On rare days, it is possible to see from Santa Cruz (California) all the way to where the distant Gabilan Range disappears over the horizon beyond Soledad. From Lighthouse State Beach, the stack tops of the Moss Landing Power Plant (as of the 1990s) seem to align with the mountain ridge near North Chalone Peak. This coincidental alignment is caused by the relative heights and distances, curvature of the Earth, and atmospheric effects.
Lighthouse State Beach, in Santa Cruz, California USA, is a well-known surfing spot called steamer lane. The locals call the adjacent beach It's beach, although I don't know why.
This lighthouse is a memorial and a surfing museum, not the navigation aid at the yacht harbor. As long as there's no fog, the view across Monterey Bay is superb.
Once in a while, a gentle easterly wind clears the coastline, and it's possible to see long distances. This tends to happen between storms in the cool. rainy season – known as winter, elsewhere. One such day, I was looking southeast across the bay. It was possible to see clearly all the way to where the Gabilan Range disappeared over the horizon along the Salinas Valley, some fifty miles away.
The year was soon after the 500th anniversary of Columbus arriving in the Western Hemisphere, and I had been thinking about historical methods for estimating the size of the Earth. A reasonably good estimate was known during the Graeco-Roman era, but at the time of Columbus, different estimates were in use. Contrary to popular mythology, educated Europeans knew that the Earth was spherical, and had an idea of its size. The most commonly used size, along with an estimate of the eastward distance to the Orient, indicated that it was impractical to get there by sailing westward. But Columbus was able to convince Spain of an erroneously small size, which would have made it practical.
Here is the view from the low cliff at Lighthouse State Beach. There is a conspicuous power plant at Moss Landing, 17 miles away, along the bay shore. The tops of the stacks seemed to touch the mountain ridge far behind them. The view is actually clearer to the naked eye, which can distinguish subtle shades.
You can't miss noticing the power plant. It is by far the largest and most conspicuous man-made object along the coast of Monterey Bay. Seen from Santa Cruz, the two stacks have greater visual separation than shown in this photo.
The ridge was in the Gabilan Range. Just to the right of the stacks was its highest local point, North Chalone Peak. On another day, I drove to the town of Gonzales, and took a photo of the ridge from nearby, from the direction aligned with Santa Cruz.
I realized that the visual alignment of stacks and mountains could provide a way to calculate the size of the Earth. The calculation turned out to be more difficult than I had imagined, since a subtle atmospheric effect must be taken into account.
On a clearer day, I went down to the beach at low tide. Using binoculars, I sketched the alignment. The reason for choosing low tide was so that my view would be from sea level, defined as the average level of high tide, without the need to swim.
I verified my line of sight using the Atlas of California.
Here are sections of USGS topographic maps, showing the sightline at Santa Cruz through the stacks at Moss Landing. At Santa Cruz, "W" marks the position of the municipal wharf.
From the beach to the power plant is 27.51 km, and from the beach to the ridge is 92.40 km. My view from sea level was at elevation zero. The topographic map shows the elevation of the peak as 3305 feet, but I am authoratatively informed that the map is wrong! The peak is only 3194 feet. I therefore reduced the ridge value to 2789 feet (850 meters).
The customer response group at Duke Energy Corp., which owned the power plant at the time, informed me that the stacks were 500 feet (really) above base elevation 32 feet, for a total elevation of 532 feet (162.2 meters) above sea level.
Atmospheric effects are important. If there were no air, then the information about heights and distances would be enough to calculate the size of the Earth. In fact, I tried it that way first, since I didn't realize how important atmospheric effects would be. The result that was not even close. To get a good result, we need to understand the refraction of light in a clear, well-mixed lower atmosphere.
On most days, a temperature inversion normally blocks or distorts the long-distance view over Monterey Bay. Typically the view is obscured by fog or mist. But even when the air is clear, an inversion distorts the path of light so that distant objects appear in atypical positions. The most severe conditions produce a type of mirage, the Fata Moraga, discussed in the Appendix.
Fortunately, the normal inversion is dissipated by the weather on days when the alignment is clearly visible. In this case, we can use a standard model of the lower atmosphere to calculate the refraction of light as it passes through air.
Surfers are familiar with wave refraction. In coastal areas, wherever the depth of ocean water is less than a few times the distances between wave crests, the speed of a wave depends upon depth. The shallower the water, the slower the wave moves. As a result, the direction of wave travel is altered as the wave approaches shore. This occurs because different parts of the same wave crest will be above water of varying depths. In fact, the excellent surfing near the lighthouse is largely due to wave refraction.
Even in the mildest climate and most temperate weather, air density decreases as height increases. As a result, the path of light is slightly and gradually bent as it passes though a long expanse of air. This "normal" mirage causes distant objects near the horizon to appear higher than they really are. The effect partially negates the appearance of objects sinking over the horizon due to Earth's curvature. The combined effect is to make Earth appear somewhat flatter than it would appear if our planet had no atmosphere.
The illustration shows light waves (imagined as water waves) traveling from right to left, but faster in the upper part. The crest will slowly curve as it moves. The direction of travel is always locally perpendicular to the crest. When a wave reaches the observer, it will appear that the source, traced backwards straight from the direction of arrival, is in the mirage location rather than in the actual location.
Imagine a tall, distant mountain range behind low foothills. These three drawings show an imaginary flat Earth with no atmosphere, a curved Earth with no atmosphere, and the actual case of curvature and atmosphere.
At right (not to scale), solid blue represents sea level. Dashed black represents a horizontal line from the observer, and red represents the line of sight from observer past the stack tops.
The first figure shows what would happen on an imaginary flat Earth, with no mirage. Given the heights of the stacks and ridge, and their distances from the observer, they would not align. The top of the ridge would appear to be well above the stacks.
The second figure accounts for the Earth's curvature, but neglects the mirage. In this case, the top of the stacks would appear well above the ridge.
With both effects, the stacks and ridge align. Dashed blue is the mirage of sea level.
We will first calculate Earth curvature. For various reasons, the Earth (geoid) is not quite spherical but that has little effect at our level of accuracy.
We will use a spherical Earth (blue line) of radius R. Its center is at point C.
Suppose an object T is located at distance D from a sea-level observer O. The horizontal plane at the observer intercepts the object at an elevation F above the object's base B. If the Earth's curvature were the only effect, then the part of the object below F would be beneath the observer's horizon.
The diagram is not to scale, so we can simplify the Pythagorean Theorem. For practical distances, R is very much larger than D or F. For Earth, average R = 6371000 meters, about 100 times the surface distances. It is a good approximation to replace (2R+F) with 2R.
This gives us a simple expression for F. It is the square of the distance, divided by twice the radius.
Curvature due to atmospheric refraction can be calculated from the physical properties of the standard atmosphere. The density of air can be calculated from its temperature and pressure. Both of the factors vary in time and place according to the geography and weather. On favorable days, the temperature and pressure profile of the lower atmosphere can be estimated in comparison to a standard.
Information from a number of tables and equations in the CRC Handbook of Chemistry and Physics was use to create this chart.
The differences in speed are less than 0.1%, but the distances are long, and the effect accumulates.
The speed of light in air is also a function of wavelength. We must use values weighted towards the red end of the spectrum, even though distant objects appear bluish. Here is why:
Imagine three identical, identically illuminated objects at various distances. In the absence of air, the only noticeable difference is that distant objects seem smaller.
Suppose the same objects are immersed in air, at nighttime. Light from an object is partially removed along its path by the surrounding air. Blue light is removed to a greater extent. The more distant the object, the darker and redder it appears. This atmospheric effect is not related to the "red shift" of Astronomy.
During the day, it is still the case that air removes light, especially blue light, along the path. But the same effects that make the sky blue cause stray light, particularly blue, to be mixed in. Distant objects appear lighter and bluish where they would otherwise be dark, and slightly darker where they would otherwise be brighter than the surrounding air. Most of the "original" light arriving from the object is reddish.
Above, the drawing imagines light waves radiating from a distant point. For red light, the distance between wave crests would be less than a millionth of a meter. The upper region is magnified at right. In a short period of time, a wave crest J1-K1 would move to J2-K2 if the speed were uniform. But since the air is less dense in the upper portion, the upper part of the wave moves faster than the lower part. So instead of arriving at J2-K2, the wave reaches J3-K3. The wave front is tilted by an angle [a]. The direction of motion is also tilted, since it is perpendicular to the wave front. From the point of view of the observer, the wave appears to originate from a higher position.
Imagine a section of the light waves. The bottom edge is at elevation H, the top is Δh higher. As the bottom edge of the light wave moves distance D, the top edge moves (D+δd) in the thinner air aloft.
Even for a specific color, the speed of light in air depends on the weather. To keep the analysis manageable yet reasonably accurate, I used data for red light at vacuum wavelength 700nm, in air at average elevation 500m above sea level. The speed of such light is close to 299713000 m/s, and its rate of change is about 7.65 m/s for each meter increase in elevation.
The ratio of distances along top and bottom is the ratio of their speeds. In meters, (D+Δd)/D = (299713000 + 7.65Δh) / 299713000. That is, (Δd/Δh) = 2.55 x 10-8 D.
The large and small triangles are nearly similar. The mirage causes an apparent height increase of Δm. Then, (Δm/D) = (Δd/Δh) = 2.55 x 10-8 D. That is, Δm = 2.55 x 10-8 D2 measured in meters.
Let's put it together. If the Earth were flat, with no mirage, then the math would be simple. For the stacks and ridge to align (as seen from sea level), it would have to be the case that the ratio of the distances (from the observer to the objects) was the same at the ratio of the heights of the objects. The ridge is 3.36 times farther than the stacks, so the ridge would have to be 3.36 times higher than the stacks. The stacks are about 162 meters above sea level, so the ridge would have to be about 544 meters high. This is not even close to the probable ridge elevation of 850 meters.
Earth's curvature lowers the stacks, and lowers the ridge even more. But the mirage partially offsets the curvature. The combined effects create apparent heights seen by the observer, which ought to align. The results are tabulated here. Taking into account the curvature and mirage, the amount of mis-alignment (for various reasons) is 12 meters at the ridge, or about 40 feet.
In the above calculation, I assumed that the radius of the Earth was already known, then determined how much the object heights were misaligned. This is historically realistic, since good means of determining dimensions were known long before there was adequate knowledge of meteorology and optics. In fact, the earliest known method for determining the size of the Earth was not significantly influenced by atmospheric optics.
But suppose we reverse history, and instead calculate the radius of the Earth assuming that all other factors are known. Then, we compare the almost-similar triangles formed with vertex at the observer, and opposite sides at the stacks and ridge. The visible height at the power plant is its true height, minus the fall-off from Earth curvature, plus the rise from atmospheric refraction. The same applies at the ridge.
Using the numerical values for distances, elevations, and the constant appearing in the refraction equation, the math is an exercise in ordinary algebra, solving for R. My result (assuming stacks and ridge are exactly aligned) is 6535370 meters, versus the actual value of 6371000 meters.
And there you have it. After all this analysis, I missed by 12 meters. That's less than 2 percent of the actual ridge elevation. Or, calculating Earth's radius from the other data, the result exceeds actual value by about 2.6%. However, this is nowhere near the accuracy that could be attained. Those of you who followed the complete mathematical analysis noted that I made simplifications to keep the geometry under control. More advanced math can account for the simplifications (such as assuming non-similar triangles to be similar). But the error from simplified math is only a couple of meters. That's not the problem.
One source of error is that I have no idea how well-aligned the stacks and ridge really are. As seen through binoculars, it looks very close. But images seen at such a distance, especially for the ridge, are not sharp. Also, the various colors of light do not produce the same mirage. I used red light at 700nm vacuum wavelengths as a reasonable guess for the average of light surviving the passage from ridge to observer.
Perhaps the most important contribution to the error was that I assumed the air profile of the "standard atmosphere," because the air was well-mixed. That's not necessarily the case, although I chose likely weather conditions. Of course, there's always the possibility that the "500 feet above ground at 32 feet elevation" stacks at Moss Landing are not really that tall. But I have no intention of finding out. There are other things to do!
REFERENCES
The U. S. Standard Atmosphere (1976) and its optical properties:
David R. Lide, Editor-in-chief. CRC Handbook of Chemistry and Physics, 80th Edition, 1999-2000. NY: CRC Press, 1999. Ch.10, p. 220; Ch. 14, p. 16-17.
Sightline Santa Cruz - ridge based on "Central California" map and USGS surveys:
Michael W. Donley, et. al. Atlas of California. Culver City CA USA: Pacific Book Center, 1979. p. 168.
Topographic map images for Santa Cruz, Moss Landing, and North Chalone Peak:
GIS Data Depot. 24K Digital Raster Graphics (DRG), United States Geological Survey topographic maps. But, Chad Moore of the National Park Service at Pinnacles National Monument informed me that the USGS map errs at North Chalone Peak. It should be 3194 ft, not 3305. I adjusted the value for the ridge accordingly, to 2789 ft rather than the map 2900.
Information about the stack heights at Moss Landing:
Customer Response team at Duke Energy Corporation, owner of the plant (circa 1999).
APPENDIX: The Fata Morgana Mirage over Monterey Bay
On most days, an atmospheric temperature inversion blankets the bay and surrounding coast. Generally, an inversion occurs when warmer air overlies cooler air. An example is the stratosphere (heated from above by UV absorption) atop the troposphere (heated by the surface). Local temperature inversions can arise in the lower troposphere if the air is cooled from the surface, or if warm air aloft arrives from elsewhere. What matters is actual air temperature at various heights, in relation to the mathematically expected decrease with height.
California coastal winds often arrive after lengthy passage over cool surface waters of the Pacific Ocean. The cooled surface air is the marine layer, which often brings fog and a temperature inversion. There is generally an abrupt air temperature change at the boundary of the marine layer.
Aside from fog, the inversion can cause some peculiar optical effects. One of them is the Fata Morgana, a type of mirage well-known in southern Italy. The name derives from the supernatural abode of Morgan Le Fay of the Arthurian (English) legends, and was probably brought to the region by Normans during the Middle Ages.
In the Monterey Bay region of California, the Fata Morgana may appear when the marine layer is very shallow, when there is no fog, and when the air above the marine layer is much warmer than below. These conditions may arise during the relatively calm period after a warm easterly wind has blown the usually-deep marine inversion layer out to sea, but before onshore breezes restore the deep inversion. During this time, only the air close to the ocean surface is cooled.
The Fata Morgana creates the appearance of tall cliffs along the shoreline, even where the cliffs would not be noticeable. Without the mirage, the apparent height of the shoreline cliffs diminishes with distance, as shown here:
Here is a magnified view of the Fata Morgana photo, showing the area near the power plant. Notice how the image just above the lines is the inverted reflection of the portion just below.
[Online at Wikimedia Commons] Gerd A.T. Mueller has a high-resolution photo of the Fata Morgana along the Norwegian coast. The circumstances are very similar to the one shown above.
The "speed of light" is a universal constant of nature in vacuum only. In any material medium, such as air, the actual speed of light depends upon wavelength and upon properties of the medium. The denser the air, the slower the speed. Even though the change in speed is very small, the effect can be noticeable.
Compare the chart of elevation versus density for the standard lower atmosphere to the chart when a severe marine layer is present. Within the marine layer, the air is much cooler and denser.
Popular books describe how the air above a hot roadway surface can act like a mirror, bending the path of light. Fata Morgana is the inverse, with cool at the bottom, and reflection from above the observer's level.
This is a type of superior mirage, caused when light paths meet an abrupt change in air density at shallow angle. The effect is like a mirror. Fata Morgana is an extreme case of the region's normal inversion. However, the mirage is generally not a simple mirror, the way I have drawn it here. The temperature boundary is rarely sharp, and even when sharp it does not reflect perfectly. In most cases, light paths are slightly bent, and the effect is to vertically stretch the appearance of the distant object. This is a more extreme case of the standard mirage, which I will explain below. In many cases, the Fata Morgana includes both raised and stretched mirage (not inverted), and an inverted top.
But on the day of the alignment, there was no inversion, and the lower atmosphere was well-mixed rather than abruptly layered. Wind from the east had blown the marine layer out to sea. The variation of temperature and pressure, as a function of altitude, was comparable to that of the standard atmosphere. Below, the image shows a section of coastline without the Fata Morgana conditions. Compare it to the one above.
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Copyright ©2000, 2005 by Robert Allgeyer. All Rights Reserved.
Original version of this essay: Mid-1990s.
Internet: At another URL, 2000 - 2005. This URL: November 11, 2005.