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Gaspard-Gustave Coriolis

Gaspard-Gustave de Coriolis

A course in two strands: the life of a frail engineer-professor of the École polytechnique, and the mechanics he left behind: work and kinetic energy, the theory of relative motion that carries his name, and the physics of billiards.

Born 21 May 1792, Paris Died 19 September 1843, Paris Compound centrifugal force 1835 Level advanced undergraduate
A puck crosses a turntable that turns anticlockwise. In the room it moves straight (dashed). Traced on the turntable, the same motion curves to the right: the compound centrifugal force \(-2m\,\boldsymbol\omega\times\mathbf v_{\text{rel}}\) at work.
Lesson 00

How to use this course

Every student of meteorology and oceanography learns the Coriolis force, and most assume that Gaspard-Gustave Coriolis studied winds or the rotation of the Earth. He did not. He was an engineer who wanted to know how much useful effect a machine delivers. That question led him to fix the modern meanings of work and kinetic energy, and then to ask how the energy balance changes when parts of a machine rotate. The force that bears his name came out of waterwheels and turbines, not weather.

Prerequisites for Part II
  • Newtonian mechanics of particles and rigid bodies: momentum, angular momentum, moment of inertia.
  • Vector calculus: cross products, line integrals, time derivatives of vectors.
  • For Lesson 08, the idea of pressure gradients in a fluid.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. The lessons follow his work: the theory of work (1829), machines and friction, energy in rotating frames (1831–1832), the compound centrifugal force (1835), its later life on the rotating Earth, billiards (1835), and turbines.

Reference

Timeline

1792Born 21 May in Paris, eldest child of Jean-Baptiste-Elzéar de Coriolis, an officer, and Marie-Sophie de Maillet. The family soon leaves Paris for Nancy.
1808Enters the École polytechnique.
1810Leaves Polytechnique for the corps of Ponts et Chaussées (bridges and roads).
1810–16Engineering service in the provinces, ending in the Vosges; serious illness.
1816After his father’s death, becomes répétiteur in analysis at the École polytechnique on Augustin-Louis Cauchy’s recommendation.
1829Du calcul de l’effet des machines: work and vis viva in their modern form. Professor of mechanics at the new École centrale des arts et manufactures.
1830Cauchy leaves France; Coriolis declines the offer of his post at Polytechnique.
1831–32Memoir on the principle of vis viva in the relative motion of machines. Teaches applied mechanics with Claude-Louis Navier at the École des ponts et chaussées.
1835Sur les équations du mouvement relatif des systèmes de corps: the compound centrifugal force. Théorie mathématique des effets du jeu de billard.
1836Succeeds Navier in the chair of applied mechanics at Ponts et Chaussées and in the mechanics section of the Académie des sciences.
1838Director of studies at the École polytechnique, succeeding Pierre Louis Dulong.
1843Health declines in the spring. Dies in Paris on 19 September, aged 51.
1844Posthumous Traité de la mécanique des corps solides.
Part I · Formation
Lesson 01

Paris and Nancy, 1792–1808

Gaspard-Gustave de Coriolis was born in Paris on 21 May 1792, the eldest of six children. His father, Jean-Baptiste-Elzéar de Coriolis, came from a noble family of Provence. He had served as an officer in the Rochambeau expedition during the American War of Independence and in 1792 was in the King’s guard. His mother, Marie-Sophie de Maillet, came from a noble family of Lorraine.

The year of his birth was the year the monarchy fell. The King’s guard was dissolved, and the family, exposed by its rank, left Paris for Nancy in Lorraine. There his father turned to industry, running a tobacco manufactory and later a wallpaper business. Coriolis grew up and went to school in Nancy, where his talent for mathematics showed early.

DiscussionCoriolis, Lazare Carnot, Gaspard Monge and many other French scientists of this period had their lives shaped by the Revolution. Compare how the Revolution affected a noble family like the Coriolis with how it opened careers to men of modest birth.

Lesson 02

Polytechnique and the bridges corps, 1808–1816

In 1808 Coriolis passed the entrance examination of the École polytechnique, near the top of his year. The school, founded in 1794, gave a common mathematical and scientific training to the future engineers of the state. Its best graduates chose the corps of Ponts et Chaussées, which built France’s roads, bridges and canals. Coriolis left the school in 1810 and entered this corps.

His years as a working engineer were spent in the provinces, and they ended in the Vosges, where he served near his family in Lorraine. His health was already fragile: a serious fever in 1812 left him weak, and illness recurred throughout his life. He missed the theoretical work of the school. The chance to return came through misfortune: his father died, the family needed support, and in 1816 a teaching post opened in Paris.

Lesson 03

Répétiteur at the École polytechnique, 1816–1829

In 1816 Augustin-Louis Cauchy recommended Coriolis for a post as répétiteur in analysis at the École polytechnique. A répétiteur held small-group sessions that complemented the professor’s lectures, so Coriolis spent the next decade teaching Cauchy’s rigorous analysis to future engineers. He used the time to experiment on friction and hydraulics, and to think about what mechanics should offer to people who built machines.

The tradition he joined had roots in Lazare Carnot’s essay on machines of 1783, which analysed machines in terms of the products of forces and displacements. In the 1820s three engineers trained at Polytechnique took up the project of a mechanics for industry: Claude-Louis Navier, Jean-Victor Poncelet and Coriolis. Poncelet showed Coriolis his work on the subject in 1824. In 1829 both published books in which the product of force and displacement was the central quantity. Coriolis gave it its name, travail, and Poncelet acknowledged his priority in the word.

DiscussionNavier, Poncelet and Coriolis all reached the concept of work at about the same time. What features of French engineering education and industry in the 1820s made the concept ripe?

Part II · The mechanics
Lesson 04

Work and vis viva (1829)

For the general reader. How much has a horse, a waterfall or a steam engine accomplished? Coriolis answered with a single number: the force exerted multiplied by the distance moved. He called it work. He then adjusted the old notion of “living force” by a factor of one half, so that the work done on a body equals exactly the change in its energy of motion. Physics has used both choices ever since.

Goals
  • Define work as a line integral and prove the work–energy theorem.
  • Understand why Coriolis put the factor \(\tfrac12\) in vis viva.
  • State the principle of transmission of work for a machine.

Definitions

Gottfried Wilhelm Leibniz had called \(mv^2\) the vis viva of a body. With this convention the energy balance of a falling body reads \(mv^2=2mgh\), and a factor of two appears in every balance. In Du calcul de l’effet des machines (1829), Coriolis proposed to call \(\tfrac12mv^2\) the vis viva. With his convention, the work of the forces equals the change in vis viva with no factor at all. He also proposed a unit, the dynamode, equal to 1000 kilogram-force metres. The word work survived; the dynamode did not.

Theorem (work–energy, Coriolis’s form)

Let a particle of mass \(m\) move along a path \(\gamma\) from time \(t_0\) to \(t_1\) under a total force \(\mathbf F\). Define the work \(W=\int_\gamma\mathbf F\cdot d\mathbf r\). Then

\[ W=\tfrac12mv_1^2-\tfrac12mv_0^2 . \]

For a system of particles, the sum of the works of all forces, internal and external, equals the change in total vis viva \(T=\sum\tfrac12m_iv_i^2\).

Proof. \(\mathbf F\cdot d\mathbf r=m\dot{\mathbf v}\cdot\mathbf v\,dt=\frac{d}{dt}\big(\tfrac12m\,\mathbf v\cdot\mathbf v\big)dt\). Integrate, and sum over particles.

Transmission of work

A machine receives work from a motor (water, animals, steam) and delivers it to a resistance (a load to lift, a millstone to turn). Over one cycle of steady operation, the vis viva returns to its starting value, so the theorem gives an exact budget.

Principle of transmission of work

For a machine running periodically, over one period

\[ W_{\text{motor}}=W_{\text{useful}}+W_{\text{friction}}+W_{\text{impacts}}, \]

where the last two terms are non-negative losses. The efficiency \(\eta=W_{\text{useful}}/W_{\text{motor}}\) is therefore at most 1, and no arrangement of gears or levers can increase it.

This is the precise content of the old engineering maxim that what is gained in force is lost in speed. It tells the engineer where to look for improvement: at friction and shocks, never at mechanisms promising more work out than in.

Exercise 4.1Express the dynamode in joules. The French cheval-vapeur was 75 kilogram-force metres per second. How many seconds does such an engine take to produce one dynamode, and how many dynamodes per hour?

Show solution
One kilogram-force metre is \(9.80665\) J, so a dynamode is \(9806.65\) J \(\approx9.8\) kJ. At 75 kgf·m/s, one dynamode takes \(1000/75\approx13.3\) s, so the engine delivers \(3600/13.3=270\) dynamodes per hour.

Exercise 4.2A pile driver drops a 300 kg ram from 2 m onto a 200 kg pile. The impact is perfectly inelastic, and ram and pile then sink 2 cm together against a constant resistance \(R\). Ignoring the weights during penetration, find \(R\) and the fraction of the work lost in the impact.

Show solution
The ram arrives with \(T=mgh=300\cdot9.81\cdot2=5886\) J. Momentum is conserved in the impact, so the common speed is \(mv/(m+M)\) and the vis viva after impact is \(T\,m/(m+M)=0.6T=3532\) J. Then \(R\cdot0.02=3532\), so \(R\approx177\) kN. The impact destroys \(M/(m+M)=40\%\) of the work, a loss that Lazare Carnot had already identified for shocks.
Lesson 05

Machines, friction and the chain of equal strength

For the general reader. Engineers of Coriolis’s time lived with friction. Every axle, sliding block and rope loses work to it. Charles-Augustin de Coulomb had found that dry friction is roughly proportional to the load. Using that rule, the work budget of Lesson 04 becomes a tool for designing inclines, screws and hoists. A related design question, which Coriolis studied in 1836, asks how to shape a hanging cable so that it is equally stressed everywhere.

Goals
  • Compute efficiencies of simple machines with Coulomb friction.
  • Derive the shape of a catenary of equal strength.
  • Find the longest span such a cable can reach.

Coulomb friction

Coulomb’s law of dry friction

When two surfaces slide, the friction force opposes the relative velocity and has magnitude \(\mu N\), where \(N\) is the normal force and \(\mu\) a coefficient depending on the materials, roughly independent of speed and contact area.

For a load of weight \(P\) pulled up a slope of angle \(\alpha\) at constant speed, the useful work over a distance \(s\) is \(Ps\sin\alpha\) and the friction loss is \(\mu Ps\cos\alpha\). The efficiency is

\[ \eta=\frac{\sin\alpha}{\sin\alpha+\mu\cos\alpha}. \]

The catenary of equal strength

A uniform cable hangs in an ordinary catenary, and its tension is largest at the supports. If the cross-section \(A\) is allowed to vary, it can be chosen so that the stress \(\sigma=T/A\) is the same everywhere. Coriolis treated this problem in 1836.

Proposition (catenary of equal strength)

Let the cable have density \(\rho\) and constant working stress \(\sigma\), and set \(c=\sigma/(\rho g)\). With the lowest point at the origin, the cable has the shape

\[ y=-c\,\ln\cos\frac xc,\qquad |x|<\frac{\pi c}{2}, \]

and its cross-section is \(A=A_0\sec(x/c)\). No such cable can span more than \(\pi c\).

Derivation. Let \(\theta\) be the angle of the cable with the horizontal. The horizontal tension \(H\) is constant, so \(T=H\sec\theta\) and \(A=T/\sigma=(H/\sigma)\sec\theta\). Vertical balance of an element of length \(ds\) gives \(d(H\tan\theta)=\rho gA\,ds\). With \(ds=\sec\theta\,dx\), this becomes \(H\sec^2\theta\,d\theta=(\rho gH/\sigma)\sec^2\theta\,dx\), so \(d\theta/dx=1/c\) and \(\theta=x/c\). Then \(dy/dx=\tan(x/c)\), which integrates to the stated curve. As \(x\to\pm\pi c/2\) the slope and section become infinite.

Exercise 5.1A cart of weight \(P\) is hauled up a 10° slope with \(\mu=0.10\). Find the efficiency. At what angle is half the work lost to friction?

Show solution
\(\eta=0.1736/(0.1736+0.0985)=0.638\). Half is lost when \(\sin\alpha=\mu\cos\alpha\), that is \(\tan\alpha=0.1\), \(\alpha\approx5.7^\circ\). Shallow slopes waste more of the work, though they need less force.

Exercise 5.2For steel with \(\rho=7850\) kg/m³ and working stress \(\sigma=500\) MPa, compute \(c\) and the maximal span of a catenary of equal strength. Compare with the span at which a uniform cable of the same steel would break under its own weight with a sag of 10% of the span.

Show solution
\(c=5\times10^8/(7850\cdot9.81)\approx6.49\) km, so the span is at most \(\pi c\approx20.4\) km. For a uniform shallow cable of span \(L\) and sag \(d\), the maximum tension is about \(\rho gAL^2/(8d)\), and breaking occurs when \(L^2/(8d)=c\). With \(d=0.1L\), \(L=0.8c\approx5.2\) km. Tapering the cable to keep the stress uniform almost quadruples the span.
Lesson 06

Energy in rotating machines (1831–1832)

For the general reader. In a waterwheel or a turbine, water moves along vanes that are themselves turning. An engineer would like to write the energy balance as seen by someone riding on the wheel. Coriolis showed in 1831 that this is possible: the usual rule that work equals change of kinetic energy still holds in the rotating frame, provided one adds the familiar centrifugal force to the real forces. Nothing else is needed for the energy balance.

Goals
  • State Coriolis’s theorem on vis viva in relative motion.
  • Derive Bernoulli’s equation in a rotating channel.
  • Apply both to a bead on a rotating rod and to a reaction wheel.

Let a frame rotate with constant angular velocity \(\boldsymbol\omega\) about a fixed axis. Write \(\mathbf r\) for the position relative to a point on the axis, and \(\mathbf v_r\) for the velocity measured in the rotating frame.

Theorem (Coriolis, 1831–1832)

In a frame rotating uniformly about a fixed axis, the change in relative vis viva \(\tfrac12mv_r^2\) of a particle equals the work, computed along the relative path, of the real forces together with the centrifugal force \(-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)\). The centrifugal force has potential \(-\tfrac12m\,|\boldsymbol\omega\times\mathbf r|^2\), so

\[ \Delta\Big(\tfrac12mv_r^2-\tfrac12m\,|\boldsymbol\omega\times\mathbf r|^2\Big)=W_{\text{real}} . \]

Proof. The relative equation of motion (Lesson 07) is \(m\dot{\mathbf v}_r=\mathbf F-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)-2m\,\boldsymbol\omega\times\mathbf v_r\). Dot with \(\mathbf v_r\). The last term vanishes because \((\boldsymbol\omega\times\mathbf v_r)\cdot\mathbf v_r=0\). For the centrifugal term, \(\frac{d}{dt}\tfrac12|\boldsymbol\omega\times\mathbf r|^2=(\boldsymbol\omega\times\mathbf r)\cdot(\boldsymbol\omega\times\mathbf v_r)=-\big(\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)\big)\cdot\mathbf v_r\), so its work is the change in \(\tfrac12m|\boldsymbol\omega\times\mathbf r|^2\). Coriolis’s 1832 memoir proved this before he had isolated the extra term; the 1835 memoir explained why it drops out.

Bernoulli in a rotating channel

For steady flow of an ideal incompressible fluid relative to the rotating frame, the same argument applied to a fluid particle gives, along a relative streamline,

\[ \frac p\rho+\tfrac12v_r^2-\tfrac12\omega^2s^2+gz=\text{const}, \]

where \(s\) is the distance to the axis. This is the starting point of the theory of turbines and centrifugal pumps.

Exercise 6.1A bead slides without friction along a straight rod rotating in a horizontal plane at constant \(\omega\) about one end. It starts at rest relative to the rod at distance \(r_0\). Find \(r(t)\) and the work done by the motor that keeps \(\omega\) constant, by the time the bead reaches \(r\).

Show solution
In the rotating frame, the rod’s normal force does no work along the rod, so \(\tfrac12\dot r^2-\tfrac12\omega^2r^2=-\tfrac12\omega^2r_0^2\), giving \(\dot r=\omega\sqrt{r^2-r_0^2}\) and \(r=r_0\cosh\omega t\). In the inertial frame the bead’s kinetic energy is \(\tfrac12m(\dot r^2+\omega^2r^2)=\tfrac12m\omega^2(2r^2-r_0^2)\), which started at \(\tfrac12m\omega^2r_0^2\). The motor’s work is the increase, \(m\omega^2(r^2-r_0^2)\): twice the work of the centrifugal force. The other half goes against the Coriolis force, which the rod must balance by pushing sideways.

Exercise 6.2A reaction wheel (Segner’s wheel) is fed with water at rest at height \(h\) above its arms, which have length \(R\) and turn at \(\omega\). Neglecting losses, find the relative exit speed \(v_r\) of the water. Show that the absolute exit speed \(v_r-\omega R\) tends to 0 as \(\omega\to\infty\), and interpret.

Show solution
Pressure is atmospheric at entry and exit, the water enters at the axis (\(s=0\)) at rest with head \(h\), and leaves at \(s=R\). Bernoulli gives \(\tfrac12v_r^2-\tfrac12\omega^2R^2=gh\), so \(v_r=\sqrt{2gh+\omega^2R^2}\). Then \(v_r-\omega R=2gh/(v_r+\omega R)\to0\). The water leaves with almost no absolute velocity, so almost all of its potential energy has gone into the wheel. The ideal efficiency tends to 1 at high speed.
Lesson 07

The compound centrifugal force (1835)

For the general reader. On a merry-go-round, a ball rolled straight outward seems to swerve sideways. Newton’s laws fail in a rotating frame unless one adds fictitious forces. Coriolis found exactly what must be added. One part is the familiar centrifugal force. The other, which he called the compound centrifugal force, pushes sideways on anything that moves relative to the rotating frame. It is the force now named after him.

Goals
  • Derive the transformation of velocity and acceleration to a rotating frame.
  • State Coriolis’s theorem on relative motion.
  • Compute the magnitudes of the terms in practical cases.

Kinematics

Let a frame with orthonormal vectors \(\mathbf e_i(t)\) rotate with angular velocity \(\boldsymbol\omega(t)\), so that \(\dot{\mathbf e}_i=\boldsymbol\omega\times\mathbf e_i\). For any vector \(\mathbf A=\sum A_i\mathbf e_i\),

\[ \left(\frac{d\mathbf A}{dt}\right)_{\text{in}}=\left(\frac{d\mathbf A}{dt}\right)_{\text{rot}}+\boldsymbol\omega\times\mathbf A . \]

Apply this twice to the position \(\mathbf r\) of a particle, measured from a point on the axis.

Theorem (Coriolis, 1835)

The absolute acceleration of a particle is

\[ \mathbf a_{\text{in}}=\mathbf a_r+\dot{\boldsymbol\omega}\times\mathbf r+\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)+2\,\boldsymbol\omega\times\mathbf v_r . \]

Hence Newton’s law in the rotating frame reads

\[ m\mathbf a_r=\mathbf F-m\,\dot{\boldsymbol\omega}\times\mathbf r-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)-2m\,\boldsymbol\omega\times\mathbf v_r . \]

The last term, the compound centrifugal force, is perpendicular to \(\mathbf v_r\) and does no work in the relative motion.

Proof. \(\mathbf v_{\text{in}}=\mathbf v_r+\boldsymbol\omega\times\mathbf r\). Differentiate in the inertial frame and use the rule again: \(\mathbf a_{\text{in}}=(\dot{\mathbf v}_r)_{\text{rot}}+\boldsymbol\omega\times\mathbf v_r+\dot{\boldsymbol\omega}\times\mathbf r+\boldsymbol\omega\times(\mathbf v_r+\boldsymbol\omega\times\mathbf r)\). Collect the two \(\boldsymbol\omega\times\mathbf v_r\) terms.

Coriolis did not separate the terms quite as textbooks now do. He regarded the sum of the ordinary and compound centrifugal forces as one effect of rotation and was interested in the compound part because it disappeared from the energy balance. The form of the acceleration had been found in special cases before him, by Leonhard Euler in 1749 and in Pierre-Simon Laplace’s tidal equations of 1778. Coriolis gave the general theorem for systems of bodies.

Exercise 7.1A child on a merry-go-round turning at one revolution every 6 s rolls a ball outward at 2 m/s relative to the platform. Find the Coriolis acceleration and the sideways drift relative to the platform after 1 s, treating the acceleration as constant.

Show solution
\(\omega=2\pi/6=1.05\) rad/s, so \(2\omega v_r=4.19\) m/s². The drift is about \(\tfrac12\cdot4.19\cdot1^2\approx2.1\) m. On a merry-go-round turning anticlockwise seen from above, the drift is to the right of the direction of throw.

Exercise 7.2A car drives at 30 m/s on a level road at latitude 45°. Compare the horizontal Coriolis acceleration with \(g\). Take \(\Omega=7.29\times10^{-5}\) s⁻¹.

Show solution
The horizontal component is \(2\Omega v\sin\varphi=2\cdot7.29\times10^{-5}\cdot30\cdot0.707=3.1\times10^{-3}\) m/s², about \(3\times10^{-4}g\). Friction of the tyres absorbs it without notice. The force matters only for motions that last hours or cover hundreds of kilometres.

Exercise 7.3Show directly from the theorem that a particle at rest in the inertial frame appears, in a frame rotating at constant \(\omega\) about \(z\), to move on a circle, and check that the centrifugal and Coriolis forces combine to give the required centripetal force.

Show solution
In the rotating frame \(\mathbf v_r=-\boldsymbol\omega\times\mathbf r\): the particle moves on a circle at speed \(\omega s\) in the opposite sense. The centrifugal force is \(m\omega^2s\) outward. The Coriolis force \(-2m\boldsymbol\omega\times\mathbf v_r=2m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)\) is \(2m\omega^2s\) inward. The net force \(m\omega^2s\) inward is exactly the centripetal force for the apparent circular motion.
Lesson 08

The rotating Earth

For the general reader. Coriolis never applied his theorem to the Earth. Others did, mostly without knowing his work. Léon Foucault’s pendulum, the large-scale winds that flow along lines of equal pressure, and the circular drift of ocean currents all follow from the same term. This lesson collects these consequences.

Goals
  • Define the Coriolis parameter and the inertial oscillation.
  • Derive the precession of the Foucault pendulum and the geostrophic wind.
  • Compute the eastward deflection of a falling body.

Use local axes at latitude \(\varphi\): \(x\) east, \(y\) north, \(z\) up. The Earth’s angular velocity is \(\boldsymbol\Omega=\Omega(0,\cos\varphi,\sin\varphi)\) with \(\Omega=7.292\times10^{-5}\) s⁻¹. Absorb the centrifugal force into effective gravity \(\mathbf g\). For horizontal motion, the horizontal Coriolis acceleration is \(-f\,\hat{\mathbf z}\times\mathbf u\) with the Coriolis parameter

\[ f=2\Omega\sin\varphi . \]
Proposition (three consequences)

(i) Inertial oscillation. A horizontal motion free of other forces satisfies \(\dot{\mathbf u}=-f\hat{\mathbf z}\times\mathbf u\): the velocity turns clockwise in the northern hemisphere with period \(2\pi/f\), on a circle of radius \(|\mathbf u|/f\).

(ii) Geostrophic balance. When the Coriolis force balances a horizontal pressure gradient, \(f\hat{\mathbf z}\times\mathbf u=-\frac1\rho\nabla p\), so \(\mathbf u=\frac{1}{\rho f}\hat{\mathbf z}\times\nabla p\): the wind blows along the isobars, with low pressure on the left in the northern hemisphere.

(iii) Foucault pendulum. For small oscillations, the plane of a pendulum turns clockwise (in the northern hemisphere) at angular rate \(\Omega\sin\varphi\), with period \(2\pi/(\Omega\sin\varphi)\), one sidereal day divided by \(\sin\varphi\).

Sketch for (iii). For small swings, write the horizontal position as a complex number \(\zeta=x+iy\). The equations become \(\ddot\zeta+2i\Omega\sin\varphi\,\dot\zeta+\omega_0^2\zeta=0\), with \(\omega_0^2=g/\ell\). Substituting \(\zeta=e^{-i\Omega\sin\varphi\,t}\eta\) and neglecting \(\Omega^2\) gives \(\ddot\eta+\omega_0^2\eta=0\): an ordinary pendulum in a frame turning at rate \(-\Omega\sin\varphi\).

Foucault showed his pendulum in Paris in 1851. William Ferrel used the deflecting effect of rotation to explain mid-latitude winds in 1856, and Christophorus Buys Ballot stated his empirical law relating wind and pressure in 1857. None of them drew on Coriolis’s paper.

Exercise 8.1Find the period of rotation of the Foucault pendulum at Paris (\(\varphi=48.85^\circ\)) and the period of inertial oscillations at 45°.

Show solution
The sidereal day is 23.93 h, so the Paris period is \(23.93/\sin48.85^\circ=23.93/0.753\approx31.8\) h. At 45°, \(f=2\Omega\sin45^\circ=1.03\times10^{-4}\) s⁻¹ and \(2\pi/f\approx6.09\times10^4\) s \(\approx16.9\) h, half a pendulum day.

Exercise 8.2At 45°, the pressure falls by 1 hPa per 100 km toward the north. With air density 1.2 kg/m³, find the speed and direction of the geostrophic wind.

Show solution
\(|\nabla p|=100\) Pa / \(10^5\) m \(=10^{-3}\) Pa/m, so \(|\mathbf u|=10^{-3}/(1.2\cdot1.03\times10^{-4})\approx8.1\) m/s. \(\nabla p\) points south, and \(\hat{\mathbf z}\times(-\hat{\mathbf y})=\hat{\mathbf x}\): the wind blows toward the east, with low pressure on its left.

Exercise 8.3A stone is dropped from rest down a 100 m shaft at latitude 45°. Neglecting air resistance, show that to first order in \(\Omega\) it lands a distance \(\tfrac13\Omega gt^3\cos\varphi\) east of the plumb line, and evaluate.

Show solution
The velocity is \(\mathbf v\approx(0,0,-gt)\). The Coriolis acceleration \(-2\boldsymbol\Omega\times\mathbf v\) has east component \(-2(\Omega_yv_z-\Omega_zv_y)=2\Omega gt\cos\varphi\). Integrating twice gives \(x=\tfrac13\Omega gt^3\cos\varphi\). With \(t=\sqrt{2h/g}=4.52\) s, \(x=\tfrac13\cdot7.29\times10^{-5}\cdot9.81\cdot92.1\cdot0.707\approx1.6\) cm.
Lesson 09

The mathematics of billiards (1835)

For the general reader. Billiards was a passion in Coriolis’s Paris, and he wrote the first full mathematical theory of the game. A ball struck off centre both slides and spins. Coriolis showed that while it slides, friction pushes it in a fixed direction, so it travels along a parabola. Once it starts rolling it goes straight. This explains the curved paths that skilled players produce with spin.

Goals
  • Find where to strike a ball so that it rolls at once.
  • Prove that a sliding ball moves on a parabola and compute its final velocity.
  • Predict the direction of the cue ball after it strikes another ball.

Model a ball of mass \(m\), radius \(R\) and moment of inertia \(I=\tfrac25mR^2\) on a horizontal table with sliding friction coefficient \(\mu\). Let \(\mathbf v\) be the velocity of the centre and \(\boldsymbol\omega\) the angular velocity. The contact point, at \(-R\hat{\mathbf z}\) from the centre, has velocity \(\mathbf u=\mathbf v+\boldsymbol\omega\times(-R\hat{\mathbf z})\). The ball rolls when \(\mathbf u=0\).

Theorem (sliding phase)

While \(\mathbf u\neq0\), the friction force \(-\mu mg\,\hat{\mathbf u}\) has constant direction, and

\[ \dot{\mathbf u}=-\tfrac72\mu g\,\hat{\mathbf u} . \]

So the centre moves with constant acceleration along an arc of a parabola. Sliding stops after time \(t_s=2|\mathbf u_0|/(7\mu g)\), when the velocity of the centre is

\[ \mathbf v_f=\mathbf v_0-\tfrac27\mathbf u_0 . \]

The ball then rolls in a straight line.

Proof. \(m\dot{\mathbf v}=-\mu mg\,\hat{\mathbf u}\) and \(I\dot{\boldsymbol\omega}=(-R\hat{\mathbf z})\times(-\mu mg\,\hat{\mathbf u})=\mu mgR\,\hat{\mathbf z}\times\hat{\mathbf u}\). Then \(\dot{\mathbf u}=\dot{\mathbf v}-R\,\dot{\boldsymbol\omega}\times\hat{\mathbf z}=-\mu g\hat{\mathbf u}-\tfrac52\mu g(\hat{\mathbf z}\times\hat{\mathbf u})\times\hat{\mathbf z}=-\tfrac72\mu g\hat{\mathbf u}\), since \((\hat{\mathbf z}\times\hat{\mathbf u})\times\hat{\mathbf z}=\hat{\mathbf u}\) for horizontal \(\hat{\mathbf u}\). So \(\hat{\mathbf u}\) never changes and \(|\mathbf u|\) decreases linearly to zero at \(t_s\). The acceleration of the centre is constant, which gives a parabola, and \(\mathbf v_f=\mathbf v_0-\mu g\,t_s\hat{\mathbf u}_0=\mathbf v_0-\tfrac27\mathbf u_0\).

Exercise 9.1A cue strikes a ball horizontally, in the vertical plane through its centre, at height \(h\) above the centre. Find \(h\) such that the ball rolls without sliding from the start. Then, for a ball struck at the centre (\(h=0\)), find the final speed and the fraction of kinetic energy lost to friction.

Show solution
An impulse \(J\) gives \(v=J/m\) and \(\omega=Jh/I=5Jh/(2mR^2)\). Rolling requires \(v=\omega R\), so \(h=\tfrac25R\), that is, at \(\tfrac75R\) above the table. For \(h=0\), \(\mathbf u_0=\mathbf v_0\) and \(v_f=\tfrac57v_0\). The final energy is \(\tfrac12mv_f^2(1+\tfrac25)=\tfrac{7}{10}m\cdot\tfrac{25}{49}v_0^2=\tfrac5{14}mv_0^2\), against \(\tfrac7{14}mv_0^2\) at the start: \(\tfrac27\) of the energy is lost.

Exercise 9.2A rolling cue ball with speed \(v\) strikes an identical object ball, at rest, at cut angle \(\phi\) (the angle between \(\mathbf v\) and the line of centres). Assume the collision is elastic and frictionless, so the cue ball keeps its spin and loses its velocity component along the line of centres. Find the cue ball’s final direction after it has resumed rolling, and the maximum deflection from its original direction.

Show solution
Let \(\hat{\mathbf n}\) be the line of centres and \(\hat{\mathbf t}\) the tangent, so \(\mathbf v=v(\cos\phi\,\hat{\mathbf n}+\sin\phi\,\hat{\mathbf t})\). After impact \(\mathbf v_0=v\sin\phi\,\hat{\mathbf t}\), and the spin still corresponds to rolling at \(\mathbf v\), so \(\mathbf u_0=\mathbf v_0-\mathbf v=-v\cos\phi\,\hat{\mathbf n}\). Then \(\mathbf v_f=v\sin\phi\,\hat{\mathbf t}+\tfrac27v\cos\phi\,\hat{\mathbf n}\). The deflection \(\theta\) from \(\mathbf v\) satisfies \(\cos\theta=(\sin^2\phi+\tfrac27\cos^2\phi)/\sqrt{\sin^2\phi+\tfrac4{49}\cos^2\phi}\). Maximizing numerically gives \(\theta_{\max}\approx33.7^\circ\) at \(\phi\approx28^\circ\). Players use the rule of thumb that a rolling cue ball deflects by about 30° over a wide range of cuts.
Lesson 10

Waterwheels and turbines

For the general reader. Coriolis’s theory of relative motion was made for machines like waterwheels and the new reaction turbines of the 1820s and 1830s. Water enters a turning wheel, slides along its vanes and leaves. How much of its energy does the wheel capture? The answer follows from the momentum and energy balances of Lessons 04 and 06, and it explains the design rules engineers still follow: enter without shock, leave without speed.

Goals
  • Derive the Euler turbine equation from the balance of angular momentum.
  • Check that it agrees with Coriolis’s energy theorem in the rotating frame.
  • Find the optimal speed of an impulse wheel.

Angular momentum balance

Consider steady flow through a wheel turning at constant \(\omega\) about a fixed axis. Water enters at radius \(r_1\) and leaves at radius \(r_2\), with mass flow rate \(\dot m\). Let \(c_{\theta1},c_{\theta2}\) be the tangential components of the absolute velocity at entry and exit.

Theorem (Euler turbine equation)

The torque the water exerts on the wheel is \(\tau=\dot m\,(r_1c_{\theta1}-r_2c_{\theta2})\), and the power delivered is

\[ P=\tau\omega=\dot m\,(U_1c_{\theta1}-U_2c_{\theta2}),\qquad U_i=\omega r_i . \]

Proof. In steady flow, the rate at which angular momentum about the axis leaves the wheel minus the rate at which it enters equals the torque exerted on the water by the wheel. That torque is \(\dot m(r_2c_{\theta2}-r_1c_{\theta1})\), and the wheel receives the opposite.

Agreement with the relative energy theorem

Write the absolute velocity as \(\mathbf c=\mathbf w+\mathbf U\), where \(\mathbf w\) is the velocity relative to the vanes. From \(w^2=c^2-2Uc_\theta+U^2\), we get \(Uc_\theta=\tfrac12(c^2+U^2-w^2)\), so

\[ \frac{P}{\dot m}=\frac{c_1^2-c_2^2}{2}+\frac{U_1^2-U_2^2}{2}+\frac{w_2^2-w_1^2}{2}. \]

The first term is the drop in absolute kinetic energy. The other two are what Coriolis’s theorem of Lesson 06 predicts along a relative streamline: the centrifugal potential and the relative kinetic energy exchange with pressure, and the Coriolis force, doing no relative work, does not appear. It is nevertheless the Coriolis force, balanced by the vanes, that transmits the torque.

Exercise 10.1An impulse wheel (a Pelton-type wheel) receives a jet of speed \(c\) at radius \(r\). The buckets move at \(U=\omega r\) and turn the relative flow through nearly \(180^\circ\) without loss, so \(w_2=w_1=c-U\) and both radii are equal. Find the power as a function of \(U\) and the optimal \(U/c\). What is the maximum efficiency?

Show solution
\(c_{\theta1}=c\) and \(c_{\theta2}=U-(c-U)=2U-c\). So \(P=\dot mU\big(c-(2U-c)\big)=2\dot mU(c-U)\), maximal at \(U=c/2\), where \(P=\tfrac12\dot mc^2\), the full kinetic energy of the jet. The efficiency is 1, and the water leaves the bucket with zero absolute velocity.

Exercise 10.2A paddle wheel driven by the impact of a stream at speed \(c\) on flat blades moving at \(U\) loses the relative velocity in a shock: the water leaves the blade moving at the blade speed. Show that the power is \(\dot mU(c-U)\) and the maximum efficiency is \(\tfrac12\). Identify the lost work with the Carnot shock loss of Lesson 04.

Show solution
\(c_{\theta1}=c\), \(c_{\theta2}=U\), same radius: \(P=\dot mU(c-U)\), maximal at \(U=c/2\) with \(P=\tfrac14\dot mc^2\), half the jet energy. Of the remaining \(\tfrac14\dot mc^2\), half leaves as the outflow’s kinetic energy \(\tfrac12\dot mU^2=\tfrac18\dot mc^2\) and half is destroyed in the inelastic shock, \(\tfrac12\dot m(c-U)^2=\tfrac18\dot mc^2\). Avoiding the shock is why curved vanes and buckets beat flat paddles.
Part III · Life and legacy
Lesson 11

Navier’s successor, 1829–1843

The year 1829 changed Coriolis’s position. His book on the effect of machines appeared, and he became professor of mechanics at the new École centrale des arts et manufactures, a private school founded that year to train engineers for industry. When Cauchy left France after the revolution of July 1830, refusing the oath to the new king, Coriolis was offered his post at the École polytechnique. He declined, wanting time for research.

In the early 1830s he taught applied mechanics alongside Claude-Louis Navier at the École des ponts et chaussées. Navier died in August 1836, and Coriolis succeeded him both in the chair and in the mechanics section of the Académie des sciences. In 1838 he succeeded the physicist Pierre Louis Dulong as director of studies at the École polytechnique, responsible for its whole curriculum. He gave up teaching applied mechanics to save his health, and contemporaries judged him an excellent director.

His health had never been good, and in the spring of 1843 it collapsed. He died in Paris on 19 September 1843, aged 51, and was buried in the Montparnasse cemetery. His Traité de la mécanique des corps solides, a revised edition of the 1829 book, appeared in 1844. His sister Cécile married the physicist Eugène Péclet.

DiscussionCoriolis declined a prestigious professorship to do research, then accepted a heavy administrative post eight years later. What did a director of studies at the École polytechnique control, and why might a researcher accept the role?

Lesson 12

Legacy: how a name reached the weather

During his life Coriolis was known for the theory of machines and the definition of work. His 1835 memoir on relative motion was read by students of mechanics but not by meteorologists. George Hadley had explained the trade winds by the Earth’s rotation in 1735, Laplace had written the rotating tidal equations in 1778, and Ferrel and Buys Ballot developed the theory of winds in the 1850s without him. His name began to appear in meteorology near the end of the nineteenth century, and the phrase “Coriolis force” became standard only in the early twentieth century, by around 1920. Today it is one of the most familiar names in geophysics.

Legacy

  • Energy: the word work in its scientific sense, the factor \(\tfrac12\) in kinetic energy, and the work–energy budget of machines.
  • Relative motion: the energy theorem in rotating frames (1831–1832) and the compound centrifugal force (1835), the basis of turbine theory and of every rotating-frame calculation in geophysics, astrophysics and engineering.
  • Billiards: the first full mechanical theory of the game, still cited by physicists of sport.
  • Education: a generation of engineers trained at the École centrale, Ponts et Chaussées and Polytechnique.
  • Honours: one of the 72 names inscribed on the Eiffel Tower; a lunar crater named in 1970; the asteroid (16564) Coriolis; a French oceanographic research vessel named in 1963.

Final projectWrite a two-page account, for a strong undergraduate, of the Coriolis force as Coriolis understood it. Start from a waterwheel or reaction turbine, derive the energy balance of Lessons 06 and 10, show where the compound centrifugal force appears and why it drops out, and end with one geophysical application from Lesson 08.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Gaspard Gustave de Coriolis, MacTutor History of Mathematics.
  2. Gaspard-Gustave de Coriolis, Wikipedia; Gaspard-Gustave Coriolis, Wikipédia (family and early career).
  3. P. Costabel, Coriolis, Gaspard Gustave de, Dictionary of Scientific Biography, vol. 3, Scribner, 1971.
  4. A. Persson, The Coriolis effect: four centuries of conflict between common sense and mathematics, History of Meteorology 2 (2005).

Science

  1. G. Coriolis, Du calcul de l’effet des machines, Carilian-Goeury, Paris, 1829.
  2. G. Coriolis, Mémoire sur le principe des forces vives dans les mouvements relatifs des machines, Journal de l’École polytechnique 13 (1832).
  3. G. Coriolis, Mémoire sur les équations du mouvement relatif des systèmes de corps, Journal de l’École polytechnique 15 (1835), 142–154.
  4. G. Coriolis, Théorie mathématique des effets du jeu de billard, Carilian-Goeury, Paris, 1835.
  5. A. Persson, How do we understand the Coriolis force?, Bulletin of the American Meteorological Society 79 (1998).
  6. Coriolis force, Wikipedia (history and standard formulas).
  7. D. G. Alciatore, The Illustrated Principles of Pool and Billiards, Sterling, 2004 (modern treatment of the 30° rule).
  8. G. K. Vallis, Atmospheric and Oceanic Fluid Dynamics, 2nd ed., Cambridge University Press, 2017 (geostrophy, inertial oscillations).