I enjoyed reading your analysis of sunrise and sunset at the time of winter solstice in Bettles, Alaska. Since the town is only a few hundred feet above sea level, your discussion was simplified. As an aircraft owner and pilot, I have seen the effect of atmospheric refraction dramatically amplified. On the ground, I am fond of pointing out jets up high, dragging contrails in bright sunlight, well after the sun has gone down on the surface. In the cockpit at 40,000 feet, still bathed in sunlight, we already see darkness on the ground with street lights coming on. Refraction is surprisingly strong at this altitude—even to an observer on a mountaintop overlooking the ocean. Can you elucidate?
I hope you will allow me to recast your question: When viewing sunset from 40,000 feet, what influences the apparent position of the sun more, the altitude or atmospheric refraction? Consider the effect of altitude first. It is common experience that the horizon expands with altitude. From the observation deck of a tall building, you can see farther than from street level (when no other buildings are in the way). From atop a tall sand dune, you can see farther out over the ocean than from the water line.
It is fairly easy to find a formula that gives the distance to the horizon as viewed from a given elevation. Refer to Figure 1. Suppose an observer at O, at elevation z above sea level, wants to know how far it is d to the horizon at H. The distance from H to the center of the earth is the earth radius R (6,371 kilometers). The distance from O to the center of the Earth is R + z. Because the line from O to H is tangent to the earth's surface, the points O, H, and the center of the earth form a right triangle with R + z as the hypotenuse. The Pythagorean Theorem relates the lengths of the three sides of a right triangle:
This formula is valid for a spherical earth with no atmosphere. The atmosphere bends light rays (refraction), and we'll deal with that effect later. If we are in an aircraft at an altitude of 40,000 feet (12.192 kilometers), we can calculate the distance to the horizon with this formula, making sure that everything is in kilometers. The answer is 394.3 kilometers (245 miles).
The line from O to F is drawn perpendicular to the line from the center of the earth to O. It is in the direction that the horizon would be if the Earth were flat. It is called the astronomical horizon. The sum of the angles α and β is thus 90°. We can easily find α from
β is thus 3.543° or 212.6′, where (′) denotes minutes of arc (60′ = 1°). β is called the dip angle, and it specifies how much the upper limb of the sun is below the astronomical horizon at sunset as viewed from 40,000 feet in an airless world. This is the effect of altitude alone. What is the effect of refraction?
An observer at O, at altitude z above the Earth's surface, views the geometric horizon H along the red line OH in an airless atmosphere. OH is tangent to the Earth's surface. The earth is assumed to be spherical with radius R. The horizon is at distance d from the observer. The line OF is in the direction of the astronomical horizon; it is perpendicular to the line from the center of the Earth to the observer. The sum the angles α and β is thus 90°, and the triangle formed by O, H, and the center of the Earth is a right triangle. β is called the dip angle.
Under normal atmospheric conditions, atmospheric density decreases with altitude. In the atmosphere, a ray of light bends downward toward higher density. Thus a ray of light coming from the horizon and observed by a pilot at 40,000 feet will bend downward, with the same sense of curvature as the earth's (convex downward). See the green curve in Figure 2, which gives the path a light ray travels from the pilot's eye to the horizon. The curvature is gentle because atmospheric density increases gradually toward the earth's surface. Contrast this with a ray of light entering a swimming pool, where the change in direction of a light ray is sudden because the density changes significantly all at once at the air/water interface. This bending of a light ray is called refraction, and it explains why objects at the bottom of a pool appear to be closer to the surface than they really are, and objects on the ground viewed obliquely from 40,000 feet appear to be somewhat closer than they really are.
Figure 2 includes the geometry from Figure 1. Because of the curvature of the green ray, a pilot can see the horizon H at a greater distance than he or she could along the straight red ray. Note that the dip angle for the green ray is less than the dip angle β for the red ray. But how much less?
This figure includes the same geometry as Figure 1 (for an airless atmosphere), but a green curve has been added, which represents the path of a light ray from the observer to the apparent horizon at H through an atmosphere whose density increases downward. The observer sees the light ray in the direction given by the dashed green line and at a dip angle from the astronomical horizon OF.
The answer depends on the density stratification, i.e., the variation in density along the ray path. To give a concrete example, consider what is called the Standard Atmosphere. It roughly approximates a globally averaged atmosphere and is used for a multitude of engineering calculations. In the Standard Atmosphere, the sea level temperature is 15°C, the sea level pressure is 1013 millibars, and there is no moisture. The troposphere extends from the surface to 11 kilometers altitude; in this layer, the temperature decreases at a rate of 6.5°C per kilometer of altitude. The stratosphere extends from 11–20 kilometers altitude. The stratospheric temperature is constant at –56.5°C. The Standard Atmosphere is also hydrostatic, that is, the pressure at a given level depends on the weight of all the overlying air. These conditions are sufficient to determine the density at each altitude, which is needed to determine the curvature of the green ray in Figure 2, given a Standard Atmosphere. That, in turn, is needed to compute the location of H, which, ultimately, can give us the dip angle . This involves lengthy calculations, but we can approximate the answer by using the diagram in Figure 3.
One can use Figure 3 to estimate the dip angle of the apparent horizon in a standard atmosphere, experienced by an observer at an altitude of 12.192 kilometers (40,000 feet), looking along the green ray in Figure 2. The average earth radius was given above as 6371 kilometers. For Figure 3, the value is R0 = 6357 kilometers, the radius at 45° latitude (the earth is not a perfect sphere). The horizontal axis gives the height above sea level z. The vertical axis is nR – R0, where n is the index of refraction in a Standard Atmosphere, and R = R0 + z. The index of refraction describes how light travels through a medium; it is the ratio of the speed of light in a vacuum to the speed of light in the medium. For ordinary water, n = 1.333, which means that light travels through a vacuum 1.333 times faster than it does through water. For air at 0°C and at a pressure of 1013 millibars, n = 1.000293, a number close to one. n decreases toward one with altitude. The lower the density, the closer n is to one. If you put a straight edge to the graph of R = nR in Figure 3; you will see that the graph is nearly linear.
Now, we must explain the meaning of the black vertical line at the right side of Figure 3. The markings on this line, with numbers from zero through 210, give the dip angles in minutes of arc for an observer at O at an altitude of z = 12.192 kilometers (40,000 feet). If the observer looks along a curved ray at any dip angle within this range, this graph tells you how close the ray will come to the earth's surface in a Standard Atmosphere. For example, draw a horizontal line from the 144-minute mark to the solid curve (for the Standard Atmosphere) as shown in Figure 3, then a vertical line down to the horizontal axis at 6,000 meters. This tells you that the closest approach to the surface of the earth by a ray, observed at a dip angle of 144′ from 40,000 feet, is 6,000 meters (19,685 feet).
A diagram that can be used to determine how close a light ray comes to the surface of the Earth, when viewed from an altitude of 12.192 km (40,000 ft) at dip angles from 0′ to 210′. The horizontal axis gives the height in meters above sea level. R0 is the radius of the Earth at 45° latitude (6357 km). R is the distance from a point in the atmosphere to the center of the Earth. The vertical axis is nR – R0, where n is the index of refraction. The solid curve is valid for R = nR in a Standard Atmosphere, for which the value of n varies with altitude. Values of dip angle in minutes of arc appear along the vertical line at right. The text explains the red lines and how to use this diagram.
We want to know the dip angle of the green ray that goes through the apparent horizon at H, the ray shown in Figure 2. We can find it by using the diagram of Figure 3 in the reverse way. Draw a vertical line up from the zero height mark (lower left of the diagram) to the lower end of the curve for R = nR. Then draw a horizontal line (dashed red) to the right until it intersects the dip angle index. The intersection occurs at a dip angle of 201′. Thus the apparent horizon, where the upper edge of the sun disappears at sunset, is at a dip angle of 201′. This compares with geometric horizon, which we computed earlier at a dip angle of 212.6′.
Finally, we can answer the original question. At 40,000 feet in a Standard Atmosphere, the effect of altitude on the dip angle of the apparent horizon is significantly larger than the effect of atmospheric refraction. The curvature of the ray between the observer and the apparent horizon changes the dip angle by just 11.6′.
Sunset occurs when the upper limb of the sun disappears below the apparent horizon. To compute the time of sunset, one must know the true position (dip angle) of the sun when it is last visible. Equivalently, we must know how much the atmosphere refracts the last ray from the sun beyond the apparent horizon (in the direction away from the observer) in a Standard Atmosphere. That number is known to be 34.5′, and most of this refraction occurs close to the earth's surface. Official sunset thus occurs when the upper edge of the sun lies 34.59 below the apparent horizon.
Weatherwise Contributing Editor THOMAS W. SCHLATTER is a retired meteorologist and volunteer at NOAA's Earth System Research Laboratory in Boulder, Colorado. Submit queries to the author at email@example.com, or by mail in care of Weatherwise, Taylor & Francis, 530 Walnut Street, Suite 850, Philadelphia, PA 19106.
I thank Andrew T. Young, an astronomer at San Diego State University and a leading expert on the green flash, for the three figures, for pointing me to his Web pages, and for helping generously with this answer. For those wanting to learn more about refraction, I highly recommend his Web page at http://www-rohan.sdsu.edu/~aty/explain/atmos_refr/astr_refr.html, which is a gateway to many other interesting Web pages.