Our analysis of hybrid corn adoption in Iowa is similar to the one we carried out for the population of California. The results are U0 = No/N* = 0.0096 and a = 1.12. Our problem of hybrid corn is easier to analyze than the problem of California's population because, from Griliches' data, we already know that N* = 100%.
Substitution of these numerical values of U0 and a into equations (23.7) provides the inflection point: ti = 4.76 (year 1937.76), Ui = 0.50, and (dU/dt)i = 0.28 per year. Substituting these same numbers into (23.6) gives the solid lines of figures 23.3(a) and 23.3(b).
TABLE 23.2
Adoption of hybrid corn in lowa, 1933–1942
Amounts are ratios of hybrid corn acreage to total corn acreage.
Year t U = N / N*
1933 0 0.01
1934 1 0.02
1935 2 0.05
1936 3 0.15
1937 4 0.30
1938 5 0.52
1939 6 0.73
1940 7 0.90
1941 8 0.97
1942 9 0.98
Source: Griliches (1960).
FIG. 23.3
Adoption of hybrid corn in lowa. The year 1933 corresponds to t = 0. (a) Arithmetic and (b) semilogarithmic plots.
Again, we can say that the arctangent-exponential equation describes the adoption of hybrid corn in Iowa rather nicely.
The Loxodrome Curve on a Sphere
We now shift the time origin of equation (23.6) to the right by an amount equal to the inflection point time, ti. In other words, t = 0 now passes through the inflection point. Consequently, (23.6) becomes
This is a remarkable answer! Does this equation look familiar? It should; it is essentially the same equation we obtained for the loxodrome curve in chapter 22. Recall that the loxodrome is a curve that maintains a constant angle with the meridians on a sphere. On a Mercator projection it appears as a straight line.
In that chapter we determined that the equation of the loxodrome passing through the point λ = 0, ø = 0 on a sphere (where λ and ø are the longitude and latitude) is
in which θ is the constant angle between the curve and the meridians.
This is a very interesting result. Comparison of equations (23.11) and (23.12) provides the following list of analogous quantities:
This analogy says that the growth of the population of California and the adoption of hybrid corn in Iowa are described by the same mathematics as the loxodrome curve on the surface of a sphere.
In other words, the latitude ø, measured from the South Pole, of a point on a loxodrome curve on a sphere corresponds exactly to the magnitude of a growing quantity (e.g., the population of California) and to the amount of new technology adoption (e.g., hybrid corn in Iowa). The longitude λ of the point on the loxodrome corresponds to the growth or adoption time.
On second thought, the correspondence or equivalence of the quantities is not that remarkable; it is really more a fortunate coincidence. We arbitrarily selected the arctangent-exponential for the growth-adoption problems and so we picked up the loxodrome analogy. Had we stayed with the logistic equation the analogy would not have appeared.
Nevertheless, the above result does bring out something that is very important; the recognition and utilization of mathematical analogies. Such analogies provide extremely useful methodologies for the applied mathematician, engineer, and scientist.
There Are All Kinds of Mathematical Analogies
Since there is an almost endless list of mathematical analogies, we shall look at only three basic kinds. However, even these three provide a large number of interesting and very helpful applications. An excellent reference on analogies is the book by Soroka (1954).
Mechanical and Electrical Circuit Analogies
The mathematical basis for these analogies is Newton's second law (for mechanical circuits) and Ohm's and Kirchhoff's laws (for electrical circuits).
The mechanical circuit features a system comprised of a mass (m), a damping chamber (c), and a spring (k). The displacement (x) and velocity (v) of the oscillating system occur as the result of an external force (F).
With precisely the same wording, the electrical circuit features a system comprised of an inductance (L), a resistance (R), and a capacitance (C). The electrical charge (q) and current (i) in the oscillating system occur as a result of an electrical voltage (E).
These mechanical and electrical circuits find exact mathematical analogies in structural engineering (oscillation of suspension bridges), naval architecture (pitching and rolling of ships), biology (prey-predator phenomena), economics (supply, demand, and investment functions), and numerous other disciplines. Currently, there is a good deal of interest in these topics in connection with nonlinear dynamics and chaos theory. MacDonald (1989) discusses the problem of “damped linear oscillators with a delayed restoring force.” Thompson and Stewart (1986) consider phenomena involving “forced nonlinear oscillators with periodic and strange attractors.”
Soap Film and Membrane Analogies
There are many very useful applications of analogies involving soap films and thin membranes.
For example, suppose that a soap film or flexible membrane covers the horizontal top of an otherwise open conduit of arbitrary cross-section. A slight air pressure beneath the film raises the film upward except along the conduit boundaries. The deflection of the film above the original datum is measured. Though it is far from obvious in our brief consideration, what has been achieved with these measurements is a solution to the Laplace equation. This equation is extremely important in many problems of electrostatics, elasticity, heat conduction, porous media flow, and hydro- and aerodynamics.
Another application of the soap film analogy relates to the branch of mathematics called the calculus of variations. As mentioned in an earlier chapter, this subject is concerned with problems involving minimum length or minimum area or minimum energy. For example, it can be shown that the “catenary of revolution” is the minimum area of a soap film connecting two coaxial circular hoops of different diameters. A good deal of useful information concerning soap bubbles and soap films is given by Isenberg (1992). Soap films are also discussed by Kappraff (1990) in connection with the problem of minimal total lengths of highway networks.
Laminar and Potential Flow Analogies
Over the years, many investigators have employed electrolytic tanks and electrically conducting paper to solve all kinds of problems dealing with so-called potential flow phenomena. A potential flow is one in which the velocity is directly proportional to the gradient of the potential (e.g., pressure).
Briefly, in an electrolytic tank (or on conducting paper), electrodes are placed along specified portions of the boundary of a shallow tank containing an electrolyte (e.g., salt water). An object (e.g., an airfoil cross-section) made of nonconducting material is placed in the center of the tank. A certain voltage is applied to one electrode; the other electrode is grounded.
A probe is then employed to measure the voltage at any point in the tank, especially near the object being studied. Connection of the points of equal voltage determines the positions of the equipotentials. Appropriate changes of the locations of the boundary electrodes and another round of voltage measurements establishes the locations of the streamlines. In this way, the entire flow field around the object is determined.
Another very useful mathematical analogy is provided by the Hele-Shaw cell. Although this analog has been known for a long time, there is considerable renewed interest in this methodology because of current interest in the subject of ground water recharge of aquifers and secondary oil recovery in petroleum reservoirs.
A Hele-Shaw cell consists of two glass plates separated by a very small gap. A highly viscous fluid occupies the gap space. Indeed, two or even three immiscible fluids, of different densities and viscosities, can be employed to provide solutions to many kinds of difficult problems. When the fluids are given different colors, the patterns created yield spectacular dynamic art.
As in the case of
electrolytic tanks, a test object (e.g., the sheet-piling configuration under a dam) is installed in the cell and appropriate boundary conditions are applied along the edges. Colored dye streams show the flow patterns. With this information, velocity and pressure distributions can be computed.
Current renewed interest in Hele-Shaw cells is due to the fact that there is exact analogy between Poiseuille's law (which governs the behavior of laminar flow between closely spaced plates) and Darcy's law (which governs the behavior of flow in porous media). This analogy has been known for a great many years. However, in recent times, applied mathematicians and petroleum engineers have become interested in problems of instability of the interface between a driving fluid and a driven fluid as they flow through the cell. Such instabilities produce complex patterns of viscous fingering and fractal structures. This is an important consideration in problems of effective oil recovery in petroleum reservoirs.
An easy to understand article about Hele-Shaw cells is provided by Walker (1987); a not-so-easy one is given by Schwartz (1988).
24
How Long Is the Seam on a Baseball?
Or, if you prefer, how long is the groove on a tennis ball? Now we all know that (a) this is not one of the great unsolved problems of mathematics and (b) you will enjoy the World Series or Wimbledon just as much never knowing. However, if you welcome an opportunity to deflate the egos of certain friends or relatives who, by their own admission, know practically everything there is to know about baseball or tennis, your moment has arrived.
We have long been aware of the fact that the seam on a baseball, the groove on a tennis ball, and the dimples on a golf ball, enormously affect the aerodynamics of ball flight. So a study of these geometrical features of sporting balls certainly relates to the very interesting problem of ball trajectories.
In the following sections, we examine a number of relationships of spherical geometry. For these endeavors, it may be helpful if you have the following items: a single-color plastic ball six to eight inches in diameter, some marker pens (two or three colors, easily erasable), a centimeter scale, and some string.
Neatly glue pieces of string onto the ball to establish an equator and four meridians spaced at 90°. Mark the poles N and S. Measure the circumference of the ball, C, or diameter, D, and compute the radius, R.
FIG. 24.1
Definition sketch for a spherical triangle.
The mathematics of our baseball seam problem is not very difficult. Mostly, it involves only spherical trigonometry; later on we use a bit of differential calculus to compute some slopes.
The following two fundamental laws are utilized in our analysis; the various symbols are defined in figure 24.1.
The law of cosines states that
and two similar equations for cos b and cos c.
The law of sines states that
A photograph of a baseball, a tennis ball, and a golf ball is shown in figure 24.2. Some of the physical features of the three types of balls are listed in table 24.1.
It will be helpful if you have, in addition to the plastic ball mentioned above, a baseball, a tennis ball, and perhaps a golf ball. You might even want to acquire a basketball.
Now a word about the all-important quantity designated “minimum latitude, ø0” in table 24.1. With your centimeter ruler or scale, carefully measure, on your baseball, the distance between the seams at the point where the seams come closest together. You should get about 3.2 centimeters. Next, measure the entire circumference of the baseball; the result is approximately 23.0 centimeters. Since there are 360° along the circumference of the ball, we define the arc angle 2ø0 = (3.2/23.0) (360) = 50.08°. Therefore, ø0= 25.04°, which we round off to ø0 = 25°. Similar measurements on a tennis ball give ø0= 40°.
FIG. 24.2
Sporting balls: base, tennis, and golf.
TABLE 24.1
In the photograph of figure 24.2, the North Pole is at the top and the South Pole at the bottom. The minimum seam gap and the minimum latitude you just now determined is at the left edge of the ball. This corresponds to longitude λ = –90°. A vertical line (not shown) halfway across the ball defines the meridian, λ = 0. The right edge is λ = +90°.
FIG. 24.3
Definition sketch for circular arcs on a sphere.
The equator, of course, is the great circle whose plane is perpendicular to the polar axis. North and south latitudes, ø, are measured from the equator along great circles passing through the poles. These are meridians. In case you have forgotten, a great circle is one whose plane passes through the sphere's center.
We Begin the Analysis of the Problem
The definition sketch for our problem is shown in figure 24.3. The North Pole, N, is at the top. Point A′, the one you just determined, is located at ø = –90, ø = ø0.
The following is very important. With baseball in hand and oriented as in figure 24.2, convince yourself that the extreme right-hand point of the seam is located at λ = 90 – ø0, ø = 0. This point is labeled A″ in figure 24.3. Equally important is the fact that the seam must pass through the point λ = 0, ø = 45. These two important features are required because the remaining portion of the baseball's entire template must be identical to the portion we are analyzing.
Here we go. With attention to figure 24.3, we are going to pass a circle—though not a great circle—through points A′ and T and another circle through points T and A″; T is the meeting point of the two circular arcs comprising the seam.
The arc radius of each of these two “small circles” is p. The center of one is located at λ = –90, ø = ø0 + ρ; the center of the other is at λ = 90 – (ø0 + ρ), ø = 0. The two centers are labeled C′ and C″.
A short pause follows. On the plastic ball you have elaborated with an equator and four meridians made of string, label with your marker pen the points A′, T, and A″. Why not use ø0 = 25, because that's a baseball. Since you know the circumference of your own plastic ball, you can easily calculate the arc distance corresponding to 25°.
Now we look at the spherical triangle C′NT, where N is the North Pole. From the law of cosines, equation (24.1), we have
By the way, in our analyses, we will frequently be using the following relationships:
Remember also that sin 0 = 0, sin 90 = 1, cos 0 = 1, and cos 90 = 0.
You should demonstrate, from equation (24.3), that the formula to compute the arc radius of the circle centered at C′ is
where sec x = 1/cos x and tan x = sin x/cos x. Substituting ø0 = 25 into (24.5) gives ρ = 47.57. This establishes the location of point C′.
FIG. 24.4
Definition sketches for (a) angle a and (b) radius r.
A similar analysis involving spherical triangle C″ NT again yields (24.5) and ρ = 47.57; this locates points C″. Now you can show points C' and C″ on your plastic ball. Point C′ is at λ = –90, ø = 72.57, and point C″ is at λ = 17.43, ø = 0.
Next, as accurately as you can, use your scale to draw circles centered at C′ and C″, with arc radius ρ = 47.57. This is easy. Just compute for your plastic ball the arc length corresponding to 47.57°. With your marker pen, locate 10 or 12 points measured from C′ and from C″ and connect these points to make the two circles.
What you have constructed should look like figure 24.3 but better, because you are in three dimensions. If you have been reasonably careful, the two circles should meet at point T. An interesting result: The curve A′TA″ is one-fourth of the baseball seam. Show it on your plastic ball with a different colored marker pen.
We Continue the Analysis of the Problem
As the next step in our analysis, we again look at spherical triangle C'NT. This time we have to analyze it twice; the first time for the case in which ø0 < 45 and the second for ø0> 45. A definition sketch for the first case is provided in figure 24.4(a).
For this analysis, we use the law of sines given by (24.2):
From this result we easily determine that the angle α* shown in figur
e 24.4(a) This tells us that the central angle, α = 180 — α*, is
For the case ø0 > 45, again using spherical triangle C′NT, we get
We note that the arc length from A′ to T is
in which, as seen in figure 24.4(b), r = R sin ρ. This equation says that the length of the circular arc from A′ to T is simply the proportion, α/360, of the circumference of a plane circle of radius r, where r = R sin ρ.
Since 360° is equal to 2π radians, equation (24.9) gives S1 = R α sin ρ. This is the length from A′ to T. By symmetry, it is also the length from T to A″. So the total length from A′ to A″ is S2 = 2S1 = 2Rα sin ρ. This is the length of the curve drawn on your plastic ball. Since this is one-fourth of the total, the length of the entire seam is
We Obtain the Answer to the Problem
We established that ø0 = 25 for a baseball and, from equation (24.5), we computed that ρ = 47.57. Therefore, sin ρ = 0.7381. Also, from (24.7), the central angle is α = 106.67° = 1.8617 radians. Substituting these numbers into (24.10) gives S/R = 10.994.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 18