The Cardioid
The word “cardioid” means a “heart-shaped curve,” so we are off to a good start. Have you always wanted to know how to construct a cardioid but were too afraid to ask? The answer: Put two large coins of the same diameter on the table. Hold one of them down firmly. Then roll the other coin around the boundary of the first without slipping. The original contact point between the two coins will trace a cardioid.
FIG. 17.4
Plots of the (a) cardioid and (b) lemniscate.
The equation of the cardioid, in a polar coordinate system, is
in which R is the radius of the coin, if you like. The entire cardioid is shown in figure 17.4(a). Now this may look like a heart but it sure doesn't look like a valentine. So we shall use only the right-hand portion and, in a moment, use something else for the left-hand portion. Figure 17.5 shows the design diagram for this one.
FIG. 17.5
Construction of a valentine with a cardioid and a lemniscate. The origin at the left refers to the lemniscate and that at the right to the cardioid.
As before, it is necessary that our cardioid-type valentine have the same overall dimensions: H = 10 cm and B = 8 cm. This requires that we compute the value of R in equation (17.11). To accomplish this, (17.8) is useful to determine the locations of the maximum x and maximum y points and the corresponding angles θ.
It turns out that the value R = 1.538 cm will force the cardioid through the point r = 4.62, θ = 120° or, equivalently, through x = –2.31, y = 4.00. Note that y = 4 is half the width of our valentine. We also determine that the plunge is h = 0.77 cm.
The portion of the cardioid we shall use is the solid line in figure 17.5; the unused portion is dashed.
The Lemniscate
Just as we attached a parabola to a circle to fabricate our first valentine, we are now going to attach a curve called a lemniscate to the cardioid to construct this one. The equation of the lemniscate is
The entire lemniscate is shown in figure 17.4(b); only the left-hand portion will be used. To simplify the algebra, we have selected a new origin for this part of the design.
The lemniscate has somewhat the shape of a leaf or a pointed circle. It looks like the trajectory of a high pop-fly to center field. If the curve is rotated about the x-axis, the resulting shape looks like a teardrop or—upside down—a hot-air balloon. In 1694, the noted Swiss mathematician Jakob Bernoulli (1654–1705) published a paper about this interesting curve. The lemniscate is a special case of another curve called the Cassinian ovals.
As before, we need to determine the value of L in equation (17.12) to assure that the final valentine will have the dimensions H = 10 cm and B = 8 cm. Again, (17.8) is helpful in determining the location of the maximum point of the lemniscate. We then match this point with the maximum point of the cardioid. The outcome is that L = 11.30 cm.
So, for our cardioid-lemniscate valentine, we have R = 1.54 cm and L = 11.30 cm. If you would like a nice problem to work on during your spare time, you can confirm that indeed H = 10 cm and B = 8 cm for this valentine.
You can also confirm, using equation (17.9), that the area of the cardioid portion of the entire valentine is Ac = 20.52 cm2 and the area of the lemniscate portion is Al = 36.24 cm2. The total area is A = 56.76 cm2.
Finally, utilizing (17.10), you might want to show that the perimeter of the cardioid portion of the entire valentine is Sc = 12.31 cm. A bit more difficult task is to establish that the perimeter of the lemniscate portion is Sl = 16.44 cm. The total perimeter is S = 28.75 cm.
Some Closing Comments
That should be sufficient, unless, of course, you want to create and analyze some other valentine configurations.
There are many references that deal with the subject of mathematical curves. Three of the best are Lawrence (1972), Lockwood (1961), and von Seggern (1990).
Finally, to help you prepare for next Valentine's Day, it is suggested that you plot, on a sheet of graph paper, the three valentines we analyzed in the preceding sections. Following this, you have two major decisions to make: (1) which valentine you prefer and (2) whether or not to go into mass production. Whatever you decide, it is recommended that you do not wait until next February to carry out this assignment.
18
Somewhere Over the Rainbow
And to Continue the Song
…bluebirds fly
Birds fly over the rainbow
Why then, oh why, can't I?
Nature is very generous. Frequently, and always gratuitously, it provides all of us with spectacular exhibitions of magnificent beauty. And surely, at or near the top of everyone's list of incredibly beautiful displays of nature would be the rainbow. For thousands of years, poets and artists, scientists and mathematicians have attempted to describe the rainbow with words and paintings, symbols and equations.
According to historical records, the noted Greek philosopher Aristotle (384–322 B.C.) made extensive studies of rainbows, trying to understand what they are and how they are created. But for many hundreds of years, the magnificent arc of color eluded mankind's efforts at quantitative description and mathematical analysis. During the long period of the Middle Ages, only the name Theodoric of Freiberg (c. 1310) stands out in connection with rainbows. The advances made by this Franco-German monk brought “rainbow theory” remarkably close to present-day points of view.
In the three hundred years following Theodoric, little progress was made by scientists and mathematicians in the analysis of rainbows. Then it all came together, as they say, in the seventeenth century. During that period, a sequence of very remarkable people developed and utilized the mathematics and physics needed to explain the phenomenon. Here is a list of these people:
Johannes Kepler, German, 1571–1630
Willebrord Snell, Dutch, 1591–1626
René Descartes, French, 1596–1650
Christiaan Huygens, Dutch, 1629–1695
Isaac Newton, English, 1643–1727
The advances made by these scientists, culminating in the work of Descartes and Newton, were so complete and accurate that we still refer to the theory of the rainbow as the Descartes–Newtonian theory. Indeed, their relatively simple geometrical optics solution to the problem explains all the main features of the rainbow; only a few minor aspects of rainbows require more complicated mathematics.
Rainbows Are in Many Places
An extremely pleasant surprise sometimes occurs when we suddenly look up or go around a corner and there is a magnificent rainbow. Even so, it is helpful to know where to look for rainbows and when to expect them. It is very simple: A bright sun needs to be at your back as you look toward a rain shower. Midmornings or earlier and midafternoons or later are best in order to catch low sun angles; this puts the bow higher in the sky.
Sometimes there are two rainbows. If so, the inner and more intense bow is called the primary rainbow. As we shall see, this bow is the arc of a circle whose radius is about 42°. Violet is on the inside of the bow and red is on the outside; all the other colors—blue, green, yellow, orange—are in between.
If we are lucky, we are able to see also the outer and less intense bow called the secondary bow. It has an angular radius of about 50°. The colors are reversed in the secondary bow; red is on the inside and violet on the outside.
However, we do not require rain showers to see rainbows. As long as we have a bright light at our back and a source of water drops in front, we can see them. For example, they appear in the spray at the base of a waterfall or near the jet of a garden hose or in the bow wave of a ship or motorboat. If you happen to be the pilot of an airliner with the sun behind you and a rainstorm ahead, maybe you will see the complete circle of the rainbow!
Some References about Rainbows
Over the years a great deal has been written about rainbows. An interesting book by Boyer (1987) gives a comprehensive history of the subject and an easy-to-understand presentation of the Decartes-Newtonian analysis of rainbows. A book by Greenle
r (1980) offers nontechnical descriptions and explanations about rainbows and includes a remarkable collection of color photographs.
An excellent survey article about the theory of rainbows is given by Nussenzveig (1977). The author includes an introduction to the more advanced topics of rainbow theory. Finally, Austin and Dunning (1988) and Minnaert (1993) are references that deal with the elementary mathematics and physics of rainbows.
Geometrical Analysis of Rainbows
Our study of rainbows begins with an analysis of the geometry involved in the phenomenon. For this, we use the principles of a branch of physics called geometrical optics. We start with figure 18.1. A ray of light from some source—for example, the sun—strikes the surface of a drop of water at point A. This ray is called the incident ray. As shown in the figure, the line A-A′, perpendicular to the surface, defines the angle i of the incident ray. Another ray, named the reflected ray, leaves the surface of the water drop at the same angle i. Thus, the angle of incidence is equal to the angle of reflection. We shall use this simple and intuitively obvious relationship throughout our analysis.
FIG. 18.1
Reflection and refraction of a light ray at an air-water interface.
However, not all of the light of the incident ray is reflected back into the air in the reflected ray. Some portion of the light is refracted into the water at an angle r, as shown in the figure. This angle is called the angle of refraction. In 1618, the Dutch scientist Willebrord Snell (1591–1626) gave the following equation relating the angle of incidence i to the angle of refraction r :
where n1 and n2 are the so-called refractive indices of air and water, respectively. The relationship of equation (18.1) is usually called Snell's law.
The numerical value of the refractive index depends on the type of medium through which the light ray passes. In a vacuum and in air, n = 1 and in water, n = 1.33. We simplify (18.1) by writing it in the form
with the understanding that we are now dealing with an air-water boundary. For example, if i = 60° and n = 1.33, then (18.2) gives r = 40.6°. We note in figure 18.1 that the reflected ray is bent toward the perpendicular A-A′.
FIG. 18.2
The primary bow. Two refractions and one reflection of incident rays in a water drop, and the corresponding emergent rays.
The Primary Rainbow
First, we examine what is called the primary rainbow. As we shall see, the primary rainbow involves two refractions and one reflection of a light ray inside a raindrop. As shown in figure 18.2, light rays from the sun strike a spherical raindrop at various distances at or above a line passing through the center of the drop. The angle of incidence ranges from i = 0° to i = 90°. At each impact point, part of the light of the ray is reflected back into the air and part is refracted into the water. For the moment, we are interested in only the latter.
As indicated in the figure, the refracted ray proceeds across the raindrop until it strikes the opposite boundary. Here a portion of the light is refracted into the air. More importantly, the remaining portion is reflected in the drop and travels to the front boundary of the drop. Once again, part of the light is reflected to another point on the inside of the drop. The remaining part is refracted into the air.
We take a closer look at the geometrical optics in figure 18.3. An incident ray strikes the drop at point A at some angle of incidence, i. Here, the ray is refracted to angle r and proceeds to point B. The ray is then reflected to point C, where it is refracted into the air. We call this the emergent ray.
FIG. 18.3
The primary bow. Path of a light ray through a raindrop.
In figure 18.3, extensions of the incident and emergent rays define a point D and an angle Ø. From the geometry of the figure, it is not hard to show that Ø expresses the total change of angle between the incident ray and the emergent ray. Its value is
A plot of this equation is shown in figure 18.4.
FIG. 18.4
The primary bow. Total angle change, Ø, as a function of the angle of incidence, i.
Next, we ask the question, is there a value of the incidence angle i that gives a minimum value of total angle change, Ø? To answer the question, it is necessary to use a bit of differential calculus. We rewrite equation (18.3) in the form
in which we have used (18.2) to relate i and r.
To determine the minimum value of Ø, equation (18.4) is differentiated with respect to i and the result is equated to zero. This gives the following answer:
where im is the value of the incidence angle giving minimum Ø. If n = 1.33, then im = 59.6°. Substituting this result into (18.4) yields Øm = 137.5°. These numerical values are identified by the solid dot on the curve in figure 18.4. A ray of light passing through a raindrop with these particular values of incidence and emergence angles is sometimes called the Descartes ray. A note in passing: In 1637, the French scientist René Descartes (1596–1650) was the first person to work out the geometrical optics of rainbows correctly.
We note from figure 18.4 that no light rays have a total angle change Ø less than about 138°. So it is apparent that all rays emerging from the raindrop are confined to a circular cone with half-angle of 180° – 138° = 42°. The apex of this cone is the eye of the observer and the axis of the cone is in the direction opposite to the sun.
Furthermore, since the curve displayed in figure 18.4 is quite flat near the minimum point, most of the emergent rays are at or near the 42° boundary of the cone. In this regard, if you would like another problem in calculus, you can show that the average value of Ø, over the entire range of i, is Ø = 151.5°. This corresponds to a cone half-angle of 28.5°.
Why Do Rainbows Have Colors?
So far so good. Perhaps we have now convinced ourselves that incident sunlight, after two refractions and one reflection within a spherical raindrop, produces emergent rays largely confined to a cone with 42° half-angle. Fine. But where does the color come from?
To answer this question we turn to results obtained by that remarkable English scientist, Isaac Newton (1643–1727). Along with the large number of other incredible advances he made, Newton was able to show, in experiments carried out in 1666, that a ray of sunlight is composed of an entire spectrum of colors. Each color of the ray has a particular numerical value of the refractive index n as the ray passes through a transparent medium such as water.
For example, Newton showed that in water, red has a refractive index n = 1.332, and at the other end of the spectrum, violet has a refractive index n = 1.344. The refractive indices of all the other colors—orange, yellow, green, blue—have numerical values between those of the red and violet extremes.
For our purpose, the rest is easy. Starting with red, we substitute n = 1.332 into equation (18.5) to compute im and then use (18.4) to determine that Øm(red) = 137.8°. This corresponds to a cone half-angle of 42.2°. At the other end of the spectrum, violet with n = 1.344 gives Øm(violet) = 139.5°, which provides a cone half-angle of 40.5°.
A quick summary of the results to here: In the primary rainbow, two refractions and one reflection of sunlight rays within a raindrop produce a bow with a cone half-angle of about 42°. In other words, the rainbow is the arc of a circle with 42° angular radius. The red color in sunlight yields a cone half-angle of 42.2° and the violet color gives a half-angle of 40.5°. The slight difference between these two angles provides the spectrum of color in a rainbow. Since the cone angle for red is larger than the cone angle for violet, it is clear that red is on the outside of the primary rainbow and violet is on the inside. The other colors are in between.
FIG. 18.5
The secondary bow. Path of a light ray through a raindrop.
The Secondary Rainbow
Our analysis of the so-called secondary rainbow is identical to that of the primary rainbow with one exception: this time there are two reflections, instead of one, of a ray of sunlight within the raindrop. So, as shown in figure 18.5, for the secondary bow a ray of light receives two refractions and
two reflections inside the drop.
In this case, application of geometrical optics leads to the expression
in which Ø is the total change in angle between the incident ray and the emergent ray. Note that this equation is slightly different from equation (18.3), which gives the total angle change for the primary rainbow.
A plot of (18.6) is shown in figure 18.6. It is apparent that again there is a minimum value of Ø. Substituting equation (18.2) into (18.6) and then differentiating with respect to i yields the result
Again, the refractive index of sunlight in water is n = 1.33. Substituting this value into (18.7) gives im = 71.9°. Using this answer in (18.6) yields the minimum value Øm = 230.1°. So, in the secondary rainbow, most of the rays are confined to a circular cone with half-angle 230° – 180° = 50°.
FIG. 18.6
The secondary bow. Total angle change, Ø, as a function of the angle of incidence, i.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 13
