Of Particular Significance

Part I, Endnote 1

  • Quote: Over two thousand years ago, Greek thinkers became experts in geometry and found clever tricks for estimating the Earth’s shape and size.

  • Endnote: The shadow that the Earth casts on the Moon during a lunar eclipse is always disk-shaped, no matter the time of day, which can be true only for a spherical planet. Earth’s size is revealed by comparing the lengths of shadows of two identical objects, separated by a known north-south distance, measured at noon on the same day.*

By the time that ancient Rome’s power was expanding, Greek scholars understood the basics of solar and lunar eclipses. Noting the relation between the phases of the Moon and the positions of the Sun and Moon, they recognized that the Moon’s light is reflected sunlight. They were aware that New Moon occurs when the Sun and Moon are on the same side of the Earth, while Full Moon occurs when they are on the opposite sides. And they knew that a lunar eclipse occurs when the Earth lies between the Sun and Moon, so that the Earth blocks the Sun’s light and casts a shadow on the Moon. This is illustrated (not to scale) in Fig. 1.

Fig. 1: (Top) When the Moon and Sun are on opposite sides of the Earth but not perfectly aligned, the Moon is full: its lit half faces the Earth. (Bottom) But when the Moon moves directly behind the Earth, it enters its shadow and is partly or completely darkened — a “lunar eclipse”. Sizes and distances are not to scale.

From these shadows, they confirmed the Earth was a sphere. Clearly, if the Earth had the shape of an X, it could cast an X-shaped shadow.

Fig. 2: Light from the Sun (orange dashed line) would cause an X-shaped Earth to cast an X-shaped shadow on the Moon.

If Earth were a flat disk like a coin, then depending on the Moon’s location in the sky, the Earth’s shadow might be circular or might be oval. The important point is the shadow of a circular disk is not always a circular disk. You can confirm this with a coin and a light bulb.

Fig. 3: If the Sun and Moon aren’t directly aligned face-on with a disk-shaped Earth, the disk’s shadow on the Moon will be an oval, not a circle.

The only shape which always creates a circular, disk-like shadow, from any angle and from any place and at any time, is a sphere, as you can confirm with a ball and a light bulb.

Fig. 4: Only a spherical Earth always casts a disk-shaped shadow, which causes the bright areas of the moon to be crescent shaped at all times during a partial lunar eclipse.

This is consistent with what is actually observed in eclipses, as in Fig. 4. The circular shadow of the Earth is shown especially clearly when sets of photos taken during an eclipse are carefully aligned.

Next, once you are aware that the Earth’s a sphere (and, as the Greeks also knew, that the Sun is far away ), it’s not hard to learn the size of the Earth. Imagine two vertical towers sitting on flat ground in two different cities. To keep things especially simple, let’s imagine one city is due north of the other, and the distance between them — call it “D” — is already known.

In each city, a person observes their tower’s shadow at exactly noon. (No clock is needed, because each person can watch the shadow over time, and noon is when the shadow is shortest.) The end of the shadow and the tower’s base and top form the points of a right-angle triangle, as in Fig. 5, whose other angles we can call ⍺ and πœƒ. For a person squatting at the shadow’s end, the angle ⍺ is easily measured: it is the angle formed by the tower’s silhouette against the sky. Because this is a right-angle triangle, the angle πœƒ is 90 degrees minus ⍺, so each observer can easily determine πœƒ for their city’s tower. We’ll call their two measurements πœƒ1 and πœƒ2. Knowing these angles and their distance D, they know everything we need to determine the size of the Earth.

Fig. 5: The right triangle formed by connecting a tower’s base, its top, and the end of its shadow at noon. Of greatest interest is the angle πœƒ.

The key observation is that the difference in these angles is the same as the difference in the latitude between the two cities. To see this, examine Fig. 6 below. The paths of sunlight (the orange dashed lines in Figs. 5 and 6) are parallel to the Earth-Sun line. [This (almost-exact) parallelism is only true because the Sun’s distance from Earth is much larger than the Earth’s size — a fact which the Greeks knew.] Meanwhile the line from the Earth’s center to the first tower (call it L1) is a continuation of the line from that tower’s base to its top. Because (a) L1 forms an angle πœƒ1 with the sunlight, as shown in Fig. 5, and (b) the sunlight line and Earth-Sun line are parallel, as shown in Fig. 6, the intersection of L1 with the Earth-Sun line is also πœƒ1! Similarly, the line from the Earth’s center to tower 2 forms the angle πœƒ2 with the Earth-Sun line. As can be seen in Fig. 6, it follows that the angle between the two lines connecting the Earth’s center to the two towers is πœƒ2 – πœƒ1 , the difference in the noon-time sun angles as seen by the two observers!

Fig. 6: The angles πœƒ1 and πœƒ2 that involve the towers’ shadows (Fig. 5) are also the angles between the Earth-Sun line and the lines connecting the Earth’s center to the two towers; the difference between the two angles is the difference between the two cities’ latitudes.

Now, however, the observers can use the fact that they know D, the distance between the cities. In particular, the distance D is to the Earth’s circumference C just as πœƒ2 – πœƒ1 is to 360 degrees (or, in radians, to 2β„Ό). In formulas

  • D/C = (πœƒ2 – πœƒ1)/360Β°

So (dividing and multiplying on both sides) the Earth’s circumference is simply

  • C = D [360Β° / (πœƒ2 – πœƒ1) ]

and since they know both D and πœƒ2 – πœƒ1 , they now know the Earth’s circumference. (Note this correctly says that if the angle were 90 degrees = 2β„Ό/4, then D would be C/4.)

Eratosthenes made this measurement (in a slightly different way) around 240 B.C.E. Reports by classical historians do not quite agree on what he found, but in the most optimistic interpretation of the historical record, he was well within 1 percent of the correct answer. And why not? Once you’ve realized what you should do, this is a relatively simple measurement; it’s one that you and a distant friend could carry out yourselves.

Note: You might prefer to see the answer in radians instead of degrees; since 360Β° = 2β„Ό radians, we can write

  • C = D [2β„Ό / (πœƒ2 – πœƒ1) ] (in radians)

and since C = 2β„ΌR, where R is the Earth’s radius, that gives us a particularly simple formula

  • R = D / (πœƒ2 – πœƒ1) ( in radians)

Note: If the two cities are not due north-south of one another, this poses no problem. Measure tower 1’s shadow at the first city’s noon, and tower 2’s shadow at the second city’s noon on the same day; then take D not to be the distance between the two cities but instead the distance between their latitude lines. Practically speaking, we can make a triangle with one side being the distance between the cities and the other two sides aligned north-south and east-west; then D is the length of the north-south line, as in Fig. 7. With this definition of D, the formulas above are still valid.

Fig. 7: For two cities that are not in a north-south line, D should be defined as the distance between their lines of latitude, and thus the length of the north-south line in a right triangle connecting them. (The triangle should be carefully drawn on the Earth’s globe, which I have not done here.)
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