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Archimedes

Archimedes of Syracuse (Ἀρχιμήδης)

A course in two strands: what can and cannot be known about the life of the Syracusan geometer, and the mathematics and physics of the treatises that survived him, from the measurement of the circle to the law of buoyancy.

Born c. 287 BC, Syracuse Died 212 BC, Syracuse Favourite theorem sphere : cylinder = 2 : 3 Level advanced undergraduate
The figure Archimedes wanted on his tomb. Sphere and circumscribed cylinder: volumes and surfaces both in the ratio 2 : 3.
Lesson 00

How to use this course

Archimedes is the most famous mathematician of antiquity and one of the least documented. No contemporary account of his life survives. What we know of him as a person comes from a few remarks in his own prefaces and from Roman and Byzantine authors writing between one and fourteen centuries after his death. His mathematics, by contrast, survives in his own words, in a dozen treatises whose proofs still hold up.

This course keeps the two kinds of evidence apart. In the biographical lessons, each story is labelled with its source and with a judgement on how far it can be trusted. In the technical lessons, the statements follow the treatises, translated into modern notation where that helps.

Prerequisites for Part II
  • Euclidean geometry: similar triangles, circles, the elements of conic sections.
  • First-year calculus, used to check Archimedes’ results (he did not have it).
  • Elementary mechanics: forces, torque, density, pressure.
  • For Lesson 11, comfort with exponents and the idea of a Pell equation.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proof sketches and exercises with folded solutions. Archimedes stated results as ratios between geometric magnitudes, never as formulas with \(\pi\). The course gives both forms.

Reference

Timeline

All dates BC are approximate unless a source fixes them. The birth year rests on a twelfth-century report that he lived 75 years.

c. 287 BCBorn in Syracuse, Sicily. His father, by his own statement, was the astronomer Phidias.
c. 270 BCHieron II takes power in Syracuse; he rules until 215 BC.
263 BCHieron allies Syracuse with Rome during the First Punic War, beginning half a century of peace for the city.
c. 260–220 BCPeriod of the surviving treatises, sent as letters to Conon of Samos, Dositheus of Pelusium and Eratosthenes of Cyrene. Exact dates and order are uncertain.
c. 216 BCDeath of Gelon, Hieron’s son, to whom The Sand Reckoner is addressed (so that work is earlier).
215 BCDeath of Hieron II. His grandson Hieronymus turns Syracuse toward Carthage.
214 BCRoman forces under Marcus Claudius Marcellus begin the siege of Syracuse. Archimedes’ engines hold off the assault from the sea.
212 BCSyracuse falls. Archimedes is killed by a Roman soldier.
75 BCCicero, quaestor in Sicily, finds the overgrown tomb marked with a sphere and cylinder.
6th c. ADEutocius of Ascalon writes commentaries; Isidore of Miletus assembles an edition in Constantinople.
c. 950 ADA Constantinople scribe copies the codex later known as the Archimedes Palimpsest.
1229 ADThe codex is scraped and overwritten as a prayer book.
1269 ADWilliam of Moerbeke translates the Greek treatises into Latin.
1544 ADFirst printed Greek edition, Basel.
1906 ADJohan Ludvig Heiberg reads the palimpsest in Constantinople and recovers The Method.
1998 ADThe palimpsest is sold at auction in New York; imaging at the Walters Art Museum follows (1999–2008).
Part I · Formation
Lesson 01

Syracuse and the problem of sources

Archimedes was born in Syracuse, the largest Greek city of Sicily, around 287 BC. The date is an inference. John Tzetzes, a Byzantine scholar of the twelfth century AD, says Archimedes lived 75 years, and the death in 212 BC is fixed by the Roman capture of the city. Nothing earlier confirms the figure.

What Archimedes says himself

The most secure biographical fact comes from Archimedes. In The Sand Reckoner he mentions the estimate of the sizes of the Sun and Moon made by Phidias, “my father”. Phidias is otherwise unknown. The remark suggests that Archimedes grew up with astronomy at home. His prefaces add a little more: the names of his friends and correspondents, his grief at Conon’s death, his habit of sending theorems to Alexandria without proofs to see who could prove them.

What later writers say

  • Polybius (2nd century BC), the nearest in time, describes the defence of Syracuse and Archimedes’ machines. His account is the most trustworthy of the narrative sources.
  • Livy (late 1st century BC) follows Polybius on the siege and gives an account of the death.
  • Cicero (1st century BC) describes Archimedes’ planetarium and his own discovery of the tomb. He calls Archimedes a man of humble station, which conflicts with the next source.
  • Plutarch (c. AD 100), in his Life of Marcellus, says Archimedes was a kinsman and friend of King Hieron II. He supplies most of the famous anecdotes, and also the claim that Archimedes despised practical mechanics, which Plutarch’s own Platonist views may have coloured.
  • Vitruvius (late 1st century BC) tells the story of the golden crown.
  • Byzantine writers such as Tzetzes and Zonaras (12th century AD) add the burning mirrors.

A biography by one Heraclides, cited by Eutocius in the sixth century AD, is lost. We do not know whether Archimedes married or had children, how he earned his living, or whether he ever left Sicily.

Exercise 1.1Rank the following claims by reliability and name the best source for each: (a) his father was an astronomer; (b) he was related to Hieron; (c) he died in 212 BC; (d) he shouted “Eureka” in the street.

Show solution
(a) Archimedes’ own statement in The Sand Reckoner: very reliable. (c) Fixed by the fall of Syracuse, reported by Polybius and Livy: reliable to within a year. (b) Plutarch only, contradicted by Cicero’s remark: uncertain. (d) Vitruvius, about two centuries later, with no earlier witness: a story, probably embellished.
Lesson 02

Letters to Alexandria

In the third century BC the centre of Greek science was Alexandria, where the Ptolemies funded the Museum and its library. Euclid had worked there a generation before. Archimedes wrote his treatises as letters addressed to Alexandrian mathematicians, which shows he knew them personally or by close correspondence. Many historians think he studied there as a young man. Diodorus Siculus says he invented the water screw in Egypt, but no source puts a date on a visit, and it is possible that all his contact was by letter.

The correspondents

  • Conon of Samos, astronomer and mathematician, was his closest friend. Archimedes sent him results first. When Conon died, Archimedes wrote that he had lost someone able to judge his work, and sent his results instead to Dositheus, who had known Conon.
  • Dositheus of Pelusium received Quadrature of the Parabola, On the Sphere and Cylinder, On Spirals and On Conoids and Spheroids.
  • Eratosthenes of Cyrene, head of the Library and measurer of the Earth, received The Method and, according to its heading, the Cattle Problem.
  • Zeuxippus received a lost work on the naming of numbers, mentioned in The Sand Reckoner.

A trap for plagiarists

In the preface to On Spirals Archimedes explains that he had earlier sent statements of theorems without proofs, and that some people in Alexandria had claimed them as their own. He admits that two of the statements were deliberately false, so that anyone claiming to have proved everything would be exposed. The passage is the closest thing we have to his voice outside mathematics: proud, careful about credit, and a little mischievous.

The inheritance

Archimedes built on two predecessors above all. Eudoxus of Cnidus had created the method of exhaustion and the theory of proportion found in Book V of Euclid’s Elements; Archimedes credits Eudoxus with the first proofs that a cone is one third of the cylinder on the same base. Euclid supplied the deductive form. Archimedes also cites Aristarchus of Samos, whose heliocentric hypothesis is known chiefly through Archimedes’ report of it.

Exercise 2.1The heading of The Method addresses Eratosthenes, and the Quadrature mentions Conon as dead. What can be concluded about the relative dates of these two works, and what cannot?

Show solution
Both postdate Conon’s death, at least for the Quadrature. Since The Method refers to results proved in other treatises (such as the sphere), it is a late work. Nothing fixes an absolute date for either, and the order of works not linked by cross-references is unknown.
Lesson 03

Hieron’s engineer: record and legend

For most of Archimedes’ adult life Syracuse was ruled by Hieron II, who allied with Rome in 263 BC and kept the city at peace for decades. Ancient writers connect several famous stories to this king. They are best read one at a time, with their sources.

The golden crown

Source: Vitruvius, De architectura, Book IX, about two centuries later. Hieron suspects that a goldsmith has mixed silver into a votive wreath. Archimedes, stepping into a full bath, sees the water overflow and realizes that the wreath’s volume can be measured by the water it displaces. He runs home shouting “Eureka” (I have found it). Assessment: the physics is sound, and the method of overflow would work in principle, but the story has no earlier witness. In 1586 the young Galileo argued in La Bilancetta that the overflow method is too crude and that Archimedes would have used a balance immersed in water, the method of Lesson 09. A Latin poem of late antiquity, the Carmen de ponderibus, already describes a balance version.

The ship and the lever

Sources: Plutarch; Athenaeus quoting Moschion; Pappus of Alexandria (4th century AD). Plutarch says Archimedes told Hieron that any weight could be moved by a given force, and then, with a compound pulley, single-handedly drew a loaded ship over dry land. Athenaeus describes the giant ship Syracusia, built for Hieron, and credits Archimedes with launching it. The boast about moving the Earth if given a place to stand appears in Pappus and later writers in varying forms. Assessment: Archimedes’ theory of the lever is real (Lesson 08); the demonstration is plausible but known only from writers centuries later.

The screw and the planetarium

Sources: Diodorus Siculus; Athenaeus; Cicero. Diodorus attributes the water-raising screw to Archimedes during a stay in Egypt. The device may be older, and no contemporary credits him with it. Cicero describes two celestial spheres taken to Rome by Marcellus, one a mechanical model that showed the motions of the Sun, Moon and planets. Pappus mentions a lost work by Archimedes, On Sphere-Making. Assessment: a planetarium by Archimedes is well attested; the screw attribution is possible but uncertain.

Exercise 3.1A crown of mass 1000 g is suspected of containing silver. Pure gold has density 19.3 g/cm³, silver 10.5 g/cm³. The crown displaces 60.0 cm³ of water. Assuming it is a gold–silver alloy with no voids, how much gold does it contain?

Show solution
Let \(m\) be the mass of gold. Then \(\frac{m}{19.3}+\frac{1000-m}{10.5}=60\). Solving, \(m\left(\frac1{10.5}-\frac1{19.3}\right)=\frac{1000}{10.5}-60\), so \(m\approx 811\) g. About 189 g of silver has replaced gold.

Exercise 3.2Test Galileo’s objection. A pure gold crown of 1000 g has volume about 51.8 cm³; the adulterated one above has 60.0 cm³. If the crown is lowered into a cylindrical vessel with an opening 20 cm across, how much higher does the water rise for the fake crown than for the real one?

Show solution
The extra volume is 8.2 cm³. The opening has area \(\pi\cdot10^2\approx314\) cm², so the extra rise is about 0.026 cm, a quarter of a millimetre, hard to see against a meniscus. A balance comparing the crown with equal mass of gold, both under water, shows a difference in buoyancy of about 8 g weight, which is easy to detect. Galileo’s point stands.
Part II · The mathematics and physics
Lesson 04

Measuring the circle

For the general reader. How long is the circumference of a circle compared with its diameter? Archimedes trapped the answer between two fractions by drawing polygons inside and outside the circle and doubling their number of sides until they had 96. His bounds, 3 10/71 and 3 1/7, are the source of the school value 22/7.

Goals
  • State the two main propositions of Measurement of a Circle.
  • Derive the polygon-doubling recurrence and run it to 96 sides.
  • See how Archimedes controlled square roots with rational bounds.
Proposition 1

The area of any circle equals the area of a right triangle in which one of the sides about the right angle equals the radius and the other equals the circumference.

In modern form, \(A=\tfrac12 rC\). With \(C=2\pi r\) this is \(A=\pi r^2\), so Proposition 1 says the same constant governs area and circumference.

Proof sketch (double reduction to absurdity). Let \(T=\tfrac12 rC\). Suppose \(A>T\). Inscribe a square and keep doubling the number of sides; each doubling removes more than half of the remaining gap between polygon and circle, so eventually an inscribed polygon \(P\) has \(A-P<A-T\), that is \(P>T\). But \(P\) is half its perimeter times its apothem, both smaller than \(C\) and \(r\), so \(P<T\). Contradiction. Suppose \(A<T\). Circumscribed polygons \(Q\) eventually satisfy \(Q-A<T-A\), so \(Q<T\); but \(Q\) is half its perimeter times \(r\), and its perimeter exceeds \(C\), so \(Q>T\). Contradiction. Hence \(A=T\).

Proposition 3

The ratio of the circumference of any circle to its diameter is less than \(3\tfrac17\) and greater than \(3\tfrac{10}{71}\).

The doubling recurrence

For a circle of radius 1, let \(a_n\) and \(b_n\) be the semiperimeters of the circumscribed and inscribed regular \(n\)-gons, so \(b_n<\pi<a_n\). Archimedes’ geometric steps (bisecting an angle and using the angle-bisector theorem) are equivalent to

\[ a_{2n}=\frac{2a_nb_n}{a_n+b_n},\qquad b_{2n}=\sqrt{a_{2n}\,b_n}, \]

starting from the hexagons \(a_6=2\sqrt3\approx3.4641\), \(b_6=3\). Four doublings give \(b_{96}\approx3.14103\) and \(a_{96}\approx3.14271\). Archimedes rounded at every step in the safe direction, always making outer bounds larger and inner bounds smaller. For \(\sqrt3\) he used, without explanation,

\[ \frac{265}{153}<\sqrt3<\frac{1351}{780}. \]

Exercise 4.1Check that \(1351^2-3\cdot780^2=1\) and \(265^2-3\cdot153^2=-2\). Explain why these identities make the bounds on \(\sqrt3\) very good, and verify that they are on the correct sides.

Show solution
\(1351^2=1\,825\,201\) and \(3\cdot780^2=1\,825\,200\), difference 1. \(265^2=70\,225\), \(3\cdot153^2=70\,227\), difference \(-2\). From \(p^2-3q^2=\varepsilon\) we get \(\frac pq-\sqrt3=\frac{\varepsilon}{q(p+q\sqrt3)}\), which is tiny when \(q\) is large and \(\varepsilon\) small. The sign of \(\varepsilon\) gives the side: \(1351/780\) is above \(\sqrt3\) and \(265/153\) below. Both fractions are convergents of the continued fraction of \(\sqrt3\), a hint of how Archimedes may have found them.

Exercise 4.2Run the recurrence from \(n=6\) to \(n=12\) by hand. Then confirm that \(3\tfrac{10}{71}<b_{96}\) and \(a_{96}<3\tfrac17\) using the values above.

Show solution
\(a_{12}=\frac{2\cdot2\sqrt3\cdot3}{2\sqrt3+3}=\frac{12\sqrt3}{2\sqrt3+3}=12(2-\sqrt3)\approx3.2154\) and \(b_{12}=\sqrt{3.2154\cdot3}\approx3.1058\). Then \(3\tfrac{10}{71}=223/71\approx3.14085<3.14103\) and \(22/7\approx3.14286>3.14271\). The margins are small, which is why every rounding had to go the safe way.
Lesson 05

Exhaustion and the parabola

For the general reader. A straight line cut across a parabola encloses a curved region. Archimedes proved that its area is exactly four thirds of a certain triangle inside it. He filled the region with ever smaller triangles whose areas form a geometric series, then proved, without any notion of a limit, that no other answer is possible.

Goals
  • State the quadrature theorem and the axiom it depends on.
  • Follow the geometric proof by successive triangles.
  • Understand the finite identity Archimedes used in place of an infinite sum.

The axiom

Archimedes stated, as an assumption in On the Sphere and Cylinder and in the Quadrature, that the excess by which the greater of two unequal magnitudes exceeds the less can, if added to itself enough times, exceed any given magnitude of the same kind. This is now called the Archimedean property. In modern terms it excludes infinitesimals, and it is what makes exhaustion proofs work.

Quadrature of the Parabola, Proposition 24

Let a chord \(QQ'\) cut off a segment of a parabola, and let \(P\) be the point of the segment where the tangent is parallel to \(QQ'\). Then the area of the segment is \(\tfrac43\) of the area of triangle \(QPQ'\).

Proof sketch

Call the area of triangle \(QPQ'\) \(T\). The chords \(QP\) and \(PQ'\) cut off two smaller segments; in each, inscribe the triangle with vertex where the tangent is parallel to the chord. Archimedes proves these two triangles together have area \(\tfrac14T\). Repeating, the \(k\)-th stage adds triangles of total area \(T/4^k\). Each inscribed triangle is more than half of the segment containing it, so the uncovered area can be made smaller than any given magnitude. Archimedes then uses the finite identity

\[ T+\frac T4+\frac T{4^2}+\cdots+\frac T{4^n}+\frac13\cdot\frac T{4^n}=\frac43T, \]

and argues by double contradiction: if the segment were larger than \(\tfrac43T\), some stage of the triangles would already exceed \(\tfrac43T\), which the identity forbids; if it were smaller, the identity shows the triangles at some stage would exceed the segment. The infinite series \(\sum 4^{-k}=\tfrac43\) is present in substance, but every step is finite.

The same treatise contains a second proof, by mechanical balancing (Propositions 6–17), a preview of Lesson 10.

Exercise 5.1Take the parabola \(y=x^2\) and the chord from \((-1,1)\) to \((1,1)\). Find \(P\), compute \(T\), and check the theorem by integration.

Show solution
The chord is horizontal, so \(P=(0,0)\). The triangle has base 2 and height 1, so \(T=1\). The segment has area \(\int_{-1}^{1}(1-x^2)\,dx=2-\tfrac23=\tfrac43=\tfrac43T\).

Exercise 5.2In the same example, find the vertex of the second-stage triangle over the chord from \((0,0)\) to \((1,1)\) and verify that the two second-stage triangles have total area \(\tfrac14T\).

Show solution
The chord has slope 1, and \(y'=2x=1\) at \(x=\tfrac12\), so the vertex is \((\tfrac12,\tfrac14)\). The triangle with vertices \((0,0),(1,1),(\tfrac12,\tfrac14)\) has area \(\tfrac12\left|1\cdot\tfrac14-1\cdot\tfrac12\right|=\tfrac18\). By symmetry the other is also \(\tfrac18\); together \(\tfrac14=\tfrac14T\).

Exercise 5.3Prove the finite identity above by induction on \(n\).

Show solution
For \(n=0\): \(T+\tfrac13T=\tfrac43T\). Assuming it for \(n\), replace \(\tfrac13\cdot\tfrac T{4^n}\) by \(\tfrac T{4^{n+1}}+\tfrac13\cdot\tfrac T{4^{n+1}}\); this is valid because \(\tfrac14+\tfrac1{12}=\tfrac13\). Archimedes states the step in exactly this geometric form.
Lesson 06

Sphere and cylinder

For the general reader. Put a ball snugly inside a can. Archimedes proved that the ball has exactly two thirds of the can’s volume and two thirds of its total surface. He was proud enough of the result to want it on his tomb, and Cicero found it there 137 years after his death.

Goals
  • State the surface and volume theorems of Book I.
  • Derive the 2 : 3 ratios.
  • Meet the cubic problem of Book II.
On the Sphere and Cylinder I, Propositions 33 and 34

(33) The surface of any sphere is four times its greatest circle. (34) Any sphere is four times the cone whose base is equal to the greatest circle of the sphere and whose height is equal to the radius of the sphere.

In modern form, \(S=4\pi r^2\) and \(V=4\cdot\tfrac13\pi r^2\cdot r=\tfrac43\pi r^3\).

Corollary

The cylinder whose base is a great circle of the sphere and whose height is the diameter is \(\tfrac32\) of the sphere, both in volume and in total surface (bases included).

Indeed the cylinder has volume \(\pi r^2\cdot2r=2\pi r^3=\tfrac32\cdot\tfrac43\pi r^3\) and surface \(2\pi r\cdot 2r+2\pi r^2=6\pi r^2=\tfrac32\cdot4\pi r^2\).

Method of proof

Archimedes inscribes and circumscribes about the sphere solids made by rotating regular polygons about a diameter. Their surfaces are sums of lateral surfaces of cones and frusta, which he computed in earlier propositions. Two postulates do the essential work: a straight line is the shortest curve between its endpoints, and of two convex curves (or surfaces) with the same ends, one enclosing the other, the enclosing one is longer. With these, exhaustion squeezes the sphere between the two families. He also proves that the surface of a spherical segment equals the circle whose radius is the straight line from the vertex of the segment to the edge of its base. In modern terms a zone of height \(h\) has area \(2\pi rh\), the same as the band of the circumscribed cylinder at the same height.

Book II: cutting a sphere in a given ratio

Book II solves problems about segments. Proposition 4 asks for a plane cutting a sphere into segments with volumes in a given ratio. Archimedes reduces it to a problem equivalent to a cubic equation and promised a solution; Eutocius, eight centuries later, found and preserved a solution by intersecting a parabola and a hyperbola, which he attributed to Archimedes.

Exercise 6.1Show that a chord from the vertex of a spherical cap of height \(h\) to the rim of its base has length \(\sqrt{2rh}\), and deduce Archimedes’ form of the zone theorem from \(2\pi rh\).

Show solution
Place the sphere’s centre at the origin and the vertex at \((0,r)\). The rim lies at height \(r-h\) with horizontal radius \(\rho\), \(\rho^2=r^2-(r-h)^2=2rh-h^2\). The chord satisfies \(\ell^2=\rho^2+h^2=2rh\). A circle of radius \(\ell\) has area \(\pi\ell^2=2\pi rh\), the cap’s surface.

Exercise 6.2A cap of height \(h\) in a sphere of radius \(r\) has volume \(\tfrac13\pi h^2(3r-h)\). Find \(h\) (for \(r=1\)) so that the cap is one third of the sphere. Why could Archimedes not solve this with ruler and compass?

Show solution
We need \(\tfrac13\pi h^2(3-h)=\tfrac13\cdot\tfrac43\pi\), i.e. \(h^3-3h^2+\tfrac43=0\). Numerically \(h\approx0.774\). The cubic \(3h^3-9h^2+4=0\) has no rational root (try \(\pm1,\pm2,\pm4,\pm\tfrac13,\dots\)), so its roots are not constructible with ruler and compass. Greek geometers handled such problems with conic sections, as Eutocius records.
Lesson 07

Spirals, conoids and spheroids

For the general reader. Archimedes defined a new curve by motion: a point moving steadily outward along a line that itself turns steadily. He found the area it sweeps and the direction of its tangent, and used the tangent to “straighten” a circle. In a companion work he computed volumes of solids made by spinning conic sections.

Goals
  • State the area and tangent theorems for the spiral.
  • See how the tangent rectifies the circle.
  • State the volume theorem for the paraboloid of revolution.

On Spirals

In polar coordinates the spiral is \(r=a\theta\). Archimedes defined it kinematically, and the definition matters: the curve is not constructible with ruler and compass, and the tangent theorem uses the composition of the two motions.

On Spirals, Propositions 18 and 24 (modern form)

(18) At the point of the spiral reached after one complete turn, the tangent meets the line through the origin perpendicular to the radius at a distance from the origin equal to the circumference of the circle of radius \(r=2\pi a\). (24) The area bounded by the first turn of the spiral and the radius at its end is one third of the circle of radius \(2\pi a\).

Check by calculus: the area is \(\tfrac12\int_0^{2\pi}a^2\theta^2\,d\theta=\tfrac{(2\pi)^3a^2}{6}=\tfrac13\pi(2\pi a)^2\). Archimedes proved it by inscribing and circumscribing sectors of circles and summing squares, using the identity \(1^2+2^2+\cdots+n^2=\tfrac{n(n+1)(2n+1)}{6}\) in geometric form.

On Conoids and Spheroids

A “right-angled conoid” is a paraboloid of revolution, an “obtuse-angled conoid” a hyperboloid of revolution, and a spheroid an ellipsoid of revolution. Archimedes computed the volumes of their segments.

On Conoids and Spheroids, Proposition 21

Any segment of a paraboloid of revolution cut off by a plane perpendicular to its axis is \(\tfrac32\) of the cone with the same base and the same axis.

Exercise 7.1Verify Proposition 18 using the polar subtangent formula: for a curve \(r(\theta)\), the tangent meets the perpendicular to the radius through the origin at distance \(r^2/r'(\theta)\).

Show solution
For \(r=a\theta\), \(r'=a\). At \(\theta=2\pi\), \(r=2\pi a\), so the distance is \((2\pi a)^2/a=2\pi\cdot(2\pi a)=2\pi r\), the circumference of the circle of radius \(r\). Since the spiral can be drawn (mechanically), this gives a straight segment equal to a circle’s circumference.

Exercise 7.2Verify Proposition 21 for the paraboloid \(z=x^2+y^2\) cut by the plane \(z=h\).

Show solution
The horizontal section at height \(z\) is a disc of area \(\pi z\), so the segment has volume \(\int_0^h\pi z\,dz=\tfrac12\pi h^2\). The cone with base radius \(\sqrt h\) and height \(h\) has volume \(\tfrac13\pi h\cdot h=\tfrac13\pi h^2\). The ratio is \(\tfrac32\).
Lesson 08

The lever and the centre of gravity

For the general reader. A small child can balance a large adult on a seesaw if the child sits far enough out. Archimedes turned this everyday fact into a theorem deduced from a few self-evident postulates, and then used it to locate the balance points of triangles and parabolic segments. This is the first mathematical physics we possess.

Goals
  • State the postulates and the law of the lever from On the Equilibrium of Planes.
  • Follow the proof for commensurable weights.
  • Locate centres of gravity of a triangle and a parabolic segment.

Postulates (Book I, selected)

  1. Equal weights at equal distances balance; equal weights at unequal distances do not, but incline toward the weight at the greater distance.
  2. If weights at certain distances balance and something is added to one of them, they no longer balance but incline toward that weight.
  3. In figures that are equal and similar, the centres of gravity are similarly placed.
Equilibrium of Planes I, Propositions 6 and 7

Two magnitudes, whether commensurable (6) or incommensurable (7), balance at distances reciprocally proportional to the magnitudes: weights \(W_1,W_2\) at distances \(d_1,d_2\) from the fulcrum balance if and only if \(W_1:W_2=d_2:d_1\), that is \(W_1d_1=W_2d_2\).

Proof idea for commensurable weights

Let \(W_1=m\,u\) and \(W_2=n\,u\) for a common unit \(u\), and let \(d_1:d_2=n:m\). Replace \(W_1\) by \(m\) unit weights spread evenly and symmetrically about its position, and \(W_2\) by \(n\) unit weights about its position, with all units equally spaced. The spacing can be chosen so that the two blocks are adjacent and the whole row of \(m+n\) equal weights is centred on the fulcrum. By the first postulate and symmetry, the row balances; since each block has its centre of gravity at the original position, the original weights balance. The incommensurable case follows by exhaustion.

Centres of gravity

Book I proves that the centre of gravity of a triangle lies on each median, hence at their intersection, two thirds of the way from each vertex. Book II finds the centre of gravity of a parabolic segment: it lies on the diameter through \(P\) and divides it so that the part toward the vertex is \(\tfrac32\) of the part toward the base.

Exercise 8.1Using the lever law, show that the centroid of a triangle with vertices \(A,B,C\) is \(\tfrac13(A+B+C)\), by placing unit weights at the vertices.

Show solution
Weights at \(B\) and \(C\) combine into weight 2 at the midpoint \(M=\tfrac12(B+C)\). Weight 1 at \(A\) and weight 2 at \(M\) balance at the point dividing \(AM\) in ratio \(2:1\) from \(A\), namely \(A+\tfrac23(M-A)=\tfrac13(A+B+C)\). For a uniform lamina one needs Archimedes’ argument that each median contains the centre of gravity; the result is the same point.

Exercise 8.2Check the parabolic-segment result for \(0\le y\le 1\), \(y\ge x^2\). Where is the centroid on the axis, and in what ratio does it divide the axis from vertex \((0,0)\) to base \((0,1)\)?

Show solution
The width at height \(y\) is \(2\sqrt y\). Area \(=\int_0^1 2\sqrt y\,dy=\tfrac43\); moment \(=\int_0^1 2y^{3/2}\,dy=\tfrac45\). So \(\bar y=\tfrac45\cdot\tfrac34=\tfrac35\). The centroid divides the axis into \(\tfrac35\) (toward the vertex) and \(\tfrac25\) (toward the base), ratio \(3:2\).

Exercise 8.3A compound pulley with \(n\) supporting rope segments lets a force \(F\) lift a load \(nF\). Plutarch’s ship might have weighed 50 tonnes on rollers with rolling resistance 2% of its weight. If Archimedes could pull with 400 N, how many supporting segments would he need, and how far would he pull the rope to move the ship 1 m?

Show solution
Resistance \(\approx0.02\cdot5\times10^4\cdot9.8\approx9800\) N, so \(n\ge 9800/400\approx25\) segments. By conservation of work the rope moves \(n\) times as far as the load: about 25 m of rope per metre of ship. Possible in principle, slow in practice.
Lesson 09

Floating bodies

For the general reader. Why does a ship float and a stone sink? Archimedes answered with a single postulate about how fluids push, from which he derived the principle that carries his name. He then asked a much harder question, which shapes float upright and which tip over, and answered it for a whole family of solids.

Goals
  • State the postulate and the buoyancy propositions of Book I.
  • Use two weighings to determine density.
  • Understand the stability problem of Book II.

The postulate

Archimedes assumes that a fluid is such that, of its parts lying evenly and continuously, the part that is pushed less is driven along by the part pushed more, and that each part is pushed by the fluid directly above it. From this he proves (Proposition 2) that the surface of a fluid at rest is a sphere with the same centre as the Earth.

On Floating Bodies I, Propositions 5–7

(5) A solid lighter than a fluid, placed in it, sinks until the weight of fluid displaced by the immersed part equals the weight of the whole solid. (6) If such a solid is forced under the surface, it is driven upward with a force equal to the difference between the weight of the fluid it displaces and its own weight. (7) A solid heavier than a fluid sinks to the bottom, and when weighed in the fluid it is lighter than its true weight by the weight of the fluid displaced.

Together these are Archimedes’ principle: a body in a fluid experiences an upward force equal to the weight of fluid displaced. For a body of density \(\rho\) weighed in air (\(W\)) and in water of density \(\rho_w\) (apparent weight \(W'\)),

\[ \frac{\rho}{\rho_w}=\frac{W}{W-W'}. \]

Stability (Book II)

Book II studies a right segment of a paraboloid of revolution floating in a denser fluid. Archimedes finds conditions on the ratio of the axis to the parameter of the parabola, and on the density, under which the segment returns to upright when tilted, and the positions of equilibrium when it does not. The analysis compares the centre of gravity of the solid with that of the displaced fluid, the idea behind the metacentre of naval architecture two thousand years later. The only Greek text of On Floating Bodies is in the palimpsest (Lesson 13).

Exercise 9.1Ice has density 0.917 g/cm³ and seawater 1.025 g/cm³. What fraction of an iceberg is submerged?

Show solution
By Proposition 5 the displaced seawater weighs as much as the whole iceberg: \(\rho_wV_{\text{sub}}=\rho V\), so \(V_{\text{sub}}/V=0.917/1.025\approx0.895\). About 89% lies below the surface.

Exercise 9.2A crown weighs 10.00 N in air and 9.40 N in water. Find its density and say whether it can be pure gold (19.3 g/cm³).

Show solution
\(\rho/\rho_w=10.00/0.60\approx16.7\), so \(\rho\approx16.7\) g/cm³. Pure gold would lose only \(10/19.3\approx0.52\) N. The crown is adulterated.
Lesson 10

The Method

For the general reader. Greek geometry presents finished proofs and hides how results were found. In one letter to Eratosthenes, Archimedes explained his secret: he imagined slicing figures into thin sections and hanging the slices on a balance. The method was lost for a thousand years and recovered only in 1906.

Goals
  • Understand the logic of the mechanical method and Archimedes’ view of its status.
  • Derive the volume of the sphere by balancing slices.
  • Apply the method to a paraboloid.

Discovery versus proof

Archimedes tells Eratosthenes that some results first became clear to him by mechanics and were afterwards proved by geometry, because investigation by the method did not furnish an actual demonstration. Having found a result mechanically, he writes, it is easier to supply the proof. He also notes that Democritus stated the cone and pyramid volumes without proof, before Eudoxus proved them. The method treats a plane figure as composed of its parallel line sections and a solid as composed of its plane sections. Archimedes knew this was not rigorous.

The sphere (Method, Proposition 2)

Place a sphere of radius \(r\) with a diameter along the \(x\)-axis from \(0\) to \(2r\). Add the cone with vertex at \(0\) and base of radius \(2r\) at \(x=2r\), and the cylinder of radius \(2r\) on the same axis from \(0\) to \(2r\). At abscissa \(x\) the three sections have areas

\[ \text{sphere: }\pi(2rx-x^2),\qquad \text{cone: }\pi x^2,\qquad \text{cylinder: }4\pi r^2. \]

Sphere and cone together give \(2\pi rx\). Hang these two sections at distance \(2r\) on the other side of a fulcrum at the origin; leave the cylinder’s section where it is, at distance \(x\). Then

\[ 2r\cdot 2\pi r x = x\cdot 4\pi r^2, \]

so every pair of slices balances. Hence the whole sphere plus cone, hung at \(2r\), balances the cylinder, whose centre of gravity is at \(r\):

\[ 2r\,(V_{\text{sph}}+V_{\text{cone}})=r\,V_{\text{cyl}},\qquad V_{\text{sph}}=\tfrac12\cdot8\pi r^3-\tfrac83\pi r^3=\tfrac43\pi r^3. \]

Archimedes adds that this result suggested to him the surface theorem: as a circle equals a triangle with base the circumference and height the radius, so a sphere should equal a cone with base the surface and height the radius.

New theorems

The preface announces two results not found elsewhere: the volume cut from a cylinder by a plane through a diameter of its base (a “hoof”) is \(\tfrac16\) of the circumscribed prism, and the solid common to two cylinders of equal radius with perpendicular axes is \(\tfrac23\) of the circumscribed cube. Both are curved solids measured by rectilinear ones, without \(\pi\).

Exercise 10.1Use the method to find the volume of the paraboloid segment \(\{x^2+y^2\le z\le h\}\) by balancing it against the cylinder of radius \(\sqrt h\) and height \(h\).

Show solution
Fulcrum at the vertex, axis along \(z\). The paraboloid slice at height \(z\) has area \(\pi z\); hang it at distance \(h\). The cylinder slice has area \(\pi h\) and stays at distance \(z\). Moments: \(h\cdot\pi z=z\cdot\pi h\). So \(h\,V_{\text{par}}=\tfrac h2V_{\text{cyl}}\), giving \(V_{\text{par}}=\tfrac12\pi h^2\), in agreement with Lesson 07.

Exercise 10.2Verify the two-cylinder theorem: the solid \(x^2+y^2\le r^2,\ x^2+z^2\le r^2\) has volume \(\tfrac{16}{3}r^3\). (Hint: slice perpendicular to \(x\).)

Show solution
At abscissa \(x\) the section is a square of side \(2\sqrt{r^2-x^2}\), area \(4(r^2-x^2)\). The volume is \(\int_{-r}^{r}4(r^2-x^2)\,dx=\tfrac{16}{3}r^3\), which is \(\tfrac23\) of the cube \((2r)^3=8r^3\). Square slices are why no \(\pi\) appears: the inscribed sphere has circular slices of area \(\pi(r^2-x^2)\), in ratio \(\pi:4\).
Lesson 11

Very large numbers

For the general reader. Some people, Archimedes wrote to Gelon, think the number of grains of sand is infinite. He set out to show that a number larger than the grains needed to fill the whole universe could be named. Along the way he recorded the heliocentric hypothesis of Aristarchus and made a careful measurement of the Sun’s apparent size.

Goals
  • Understand the system of orders and periods in The Sand Reckoner.
  • Reproduce the estimate of \(10^{63}\) grains.
  • Meet the Cattle Problem and its Pell equation.

Orders and periods

Greek counting went up to a myriad, \(10^4\), and a myriad myriads, \(10^8\). Archimedes called numbers up to \(10^8\) the first order, took \(10^8\) as the unit of the second order (up to \(10^{16}\)), and so on up to the \(10^8\)-th order, ending at \(P=(10^8)^{10^8}\). All these form the first period. Then \(P\) is the unit of the second period, and the construction repeats for \(10^8\) periods. The last number is

\[ P^{10^8}=10^{8\cdot10^{16}}. \]
The law of exponents (Sand Reckoner)

If numbers are in continued proportion from 1, as \(1,A_1,A_2,\dots\) with \(A_n=A^n\), then the product of the \(m\)-th and \(n\)-th terms is the \((m+n)\)-th term: \(A^mA^n=A^{m+n}\).

The estimate

Archimedes assumed generously large values: the Earth’s circumference at most 3 million stadia, the Sun at most 30 times the Moon, a universe (the sphere of the fixed stars under Aristarchus’ hypothesis) of diameter less than \(10^{14}\) stadia. A poppy seed holds at most \(10^4\) grains, 40 poppy seeds side by side span a finger-breadth, and a stadium is less than \(10^4\) finger-breadths. He measured the Sun’s apparent diameter with a rod and disc at sunrise and bounded it between \(\tfrac1{200}\) and \(\tfrac1{164}\) of a right angle, correcting for the size of the eye. The result: fewer than \(10^{63}\) grains, which he expressed as a thousand myriad units of the eighth order.

The Cattle Problem

A Greek epigram of 44 lines, found by Gotthold Ephraim Lessing in the library at Wolfenbüttel and published in 1773, presents itself as a problem Archimedes sent to Eratosthenes. It asks for the numbers of bulls and cows of four colours in the herd of the Sun god, subject to seven linear conditions. The smallest solution of this part has 50,389,082 cattle. Two further conditions, that white and black bulls together form a square number and yellow and dappled bulls a triangular number, lead to the Pell equation

\[ x^2-410\,286\,423\,278\,424\,y^2=1 \qquad(410\,286\,423\,278\,424=4\,729\,494\cdot9314^2). \]

A. Amthor showed in 1880 that the smallest herd then has about \(7.76\times10^{206544}\) animals. All 206,545 digits were first computed in 1965 by H. C. Williams, R. A. German and C. R. Zarnke. Whether Archimedes himself composed the problem is debated; whoever did knew that it was far beyond computation.

Exercise 11.1In which order of the first period does \(10^{63}\) lie, and why is it “a thousand myriad units of the eighth order”?

Show solution
The \(k\)-th order runs from \(10^{8(k-1)}\) up to \(10^{8k}\). Since \(56\le63<64\), \(10^{63}\) is in the eighth order, whose unit is \(10^{56}\). Then \(10^{63}=10^7\cdot10^{56}\), and \(10^7=1000\cdot10^4\) is a thousand myriads.

Exercise 11.2Show that \(N\) is triangular, \(N=\tfrac{k(k+1)}2\), if and only if \(8N+1\) is a perfect square. Explain why combining this with a square condition leads to a Pell-type equation.

Show solution
\(8\cdot\tfrac{k(k+1)}2+1=4k^2+4k+1=(2k+1)^2\). Conversely, if \(8N+1=s^2\) then \(s\) is odd, \(s=2k+1\), and \(N=\tfrac{k(k+1)}2\). The linear conditions make both bull counts fixed multiples of a parameter \(t\); requiring \(at=u^2\) and \(8bt+1=s^2\) eliminates \(t\) to give \(s^2-Du^2=1\) for a fixed \(D\).
Part III · Life and legacy
Lesson 12

The siege and the death

Hieron II died in 215 BC, in the middle of the Second Punic War. His grandson and successor Hieronymus abandoned the Roman alliance for Carthage and was soon murdered, and Syracuse fell under leaders hostile to Rome. In 214 BC a Roman army and fleet under Marcus Claudius Marcellus attacked the city.

The engines

Polybius, Livy and Plutarch agree that Archimedes’ machines stopped the assault. Catapults of graded ranges struck ships at every distance. Cranes dropped heavy stones through the decks of ships that came close to the walls. The best-attested device, called the claw, gripped a ship’s prow with an iron hook, lifted it by a counterweighted beam and let it fall, capsizing or swamping it. The Romans gave up the direct attack and settled into a blockade.

The story that Archimedes set the fleet on fire with mirrors is late. Lucian, in the second century AD, says Archimedes burned the ships but mentions no mirrors. Galen, in the same century, is the first to mention burning mirrors. Anthemius of Tralles (sixth century) discussed how such mirrors might be built, and Tzetzes and Zonaras (twelfth century) tell the full story. Polybius and Livy say nothing of them. Descartes rejected the story, and modern trials have not shown that it would work against moving ships at the distances involved.

The death

In 212 BC the Romans entered the city, Livy says during a festival when the guard was lax. Archimedes was killed in the sack. The sources differ on how.

  • Livy: Archimedes, intent on figures he had drawn in the dust, was killed by a soldier who did not know who he was. Marcellus grieved and saw to his burial.
  • Plutarch gives three versions: a soldier ordered him to come to Marcellus, he refused until he had finished his problem, and the soldier killed him; or a soldier came to kill him outright; or soldiers thought the instruments he was carrying to Marcellus were gold.
  • Valerius Maximus (1st century AD) has him beg the soldier not to disturb his drawing. The familiar Latin phrase about his circles is a later polish of this; it is not found in the ancient sources.

All versions agree that Marcellus had ordered him spared.

The tomb

Plutarch says Archimedes asked his friends to place on his tomb a cylinder enclosing a sphere, with the ratio of the containing solid to the contained. In 75 BC Cicero, serving as quaestor in Sicily, searched for it near the Agrigentine gate. The Syracusans denied it existed. Cicero found a small column overgrown with brambles, carrying the sphere and cylinder, and had it cleared. He tells the story in the Tusculan Disputations with some pride that a Roman had rescued the monument of the city’s greatest mind. The site is now lost.

Exercise 12.1A claw lifts the prow of a 40 m ship by raising one end with a beam pivoted on the wall. If the effective lifting arm of the beam outside the wall is 15 m and the counterweight arm inside is 5 m, what counterweight balances a 30 kN lift at the prow? Comment on the role of Lesson 08.

Show solution
By the lever law \(W\cdot5=30\cdot15\), so \(W=90\) kN, about 9 tonnes. A counterweight of this size on a short arm can raise a ship’s bow, and releasing it drops the ship. The numbers are illustrative; the sources give no dimensions.
Lesson 13

Manuscripts, palimpsest, legacy

A thin line of transmission

Archimedes was admired in antiquity but little read. Heron, Pappus and Theon quoted him; his treatises were hard, and only a few circulated. In the sixth century AD Eutocius of Ascalon wrote commentaries on On the Sphere and Cylinder, Measurement of a Circle and On the Equilibrium of Planes, and Isidore of Miletus, an architect of Hagia Sophia, assembled the works in Constantinople. Three Byzantine codices, known to scholars as A, B and C, carried them into the Middle Ages. Arabic scholars, among them Thābit ibn Qurra, translated several works and preserved others now known only in Arabic, such as the Book of Lemmas with its arbelos theorems and a construction of the regular heptagon attributed to Archimedes.

Codices A and B passed to the West. In 1269 the Flemish Dominican William of Moerbeke translated them into Latin, including On Floating Bodies from B. Both Greek manuscripts were later lost, though copies of A survive. The first printed Greek edition appeared in Basel in 1544, and Federico Commandino’s Latin translations followed. Galileo, Kepler, Torricelli, Cavalieri and their contemporaries learned to measure curved figures from these editions; the infinitesimal methods of the seventeenth century grew from reading Archimedes.

The palimpsest

Codex C was copied in Constantinople around 950. On 13 April 1229 a priest in or near Jerusalem scraped its pages, folded them, and wrote a prayer book over them. The volume stayed in the Mar Saba monastery and later in Constantinople. In 1906 the Danish philologist Johan Ludvig Heiberg, alerted by an 1899 catalogue entry, examined it with a magnifying glass and photographs and found under the prayers seven Archimedean treatises: the only Greek text of On Floating Bodies, part of the Stomachion, and the otherwise unknown Method. He published them in his second edition (1910–1915).

The manuscript vanished in the 1920s, was damaged by mould, and had forged illuminations painted over four pages. It reappeared in 1998, when it was sold at auction in New York for about two million dollars to an anonymous buyer, who lent it to the Walters Art Museum in Baltimore. From 1999 to 2008 a team led by William Noel used multispectral imaging and X-ray fluorescence to read nearly all the hidden text. Reviel Netz and Nigel Wilson produced a new transcription. The Stomachion turned out to concern, in Netz’s reading, the number of ways fourteen pieces can be arranged into a square, possibly the earliest combinatorics in Greek mathematics.

Legacy

  • Mathematics: exhaustion arguments that anticipate the Riemann integral; the first rigorous computations of areas, volumes and surfaces of curved figures; rational bounds for \(\pi\); a positional-like system for large numbers.
  • Physics: the founding texts of statics and hydrostatics, with physical laws deduced from postulates.
  • Method: The Method is the earliest surviving text in which a mathematician explains how discovery differs from proof.
  • Honours: the Fields Medal bears his portrait and a Latin line urging one to rise above oneself and grasp the world. A lunar crater bears his name. The first screw-propelled steamship, the SS Archimedes (1839), was named after him.

Final projectChoose one result from Part II. Write two pages for a strong undergraduate presenting Archimedes’ proof in his own terms (ratios, exhaustion, no \(\pi\)) alongside the modern calculus derivation, and explain what the ancient proof needs that the modern one takes for granted.

Reference

Connections

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

Influenced by

Influenced

Colleagues and collaborators

Friends

  • Conon of Samos

Family

  • Phidias

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography and transmission

  1. J. J. O’Connor and E. F. Robertson, Archimedes of Syracuse, MacTutor History of Mathematics.
  2. Archimedes, Wikipedia (with references to the ancient sources).
  3. Archimedes Palimpsest, Wikipedia.
  4. Plutarch, Life of Marcellus 14–19; Livy, History of Rome 24.34 and 25.31; Polybius, Histories, Book 8; Cicero, Tusculan Disputations 5.64–66; Vitruvius, De architectura 9, preface.
  5. E. J. Dijksterhuis, Archimedes, Munksgaard, 1956; reprinted Princeton University Press, 1987.
  6. R. Netz and W. Noel, The Archimedes Codex, Weidenfeld & Nicolson, 2007.
  7. M. Clagett, Archimedes in the Middle Ages, 5 vols., 1964–1984.

Texts and mathematics

  1. J. L. Heiberg (ed.), Archimedis opera omnia cum commentariis Eutocii, 2nd ed., Teubner, 1910–1915.
  2. T. L. Heath, The Works of Archimedes, Cambridge, 1897; supplement The Method of Archimedes, 1912.
  3. R. Netz, The Works of Archimedes, vol. 1: The Two Books On the Sphere and the Cylinder, Cambridge University Press, 2004.
  4. The Sand Reckoner and Archimedes’s cattle problem, Wikipedia.
  5. H. W. Lenstra Jr., Solving the Pell equation, Notices of the AMS 49 (2002).
  6. H. C. Williams, R. A. German and C. R. Zarnke, Solution of the cattle problem of Archimedes, Mathematics of Computation 19 (1965).
  7. S. Stein, Archimedes: What Did He Do Besides Cry Eureka?, Mathematical Association of America, 1999.