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Claudius Ptolemy

Klaudios Ptolemaios (Κλαύδιος Πτολεμαῖος) · Claudius Ptolemaeus

A course in two strands: the little that is known of an Alexandrian scholar of the second century, and the mathematical astronomy, trigonometry, cartography and optics with which he organized the sky and the known world for fourteen hundred years.

Born c. AD 100, Egypt (place unknown) Died c. AD 160s–170s, probably Alexandria Book Mathematical Syntaxis (the Almagest), c. AD 150 Level advanced undergraduate
Ptolemy’s model for Mars, computed with his parameters (deferent 60, eccentricity 6, epicycle 39;30). The epicycle’s centre moves uniformly as seen from the equant Q. The traced path over 780 days shows one retrograde loop.
Lesson 00

How to use this course

Claudius Ptolemy wrote the textbook of mathematical astronomy that was used, copied, translated and argued with from the second century to the seventeenth. The Almagest is the only complete treatise of Greek astronomy to survive. It contains the oldest surviving table of trigonometric values, the oldest surviving star catalogue, and models for the Sun, Moon and planets that predicted positions to within a few tenths of a degree for centuries. Ptolemy also wrote the Geography, the founding text of mapmaking with coordinates, and works on optics, music theory and astrology.

About the man himself almost nothing is known. This course keeps the two kinds of evidence apart. Part I says what the sources are and how far each can be trusted. Part II follows the Almagest and the Geography lesson by lesson, in modern notation, with exercises that use Ptolemy’s own numbers.

Prerequisites for Part II
  • Plane geometry and trigonometry; the inscribed-angle theorem.
  • Vectors in three dimensions, for spherical astronomy.
  • Complex numbers in polar form, \(re^{i\theta}\), used to compose circular motions.
  • Some familiarity with the night sky: the ecliptic, the equinoxes, the phases of the Moon.

Ptolemy computed in base 60. A number such as 2;30 means \(2+\tfrac{30}{60}=2.5\), and 1;2,50 means \(1+\tfrac{2}{60}+\tfrac{50}{3600}\). The course uses this notation when quoting his values. The companion course on Nicolaus Copernicus begins where this one ends, with the Ptolemaic inheritance as a sixteenth-century reader received it; the course on Aristarchus of Samos treats one of the earlier astronomers whose work Ptolemy cited.

Reference

Timeline

Ptolemy’s own dates are inferred from the observations he reports. Dates for earlier astronomers are those the Almagest gives or modern estimates.

747 BC26 February: epoch of the era of Nabonassar, the reference date from which the Almagest counts time.
721–720 BCBabylonian lunar eclipses, the oldest observations Ptolemy uses.
280 BCSummer solstice observed by Aristarchus of Samos, cited in Almagest III.
c. 147–127 BCObservations of Hipparchus, on whose solar and lunar theory and star work Ptolemy builds.
AD 98Menelaus of Alexandria observes lunar occultations of stars in Rome; Ptolemy uses them to test precession.
c. AD 100Ptolemy born in Roman Egypt. Place and date unknown.
AD 12726 March: earliest of Ptolemy’s own dated observations, at Alexandria.
AD 1412 February: latest dated observation in the Almagest.
AD 146/147The Canobic Inscription, a list of astronomical parameters, is dated to the tenth year of Antoninus Pius.
c. AD 150The Almagest completed.
after 150Handy Tables, Planetary Hypotheses, Tetrabiblos, Geography, Optics, Harmonics; order partly uncertain, with the Geography later than the Almagest.
c. 160s–170sDies, probably at Alexandria. No ancient source gives the year.
4th c.Commentaries by Pappus (fragments survive) and Theon of Alexandria, who also edits the Handy Tables.
9th c.Arabic translations of the Almagest in Baghdad under the Abbasid caliphs.
c. 1154Eugenius of Palermo translates the Optics into Latin from Arabic.
12th c.Two Latin translations of the Almagest: one from Greek in Sicily, one from Arabic by Gerard of Cremona at Toledo (finished 1175).
1406–07Jacopo Angeli translates the Geography into Latin.
1477First printed Geography with maps, Bologna.
1496The Epitome of the Almagest by Peurbach and Regiomontanus printed at Venice.
1543Copernicus’s De revolutionibus, modelled book by book on the Almagest.
1898–1903J. L. Heiberg’s critical edition of the Greek text.
1977Robert R. Newton’s The Crime of Claudius Ptolemy accuses him of fabricating observations.
2022Fragments of Hipparchus’s star catalogue are read in a palimpsest, showing that Ptolemy did not simply copy it.
Part I · Formation
Lesson 01

Alexandria and the problem of sources

No ancient biography of Ptolemy survives, and probably none was written. What can be said rests on three kinds of evidence: his name, the dates of his observations, and remarks by much later authors.

The name

“Ptolemaios” was a Greek name common in Egypt, where the Ptolemaic dynasty had ruled from 323 to 30 BC. It does not imply royal descent. The ninth-century astrologer Abu Ma’shar made him a member of the royal family, an error repeated in medieval portraits that show him crowned. “Claudius” is a Roman family name. It indicates that Ptolemy was a Roman citizen, probably because an ancestor received citizenship under the emperor Claudius or Nero, as Gerald Toomer suggested. He wrote in Greek, and he was either a Greek or a Hellenized Egyptian living in the Roman province of Egypt.

The dates

The Almagest reports observations that Ptolemy says he made himself, dated precisely by the Egyptian calendar. The earliest is from 26 March AD 127 and the latest from 2 February AD 141. All were made at Alexandria, and no evidence places him anywhere else. The Canobic Inscription, a list of parameters said to have been set up at Canopus near Alexandria, is dated to the tenth year of Antoninus Pius, AD 146/147. Its values are earlier versions of those in the Almagest, so the book was finished around AD 150. Since several major works followed it, he probably lived into the 160s or 170s. A late scholiast says he flourished under Hadrian and lived into the reign of Marcus Aurelius (161–180).

Later reports

  • Olympiodorus (6th century) says Ptolemy worked for forty years in a part of the temple at Canopus. Toomer judged this probably a fictional elaboration of the Canobic Inscription.
  • Theodore Meliteniotes (c. 1360) says he was born at Ptolemais Hermiou in Upper Egypt. The claim appears more than a thousand years after Ptolemy’s death and has no earlier support.
  • Arabic biographical works give a physical description and anecdotes. They are late and generally discounted.

We do not know his family, whether he taught, who paid for his work, or exactly when he died.

Exercise 1.1Rank these claims by reliability and name the evidence for each: (a) Ptolemy observed at Alexandria in the 130s; (b) he was born in Ptolemais Hermiou; (c) he was a Roman citizen; (d) he was a king of Egypt.

Show solution
(a) is the most secure: his own dated observations in the Almagest. (c) is a strong inference from the name Claudius, which only a citizen would bear. (b) rests on a single fourteenth-century writer and is unsupported. (d) is a medieval confusion with the Ptolemaic dynasty and is false; the last Ptolemaic ruler died in 30 BC.
Lesson 02

Eight centuries of observations

Ptolemy did not start from nothing. Alexandria had been a centre of Greek science since the third century BC, and its library and scholarly community gave him access to a long record of observations and theories. The Almagest is our main source for much of this earlier work, because most of it is otherwise lost.

Babylonian records

Ptolemy counts time in Egyptian years of 365 days from the accession of the Babylonian king Nabonassar, 26 February 747 BC. For his lunar theory he uses Babylonian eclipse reports from 721 and 720 BC, which reached Greek astronomers through Hipparchus. These long baselines let small errors in mean motions accumulate to measurable size, which is why ancient period values were so accurate.

Hipparchus

Hipparchus of Nicaea observed at Rhodes and elsewhere between about 147 and 127 BC. Only one minor work of his survives. Through Ptolemy we know that he built the eccentric model of the Sun, an epicycle model of the Moon, a method for finding the Moon’s distance from parallax, a table of chords, and a catalogue of star positions; and that he discovered the precession of the equinoxes. Ptolemy treats him with great respect, calls him a lover of truth, and adopts many of his parameters unchanged. Hipparchus had not produced satisfactory planetary models; Ptolemy says he collected observations and showed that existing hypotheses did not fit them.

Other predecessors

  • Apollonius of Perga (c. 200 BC) proved the equivalence of eccentric and epicyclic models and found the condition for a planet’s stationary points (Lesson 08).
  • Menelaus of Alexandria (c. AD 100) developed spherical trigonometry in his Sphaerica; Ptolemy’s spherical astronomy rests on his theorem (Lesson 05).
  • Theon “the mathematician”, whose planetary observations from the early second century Ptolemy uses, is probably Theon of Smyrna and may have been his teacher. Most of Ptolemy’s early works are dedicated to one Syrus, otherwise unknown.

Exercise 2.1Ptolemy measured the length of the tropical year by comparing his own equinox observations with those of Hipparchus about 285 years earlier. Suppose each equinox time is uncertain by 6 hours. What is the resulting uncertainty in the year length, in minutes?

Show solution
The error in the interval is at most 12 hours, spread over 285 years: \(720/285\approx2.5\) minutes per year. Ptolemy’s value, \(365\tfrac14-\tfrac1{300}\) days, is about 6 minutes too long. So his error is larger than this estimate allows, which is one of the puzzles behind the charge that he adjusted his equinox observations (Lesson 12).

Exercise 2.2How many days separate the epoch of Nabonassar from the same calendar point 868 Egyptian years later, roughly when Ptolemy wrote? Why was the Egyptian year convenient for astronomers?

Show solution
\(868\times365=316\,820\) days. The Egyptian year had no leap days, so every year had exactly 365 days and intervals could be computed by simple multiplication. Its drift against the seasons (one day every four years) did not matter for computation.
Lesson 03

The plan of the Almagest

Ptolemy called his work the Mathematical Syntaxis (compilation or treatise). Later Greek readers called it the Great Syntaxis; the superlative form became al-majisṭī in Arabic and Almagest in Latin. Its thirteen books proceed in strict logical order: each part uses only what has been established before.

  • I–II: the arrangement of the cosmos, the table of chords, spherical astronomy and its applications to rising times, day lengths and shadows.
  • III: the length of the year and the theory of the Sun.
  • IV–V: the theory of the Moon; parallax; the distances of the Sun and Moon.
  • VI: eclipses.
  • VII–VIII: the fixed stars, precession, and the star catalogue.
  • IX–XIII: the planets: longitude models, stations and retrogradations, latitudes, and visibility.

Philosophy and method

In the preface Ptolemy follows Aristotle in dividing theoretical philosophy into theology, mathematics and physics. He argues that only mathematics gives firm knowledge, since theology deals with something invisible and physics with matter that is unstable and unclear. Astronomy, as the mathematics of eternal and unchanging things, is therefore the most certain of sciences. A short epigram attached to the work in the manuscripts, accepted by many scholars as his, expresses the feeling of the mortal astronomer raised up by contemplating the stars.

Book I also argues that the heavens are spherical and move as a sphere, that the Earth is spherical, lies at the centre and is a point compared with the heavens, and does not move. Against a rotating Earth he raises the objections discussed in Lesson 08 of the Copernicus course: objects not attached to the Earth would be left behind, and clouds and thrown bodies would drift westward. He grants that a daily rotation of the Earth would simplify the description of the heavens, and rejects it on physical grounds.

The method is hypothetical and quantitative. For each body Ptolemy chooses a model made of uniform circular motions, derives its parameters from a small number of selected observations, and builds tables from which anyone can compute positions for any date. He openly simplifies, and he states that different models may fit the same phenomena.

Exercise 3.1Ptolemy argues that the Earth must be at the centre because, from every place on Earth, the horizon divides the sky into two equal halves: six signs of the zodiac are always above it. Explain the geometry. Does the argument rule out an Earth in orbit around the Sun?

Show solution
If the observer were far from the centre of the celestial sphere, the horizon plane would cut the sphere off-centre, and more than half the zodiac would sometimes be visible and sometimes less. The argument shows only that the Earth’s displacement from the centre is negligible compared with the radius of the stars. A heliocentric astronomer accepts that conclusion, so the argument does not decide between the systems. Copernicus used exactly this point in his fourth postulate.
Part II · The astronomy and mathematics
Lesson 04

The table of chords

For the general reader. To compute where a planet will appear, an astronomer must turn angles into lengths and back. Greek astronomers did this with the chord: the straight line joining the ends of an arc. Ptolemy built a table of chords for every half degree from \(\tfrac12^\circ\) to \(180^\circ\), using a handful of geometric theorems and one clever inequality. It is the oldest trigonometric table that survives.

Goals
  • Relate chords to sines and use Ptolemy’s base-60 values.
  • Prove Ptolemy’s theorem on cyclic quadrilaterals and derive the addition and half-angle rules.
  • Follow Ptolemy’s bound for the chord of \(1^\circ\), and recover his value of \(\pi\).

Chords

Ptolemy takes a circle of diameter 120 parts, radius \(R=60\). The chord of a central angle \(\alpha\) is

\[ \operatorname{crd}\alpha=2R\sin\frac{\alpha}{2}=120\sin\frac{\alpha}{2}. \]

Some chords follow from regular polygons: \(\operatorname{crd}60^\circ=60\) (hexagon), \(\operatorname{crd}90^\circ=60\sqrt2\) (square), \(\operatorname{crd}120^\circ=60\sqrt3\) (triangle). From the construction of the decagon and pentagon in Euclid’s Elements XIII, Ptolemy obtains \(\operatorname{crd}36^\circ=37;4,55\) and \(\operatorname{crd}72^\circ=70;32,3\). Supplementary chords satisfy \(\operatorname{crd}^2\alpha+\operatorname{crd}^2(180^\circ-\alpha)=120^2\), by Thales’ theorem.

Theorem (Ptolemy)

If \(ABCD\) is a quadrilateral inscribed in a circle, then \(AC\cdot BD=AB\cdot CD+AD\cdot BC\).

Proof. Choose \(E\) on the diagonal \(AC\) with \(\angle ABE=\angle DBC\). Angles \(\angle BAE=\angle BAC\) and \(\angle BDC\) stand on the same arc \(BC\), so they are equal, and triangles \(ABE\) and \(DBC\) are similar: \(AE/AB=DC/DB\), so \(AE\cdot BD=AB\cdot CD\). Adding \(\angle EBD\) to both sides of \(\angle ABE=\angle DBC\) gives \(\angle ABD=\angle EBC\); also \(\angle ADB=\angle ACB\) (same arc \(AB\)). So triangles \(ABD\) and \(EBC\) are similar: \(EC/BC=AD/BD\), so \(EC\cdot BD=AD\cdot BC\). Adding, \(AC\cdot BD=AB\cdot CD+AD\cdot BC\).

Applied with one side a diameter, the theorem gives the chord of a difference of arcs, and hence (Lesson 10 of the Copernicus course) the addition formula for sines. Ptolemy also proves the half-angle rule

\[ \operatorname{crd}^2\frac{\alpha}{2}=60\,\big(120-\operatorname{crd}(180^\circ-\alpha)\big), \]

which in modern form is \(\sin^2(\alpha/4)=\tfrac12(1-\cos(\alpha/2))\). From \(\operatorname{crd}72^\circ\) and \(\operatorname{crd}60^\circ\) he gets \(\operatorname{crd}12^\circ\), and by halving, \(\operatorname{crd}6^\circ,3^\circ,1\tfrac12^\circ,\tfrac34^\circ\). But \(1^\circ\) cannot be reached by halving and adding from these.

The chord of one degree

Ptolemy closes the gap with an inequality that Aristarchus had used and Euclid’s methods prove (see the Aristarchus course, Lesson 06):

Lemma (Aristarchus’s inequality, chord form)

If \(0<\beta<\alpha\le180^\circ\), then \(\dfrac{\operatorname{crd}\alpha}{\operatorname{crd}\beta}<\dfrac{\alpha}{\beta}\).

In modern terms, \(\sin x/x\) is decreasing on \((0,\pi/2]\). With \(\operatorname{crd}\tfrac34^\circ=0;47,8\) and \(\operatorname{crd}1\tfrac12^\circ=1;34,15\):

\[ \operatorname{crd}1^\circ<\tfrac43\operatorname{crd}\tfrac34^\circ=1;2,50,40,\qquad \operatorname{crd}1^\circ>\tfrac23\operatorname{crd}1\tfrac12^\circ=1;2,50 . \]

So, to two sexagesimal places, \(\operatorname{crd}1^\circ=1;2,50\), and \(\operatorname{crd}\tfrac12^\circ=0;31,25\). Repeated use of the sum and difference rules then fills the table at half-degree steps. A third column gives one thirtieth of the difference between successive entries, the increment per minute of arc, for linear interpolation.

The value for \(1^\circ\) also fixes \(\pi\). A polygon of 360 sides each \(1;2,50\) has perimeter \(360\times1;2,50=377;0\) in units where the diameter is 120, so Ptolemy states \(\pi\approx3;8,30=3\tfrac{17}{120}\approx3.14167\).

Exercise 4.1Prove the chord form of Aristarchus’s inequality from the fact that \(\sin x/x\) is decreasing on \((0,\pi/2]\). Then check Ptolemy’s two bounds for \(\operatorname{crd}1^\circ\) numerically, and compare with the true value \(120\sin\tfrac12^\circ\).

Show solution
\(\operatorname{crd}\alpha/\alpha=120\sin(\alpha/2)/\alpha\) is a decreasing function of \(\alpha\), so for \(\beta<\alpha\), \(\operatorname{crd}\alpha/\alpha<\operatorname{crd}\beta/\beta\). Numerically \(\tfrac23\times1.57083=1.04722\) and \(\tfrac43\times0.78556=1.04741\). The true value is \(1.047184\), which is \(1;2,49,51\): Ptolemy’s 1;2,50 is correct in both places he keeps.

Exercise 4.2Use Ptolemy’s theorem with \(AD\) a diameter of length 120, \(B\) and \(C\) on the same semicircle, to show that \(\operatorname{crd}(\alpha-\beta)\cdot120=\operatorname{crd}\alpha\,\operatorname{crd}(180^\circ-\beta)-\operatorname{crd}\beta\,\operatorname{crd}(180^\circ-\alpha)\), where \(\alpha=\) arc \(AC\), \(\beta=\) arc \(AB\). Use it to compute \(\operatorname{crd}12^\circ\).

Show solution
In \(ABCD\) (in order around the circle) the sides are \(AB=\operatorname{crd}\beta\), \(BC=\operatorname{crd}(\alpha-\beta)\), \(CD=\operatorname{crd}(180^\circ-\alpha)\), \(DA=120\), and the diagonals \(AC=\operatorname{crd}\alpha\), \(BD=\operatorname{crd}(180^\circ-\beta)\). Ptolemy’s theorem \(AC\cdot BD=AB\cdot CD+BC\cdot DA\) rearranges to the stated rule. With \(\alpha=72^\circ,\beta=60^\circ\): \(\operatorname{crd}108^\circ=\sqrt{14400-70.534^2}=97.082\) and \(\operatorname{crd}120^\circ=103.923\), so \(\operatorname{crd}12^\circ=(70.534\times103.923-60\times97.082)/120=12.543\), i.e. 12;32,36. The true value \(120\sin6^\circ=12.5434\).

Exercise 4.3Convert Ptolemy’s \(\pi=3;8,30\) to a fraction and a decimal, and compare its error with Archimedes’ bounds \(3\tfrac{10}{71}<\pi<3\tfrac17\).

Show solution
\(3+\tfrac{8}{60}+\tfrac{30}{3600}=3+\tfrac{17}{120}=\tfrac{377}{120}=3.141\overline{6}\). The error is \(+7.4\times10^{-5}\). Archimedes’ bounds are \(3.14085\) and \(3.14286\), so Ptolemy’s value lies between them and is much closer to \(\pi\) than either.
Lesson 05

The sphere: ecliptic, equator, climata

For the general reader. The Sun moves along a great circle of the sky, the ecliptic, tilted to the celestial equator. Whether it is high or low at noon, how long the day lasts, and when a star rises all depend on that tilt and on the observer’s latitude. Ptolemy computed all these with Menelaus’s theorem on spherical triangles, and used the length of the longest day to divide the inhabited world into zones.

Goals
  • Know Ptolemy’s value of the obliquity and how he measured it.
  • Derive declination and right ascension from ecliptic longitude.
  • Compute the length of the longest day at a given latitude, and reconstruct Ptolemy’s climata.

The obliquity

Ptolemy measured the noon altitude of the Sun at the two solstices with a meridian instrument. He reports that the arc between the solstices always lies between \(47;40^\circ\) and \(47;45^\circ\), close to the ratio \(11/83\) of a full circle that Eratosthenes found and Hipparchus used. Half of \(11/83\) of \(360^\circ\) is \(23;51,20^\circ\), the value he adopts for the obliquity \(\varepsilon\). The true value in his time was about \(23;41^\circ\).

Menelaus’s theorem

Theorem (Menelaus, spherical form)

Let two great-circle arcs \(AE\) and \(AD\) be cut by two further great-circle arcs \(BE\) and \(CD\) that cross at \(F\), with \(B\) on \(AD\) and \(C\) on \(AE\). Then

\[ \frac{\sin CE}{\sin CA}=\frac{\sin FE}{\sin FB}\cdot\frac{\sin DB}{\sin DA}. \]

Ptolemy proves it from the plane version (a transversal across a triangle) by projecting from the centre of the sphere. Every rule of spherical astronomy in the Almagest comes from repeated use of this one theorem. Today we use vectors instead. Put the equator in the \(xy\)-plane and the vernal equinox on the \(x\)-axis. The Sun at ecliptic longitude \(\lambda\) has direction

\[ \mathbf s=(\cos\lambda,\ \sin\lambda\cos\varepsilon,\ \sin\lambda\sin\varepsilon), \]

from which

\[ \sin\delta=\sin\varepsilon\sin\lambda,\qquad \tan\alpha=\cos\varepsilon\tan\lambda, \]

for the declination \(\delta\) and right ascension \(\alpha\). Ptolemy tabulates both.

Day length and climata

Proposition (half-day arc)

At latitude \(\varphi\), a body of declination \(\delta\) with \(|\tan\varphi\tan\delta|<1\) stays above the horizon for the hour angle \(2H\), where

\[ \cos H=-\tan\varphi\tan\delta . \]

The longest day, at the summer solstice, lasts \(2H/15^\circ\) hours with \(\delta=\varepsilon\).

Derivation. The altitude \(h\) satisfies \(\sin h=\sin\varphi\sin\delta+\cos\varphi\cos\delta\cos H\). Setting \(h=0\) gives the formula. Refraction and the size of the Sun’s disc are ignored, as they were by Ptolemy.

In Almagest II Ptolemy lists parallels at which the longest day increases by a quarter hour or half hour, from the equator (12 hours) northward. These zones, the climata, served as a latitude coordinate for geography. The parallel of Rhodes, at \(36^\circ\), has a longest day of \(14\tfrac12\) hours.

Exercise 5.1With \(\varepsilon=23;51,20^\circ\), compute the longest day at Alexandria (\(\varphi=31^\circ\)) and at Rhodes (\(\varphi=36^\circ\)).

Show solution
\(\tan\varepsilon=0.4424\). Alexandria: \(\cos H=-0.6009\times0.4424=-0.2658\), \(H=105.4^\circ\), longest day \(14.05\) hours. Rhodes: \(\cos H=-0.7265\times0.4424=-0.3214\), \(H=108.75^\circ\), longest day \(14.50\) hours, as Ptolemy states.

Exercise 5.2The Geography takes the inhabited world from about \(16;25^\circ\) south to \(63^\circ\) north (Thule). Show that these are the parallels where the longest day is 13 hours and 20 hours.

Show solution
Solve \(\tan\varphi=-\cos H/\tan\varepsilon\). For 13 hours, \(H=97.5^\circ\): \(\tan\varphi=0.1305/0.4424=0.2950\), \(\varphi=16.43^\circ\approx16;26^\circ\). For 20 hours, \(H=150^\circ\): \(\tan\varphi=0.8660/0.4424=1.9576\), \(\varphi=62.94^\circ\approx63^\circ\). Ptolemy defined his limits by day length, not by measured latitude.

Exercise 5.3Find the right ascension of the Sun at longitude \(\lambda=30^\circ\) (the beginning of Taurus). Why do the signs of the zodiac take different times to rise?

Show solution
\(\tan\alpha=\cos23.86^\circ\tan30^\circ=0.9146\times0.5774=0.5280\), so \(\alpha=27.84^\circ\). Equal arcs of the ecliptic correspond to unequal arcs of the equator, and the equator turns uniformly; at a given latitude the horizon’s tilt adds a further inequality. Ptolemy tabulates these rising times for each clima.
Lesson 06

The Sun and the eccentric

For the general reader. The seasons are not equal: from the spring equinox to the summer solstice takes about three days longer than from the autumn equinox to the winter solstice. Hipparchus explained this by placing the Earth slightly off the centre of the Sun’s circle. Ptolemy confirmed his numbers and turned them into tables. The model is simple, and it is a good first example of how a Greek astronomer fitted a circle to observations.

Goals
  • Set up the eccentric solar model and its equation of centre.
  • Derive the eccentricity and apogee from the lengths of two seasons.
  • Understand Ptolemy’s year length and its error.

The model

The Sun moves uniformly, with mean motion \(\bar n=360^\circ/\text{year}\), on a circle of radius 60 whose centre \(C\) is displaced from the Earth \(O\) by \(e\) toward the apogee, at longitude \(\lambda_A\). With mean anomaly \(M\) measured from apogee, the Sun’s position relative to the Earth is \(z=e+60e^{iM}\) (taking the apogee direction as real axis), and its true longitude is \(\lambda=\lambda_A+M-q(M)\), where the equation of centre is

\[ q(M)=\arg\big(1+\tfrac{e}{60}e^{-iM}\big)^{-1},\qquad \tan q=\frac{e\sin M}{60+e\cos M}. \]

Here \(\arg z=M-q\). The maximum of \(q\) is \(\arcsin(e/60)\), reached when the Sun is a quarter circle from apogee as seen from the Earth.

Proposition (eccentricity from two seasons)

Let the Sun take \(t_1\) days from vernal equinox to summer solstice and \(t_2\) days from summer solstice to autumnal equinox, and let the year be \(T\). Put \(\theta_k=360^\circ t_k/T\), and

\[ \delta_1=\tfrac12(\theta_1+\theta_2-180^\circ),\qquad \delta_2=\theta_1-90^\circ-\delta_1 . \]

Then the centre of the eccentric lies at \(\big(\tfrac{e}{60}\cos\lambda_A,\ \tfrac{e}{60}\sin\lambda_A\big)=(\sin\delta_2,\ \sin\delta_1)\) in units of the radius, with the equinox along the \(x\)-axis.

Derivation. The line of equinoxes and the line of solstices pass through \(O\) and are perpendicular. The Sun travels the arc \(\theta_1+\theta_2\) of its own circle while passing from one equinox to the other, so the equinox chord lies at distance \(\sin\delta_1\) from \(C\). Similarly, the arc \(\theta_1\) from equinox to solstice exceeds a quadrant by \(\delta_1+\delta_2\), and the solstice chord lies at distance \(\sin\delta_2\) from \(C\). These two distances are the coordinates of \(C\) relative to \(O\).

Hipparchus took \(t_1=94\tfrac12\) and \(t_2=92\tfrac12\) days. Ptolemy says his own observations confirm them and adopts \(e=2;30\) and \(\lambda_A=65;30^\circ\) (\(5;30^\circ\) of Gemini). His year is \(365\tfrac14-\tfrac1{300}\) days, \(365;14,48\) days, which he also took from Hipparchus and claimed to confirm. It is too long by about 6 minutes. Because the solar theory is the basis for the lunar and planetary theories, this error propagates through the whole Almagest.

Exercise 6.1Using \(t_1=94\tfrac12\), \(t_2=92\tfrac12\), \(T=365\tfrac14\), compute \(e\) (in parts of 60) and \(\lambda_A\). Compare with Ptolemy’s 2;30 and \(65;30^\circ\).

Show solution
\(\theta_1=93.142^\circ\), \(\theta_2=91.170^\circ\), so \(\delta_1=2.156^\circ\), \(\delta_2=0.986^\circ\). Then \(\sin\delta_1=0.03762\), \(\sin\delta_2=0.01720\), so \(e/60=0.04137\), \(e=2.48\approx2;29\), and \(\tan\lambda_A=0.03762/0.01720\) gives \(\lambda_A=65.4^\circ\). Rounding gives Ptolemy’s values.

Exercise 6.2Find the greatest equation of centre in Ptolemy’s solar model. How long are autumn and winter in the model?

Show solution
\(\arcsin(2.5/60)=2.39^\circ\approx2;23^\circ\). Autumn and winter together take \(365.25-187=178.25\) days. Computing the mean anomaly at each equinox and solstice, the model gives about \(88\tfrac18\) days for autumn and \(90\tfrac18\) for winter. Ptolemy quotes only the two seasons he used.

Exercise 6.3Show that the eccentric model is equivalent, in direction, to Kepler’s ellipse to first order in the eccentricity, provided the eccentric’s \(e/60\) equals twice the orbital eccentricity. The Earth’s orbital eccentricity in Ptolemy’s time was about 0.017. Compare with \(2.5/60\).

Show solution
Kepler’s equation of centre is \(2\epsilon\sin M+O(\epsilon^2)\); the eccentric gives \((e/60)\sin M+O(e^2)\). They agree when \(e/60=2\epsilon=0.0348\), i.e. \(e\approx2;5\). Ptolemy’s 2;30 is too large by about 20 percent, a consequence of the season lengths inherited from Hipparchus.
Lesson 07

The Moon, parallax and distances

For the general reader. The Moon is the hardest body to model and the only one whose distance an ancient astronomer could measure directly, because it is close enough to shift against the stars as the observer moves. Ptolemy found a new irregularity in its motion, built a model to capture it, and measured the distances of the Moon and the Sun in Earth radii.

Goals
  • Describe the first lunar anomaly and its epicycle.
  • Understand the second anomaly and Ptolemy’s crank mechanism.
  • Compute the Moon’s distance from parallax and the Sun’s from the eclipse diagram.

Two anomalies

The Moon’s speed varies with a period of about 27.55 days, the anomalistic month. Hipparchus represented this first anomaly with an epicycle. Ptolemy fixes its radius at \(5;15\) parts of 60, which gives a greatest equation of \(\arcsin(5.25/60)=5;1^\circ\). This model works at new and full moon, where eclipses test it.

Ptolemy then found that at the quarters the Moon could depart from the model by up to about \(2;40^\circ\) more. His fix moves the epicycle closer to the Earth at the quarters: the epicycle’s centre rides on an eccentric deferent whose centre revolves around the Earth, so that the deferent’s centre is at distance \(10;19\) from the Earth and the epicycle comes nearest twice a month, at the quarters. The greatest equation then rises to \(7;40^\circ\). He also added a small correction, reckoning the Moon’s motion on the epicycle from a direction aimed at a point opposite the deferent’s centre, on the far side of the Earth. The second anomaly is the inequality now called evection.

Proposition (the flaw in the crank)

In Ptolemy’s lunar model with epicycle radius \(5;15\) and crank \(10;19\), the Moon’s distance from the Earth ranges from \(60+5;15=65;15\) to \(60-2(10;19)-5;15=34;7\) parts. So the Moon’s apparent diameter should vary by a factor of about 1.9.

The Moon’s observed diameter varies by only about 12 percent. Ptolemy’s model gives good longitudes and wrong distances, and he does not discuss the discrepancy. Copernicus made the point a main criticism, and his own lunar model, equivalent to that of Ibn al-Shatir, removed it.

Parallax and the Moon’s distance

An observer on the Earth’s surface sees the Moon displaced from where an observer at the centre would see it. If the Moon is at distance \(D\) Earth radii and at true zenith distance \(z\), the parallax \(p\) satisfies

\[ \sin p=\frac{\sin z'}{D}\approx\frac{\sin z}{D}, \]

where \(z'\) is the apparent zenith distance. Ptolemy measured the Moon’s zenith distance at meridian transit with his parallactic instrument and compared it with the position the model predicts for an observer at the Earth’s centre. He found the Moon’s distance at the time of observation to be about 40 Earth radii and, reducing through the model, a mean distance at new and full moon of 59 Earth radii. The modern mean value is about 60.3.

The Sun from the eclipse diagram

Ptolemy then uses the eclipse method of Aristarchus and Hipparchus, explained in detail in Lesson 08 of the Aristarchus course. With the Moon at 64;10 Earth radii at greatest distance, an apparent radius of the Sun and Moon of \(0;15,40^\circ\), and an Earth’s shadow \(2\tfrac35\) times the Moon’s radius at that distance, he finds the Sun at 1210 Earth radii. The true value is about 23,500. The result is very sensitive to the inputs, because the Sun’s parallax comes out as a small difference of much larger angles.

Exercise 7.1With the eclipse relation \(p_{\text{Moon}}+p_{\text{Sun}}=\rho_{\text{shadow}}+\rho_{\text{Sun}}\) (horizontal parallaxes and apparent radii), take \(D_{\text{Moon}}=64;10\), \(\rho_{\text{Sun}}=0;15,40^\circ\) and \(\rho_{\text{shadow}}=2.6\,\rho_{\text{Sun}}\). Find the Sun’s distance in Earth radii.

Show solution
\(p_{\text{Moon}}=1/64.167\) rad \(=53.58'\). \(\rho_{\text{Sun}}=15.67'\), \(\rho_{\text{shadow}}=40.73'\). So \(p_{\text{Sun}}=40.73'+15.67'-53.58'=2.82'\), and \(D_{\text{Sun}}=3437.7/2.82\approx1220\). Ptolemy, rounding differently, gets 1210. Note that \(p_{\text{Sun}}\) is the small difference of large quantities: a change of \(1'\) in any input changes the answer by a third.

Exercise 7.2Verify that an epicycle of radius 5;15 on a deferent of 60 gives a greatest equation of 5;1°. With the crank at quadrature, where the epicycle’s centre is at \(60-2(10;19)\) from the Earth, what is the greatest equation?

Show solution
\(\arcsin(5.25/60)=5.02^\circ\approx5;1^\circ\). At quadrature the centre distance is \(60-20.633=39.367\), and \(\arcsin(5.25/39.367)=7.66^\circ\approx7;40^\circ\), Ptolemy’s value.
Lesson 08

The planets and the equant

For the general reader. The planets show two irregularities: they move faster in some parts of the zodiac than in others, and they periodically stop and go backward. Ptolemy handled the second with an epicycle and the first with an eccentric circle viewed from a special point, the equant. The equant made his planetary models accurate, and it made them philosophically troubling for Islamic and European astronomers for more than a thousand years.

Goals
  • Write Ptolemy’s outer-planet model in complex form.
  • Derive Apollonius’s condition for stationary points.
  • See why the equant succeeds, and why Copernicus wanted to remove it.

The model for an outer planet

Put the Earth \(O\) at the origin. The deferent, radius 60, has centre \(C\) at distance \(e\) along the apse line. The equant \(Q\) lies at \(2e\) on the same line. The epicycle’s centre \(K\) moves on the deferent so that the angle \(M\) at \(Q\) increases uniformly. The planet moves on the epicycle, radius \(r\), so that its radius vector stays parallel to the direction from the Earth to the mean Sun:

\[ K=2e+\rho(M)\,e^{iM},\quad \rho(M)=-e\cos M+\sqrt{60^2-e^2\sin^2M},\qquad P=K+r\,e^{i\bar\lambda_\odot}. \]

For Mars Ptolemy takes \(e=6\) and \(r=39;30\). The ratio \(r/60\) is the ratio of the Earth’s orbit to Mars’s in the heliocentric picture: the table in Lesson 06 of the Copernicus course lists all five.

Theorem (Apollonius on stationary points)

In the concentric epicycle model with deferent radius \(R\), epicycle radius \(r\), angular speed of the epicycle’s centre \(\omega_1\) and of the planet on the epicycle relative to the line from the Earth to the centre \(\omega_2\), the planet is stationary when it lies on a line \(OST\) through the Earth that cuts the epicycle at \(S\) (near) and \(T\) (far) with

\[ \frac{\tfrac12 ST}{OS}=\frac{\omega_1}{\omega_2}. \]

In heliocentric language, with the Earth on a circle of radius 1 and the planet on a circle of radius \(a\) with angular speeds \(n_E\) and \(n\), the geocentric vector is \(w=ae^{i\theta}-e^{i\theta_E}\), and the planet is stationary when \(\operatorname{Im}(\dot w\,\bar w)=0\). Writing \(\psi=\theta_E-\theta\) for the angle at the Sun,

\[ \operatorname{Im}(\dot w\bar w)=a^2n+n_E-a(n+n_E)\cos\psi=0\quad\Longrightarrow\quad \cos\psi=\frac{a^2n+n_E}{a(n+n_E)} . \]

With Kepler’s third law \(n=n_Ea^{-3/2}\) this simplifies to \(\cos\psi=\sqrt a/(1-\sqrt a+a)\). Ptolemy computed stations geometrically in Almagest XII and tabulated them.

Why the equant works

Lesson 07 of the Copernicus course expands the equant model in powers of \(e/60\). To first order it agrees with motion on a Keplerian ellipse with orbital eccentricity \(e/60\), focus at the Earth, obeying the area law. The bisection of the eccentricity (equant at \(2e\), centre at \(e\)) is exactly what makes both the direction and the distance right to that order. Ptolemy found it empirically, from observations of Mars at opposition and of the size of retrograde arcs. It violated the principle that each celestial motion is uniform about its own centre. Ptolemy acknowledges in Almagest IX that his methods are sometimes contrary to the general principles, and defends them by their agreement with the phenomena.

Exercise 8.1For Mars take \(a=1.524\). Find the heliocentric angle \(\psi\) at the stations, and the number of days from the first station to opposition. Use \(T_E=365.25\) and \(T_M=686.98\) days.

Show solution
\(\sqrt a=1.2345\), \(\cos\psi=1.2345/(1-1.2345+1.524)=0.9573\), so \(\psi=16.8^\circ\). The angle closes at \(360^\circ(1/365.25-1/686.98)=0.4616^\circ\) per day, so the first station falls \(16.8/0.4616\approx36\) days before opposition, and the retrograde arc lasts about 73 days.

Exercise 8.2In Ptolemy’s Mars model (\(e=6\), deferent 60), with the epicycle centre at \(M=90^\circ\) from the apse line as seen from the equant, find the angle between the epicycle centre’s direction seen from the equant and from the Earth.

Show solution
\(\rho(90^\circ)=\sqrt{3600-36}=59.70\), so \(K=12+59.70i\). Seen from the Earth its direction is \(\arctan(59.70/12)=78.63^\circ\); from the equant, \(90^\circ\). The difference, \(11.4^\circ\), is the eccentric part of Mars’s equation. It is about \(2e/60\) radians, twice what a simple eccentric would give.

Exercise 8.3Show that in Ptolemy’s outer-planet model, at mean opposition (mean Sun opposite the planet’s mean longitude, reckoned from the equant) the planet is at the perigee of its epicycle as reckoned from the equant. Why does this make the planet brightest in the middle of its retrograde arc?

Show solution
At mean opposition the mean Sun’s direction is opposite to the direction \(QK\). The epicycle radius is parallel to the Earth–mean-Sun line, so it points from \(K\) back along \(KQ\): the planet is at the mean perigee, close to the true perigee on the line \(KO\). Retrogradation is centred there (Exercise 8.1), and the planet is then nearest the Earth, hence brightest. In heliocentric terms, opposition is closest approach.
Lesson 09

Fixed stars and precession

For the general reader. Books VII and VIII of the Almagest contain a catalogue of 1,022 stars in 48 constellations, with a position and a brightness for each. Ptolemy also measured how far the stars had shifted relative to the equinoxes since Hipparchus. His catalogue was the standard for more than a millennium, and the question of where its numbers came from has been argued over for four centuries.

Goals
  • Understand ecliptic coordinates and the effect of precession on them.
  • Evaluate Ptolemy’s precession constant.
  • Weigh the evidence in the debate over the catalogue’s origin.

The catalogue

Each entry gives a description of the star’s place in its constellation figure, its ecliptic longitude and latitude, and a magnitude from 1 (brightest) to 6 (faintest visible), sometimes refined as “greater” or “less”. The 48 constellations are 12 of the zodiac, 21 northern and 15 southern; most survive among the 88 modern constellations. The catalogue’s positions are given for the first year of Antoninus, AD 137.

Precession

Because ecliptic latitudes do not change under precession, while longitudes all increase at the same rate, a catalogue in ecliptic coordinates can be moved to a new date by adding a constant to every longitude.

Proposition (precession in ecliptic coordinates)

If the equinox moves westward along the ecliptic at a uniform rate \(k\), then every star’s ecliptic longitude increases by \(k\,\Delta t\) in time \(\Delta t\), and its ecliptic latitude is unchanged. Right ascension and declination change by amounts that depend on the star’s position.

Ptolemy compared his observations of stars such as Regulus and Spica with those of Hipparchus and of Timocharis (early third century BC). He found that longitudes had increased by about \(2;40^\circ\) in the 265 years since Hipparchus, and adopted Hipparchus’s lower limit as the rate: \(1^\circ\) in 100 years, \(36''\) per year. The true rate is about \(50''\) per year, \(1^\circ\) in about 72 years.

The debate

In the sixteenth century Tycho Brahe found that Ptolemy’s longitudes were systematically about \(1^\circ\) too small for AD 137. Tycho and later Jean-Baptiste Delambre suggested the catalogue was Hipparchus’s, moved forward with Ptolemy’s too-slow precession. Robert R. Newton, in 1977, accused Ptolemy of fabricating observations throughout the Almagest. Others, including Owen Gingerich and Bernard Goldstein, argued that Ptolemy selected and adjusted observations to illustrate models he believed correct, a practice that was not regarded as fraud in his time. In 2022 fragments of Hipparchus’s own catalogue were recovered from a palimpsest manuscript by multispectral imaging. Their coordinates differ from Ptolemy’s in ways that show he did not simply copy Hipparchus, though the relation between the two catalogues is still studied.

Exercise 9.1Suppose Ptolemy took a position measured by Hipparchus around 128 BC and added his precession of \(1^\circ\) per century for the roughly 265 years to AD 137. With the true rate of \(50.3''\) per year, how large is the resulting error in longitude? Compare with Tycho’s finding.

Show solution
Ptolemy adds \(2.65^\circ\); the true shift is \(265\times50.3''=13\,330''=3.70^\circ\). The longitudes would be about \(1.05^\circ\) too small, matching the systematic error Tycho found. The agreement is suggestive but does not prove copying, because an error in the equinox (Exercise 2.1) shifts all longitudes in the same way.

Exercise 9.2Modern magnitudes are defined so that 5 magnitudes correspond to a factor of 100 in brightness. What brightness ratio corresponds to Ptolemy’s range from 1 to 6? Why can Ptolemy’s magnitude scale be compared with the modern one at all?

Show solution
\(100^{5/5}=100\). The modern scale was deliberately calibrated in the nineteenth century to agree roughly with the Greek classes. Perceived brightness is approximately logarithmic in intensity, so equal steps of the eye’s judgement correspond to roughly equal brightness ratios.
Lesson 10

Mapping the world; measuring refraction

For the general reader. In the Geography Ptolemy proposed something new: to describe every place by two numbers, latitude and longitude, and to draw the curved surface of the Earth on a flat sheet by a stated mathematical rule. In the Optics he measured how light bends when it passes from air into water. Both works show the same method as the Almagest: geometry, applied to measurements, organized into tables.

Goals
  • Reconstruct Ptolemy’s first projection from his conditions.
  • Understand the consequences of his small value for the Earth’s circumference.
  • Analyse the refraction table and compare it with Snell’s law.

The Geography

The work has eight books. Book I discusses method and criticizes the lost atlas of Marinus of Tyre. Books II–VII list about 8,000 places, most with coordinates in degrees. Book VIII describes 26 regional maps. Longitude runs from a prime meridian through the Fortunate (Blessed) Isles in the Atlantic, eastward through \(180^\circ\) to the land of the Seres and Sinae. Ptolemy takes the Earth’s circumference as 180,000 stadia, 500 stadia per degree, against Eratosthenes’ 252,000 (700 per degree).

Marinus had drawn parallels and meridians as straight perpendicular lines, with the scale correct on the parallel of Rhodes, where a degree of longitude is about \(\cos36^\circ\approx\tfrac45\) of a degree of latitude. Ptolemy criticized this and proposed a better construction.

Ptolemy’s first projection (conditions)
  1. Meridians are straight lines through a common point \(P\), the image of the pole.
  2. Parallels are circular arcs centred at \(P\), equally spaced along every meridian (distances along meridians are true).
  3. Lengths along the parallel of Thule (\(63^\circ\)) and the equator are in their true ratio \(\cos63^\circ:1\).

Construction. Let \(\rho(\varphi)\) be the radius of the arc for latitude \(\varphi\). By (2), \(\rho(\varphi)=\rho_0-\varphi\) in units of one degree of latitude. By (1) and (3), \(\rho(63)/\rho_0=\cos63^\circ\approx0.454\). So \(\rho_0(1-0.454)=63\) and \(\rho_0\approx115\), \(\rho(63)\approx52\), \(\rho(36)\approx79\). Ptolemy gives exactly these numbers. The angular width of each meridian step is then chosen to make the scale true on the parallel of Rhodes. His second projection, which also curves the meridians,, approximates what later became the Bonne projection.

The Optics

The Optics survives only in a Latin translation, made around 1154 by Eugenius of Palermo from a lost Arabic version, and lacks its first book and the end of the fifth. It treats vision by rays from the eye, colour, reflection in plane and curved mirrors, and refraction. In Book V Ptolemy describes an experiment with a graduated disc half immersed in water and tabulates the angle of refraction for angles of incidence from \(10^\circ\) to \(80^\circ\). For air to water his values are:

Incidence10°20°30°40°50°60°70°80°
Ptolemy8°15½°22½°29°35°40½°45½°50°
Snell, \(n=1.333\)7.5°14.9°22.0°28.8°35.1°40.5°44.8°47.6°

The first differences of Ptolemy’s column are \(8,7\tfrac12,7,6\tfrac12,\dots\), with constant second differences of \(\tfrac12^\circ\). The table therefore looks fitted to a quadratic rule, probably adjusted from a few measured values. The law of sines for refraction was found by Ibn Sahl in the tenth century and again by Willebrord Snell and René Descartes in the seventeenth.

Exercise 10.1Ptolemy places the eastern edge of the known world \(180^\circ\) east of the Fortunate Isles. If his distances in stadia were roughly right but he converted them at 500 stadia per degree, by what factor are his longitude differences stretched relative to a conversion at Eratosthenes’ 700? What would such a map suggest to a navigator like Columbus?

Show solution
The factor is \(700/500=1.4\). A modern study of 80 cities in the Geography finds a systematic stretch of about 1.43 in longitude. Eurasia spans too many degrees, so the ocean between western Europe and eastern Asia appears correspondingly narrower. Columbus’s case for a short westward route drew on estimates of this kind.

Exercise 10.2Show that Ptolemy’s refraction values satisfy \(r(i)=\tfrac{i}{10}\big(8.25-\tfrac{i}{40}\big)\) (degrees). Using \(\sin i/\sin r\), compute the implied index of refraction at \(i=30^\circ\) and \(i=80^\circ\). Which part of the table agrees best with Snell’s law?

Show solution
At \(i=10\): \(1\times(8.25-0.25)=8\); at \(i=50\): \(5\times7=35\); at \(i=80\): \(8\times6.25=50\); the other entries check similarly. At \(30^\circ\), \(\sin30^\circ/\sin22.5^\circ=1.31\); at \(80^\circ\), \(\sin80^\circ/\sin50^\circ=1.29\). The middle of the table agrees best (exact at \(60^\circ\)). Large angles, where refraction is hardest to read, deviate most.

Exercise 10.3In the first projection, choose the spacing of the meridians so that the scale is true on the parallel of Rhodes. Find the scale on the parallel of Thule and on the equator, relative to the true globe.

Show solution
If the parallels at radius \(\rho\) are drawn with angular step \(\kappa\) per degree of longitude, the map length of one degree on a parallel is \(\kappa\rho\). Choosing \(\kappa\) so that \(\kappa\rho(36)=\cos36^\circ\), the scale at Thule is \(\kappa\cdot52=0.809\times52/79=0.533\) against the true \(\cos63^\circ=0.454\), and at the equator \(0.809\times115/79=1.18\) against 1. Ptolemy accepted errors of this size away from the central parallel; condition (3) controls only their ratio.
Part III · Life and legacy
Lesson 11

The later works

After the Almagest Ptolemy continued to revise his astronomy and extended his methods to other subjects. Internal references fix a partial order: the Tetrabiblos, Handy Tables and Planetary Hypotheses all cite the Almagest, and the Geography is later still.

  • Handy Tables. The tables of the Almagest, rearranged for practical use and with some parameters changed. They survive in the revision by Theon of Alexandria and became the model for the Arabic and Latin astronomical tables (zījes) of later centuries.
  • Planetary Hypotheses. Two books presenting the models as physical bodies, nested so that each body’s greatest distance equals the least distance of the next. From the Moon’s distance this fixes the size of the whole cosmos: the Sun at a mean of about 1,210 Earth radii, and the sphere of the fixed stars at about 20,000. Here Mercury is placed between the Moon and Venus, an order left open in the Almagest. The scheme dominated medieval cosmology.
  • Tetrabiblos. Four books on astrology, which Ptolemy treated as the second, less certain part of the study of the heavens: predicting earthly effects from celestial configurations. It was the most widely read of his works for centuries.
  • Harmonics. Three books on musical theory. Ptolemy argues that musical intervals should be defined by ratios of whole numbers, as the Pythagoreans held, but that the ratios must be tested by the ear, using a monochord. He gives his own divisions of the tetrachord, such as the “tense diatonic” with steps \(\tfrac{16}{15},\tfrac98,\tfrac{10}{9}\).
  • Smaller works: the Analemma, on orthographic projection for sundials; the Planisphaerium, on stereographic projection of the sky, the mathematics behind the astrolabe; the Phaseis, on the risings of stars; and On the Criterion, a short philosophical essay on knowledge.

Ptolemy probably died in the 160s or 170s. The place is presumed to be Alexandria; no source records it, and no tomb or monument is known.

Exercise 11.1Show that the three steps of Ptolemy’s tense diatonic multiply to a perfect fourth, \(\tfrac43\). Which two steps are equal to the major and minor whole tones of later just intonation?

Show solution
\(\tfrac{16}{15}\cdot\tfrac98\cdot\tfrac{10}9=\tfrac{1440}{1080}=\tfrac43\). The steps \(\tfrac98\) and \(\tfrac{10}9\) are the major and minor tones; \(\tfrac{16}{15}\) is the diatonic semitone. This tuning reappears in Renaissance theory as the syntonic diatonic, named after Ptolemy.

Exercise 11.2In the nested-sphere scheme, the Moon’s sphere extends to 64 Earth radii and Mercury’s epicycle-plus-deferent model has a ratio of greatest to least distance of about 88:34. Find the outer radius of Mercury’s sphere.

Show solution
Mercury’s least distance equals the Moon’s greatest, 64. Its greatest is \(64\times88/34\approx166\) Earth radii. Continuing outward the same way, Venus’s sphere reaches about 1,079 Earth radii, close to the Sun’s least distance of about 1,160. Ptolemy took the near fit as support for placing Mercury and Venus below the Sun.
Lesson 12

Fourteen centuries of Ptolemy

The Almagest made earlier Greek astronomy obsolete, and copies of the older works ceased to be made. In late antiquity it was the subject of commentaries by Pappus and by Theon of Alexandria, whose daughter Hypatia may have worked on its text.

Arabic astronomy

The Almagest was translated into Arabic several times in the ninth century in Baghdad. Astronomers of the Islamic world re-measured its parameters (al-Battani found the obliquity smaller and the solar apogee moved), and they criticized its foundations. Ibn al-Haytham, in the eleventh century, wrote Doubts concerning Ptolemy, objecting that the equant could not be realized by real bodies. Astronomers at Maragha and later Ibn al-Shatir in Damascus built equant-free models; Lesson 07 of the Copernicus course follows their devices into De revolutionibus.

Latin Europe

In the twelfth century the Almagest was translated into Latin twice, from Greek in Sicily and from Arabic by Gerard of Cremona at Toledo, finished in 1175. The Tetrabiblos was translated by Plato of Tivoli in 1138. Georg Peurbach and Regiomontanus prepared an Epitome of the Almagest, printed in 1496, from which Copernicus learned Ptolemaic astronomy. The Geography, rediscovered in Byzantium around 1300 and translated into Latin in 1406–07, transformed European mapmaking: its coordinate grid became standard, and its printed maps (from 1477) circulated widely in the age of exploration.

Assessment

  • Astronomy: Copernicus kept Ptolemy’s method, his order of presentation, and most of his devices, while moving the Earth. Kepler’s ellipse, which finally replaced the equant, agrees with it to first order. The models of the Almagest remained the most accurate available until Tycho Brahe and Kepler.
  • Mathematics: the chord table and the systematic use of Menelaus’s theorem made trigonometry a computational tool. Ptolemy’s theorem remains a standard result of Euclidean geometry.
  • Geography and optics: the idea of mapping by coordinates and by a defined projection, and one of the earliest recorded series of physical measurements, the refraction table.
  • Integrity: the charge of fraud has not been sustained in its strong form. Most historians conclude that Ptolemy selected, rounded and sometimes adjusted observations to fit models derived from a larger body of data, without the modern distinction between raw data and fitted values.
  • Names: large craters on the Moon and on Mars bear his name.

Final projectChoose one body: the Sun, the Moon or Mars. Using Ptolemy’s parameters from this course, compute its position for a date of your choice and compare with a modern ephemeris. Then write two pages explaining which part of the error comes from the model and which from the parameters, and how Copernicus or Kepler would each have reduced it.

Reference

Connections

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

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Reference

Sources and further reading

Biography and context

  1. J. J. O’Connor and E. F. Robertson, Claudius Ptolemy, MacTutor History of Mathematics.
  2. G. J. Toomer, Ptolemy, Dictionary of Scientific Biography, vol. 11, Scribner, 1975.
  3. Ptolemy, Wikipedia (with extensive references).
  4. A. Jones (ed.), Ptolemy in Perspective, Springer, 2010.
  5. O. Neugebauer, A History of Ancient Mathematical Astronomy, Springer, 1975.

Works and mathematics

  1. G. J. Toomer, Ptolemy’s Almagest, translation and notes, Duckworth, 1984; Princeton University Press, 1998.
  2. Almagest and Ptolemy’s table of chords, Wikipedia.
  3. O. Pedersen, A Survey of the Almagest, Odense, 1974; revised edition, Springer, 2011.
  4. J. Evans, The History and Practice of Ancient Astronomy, Oxford University Press, 1998.
  5. J. L. Berggren and A. Jones, Ptolemy’s Geography: An Annotated Translation of the Theoretical Chapters, Princeton University Press, 2000.
  6. Geography (Ptolemy) and Optics (Ptolemy), Wikipedia.
  7. A. M. Smith, Ptolemy’s Theory of Visual Perception: An English Translation of the Optics, American Philosophical Society, 1996.
  8. R. R. Newton, The Crime of Claudius Ptolemy, Johns Hopkins University Press, 1977.
  9. R. Fitzpatrick, A Modern Almagest, University of Texas, open text, 2010.