How Did Eratosthenes Measure the Earth?

Shadows farther and farther north

Stay with 21 June at noon, when the Sun is overhead at Ranchi. Imagine friends with 1 m sticks farther and farther due north of Ranchi. Our script worked out, for the real Earth, the Sun’s angle from straight up at each place, and the shadow:

Distance north of Ranchi	0 km	480 km (Bhaktapur)	1,000 km	2,000 km	3,000 km	5,000 km
Sun’s angle from straight up	0∘°	4.3∘°	9.0∘°	18.0∘°	27.0∘°	45.0∘°
Shadow of a 1 m stick	0 cm	7.6 cm	15.8 cm	32.5 cm	50.9 cm	100 cm

(Asha measured 7.4 cm, not 7.6 cm. There is a reason, and we come to it.)

Look at the two rows separately.

• a.The angle grows in step with the distance: 1,000 km gives 9∘°, 2,000 km gives 18∘°, 3,000 km gives 27∘°. Every extra 1,000 km north adds the same 9∘°. Twice the distance, twice the angle.
• b.The shadow doesn’t. 1,000 km gives 15.8 cm, but 2,000 km gives more than twice that, 32.5 cm, and 5,000 km gives a whole meter. The shadow grows faster and faster.

So the angle, not the shadow, is the thing that keeps in step with the distance. Keep that in mind: it is the key to the puzzle.

Two ideas from Grade 8 do all the work: the angles where a line crosses two parallel lines, and proportion.

The Sun’s rays are parallel

The Sun is about 150 million km away. Light from it reaches Ranchi and Bhaktapur along two lines that are, for every purpose here, parallel: aimed at two places 480 km apart from 150 million km away, they differ in direction by less than a five-thousandth of a degree.

So on a flat Earth, every stick would cast the same shadow at the same moment. Different shadows mean the ground itself is tilted differently from place to place: the Earth is curved.

A stick points away from the center

A stick standing straight up, as a plumb line shows, points straight away from the center of the Earth. Make Asha’s stick longer and longer downward, in your mind, through the ground, and it would reach the center. The same for Rohan’s stick in Ranchi, and there the Sun’s ray runs straight down the stick and on towards the center too, which is why it has no shadow.

This is how Eratosthenes saw it, 2,200 years ago, in Egypt: at noon on midsummer’s day the Sun was overhead at Syene (now Aswan), in the south, but a stick at Alexandria, in the north, showed the Sun about 7.2∘° from straight up.

Look at Figure 22.1. There are two parallel lines: the ray that grazes the top of Alexandria’s stick, and the ray down Syene’s stick to the center. The line from the center up through Alexandria’s stick crosses both of them: it is a transversal. Where a transversal crosses two parallel lines, the alternate angles (on opposite sides of the transversal, between the parallels, in a Z shape) are equal.

One of them is the angle at the top of the stick, between the stick and the ray going on to the shadow’s tip: the angle the shadow shows. The other is the angle at the center of the Earth between the two cities. So:

The angle a shadow makes at the top of its stick is the angle at the Earth’s center between the stick’s place and the place where the Sun is overhead.

That is why the angle in the table keeps in step with the distance. An angle at the center of a circle is called a central angle, and the piece of the circle’s edge between its two arms is an arc. Double the arc, and you double the central angle; take a third of the arc, and you get a third of the angle. (This needs both cities on one north–south line, so that they and the center lie in the one slice of the Earth that the noon Sun shines along. Ranchi is only about 10 km west of due south of Bhaktapur.)

A share of the whole circle

Now proportion. The whole way round a circle is 360∘°, so a central angle tells you what share of the circle the arc is:

$$\frac{\text{arc}}{\text{all the way round}} = \frac{\text{central angle}}{360^\circ}$$

A quarter turn, 90∘°, is a quarter of the circle; 1∘° is a 360th. Eratosthenes’ 7.2∘° is $\tfrac{7.2}{360} = \tfrac{1}{50}$ of a full turn. So the 5,000 stadia (a Greek unit of distance) between Syene and Alexandria was a fiftieth of the way round, and the whole Earth was

$$50 \times 5{,}000 = 250{,}000 \text{ stadia.}$$

Asha measured a shadow, not an angle. How does 7.4 cm become degrees? Two ways, both with what you already know.

By scale drawing

Draw it smaller, as in Grade 8’s scale drawings. A 1 m stick and a 7.4 cm shadow, drawn at a tenth of the size, become a line 10 cm tall with 7.4 mm along the ground at its foot. Join the top of the line to the end of the 7.4 mm, and measure the angle at the top with a protractor.

The trouble is that it is about 4∘°, and a school protractor reads to the nearest degree. Is it 4∘° or 4.5∘°? Our script tried both: 4∘° gives $\tfrac{360}{4} \times 480 = 43{,}200$ km; 4.5∘° gives 38,400 km. Half a degree, a mark you can hardly see on the protractor, changes the answer by 4,800 km. A small angle must be measured very carefully, because everything is multiplied by 360 divided by it.

The stick is a little Earth

There is a neater way, with no protractor. Swing the stick’s length round its top like the hand of a clock: its foot draws a circle of radius 1 m. The whole way round that circle is $2 \times \pi \times 100 \approx 628$ cm (the circumference of a circle, from Grade 8), so 1∘° of it is $628 \div 360 \approx 1.745$ cm, about 1.75 cm.

For a small angle, the arc swung from the stick’s foot ends almost exactly at the shadow’s tip, as Figure 22.2 shows. So Asha’s 7.4 cm shadow is about $7.4 \div 1.745 \approx 4.24$ of those one-degree steps: 4.24∘°. (Our script found the arc and the shadow differ by 0.2% at this angle, far less than the shadow’s blurry edge; below.)

Then the Earth:

$$\frac{360}{4.24} \times 480 \text{ km} \approx 84.9 \times 480 \text{ km} \approx 40{,}800 \text{ km.}$$

And here is a shortcut, from the same picture. The two angles in Figure 22.2 are equal, so each arc is the same share of its own radius: 7.4 cm is $\tfrac{7.4}{100}$ of the stick, so 480 km is $\tfrac{7.4}{100}$ of the Earth’s radius. The Earth’s radius is

$$480 \times \frac{100}{7.4} \approx 6{,}490 \text{ km},$$

and once round the Earth is $2 \times \pi \times 6{,}490 \approx 40{,}800$ km again. That is the answer the widget found. The stick is a little Earth: its shadow is to its height as the distance between the cities is to the radius of the Earth.

The missing 2%

Asha’s 40,800 km is 2% more than the real 40,008 km. Where did it go wrong? Rohan’s stick was not quite shadowless. On 21 June the Sun is overhead at noon on the Tropic of Cancer, which runs about 8 km north of Ranchi. Rohan, 8 km south of it, sees the Sun a tiny 0.08∘° to his north, and a 1 m stick casts a shadow 1.3 mm long, pointing south: too short to notice, but it means the ground between the cities has turned by $4.24 + 0.08 = 4.32°$, not 4.24∘°. Then

$$\frac{360}{4.32} \times 480 \text{ km} = 40{,}000 \text{ km.}$$

Two school sticks, a tape measure, a map and some care: the size of the Earth, right to within a fraction of a percent.

How long is a degree?

If once round the Earth is 40,008 km, one degree of it is $40{,}008 \div 360 \approx 111$ km, and Bhaktapur, 27.7∘° north of the equator, is about $27.7 \times 111 \approx 3{,}075$ km from it.

The twist: two shadows can’t tell a flat Earth

Eratosthenes’ argument rests on two assumptions: the Earth is round, and the Sun is so far away that its rays are parallel. What if both were false? Suppose the Earth were flat, and the Sun close above it, straight above Ranchi.

Then the stick in Ranchi has no shadow, and the stick in Bhaktapur, 480 km away, has a shadow because the Sun’s rays reach it slanting. For a 7.4 cm shadow, the slant must be 7.4 cm across for every 100 cm down; so the Sun must be $480 \times \tfrac{100}{7.4} \approx 6{,}490$ km up, the same number as the Earth’s radius, by the same similar triangles. The two shadows are explained just as well: the same two shadows, in two different worlds.

So two measurements can never tell the two worlds apart, but a third can, and measurements at many places, the Earth’s round shadow on the Moon in an eclipse, and ships rising hull first over the horizon all agree with a round Earth and a far Sun. The lesson is about mathematics, though: a calculation is only as good as its assumptions. Eratosthenes’ method works because its two big assumptions, a round Earth and a far-off Sun, are true.

At Science Playground, the Analemmatic Sundial is set out for Kathmandu, and you are its stick: your shadow tells the time. At noon by the Sun it points due north and is at its shortest, the moment Asha needed.

• What Is the Shortest Way to the River and Back? Equal angles again, this time making the shortest path, the way light bounces off a mirror.
• Why Do Flights to America Fly Near the North Pole? (Grades 9–12) Now that you know the Earth’s size: the shortest way between two cities on a round Earth.
• How Did Madhava Find π Without a Calculator? (Grades 9–12) The number in every circle you measured here, worked out to many places in Kerala six hundred years ago.
• Why Is It Cold in Winter? (Science, Going Further 6–8) Why the noon Sun is overhead on the Tropic of Cancer in June, and lower in the sky in December.
• Grade 10 Mathematics, Heights and distances: the tangent, which turns a shadow into an exact angle, even a big one.