On quite another topic and as we established in chapter 1 the perimeter p of a regular five-pointed star of radius R is . The dimensions of playing cards are not greatly different from the ratio ø. Most books seem to have this proportion, and virtually all the flags of the world have length:height ratios close to ø:1.
The golden ratio or golden number, ø, appears frequently in other quite unexpected places. Quite a few of these are discussed by Huntley (1970); here is one more. Consider a square of unit side length as shown in figure 9.3. Another square of side length x is removed from the first square. The center of gravity of the L-shaped area is at 0. What is the value of x?
This problem, proposed by Lord (1995), is solved as follows. We compute the torques (or moments) about axis 00' (or axis 00"). Equating the clockwise torques to the counterclockwise gives
FIG. 9.3
Definition sketch for a balanced area.
which yields
x2 + x – 1 = 0.
The solution to this quadratic equation is . This is the length of the rectangular leg; its width, of course, is (1 – x). A bit of algebra shows that the length-width ratio x/ . Thus, the two rectangular legs of the L-shaped area are golden rectangles: an interesting result. You might want to test this answer by very carefully cutting the L-shaped area from a piece of stiff cardboard and balancing it on a needle at point O.
Finally, as illustrated in figure 9.4, suppose we start with a relatively large golden rectangle of length L and width H. We mark off the corresponding square of side length L – H. This leaves another but smaller golden rectangle. We again mark off the corresponding square. This process is repeated until we begin to approach the pole, O. Then a smooth curve is drawn through the corners of each of the squares. It turns out that this curve is what is called an equiangular spiral or logarithmic spiral. It is surely one of the most beautiful curves in all of mathematics. Let us take a closer look at it.
The Golden Section and the Logarithmic Spiral
In the branch of mathematics called analytic geometry, two basic systems are utilized to display and mathematically describe plane curves. One is the rectangular coordinate system, in which any point p is specified by its rectangular (x, y) coordinates. The other is the polar coordinate system, in which a point is identified by its polar (r, θ) coordinates. We shall use polar coordinates in our present analysis.
FIG. 9.4
The golden rectangle: a geometrical basis for the logarithmic spiral curve.
The pole O and a portion of the curve illustrated in figure 9.4 are shown in figure 9.5. The equation of the logarithmic spiral is
in which (r, θ) are the polar coordinates, ro is the value of r when θ = 0, and α is the angle between the radius vector r and the tangent to the curve at any point P; the symbol cot α is the cotangent a = 1/tangent α. The fact that this angle is constant all along the curve is the reason that it is sometimes called the equiangular spiral.
It is not difficult to show, using equation (9.8), that the length of the logarithmic spiral is
where sec α = secant α = 1/cosine α.
There are numerous interesting features of this beautiful curve. Perhaps the most noteworthy is that the shape of the curve does not change as it increases in size. In nature, this remarkable property is vividly displayed in the Nautilus sea shell. As seen in figure 9.6, this shell has the shape of a logarithmic spiral. Other kinds of sea shells, the horns of various types of animals, and growth patterns of sunflowers and other plants all have these spiral shapes. These and related topics are discussed by Huntley (1970) and by Thompson (1961).
FIG. 9.5
Definition sketch for the logarithmic spiral.
We conclude our brief look at logarithmic spirals with a numerical example involving naval warfare tactics.
EXAMPLE: A DESTROYER INTERCEPTS A SUBMARINE. A destroyer sights an enemy submarine at a distance L = 4.0 nautical miles. Knowing that it has been sighted, the submarine quickly dives and heads off at maximum velocity Us = 10 knots, in a certain direction θ. The velocity of the destroyer is Ud = 30 knots. Assuming that the submarine's velocity and direction do not change, what should be the destroyer's tactics to assure an interception regardless of the direction the submarine takes?
Our example is illustrated in figure 9.7. The points D and S identify the locations of the destroyer and the submarine at the moment of initial visual contact. Now the submarine, initially at point S, immediately dives after visual contact and proceeds along some constant course, θ, at a velocity Us = 10 knots. The destroyer, initially at D, should head directly for point S at a velocity Ud = 30 knots and maintain that course for 3.0 miles, that is, 6 minutes. (Why? Hint: if the submarine were to select the course θ = 0, (that is, headed directly for the destroyer, interception would occur at point C.)
FIG. 9.6
A logarithmic spiral in nature: the Nautilus sea shell.
When the destroyer reaches point C, it should make a sudden radical course change and head initially in the direction θ = α = arccos(Us/ Ud) = 70.5°. Thereafter, the destroyer should follow the continuously changing course defined by the logarithmic spiral of equation (9.8) in which ro = 1.0 miles and cotα: = 0.3536. The spiral course of the destroyer is shown in figure 9.7.
It should be apparent that the destroyer will intercept the submarine regardless of the direction θ of the latter. You can check your calculations by using equation (9.9) to compute the distances covered by the destroyer. With this information, you can calculate interception times. A typical result: If the submarine were to select a course θ = 270°, interception would occur at a distance 5.3 miles from the original position of the submarine, 32 minutes after initial contact. The destroyer would have traveled a distance of 15.9 miles.
FIG. 9.7
The logarithmic spiral. Example: Naval warfare tactics.
A reminder: When computing, make certain you express θ in radians, not degrees. Recall that 1.0 radian = 360/2π= 57.3°.
Examining Hailstone Numbers
After investigating two quite ancient sequences of numbers—the prime numbers and the Fibonacci numbers—we now turn to a sequence that is relatively new: the so-called hailstone numbers.
These numbers are generated by an extremely simple mathematical process in what is called the 3N + 1 problem; it is also known as the Collatz problem and the Syracuse problem. By whatever name, it seems to have started in the 1930s, went away for a while, and then came back with renewed vigor in the early 1970s. For quite a few years it attracted the interest and efforts of many mathematicians in numerous American universities. Indeed, the joke went around in those days that the 3N + 1 problem was planted in mathematics departments by enemy agents in a diabolical attempt to divert mathematicians from serious important research. To date, the basic problem has not been solved, so people are still working on it though perhaps not as vigorously as they were before.
In any event, here is a description of the 3N + 1 problem and how it generates hailstone numbers. Select a positive integer, preferably small at the outset, for instance, less than 10 or 20. If it is an odd number, multiply it by three and add one (that's the 3N + 1 thing); if it is even, divide it by two. Keep repeating this process until you cannot go any further. Let's try a few numbers and see what happens.
a. Start with N = 3. Then, applying the rules (if odd 3N + 1; if even N/2) we generate the sequence 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4,
b. Start with N = 5. We obtain 5, 16, 8, 4, 2, 1, 4,
c. Start with N = 7. We obtain 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1,
So far, we have tried three starting numbers: 3, 5, and 7. In each case, our computations terminated when we entered a 4, 2, 1, 4, 2, 1 cycle. For the N = 7 case, we reached a maximum value of 52 and went through 16 computational steps (i.e., a path length of 16) before we reached the 421421 endless loop. Does this always happen regardless of the magnitude of the starting number? Perhaps we should begin with a larger number.<
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d. Start with N = 25. We obtain the sequence 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. That's more like it! This time we reached a maximum value of 88 and a path length of 23: an improvement but not good enough.
FIG. 9.8
The hailstone numbers with starting number N = 27.
e. Start with N = 27. Well! What happened? As you will want to confirm, in this case the maximum value of the sequence was an amazing 9,232 and the path length was 111. A plot of this N = 27 sequence is shown in figure 9.8. We note that the plot displays several intermediate maximum values, notably 1,780 and 7,288, before it reaches the peak of 9.232 and then plunges to N = 1. Indeed, the entire plot of figure 9.8 looks like the great Wall Street crash of October 1929.
Incidentally, the reason these sequences are called “hailstone numbers” is because they behave something like the hail particles we associate with those enormous cumulonimbus clouds we call thunderheads. Small-diameter hail particles are caught in strong updrafts of air and are carried to great heights, growing in size as they do so. At high altitudes, these hail particles tumble out of the main current of the updraft and begin to fall, still growing in size. Near the base of the massive cloud they are again transported upward. This rising and falling pattern is repeated a few times. Finally, the hail particles become so large and heavy that they fall to the earth—something like our hailstone numbers going up and down several times before finally crashing to N = 1.
Up to here we have demonstrated that maximum values and path lengths of hailstone sequences depend somewhat on the magnitude of the starting number. More importantly, regardless of where the sequence starts, invariably it seems to get caught in the 421421 cycle and falls to N = 1.
We wonder if this is always the behavior. Does the sequence always terminate with 4, 2, 1, or could it, for some starting numbers and possibly after many ups and downs, nevertheless go off to infinity? Mathematicians have not yet been able to answer this question; the general solution remains an unsolved problem. Researchers at the University of Tokyo have tested every starting number up to 1.2 trillion; the sequence produced by every one of them sooner or later enters the 421 loop and ends up at N = 1. However, the conjecture that this always happens, regardless of the magnitude of the starting number, has not yet been theoretically established. For our general information, the paths lengths and successively larger maximum values of sequences corresponding to values of starting numbers N up to 100,000 are listed in table 9.3.
The hailstone number sequence: what a pleasant and easy problem to define and to carry out mathematically! Yet, so far, the 3N + 1 conjecture has eluded rigorous proof by mathematicians. In the meantime, the problem nevertheless provides a very interesting exercise in arithmetic and point plotting for students at all levels.
Here are four references dealing with the 3N + 1 problem and hailstone numbers. The first reference is the easiest: the paper by Bruce (1978) entitled “Crazy Roller Coasters.” Then come Hayes (1984) and Lines (1990). The last reference, the most difficult, is by Lagarias (1985), with the title “The 3x + 1 Problem and Its Generalizations.” The author presents a short history of the problem and includes seventy references, most of which are not easy reading unless you're a real mathematician.
TABLE 9.3
10
A Fast Way to Escape
Take a tennis ball and toss it straight upward. It goes to a certain height and falls back to your hands. Keep tossing or throwing the ball, each time with greater velocity. Observe that each time the ball goes higher and the total flight time is longer. An amazing observation! Anything else new? Most likely; we shall get to that shortly.
However, before going further we do magic: we remove all the air in the world. The resulting perfect vacuum greatly simplifies our mathematical analysis because now we can neglect the effect of air resistance. In this case, as we shall see, the height ym to which our tennis ball ascends is , where Uo is the velocity of the ball as it leaves your hand. The quantity g is the acceleration due to gravity; more precisely, it is the force exerted by gravity per unit mass of an object. In the metric system of units, g has the approximate value of 9.82 newtons of force per kilogram of mass or, equivalently, 9.82 meters per second squared; that is, g = 9.82 m/s2. In the English or engineering system of units, g = 32.2 ft/s2.
Now g is not really constant. At the earth's equator, g = 9.780 m/s2 and at the poles, g = 9.832 m/s2. The smaller value of g at the equator is due to the centrifugal force created by the rotation of the earth. As we shall see, g also depends very much on the distance from the center of the earth. However, for the moment, with a subscript to emphasize its constant value, we take g0 = 9.82 m/s2.
We begin our analysis with an introduction of Newton's famous laws of motion. These remarkable relationships were devised by the English mathematician, Issac Newton (1643-1727), one of the greatest mathematicians of all time. The three laws are stated as follows:
First law: An object at rest or moving at constant velocity will remain so unless an external force is applied
Second law: The acceleration of an object is proportional to the overall force applied on the object divided by the mass of the object
Third law: Every action (force) has an equal and opposite reaction (force)
Newton's second law is now employed in our tennis ball throwing experiment. We have
in which is the summation of all the forces acting on the ball, m is its mass, and a is its acceleration. By definition, we also have the relationships
These last two expressions indicate that the acceleration a is the rate of change of velocity U with time and that the velocity is the rate of change of vertical distance y with time.
If we neglect the effect of air resistance, the only force acting on the ball is its own weight. That is, . The minus sign is employed because the gravitational force acts in the downward (negative y) direction. So, utilizing equation (10.1) and the first relationship of (10.2), we obtain
which becomes
The second expression given in (10.4) is now in proper form for utilizing integral calculus. The quantities in the lower limits of the integrals describe the so-called “initial condition”: U = Uo when t = 0. Carrying out the integration yields
Utilizing the second relationship of (10.2) gives
The lower limits of the integrals indicate that y = 0 when t = 0. Integrating equation (10.6) gives
Using equations (10.5) and (10.7) to eliminate the time variable t provides the relationship
At the instant the tennis ball is at its highest point, y = ym and the velocity U = 0. So from (10.8), or, alternatively, . For example, if the observed height ym = 10 m, then the initial velocity U0 = 14.0 m/s.
We note that if we neglect the effect of air resistance and assume that gravity is constant, it is easy to calculate the maximum height of our tennis ball. Clearly, the faster we throw the ball the higher it goes. However, we recall that the gravitational force g is not really constant. This leads us to another of Isaac Newton's remarkable contributions: his famous law of gravitation,
in which F is the force of attraction between two bodies of masses m and M, whose centers are separated by a distance r. The quantity G is the gravitational constant. In the metric system of units, G = 6.673 X 10-11 newton m2/kg2.
We shall consider m to be the mass of the tennis ball and M the mass of the earth. If R is the radius of the earth and the force F is the weight of our ball at the earth's surface, then F = mg0. Consequently, from equation (10.9) we obtain
where r is the distance of the ball from the earth's center. This equation says that the weight of the ball is inversely proportional to the square of its distance from the center of the earth. So it follows from equation (10.10) that g = g0R2/r2 describes the way in which g changes when r > R (i.e., the ball outside the earth). Incidentally, r = R + y, where y, of course, is the vertical distance above the earth's surface. By the way, shortly we shall present an expres
sion for g when r < R (i.e., the ball is inside the earth). Analyses dealing with these matters are presented by Boas (1983).
Returning to Newton's second law, equation (10.1), we have
Now, since U = dr/dt, we can write dU/dt = (dU/dr)(dr/dt) = U(dU/dr). Also, since g = g0R2/r2, (10.11) becomes
The lower limits of the integral state that when r = R (i.e., on the earth's surface), the “launch velocity” is U0. Integrating this expression gives
Recall that when the gravitational force is constant, g = g0, the velocity of the tennis ball becomes zero at a certain height . The ball then falls back to earth. However, when the gravitational force changes according to the relationship g = g0R2/r2, quite a different answer is obtained. We ask the question: can we toss the tennis ball or, more realistically, launch a projectile or space vehicle with sufficiently large velocity U0 that it simply does not return? In other words, can the launch velocity be made so great that the ball or projectile actually escapes from the earth's gravity?
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