WHEN THE SUN SETS OVER open water, its reflection can be oval or columnar. What determines the shape? Lorne Whitehead of TIR Systems, Ltd., in Vancouver has looked into the matter and reports (in a manuscript he sent to me) that the shape is related to the maximum slope of the waves on the water. By photographing the reflection one can compute the maximum slope without having to go out on the water.

To follow Whitehead's work I examine first what determines the height and width of the reflection region you observe. For simplicity assume that the sun is a point source of light. (It actually occupies about .5 degree of arc in the field of view.) Assume also that the water surface is smooth.

Suppose the water surface is flat. Since the sun is so distant, the rays arrive at the water parallel to one another. They reflect from the water and some of them reach your eye. You perceive a reflection of the sun as if the surface of the water were a mirror laid flat in front of you.

The reflection of a light ray from a surface is usually described in terms of angles measured with respect to what is called the normal: a line perpendicular to the surface at the point of reflection. The ray is incident at a certain angle measured with respect to the normal. The reflected ray forms the same angle. This rule applies even if the surface is tilted or curved. Wherever a ray reflects from the surface, mentally construct a normal perpendicular to the surface at that point.

When the water surface is flat, a single spot on the surface reflects rays toward you. One way to determine where that spot is in your field of view is to imagine that you are viewing the scene without the benefit of depth perception, as if you were looking at a photograph.

This flat representation is called a projection plane [Figure 3]. The horizon lies across the plane. The spot reflecting light rays to you lies below the horizon on the projection plane. That is where you see the sun's image. When the water surface is flat, the image on the projection plane is as far below the horizon as the sun is above it. As the sun sets the image rises toward the horizon. As the sun dips below the horizon the image disappears at the horizon.

When the water surface is curved by smooth waves, the sides of the waves reflect the rays according to the simple rule for the angles in a reflection. Now many spots on the water surface reflect rays to you, however, and their locations constantly vary as the waves reshape the water surface while moving in every possible direction. At any instant you see many images of the sun on the projection plane. The reflection region associated with a setting sun is the composite of those many images.

In order to determine how the waves spread the images of the sun I first investigated how an image moves on a projection plane when a reflecting surface is uniformly tilted in one direction. Having taped a pencil upright on a table, I laid a flat, rectangular mirror in front of the pencil and positioned my eyes near the tabletop. The far edge of the mirror, which was at the base of the pencil, functioned as the horizon.

Ignoring any sense of depth, I imagined the pencil and its image to be on a projection plane. To measure distances on the plane I used a transparent ruler, keeping it and my head stationary while I noted where on the ruler various parts of the projection plane were aligned. For example, in my field of view the pencil point was three centimeters above the far edge of the mirror and the image of the point was three centimeters below it, according to the rule for flat surfaces.

I then raised the near edge of the mirror, keeping the far edge on the table. The spot reflecting the pencil point moved toward me along the surface of the mirror. On the projection plane the image of the point shifted downward.

Next I tilted the mirror toward me. This time the spot reflecting the pencil point moved along the surface away from me. On the projection plane the image of the point moved upward.

My demonstration is not entirely faithful to the reflection of rays from the sun by a water surface because the, rays arriving at the mirror are not parallel. Still, the general shift of the image on the projection plane is similar. The image of the sun moves downward on the plane as the water surface tilts away from the observer and upward when the surface tilts toward the observer. The slope of the surface determines the extent of the shift.

The slope can be defined as the tangent of the angle through which the surface is tilted. For example, if the tilt of the surface is 16.7 degrees with respect to the horizontal, the slope is .3 (a dimensionless number). The slope at a point on a curved surface is that of a line drawn tangent to the surface l at that point.

A vertical axis on the projection plane can also be marked off in terms of slope. For example, suppose the sun is 38.7 degrees above the horizon in your field of view. On the projection plane the horizon is at zero height and the sun is in a position that is the tangent of 38.7 degrees, or .8, above it. If the water surface is flat, the image of the sun is .8 below the horizon on the projection plane.

Imagine that the water surface is initially flat and then uniformly tilts away from you until it has a final slope of .3. The image of the sun moves downward on the projection plane by twice the final slope of the surface, that is, by .6. If instead the water surface uniformly tilts toward you with a slope of .3, the image of the sun moves upward on the projection plane by twice the slope of the surface, or .6.

A natural water surface is seldom tilted uniformly. Usually it has a range of slopes because of waves. Whitehead's point is that the maximum slope of the waves determines the dimensions of the sun's reflection region on the water. To follow his argument first assume that the water surface is flat and the sun and its image are .8 above and below the horizon on the projection plane. Now examine how the waves alter the reflections from the spot that creates the normal image of the sun. Sometimes the water surface ; there is flat and you see the image. At other times the waves tilt the surface and eliminate the image. Thus the image appears and disappears as waves sweep through the corresponding spot on the water.

Next imagine a spot on the surface that is slightly higher on the projection plane. If the surface were constantly flat, you would never see the sun's image, but since waves move through it, the spot sometimes tilts toward you enough so that you can see a reflection of the sun. Again the image appears and disappears as waves move through the spot.

Now think of a spot on the water whose image on the projection plane is lower than the normal image. Sometimes the water surface tilts away from you enough for you to see the sun's image there.

How far above and below the normal position can you see images of the sun? The limits are set by the maximum slope of the largest waves. Suppose the waves have a maximum slope of .3 (16.7 degrees). The highest image position on the projection plane is above the normal position by a distance that is twice the maximum slope of the waves, that is, by a distance of .6. Since in my example the normal position is .8 below the horizon on the projection plane, the image position is .2 below the horizon.

Similarly, the lowest image position is below the normal position by a distance that is twice the maximum slope of the waves, placing the image at a position of 1.4 on the vertical axis of the projection plane. In this example you can see images of the sun throughout the range from .2 to 1.4 on the vertical axis. The images are not constant, because the slopes of the water surface constantly change. Since the visual system averages brightness over time, you perceive a fairly constant illumination. A photograph made with a high shutter speed, however, would reveal areas within the region where there is no image of the sun.

As the sun sinks, the reflection region on the water moves toward the horizon. In terms of the projection plane the region ascends. In my example the far end of the reflection region reaches the horizon when the sun has a slope of .6 in your field of view, which is about 31 degrees. Eventually the region disappears into the horizon.

The vertical extent of the reflection region on the projection plane depends on the maximum slope of the waves. If the maximum slope increases, the vertical extent increases by twice as much on the projection plane. This relation enables you to measure the maximum slope of the waves. Photograph the scene before the reflection region reaches the horizon. Also measure the angle between the horizon and the sun. Place over the photograph a sheet of clear plastic on which the projection plane is to be marked. Trace the horizon and mark the sun as a point.

Compute the tangent of the sun's angle with respect to the horizon. Suppose the angle is 38.7 degrees (a tangent of .8) when you make the photograph. Use this result to scale the vertical axis on the projection plane. Mark the sun's position on the photograph as .8 above the horizon. With a ruler measure the number of centimeters between the horizon and the sun in the photograph. Suppose the distance is four centimeters. Each centimeter on the ruler corresponds, then, to a distance of .2 on the vertical axis on the projection plane. Scale the axis with this relation.

Next determine the extent of the reflection region along the axis. Suppose it stretches from .5 to 1.1. Since the sun is .8 above the horizon, the normal position of the sun's image must be .8 below the horizon. The lowest point of the reflection region is therefore .3 below the normal position. Since the distance along the axis from the normal position to the lowest point of the region is twice the maximum slope of the waves, the maximum slope must be .15, which corresponds to 8.5 degrees.

Whitehead has also explained the width of the reflection region on the projection plane. To follow his explanation I returned to my pencil and mirror. This time I placed the right-hand edge of the mirror along a line between me and the base of the pencil, with the far-right corner of the mirror touching the base. When I tilted the mirror to the right, the image of the pencil on the projection plane rotated about the base of the pencil by an angle that was twice the tilt of the mirror. When the tilt of the mirror was 45 degrees, the image rotated through 90 degrees and The image of the pencil point can be found on the projection plane by superposing a line that extends across the plane from the image to the pencil point. The line tilts from the vertical axis by as much as the mirror is tilted. Also superposed on the projection plane is the far edge of the mirror. On the plane the image is as far from the edge of the mirror as the actual pencil point is.

Whitehead employs an analogous construction on a projection plane to determine the width of the reflection region on the water surface [see middle illustration on opposite page]. Suppose the waves have a maximum slope of .3 (16.7 degrees). On the projection plane draw through the position of the sun a line tilted to the right of the vertical by 16.7 degrees. Extend the line toward the bottom left of the projection plane.

Now draw a second line that is perpendicular to the first one and passes through the point on the horizon directly below the sun. This line corresponds to the far edge of the mirror Along the first line measure the distance between the sun and the intersection of the first and second lines Starting at the intersection, measure off an equal distance along the first line toward the bottom left of the projection plane. The point you reach is the position of the sun's image in a mirror tilted to the right by 16.7 degrees.

Repeat the entire procedure on the right-hand side of the sun. You now have two lines that diverge toward the bottom of the projection plane. The reflection region lies between those lines and coincides with them at the points corresponding to rightward and leftward-tilted mirrors. The angle between the lines is twice the maximum slope of the waves.

If you add to the projection plane the highest and lowest points of the reflection region according to my previous explanation, you can sketch the entire reflection region by drawing a smooth curve connecting the extreme points [see Figure 5]. If the maximum wave slope increases, the angle between the diverging lines becomes larger and the reflection region grows both wider and higher on the projection plane. When the sun is high, the region is oval, smaller in width than in height. As the sun descends, the region slips toward the horizon, becoming narrower because it is constrained by the diverging lines. The oval may then seem to be a long column that stretches toward the horizon. Because Whitehead and I have treated the sun as a point source of light, the upper end of the column should be a point. Since the sun actually occupies about .5 degree in your field of view, however, the upper end of the column cannot be narrower than .5 degree.

The fact that the column has unequal width and height is surprising when one recalls that the reflections are from waves moving in every direction. The column is also surprising in seeming to be long. Actually, however, it occupies only a small angle on the projection plane. Some of its apparent length derives from a perception of depth in the scene: the column looks as though it stretches over a long distance to the horizon. Part of the illusion is also due to a misinterpretation of angles near the horizon: the low sun and the upper end of the column look large because the horizon seems to be distant. (The same phenomenon accounts for the apparent enlargement of the moon when it is near the horizon.)

You might enjoy studying reflection regions produced by the sun or the moon under other circumstances. In some cases you might be able to detect surface currents on the water if the maximum slope of the waves in a current differs from the slope of the waves in the surrounding water. Does the reflection region then become asymmetric or otherwise distorted? What happens to it when the waves move uniformly in one direction? How does the statistical distribution of wave slopes alter Whitehead's calculations of the shape of the reflection region? What happens to the shape of the region and the distribution of the light when the waves become so large that they are no longer sinusoidal?

Last August I described a number of arrangements for stacking dominoes. One of them called for a series of dominoes to be laid one on another so as to extend beyond the edge of a table. Each domino should have a broad face down, with its long dimension perpendicular to the edge of the table. The aim is to find the maximum extension. Of the overhanging stack for a given number of dominoes.

Several readers (Eugene Wall of the Aspen Systems Corporation in Rockville, Md.; R. F. Tindall of Cambridge, England; David Callway of Fort Collins, Colo., and Wayne Fullerton of Houston) pointed out that the mathematical series I gave to describe the overhang can be written in terms of Euler's constant (.57722). The horizontal distance from the table's edge to the outer edge of the nth domino is half a domino length multiplied by the sum of Euler's constant and the digamma function for n + 1. (The digamma function is the logarithmic de rivative of the common gamma function.) When n is large, the digamma function can be approximated as the subtraction of 1 /(2n + 2) from the natural logarithm of + 1.

When formulas derived from one are employed to compute the number of dominoes needed for an overhang of 50 domino lengths, one finds that 1.5 X 10⁴³ dominoes are needed, not 1.5 X 10⁴⁴ as I stated.

Is there a way to stack dominoes so as to build a large overhang more economically? Tindall and Hans-Hellmuth Cuno of Waldetzenberg in West Germany suggested that a given overhang can be achieved with fewer dominoes by counterbalancing. In their scheme each new domino is positioned so that its midpoint is above the outer edge of the domino just below it. Such an arrangement would of course make the lower domino unstable. Stability is restored by putting a counterweight domino on the inner edge of the lower one [see above].

For example, to achieve an overhang of three domino lengths, align six dominoes in steps from the edge of the table, each with its midpoint over the outer edge of the domino (or table) just below it. To provide counterbalance 31 dominoes are stacked above the inner edge of the domino on the table, 15 above the second domino, seven above the third, three above the fourth and one above the fifth. The sixth and outermost domino requires no counterbalancing. All told 63 dominoes are required for this arrangement, easily beating the 227 dominoes required in my stacking scheme.

E. James Morton of the John Hancock Mutual Life Insurance Company in Boston pointed out that some clever stacking schemes for four rectangular objects such as dominoes have been published by Stephen Ainley (The Mathematical Gazette, Vol. 63, No. 426, page 272; December, 1979). Each domino's long dimension must be perpendicular to the edge of the table. If the dominoes are stacked in the way I described last August, their maximum overhang is 1 1/24 times their long dimension. The first two of Ainley's schemes, which are depicted in the illustration below, generate an overhang 1 1/8 times the long dimension. (The fractions shown are in terms of the domino's long dimension.) One or two dominoes serve as a counterbalance.

In a third arrangement counterbalancing is achieved by a domino on the inner edge of the lowest domino and another domino on top of the stack.

The top domino is positioned directly above the one at the bottom of the stack. This scheme yields an overhang 11/6 times the long dimension.

The winner of the balancing act is the last arrangement shown in the illustration. Note that the topmost domino is not exactly above the bottom one. This arrangement gives an overhang approximately 1.1679 times a domino's long dimension, barely beating out the previous arrangement.

Readers interested in such puzzles may enjoy a note published by R. E. Scraton (The Mathematical Gazette, Vol. 64, No. 429, pages 202-203; October, 1980). He shows how a complex stacking of 28 dominoes, each two inches by one inch by half an inch, can be built with an overhang of almost eight inches. If the stack is rotated so that the diagonals of the dominoes are at right angles to the table's edge, the overhang is almost 8.9 inches.

Bibliography

THE POLARIZATION OF LIGHT AT SEA. E. O. Hulburt in Journal of the Optical Society of America, Vol. 24, No. 2, pages 35-42; February, 1934.

REFLECTION OF LIGHT. M. Minnaert in The Nature of Light and Colour in the Open Air. Dover Publications, Inc., 1954.

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