiss mathematician Johann Bernoulli (1667-1748) posed the following problem:
Due to the force of gravity, a particle, initially at rest, slides without friction along a curve in a vertical plane from point P to point Q. What should be the shape of the curve to assure minimum transit time between the two points?
As was the custom in those days, mathematicians throughout Europe competed for the honor to provide answers to such problems. One solution to this problem was given by Jakob Bernoulli (1654-1705), the older brother of Johann, and another was submitted by the well-known German mathematician and coinventor of calculus, Gottfried Leibniz (1646-1716). The famous English mathematician and other coinventor of calculus, Isaac Newton (1643-1727), also solved the problem. Indeed it is said that the day after he received the problem, Newton sent the solution to a friend.
The answer to this intriguing problem is that a curve called the brachistochrone is the curve that provides quickest descent. Indeed, the name itself is derived from the Greek words brachystos (shortest) and chronos (time). As Newton pointed out, this curve is the same thing as the cycloid, which is the path described by a point on the periphery of a circle of radius a as it moves along a straight line.
In parametric form, the cycloid is defined by the equations
where θ is the parameter, whose value ranges from θ = 0 to θ = .
To obtain the answer expressed by equations (20.7), we utilize the branch of mathematics called the calculus of variations. This provides a method that mathematicians employ to determine the forms of functions providing maximum or minimum values of things. In the brachistochrone problem, minimum time is the quantity we are examining. In other problems, minimum length, minimum area, minimum energy, or something else might be the variable of interest. A suggested reference on the subject is the fairly easy to read book by Forray (1986).
Since the main subject of this chapter is mathematical mountains, it is appropriate that we construct a brachistochrone mountain. To do so, we need to modify (20.7) to the form
where (r, z) are, respectively, the radial and vertical coordinates of the profile of our axially symmetric mountain. The coordinates of the top of the mountain are (0, z0) and the coordinates of the base are (r*,0). It is easy to show that z0 = 2a and r* = a, where, again, a is the radius of the circle that generates the cycloid.
If we like, we can eliminate the parameter θ by using the second of equations (20.8) to obtain an expression for θ and substituting the answer into the first equation. The result is
This is an alternative form of the brachistochrone curve. A plot of (20.8)—or the equivalent (20.9)—is displayed in figure 20.3.
FIG. 20.3
A brachistochrone mountain of height z0 and radius r*.
Can you show, using either (20.8) or (20.9), that the slope of this curve, dz/dr, is infinite at the top of the mountain (0, z0) and zero at the base (r*, 0)? Accordingly, we would say that a cusp exists in the curve at the top of the mountain.
Now we can calculate a few things about our brachistochrone mountain.
Arc Length
Starting with the relationship ds2 = dr2 + dz2 and utilizing the parametric relationships of equations (20.8), it is not difficult to establish that the length of the curve from the top of the mountain to the base is S = 2z0.
Time of Descent
The time of descent of an object sliding from the top to the base is given by the equation
An interesting property of the cycloid is that the time of descent from any point (r, z) along the profile to the base (r*, 0) is always this same value. The noted Dutch scientist Christiaan Huygens (1629-1695) discovered this feature of the cycloid and utilized it in his tautochrone clock. In such a clock the period of the pendulum is constant regardless of the amplitude.
An object falling freely in a vacuum has the descent time . Show that an object sliding without friction along a straight line between the top of the mountain (0, z0) and the base (r*,0) has the descent time
At this point it may be helpful to introduce some numbers. Suppo
se you are ready to go on your frictionless skis at the top of our snow-covered brachistochrone mountain of height z0 = 100 m, and radius r* = (/2)z0 = 157 m. As we established above, the arc length of your trip from the top to the base is S = 2z0 = 200 m.
From (20.10), the time of your descent is = 7.09 s. Accordingly, your average velocity is = S/T = 200/7.09 = 28.2 m/s. Compare this to your velocity at the instant you reach the base,
If somehow you could have gone in a straight line from the top to the base, the distance would have been 186 m. From equation (20.11), your time of descent would have been T = 8.40 s with an average velocity = 186/8.40 = 22.1 m/s. So even though the straight-line path is shorter than the brachistochrone path (186 meters vs. 200 meters), the straight route takes longer (8.40 seconds vs. 7.09 seconds).
You might want to examine the times of descent T, arc lengths S, and average velocities , of some other descent curves. Parabolic, circular, and sinusoidal paths would be interesting cases to consider.
FIG. 20.4
The Mayon Volcano in the Philippines. (Photograph provided by Embassy of the Philippines, Washington.)
Numerous presentations about the history of the brachistochrone—and many other stories concerning the development of mathematics—are given by Struik (1967) and by Dunham (1990).
The Mayon Volcano in the Philippines
Without doubt the most perfectly symmetrical mountain in the world is beautiful Mayon Volcano in the province of Albay in the southern part of Luzon in the Philippines. This peak, shown in figure 20.4, has an elevation of 2,460 meters and covers an area of over 300 square kilometers. It has a long history of eruption. The first recorded event was in 1616. There have been over fifty eruptions since then, including one in 1814 that completely buried the town of Cagsawa with substantial loss of life. There was intense activity of the volcano in 1993.
FIG. 20.5
The profile of Mayon Volcano based on a hyperbolic sine equation.
This mountain is so symmetrical that, over the years, many mathematically inclined people have attempted to describe its shape in terms of mathematical equations. Among the first to do so was Becker (1905), who decided the peak looked like a hyperbolic sine curve. Specifically, he employed the equation
where the scaling constant c has the numerical value c = 800 m. Others have matched the volcano's lovely profile to exponential curves, cosine curves, and logarithmic curves.
To conclude our chapter on mathematical mountains, we take a closer look at the hyperbolic sine relationship expressed by equation (20.12). The elevation of Mayon Volcano is z0 = 2,460 m. Substituting this quantity and the value c = 800 m into (20.12), we compute the profile z = ƒ(r). The result, shown in figure 20.5, is displayed to the same scale as the photograph in figure 20.4.
PROBLEM 1. It is appropriate that we close this chapter with the assignment that you compute the hypsometric curve for (a) a brachistochrone mountain and (b) a hyperbolic sine mountain.
PROBLEM 2. Finally, here is a little problem designed to help prepare us for the next chapter. A hunter leaves his home one morning and goes straight south for one mile. Then he turns 90° and goes straight east for one mile. At that point he sees a bear and takes a shot at it. Wisely, he decides to go directly home. When he gets there, he determines that his entire trip was exactly three miles.
What color was the bear?
21
Moving Continents from Here to There
What Color Was the Bear?
It turns out the bear was white, because it was a polar bear, because the hunter lived right at the North Pole, because only at the North Pole could one carry out the geometrical journey described in the question. The crux of the matter, as shown in figure 21.1, is that we are dealing with a spherical triangle, not a plane triangle. You might want to examine this geometry on your globe of the world or your basketball.
When dealing with problems involving distances on the earth's surface, our computations are simplified if we use nautical miles, instead of statute miles or kilometers. The nautical mile is explained in the following paragraph.
We assume that the earth is a perfect sphere. Lines of constant longitude passing through the North and South Poles—the great circles called meridians—can be divided into 360 degrees; each degree consists of 60 minutes of latitude change. By definition, the arc length along a meridian corresponding to one minute of latitude difference is one nautical mile. The maritime unit of velocity, the knot, is equivalent to one nautical mile per hour.
Clearly, the circumference of the world is C = (360)(60) = 21,600 nautical miles. It can be established that 1.0 nautical mile = 1.1510 statute miles = 1.8520 kilometers. On this basis, the radius of the earth is R = C/2π = 3,437.7 nautical miles = 3,956.2 statute miles = 6,366.7 kilometers.
FIG. 21.1
The journey of the hunter who lived at the North Pole
How far “around the world” did the hunter go on the west-to-east (B to C) leg of his journey just before he spotted the bear? It is helpful to refer to figure 21.2 to answer this question.
FIG. 21.2
Definition sketch for spherical geometry involving the North Pole.
It is seen in the figure that sin θ = r/R. On his west-to-east segment, the hunter was walking along the line of latitude ø = 89°59'N. Accordingly, θ=1/60°. Letting C*=2πr be the circumference of this latitude circle, we obtain C*=C sinθ = 21,600 sin(1/60) = 6.2832 nautical miles. Since the hunter covered a west-to-east distance of 1.0 nautical miles, it follows that he went (1.0/6.2832)(360°) = 57.296° around the world.
Suppose the hunter, as he left his home and went straight south on leg A to B, was in fact headed directly toward San Francisco, California (longitude λ = 122°W). Then just after he took a shot at the bear and went straight north toward home on leg C to A, he was headed directly away from Moncton, New Brunswick (λ = 65°W). It's a small world.
Wherein We Construct an Enormous Mountain at the North Pole
The main reason we came to the North Pole was to supervise the moving of the earth's entire land mass to the region of the North Pole. The authorities inform us that it is necessary to arrange the land mass so that it is identical to the axially symmetric peak shown in figure 20.2. We put Mount Everest (elevation 8,847 m) at the center, on the polar axis.
Bear in mind that we also move Antarctica and Greenland, with all their ice caps and glaciers, to the pole. So we shall have a truly beautiful ice- and snow-capped mountain.
The total area of the world's above-sea-level land mass is Ac = 1.49 × 108 km2. This corresponds to a radius of 6,887 kilometers on a plane surface. However, we want the land mass to fit symmetrically onto a spherical surface. With its center at the North Pole, to what latitude will our big circular mountain extend southward?
With reference to figure 21.2, the area of a spherical cap is Ac = 2πRh. Using Ac = 1.49 × 108 km2 and R = 6.370 × 103 km, we obtain h = 3,723 km. From the figure, cos θ = (R – h)/R. This gives θ = 65.45° and so the latitude at the edge of our circular land mass is ø = 90° – θ = 24°33'N. The tropic of Cancer is located at ø = 23°27'N; consequently, the land mass extends almost to the tropics. Some cities at or near the edge of our circular island are Key West, Florida (24°33'), Honolulu, Hawaii (21°18'), Brownsville, Texas (25°54'), Riyadh, Saudi Arabia (24°40'), Karachi, Pakistan (24°52'), and Taipei, Taiwan (25°03').
What is the length of the shoreline of the circular island? If the land mass were on a plane surface the circumference would be C* = 2π(6,887) = 43,272 km. However, on a spherical surface, C* = 2πr = 2πRsin θ = 36,406 km. Clearly, this is the circumference of a circle located at latitude ø = 24°33'.
Obviously, the remaining and larger area of the world is covered by water. Thank goodness the two areas—lands and oceans—are not equal. Otherwise it would be necessary to resolve the following troublesome dilemma:
If the entire northern hemisphere were land and if the entire southern hemisphere were water, would we have an island in the
center of a large body of water or a lake in the center of a large mass of land?
Freezing Water and Melting Ice
During the past several million years, the climate of the earth has alternated between cold periods resulting in ice formation and glacial advance and warm periods causing ice melting and glacial retreat.
At the height of the most recent “ice age”—about twenty thousand years ago—vast sheets of ice covered much of Alaska, all of Canada, Greenland, Iceland, and the British Isles, and a good deal of the northern regions of Europe and Asia. In the United States, enormous ice sheets and glaciers extended over much of the North Central, Middle Atlantic, and New England states.
It has been estimated by Denton and Hughes (1981) that the total volume of ice during the ice age was about 90 million cubic kilometers—three times more than we presently have, mostly in Antarctica and Greenland. That enormous volume would provide an ice blanket 4,800 meters thick, about three miles, over Alaska, Canada, and the forty-eight states of the United States.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 15
