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Christopher Wren

Sir Christopher Wren PRS

A course in two strands: the life of the Savilian professor of astronomy who became the architect of St Paul’s and of London rebuilt, and the geometry, mechanics and astronomy he did before the Fire and carried into his buildings.

Born 20 October 1632 (O.S.), East Knoyle, Wiltshire Died 25 February 1723 (O.S.), London Arc of the cycloid 1658 Level advanced undergraduate
Wren’s theorem (1669). Every line drawn here is straight, yet together they sweep out a curved hyperboloid. Each family joins a point of the top circle to a point of the bottom circle a fixed angle further round.
Lesson 00

How to use this course

Christopher Wren is remembered as England’s greatest architect. Until he was past thirty he was something else: a mathematician and astronomer, professor at Gresham College and then Savilian Professor of Astronomy at Oxford, a founder of the Royal Society, and in Isaac Newton’s judgement one of the leading geometers of his day. This course treats the mathematics first. It then follows his science into the buildings, where statics, geometry and the mechanics of masonry decided what could stand.

Prerequisites for Part II
  • Single-variable calculus, including parametric curves and arc length.
  • Analytic geometry of conics and quadric surfaces.
  • First-year mechanics: momentum, energy, circular motion, static equilibrium.
  • For Lesson 09, the idea of forces per unit length in a thin shell is developed in the lesson.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs or derivations, and exercises with folded solutions. Wren’s own proofs were geometric, in the style of Archimedes. The lessons give modern versions and say where they differ from his. Dates before 1752 are in the Old Style (Julian) calendar used in England at the time.

Reference

Timeline

1632Born 20 October at East Knoyle, Wiltshire, son of the rector, Christopher Wren the elder, and Mary Cox.
1635His father becomes Dean of Windsor.
c. 1641–46Westminster School under Richard Busby. The Civil War disrupts the Royalist family; William Holder, his brother-in-law, tutors him in mathematics.
c. 1647Assists the anatomist Charles Scarburgh, making models of muscles.
c. 1650Wadham College, Oxford, under John Wilkins. BA 1651, MA 1653.
1653Fellow of All Souls College, Oxford.
1657Professor of Astronomy at Gresham College, London.
1658Rectifies the cycloid; solves and poses Kepler’s problem; Gresham lectures on Saturn.
1659John Wallis publishes Wren’s cycloid results in his treatise on the cycloid.
166028 November: after Wren’s Gresham lecture, the meeting that founds the Royal Society.
1661Savilian Professor of Astronomy, Oxford (to 1673). Presents Charles II with a relief globe of the Moon.
1663–64Pembroke College Chapel, Cambridge; designs the Sheldonian Theatre, Oxford (built 1664–69).
1665July: visits Paris, his only journey abroad; meets Gian Lorenzo Bernini.
1666September: the Great Fire. Within days, presents a plan for a new City to Charles II. Commissioner for rebuilding.
166817 December: his law of collision read to the Royal Society (printed January 1669).
1669Surveyor of the King’s Works. Marries Faith Coghill. Paper on the ruled hyperboloid and a lens-grinding engine.
1671–77The Monument to the Great Fire, with Robert Hooke.
1673The Great Model of St Paul’s. Knighted 14 November. Resigns the Savilian chair.
1675Warrant Design approved; first stone of St Paul’s laid in June. Royal Observatory, Greenwich, built 1675–76. Faith Wren dies.
1677Marries Jane Fitzwilliam (died 1680).
1680–82President of the Royal Society.
1684January: with Hooke and Edmond Halley, discusses the inverse-square law; offers a prize for a proof.
1685–1702Member of Parliament at intervals (Plympton Erle, New Windsor, Weymouth and Melcombe Regis).
1696Surveyor of Greenwich Hospital (to 1716).
170826 October: the last stone placed on the lantern of St Paul’s.
1711Parliament declares St Paul’s complete.
171826 April: dismissed as Surveyor of the King’s Works.
1723Dies 25 February, aged 90. Buried 5 March in the crypt of St Paul’s.
Part I · Formation
Lesson 01

East Knoyle, Windsor, Westminster, 1632–1649

Christopher Wren was born on 20 October 1632 at East Knoyle in Wiltshire, where his father, also Christopher, was rector. His mother was Mary Cox. He was a sickly child. In 1635 his father became Dean of Windsor, and the boy grew up partly in the precincts of Windsor Castle, among a family of High Church clergy close to the court. His uncle Matthew Wren was Bishop of Ely.

The Civil War broke the family’s world. Matthew Wren was imprisoned in the Tower of London by Parliament in 1642 and stayed there until 1660. The deanery at Windsor was plundered, and Wren’s father retired to the house of his son-in-law William Holder at Bletchingdon in Oxfordshire. Holder, a clergyman with a strong interest in mathematics and music theory, tutored the boy and gave him his first serious mathematics.

Wren attended Westminster School under its formidable headmaster, Richard Busby, probably from about 1641 to 1646. After leaving he was taken up by the physician Charles Scarburgh, a mathematician as well as an anatomist, who lectured on anatomy and translated Euclid. Wren assisted him and made pasteboard models showing how the muscles act, his first scientific work. He was already devising instruments, among them a pen for writing two copies at once.

Lesson 02

Wadham and All Souls, 1649–1657

Around 1650 Wren entered Wadham College, Oxford, as a gentleman commoner. Its Warden, John Wilkins, had made it the centre of a club of experimental philosophers that met in his lodgings: the mathematicians Seth Ward, Savilian Professor of Astronomy, and John Wallis, Savilian Professor of Geometry; the physicians Thomas Willis and William Petty; and, from the mid-1650s, Robert Boyle. Wren took his BA in 1651 and his MA in 1653, and in 1653 he was elected a Fellow of All Souls College, where he lived until 1657.

He was the youngest member of the club and its most versatile. With Ward he observed through the telescopes at Wadham. He worked with Paul Neile on a 35-foot telescope and helped Wallis with a longer one. In anatomy he drew the plates of the brain for Willis’s Cerebri anatome (1664), and he injected liquids into the veins of living dogs with a quill and bladder, among the first experiments of intravenous injection. Boyle reported the experiments and credited Wren with the idea.

The friends he made at Oxford stayed with him for life: Boyle, Wallis, Willis, and the young Robert Hooke, then Boyle’s assistant. Hooke would later be his partner in the rebuilding of London.

Lesson 03

Professor of astronomy, 1657–1665

In 1657, aged 24, Wren became Professor of Astronomy at Gresham College in London. His inaugural lecture, which survives, praised London as a city with a taste for mathematics and set out a programme of experimental and observational astronomy. In the next two years he did the mathematics for which he is remembered. In 1658 he found the length of the cycloid (Lesson 04). The same year an anonymous challenge on the ellipse came from France, possibly from Blaise Pascal. Wren solved it and sent back a counter-challenge, Kepler’s problem, which he had solved himself (Lesson 05). In lectures at Gresham he offered a theory of the strange appearances of Saturn, and gave it up gracefully when he learned of Christiaan Huygens’s ring (Lesson 08).

On 28 November 1660, after Wren’s lecture at Gresham, twelve of those present resolved to found a college for the promotion of physico-mathematical experimental learning. This became the Royal Society, chartered in 1662. In 1661 Wren succeeded Ward as Savilian Professor of Astronomy at Oxford. That year he made for Charles II a globe of the Moon, ten inches across, showing its visible face in relief. For the Society he proposed instruments by the dozen: a tipping-bucket rain gauge in 1662, a self-recording weather clock in 1663, and a double telescope with a scale for measuring distances.

The turn to architecture

Architecture was then a branch of applied mathematics open to a gentleman with geometry. In 1663 his uncle Matthew commissioned a chapel for Pembroke College, Cambridge, and Archbishop Gilbert Sheldon asked him to design a theatre for university ceremonies at Oxford (Lesson 09). He was also consulted about repairs to the decaying old St Paul’s Cathedral. In July 1665, with plague in London, he went to Paris, his only journey abroad. He studied the new buildings, met the sculptor-architect Gian Lorenzo Bernini, and came home with books and engravings. Fourteen months later London burned.

Part II · The mathematics
Lesson 04

The length of the cycloid

For the general reader. Mark a point on the rim of a rolling wheel and it traces a row of arches called cycloids. In the 1650s the best mathematicians in Europe competed to measure them. Wren found the length of an arch: exactly four times the diameter of the wheel. It was one of the first curves whose length was found exactly, and Wallis published it in 1659 with full credit to Wren.

Goals
  • Parametrize the cycloid and compute its arc length.
  • State Wren’s theorem in its geometric form and prove it.
  • Find the area under the arch.

The curve

A circle of radius \(r\) rolls along the \(x\)-axis. After it has turned through angle \(\theta\), its centre is at \((r\theta, r)\), and the marked point, which started at the origin, is at

\[ x = r(\theta-\sin\theta), \qquad y = r(1-\cos\theta), \qquad 0\le\theta\le2\pi. \]

Galileo had named the curve and tried to find its area by weighing paper cut-outs. Gilles Personne de Roberval and Evangelista Torricelli found the area. In 1658 Pascal, writing as A. Dettonville, offered prizes for harder problems on the cycloid. Wallis competed. Wren sent Pascal his rectification, which Wallis printed in his Tractatus duo of 1659 as the first finding of a straight line equal to the curve.

Theorem (Wren, 1658)

The arc of a cycloid measured from its vertex to any point \(P\) is twice the chord of the generating circle drawn from the vertex parallel to the tangent at \(P\). In particular, a whole arch has length \(8r\), four times the diameter of the generating circle.

Proof

Differentiate: \(\dot x=r(1-\cos\theta)=2r\sin^2(\theta/2)\) and \(\dot y=r\sin\theta=2r\sin(\theta/2)\cos(\theta/2)\). Hence

\[ \frac{ds}{d\theta}=\sqrt{\dot x^2+\dot y^2}=2r\sin\frac\theta2\sqrt{\sin^2\frac\theta2+\cos^2\frac\theta2}=2r\sin\frac\theta2, \qquad 0\le\theta\le2\pi. \]

Integrating from the vertex \(\theta=\pi\) to \(\theta\) gives arc length \(s=4r\,|\cos(\theta/2)|\). Now place the generating circle with its top at the vertex of the arch. The tangent direction at parameter \(\theta\) is \((\sin(\theta/2),\cos(\theta/2))\), making angle \(\pi/2-\theta/2\) with the horizontal. A chord from the top of a circle of diameter \(2r\) making angle \(\psi\) with the horizontal tangent there has length \(2r\sin\psi\); with \(\psi=\pi/2-\theta/2\) that is \(2r\cos(\theta/2)\). So \(s=2\times\text{chord}\). At \(\theta=0\), \(s=4r\), and the full arch is \(8r\). Wren’s own proof was geometric, a method of exhaustion in the manner of Archimedes, with no coordinates and no integrals.

Exercise 4.1Show that the area under one arch of the cycloid is \(3\pi r^2\), three times the area of the generating circle.

Show solution
\(A=\int y\,dx=\int_0^{2\pi}r(1-\cos\theta)\cdot r(1-\cos\theta)\,d\theta=r^2\int_0^{2\pi}(1-2\cos\theta+\cos^2\theta)\,d\theta=r^2(2\pi-0+\pi)=3\pi r^2\).

Exercise 4.2A wheel of diameter 0.70 m rolls without slipping for one full turn. How far does a stone stuck in its tread travel? Compare with how far the axle travels.

Show solution
The stone traces one arch: \(4d=2.80\) m. The axle moves one circumference, \(\pi d\approx2.20\) m. The stone goes 27% farther, because it rises and falls as well as moving forward.

Exercise 4.3Find the arc length from the vertex to the point where the tangent makes 30° with the horizontal, using Wren’s chord rule only.

Show solution
The chord from the top of the generating circle parallel to the tangent makes 30° with the horizontal tangent at the top, so its length is \(2r\sin30^\circ=r\). The arc is twice this: \(2r\). (Check: \(\psi=30^\circ\) means \(\theta/2=60^\circ\), and \(4r\cos60^\circ=2r\).)
Lesson 05

Kepler’s problem

For the general reader. Johannes Kepler found that a planet sweeps out equal areas in equal times. To predict where a planet will be on a given date, you must cut an ellipse into a piece of a given area by a line from the Sun. Kepler could not do it exactly and doubted anyone could. Wren gave a geometrical construction in 1658 and sent the problem to France as a challenge. The equation behind it, Kepler’s equation, cannot be solved by algebra, which is why Wren needed a special curve.

Goals
  • Derive Kepler’s equation from the area law.
  • See why a cycloid-type curve solves it.
  • Solve it numerically by Newton’s method.

The problem as Wren posed it

Given a semicircle of radius \(a\) and a point \(S\) on its diameter at distance \(ae\) from the centre \(C\), draw a line from \(S\) to the arc that cuts off a given fraction of the area of the semicircle. Because an ellipse is a circle compressed in one direction, which scales all areas by the same factor \(b/a\), this is equivalent to cutting an ellipse from its focus.

Proposition (Kepler’s equation)

Let \(A\) be the end of the diameter beyond \(S\), and \(P\) the point of the circle with \(\angle ACP=E\). The area bounded by \(SA\), the arc \(AP\) and \(PS\) equals \(\tfrac12a^2(E-e\sin E)\). Hence the line \(SP\) cuts off the fraction \(M/\pi\) of the semicircle exactly when

\[ M = E - e\sin E. \]

Proof. The region is the circular sector \(CAP\), of area \(\tfrac12a^2E\), minus the triangle \(CSP\). That triangle has base \(CS=ae\) and height \(a\sin E\), so area \(\tfrac12a^2e\sin E\). In astronomy \(E\) is the eccentric anomaly, \(e\) the eccentricity of the orbit and \(M=2\pi t/T\) the mean anomaly, proportional to time since perihelion.

Why a cycloid

Roll a circle of radius \(a\) along a line and follow a point fixed at distance \(ae\) from the centre, with \(e<1\). Its horizontal position after a turn through \(\theta\) is \(a(\theta-e\sin\theta)\). This is a curtate cycloid. Solving Kepler’s equation for \(E\) is the same as finding where this curve reaches the horizontal position \(aM\). Wren, who had just mastered the cycloid, solved the problem with a construction of this kind, and Wallis printed it in 1659. Newton later used the same idea in the Principia, Book I, Proposition 31, and also gave an approximation scheme.

Theorem (uniqueness)

For \(0\le e<1\) the function \(f(E)=E-e\sin E\) is strictly increasing with \(f'(E)=1-e\cos E\ge1-e>0\). So for every \(M\) there is exactly one \(E\), and Newton’s iteration \(E_{k+1}=E_k-\dfrac{E_k-e\sin E_k-M}{1-e\cos E_k}\) converges quadratically from \(E_0=M\) for moderate \(e\).

Exercise 5.1Take \(e=0.2\), \(M=1\) radian. Perform Newton’s iteration from \(E_0=1\) to six decimals.

Show solution
\(E_1=1-\dfrac{1-0.2\sin1-1}{1-0.2\cos1}=1+\dfrac{0.168294}{0.891940}=1.188683\). \(E_2=1.185325\), \(E_3=1.185324\). Converged: \(E\approx1.185324\). The number of correct digits roughly doubles each step.

Exercise 5.2In Wren’s semicircle form, \(S\) lies halfway from the centre to the end \(A\) (\(e=\tfrac12\)). At what angle \(E=\angle ACP\) must \(P\) lie for \(SP\) to cut off one quarter of the semicircle, measured from \(A\)?

Show solution
One quarter of the semicircle means \(M/\pi=\tfrac14\), \(M=\pi/4\). Solve \(E-\tfrac12\sin E=0.785398\). Newton from \(E_0=0.7854\): \(E_1\approx1.3323\), \(E_2\approx1.2631\), \(E_3\approx1.2617\). So \(E\approx1.2617\) rad \(\approx72.3^\circ\). It is greater than 45° because \(S\) sits near \(A\), so the region near \(A\) is thin.

Exercise 5.3Show that for small \(e\), \(E\approx M+e\sin M+\tfrac12e^2\sin2M\).

Show solution
Write \(E=M+e\,u_1+e^2u_2+O(e^3)\). Then \(e\sin E=e\sin M+e^2u_1\cos M+O(e^3)\). Matching powers in \(E=M+e\sin E\): \(u_1=\sin M\) and \(u_2=u_1\cos M=\sin M\cos M=\tfrac12\sin2M\).
Lesson 06

The ruled hyperboloid and the perfect lens

For the general reader. Bundle straight sticks between two hoops, twist one hoop, and the sticks form a waisted surface like a cooling tower. Wren showed that this surface is a hyperboloid. He published the result in 1669 with a practical aim: lenses with hyperbolic surfaces focus light perfectly, and a surface made of straight lines could be ground by a straight-edged tool.

Goals
  • Prove that the hyperboloid of one sheet is doubly ruled.
  • Find the waist of a twisted bundle of lines.
  • Derive the hyperbolic lens from Fermat’s principle.
Theorem (Wren, 1669)

The hyperboloid of revolution of one sheet, \(\dfrac{x^2+y^2}{a^2}-\dfrac{z^2}{c^2}=1\), is generated by rotating about the \(z\)-axis a straight line that does not meet the axis and is not parallel to it. Through every point of the surface pass exactly two straight lines lying in the surface.

Proof

The line \(\ell(t)=(a,\ at,\ ct)\) satisfies \(\dfrac{a^2+a^2t^2}{a^2}-\dfrac{c^2t^2}{c^2}=1\) for all \(t\), so it lies on the surface; rotating it gives the whole surface, because every height \(z=ct\) is reached at radius \(a\sqrt{1+t^2}\). The line \((a,-at,ct)\) gives the second family. For the factorized form, write the equation as

\[ \Big(\frac xa-\frac zc\Big)\Big(\frac xa+\frac zc\Big)=\Big(1-\frac ya\Big)\Big(1+\frac ya\Big). \]

For each \(\lambda\) the two planes \(\frac xa-\frac zc=\lambda\big(1-\frac ya\big)\) and \(\lambda\big(\frac xa+\frac zc\big)=1+\frac ya\) meet in a line on the surface; replacing \(1\mp y/a\) by \(1\pm y/a\) gives the other family. Lines of the same family are skew; lines of different families meet or are parallel.

The hyperbolic lens

René Descartes had shown in 1637 that a lens with a hyperbolic surface brings parallel light to an exact focus, free of the spherical aberration of ordinary lenses. Nobody could grind such a surface. Wren’s idea was to rotate the glass against a tool shaped by straight edges laid along the rulings. His paper in the Philosophical Transactions for November 1669 described an engine for doing so. There is no clear evidence that it was built.

Proposition (Descartes)

Rays travelling parallel to the axis inside glass of index \(n>1\) leave through a surface into air and meet at a point \(F\) exactly when the surface is a hyperboloid of revolution with focus \(F\) and eccentricity \(e=n\).

Proof by Fermat’s principle. All rays must have equal optical path to \(F\). Measure \(z\) along the axis toward \(F\) from a reference plane in the glass. A ray meeting the surface at a point at distance \(\rho\) from \(F\) has optical path \(nz+\rho\), so \(\rho=K-nz\). The quantity \(K/n-z\) is the distance from the point to a fixed plane, so \(\rho=n\cdot(\text{distance to a plane})\): a conic with focus \(F\) and eccentricity \(n\). Since \(n>1\) it is a hyperbola.

Exercise 6.1Straight rods join each point of a circle of radius \(R\) at height \(+h\) to the point a fixed angle \(\varphi\) further round on a circle of radius \(R\) at height \(-h\). Find the radius of the waist.

Show solution
By symmetry the narrowest point of each rod is its midpoint, at height 0. The midpoint of the chord between angles \(\alpha\) and \(\alpha+\varphi\) on a circle of radius \(R\) lies at distance \(R\cos(\varphi/2)\) from the centre. So the waist radius is \(R\cos(\varphi/2)\). With \(\varphi=180^\circ\) the rods pass through the axis and the surface degenerates to a double cone.

Exercise 6.2For crown glass with \(n=1.5\), find the angle between the axis of the hyperbolic lens surface and the asymptotes of the hyperbola.

Show solution
For a hyperbola, \(b/a=\sqrt{e^2-1}=\sqrt{1.25}\approx1.118\). The asymptotes make angle \(\arctan(b/a)\approx48.2^\circ\) with the axis. Far from the vertex the surface approaches a cone of that half-angle.
Lesson 07

The law of collision

For the general reader. What happens when two hard balls collide? Descartes had published rules that were mostly wrong. In 1668 the Royal Society asked for the true laws, and three answers arrived within weeks, from Wallis, from Huygens and from Wren. Wren’s rule for hard, springy balls was short and correct. Seen from their common centre of gravity, the two balls simply bounce back with their speeds unchanged.

Goals
  • State Wren’s rule in modern terms and derive the final velocities.
  • Show that it conserves momentum and kinetic energy.
  • Test Descartes’s rules against it.

Wren’s rule

Wren’s paper, Lex naturae de collisione corporum, was read on 17 December 1668 and printed in the Philosophical Transactions in January 1669, next to Wallis’s. Wren said he had found the law around 1660 and that Lawrence Rooke, who died in 1662, had confirmed it by experiment. Wren compared the collision to a balance with its fulcrum at the common centre of gravity: the velocities of the bodies relative to that point, which he called their proper velocities, are inversely as the bodies, like weights balancing on a lever.

Theorem (perfectly elastic collision, Wren and Huygens 1668–69)

Two bodies of masses \(m_1,m_2\) move along a line with velocities \(u_1,u_2\). Let \(V=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\) be the velocity of their centre of gravity. After a perfectly elastic collision each velocity relative to the centre of gravity is reversed:

\[ v_i = 2V-u_i,\qquad\text{so}\qquad v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2},\quad v_2=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}. \]

Check. Momentum: \(m_1v_1+m_2v_2=2(m_1+m_2)V-(m_1u_1+m_2u_2)=m_1u_1+m_2u_2\). Energy: in the centre-of-gravity frame each speed is unchanged, so the kinetic energy there is unchanged, and the energy in any frame is that energy plus \(\tfrac12(m_1+m_2)V^2\). Also \(v_2-v_1=-(u_2-u_1)\): the bodies separate as fast as they approached, which Huygens made an axiom. Newton, in the Scholium to the laws of motion in the Principia, cited Wren, Wallis and Huygens as the leading geometers of the age for these results, and described experiments with pendulums that confirmed them.

Exercise 7.1A 3 kg ball moving at 4 m/s strikes a 1 kg ball at rest, elastically. Find the final velocities and check that kinetic energy is conserved.

Show solution
\(V=(12+0)/4=3\) m/s. \(v_1=6-4=2\) m/s, \(v_2=6-0=6\) m/s. Energy before \(\tfrac12\cdot3\cdot16=24\) J; after \(\tfrac12\cdot3\cdot4+\tfrac12\cdot1\cdot36=6+18=24\) J.

Exercise 7.2Descartes’s fourth rule says that a body at rest cannot be moved by a smaller body striking it, whatever its speed; the smaller body bounces back. Use Wren’s rule to show this is false, and find the speed given to a 2 kg ball at rest by a 1 kg ball arriving at 3 m/s.

Show solution
\(V=3/3=1\) m/s. \(v_1=2-3=-1\) m/s (the small ball does bounce back) and \(v_2=2-0=2\) m/s. The larger body is set moving at 2 m/s. In Wren’s rule the struck body always moves off with \(2V>0\) whenever anything hits it.

Exercise 7.3Show that when \(m_1=m_2\) the bodies exchange velocities, and explain Newton’s cradle.

Show solution
With \(m_1=m_2\), \(V=\tfrac12(u_1+u_2)\), so \(v_1=u_2\) and \(v_2=u_1\). In a row of equal balls, each collision passes the incoming velocity along to the next ball and leaves the striker at rest, so only the last ball swings out.
Lesson 08

Saturn, the Moon and the inverse square

For the general reader. As an astronomer Wren worked on the questions of his day: the changing shape of Saturn, the mountains of the Moon, the paths of comets, and what force holds the planets in orbit. He was early to see that Kepler’s laws imply a force falling off as the square of the distance. He could not prove that such a force gives elliptical orbits; that took Newton.

Goals
  • Explain the changing aspect of Saturn’s rings.
  • Derive the inverse-square law from Kepler’s third law for circular orbits.
  • Use angular measurement with a micrometer.

Saturn

Through small telescopes Saturn seemed to change shape over the years, at times showing two “handles” and at times none. In Gresham lectures in spring 1658 Wren proposed an elliptical corona attached to the planet at two points, and wrote it up in a manuscript, De corpore Saturni, with a copper model. Early in 1659 he learned of Huygens’s theory of a thin flat ring, nowhere touching the planet and inclined to the orbit. Wren accepted it at once, saying he preferred it to his own, and did not publish.

Proposition (aspect of a ring)

A thin circular ring of radius \(R\), seen from a distant observer at elevation \(B\) above its plane, appears as an ellipse with semi-axes \(R\) and \(R\sin B\). Saturn’s rings lie in its equator, tilted about \(26.7^\circ\) to its orbit, so \(|B|\) as seen from near the Sun ranges from 0 to about \(26.7^\circ\) and passes through 0 twice in each 29.5-year orbit.

Kepler’s third law and the inverse square

Huygens published in 1673 that a body moving uniformly in a circle of radius \(r\) with period \(T\) needs an inward force \(F=4\pi^2mr/T^2\). Kepler’s third law says \(T^2=kr^3\) with the same \(k\) for all planets. Then

\[ F=\frac{4\pi^2mr}{kr^3}=\frac{4\pi^2}{k}\cdot\frac{m}{r^2}. \]

By the early 1680s Wren, Hooke and Halley each knew this step. In January 1684, after a Royal Society meeting, Hooke claimed he could derive all the motions of the planets from an inverse-square attraction. Wren offered a book worth forty shillings to whichever of them produced a proof within two months. No proof came. In August Halley asked Newton, who said the orbit is an ellipse and in November sent De motu corporum in gyrum, the seed of the Principia. Newton acknowledged Wren, Hooke and Halley in the Principia as having inferred the inverse-square law from Kepler’s third law.

Exercise 8.1Saturn’s ring system spans about 270 000 km at its outer edge (radius 135 000 km); Saturn’s equatorial radius is 60 300 km. For what elevations \(B\) does the ring’s apparent minor semi-axis fall below the planet’s radius, so that the rings no longer extend above and below the disc?

Show solution
Need \(135\,000\sin B<60\,300\), so \(\sin B<0.447\), \(B<26.5^\circ\). Since \(B\) never exceeds about \(26.7^\circ\), the rings almost always project within the disc’s height. They are thin enough to vanish from 1650s telescopes near \(B=0\), and the changing \(B\) produced the shapes that puzzled Wren.

Exercise 8.2Jupiter’s moons Io and Callisto have orbital radii 421 700 km and 1 882 700 km, and Io’s period is 1.769 days. Predict Callisto’s period, and the ratio of the accelerations toward Jupiter at the two orbits.

Show solution
\(T_C=1.769\times(1\,882\,700/421\,700)^{3/2}=1.769\times(4.4646)^{1.5}=1.769\times9.433\approx16.69\) days (observed 16.69). The acceleration ratio is \((r_C/r_I)^{-2}=1/19.93\approx0.050\).

Exercise 8.3Wren used a micrometer in the eyepiece of his telescope to measure lunar features. A crater subtends 30 arcseconds at a mean Earth–Moon distance of 384 400 km. Estimate its diameter.

Show solution
\(30''=30/206\,265\ \mathrm{rad}=1.454\times10^{-4}\) rad. Diameter \(\approx384\,400\times1.454\times10^{-4}\approx56\) km.
Lesson 09

The mechanics of roofs, arches and domes

For the general reader. Wren came to architecture as a geometer and mechanic. His first large building, the Sheldonian Theatre in Oxford, needed a roof wider than any available timber. His greatest, St Paul’s, needed a dome that would look grand from the street, look right from inside, and carry a stone lantern of about 850 tons. He solved the dome with three shells, one inside another, and the logic of each can be followed with the statics of this lesson.

Goals
  • Find the line of thrust of an arch and why it must lie inside the masonry.
  • Compute the hoop forces in a hemispherical dome and see why iron chains were needed.
  • Find the outward thrust at the base of a cone carrying a lantern.

The Sheldonian roof

The theatre, built 1664–69 and modelled on engravings of the Theatre of Marcellus in Rome, is D-shaped, with a roof spanning over 70 feet. No English timber was long enough. Wren used a scheme that John Wallis had studied for a “geometrical flat floor”, in which short beams rest on one another in a grid so that each is supported by its neighbours, combined with trusses built up from shorter pieces and joined with iron. For many years it was celebrated as the widest unsupported floor in England.

Arches and the line of thrust

Masonry resists compression well and tension hardly at all. A chain hanging under its own weight is in pure tension. Turned upside down, the same shape is in pure compression: the principle Hooke announced in 1675, and that Wren and Hooke discussed for the dome of St Paul’s. For a uniform chain the shape is the catenary \(y=a\cosh(x/a)\). An arch stands if a line of thrust, a funicular curve for its actual loads, can be drawn inside the thickness of the masonry.

Theorem (membrane forces in a spherical dome)

A thin hemispherical shell of radius \(R\) and weight \(w\) per unit area, supported vertically at its base, carries under self-weight a meridional force \(N_\phi=-\dfrac{Rw}{1+\cos\phi}\) (compression) and a hoop force \(N_\theta=Rw\Big(\dfrac{1}{1+\cos\phi}-\cos\phi\Big)\) per unit length, where \(\phi\) is the angle from the crown. The hoop force is compressive near the crown and becomes tensile below \(\phi_0\), where \(\cos\phi_0=\tfrac{\sqrt5-1}{2}\), \(\phi_0\approx51.8^\circ\).

Derivation. The cap above angle \(\phi\) has weight \(2\pi R^2w(1-\cos\phi)\). It is held by the meridional force acting around the circle of radius \(R\sin\phi\), with vertical component \(-N_\phi\sin\phi\cdot2\pi R\sin\phi\). Equating gives \(N_\phi=-Rw(1-\cos\phi)/\sin^2\phi=-Rw/(1+\cos\phi)\). Normal equilibrium of a shell element gives \(N_\phi+N_\theta=-wR\cos\phi\), which yields \(N_\theta\). Setting \(N_\theta=0\) gives \(\cos^2\phi+\cos\phi-1=0\).

St Paul’s

The inner dome, which rises to about 214 feet above the floor, is shaped to look right from inside. The lead-covered outer dome, on a timber frame, is shaped to look right from the city. Neither carries the lantern. Between them a cone of brick, 18 inches thick, rises from the drum to the lantern and carries it, together with the timbers of the outer dome. Iron chains set in the masonry resist the outward spreading at the base, the tension the theorem above predicts. Structural engineers, notably Jacques Heyman, have used the same thrust-line reasoning to explain why the dome stands.

Exercise 9.1A chain hangs between supports 20 m apart in the shape \(y=a\cosh(x/a)\) with \(a=10\) m. Find its sag and length.

Show solution
Sag \(=a(\cosh(10/a)-1)=10(\cosh1-1)=10(0.5431)\approx5.43\) m. Length \(=2a\sinh(10/a)=20\sinh1\approx23.50\) m. Inverted, this is a pure-compression arch of span 20 m and rise 5.43 m for a load distributed uniformly along the arch.

Exercise 9.2A masonry hemisphere of radius 15 m and thickness 0.45 m (density 1800 kg/m³) rests on vertical supports. Find the hoop force per unit length at the base, \(\phi=90^\circ\), and the total hoop tension carried across a meridional cut through the tensile band \(\phi_0\le\phi\le90^\circ\). This is the tension that chains must supply if the masonry carries none.

Show solution
\(w=1800\times9.81\times0.45\approx7946\ \mathrm{N/m^2}\). At \(\phi=90^\circ\), \(N_\theta=Rw\,(1-0)=15\times7946\approx1.19\times10^5\) N/m, in tension. The total over the band is \(\int_{\phi_0}^{\pi/2}N_\theta\,R\,d\phi=R^2w\big[\tan\tfrac\phi2-\sin\phi\big]_{\phi_0}^{\pi/2}=R^2w\,\big(0-(0.4859-0.7862)\big)\approx0.300\,R^2w\). With \(R^2w=1.79\times10^6\) N this is about \(5.4\times10^5\) N, some 55 tonnes-force.

Exercise 9.3A cone of semi-vertical angle \(\alpha\) carries a lantern of weight \(W\) at its apex. Ignoring the cone’s own weight, show that the total ring tension at the base is \(T=\dfrac{W\tan\alpha}{2\pi}\). Evaluate for \(W=850\) tonnes-force and an illustrative \(\alpha=20^\circ\).

Show solution
Let the thrust along the generators be \(N\) per unit length at base radius \(r\). Its vertical component balances the load: \(N\cos\alpha\cdot2\pi r=W\). The horizontal outward component per unit length is \(H=N\sin\alpha=\dfrac{W\tan\alpha}{2\pi r}\). A ring of radius \(r\) under outward load \(H\) per unit length carries tension \(T=Hr=W\tan\alpha/(2\pi)\), independent of \(r\). With \(\alpha=20^\circ\): \(T=850\times0.364/6.283\approx49\) tonnes-force. A steeper cone (smaller \(\alpha\)) needs a weaker chain.
Lesson 10

The logarithmic spiral and the growth of shells

For the general reader. A snail shell grows by adding new material at its open end while keeping its shape. Wren noticed that this kind of growth produces a logarithmic spiral, a curve that looks the same at every scale, and he suggested that shells are spirals of this kind wound in three dimensions. The idea was taken up again by D’Arcy Thompson in 1917 and is now standard in the mathematics of biological form.

Goals
  • Characterize the logarithmic spiral by its constant angle and its self-similarity.
  • Compute its arc length, a result of Torricelli’s that Wren knew.
  • Model a shell by a spiral with expanding cross-section.

The curve

In polar coordinates the logarithmic spiral is \(r=a\,e^{b\theta}\) with \(a>0\), \(b\neq0\). Its tangent makes the same angle \(\alpha\) with the radius at every point, with \(\cot\alpha=b\). For this reason it is also called the equiangular spiral.

Theorem (characterizations)

For a smooth curve \(r=f(\theta)\) with \(f>0\), the following are equivalent: (1) the angle between the tangent and the radius vector is constant; (2) \(f(\theta)=a\,e^{b\theta}\); (3) the curve is carried onto itself by every similarity \(z\mapsto\lambda e^{i\beta}z\) with \(\lambda=e^{b\beta}\), a rotation combined with a matching dilation about the pole.

Proof. The angle \(\alpha\) between radius and tangent satisfies \(\cot\alpha=f'(\theta)/f(\theta)\). It is constant exactly when \((\ln f)'\) is constant, that is, when \(f=a e^{b\theta}\). For (3), rotating by \(\beta\) sends \(r=ae^{b\theta}\) to \(r=ae^{b(\theta-\beta)}=e^{-b\beta}\,ae^{b\theta}\), which is the original curve scaled by \(e^{-b\beta}\); so a rotation followed by the dilation \(e^{b\beta}\) returns the curve to itself. Conversely, invariance under a continuous family of such similarities forces \(f(\theta+\beta)=e^{b\beta}f(\theta)\) for all \(\beta\), whence \(f=ae^{b\theta}\).

Arc length

With \(ds^2=dr^2+r^2d\theta^2\) and \(dr=br\,d\theta\), we get \(ds=\sqrt{1+b^2}\,r\,d\theta=\dfrac{\sqrt{1+b^2}}{b}\,dr\). Therefore

\[ s(\theta_1\to\theta_2)=\frac{\sqrt{1+b^2}}{b}\,\big(r_2-r_1\big)=\frac{r_2-r_1}{\cos\alpha}. \]

As \(\theta\to-\infty\) (for \(b>0\)) the spiral winds infinitely often about the pole, yet the length from any point to the pole is finite, \(r/\cos\alpha\). Torricelli found this in the 1640s; it was, with Wren’s cycloid, among the first rectifications of curves.

Shells

Let a closed cross-section (the opening of the shell) be carried along a logarithmic spiral while it grows in proportion to its distance from the axis, and let the whole spiral also move along the axis in proportion to that distance. The resulting surface is invariant under a one-parameter group of similarities, so every whorl is a scaled copy of the last. This is the content of Wren’s suggestion, made precise by Henry Moseley in 1838 and by D’Arcy Thompson in On Growth and Form.

Exercise 10.1A nautilus shell’s whorl grows by a factor of about 3 in radius per full turn. Find \(b\) and the constant angle \(\alpha\).

Show solution
\(e^{2\pi b}=3\) gives \(b=\ln3/(2\pi)\approx0.1748\). Then \(\cot\alpha=0.1748\), so \(\alpha\approx80.1^\circ\). The curve is close to a circle locally, crossing each radius at just under a right angle.

Exercise 10.2For the spiral of Exercise 10.1 with outer radius 10 cm, how long is the whole spiral from the outer end to the pole?

Show solution
\(s=r/\cos\alpha=r\sqrt{1+b^2}/b=10\times\sqrt{1.0306}/0.1748\approx10\times5.808\approx58.1\) cm, finite despite infinitely many turns.
Part III · Life and legacy
Lesson 11

Surveyor: London rebuilt, 1666–1711

The Great Fire began on 2 September 1666 and burned for four days, destroying most of the medieval City, 87 parish churches and old St Paul’s. Within days Wren presented Charles II with a plan for a new City of broad straight streets and piazzas. It was not adopted; property owners rebuilt on the old lines. Wren was named one of the commissioners for rebuilding, with Hooke as one of the City’s surveyors. In 1669 he became Surveyor of the King’s Works, the post that made him the chief architect of the Crown, and in 1670 the rebuilding of the City churches began under his office. Fifty-one were rebuilt to designs from his office, each a new solution to an awkward medieval site.

The Monument to the Great Fire, a Doric column 202 feet tall standing 202 feet from the site in Pudding Lane where the fire began, was built from 1671 to 1677. Hooke drew the design under Wren’s authority. Its hollow shaft was meant also to serve as a zenith telescope and a laboratory for pendulum and pressure experiments, but the traffic shook it too much.

St Paul’s

For the cathedral Wren made several designs. The Great Model of 1673, a vast oak model of a centralized church under a great dome, was his favourite; the clergy wanted a traditional long nave. The Warrant Design of 1675, with a Latin-cross plan and a curious spire on a shallow dome, received royal approval, with permission for ornamental changes, which Wren used to transform it. The first stone was laid in June 1675. The last stone on the lantern was set on 26 October 1708, and Parliament declared the building complete in 1711. Wren lived to see it finished.

Science, Society and family

Wren resigned the Savilian chair in 1673 and was knighted on 14 November that year. He designed the Royal Observatory at Greenwich in 1675–76 for John Flamsteed, the first Astronomer Royal. He served as President of the Royal Society from 1680 to 1682, after Boyle declined. He married Faith Coghill in 1669; she died in September 1675. He married Jane Fitzwilliam in 1677; she died in 1680. He had four children, of whom his son Christopher later compiled the family memoirs. He sat in Parliament at intervals between 1685 and 1702. His office trained a generation of architects, Nicholas Hawksmoor foremost among them. Among his other buildings are the Library of Trinity College, Cambridge (1676–84), Tom Tower at Christ Church, Oxford, and the Royal Hospital, Chelsea (1682–92).

DiscussionWren’s 1666 plan for London was never built. Using his plan and a map of the City as rebuilt, discuss what a geometrically rational street plan would have gained and what it would have cost the citizens whose property lines it ignored.

Lesson 12

Old age, death, legacy

In 1696 Wren became Surveyor of the new Royal Hospital for Seamen at Greenwich, which he laid out so as to keep open the view of the Queen’s House from the river; he resigned in 1716. From 1699 until his death he was Surveyor of Westminster Abbey. Under Queen Anne and George I, younger rivals and politics turned against him, and on 26 April 1718, aged 85, he was dismissed as Surveyor of the King’s Works after almost half a century.

He spent his last years at his house at Hampton Court and in London, reportedly still working on problems of longitude. He died on 25 February 1723, aged 90, after catching a chill, at his house in St James’s, London. He was buried on 5 March in the crypt of St Paul’s. His son’s Latin epitaph above the grave ends with the line familiar to every visitor: Lector, si monumentum requiris, circumspice, reader, if you seek his monument, look around you.

Legacy

  • Mathematics: the rectification of the cycloid, among the first exact lengths of a curve; a solution of Kepler’s problem; the ruled hyperboloid, now used for cooling towers and lattice towers.
  • Mechanics and astronomy: the correct law of elastic collision, the inverse-square step from Kepler’s third law, and the 1684 challenge that set Halley on the road to Newton.
  • Science as an institution: a founder and President of the Royal Society, and the lecture after which it was founded.
  • Architecture: St Paul’s, the City churches, Greenwich, Chelsea and Trinity Library, buildings in which geometry and statics are as visible as style. His portrait appeared on the Bank of England £50 note from 1981 to 1996.

Final projectChoose one of Wren’s buildings. Identify one structural problem it posed (a span, a dome, a tower), model it with the methods of Lesson 09, and explain how Wren’s solution relates to the mathematics he knew from his scientific work.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Christopher Wren, MacTutor History of Mathematics.
  2. Christopher Wren, Wikipedia.
  3. J. A. Bennett, The Mathematical Science of Christopher Wren, Cambridge University Press, 1982.
  4. L. Jardine, On a Grander Scale: The Outstanding Career of Sir Christopher Wren, HarperCollins, 2002.
  5. K. Downes, The Architecture of Wren, Granada, 1982.
  6. Wren, Christopher, in the Biographical Encyclopedia of Astronomers (via MacTutor).
  7. C. Wren (son) and S. Wren, Parentalia, or Memoirs of the Family of the Wrens, London, 1750.

Mathematics, science and structure

  1. S. Hart, Sir Christopher Wren: Mathematician, Gresham College lecture, 2023 (transcript).
  2. J. Wallis, Tractatus duo, prior de cycloide, Oxford, 1659 (contains Wren’s rectification).
  3. C. Wren, Lex naturae de collisione corporum, Phil. Trans. 3 (1669), 867–868. Royal Society, manuscript in Making Science.
  4. C. Wren, on the generation of the hyperbolical cylindroid and an engine for grinding hyperbolical glasses, Phil. Trans. 4 (1669).
  5. Royal Society, The handles of Saturn (Wren’s Saturn manuscript).
  6. J. Heyman, The Stone Skeleton, Cambridge University Press, 1995.
  7. St Paul’s Cathedral, Sheldonian Theatre and Monument to the Great Fire of London, Wikipedia.