Eratosthenes Measures the Earth: Shadows, Angles and Stated Assumptions

How Eratosthenes, as Cleomedes reports him, turned a shadow angle and a distance into the size of the Earth, with every assumption listed and the uncertain length of the stade stated plainly.

Eratosthenes of Cyrene (about 276 to 194 BC) was a scholar and librarian at Alexandria. By later report, he found the size of the Earth from a shadow. His book, On the Measurement of the Earth, is lost. The fullest account is Cleomedes' Caelestia I.7 (I.10 in older editions), by an author of uncertain date. Some historians doubt that this tidy version is what Eratosthenes actually did. The method and the reported figures are Established; the accuracy in kilometres is Aporetic (unresolved).

What Cleomedes reports

Syene (modern Aswan), says Cleomedes, lies under the summer tropic, where the noon Sun stands straight overhead at the June solstice. Then the pointers (gnomons) of sundials there throw no shadow. At Alexandria, further north, they do. In a bowl-shaped sundial the shadow marks an arc. Eratosthenes found it to be one fiftieth of the bowl's circle: 360 / 50 = 7.2 degrees. With the cities taken as 5,000 stades apart (a Greek length), the whole circle is 50 × 5,000 = 250,000 stades.

Strabo (Geography II.5.7) and Pliny (Natural History II.247) give 252,000 stades. Strabo mentions sixty divisions of the circle and, following Hipparchus, 360 parts of 700 stades. Since 252,000 gives 4,200 per sixtieth and 700 per degree, some modern scholars conjecture that it was adjusted to divide evenly. No ancient source checked here gives the reason.

The well. Many retellings add a well at Syene, lit to the bottom at noon. Cleomedes does not. Pliny (II.183) and Strabo (XVII.1.48) describe one without naming Eratosthenes. Pliny says it was made to test the missing shadow.

The geometry

Take the Sun's rays at both cities to be parallel. The ray at Syene falls straight down, along the line to the Earth's centre. Extended down, the upright gnomon at Alexandria also reaches the centre, so its line crosses both parallel rays. A line falling on parallels makes equal alternate angles (Elements I.29, which the angle-sum proof also uses). So the shadow angle at Alexandria equals the angle at the centre between the cities. The shadow angle is also the Sun's zenith angle: its distance from straight overhead. Angles at a circle's centre have the ratio of the arcs they stand on (VI.33), so angle / 360 = distance / circumference, or:

circumference = distance × 360 / angle

The Eratosthenes lab runs the measurement, and The Sky as a Sphere explains the axial tilt behind the tropics.

Every assumption on the table

Cleomedes lists five assumptions: both cities on one meridian (the north-south line through a place), 5,000 stades apart, parallel rays, and the two theorems. The method also needs Syene on the tropic, a correct reading and a round Earth. Modern data show how far each held.

Parallel rays. The Sun is about 149.6 million km away and the cities less than 850 km apart, so the rays differ in direction by about 0.0003 degrees. The Sun's disc, about 0.52 degrees wide in June, blurs every shadow edge.
Syene on the tropic. The tropic's latitude equals the axial tilt: about 23.72 degrees around 240 BC, by J. Laskar's 1986 formula. Aswan is at 24.09 degrees north, about 0.36 degrees (40 km) north of the tropic. The Sun's disc came within about 0.1 degrees of the zenith, so shadows were very short, not nil.
Same meridian. Aswan is 3.0 degrees of longitude east of Alexandria. The angle measures only the north-south separation, about 788 km; the surface distance is about 842 km, 7 per cent more.
The angle. The modern latitude difference is 7.11 degrees, so 7.2 looks excellent. But with Syene 0.36 degrees north of the tropic, a perfect reading at Alexandria would have been about 7.47 degrees. The reported 7.2 is 0.27 below that, about the Sun's radius, so two errors partly cancel.
A round Earth. The Earth is slightly flattened, but its circumferences through the poles and round the equator (40,008 and 40,075 km) differ by only 0.17 per cent.

The stade problem

A stade (stadion) was a Greek length with no single agreed value. Much of the modern debate starts from Pliny (Natural History XII.53): by Eratosthenes' reckoning a schoenus, an Egyptian measure, is 40 stades, "that is, five miles". With eight stades to a Roman mile (1,000 paces, about 1.48 km), the stade is about 185 m. That fits Pliny's rule (II.85) of 125 paces to the stade. Rawlins (1982), Engels (1985) and others adopt 185 m. Hultsch (1882) and others instead took a schoenus of 12,000 Egyptian royal cubits of about 0.525 m. That makes the stade 300 cubits, or 157.5 m. Gulbekian (1987) proposed 166.7 m, and some have argued for about 148 m. All are known here from secondary reports.

157.5 m. 250,000 stades: 39,375 km (1.6% low). 252,000 stades: 39,690 km (0.8% low).
166.7 m. 250,000 stades: 41,675 km (4.2% high). 252,000 stades: 42,008 km (5.0% high).
185 m. 250,000 stades: 46,250 km (15.6% high). 252,000 stades: 46,620 km (16.5% high).

Percentages are against the circumference through the poles, 40,008 km, since the method measures along a meridian. For 250,000 stades the result is within 2 per cent only if the stade is between about 156.8 and 163.2 m.

Aporetic "He was accurate to within 2 per cent" is not defensible as stated. It picks one stade from a disputed range, and a stade chosen because it fits cannot then confirm the fit. It ignores the 252,000 figure. It rests on round inputs, 1/50 and 5,000, and on errors that partly cancel. No source checked says how the 5,000 was obtained. What is defensible: the method is sound and the result is the right size. Across the table, the error runs from under 1 to over 16 per cent. Nobody can now fix the unit.

Engineers meet a milder version of the stade problem, because each candidate unit has an exact size. A 500 GB restore at "100 MB/s" needs about 83 minutes of transfer. If that MB means 1,048,576 bytes, not a million, it needs about 79.

Repeat it with two cities and a stick

Choose two cities on nearly the same meridian, several hundred kilometres or more apart, both north of the Tropic of Cancer. Take their north-south distance D from a map, not from latitudes, which would be circular.
On one day, at each city's local solar noon (the shortest shadow), measure the shadow s of a vertical stick of height h on level ground. Published noon Sun elevations will serve instead.
Each zenith angle is z = arctan(s / h), or 90 degrees minus the elevation. Take the difference, Δz.
The circumference is D × 360 / Δz.

With invented readings: a 1.00 m stick gives noon shadows of 0.50 m and 1.30 m, so z = 26.57 and 52.43 degrees, and Δz = 25.87 degrees. For cities 2,870 km apart, the circumference is 2,870 × 360 / 25.87, about 39,900 km.

Here, misreading both shadows by 5 mm shifts Δz by at most about 1.3 per cent. On a Δz of 7.2 degrees, a quarter of a degree is 3.5 per cent. Write the assumptions as a list with a size for each, as in Quantitative Reasoning, and give the answer as a range.

Where this fits

In Nicomachus's division of quantity, geometry studies magnitude at rest and astronomy magnitude in motion. Reading this measurement as geometry applied to astronomy is an interpretation, not his claim. Testing models against observation continues in Models of the Heavens; Geometry, Euclid and Trigonometry covers the Euclid.

Try this

Recompute 360 / 50, 50 × 5,000 and 252,000 / 360. Multiply 250,000 stades by 157.5 m and by 185 m and compare each with 40,008 km.
Find the stade lengths that would make "within 2 per cent" true for 250,000 stades, then give two reasons why that does not show Eratosthenes was that accurate.
Repeat the measurement with two cities, as above. Write your assumptions first and give a range.