In 240 BC, the Greek astronomer, geographer, mathematician, music theorist, and librarian Eratosthenes of Cyrene (c. 276 BC – c. 195/194 BC) calculated the circumference of the Earth without even leaving Egypt. Here’s how:

## Eratosthenes’ method to calculate the Earth’s circumference

Eratosthenes knew that at local noon on the summer solstice (at the time of the longest day, about 21 June in the northern hemisphere) in Syene (modern Aswan, Egypt), the Sun was directly overhead – Syene was in fact slightly north of the tropic, though (see notes 1). Local noon is – technically when the sub-solar point is somewhere over your meridian, it’s noon for you. So, on that day, Syene is the sub-solar point of Earth (the sub-solar point on a planet is where its sun is perceived to be directly overhead). To learn more about the local noon and the subsolar point, see the article titled “How Earth Moves“.

Related: How Earth moves?

He knew that the Sun was directly overhead in Syene because the shadow of someone looking down a deep well at that time in Syene blocked the reflection of the Sun on the water. He also knew that, even on the Summer Solstice, in his home city of Alexandria, which is further north than Syene, the Sun was never directly overhead. In Alexandria, the angle of elevation of the Sun would be 83° or 7° south of the zenith on June, 21.

Assuming that Alexandria was due north of Syene (in fact, Alexandria is on a more westerly longitude) he concluded, using geometry of parallel lines, that the distance from Alexandria to Syene must be 7/360 of the total circumference of the Earth (see the image below).

So, if one would know the distance between Syene and Alexandria, s/he could calculate the circumference of the Earth. 360/7 is close to 1/50th of a circle, so Eratosthenes concluded that the Earth’s circumference was fifty times that distance.

C = 360/7 x d (C-> circumference of Earth, d -> the distance between Syene and Alexandria)

The distance between the cities was known from caravan travelings to be about 5,000 stadia (see notes 2). Carl Sagan even says that Eratosthenes paid a man to walk and measure the distance (see the video below).

Today, we know that the equatorial circumference of Earth is 40,075.017 km (24,901.461 mi). If Eratosthenes used the Olympic stadion of 176.4 m, which would imply a circumference of 44,100 km, an error of 10%. If he used the 184.8 m Italian stadion which became (300 years later) the most commonly accepted value for the length of the stade, the calculation gives a circumference of 46,100 km, an error of 15%.

Please note that, Eratosthenes made two important assumptions, neither of which is perfectly accurate:

1. That the distance between Alexandria and Syene was 5000 stadia,
2. That the Earth was a perfect sphere.

Seventeen hundred years after Eratosthenes’ death, while Christopher Columbus studied what Eratosthenes had written about the size of the Earth, he chose to believe, based on a map by the Italian astronomer, mathematician, and cosmographer Paolo dal Pozzo Toscanelli (1397 – 10 May 1482), that the Earth’s circumference was one-third smaller. Had Columbus set sail knowing that Eratosthenes’ larger circumference value was more accurate, he would have known that the place that he made landfall was not Asia, but rather a New World.

Another video, an excerpt from Carl Sagan’s “Cosmos: A Personal Voyage” (a thirteen-part television series written by Carl Sagan, Ann Druyan, and Steven Soter, with Sagan as the presenter).

## Notes

1. On the summer solstice at local noon on the Tropic of Cancer, the Sun would appear at the zenith, directly overhead (sun elevation of 90°).
2. The stadion (Latin: stadium, plural: stadia) was a distance unit that was often used in ancient times. However, not everybody used a stadion of the same length. Some claim Eratosthenes used the Olympic stadion of 176.4 m, which would imply a circumference of 44,100 km, an error of 10%, but the 184.8 m Italian stadion became (300 years later) the most commonly accepted value for the length of the stadion, which implies a circumference of 46,100 km, an error of 15% That is still pretty good!

## Sources

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