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.
At a glance · about 2 minutes
From shadow to circumference
- Two shadowsSolstice noon: none at Syene, one fiftieth of a circle at Alexandria.
- AssumptionsSame meridian, 5,000 stades apart, parallel rays.
- Alternate anglesShadow angle equals the angle at the Earth's centre (Euclid I.29).
- Scale upOne fiftieth of the circle is 5,000 stades, so 250,000 in all (VI.33).
- Stade unitThe stade's length is disputed, so kilometres are uncertain.
In brief
Eratosthenes, as Cleomedes reports him, compared noon shadows at Syene and Alexandria. A shadow angle of one fiftieth of a circle and an assumed distance of 5,000 stades gave a circumference of 250,000 stades. Every step rests on a stated assumption, and the length of the stade is so uncertain that his accuracy in kilometres cannot be fixed.
Key ideas
- Eratosthenes, as Cleomedes reports him, turned one shadow angle (one fiftieth of a circle, 7.2 degrees) and an assumed distance of 5,000 stades into a circumference of 250,000 stades, using parallel rays, alternate angles (Euclid I.29) and arcs proportional to angles at the centre (VI.33).
- Every step rests on an assumption, and some were only roughly true. Syene lay about 0.36 degrees north of the tropic and 3 degrees of longitude east of Alexandria. A perfect reading at Alexandria would have been about 7.47 degrees, so the reported 7.2 was 0.27 low, and the two errors partly cancel.
- The stade has no agreed length, so the result in kilometres is uncertain: 250,000 stades is about 39,400 km at 157.5 m and 46,250 km at 185 m, and 'within 2 per cent' holds only for stades of about 157 to 163 m. State the range, not one figure.
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Eratosthenes of Cyrene (about 276 to 194 BC) was a scholar and librarian at Alexandria who, by later report, found the size of the Earth from the length of a shadow. His book On the Measurement of the Earth is lost. The fullest account is in Cleomedes' Caelestia I.7 (I.10 in older editions), by an author of uncertain date, and some historians doubt that this tidy version is what Eratosthenes actually did. This page lists the method's assumptions and asks how far to trust the answer. 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, and Eratosthenes found it to be one fiftieth of the bowl's circle: 360 / 50 = 7.2 degrees. With the distance from Syene to Alexandria taken as 5,000 stades (a Greek length; see below), 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. It pictures the overhead Sun; it is not Cleomedes' evidence.
The geometry
Take the Sun to be so far away that its rays at both cities are 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, the Sun's zenith angle (its distance from straight overhead), equals the angle at the centre between the cities. 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: Days, Seasons and the Phases of the Moon 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. He also takes Syene to lie on the tropic, the reading to be correct and the Earth to be round. 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, changing by only 0.01 degrees in Eratosthenes' lifetime. 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 off the tropic, a perfect reading at Alexandria would have been about 7.47 degrees: Syene's position added 0.36 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", though some gave it 32. With eight stades to a Roman mile (1,000 paces, about 1.48 km), which fits Pliny's rule (II.85) of 125 paces to the stade, the stade is about 185 m, the value adopted by Rawlins (1982), Engels (1985) and others. With a schoenus of 12,000 Egyptian royal cubits of about 0.525 m, it is 300 cubits, or 157.5 m, as Hultsch (1882) and others took it. Gulbekian (1987) proposed 166.7 m, and some have argued for about 148 m. All are known here from secondary reports.
| Stade | 250,000 stades | 252,000 stades |
|---|---|---|
| 157.5 m | 39,375 km (1.6% low) | 39,690 km (0.8% low) |
| 166.7 m | 41,675 km (4.2% high) | 42,008 km (5.0% high) |
| 185 m | 46,250 km (15.6% high) | 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. A 157.5 m stade makes 5,000 stades match the north-south distance and about 168 m the surface distance; no source checked says how the 5,000 was obtained. What is defensible: the method is sound and the result is the right size, with an error from under 1 to over 16 per cent across the table. Nobody can now fix the unit.
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.
The Sun's width blurs the shadow's end by about 1 cm at s = 0.5 m and 2.5 cm at s = 1.3 m. A 5 mm misreading shifts z by 0.23 degrees at s = 0.5 and 0.11 degrees at s = 1.3, together at most about 1.3 per cent of Δz. 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 with Stated Assumptions, and give the answer as a range.
Where this fits
In Nicomachus's division (The Order of the Seven Liberal Arts: Substance, Quantity and the Mind), geometry studies magnitude at rest and astronomy magnitude in motion. Reading this measurement as geometry applied to astronomy is an interpretation, not his claim. Models of the Heavens: Ptolemy, Copernicus, Tycho and Kepler continues testing models against observation; 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.
Further reading
- Cleomedes, Caelestia I.7 (older I.10): Bowen and Todd, Cleomedes' Lectures on Astronomy (2004), or Heath's translation in Cohen and Drabkin, A Source Book in Greek Science (1948).
- Strabo, Geography II.5 and XVII.1.48 (Project Gutenberg); Pliny, Natural History II.183, II.247 and XII.53 (Latin, LacusCurtius).
- Euclid, Elements I.29 and VI.33, in D. E. Joyce's online edition.
- MacTutor, "Eratosthenes of Cyrene"; I. Tupikova, "A common-sense approach to the problem of the itinerary stadion" (2022, open access).
Check yourself
Answer each question from memory before you open it. Retrieval, not rereading, is what makes learning last (see the weekly loop). Grade yourself honestly and the study dashboard will bring each card back just before you'd forget it.
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recall
According to Cleomedes, which observations and which distance did Eratosthenes combine, and what circumference did they give?
Show answer
At noon on the summer solstice, sundial pointers at Syene throw no shadow, while at Alexandria the shadow marks an arc one fiftieth of the bowl's circle (7.2 degrees). Taking the two cities as 5,000 stades apart, the whole circle is 50 × 5,000 = 250,000 stades.
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recall
Which ancient writers give 252,000 stades rather than 250,000, and what is one suggested reason for the difference?
Show answer
Strabo (Geography II.5.7) and Pliny (Natural History II.247). Some modern scholars conjecture that the figure was adjusted to divide evenly: 252,000 gives 4,200 stades for each sixtieth of the circle and 700 for each degree. No ancient source checked for the page gives the reason.
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explain
Why does the shadow angle at Alexandria equal the angle at the Earth's centre between the two cities?
Show answer
The Sun's rays are taken as parallel. The ray at Syene falls straight down, along the line to the Earth's centre, and the upright gnomon at Alexandria, extended down, also reaches the centre. So the gnomon's line crosses both parallel rays, and a line falling on parallels makes equal alternate angles (Euclid I.29). The shadow angle at the gnomon's tip therefore equals the angle at the centre.
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apply
Two cities lie 1,000 km apart on one meridian. On the same day the Sun's noon zenith angles at them differ by 9.0 degrees. What circumference follows, and how much does a quarter-degree reading error matter?
Show answer
1,000 × 360 / 9.0 = 40,000 km, assuming one meridian, parallel rays and a true north-south distance. A quarter of a degree is about 2.8 per cent of 9.0 degrees, so the honest answer is a range, roughly 38,900 to 41,100 km.
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connect
The page sends you to 'Quantitative Reasoning with Stated Assumptions' when it asks you to list your assumptions with a size for each. What do the two pages share, and how does the stade show it?
Show answer
Both treat a number as only as good as the assumptions beneath it. Eratosthenes' arithmetic is exact (50 × 5,000), but the result in kilometres depends on a stade nobody can now fix, from about 39,400 km at 157.5 m to 46,250 km at 185 m. So list the assumptions with a size for each, and give the answer as a range.
Open questions
No answer key. Think them through, write a paragraph, or argue them out with someone; the aim is a better question.
- Cleomedes states his assumptions before he calculates (Caelestia I.7), and some were only roughly true. Is a method trustworthy because its assumptions are stated, or only once they have been tested?
- The result may owe its closeness to the truth partly to errors that cancel. Should we admire the method, the result, or neither? What would each judgement leave out?
- Cleomedes reports a result in a unit nobody can now fix (Caelestia I.7). Does a measurement whose unit is lost still count as knowing the Earth's size, and what would you need to know to say so?
Sources
- Cleomedes. Caelestia (On the Circular Motions of the Celestial Bodies) I.7 in Todd's numbering (used by Bowen and Todd), I.10 in Ziegler's older numbering. T. L. Heath's translation in Greek Astronomy (London, 1932), as reproduced in Cohen and Drabkin, A Source Book in Greek Science (1948), pp. 149-153, and online at roger-pearse.com/weblog/2017/04/15/cleomedes-how-big-is-the-earth/.
- Bowen, A. C. and Todd, R. B. Cleomedes' Lectures on Astronomy: A Translation of The Heavens (University of California Press, 2004), pp. 81-84, as quoted at nonagon.org/ExLibris/eratosthenes-measures-earth; chapter number confirmed by the review in Bryn Mawr Classical Review 2004.08.06.
- Strabo. Geography II.5.7, II.5.34 and XVII.1.48. Translated by H. C. Hamilton and W. Falconer (Bohn, 1854-57), Project Gutenberg eBooks 44884 (vol. 1) and 44886 (vol. 3).
- Pliny the Elder. Natural History II.85, II.183, II.247 and XII.53. Latin text at penelope.uchicago.edu (LacusCurtius).
- Euclid. Elements I.29 and VI.33. Text and commentary by D. E. Joyce, Clark University (mathcs.clarku.edu/~djoyce/elements).
- O'Connor, J. J. and Robertson, E. F. 'Eratosthenes of Cyrene'. MacTutor History of Mathematics (mathshistory.st-andrews.ac.uk/Biographies/Eratosthenes/).
- Tupikova, I. 'A common-sense approach to the problem of the itinerary stadion'. Archive for History of Exact Sciences 76 (2022), 319-361, open access at link.springer.com/article/10.1007/s00407-022-00287-6. Secondary; used for who proposed the 157.5 m, 148 m and 185 m stades.
- Walkup, N. 'Eratosthenes and the Mystery of the Stades' (student survey, home.adelphi.edu/~bradley/HOMSIGMAA/Walkup.pdf). Secondary; used for Rawlins's 185 m statement, the Roman-mile conversion and the conjecture about 252,000.
- NASA Goddard Space Flight Center. Earth Fact Sheet and Sun Fact Sheet (nssdc.gsfc.nasa.gov/planetary/factsheet/). Radii, obliquity, distance and the Sun's apparent diameter.
- Laskar, J. (1986), polynomial for the mean obliquity of the ecliptic, as printed in the Wikipedia article 'Axial tilt'.
- Wikipedia infoboxes for Alexandria and Aswan (coordinates).