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Gilles Personne de Roberval

Gilles Personne (or Personier), sieur de Roberval

A course in two strands: the life of a Picard farmer’s son who held a royal chair in Paris for forty years by winning it in open contest, and the cycloid, the kinematic tangents, the indivisibles and the balance that he kept, for the most part, to himself.

Born 10 August 1602, near Senlis Died 27 October 1675, Paris Ramus chair won 24 June 1634 Level advanced undergraduate
The roulette. The tracing point moves with the centre and turns about it; the diagonal of the two motions is the tangent, and it passes through the top of the wheel. Dashed: Roberval’s companion curve.
Lesson 00

How to use this course

Gilles Personne de Roberval was the one professional mathematician in the Paris circle of Marin Mersenne, the colleague of Fermat, the Pascals and Descartes, and the enemy of the last. He found the area under the cycloid, a general way to draw tangents by composing motions, and the quadrature of \(y=x^n\) and of the sine, all in the 1630s. He published almost none of it. His name survives today mostly on a kitchen scale.

Prerequisites for Part II
  • Single-variable calculus: integrals, parametric curves, arc length, volumes of revolution.
  • Vectors in the plane: addition, the parallelogram rule.
  • Elementary statics: forces, moments, equilibrium.
  • For Lesson 09, the idea of air pressure and of a pendulum.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, sketches of Roberval’s methods, and exercises with folded solutions. Many of his results survive only in the posthumous Divers ouvrages of 1693 and in letters, so dates are often given as “before” a letter in which a result is mentioned. The course follows the dates established by Kokiti Hara in the Dictionary of Scientific Biography.

Reference

Timeline

1602Born 10 August at Noël-Saint-Martin, near Villeneuve-sur-Verberie and the village of Roberval (Oise), to Pierre Personne and Jeanne Le Dru, farmers.
c. 1616At about 14, begins Latin and mathematics with the parish priest of Rhuis.
1620sTravels through France, earning his living by teaching mathematics; meets Fermat in Bordeaux.
1627At the siege of La Rochelle, studying fortification and ballistics.
1628Arrives in Paris; joins the circle of Mersenne. Takes the name “de Roberval”. Mersenne draws his attention to the cycloid.
1632Professor of philosophy at the Collège de Maître Gervais.
163424 June: wins the competition for the Ramus chair of mathematics at the Collège Royal.
by 1636Area of the cycloid and volume of its solid of revolution about the base; tangents by composition of motions; quadrature of \(y=x^n\). Traité de mécanique published.
1637–38With Fermat, in dispute with Descartes over tangents and the Géométrie.
1642–43Compares the arc lengths of the spiral and the parabola.
1643October: Torricelli writes to him on the cycloid. A priority dispute follows the publication of Torricelli’s Opera geometrica in 1644.
1644Aristarchi Samii de mundi systemate, a heliocentric system with universal attraction.
1646Dispute with Descartes over the centre of oscillation of a compound pendulum.
1647–48Two Narrationes on the vacuum experiments; the carp-bladder experiment; apparatus for the “vacuum within a vacuum”.
1655At the end of the year succeeds Gassendi in a second mathematics chair at the Collège Royal.
1658Pascal’s Histoire de la roulette credits him with the first quadrature of the cycloid.
1666Founding member of the Académie royale des sciences.
166921 August: presents the balance that bears his name to the Académie.
1675Dies in Paris on 27 October, aged 73.
1693Divers ouvrages de M. de Roberval published by the Académie.
Part I · Formation
Lesson 01

A harvest field in the Oise, 1602–1627

Gilles Personne was born on 10 August 1602 at Noël-Saint-Martin, a hamlet of Villeneuve-sur-Verberie near Senlis, north-east of Paris. His parents, Pierre Personne and Jeanne Le Dru, worked the land. According to family tradition his mother gave birth in the fields during the harvest. He had brothers and sisters; only his sister Marie outlived him.

At about 14 he began to study Latin and mathematics with the parish priest of Rhuis, a chaplain of the queen mother Marie de’ Medici. He probably also learned some Greek. There was no question of a university. Instead, by his early twenties he left home and travelled through France, supporting himself by giving lessons in mathematics and seeking out university teachers to discuss harder questions. In Bordeaux he met Pierre de Fermat; the two would correspond for years.

In September 1627 he was at the siege of La Rochelle, the Huguenot stronghold besieged by the army of Louis XIII. There he studied fortification and the motion of projectiles. Descartes was present at the same siege.

Lesson 02

Paris and the Mersenne circle, 1628–1634

In 1628 he arrived in Paris and found his way into the circle of the Minim friar Marin Mersenne. Its members met to read papers and propose problems: Claude Mydorge, Claude Hardy, Étienne Pascal and later his son Blaise. Mersenne valued him highly. Around this time he added “de Roberval” to his name, after the village near his birthplace, with local permission.

Mersenne collected open problems and passed them on. One of them was the cycloid, the curve traced by a point on the rim of a rolling wheel. Galileo had named it and had tried to find its area by weighing cut-out shapes. Mersenne put the problem to Roberval in 1628. Roberval later called the curve the trochoid, from the Greek for wheel; Pascal’s French name roulette also stuck. Lesson 05 tells what he found.

Before 1632 he had also studied the symbolic algebra of François Viète. He wrote a short treatise on the theory of equations, De recognitione aequationum, probably before the publication of Descartes’s Géométrie in 1637, though it goes only as far as the basics. In 1632 he was appointed professor of philosophy at the Collège de Maître Gervais in Paris, where he lived for the rest of his life.

Lesson 03

The Ramus chair and the habit of secrecy

The philosopher Petrus Ramus, killed in the massacre of St Bartholomew’s Day in 1572, had endowed a chair of mathematics at the Collège Royal, the royal foundation that is now the Collège de France. Its terms were unusual. The chair was filled by public competition, and every three years the holder had to defend it against challengers. The holder set problems, and anyone who solved them better could claim the chair.

On 24 June 1634 Roberval won the competition. He then held the chair for 41 years, until his death, defending it every three years. The likely consequence, argued by historians since the eighteenth century, is his secrecy. A method that only he knew was a weapon for the next competition. He let results be known through Mersenne and in letters, kept his methods hidden, and published very little. In his lifetime he printed only two works: a short Traité de mécanique (1636), which appeared in Mersenne’s Harmonie universelle, and the astronomical treatise attributed to Aristarchus (1644).

The price was credit. Bonaventura Cavalieri published his indivisibles in 1635 and his name became attached to them; Evangelista Torricelli published the cycloid in 1644. Roberval spent much of the rest of his career claiming priority, often angrily. Contemporaries described his temper as difficult, and he had a reputation for meanness, though he helped to provide dowries for his relatives. He lived austerely and never married.

At the end of 1655, on the death of Pierre Gassendi, he was also appointed to Gassendi’s mathematics chair at the Collège Royal. This one did not require competition, and he held both chairs until his death.

Part II · The mathematics and mechanics
Lesson 04

Indivisibles and sums of powers

For the general reader. Roberval, like Cavalieri, found areas by cutting figures into very many thin strips. He said he took the idea from Archimedes, not from Cavalieri, whose book he had not seen. His strips were thin but not quite zero: he thought of a line as made of many small lines, and a surface as made of many small surfaces. This made his method arithmetical. Areas became sums of numbers, and the key was a formula for sums of powers.

Goals
  • Understand Roberval’s view of indivisibles as small elements.
  • Prove the power-sum inequalities and deduce \(\int_0^a x^n\,dx=a^{n+1}/(n+1)\).
  • See how the same method gives areas by comparing curves.

The Traité des indivisibles

The Traité des indivisibles, printed in 1693 from his papers, explains that one may treat a line as composed of infinitely many lines, a surface of infinitely many surfaces, a solid of infinitely many solids, as long as the elements are homogeneous with the whole. In practice he divided a segment into \(N\) equal parts, built figures on the parts, computed, and let \(N\) grow. Hara dates his quadrature of the higher parabolas to about 1636. In a letter to Fermat he stated the inequalities that make it work.

Lemma (Roberval)

For all positive integers \(n\) and \(N\),

\[ 1^n+2^n+\cdots+(N-1)^n \;<\; \frac{N^{n+1}}{n+1} \;<\; 1^n+2^n+\cdots+N^n . \]
Theorem (Roberval, Fermat, Cavalieri, 1630s)

For every positive integer \(n\), the region under \(y=x^n\) between \(x=0\) and \(x=a\) has area \(\dfrac{a^{n+1}}{n+1}\), that is, \(\dfrac1{n+1}\) of the rectangle \(a\times a^n\).

Proof

Divide \([0,a]\) into \(N\) equal parts of width \(a/N\). The region is squeezed between the inscribed and circumscribed stacks of rectangles:

\[ \frac aN\sum_{i=0}^{N-1}\Big(\frac{ia}N\Big)^n \;<\; \text{Area} \;<\; \frac aN\sum_{i=1}^{N}\Big(\frac{ia}N\Big)^n . \]

Both bounds equal \(\frac{a^{n+1}}{N^{n+1}}\) times a power sum. By the lemma, the lower bound is less than \(\frac{a^{n+1}}{n+1}\) and the upper bound greater, and the bounds differ by \(\frac{a^{n+1}}{N}\). Since this is true for every \(N\), the area is exactly \(\frac{a^{n+1}}{n+1}\). This is an exhaustion argument in the Greek style with an arithmetical engine, which is why Roberval could honestly say he followed Archimedes.

To prove the lemma, expand \((k+1)^{n+1}-k^{n+1}=(n+1)k^n+\binom{n+1}{2}k^{n-1}+\cdots+1\). This is greater than \((n+1)k^n\). By the mean value theorem, or by the identity \(b^{n+1}-c^{n+1}=(b-c)\sum b^jc^{n-j}\), it is less than \((n+1)(k+1)^n\). Summing over \(k=0,\dots,N-1\) telescopes the left side to \(N^{n+1}\).

Exercise 4.1Verify the lemma for \(n=2\), \(N=4\), and find the exact formula for \(\sum_{i=1}^N i^2\) from the identity \((k+1)^3-k^3=3k^2+3k+1\).

Show solution
\(1+4+9=14<\tfrac{64}3\approx21.33<14+16=30\). Summing the identity from \(k=1\) to \(N\): \((N+1)^3-1=3\sum k^2+\tfrac{3N(N+1)}2+N\), so \(\sum_{k=1}^N k^2=\tfrac{N(N+1)(2N+1)}6\).

Exercise 4.2Using the theorem, find the area enclosed between the curves \(y=x^2\) and \(y=x^3\) on \([0,1]\), and the ratio in which the curve \(y=x^3\) divides the region under \(y=x^2\).

Show solution
Under \(y=x^2\): \(\tfrac13\). Under \(y=x^3\): \(\tfrac14\). Between them: \(\tfrac1{12}\). The curve \(y=x^3\) divides the region under \(y=x^2\) in the ratio \(\tfrac14:\tfrac1{12}=3:1\).
Lesson 05

The roulette: area and solids

For the general reader. Roll a coin along a ruler and follow a point on its rim. The point traces a row of arches, the cycloid. Galileo guessed, by weighing paper models, that one arch encloses about three times the area of the coin. Roberval proved that it is exactly three times, by inventing a second curve that splits the arch into pieces he could measure. The second curve, his “companion of the cycloid”, is the first sine curve drawn in the history of mathematics.

Goals
  • Parametrize the cycloid and its companion curve.
  • Prove that one arch has area three times the generating circle.
  • Compute the volume of the solid generated by rotating the arch about its base.

The curves

Let a circle of radius \(a\) roll along the \(x\)-axis, and let \(P\) be the point of the rim that starts at the origin. After the circle has turned through angle \(t\), its centre is at \((at,a)\), and

\[ P(t)=\big(a(t-\sin t),\ a(1-\cos t)\big),\qquad 0\le t\le2\pi . \]

Roberval’s companion point \(Q(t)\) has the same height as \(P\) but sits directly below the centre of the circle’s horizontal position:

\[ Q(t)=\big(at,\ a(1-\cos t)\big). \]

So \(Q\) traces \(y=a(1-\cos(x/a))\), a sine curve. The horizontal segment \(QP\) has length \(a\sin t\), which is exactly the half-chord of the rolling circle at height \(y\).

Theorem (Roberval, before the end of 1636)

One arch of the cycloid generated by a circle of radius \(a\) encloses, with its base, the area \(3\pi a^2\), three times the area of the generating circle. Rotated about its base, the arch generates a solid of volume \(5\pi^2a^3\).

Roberval’s argument for the area

Take half an arch, \(0\le t\le\pi\), inside the rectangle of width \(\pi a\) and height \(2a\), area \(2\pi a^2\).

  1. Between the cycloid and the companion curve, the horizontal segment at height \(y\) equals the half-chord of the generating circle at that height. By the principle of indivisibles, the region between them has the area of a half-disk, \(\tfrac12\pi a^2\).
  2. The companion curve passes through the centre of the rectangle and is symmetric about it, so it cuts the rectangle into two equal parts of area \(\pi a^2\).
  3. The region under the half-arch is therefore \(\pi a^2+\tfrac12\pi a^2=\tfrac32\pi a^2\), and the whole arch \(3\pi a^2\).

In modern terms the calculation is \(\int_0^{2\pi}a(1-\cos t)\cdot a(1-\cos t)\,dt=a^2\int_0^{2\pi}(1-2\cos t+\cos^2t)\,dt=a^2(2\pi+\pi)=3\pi a^2\).

Roberval told Descartes the result. Descartes answered that it was a pretty result but would give no trouble to any moderately skilled geometer, and set Roberval to find the tangent instead. Roberval also extended his results to the general trochoid, traced by a point inside or outside the rim, and between 1644 and 1645 found the volume of the solid generated by rotating the arch about its axis of symmetry (Lesson 10).

Exercise 5.1Verify the volume \(5\pi^2a^3\) of the solid generated by rotating one arch about its base.

Show solution
By disks, \(V=\pi\int y^2\,dx=\pi\int_0^{2\pi}a^2(1-\cos t)^2\cdot a(1-\cos t)\,dt=\pi a^3\int_0^{2\pi}(1-\cos t)^3\,dt\). Expand: \(1-3\cos t+3\cos^2t-\cos^3t\). Over a period the odd powers of cosine integrate to 0 and \(\int\cos^2=\pi\), so the integral is \(2\pi+3\pi=5\pi\). Hence \(V=5\pi^2a^3\).

Exercise 5.2Check step 1 of Roberval’s argument by computing \(\int_0^{2a}|QP|\,dy\) over the half-arch.

Show solution
\(|QP|=a\sin t\) and \(dy=a\sin t\,dt\). So \(\int_0^{\pi}a\sin t\cdot a\sin t\,dt=a^2\cdot\tfrac\pi2=\tfrac12\pi a^2\), a half-disk.

Exercise 5.3Find the centroid height \(\bar y\) of the region under one arch, and use the Pappus–Guldin theorem to check the volume in Exercise 5.1.

Show solution
\(\bar y=\frac1A\int\tfrac12y^2\,dx=\frac{1}{3\pi a^2}\cdot\tfrac12a^3\cdot5\pi=\tfrac56a\). Pappus–Guldin: \(V=2\pi\bar y\,A=2\pi\cdot\tfrac56a\cdot3\pi a^2=5\pi^2a^3\).
Lesson 06

Tangents by composition of motions

For the general reader. Swing a stone on a string and let go: it flies off along the tangent to the circle. Roberval made this into a method. Regard a curve as the path of a moving point. If the motion is a combination of simpler motions, add their velocities like forces, by the parallelogram rule. The result points along the tangent. He developed the method before 1636 and taught it from 1639 to 1644; his pupil François du Verdus wrote up the lessons, published in 1693 as Observations sur la composition des mouvemens.

Goals
  • State the kinematic principle of tangents.
  • Find the tangents to the cycloid, the parabola and the ellipse.
  • Compare the method with the derivative.
Principle (Roberval)

Let a curve be traced by a point whose motion is composed of motions with velocities \(\mathbf v_1,\dots,\mathbf v_k\) at a given instant. Then the tangent at that point has the direction of \(\mathbf v_1+\cdots+\mathbf v_k\).

In modern terms, if \(\gamma(t)=\gamma_1(t)+\gamma_2(t)\) then \(\gamma'(t)=\gamma_1'(t)+\gamma_2'(t)\), and \(\gamma'(t)\) is tangent to \(\gamma\) when it is non-zero. The principle is the linearity of the derivative, used before the derivative existed. The difficulty, which critics noted, is to choose the component motions correctly and to know their relative speeds.

The cycloid

The tracing point moves with the centre of the wheel, velocity \(a\omega\) horizontally, and turns about the centre, velocity \(a\omega\) along the circle. The two components have equal length, so their sum bisects the angle between them. A short calculation shows that the bisector passes through the highest point of the rolling circle; equivalently, the normal passes through the point of contact with the ground.

Conics

A point on an ellipse moves so that the sum of its distances to the foci is constant. As it moves, it recedes from one focus exactly as fast as it approaches the other. So its velocity has equal components along the two focal radii, one away from \(F_1\), one toward \(F_2\), and the tangent bisects the angle between \(\overrightarrow{F_1P}\) and \(\overrightarrow{PF_2}\), the external angle at \(P\) of the triangle \(F_1PF_2\). For the parabola the same reasoning uses the focus and the directrix.

Exercise 6.1With \(P(t)=(a(t-\sin t),a(1-\cos t))\), show that \(P'(t)\) is parallel to the vector from \(P\) to the top of the rolling circle, \(T=(at,2a)\).

Show solution
\(P'(t)=a(1-\cos t,\ \sin t)\). \(T-P=a(\sin t,\ 1+\cos t)\). The cross product is \(a^2[(1-\cos t)(1+\cos t)-\sin t\sin t]=a^2[1-\cos^2t-\sin^2t]=0\). So they are parallel.

Exercise 6.2For the parabola with focus \(F\) and directrix \(d\), use Roberval’s method to show that the tangent at \(P\) bisects the angle between \(PF\) and the perpendicular from \(P\) to \(d\).

Show solution
The defining property is \(|PF|=\mathrm{dist}(P,d)\). As \(P\) moves, these two distances change at the same rate. So the velocity has equal components along the unit vector \(\mathbf u\) from \(F\) toward \(P\) and along the unit vector \(\mathbf w\) perpendicular to \(d\), pointing away from it: \(\mathbf v\cdot\mathbf u=\mathbf v\cdot\mathbf w\), i.e. \(\mathbf v\cdot(\mathbf u-\mathbf w)=0\). The tangent is perpendicular to \(\mathbf u-\mathbf w\), hence along \(\mathbf u+\mathbf w\), the bisector of the angle between them. This is the reflective property of the parabola.

Exercise 6.3Show that the method needs care: the Archimedean spiral \(r=c\theta\) is traced by a point moving outward along a rotating ray at speed \(c\omega\) while the ray turns at rate \(\omega\). Find the angle \(\psi\) between the tangent and the radius at distance \(r\).

Show solution
The radial component of velocity is \(c\omega\) and the transverse component is \(r\omega\). Their sum makes angle \(\psi\) with the radius where \(\tan\psi=\frac{r\omega}{c\omega}=\frac rc=\theta\). The trap is to forget that the transverse speed grows with \(r\): the two motions are not uniform in the same sense, and getting their ratio right is the whole difficulty.
Lesson 07

The sine, and the lengths of curves

For the general reader. The companion curve of the cycloid led Roberval to the area under a sine curve, one of the first integrals of a trigonometric function. He also compared lengths of curves, a question the Greeks had abandoned as hopeless. He showed that a spiral and a parabola can have equal length, and he claimed, before 1640, to know the length of the cycloid.

Goals
  • Compute \(\int\sin\) as Roberval did, by indivisibles on the circle.
  • Prove the equality of arc lengths of the Archimedean spiral and a parabola.
  • Find the length of the cycloid arch.
Theorem (Roberval, 1630s)

For \(0\le\theta\le\pi\), \(\displaystyle\int_0^\theta\sin\varphi\,d\varphi=1-\cos\theta\). In particular one arch of the sine curve \(y=\sin x\) has area 2.

Roberval’s idea, reconstructed. Divide the arc of the unit circle from angle 0 to \(\theta\) into equal small arcs \(d\varphi\). Each small arc, almost straight, has a horizontal projection of length \(\sin\varphi\,d\varphi\) onto the horizontal diameter, by similar triangles. The projections tile the segment of the diameter from \(\cos\theta\) to 1. So the sum of the sines times the arcs equals \(1-\cos\theta\). He used this to solve a problem he liked to set: to trace on a right circular cylinder, with a single motion of a compass, a region equal in area to a given square.

Theorem (Roberval, winter 1642–43)

The arc of the Archimedean spiral \(r=c\theta\) from \(\theta=0\) to \(\theta=\Theta\) has the same length as the arc of the parabola \(y=x^2/(2c)\) from \(x=0\) to \(x=c\Theta\).

Proof. In polar coordinates, \(ds^2=dr^2+r^2d\theta^2\), so along the spiral \(ds=c\sqrt{1+\theta^2}\,d\theta\). Along the parabola, \(ds=\sqrt{1+x^2/c^2}\,dx\). Substituting \(x=c\theta\) gives the same element \(c\sqrt{1+\theta^2}\,d\theta\). Roberval obtained this by comparing the motions that generate the two curves.

Theorem (rectification of the cycloid)

One arch of the cycloid generated by a circle of radius \(a\) has length \(8a\), four times the diameter of the circle.

Roberval claimed this before 1640, by reducing it to an integral of the sine; Hara notes there is no independent evidence of the claim. Christopher Wren found a proof in 1658 in response to Pascal’s challenge, and Torricelli had a version around 1645. The proof is Exercise 7.2.

Exercise 7.1Compute the area under \(y=a(1-\cos(x/a))\) on \([0,\pi a]\) and confirm step 2 of Lesson 05.

Show solution
\(\int_0^{\pi a}a(1-\cos(x/a))\,dx=\pi a^2-a^2[\sin(x/a)]_0^{\pi a}=\pi a^2\): half the rectangle of area \(2\pi a^2\).

Exercise 7.2Prove that one arch of the cycloid has length \(8a\).

Show solution
\(|P'(t)|^2=a^2[(1-\cos t)^2+\sin^2t]=2a^2(1-\cos t)=4a^2\sin^2(t/2)\). So \(|P'(t)|=2a\sin(t/2)\) on \([0,2\pi]\), and the length is \(\int_0^{2\pi}2a\sin(t/2)\,dt=[-4a\cos(t/2)]_0^{2\pi}=8a\). The reduction to an integral of a sine is exactly the step Roberval claimed.

Exercise 7.3Compute the length of the first turn of the spiral \(r=\theta\) (\(0\le\theta\le2\pi\)) in closed form.

Show solution
\(L=\int_0^{2\pi}\sqrt{1+\theta^2}\,d\theta=\tfrac12\big[\theta\sqrt{1+\theta^2}+\ln(\theta+\sqrt{1+\theta^2})\big]_0^{2\pi}\approx\tfrac12(6.2832\times6.3623+2.5373)\approx21.26\). This equals the length of \(y=x^2/2\) from 0 to \(2\pi\).
Lesson 08

Statics: inclined planes and the balance

For the general reader. Roberval’s first publication, the Traité de mécanique of 1636, asked how much effort holds a weight on a slope, and answered by showing how forces pulling in different directions combine. Thirty-three years later he showed the Académie a balance with an odd property: the weighing comes out the same wherever you put the load on the pan. Versions of it are still in shops and kitchens.

Goals
  • Derive the law of the inclined plane and the composition of forces.
  • Explain the Roberval balance by moments and by virtual work.

The inclined plane and two cords

The treatise studies a weight held on an inclined plane by a force, then the same weight hanging from two cords. Following a route close to that of Simon Stevin, and reducing the inclined plane to a balance in the manner of Galileo, Roberval obtained what is now the parallelogram law for forces. He also gave a clear account of the pressure that a body exerts on the plane supporting it.

Inclined plane

A weight \(W\) rests on a smooth plane inclined at angle \(\alpha\). A force \(F\) along the plane holds it in equilibrium if and only if \(F=W\sin\alpha\). The plane then exerts a normal reaction \(N=W\cos\alpha\).

The balance

On 21 August 1669 Roberval presented to the Académie the plan of a balance whose pans sit above the beam. Two parallel horizontal beams, pivoted at their middles on a central post, are joined at each end by a vertical rod that carries a pan. The four bars form a parallelogram on each side, so the rods stay vertical and the pans stay level as the beams tip.

Theorem (Roberval balance)

In an ideal frictionless Roberval balance, the condition for equilibrium is equality of the two loads, independent of where each load sits on its pan.

Virtual-work proof. Tip the beams through a small angle \(\delta\theta\). Each vertical rod is translated without rotation, so every point of the left pan, wherever it is, rises by the same \(\ell\,\delta\theta\), and every point of the right pan falls by \(\ell\,\delta\theta\), where \(\ell\) is the half-length of the beams. The work done by gravity is \((W_2-W_1)\ell\,\delta\theta\), which vanishes for all \(\delta\theta\) exactly when \(W_1=W_2\). Hara notes that Roberval’s own analysis contains this idea of “moment” as weight times displacement, which is an early form of the principle of virtual work.

Exercise 8.1A weight \(W\) on a smooth plane inclined at \(\alpha\) is held by a force \(F\) making angle \(\beta\) with the plane (pulling up and away). Find \(F\) and the normal reaction.

Show solution
Along the plane: \(F\cos\beta=W\sin\alpha\), so \(F=\frac{W\sin\alpha}{\cos\beta}\). Normal: \(N+F\sin\beta=W\cos\alpha\), so \(N=W\cos\alpha-W\sin\alpha\tan\beta\). The force is least when \(\beta=0\).

Exercise 8.2A load \(W\) sits on the left pan of a Roberval balance at horizontal distance \(d\) outside the vertical rod. The upper and lower beams are a vertical distance \(h\) apart. Find the horizontal forces that the rod exerts on the two beams, and explain why they do not affect the balance.

Show solution
The load exerts on the rod a downward force \(W\) and a moment \(Wd\). The moment is balanced by horizontal forces \(\pm Wd/h\) from the beams: the upper beam pulls the rod inward with \(Wd/h\) and the lower one pushes outward with \(Wd/h\). These forces act along the beams, through their central pivots, so they have no moment about the pivots. Only the vertical force \(W\), at arm \(\ell\), turns the beams.
Lesson 09

The vacuum, the pendulum and the system of the world

For the general reader. In the 1640s Roberval took part in three debates in physics. He helped interpret the new mercury tube of Torricelli, with an experiment using a fish’s bladder. He argued with Descartes about where the “centre of oscillation” of a swinging body lies. And under the name of an ancient Greek astronomer he published a sun-centred universe held together by mutual attraction.

Goals
  • Explain the barometer and Roberval’s carp-bladder experiment.
  • Compute centres of oscillation of compound pendulums.
  • Place the Aristarchus of 1644 between Kepler and Newton.

The vacuum

Torricelli’s experiment of 1644 left an apparently empty space above a column of mercury. In Rouen and Paris, Blaise Pascal repeated it with tubes of many shapes. Roberval wrote two Latin reports, or Narrationes. The first, dated 20 September 1647, described Pascal’s experiments and his own, and concluded that the space, if not empty, did not contain the substances the philosophers proposed. The second, from 1648, explained the suspension of the mercury by the pressure of the outside air. He placed a carp’s bladder, almost emptied of air and tied off, in the space at the top of the tube, and it swelled: the little air shut inside it expanded, so the space around it pressed on it hardly at all. His apparatus was the prototype of Pascal’s experiment of the “vacuum within a vacuum”. He avoided declaring whether a true vacuum exists, but he rejected Descartes’s identification of space with matter.

Hydrostatic balance

A column of liquid of density \(\rho\) and height \(h\), closed at the top by a near-vacuum, is held by atmospheric pressure \(p_0\) when \(p_0=\rho g h\). For mercury (\(\rho\approx13\,600\ \mathrm{kg/m^3}\)) at sea level (\(p_0\approx101\ \mathrm{kPa}\)), \(h\approx0.76\ \mathrm m\).

The compound pendulum

Mersenne asked for the length of a simple pendulum that swings in step with a given rigid body, the body’s centre of oscillation. In 1646 Descartes and Roberval argued over it. Descartes had the better idea of what the centre of oscillation is but neglected the directions of the forces on the parts of the body; Roberval corrected this but doubted that the point he found was exactly right. Christiaan Huygens settled the problem in 1673.

Theorem (Huygens, 1673; modern form)

A rigid body of mass \(m\), swinging about a horizontal axis at distance \(d\) from its centre of mass, with moment of inertia \(I\) about the axis, oscillates for small amplitudes like a simple pendulum of length \(L=I/(md)\).

Aristarchus

In 1644 Roberval published Aristarchi Samii de mundi systemate, presented as a work of the Greek astronomer Aristarchus of Samos, recovered from an Arabic manuscript, with notes. The text was his own. It described a heliocentric system and grounded it on a universal attraction between all parts of matter, an idea he drew from Kepler. In the dedication he wrote that perhaps all three systems, those of Ptolemy, Tycho and Aristarchus, were false and the true one unknown, though that of Aristarchus seemed the simplest. The disguise gave cover in the years after Galileo’s condemnation.

Exercise 9.1Water has density \(1000\ \mathrm{kg/m^3}\). How tall would a water barometer be? A bladder holding \(0.1\ \mathrm{cm^3}\) of air at atmospheric pressure (760 mmHg) is carried into a space where the pressure is 10 mmHg. Assuming Boyle’s law (stated two decades later), what volume would the air occupy?

Show solution
\(h=p_0/(\rho g)\approx101\,000/(1000\times9.81)\approx10.3\ \mathrm m\). Boyle: \(pV\) constant, so \(V\approx0.1\times760/10=7.6\ \mathrm{cm^3}\), a 76-fold expansion: a nearly flat bladder becomes visibly inflated, as Roberval observed.

Exercise 9.2Find the centre of oscillation of a uniform rod of length \(\ell\) hung from one end, and of a uniform disk of radius \(R\) hung from a point of its rim, swinging in its own plane.

Show solution
Rod: \(I=\tfrac13m\ell^2\), \(d=\ell/2\), so \(L=\tfrac23\ell\). Disk: \(I=\tfrac12mR^2+mR^2=\tfrac32mR^2\), \(d=R\), so \(L=\tfrac32R\). In both cases the centre of oscillation lies below the centre of mass, which is what made the problem hard for Descartes and Roberval.
Lesson 10

Trochoids and the solid about the axis

For the general reader. A point on a wheel’s spoke, inside the rim, traces a wavy curve that never touches the ground; a point on a train wheel’s flange, outside the rim, traces loops. Roberval called all these curves trochoids and extended his cycloid results to them. Between May 1644 and October 1645 he also solved a harder problem: the volume of the solid made by spinning the cycloid arch about its axis of symmetry, the vertical line through its top. He set his results down in a piece titled Ad trochoidem, ejusque solida.

Goals
  • Parametrize the general trochoid and find its area by the companion-curve method.
  • Compute the volume of the cycloidal solid about the axis by shells.

The general trochoid

Let a circle of radius \(a\) roll on the \(x\)-axis and follow a point at distance \(b\) from its centre. The point traces

\[ P(t)=\big(at-b\sin t,\ a-b\cos t\big). \]

For \(b<a\) the curve is a curtate trochoid, for \(b=a\) the cycloid, for \(b>a\) a prolate trochoid with loops. The companion point \(Q(t)=(at,\ a-b\cos t)\) lies at the same height, and the horizontal gap \(|QP|=b\sin t\) is the half-chord at that height of the circle of radius \(b\) described by the point about the centre.

Theorem (area of the trochoid)

For \(b\le a\), the region between one arch of the trochoid (\(0\le t\le2\pi\)) and the line \(y=a-b\) through its lowest points has area \(2\pi ab+\pi b^2\).

Proof by the companion curve. Between the trochoid and its companion, the horizontal segments are the half-chords of a circle of radius \(b\), so over the two halves of the arch the regions between them add up to a full disk, \(\pi b^2\). Measured from the line \(y=a-b\), the companion \(y=b(1-\cos(x/a))\) has area \(2\pi ab\) under it on \([0,2\pi a]\), half of the rectangle \(2\pi a\times2b\) by central symmetry. With \(b=a\) the total is \(3\pi a^2\), as in Lesson 05.

The solid about the axis

Theorem (Roberval, 1644–45)

The region under one arch of the cycloid of radius \(a\), rotated about the vertical line through the top of the arch, generates a solid of volume

\[ V=\pi a^3\Big(\frac{3\pi^2}{2}-\frac83\Big). \]

Derivation by shells. The half-arch \(0\le t\le\pi\) sweeps the whole solid. A vertical strip at abscissa \(x\), of height \(y\), lies at distance \(\pi a-x\) from the axis and sweeps a cylindrical shell of volume \(2\pi(\pi a-x)\,y\,dx\). With \(x=a(t-\sin t)\), \(y=a(1-\cos t)\), \(dx=a(1-\cos t)\,dt\):

\[ V=2\pi a^3\int_0^{\pi}\big(\pi-t+\sin t\big)(1-\cos t)^2\,dt . \]

Expand \((1-\cos t)^2=\tfrac32-2\cos t+\tfrac12\cos2t\) and use \(\int_0^\pi(\pi-t)\,dt=\tfrac{\pi^2}{2}\), \(\int_0^\pi(\pi-t)\cos t\,dt=2\), \(\int_0^\pi(\pi-t)\cos2t\,dt=0\), and \(\int_0^\pi\sin t(1-\cos t)^2dt=\tfrac83\). The first part gives \(\tfrac32\cdot\tfrac{\pi^2}2-2\cdot2=\tfrac{3\pi^2}4-4\). So \(V=2\pi a^3\big(\tfrac{3\pi^2}{4}-4+\tfrac83\big)=\pi a^3\big(\tfrac{3\pi^2}{2}-\tfrac83\big)\approx38.13\,a^3\).

Exercise 10.1Verify the integral \(\int_0^\pi(\pi-t)\cos t\,dt=2\) and \(\int_0^\pi\sin t\,(1-\cos t)^2\,dt=\tfrac83\).

Show solution
Substitute \(u=\pi-t\): \(\int_0^\pi u\cos(\pi-u)\,du=-\int_0^\pi u\cos u\,du=-[u\sin u+\cos u]_0^\pi=-(-1-1)=2\). For the second, put \(w=1-\cos t\), \(dw=\sin t\,dt\), \(w\) from 0 to 2: \(\int_0^2w^2\,dw=\tfrac83\).

Exercise 10.2Check the trochoid area formula directly from \(\int(y-(a-b))\,dx\) for \(b\le a\).

Show solution
\(y-(a-b)=b(1-\cos t)\), \(dx=(a-b\cos t)\,dt\). So the area is \(\int_0^{2\pi}b(1-\cos t)(a-b\cos t)\,dt=b\int_0^{2\pi}\big(a-(a+b)\cos t+b\cos^2t\big)dt=b(2\pi a+\pi b)=2\pi ab+\pi b^2\).

Exercise 10.3For which \(b\) does the trochoid have self-intersections, and why does the area formula then need care?

Show solution
\(x'(t)=a-b\cos t\) changes sign exactly when \(b>a\): the point moves backwards near the bottom of each turn and the curve forms loops. The integral \(\int(y-(a-b))\,dx\) then counts the loop regions with a negative sign, so it gives a signed area, not the area of the region enclosed by the outer boundary.
Part III · Life and legacy
Lesson 11

Quarrels: Descartes, Torricelli, Pascal

Descartes

Descartes was Roberval’s “irreconcilable adversary”, in Hara’s phrase. In 1637–38, after the publication of the Discours de la méthode with the Géométrie and the Dioptrique, Roberval sided with Fermat and Jean de Beaugrand against Descartes in a long dispute, conducted through Mersenne, over methods for tangents and maxima. Descartes challenged Roberval to find the tangent to the cycloid; Roberval did not produce it at once, while Fermat and Descartes did. The curve now called the folium of Descartes entered the same exchanges; Roberval thought it formed four petals and called it the fleur de jasmin, but only one loop is real. They quarrelled again in 1646 over the compound pendulum, and in 1647 over the vacuum. Jacqueline Pascal described in a letter of 25 September 1647 a meeting at which Descartes visited her brother and Roberval defended Blaise’s experiments against him with some heat.

Torricelli

In October 1643 Torricelli wrote to Roberval about the cycloid. In 1644 his Opera geometrica published the area of the cycloid and its tangent. Roberval, who had the area by 1636 but had not published it, accused Torricelli of taking the results from papers sent from Paris to Galileo’s circle. Modern historians generally find that Torricelli worked independently. The dispute was unresolved at Torricelli’s death in 1647. Their correspondence also touched mechanics: in a 1647 letter Roberval outlined a treatise in eight books on forces, the balance, cords, machines, compound motions and the centre of percussion, which was never published.

Pascal

Roberval had known Blaise Pascal since Blaise was a boy in Mersenne’s circle and supported him in the vacuum controversy. In 1658 Pascal set his challenge problems on the cycloid under the name Amos Dettonville, and his Histoire de la roulette (10 October 1658) took Roberval’s side against Torricelli on the priority for the area. Roberval was among the judges who awarded no prize. Pascal also showed in 1659 that arcs of a generalized cycloid equal arcs of an ellipse, a result Roberval may have known before August 1648.

Exercise 11.1The folium of Descartes is \(x^3+y^3=3axy\). Parametrize it by \(y=tx\), find the tangent at the point where \(t=1\), and explain why Roberval’s four petals are impossible.

Show solution
Substituting \(y=tx\): \(x=\frac{3at}{1+t^3}\), \(y=\frac{3at^2}{1+t^3}\). At \(t=1\) the point is \((\tfrac32a,\tfrac32a)\). By symmetry in \(x\leftrightarrow y\), the tangent there is perpendicular to \(y=x\): \(x+y=3a\). For the loop in the third quadrant one would need \(x,y<0\), but then \(x^3+y^3<0<3axy\) (for \(a>0\)), so no point of the curve lies there. The only loop is in the first quadrant.

DiscussionRoberval protected his chair by keeping methods secret and lost priority for most of his discoveries. Torricelli and Cavalieri published and gained it. Was Roberval’s choice rational under the rules of the Ramus chair? What does the episode suggest about how institutions shape credit in science?

Lesson 12

The Académie, death and legacy

On 22 December 1666 the Académie royale des sciences, founded under Jean-Baptiste Colbert, held its first meeting in the king’s library. Roberval was one of the founding mathematicians, with Christiaan Huygens, Pierre de Carcavi, Adrien Auzout, Bernard Frénicle de Bessy and Jacques Buot. He received a royal pension and was active in the Académie’s early years, presenting his balance there in 1669 and drafting around the same time a plan for a book of mechanics founded on composed motions. He also contributed on optics to Mersenne’s posthumous L’optique et la catoptrique (1651), with a study of the oval curves of Descartes’s Dioptrique.

He died in Paris on 27 October 1675, aged 73, still holding both chairs at the Collège Royal. A considerable sum of money was found in his rooms. He left his papers to the Académie, which published a selection in 1693 as Divers ouvrages de M. de Roberval: the treatises on indivisibles, on the composition of motions and tangents, on the trochoid and on equations, and letters.

Legacy

  • Calculus before calculus. The quadrature of \(y=x^n\) and of the sine, the cycloid’s area and solids, and the comparison of arc lengths place him with Cavalieri, Fermat and Torricelli among the immediate precursors of Newton and Leibniz.
  • Kinematic geometry. Tangents by composition of motions anticipate Newton’s fluxions in spirit and are standard today as the rule that the velocity of a sum is the sum of the velocities.
  • The sine curve. His companion curve of the cycloid is the first appearance of the graph of the sine.
  • Mechanics. The parallelogram of forces in 1636 and the balance of 1669, with its early form of the principle of virtual work.
  • Physics. The interpretation of the barometer by air pressure, and a universal attraction stated four decades before the Principia.
  • Names. The Roberval balance, the lunar crater Roberval, and the town of Roberval in Quebec, named after a different man (Jean-François de La Rocque de Roberval).

Final projectWrite a two-page history of the cycloid from Galileo to Johann Bernoulli’s brachistochrone. Include Roberval’s area proof in full, a tangent construction, the arc length, and a paragraph on the priority disputes.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Gilles Personne de Roberval, MacTutor History of Mathematics.
  2. K. Hara, Roberval, Gilles Personne (or Personier) de, Dictionary of Scientific Biography.
  3. Gilles de Roberval and Gilles Personne de Roberval, Wikipedia (English and French).
  4. Linda Hall Library, Scientist of the Day: Gilles Personne de Roberval.
  5. L. Auger, Un savant méconnu : Gilles Personne de Roberval (1602–1675), Blanchard, Paris, 1962.
  6. D. J. Sturdy, Science and Social Status: The Members of the Académie des Sciences 1666–1750, Boydell, 1995.
  7. Museo Galileo, Roberval and the vacuum experiments.

Mathematics and mechanics

  1. Divers ouvrages de M. de Roberval, Académie royale des sciences, Paris, 1693.
  2. G. P. de Roberval, Aristarchi Samii de mundi systemate, Paris, 1644.
  3. MacTutor, Cycloid and Folium of Descartes.
  4. B. Hughes, The companion curves of Gilles Personne de Roberval (2002).
  5. P. R. Wolfson, The crooked made straight: Roberval and Newton on tangents, American Mathematical Monthly 108 (2001).
  6. J. Itard, La lettre de Torricelli à Roberval d’octobre 1643, Revue d’histoire des sciences (1975).
  7. V. Jullien, Descartes–Roberval, une relation tumultueuse (1998).
  8. Roberval balance, Wikipedia.
  9. E. A. Whitman, Some historical notes on the cycloid, American Mathematical Monthly 50 (1943).