Lesson 00

How to use this course

Eratosthenes is remembered for two things that every school pupil meets: a way of finding prime numbers and a measurement of the Earth. Both reach us only through later writers. Not one of his scientific books survives. What we have are summaries, quotations and criticisms by authors who lived between one and eight centuries after him: Strabo, Cleomedes, Nicomachus, Theon of Smyrna, Ptolemy, Pappus and Eutocius.

The course therefore keeps two kinds of statement apart. Biographical claims are labelled with their source, usually the Byzantine lexicon called the Suda or the geographer Strabo. Technical lessons present the methods as the later reports describe them, then rebuild them in modern notation and test them against modern data.

Prerequisites for Part II

Euclidean geometry: parallel lines, similar triangles, the angle properties of circles.
Elementary number theory: primes, divisibility, the floor function.
Trigonometry on the sphere at the level of latitude, longitude and the Sun’s declination.
For Lessons 04 and 08, a first acquaintance with sums over primes and with continued fractions is helpful but not required.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Eratosthenes did not use degrees: he divided the circle into sixty parts, and he expressed results as ratios and fractions of the whole circle. The course gives both his form and ours.

Reference

Timeline

All dates are BC unless marked AD, and almost all are approximate. The birth year follows the Suda, which places it in the 126th Olympiad (276–273 BC); older scholarship proposed dates as early as 296 or 284 BC.

c. 276	Born in Cyrene, a Greek city in what is now Libya, son of Aglaos (Suda).
late 260s	Goes to Athens. Studies with the Stoic Ariston of Chios, the Academic Arcesilaus and the Cynic Bion of Borysthenes; Strabo also calls him a pupil of Zeno of Citium.
246	Ptolemy III Euergetes becomes king of Egypt.
c. 245	Summoned from Athens to Alexandria, aged about thirty. Becomes tutor to the king’s son, the future Ptolemy IV.
c. 240s–230s	Becomes head of the Library of Alexandria, succeeding Apollonius of Rhodes; the exact year is unknown.
238	The Canopus Decree of the Egyptian priests orders a leap day every fourth year. Later reports credit Eratosthenes with the same calendar rule.
before 212	Archimedes addresses The Method to him, and, according to its heading, the Cattle Problem.
c. 240–200	Period of his scientific works: On the Measurement of the Earth, Geographika in three books, Chronographiai, Olympic Victors, the mesolabe and its dedication to the king.
221	His pupil becomes King Ptolemy IV Philopator.
204	Death of Ptolemy IV; Queen Arsinoe III is murdered. Eratosthenes later writes a memoir, Arsinoe.
c. 200	Aristophanes of Byzantium, named by the Suda as his pupil, becomes head of the Library.
c. 194	Dies in Alexandria, aged 80, 81 or 82 according to different sources. The Suda says he went blind and starved himself.
2nd c.	Hipparchus writes three books Against the Geography of Eratosthenes.
c. AD 20	Strabo’s Geography, the main surviving source for Eratosthenes’ geography.
c. AD 100	Nicomachus of Gerasa describes the sieve in his Introduction to Arithmetic.
AD 1772	Samuel Horsley publishes “Κόσκινον Ἐρατοσθένους, or, The Sieve of Eratosthenes” in the Philosophical Transactions.

Part I · Formation

Lesson 01

Cyrene, Athens and the sources

Eratosthenes was born around 276 BC in Cyrene, the leading Greek city of North Africa, then under the rule of the Ptolemies of Egypt. The Suda, a Byzantine encyclopedia of the tenth century AD that drew on lost Hellenistic reference works, names his father as Aglaos and lists his teachers. Nothing else is known of his family. Cyrene was a city of learning: the poet Callimachus came from there, as did the philosopher Carneades a generation later.

Teachers

The Suda names the grammarian Lysanias of Cyrene, and calls Eratosthenes a pupil of Callimachus. Modern scholars doubt that Callimachus formally taught him, since Callimachus had left for Alexandria early, but the two Cyreneans were certainly connected later. In the late 260s Eratosthenes went to Athens, where he spent roughly twenty years. Strabo, writing two centuries later, lists his Athenian masters: the Stoic Ariston of Chios, the head of the Academy Arcesilaus of Pitane, and the Cynic Bion of Borysthenes. Strabo also calls him a pupil of Zeno of Citium, the founder of Stoicism, which is chronologically awkward because Zeno died around 262 BC.

Strabo’s tone is critical. He says Eratosthenes moved between schools, belonging fully to none, and treats his philosophy as amateur. The judgement matters because Strabo is our main witness to his geography too. Throughout this course we read Strabo as an informed but hostile reporter.

The shape of the evidence

The Suda (10th century AD): birth Olympiad, father, teachers, nicknames, pupils, blindness and death by starvation.
Strabo, Geography (c. AD 20): over a hundred fragments of the Geographika, the Athenian teachers, many criticisms.
Cleomedes, a Stoic teacher probably of the Roman imperial period: the simplified account of the Earth measurement.
Eutocius of Ascalon (6th century AD): a letter to King Ptolemy on doubling the cube, with an epigram, attributed to Eratosthenes.
Nicomachus of Gerasa and Theon of Smyrna (1st–2nd centuries AD): the sieve and extracts from the Platonicus.
Archimedes, the only contemporary witness: the preface to The Method, addressed to him.

Exercise 1.1Strabo calls Eratosthenes a pupil of Zeno of Citium. Zeno died around 262 BC, and Eratosthenes was born around 276 BC and reached Athens in the late 260s. What does this suggest about the claim, and what alternative readings are possible?

Show solution

Eratosthenes would have been about fourteen when Zeno died, and could have overlapped with him in Athens for a year at most. Formal study is unlikely. Strabo may use “pupil” loosely for someone formed in Zeno’s school, through Ariston; or the birth date may be earlier than the Suda suggests, as some nineteenth-century scholars argued. The tension is one reason the birth year is uncertain.

Lesson 02

Called to Alexandria

In 246 BC Ptolemy III Euergetes succeeded his father as king of Egypt. Soon afterwards he summoned Eratosthenes from Athens, when Eratosthenes was about thirty. Ancient reports connect the invitation with Callimachus, the great scholar-poet at court, and with Queen Berenice II, herself from Cyrene. His first post was as tutor to the king’s son, the future Ptolemy IV Philopator.

The Museum and the Library

Ptolemy I had founded the Museum, a community of scholars supported by the crown, and the Library beside it. A list of librarians preserved on a papyrus from Oxyrhynchus places Eratosthenes after Apollonius of Rhodes and before Aristophanes of Byzantium; Zenodotus of Ephesus had been the first. Eratosthenes thus became the third head of the Library, at a date between the 240s and the 230s. Callimachus, who catalogued the collection in his Pinakes, never held the post.

The Library gave Eratosthenes what no other place could: reports of distances from royal surveyors and travellers, records of the Ptolemaic expeditions up the Nile and along the Red Sea, the observations of earlier astronomers, and lists of Olympic victors and kings. His later work in geography and chronology is unthinkable without this archive.

The letter from Archimedes

Archimedes of Syracuse addressed The Method of Mechanical Theorems to Eratosthenes. In the preface he calls him a serious scholar and a man eminent in philosophy, and explains that he is sending him a way of discovering results by mechanics, which he thinks Eratosthenes will appreciate. The Cattle Problem, an epigram found in 1773, also presents itself as a problem sent by Archimedes to Eratosthenes for the mathematicians of Alexandria. These two documents are the only contemporary evidence for Eratosthenes as a mathematician, and they show that the best geometer of the age treated him as an equal.

Exercise 2.1The Oxyrhynchus list gives the order of librarians but no dates. The ancient sources say Eratosthenes was summoned by Ptolemy III (from 246 BC) and that Aristophanes of Byzantium took over around 200 BC. If Eratosthenes died around 194 BC, what does this suggest about how his tenure ended?

Show solution

If Aristophanes became librarian around 200 BC while Eratosthenes lived until about 194 BC, Eratosthenes must have given up the post some years before his death. This fits the Suda’s report that he went blind in old age. The dates are approximate enough that the inference is suggestive, not certain.

Lesson 03

Pentathlos and Beta

According to Suetonius, Eratosthenes was the first to call himself a philologos, a lover of learning in every form. His output ranged across mathematics, astronomy, geography, chronology, literary criticism, philosophy and poetry. Admirers called him Pentathlos, the all-round athlete of the five-event contest. Critics, the Suda says, called him Beta, the second letter, because he came second to the specialists in every field. Some modern scholars suggest the nickname may simply refer to his place in the sequence of librarians, or to the Library room where he worked; there is no way to decide.

The works

Mathematics: the Platonicus, on the mathematics in Plato, known through Theon of Smyrna; On Means, listed by Pappus among the works of the “Treasury of Analysis”; the letter and epigram on doubling the cube; the sieve.
Geodesy and astronomy: On the Measurement of the Earth, known from Cleomedes, Heron and Galen; work on the obliquity of the ecliptic reported by Ptolemy; a calendar rule.
Geography: the Geographika in three books, the first work with that title and the source of the word “geography”.
Chronology: Chronographiai and Olympic Victors.
Literature: On the Old Comedy, in at least twelve books; the poems Hermes, Erigone and Anterinys (or Hesiod); the memoir Arsinoe.

A surviving collection of star myths, the Catasterismi, circulates under his name. Scholars regard it as a later epitome at best, loosely connected with a lost work of his.

A method across fields

The fragments share a single habit of mind: measure what can be measured, fix a frame of reference, and refuse stories that cannot be tested. He used parallels and meridians for space and the Olympiad cycle for time. He argued that Homer’s geography was poetic invention, saying, as Strabo reports, that one would find the scene of Odysseus’ wanderings when one found the cobbler who stitched the bag of the winds. In the Platonicus he made ratio and proportion the common thread of the mathematical sciences.

Exercise 3.1Sort the list of works above into three groups: (a) survives in its own words; (b) known from substantial later summaries or quotations; (c) known by title or a few lines only. What does the result imply for any assessment of Eratosthenes as a mathematician?

Show solution

(a) Nothing, except perhaps the epigram on the cube quoted by Eutocius, whose authenticity is debated, and fragments of verse. (b) The Earth measurement (Cleomedes), the Geographika (Strabo), the mesolabe (Eutocius, Pappus), the sieve (Nicomachus), parts of the Platonicus (Theon). (c) On Means, Chronographiai, On the Old Comedy, most poems. His mathematics is therefore judged through intermediaries who chose what to report; the “Beta” verdict may partly reflect what was copied, not what he wrote.

Part II · The mathematics and sciences

Lesson 04

The sieve

For the general reader. To find all primes up to a hundred, write the numbers down, then cross out every multiple of 2 after 2, every multiple of 3 after 3, and so on. What survives are the primes. The method is named after Eratosthenes, and its descendants remain central tools in modern number theory.

Goals

State the sieve precisely and prove that it works.
Turn the sieve into an exact counting formula by inclusion and exclusion.
Estimate the work it costs, using sums over primes.

The source

The earliest description is in the Introduction to Arithmetic of Nicomachus of Gerasa, written around AD 100. Nicomachus names the method after Eratosthenes and calls it a sieve (koskinon). His version lists only the odd numbers and strikes out multiples of each odd number in turn, by counting places in the list. Some historians argue that Eratosthenes named the procedure rather than invented it.

The algorithm and its correctness

Fix \(N\ge2\). Start with the list \(2,3,\dots,N\). Repeatedly take the smallest number \(p\) not yet used as a sieving number and not struck out, and strike out \(p^2, p^2+p, p^2+2p,\dots\) up to \(N\). Stop once \(p^2>N\).

Proposition

Every composite \(n\le N\) has a prime factor \(p\le\sqrt N\). Consequently the numbers that survive the sieve by the primes \(p\le\sqrt N\) are exactly the primes in \((\sqrt N,N]\), and together with those sieving primes they are all the primes up to \(N\).

Proof. If \(n=ab\) with \(1<a\le b\), then \(a^2\le ab=n\le N\), so \(a\le\sqrt N\), and any prime factor of \(a\) is at most \(\sqrt N\). Each \(p\) used is prime, because it survived sieving by all smaller primes. A composite \(n\) with least prime factor \(p\) satisfies \(n\ge p^2\) and is a multiple of \(p\), so it is struck out. A prime is never struck out, because all struck numbers are proper multiples of a smaller number.

Counting by inclusion and exclusion

Let \(P\) be the product of the primes \(p\le\sqrt N\) and \(\mu\) the Möbius function. Counting the integers in \([1,N]\) with no prime factor up to \(\sqrt N\) gives

\[ \pi(N)-\pi(\sqrt N)+1=\sum_{d\mid P}\mu(d)\left\lfloor\frac Nd\right\rfloor, \]

where \(\pi(x)\) counts primes up to \(x\) and the \(+1\) accounts for the integer 1. The identity is often credited to Adrien-Marie Legendre. Its error terms grow quickly with the number of primes in \(P\), and controlling them is the starting point of modern sieve theory.

The cost

Sieving by \(p\) touches about \(N/p\) entries, so the total work is about \(N\sum_{p\le\sqrt N}1/p\). By Mertens’ theorem \(\sum_{p\le x}1/p=\log\log x+M+o(1)\) with a constant \(M\approx0.2615\), so the sieve uses \(O(N\log\log N)\) operations and \(O(N)\) memory. Splitting \([1,N]\) into blocks of length about \(\sqrt N\) (the segmented sieve) cuts the memory to \(O(\sqrt N)\).

Exercise 4.1Run the sieve to \(N=100\). Which primes do you sieve by, and how many primes survive?

Show solution

Since \(\sqrt{100}=10\), sieve by 2, 3, 5, 7. The survivors are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97: twenty-five primes. The next prime, 11, has \(11^2=121>100\), so it strikes nothing.

Exercise 4.2Check the inclusion–exclusion formula for \(N=100\).

Show solution

Here \(P=2\cdot3\cdot5\cdot7=210\). The sum is \(100-(50+33+20+14)+(16+10+7+6+4+2)-(3+2+1+0)+0=100-117+45-6=22\). The single terms use \(d=2,3,5,7\); the pairs \(d=6,10,14,15,21,35\); the triples \(d=30,42,70,105\); and \(d=210\). The left side is \(\pi(100)-\pi(10)+1=25-4+1=22\).

Exercise 4.3Explain why striking out may start at \(p^2\) rather than \(2p\), and compute how many strikes the sieve to 100 makes with this rule, counting repeats.

Show solution

A multiple \(kp\) with \(k<p\) has a prime factor of \(k\) smaller than \(p\), so it was struck earlier. Strikes: for 2, the even numbers 4 to 100: 49; for 3, the multiples 9, 12, …, 99: 31; for 5, 25, 30, …, 100: 16; for 7, 49, 56, …, 98: 8. Total 104 strikes for 74 composites; the 30 repeats are numbers with two or more prime factors below 10 that are struck more than once.

Lesson 05

Two mean proportionals and the mesolabe

For the general reader. To double the volume of a cube you must multiply its side by the cube root of 2. Greek geometers could not construct this length with ruler and compass, and today we know it is impossible. Eratosthenes built a sliding instrument that finds it mechanically, set up a bronze model in Alexandria, and wrote a poem about it.

Goals

Reduce doubling the cube to two mean proportionals.
Prove that the mesolabe finds them.
Compare it with Menaechmus’ solution by conics and with the modern impossibility proof.

The problem and its history

Eutocius, commenting on Archimedes’ On the Sphere and Cylinder in the sixth century AD, collected a dozen ancient solutions to the problem of finding two mean proportionals. One is introduced by a letter to King Ptolemy in Eratosthenes’ name. The letter tells how Hippocrates of Chios reduced doubling the cube to finding two mean proportionals, how the Delians were told by an oracle to double an altar, and how Archytas, Eudoxus and Menaechmus found theoretical solutions that could not be carried out by hand. Many scholars regard the letter as a later composition built on genuine material; the closing epigram and the description of the instrument are generally accepted as derived from Eratosthenes.

Reduction (Hippocrates of Chios)

Let \(a,b>0\). If \(x,y\) satisfy \(a:x=x:y=y:b\), then \(x^3=a^2b\). In particular, for \(b=2a\), the cube on \(x\) is double the cube on \(a\).

Proof. Let \(r=x/a\). Then \(y=xr=ar^2\) and \(b=yr=ar^3\). So \(r^3=b/a\) and \(x^3=a^3\cdot b/a=a^2b\).

The mesolabe

Eutocius and Pappus describe a frame holding three equal rectangular plates that slide along a pair of parallel rails, each able to pass behind the next. On each plate a diagonal is drawn from the upper left corner to the lower right corner. Model it as follows. Let \(AE\) be the left edge of the first plate, of length \(a\), with \(E\) on the lower rail. Let \(F\), \(G\), \(H\) be the lower right corners of plates 1, 2, 3, so that \(AF\) is the diagonal of plate 1. Slide plate 2 partly behind plate 1, and plate 3 partly behind plate 2. Let \(B\) be the point where the diagonal of plate 2 crosses the right edge of plate 1, and \(C\) the point where the diagonal of plate 3 crosses the right edge of plate 2. Then \(BF\) and \(CG\) are vertical, the diagonal of plate 2 ends at \(G\) and that of plate 3 at \(H\). Mark \(D\) on the right edge of plate 3 with \(DH=b<a\), and slide the plates until \(A\), \(B\), \(C\), \(D\) lie on one straight line.

Theorem (Eratosthenes’ mesolabe)

If \(A,B,C,D\) are collinear, then \(AE:BF=BF:CG=CG:DH\). So \(BF\) and \(CG\) are two mean proportionals between \(a\) and \(b\).

Proof. If \(AE=DH\) there is nothing to prove, so let the line \(ABCD\) meet the base at \(K\). The verticals are parallel, so triangles \(KAE\), \(KBF\), \(KCG\), \(KDH\) are similar and

\[ \frac{AE}{BF}=\frac{KA}{KB}=\frac{KE}{KF},\qquad \frac{BF}{CG}=\frac{KB}{KC}=\frac{KF}{KG}. \]

The diagonals of congruent plates are parallel, so \(AF\parallel BG\parallel CH\). Because \(AF\parallel BG\), triangles \(KAF\) and \(KBG\) are similar, so \(KA/KB=KF/KG\). Combining, \(AE/BF=KF/KG=BF/CG\). The same argument with \(BG\parallel CH\) gives \(BF/CG=CG/DH\). With \(n\) plates the same instrument gives \(n-1\) means, as the letter claims.

Ratio, means and musical scales

For the general reader. Greek mathematicians thought of the average of two numbers in several ways, each defined by a proportion. The same proportions describe musical intervals: an octave is the ratio 2 to 1, a fifth 3 to 2. Eratosthenes wrote a book on means and divided the musical fourth into small steps in his own way.

Goals

Define the arithmetic, geometric and harmonic means by proportions and prove their order.
Read the “perfect proportion” 6, 8, 9, 12 musically.
Check Eratosthenes’ divisions of the tetrachord as reported by Ptolemy.

What the sources say

Theon of Smyrna, in the second century AD, reports from the Platonicus that Eratosthenes treated ratio as the source of proportion, and proportion as the bond of the mathematical sciences; he also regarded the unit as the starting point of number. Pappus lists Eratosthenes’ On Means among the classical books of geometrical analysis, but its contents are lost. Ptolemy’s Harmonics (Book II) preserves his divisions of the tetrachord.

The three classical means

Let \(0<a<c\). A number \(b\) with \(a<b<c\) is the arithmetic, geometric or harmonic mean of \(a\) and \(c\) when, respectively,

\[ \frac{b-a}{c-b}=\frac aa,\qquad \frac{b-a}{c-b}=\frac ab,\qquad \frac{b-a}{c-b}=\frac ac. \]

Equivalently \(A=\tfrac{a+c}2\), \(G=\sqrt{ac}\), \(H=\tfrac{2ac}{a+c}\). Then \(H<G<A\) and \(G^2=AH\).

Proof. Solving the defining proportions gives the closed forms. \(AH=\tfrac{a+c}2\cdot\tfrac{2ac}{a+c}=ac=G^2\), and \(A-G=\tfrac12(\sqrt c-\sqrt a)^2>0\). Since \(G\) is the geometric mean of \(A\) and \(H\) and \(A>G\), we get \(H<G\).

The musical proportion

Between 6 and 12, the arithmetic mean is 9 and the harmonic mean is 8. The four numbers 6, 8, 9, 12 contain the octave \(12:6\), the fifths \(12:8=9:6\), the fourths \(12:9=8:6\), and the whole tone \(9:8\), the difference between a fifth and a fourth. The fourth, \(4:3\), was divided into three steps to form a tetrachord, in three “genera”.

Eratosthenes’ tetrachords (Ptolemy, Harmonics II)

Diatonic: \(\tfrac98\cdot\tfrac98\cdot\tfrac{256}{243}\), the Pythagorean division. Chromatic: \(\tfrac65\cdot\tfrac{19}{18}\cdot\tfrac{20}{19}\). Enharmonic: \(\tfrac{19}{15}\cdot\tfrac{39}{38}\cdot\tfrac{40}{39}\). Each product equals \(\tfrac43\).

In the chromatic and enharmonic genera the two small steps arise by splitting an interval \(\tfrac{n+1}{n-1}\) at the arithmetic mean of its terms: \(\tfrac{20}{18}=\tfrac{20}{19}\cdot\tfrac{19}{18}\) and \(\tfrac{40}{38}=\tfrac{40}{39}\cdot\tfrac{39}{38}\). This is proportion theory applied to the ear, in the spirit Theon describes.

Exercise 6.1Verify that each of Eratosthenes’ three tetrachords multiplies to \(4/3\).

Show solution

Diatonic: \(\tfrac{81}{64}\cdot\tfrac{256}{243}=\tfrac{81\cdot256}{64\cdot243}=\tfrac{4}{3}\). Chromatic: \(\tfrac65\cdot\tfrac{19}{18}\cdot\tfrac{20}{19}=\tfrac65\cdot\tfrac{20}{18}=\tfrac{120}{90}=\tfrac43\). Enharmonic: \(\tfrac{19}{15}\cdot\tfrac{40}{38}=\tfrac{760}{570}=\tfrac43\).

Exercise 6.2Show that the octave \(2:1\) cannot be divided into two equal rational intervals, and relate this to the geometric mean.

Show solution

Two equal intervals of ratio \(r\) give \(r^2=2\), so \(r=\sqrt2\), the geometric mean of 1 and 2, which is irrational. Equal division of rational intervals requires geometric means, which is why the mean-proportional problem of Lesson 05 also has a musical side.

Exercise 6.3For \(a=6\), \(c=12\) check \(G^2=AH\) and locate \(G\) between 8 and 9.

Show solution

\(A=9\), \(H=8\), \(AH=72=6\cdot12\), so \(G=\sqrt{72}\approx8.485\). The geometric mean splits the octave into two equal halves, lying between the harmonic and arithmetic means.

Lesson 07

Measuring the Earth

For the general reader. On midsummer noon at Syene, near modern Aswan, the Sun stood overhead. At Alexandria, about 5,000 stadia to the north, a vertical rod still cast a short shadow. From the angle of that shadow and the distance between the cities, Eratosthenes worked out the size of the whole Earth.

Goals

Reconstruct the argument reported by Cleomedes and prove its key step.
Analyse the sensitivity of the result to each measurement.
Understand the debate over the length of the stadion.

The report

Eratosthenes’ book On the Measurement of the Earth is lost. The fullest account is in Cleomedes’ On the Circular Motions of the Heavenly Bodies, which says it simplifies the original. Its assumptions are these: Syene lies on the summer tropic, so the Sun is at the zenith there at noon on the solstice; Syene and Alexandria lie on the same meridian; the distance between them is 5,000 stadia; the Sun is so distant that its rays reaching the Earth are parallel. Cleomedes says the shadow was read in a skaphe, a hemispherical bowl sundial, and that it measured one fiftieth of a full circle.

Proposition (the shadow angle)

Let \(O\) be the Earth’s centre, \(S\) a point where the Sun is at the zenith and \(A\) another point on the same meridian. If the Sun’s rays are parallel, the angle between a vertical gnomon at \(A\) and the ray through its tip equals \(\angle AOS\). Hence

\[ C=s\cdot\frac{2\pi}{\theta}=s\cdot\frac{360^\circ}{\theta}, \]

where \(s\) is the arc \(AS\), \(\theta=\angle AOS\) and \(C\) is the circumference.

Proof. The ray at \(S\) points along \(OS\). The gnomon at \(A\) lies along \(OA\). The ray through the gnomon’s tip is parallel to \(OS\), and the line \(OA\) is a transversal of the two parallel lines, so the alternate angles are equal (Euclid, Elements I.29). Arc length is proportional to central angle (Elements VI.33).

With \(\theta=\tfrac1{50}\) of a circle, that is \(7^\circ12'\), and \(s=5000\) stadia, \(C=250{,}000\) stadia. Later writers (Strabo, after Hipparchus; Pliny) give 252,000 stadia. The adjusted number is divisible by 60, matching Eratosthenes’ division of the circle into sixtieths, and gives 700 stadia to one degree once degrees were adopted. Whether he refined the measurement or simply rounded is unknown.

How good was it?

The answer depends on the stadion, whose length varied by region and period. With the Attic-Olympic stadion of about 176–185 m, 252,000 stadia is 44,000–47,000 km, about 10–15 percent too large. With a stadion of about 157.5 m, sometimes inferred from Pliny, it is about 39,700 km, within a few percent of the true meridian circumference of 40,008 km. Neither value is securely tied to Eratosthenes. Two of his assumptions are also off: Syene lies at about \(24.1^\circ\) N, slightly north of the tropic of his day (about \(23.7^\circ\)), and about \(3^\circ\) east of Alexandria. The latitude difference between the two cities is about \(7.1^\circ\), so his angle was close.

Exercise 7.1A gnomon of height 1 m stands at Alexandria at solstice noon, with the Sun \(7^\circ12'\) from the zenith. How long is its shadow? What does this suggest about why a bowl sundial was preferred?

Show solution

Shadow length \(=\tan7.2^\circ\approx0.126\) m, about 12.6 cm. The fuzzy edge of the shadow, caused by the Sun’s half-degree disc, is a large fraction of so short a length. In a hemispherical bowl the shadow of the pointer falls directly on a scale of arc, so the reading is the angle itself as a fraction of the circle.

Exercise 7.2Show that \(\dfrac{\Delta C}{C}\approx\dfrac{\Delta s}{s}-\dfrac{\Delta\theta}{\theta}\). If the angle is in error by \(0.25^\circ\), the Sun’s angular radius, what is the relative error in \(C\)?

Show solution

From \(\ln C=\ln s-\ln\theta+\text{const}\), differentiate: \(dC/C=ds/s-d\theta/\theta\). With \(\theta=7.2^\circ\), \(0.25/7.2\approx3.5\%\). An error of a quarter-degree in reading the shadow moves the circumference by about 3.5 percent, as much as many proposed stadion corrections.

Exercise 7.3Use modern data: Alexandria \(31.20^\circ\) N, Syene \(24.09^\circ\) N, mean length of one degree of latitude about 111.1 km. Find the meridian distance between their parallels and the stadion length that would make “5,000 stadia” exact.

Show solution

The latitude difference is \(7.11^\circ\), so the meridian distance is about \(7.11\times111.1\approx790\) km. Then 5,000 stadia would be 790 km, a stadion of about 158 m. The actual road distance is longer, and the true great-circle distance is about 843 km because Syene lies to the east.

Exercise 7.4From \(C=252{,}000\) stadia, find the radius of the Earth in stadia, taking \(\pi\approx\tfrac{22}{7}\) as Archimedes’ upper bound suggests.

Show solution

\(R=C/(2\pi)\approx252{,}000\cdot\tfrac{7}{44}\approx40{,}091\) stadia; with \(\pi\approx3.1416\), \(R\approx40{,}107\) stadia. The convenient divisibility of 252,000 by 7 makes Archimedes’ bound give a clean figure.

Lesson 08

Tropics, zones and the obliquity

For the general reader. The Sun’s noon height changes through the year because the Earth’s axis is tilted. Measuring the noon Sun on the longest and shortest days gives both the tilt and the observer’s latitude. Ptolemy reports a ratio for the tilt, 11 to 83, connected with Eratosthenes, and Eratosthenes used the tilt to divide the globe into climate zones.

Goals

Derive latitude and obliquity from solstitial noon altitudes.
Understand the ratio \(11/83\) and the debate about its origin.
Reconstruct Eratosthenes’ zones in sixtieths of a circle.

Proposition

At latitude \(\varphi\) (northern hemisphere, outside the tropics), the noon altitude of the Sun at the summer and winter solstices is \(h_s=90^\circ-\varphi+\varepsilon\) and \(h_w=90^\circ-\varphi-\varepsilon\), where \(\varepsilon\) is the obliquity of the ecliptic. Hence

\[ \varepsilon=\tfrac12(h_s-h_w),\qquad \varphi=90^\circ-\tfrac12(h_s+h_w). \]

Derivation. At noon the Sun is on the meridian, at declination \(\delta\), and its altitude is \(90^\circ-\varphi+\delta\). At the solstices \(\delta=\pm\varepsilon\). Adding and subtracting the two equations gives the result. The arc between the two noon positions is \(2\varepsilon\), the angular distance between the tropics.

The ratio 11/83

In the Almagest (I.12), Ptolemy reports his own measurement of the arc between the tropics as between \(47^\circ40'\) and \(47^\circ45'\), and says that this agrees closely with the ratio of 11 to 83 of the whole meridian, which Eratosthenes found and Hipparchus also used. Then \(2\varepsilon=\tfrac{11}{83}\cdot360^\circ\approx47^\circ42'40''\) and \(\varepsilon\approx23^\circ51'20''\). The true obliquity around 230 BC was about \(23^\circ43'\). Historians disagree on whether Eratosthenes himself stated \(11/83\), or used a rounder value such as \(24^\circ\), with \(11/83\) a later refinement.

Zones in sixtieths

Eratosthenes divided the meridian circle into 60 parts, each 4,200 stadia for a circumference of 252,000. A reconstruction from Strabo’s figures places the tropic at 4 sixtieths (\(24^\circ\)) from the equator, gives each temperate zone a width of 5 sixtieths and each frigid zone a radius of 6 sixtieths around the pole: \(4+5+6=15\) sixtieths make a quarter circle of 63,000 stadia.

Exercise 8.1Compute the continued fraction of \(11/83\) and its convergents. Which convergent corresponds to an obliquity of exactly \(24^\circ\)?

Show solution

\(83/11=7+6/11\), \(11/6=1+5/6\), \(6/5=1+1/5\), so \(11/83=[0;7,1,1,5]\). The convergents are \(1/7,\ 1/8,\ 2/15,\ 11/83\). The convergent \(2/15\) of a circle is \(48^\circ\), an obliquity of \(24^\circ\). So \(24^\circ\) and \(11/83\) are successive approximations from the same expansion, which is why some historians link \(11/83\) to anthyphairesis, the Greek form of the Euclidean algorithm.

Exercise 8.2Assume, as Cleomedes does, that Syene lies on the tropic, and that the summer noon Sun at Alexandria is \(7^\circ12'\) from the zenith. With \(\varepsilon=23^\circ51'\), what latitude does this give Alexandria? Compare with \(31^\circ12'\).

Show solution

Zenith distance at summer noon is \(\varphi-\varepsilon\), so \(\varphi=23^\circ51'+7^\circ12'=31^\circ03'\), nine minutes of arc south of the true value. The two errors (obliquity too large, zenith distance slightly too large) partly cancel.

Exercise 8.3In the zone scheme, convert the boundaries into degrees and stadia from the equator, using 252,000 stadia for the circumference.

Show solution

One sixtieth is \(6^\circ\) and 4,200 stadia. Tropic: 4 sixtieths, \(24^\circ\), 16,800 stadia. Boundary of the frigid zone: 9 sixtieths, \(54^\circ\), 37,800 stadia. Pole: 15 sixtieths, \(90^\circ\), 63,000 stadia.

Lesson 09

A gridded map of the world

For the general reader. Earlier Greek maps were sketches. Eratosthenes drew the known world on a grid of lines running east–west and north–south, placed by measured distances and by the length of the longest day. He also believed the world was a globe on which one could, in principle, sail west from Spain to India.

Goals

Describe the main parallel and meridian of Eratosthenes’ map and its dimensions.
Compute lengths along parallels on a sphere.
Use the longest day to find latitude, the method Hipparchus pressed against him.

The frame

The Geographika survives only in fragments, mostly in Strabo and Pliny. Its first book reviewed earlier geography, including Homer; the second set out the mathematical foundations, including the size of the Earth; the third drew the map. Its frame was a main parallel running from the Pillars of Heracles (Gibraltar) through the Strait of Messina, Rhodes and the Gulf of Issus, along the Taurus range to the eastern mountains, and a main meridian through Meroë, Syene, Alexandria, Rhodes and the mouth of the Borysthenes (Dnieper). The inhabited world (oikoumene) measured about 38,000 stadia from north to south and about 78,000 from west to east, twice as long as wide. For the regions east of the Mediterranean he divided the land into irregular quadrilaterals called sphragides, “seals”, and estimated their sides.

Proposition (parallels on a sphere)

On a sphere of circumference \(C\), the parallel of latitude \(\varphi\) is a circle of circumference \(C\cos\varphi\). An east–west distance \(d\) along it spans a longitude difference \(\Delta\lambda=360^\circ\cdot d/(C\cos\varphi)\).

The parallel at latitude \(\varphi\) is a circle in a plane perpendicular to the axis, with radius \(R\cos\varphi\), and longitude is the angle around the axis.

The longest day

Greek geographers described latitudes by the length of the longest day, the klima. On the day of the summer solstice the Sun’s hour angle at sunset satisfies \(\cos H_0=-\tan\varphi\tan\varepsilon\), and the length of daylight is \(2H_0/15^\circ\) hours. Hipparchus, in his three books Against the Geography of Eratosthenes, objected that Eratosthenes placed too many points by travellers’ distances rather than by such astronomical observations.

Exercise 9.1Rhodes lies at about \(36^\circ\) N. With \(C=252{,}000\) stadia, find the circumference of the parallel of Rhodes, and the longitude span of an inhabited world 78,000 stadia long along it.

Show solution

\(252{,}000\cos36^\circ\approx203{,}900\) stadia. Then \(\Delta\lambda\approx360^\circ\cdot78{,}000/203{,}900\approx138^\circ\). The remaining \(222^\circ\) of the parallel is unknown sea, which is why Eratosthenes could say that, but for the size of the Atlantic, one might sail along the same parallel from Iberia to India.

Exercise 9.2Find the length of the longest day at Rhodes (\(\varphi=36^\circ\), \(\varepsilon=23^\circ51'\)), ignoring refraction and the Sun’s disc. The ancient value was \(14\tfrac12\) hours.

Show solution

\(\tan36^\circ\approx0.7265\), \(\tan23.85^\circ\approx0.4421\), product \(0.3212\). So \(\cos H_0=-0.3212\), \(H_0\approx108.7^\circ\), and the day lasts \(2\cdot108.7/15\approx14.5\) hours.

Exercise 9.3If the inhabited world spans 38,000 stadia north to south, how many degrees of latitude is that with 700 stadia per degree? With a southern limit near \(12^\circ\) N, where is its northern edge?

Show solution

\(38{,}000/700\approx54^\circ\). From about \(12^\circ\) N this reaches about \(66^\circ\) N, close to the Arctic Circle, the region where Eratosthenes, following the reports of Pytheas, placed Thule.

Lesson 10

Chronology and the calendar

For the general reader. Greek cities each counted years in their own way, by local officials or kings. Eratosthenes built a single timeline for Greek history, counted in Olympiads, the four-year cycles of the Olympic games. He dated the fall of Troy to what we call 1184/3 BC, a date ancient historians used for centuries.

Goals

Convert Olympiad dates to years BC.
Reconstruct the intervals of Eratosthenes’ chronology.
Compute calendar drift and the effect of a four-year leap rule.

The system

The Chronographiai and Olympic Victors are lost, but later chronographers, especially Apollodorus of Athens, followed them. Eratosthenes fixed a sequence of epochs by counting intervals: fall of Troy 1184/3 BC; return of the Heraclidae (Dorian migration) 1104/3; first Olympiad 776; Xerxes’ crossing into Greece 480/79; start of the Peloponnesian War 432/1. Earlier dates rested on generations and king lists; later ones on recorded Olympic victors.

Conversion rule

Olympiad \(n\), year \(k\) (\(k=1,2,3,4\)) begins in midsummer of the year

\[ Y=776-4(n-1)-(k-1)\ \text{BC}, \]

and runs to midsummer of the next year, written \(Y/(Y-1)\) BC.

The calendar

The Egyptian civil year had exactly 365 days, so it slipped against the seasons by one day every four years. In 238 BC the priests meeting at Canopus decreed an extra day every fourth year in honour of Ptolemy III. Later sources credit Eratosthenes with a year of 365¼ days, and it is plausible that the court astronomer and librarian was involved, though no ancient text says so. The reform did not last; a fixed Egyptian year with a leap day was established only under Augustus.

Exercise 10.1Check that Xerxes’ invasion (Olympiad 75, year 1) falls in 480/79 BC and the start of the Peloponnesian War (Olympiad 87, year 1) in 432/1 BC.

Show solution

\(776-4\cdot74=776-296=480\). \(776-4\cdot86=776-344=432\). Both are first years of their Olympiads.

Exercise 10.2How many years separate the fall of Troy from the first Olympiad in Eratosthenes’ scheme? Express the interval from the return of the Heraclidae to the first Olympiad in generations of about 33 years.

Show solution

\(1184-776=408\) years, or 407 counting from 1183. From 1104 to 776 is 328 years, about ten generations of 33 years, the kind of genealogical reckoning used for the early period.

Exercise 10.3The solar year is about 365.2422 days. How many years does it take for a 365-day calendar to drift a full year? How fast does a calendar with a leap day every fourth year drift, and in which direction?

Show solution

Drift is 0.2422 days per year, so a full cycle takes \(365.2422/0.2422\approx1508\) tropical years (the Egyptian “Sothic” cycle of 1460 years uses \(\tfrac14\) day and the star Sirius instead). With 365.25 days the calendar runs ahead of the Sun by 0.0078 days a year, about one day in 128 years, the error the Gregorian reform corrected in 1582.

Part III · Life and legacy

Lesson 11

Old age, critics and successors

Eratosthenes remained in royal favour through the reign of his pupil Ptolemy IV (221–204 BC). He wrote a memoir, Arsinoe, about Ptolemy IV’s queen and sister, who was murdered in the palace intrigues after the king’s death; Athenaeus quotes from it. The Suda names his pupils Aristophanes of Byzantium, who succeeded him at the Library and became the greatest Homeric scholar of the age, Mnaseas of Patara, Menander and Aristis.

Blindness and death

According to the Suda, Eratosthenes lost his sight in old age and, in despair, starved himself to death, at the age of 80. Other ancient sources give his age at death as 81 (Censorinus) or 82 (a work attributed to Lucian). He died in Alexandria around 194 BC. An epigram by Dionysius of Cyzicus, preserved in the Greek Anthology, mourns that he was buried not in Cyrene but on the shore of Egypt.

The critics

Hipparchus of Nicaea, the greatest astronomer of the next century, wrote three books against the Geographika. He accepted Eratosthenes’ method but demanded that positions rest on astronomical observations of latitude and longitude, not on travellers’ distances. He adopted Eratosthenes’ value for the Earth’s circumference and worked with 700 stadia to a degree. Polybius and Strabo criticized details of the map. Strabo nonetheless followed him closely on the regions east of the Mediterranean, and thereby preserved most of what we know.

Exercise 11.1Hipparchus objected that one should not combine distances measured along roads with positions fixed by the stars. Give a concrete geometrical reason why road distances between two cities cannot directly give their difference of latitude.

Show solution

A road distance measures path length, which exceeds the great-circle distance, and the great-circle distance mixes north–south and east–west components. Only the meridian component converts to latitude. Alexandria and Syene illustrate this: they differ by about \(3^\circ\) in longitude, so the 5,000 stadia overstate the meridian arc. Latitude measured from the Sun or the length of day avoids the problem.

Lesson 12

Legacy

The size of the Earth after Eratosthenes

A century later Posidonius estimated the circumference by a different route: the star Canopus just grazes the horizon at Rhodes but rises to about a forty-eighth of a circle above it at Alexandria. Cleomedes reports a result of 240,000 stadia; Strabo reports a smaller figure of 180,000 stadia attributed to Posidonius. Ptolemy’s Geography adopted the smaller value, which made Asia seem to reach far around the globe. Christopher Columbus, relying on low estimates of this kind, expected Asia within reach across the Atlantic. Eratosthenes’ larger and more accurate figure would have told him otherwise.

The sieve after Eratosthenes

Through Nicomachus and Boethius the sieve entered the medieval arithmetic curriculum. In 1772 Samuel Horsley published a reconstruction of it in the Philosophical Transactions. In the twentieth century the sieve became a theory: Viggo Brun’s sieve (1915–1920) showed that the sum of the reciprocals of the twin primes converges, and Atle Selberg’s sieve of the 1940s gave sharp, flexible upper bounds. Sieve methods lie behind Chen Jingrun’s theorem on even numbers as a prime plus a product of at most two primes, and behind Yitang Zhang’s 2013 proof of bounded gaps between primes.

Honours

The lunar crater Eratosthenes, and the Eratosthenian period of lunar geology named after it.
The Eratosthenes Seamount in the eastern Mediterranean.
The Earth measurement is repeated every year by school classes around the world, who share shadow readings at noon on the equinox to compute the circumference.

Exercise 12.1Posidonius’ method: Canopus grazes the horizon at Rhodes and stands \(1/48\) of a circle high at Alexandria, assumed 5,000 stadia away. What circumference follows? If the distance is replaced by 3,750 stadia?

Show solution

\(48\cdot5000=240{,}000\) stadia, and \(48\cdot3750=180{,}000\) stadia. The method is the same as Eratosthenes’ with a star replacing the Sun; the angle is the difference of latitude. The true arc between Rhodes and Alexandria is about \(5^\circ14'\), not \(7^\circ30'\): horizon observations of a grazing star are badly affected by refraction.

Exercise 12.2Two classes, one at latitude \(45.0^\circ\) N and one at \(40.5^\circ\) N on the same meridian, 500 km apart, measure the noon Sun at the equinox. What zenith angles do they find, and what circumference do they compute?

Show solution

At the equinox the Sun is overhead at the equator, so the noon zenith angle equals the latitude: \(45.0^\circ\) and \(40.5^\circ\). The difference is \(4.5^\circ=1/80\) of a circle, so \(C=80\cdot500=40{,}000\) km.

Final projectChoose one of Eratosthenes’ results from Part II. Write two pages for a strong undergraduate that set out what the ancient sources actually report, reconstruct the argument in modern notation, and assess which parts of the result are his, which are the reporter’s, and which are modern inference.

Reference

Connections

People linked to Eratosthenes across this collection. A name without a link has no page here yet. See Eratosthenes in the network of all 337 lives →

Teachers and advisors

Arcesilaus of Pitane
Ariston of Chios
Bion of Borysthenes
Lysanias of Cyrene

Students

Aristophanes of Byzantium
Mnaseas of Patara
Ptolemy IV Philopator

Influenced by

Euclid
Callimachus
Pytheas of Massalia
Zeno of Citium

Influenced

Atle Selberg
Claudius Ptolemy
Hipparchus
Nicomachus of Gerasa
Posidonius
Strabo

Colleagues and collaborators

Archimedes

Rivals

Hipparchus

Family

Aglaos

Also mentioned in this life

Fields

geography 5 lives
geodesy 12 lives
number theory 45 lives
geometry 32 lives
astronomy 50 lives

On CoursesHub

Astrophysics for Beginners

Reference

Sources and further reading

Biography and sources

J. J. O’Connor and E. F. Robertson, Eratosthenes of Cyrene, MacTutor History of Mathematics.
D. R. Dicks, Eratosthenes, Dictionary of Scientific Biography.
Eratosthenes, Wikipedia (with references to the ancient testimonia).
D. W. Roller, Eratosthenes’ Geography: Fragments Collected and Translated, with Commentary, Princeton University Press, 2010.
K. Geus, Eratosthenes von Kyrene: Studien zur hellenistischen Kultur- und Wissenschaftsgeschichte, Beck, 2002.
P. M. Fraser, Ptolemaic Alexandria, 3 vols., Oxford, 1972.
Strabo, Geography, Books 1–2, English translation by H. L. Jones (Loeb, on LacusCurtius).

Mathematics and science

A. C. Bowen and R. B. Todd, Cleomedes’ Lectures on Astronomy, University of California Press, 2004.
T. L. Heath, A History of Greek Mathematics, 2 vols., Oxford, 1921 (doubling the cube, the mesolabe, the Earth measurement).
W. R. Knorr, Textual Studies in Ancient and Medieval Geometry, Birkhäuser, 1989 (the Eutocius letter).
The Duplicators, Part I: Eutocius’ collection of cube duplications, MAA Convergence.
Sieve of Eratosthenes and Earth’s circumference, Wikipedia.
S. Horsley, Κόσκινον Ἐρατοσθένους, or, The Sieve of Eratosthenes, Philosophical Transactions 62 (1772).
J. Friedlander and H. Iwaniec, Opera de Cribro, AMS Colloquium Publications 57, 2010 (modern sieve theory).
G. J. Toomer, Ptolemy’s Almagest, Duckworth, 1984 (I.12 on the obliquity).
A. Barker, Greek Musical Writings, vol. 2, Cambridge, 1989 (Ptolemy’s tables of tetrachords).

How to use this course

Timeline

Cyrene, Athens and the sources

Teachers

The shape of the evidence

Called to Alexandria

The Museum and the Library

The letter from Archimedes

Pentathlos and Beta

The works

A method across fields

The sieve

The source

The algorithm and its correctness

Counting by inclusion and exclusion

The cost

Two mean proportionals and the mesolabe

The problem and its history

The mesolabe

Other solutions, and impossibility

Ratio, means and musical scales

What the sources say

The musical proportion

Measuring the Earth

The report

How good was it?

Tropics, zones and the obliquity

The ratio 11/83

Zones in sixtieths

A gridded map of the world

The frame

The longest day

Chronology and the calendar

The system

The calendar

Old age, critics and successors

Blindness and death

The critics

Legacy

The size of the Earth after Eratosthenes

The sieve after Eratosthenes

Honours

Connections

Teachers and advisors

Students

Influenced by

Influenced

Colleagues and collaborators

Rivals

Family

Also mentioned in this life

Fields

On CoursesHub

Sources and further reading

Biography and sources

Mathematics and science