Astronomy · Round I · Foundations · Measuring the heavens

Reproduce Eratosthenes' measurement of the Earth

I can reproduce Eratosthenes' calculation of the Earth's circumference from a shadow angle and a stated distance, list the assumptions it needs with a size for each, and say why its accuracy in kilometres is uncertain.

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Round IReproduce Eratosthenes' measurement of the Earth (this page)
Round IIDistances to the Moon and the SunComing
Round IIIStellar parallax and the scale of the universeComing

1 Learn

You will be able to

Compute the circumference as the distance times 360 divided by the angle, as Cleomedes reports it: 7.2 degrees (one fiftieth of a circle) and 5,000 stades give 250,000 stades, where Strabo and Pliny report 252,000.
Explain, with parallel rays and Elements I.29, why the shadow angle at Alexandria equals the angle at the Earth's centre between the two cities.
List the assumptions of the method with the size of each departure from the ideal, such as Syene lying about 0.36 degrees north of the tropic in Eratosthenes' time.
Convert the result to kilometres for stades of 157.5 m and 185 m, compare it with 40,008 km, and say why 'accurate to within 2 per cent' is not defensible as stated.

2 Practise

Answer each card from memory, then grade yourself honestly. The cards join your review deck and come back just before you would forget them.

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.
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.
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.
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.
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.
explain
Why can nobody say without qualification how many kilometres Eratosthenes measured, or that he was within 2 per cent?

Show answer
The length of the stade is uncertain. At about 157.5 metres his 250,000 stades is 39,375 km, 1.6 per cent below 40,008 km. At 185 metres it is 46,250 km, 15.6 per cent above.
apply
Two places lie on the same north-south line, 800 km apart. At noon on the same day the Sun is overhead at one, and at the other the Sun's rays make an angle of 7.2 degrees with an upright gnomon. What circumference does the method of Eratosthenes give?

Show answer
The angle at the Earth's centre is also 7.2 degrees, one fiftieth of a circle, so the circumference is 800 × 360 ÷ 7.2 = 40,000 km.

3 Prove it: the mastery check

5 questions drawn from a pool of 11. The pass mark is 80%. There is no time limit, and you can retake it with new questions; your best result counts.

4 Prove it: the performance task

Repeat the measurement and state its limits

Choose any two cities on nearly the same meridian (within about 2 degrees of longitude), at least 500 km apart and both north of the Tropic of Cancer. They need not be places you have been. Part A. From a map's scale or a distance tool, take the distance D between them in kilometres. Do not work it out yourself from their latitudes and a length for a degree: that would build in the size of the Earth you are trying to find. Part B. For one date, take each city's noon Sun elevation from a published source such as an almanac or a sun-position calculator, or use noon shadows of a vertical stick measured in both cities on the same day. Work out each zenith angle (90 degrees minus the elevation, or arctan of shadow over height), the difference, and the circumference D × 360 / difference. Part C. List at least four assumptions, each with a size or a bound, then give your circumference as a range and compare it with 40,008 km. Modern maps and distance tools are themselves built on a known size of the Earth, and published elevations follow from the cities' latitudes: add one sentence on what your result can show and what it cannot. Part D. In two or three sentences, convert Eratosthenes' 250,000 stades to kilometres with a stade of 157.5 m and with one of 185 m, and say why his result cannot be given as one figure in kilometres.

What to hand in: A one-page worked table (the two cities, D, the two zenith angles, their difference, the circumference, the assumptions with sizes and the range), a sentence on what the result can show, and a closing paragraph of two or three sentences. Save it on this page or download it.

Check your work against the rubric The two cities lie on nearly the same meridian, at least 500 km apart and north of the Tropic of Cancer, and D comes from a map or distance tool, not worked out by hand from the cities' latitudes. Each zenith angle is 90 degrees minus the noon elevation (or arctan of shadow over height), and the circumference is shown as D × 360 / difference with the numbers written out. At least four assumptions are listed (for example parallel rays, one meridian, the distance, a round Earth, the reading error), and each has a size or a bound. The answer is a range and not a single figure, it is compared with about 40,008 km, the largest source of uncertainty is named, and one sentence says that because the distance (and any published elevations) already rest on a known size of the Earth, agreement checks the method rather than measuring the Earth afresh. The closing paragraph says that 250,000 stades gives 39,375 km for a stade of 157.5 m and 46,250 km for a stade of 185 m, that nobody can now fix the unit, and that 'within 2 per cent' is therefore not defensible as stated.

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5 Discuss: the seminar

Read the text, then think, write or talk through the question with someone. There is no answer key: the aim is a better question.

Astronomy · Round 1

Aristotle, On the Heavens (De Caelo), II.14 (Bekker 296a24-298a20; the arguments for a spherical Earth run from 297a8)J. L. Stocks's Oxford translation on the MIT Internet Classics Archive. The page carries Book II in full without Bekker numbers; chapter 14 is 'Part 14', the last on the page. Aristotle argues for a spherical Earth from the curved shadow in lunar eclipses and from the stars that change as one travels north or south, and cites the mathematicians' circumference of 400,000 stades.
Cleomedes, On the Circular Motion of the Celestial Bodies (Caelestia), Book I, chapter 10 in Ziegler's numbering (chapter 7 in Todd's numbering)T. L. Heath's English translation of the chapter, reproduced on a freely readable page; it also gives Posidonius's method. Cleomedes reports the measurement of Eratosthenes (about 276 to 194 BC). The length of the stade is uncertain, so the result in kilometres is too. Bowen and Todd (2004) is the standard translation but is not free.

Aristotle argues that the Earth is a sphere from the curved shadow in an eclipse and from the stars that change as you travel, and Cleomedes reports that Eratosthenes found its size from one shadow angle and one distance. Which of these arguments depends most on what we already assume, and which could you check yourself this week?

Eratosthenes needs Syene and Alexandria on one meridian and the two lines to the Sun to be parallel. Cleomedes states both. Does his reason for the parallel lines convince you, and what would go wrong if they were not parallel?
Cleomedes also reports Posidonius, who used the star Canopus, on the horizon at Rhodes and one forty-eighth of a circle above it at Alexandria. Compare the two methods: which hypotheses do they share, and which does each add?
Aristotle gives 400,000 stades as the figure of unnamed mathematicians. How much weight can a figure carry when neither its method nor its unit is given?

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