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Simon Stevin

Simon Stevin of Bruges (Simon Stevinus Brugensis)

A course in two strands: the life of the Flemish clerk who became engineer to the Dutch Republic, and the mathematics he gave it: decimal fractions, real numbers, statics, hydrostatics and equal temperament.

Born 1548, Bruges Died 1620, The Hague De Thiende Leiden, 1585 Level undergraduate → advanced undergraduate
The clootcrans: fourteen equal balls on a closed string. If the chain moved, it would move forever. It cannot, so four balls on the long side balance two on the short side.
Lesson 00

How to use this course

Simon Stevin worked at the hinge between Renaissance learning and the new science. He wrote for merchants, surveyors and soldiers, in Dutch, and he insisted that every theory earn its keep in practice. The results were large: decimal arithmetic for everyday use, an account of number in which irrationals are as good as integers, the first sound statics since Archimedes, the hydrostatic paradox, and an early correct statement of equal temperament.

Prerequisites for Part II
  • School algebra, geometric series and logarithms.
  • Vectors in the plane and the idea of a force as a vector.
  • First-year calculus for Lessons 06 and 09 (limits, the intermediate value theorem, a simple integral).
  • No music theory is assumed in Lesson 10.

Each technical lesson opens with a paragraph for the general reader, then states results precisely in modern notation, sketches Stevin’s argument, and ends with exercises with folded solutions. Stevin’s own notation was very different; where it matters, the lesson shows both. Many details of his early life are unknown, and the course says so where the sources disagree.

Reference

Timeline

1548Born in Bruges, in the Habsburg Netherlands, to Anthuenis Stevin and Cathelijne van der Poort. Exact date unknown.
c. 1571–77Bookkeeper and cashier for a firm in Antwerp; travels in Prussia, Poland and Scandinavia.
1577–81Clerk in the tax administration at Bruges.
1581Moves to Leiden in the northern provinces; attends the Latin school.
1582Tafelen van Interest, published tables of compound interest.
158316 February: enrols at Leiden University. Problemata geometrica.
1585De Thiende and its French version La Disme; L’Arithmétique; Dialectike ofte Bewysconst.
1586De Beghinselen der Weeghconst, De Weeghdaet, De Beghinselen des Waterwichts. Falling-body test at Delft with Jan Cornets de Groot.
1590Vita Politica, Het Burgherlick leven, on civic duty.
1594De Stercktenbouwing, on fortification.
1599De Havenvinding, on finding a ship’s position by the compass declination.
c. 1600Sails his land yacht on the beach from Scheveningen toward Petten with Prince Maurice.
1600January: the Dutch-language engineering course (Duytsche Mathematique) opens at Leiden on a plan Stevin drew up.
c. 1604Quartermaster-general of the army of the States-General.
1605–08Wisconstighe Ghedachtenissen, his collected mathematical memoirs, including De Hemelloop (1608), a defence of Nicolaus Copernicus.
1610 or 1614Marries Catharina Krai (spellings vary); four children, including Hendrick.
1612Buys a house on the Raamstraat in The Hague.
1617Castrametatio and Nieuwe Maniere van Stercktebou door Spilsluysen.
1620Dies, probably in February, in The Hague (some sources say Leiden).
1634Les Œuvres mathématiques de Simon Stevin, French collected edition, Leiden.
Part I · Formation
Lesson 01

Bruges, Antwerp and the counting house, 1548–1581

Simon Stevin was born in Bruges in 1548. His mother, Cathelijne van der Poort, came from a burgher family of Ypres; his father, Anthuenis Stevin, is thought to have been a younger son of a mayor of Veurne. His parents were not married. His mother later married Joost Sayon, who traded in carpets and silk, and the household was Calvinist. That fact shaped his life, because the southern Netherlands were then under the Catholic rule of Philip II of Spain, and the revolt of the northern provinces began when Stevin was a young man.

Nothing is known of his schooling. What is known is that he learned commerce well. Around 1571 he left Bruges and worked as a bookkeeper and cashier in Antwerp, then the great trading city of northern Europe. He later wrote of travels in Prussia, Poland, Norway and other northern lands in these years. By 1577 he was back in Bruges as a clerk in the administration of taxes for the Brugse Vrije, the rural district around the city.

These years explain his later style. He thought like an accountant: he wanted methods that were exact, checkable and fast, and he had no patience for Latin learning that could not be put to work. He would later recommend double-entry bookkeeping for the finances of a prince’s household, a radical idea for public administration.

Lesson 02

Leiden and the first books, 1581–1586

In 1581, as the Spanish reconquest of Flanders gathered pace, Stevin moved north to Leiden in Holland. Its university had been founded in 1575 as a reward to the city for surviving a Spanish siege. Stevin first attended the Latin school, then on 16 February 1583 enrolled at the university as Simon Stevinus Brugensis, at about 35. He appears in the registers until 1590 and never took a degree.

He published at astonishing speed. The Tafelen van Interest (1582) put into print tables that bankers had treated as trade secrets. The Latin Problemata geometrica (1583) dealt with polyhedra and geometric constructions. In 1585 came De Thiende, a short booklet on decimal fractions, issued in Dutch and in French as La Disme; the long French L’Arithmétique, on numbers and equations; and a Dutch textbook of logic. In 1586 he published the three works on statics and hydrostatics that are his main contribution to physics.

Almost everything after 1583 was written in Dutch. Stevin argued, with characteristic confidence, that Dutch was the best language for science because its many one-syllable roots let new compound words explain themselves. He coined or popularized much of the Dutch scientific vocabulary. Words he coined or promoted, such as wiskunde (mathematics), meetkunde (geometry) and sterrenkunde (astronomy), are still standard Dutch.

Lesson 03

Mills, sluices and a prince, 1586–1600

In Leiden Stevin met Maurice of Nassau, second son of William of Orange. After William’s assassination in 1584, Maurice became stadholder of Holland and Zeeland and, soon, commander of the Dutch army. Stevin became his tutor in mathematics and his close adviser. Several of Stevin’s later books grew from lessons written for the prince.

The Dutch problem was water. Stevin received a patent in 1586 for improvements to drainage mills and worked on locks and sluices. In Delft he collaborated with Jan Cornets de Groot, a lawyer and the father of the jurist Hugo Grotius. De Groot helped Stevin obtain permission to modify the town’s mills and invested in mills built to his design. Around 1586 the two men dropped two lead balls, one about ten times heavier than the other, some thirty feet from the tower of the Nieuwe Kerk onto a board, and heard them land together. Stevin reported the result in his 1586 writings on weighing and hydrostatics, several years before Galileo Galilei began writing on the subject.

Military engineering followed. De Stercktenbouwing (1594) adapted Italian fortification to Dutch conditions. Stevin also devised a defensive system of controlled flooding by opening chosen sluices in the dikes, an idea the Republic used for centuries. Around 1600 he built a wind-driven land yacht, which carried Maurice and some 26 passengers along the beach from Scheveningen toward Petten.

Part II · The mathematics
Lesson 04

Tables of interest

For the general reader. Money paid in the future is worth less than money paid today. Before 1582, the tables that let a merchant compare such payments were closely guarded. Stevin published them, with clear rules and worked examples, and in doing so wrote one of the first textbooks of financial mathematics.

Goals
  • Express compound interest and present value in modern notation.
  • Understand how a table of discount factors answers annuity problems.
  • See why Stevin scaled his tables by a large power of ten.

Rates as “the penny”

Stevin’s contemporaries quoted interest as den penninck \(n\): one unit of interest per \(n\) units of principal per year. At the 16th penny, 16 grows to 17 in a year, a rate of \(1/16=6.25\%\). His tables list, for each rate and each number of years, the present value of a sum due in the future, scaled to an integer base of \(10^7\) so that only whole numbers appear.

Present value and annuities

At interest rate \(r\) per period, a sum \(S\) due after \(k\) periods has present value \(S\,v^k\) with \(v=\frac{1}{1+r}\). An annuity paying \(S\) at the end of each of \(N\) periods has present value

\[ A_N = S\sum_{k=1}^{N} v^k = S\,\frac{v\,(1-v^N)}{1-v} = S\,\frac{1-(1+r)^{-N}}{r}. \]

For the penny \(n\), \(r=1/n\) and \(v=\frac{n}{n+1}\). Stevin’s tables gave the column \(10^7 v^k\) and, beside it, the running sums \(10^7\sum_{j\le k}v^j\). Every problem of loans, rents and instalments then reduces to looking up two numbers and doing a rule of three. The method, a table of discount factors plus cumulative sums, is the one actuaries used until computers arrived.

Stevin also took care over the difference between simple and compound interest, and solved the inverse problems: given payments and a price, find the rate. With tables, he did this by interpolation between columns.

Exercise 4.1At the 16th penny, what is the present value of 1000 due in three years? Give the entry that would stand in a table with base \(10^7\).

Show solution
\(v=16/17\), so \(1000\,v^3=1000\cdot 4096/4913\approx 833.71\). The table entry would be \(10^7\cdot(16/17)^3\approx 8\,337\,065\).

Exercise 4.2A rent of 100 is paid at the end of each year for 5 years, at the 16th penny. Find its present value, and check your answer with the closed formula.

Show solution
Summing \(100\,v^k\) for \(k=1,\dots,5\) with \(v=16/17\) gives about \(418.39\). The formula with \(r=1/16\): \(100\cdot\frac{1-(17/16)^{-5}}{1/16}=1600\,(1-(16/17)^5)\approx1600\cdot0.26149\approx418.39\).

Exercise 4.3Show that a perpetual rent of 1 per year at the penny \(n\) is worth exactly \(n\). Why does this explain the name of the rate?

Show solution
\(\sum_{k\ge1}v^k=\frac{v}{1-v}=\frac{n/(n+1)}{1/(n+1)}=n\). Quoting a rate as “the penny \(n\)” says that a perpetual income of 1 costs \(n\): the price is \(n\) years’ purchase.
Lesson 05

De Thiende: decimal fractions

For the general reader. Stevin did not invent decimal fractions. Mathematicians in the Islamic world, among them al-Uqlidisi in the tenth century and Jamshid al-Kashi in the fifteenth, had used them. What Stevin did was to argue, in 29 pages anyone could read, that all everyday calculation should be done with them, and that weights, measures and coins should be divided by ten. He addressed the book to stargazers, surveyors, carpet-makers, wine-gaugers and mint-masters.

Goals
  • Read Stevin’s decimal notation and translate it.
  • See why the four operations on decimals reduce to those on integers.
  • Understand his proposal for decimal measures.

Notation

Stevin wrote the unit position as ⓪, tenths as ①, hundredths as ②, and so on, each digit followed by its circled order. His standard example is

27 ⓪ 8 ① 4 ② 7 ③  means  \(27+\tfrac{8}{10}+\tfrac{4}{100}+\tfrac{7}{1000}=27.847\).

He called the orders commencement, prime, seconde, tierce, by analogy with sexagesimal minutes and seconds. The decimal point emerged from his notation within a generation: John Napier used a point in his work on logarithms, and Christopher Clavius used decimals in astronomical tables.

The rules of De Thiende

Write \(x=\sum_{i} a_i 10^{-i}\) and \(y=\sum_j b_j 10^{-j}\) with finitely many digits. Then addition and subtraction act digit by digit on aligned orders with carries, and

\[ xy=\sum_{i,j}a_ib_j\,10^{-(i+j)}, \]

so the order of a product is the sum of the orders. Division is long division of integers, with the order of the quotient equal to the difference of orders, continued to as many places as needed.

The point of the rule for products is that the decimal point never needs to be tracked digit by digit. One multiplies integers and then counts orders. Stevin stressed that when a division does not terminate, as with \(\frac13=0.333\ldots\), one stops at the precision the application requires. This was the first explicit statement that approximation, not exact fractions, is the right goal of practical arithmetic.

The appendix

The second part of De Thiende applies decimals to surveying, measuring tapestry, gauging wine casks, astronomy and coinage. It ends by proposing that the government adopt decimal subdivisions of all units. The proposal waited two centuries, for the French metric system of the 1790s. Robert Norton’s English translation, Disme (1608), is reported to have influenced Thomas Jefferson’s plan for American decimal coinage.

Exercise 5.1Multiply 3 ⓪ 7 ① 5 ② by 4 ⓪ 2 ① using Stevin’s rule, and write the answer in his notation.

Show solution
As integers, \(375\times42=15750\). The orders are 2 and 1, so the product has order 3: \(15.750\), that is 15 ⓪ 7 ① 5 ② 0 ③. Check: \(3.75\times4.2=15.75\).

Exercise 5.2Show that a fraction \(p/q\) in lowest terms has a terminating decimal expansion if and only if \(q\) has no prime factors other than 2 and 5. What does this mean for Stevin’s advice to stop at a chosen order?

Show solution
If \(p/q=m/10^k\) then \(q\mid10^k\), so \(q=2^a5^b\). Conversely if \(q=2^a5^b\), multiply top and bottom by \(2^{c-a}5^{c-b}\) with \(c=\max(a,b)\) to get denominator \(10^c\). For every other \(q\) the expansion is infinite (and periodic), so truncation is unavoidable, and the error after order \(k\) is below \(10^{-k}\).
Lesson 06

Numbers, equations and approximation

For the general reader. For the Greeks, “number” meant a whole number. Lengths like \(\sqrt2\) were magnitudes, not numbers. Stevin rejected this. In L’Arithmétique (1585) he declared that the unit is a number, that roots are numbers, that negative numbers deserve the same rights, and that there are no “absurd” or “irrational” numbers at all. With decimals, any such number can be approached as closely as one likes.

Goals
  • State Stevin’s view of number as a continuum of magnitudes.
  • Understand his decimal search for roots of polynomials.
  • Connect it to the intermediate value theorem.

A continuum of numbers

Stevin compared number to water: as a continuous quantity of water stays water when divided, so the continuous quantity of number stays number. The theory to support this view came only with Richard Dedekind and Georg Cantor in the 1870s, but Stevin’s infinite decimal expansions are already a concrete model of the real line.

He also unified the cases of the quadratic equation. Earlier algebraists treated \(x^2+bx=c\), \(x^2=bx+c\) and \(x^2+c=bx\) separately, because coefficients had to be positive. Stevin accepted negative coefficients and handled them by one rule. He used the signs \(+\) and \(-\), wrote radicals with \(\surd\), and indicated powers with circled exponents, an anticipation of index notation.

Roots digit by digit

For a polynomial equation \(p(x)=0\) of any degree, Stevin described a procedure. Find an integer \(a\) with \(p(a)\) and \(p(a+1)\) of opposite sign. Then try \(a+\tfrac{1}{10},a+\tfrac{2}{10},\dots\) to find the tenth at which the sign changes; then hundredths, and so on. Each step fixes one more decimal of a root.

Theorem (intermediate value property for polynomials)

Let \(p\) be a polynomial with real coefficients and \(a<b\) real numbers with \(p(a)<0<p(b)\). Then there is \(c\in(a,b)\) with \(p(c)=0\). Stevin’s decimal search produces nested intervals \([a_k,a_k+10^{-k}]\) on which \(p\) changes sign, and \(c=\lim a_k\).

Proof sketch. At each stage, divide the current interval into ten equal parts. Since \(p\) changes sign between the ends, it changes sign (or vanishes) between the ends of some part. The left endpoints \(a_k\) form a non-decreasing sequence bounded by \(b\), so in the reals they converge to some \(c\). Because \(p\) is continuous, \(p(c)=\lim p(a_k)\le0\) and \(p(c)=\lim p(a_k+10^{-k})\ge0\), so \(p(c)=0\). The completeness used in the second step is exactly what Stevin took for granted when he called every decimal expansion a number. A rigorous proof had to wait for Bernard Bolzano in 1817.

Exercise 6.1Use Stevin’s method to find the root of \(x^3-2x-5=0\) to three decimals.

Show solution
\(p(2)=-1\), \(p(3)=16\): the root is in \((2,3)\). \(p(2.0)=-1\), \(p(2.1)=0.061\): in \((2.0,2.1)\). \(p(2.09)\approx-0.051\), \(p(2.10)=0.061\): in \((2.09,2.10)\). \(p(2.094)\approx-0.0062\), \(p(2.095)\approx0.0050\). So the root is \(2.094\ldots\) (in fact \(2.0946\ldots\)).

Exercise 6.2Show that \(x^3=300x+33915024\) has the integer root 324, by locating it digit by digit starting from the hundreds.

Show solution
Let \(p(x)=x^3-300x-33915024\). \(p(300)=27\,000\,000-90\,000-33\,915\,024<0\) and \(p(400)=64\,000\,000-120\,000-33\,915\,024>0\), so the hundreds digit is 3. \(p(320)=32\,768\,000-96\,000-33\,915\,024<0\), \(p(330)=35\,937\,000-99\,000-33\,915\,024>0\): tens digit 2. Finally \(324^3=34\,012\,224\) and \(300\cdot324=97\,200\), so \(p(324)=0\).

Exercise 6.3How many evaluations of \(p\) does decimal search need, at most, to fix \(k\) decimals after the integer part is known? Compare with bisection.

Show solution
At most 9 new evaluations per digit (the tenth is implied by the sign at the right end), so at most \(9k\). Bisection gains a factor 2 per evaluation and needs about \(k\log_2 10\approx3.32k\) evaluations. Stevin’s scheme is slower but produces the decimal digits directly.
Lesson 07

The inclined plane and the wreath

For the general reader. How hard must you pull to hold a cart on a slope? Stevin answered with a thought experiment so elegant he put it on the title page of his statics. Hang a loop of equal balls over a triangular ridge. If the balls on one side outweighed those on the other, the loop would slide forever, a perpetual motion. That is absurd, so they balance. The motto he printed beside it says a wonder is no wonder.

Goals
  • Reconstruct the wreath argument.
  • Derive the law of the inclined plane in modern form.
  • See the impossibility of perpetual motion used as a physical axiom.

The argument

Take a triangular prism with horizontal base and two sloping faces of lengths \(\ell_1\) and \(\ell_2\), with \(\ell_1=2\ell_2\) in Stevin’s figure. Drape over it a closed string of 14 equal, equally spaced balls, so that 4 lie on the long face, 2 on the short face, and 8 hang below in a symmetric festoon.

  1. The chain cannot move by itself. If it did, each new position would look exactly like the old one, so it would keep moving forever, which is impossible.
  2. The hanging festoon is symmetric. Cut it away and the remainder is still in equilibrium.
  3. So 4 balls on the long face balance 2 on the short face. The weights along each face are proportional to the face lengths.
Theorem (Stevin, Weeghconst, 1586)

Two weights \(W_1,W_2\) on smooth inclined planes of the same height \(h\) and lengths \(\ell_1,\ell_2\), joined by a string over the top, are in equilibrium when \(W_1/W_2=\ell_1/\ell_2\). Equivalently, the force along a smooth plane inclined at angle \(\theta\) needed to hold a weight \(W\) is \(W\sin\theta\).

The equivalence follows from \(\sin\theta_i=h/\ell_i\): the condition \(W_1\sin\theta_1=W_2\sin\theta_2\) is \(W_1/\ell_1=W_2/\ell_2\). Taking the second plane vertical (\(\ell_2=h\)) gives the force needed to hold \(W_1\) on its slope. A medieval treatise attributed to Jordanus de Nemore already contained the law, argued by a form of virtual work. Stevin’s derivation from the impossibility of perpetual motion is his own, and it became the classic one.

A principle, not a measurement

Stevin did not measure anything. He used the impossibility of perpetual motion as an axiom, and drew a quantitative law from symmetry. The same reasoning reappears in the principle of virtual work and in energy conservation. Ernst Mach, in his history of mechanics, singled out the wreath as a model of how physical insight works.

Exercise 7.1A cart of mass 500 kg rests on a smooth ramp rising 1 m over a sloping length of 8 m. What force along the ramp holds it? Take \(g=9.8\,\mathrm{m\,s^{-2}}\).

Show solution
\(\sin\theta=1/8\), so \(F=500\cdot9.8/8=612.5\) N, one eighth of the weight.

Exercise 7.2Show that the wreath argument works for any triangle, not just a 2:1 ratio, provided the chain has uniform weight per unit length. What is the role of the base being horizontal?

Show solution
With uniform density \(\lambda\), the weights on the faces are \(\lambda\ell_1\) and \(\lambda\ell_2\). The hanging part is symmetric about the vertical through its lowest point only if both ends of the festoon are at the same height, which is why the base corners must be level. Removing it leaves \(\lambda\ell_1\) balancing \(\lambda\ell_2\), so \(W_1/W_2=\ell_1/\ell_2\) for any lengths.
Lesson 08

The triangle of forces

For the general reader. When a lamp hangs from two cords, how is its weight shared? Stevin found the rule. Draw a triangle whose sides are parallel to the two cords and to the vertical. Its side lengths are in the same ratios as the tensions and the weight. This is the first clear statement of what is now called vector addition of forces.

Goals
  • State the triangle of forces and its equivalent forms.
  • Derive it from the inclined plane.
  • Solve problems with two and three cords.
Theorem (triangle of forces)

Three non-parallel forces acting at a point are in equilibrium if and only if they can be represented, in magnitude and direction, by the sides of a triangle taken in order. Equivalently, \(\vec F_1+\vec F_2+\vec F_3=\vec 0\), and then (Lami’s form)

\[ \frac{F_1}{\sin\alpha_1}=\frac{F_2}{\sin\alpha_2}=\frac{F_3}{\sin\alpha_3}, \]

where \(\alpha_i\) is the angle between the two forces other than \(\vec F_i\).

From the inclined plane. A weight \(W\) resting on a smooth plane at angle \(\theta\) and held by a force along the plane feels three forces: \(W\) downward, the hold \(W\sin\theta\) along the plane, and the normal reaction \(N\) perpendicular to it. Stevin observed that a triangle with sides vertical, along the plane and perpendicular to it is similar to the triangle of the plane itself, and that the ratios \(W:W\sin\theta:W\cos\theta\) match its sides. Replacing the plane by a cord pulling in the direction of \(N\) gives the general rule. Stevin also drew the resolution of a single force into two components, which is the parallelogram rule in disguise. Pierre Varignon and Isaac Newton stated it systematically a century later.

Exercise 8.1A weight of 100 N hangs from two cords making angles of \(30^\circ\) and \(60^\circ\) with the horizontal. Find the tensions.

Show solution
Horizontal: \(T_1\cos30^\circ=T_2\cos60^\circ\). Vertical: \(T_1\sin30^\circ+T_2\sin60^\circ=100\). From the first, \(T_2=\sqrt3\,T_1\); then \(T_1(\tfrac12+\tfrac32)=100\), so \(T_1=50\) N and \(T_2=50\sqrt3\approx86.6\) N. The cords meet at right angles, and the triangle of forces is a 30–60–90 triangle with hypotenuse 100.

Exercise 8.2Show that a symmetric pair of cords at angle \(\varphi\) below the horizontal supporting a weight \(W\) each carry tension \(W/(2\sin\varphi)\). What happens as \(\varphi\to0\), and what does this say about a washing line pulled taut?

Show solution
Horizontal components cancel; vertical: \(2T\sin\varphi=W\). As \(\varphi\to0\), \(T\to\infty\): no finite tension can make a loaded line perfectly straight.
Lesson 09

Hydrostatics and the paradox

For the general reader. Fill a wide jar and a narrow tube with water to the same height, with bottoms of the same area. The water in the jar weighs far more, yet the force on the bottom is the same. Stevin proved this in his Waterwicht (1586). The pressure of a liquid depends only on depth, not on the shape of the vessel. He also computed the force on a sloping or vertical wall, using a method of upper and lower sums that anticipates the integral.

Goals
  • State and prove the hydrostatic paradox with Stevin’s solidification principle.
  • Compute the force on a vertical wall by his bounding method.
  • Relate it to Archimedes’ principle.

The solidification principle

Stevin’s key axiom: any part of a fluid at rest can be imagined to become solid, with the same weight, without disturbing the equilibrium. Imagine a vessel of any shape filled to height \(h\) above a horizontal bottom of area \(A\). Solidify all the water except the vertical column standing on the bottom. Nothing changes, so the force on the bottom is the weight of that column.

Theorem (hydrostatic paradox)

For a liquid of density \(\rho\) at rest under gravity \(g\), the force on a horizontal bottom of area \(A\) at depth \(h\) below the free surface is \(\rho g h A\), whatever the shape of the vessel. The pressure at depth \(h\) is \(p=\rho g h\) (above atmospheric).

The same principle gives Archimedes’ law: solidify a portion of liquid shaped like a submerged body; it floats in equilibrium, so the surrounding liquid pushes up on it with a force equal to its weight.

Force on a vertical wall

Take a vertical rectangular wall of width \(b\), submerged from depth 0 to \(h\). Divide it into \(n\) horizontal strips of height \(h/n\). The pressure on strip \(k\) lies between \(\rho g\frac{(k-1)h}{n}\) and \(\rho g\frac{kh}{n}\). So the total force \(F\) satisfies

\[ \rho g b\frac{h^2}{n^2}\sum_{k=1}^{n}(k-1)\;\le\;F\;\le\;\rho g b\frac{h^2}{n^2}\sum_{k=1}^{n}k, \qquad\text{that is}\qquad \rho g b h^2\,\frac{n-1}{2n}\le F\le\rho g b h^2\,\frac{n+1}{2n}. \]

Both bounds tend to \(\tfrac12\rho g b h^2\), and Stevin concluded that the force equals half that on a horizontal bottom of the same area at depth \(h\). In modern terms he evaluated \(\int_0^h \rho g b\,y\,dy\) by squeezing it between Riemann sums.

Exercise 9.1A lock gate 6 m wide holds water 4 m deep. Find the total force and the depth at which it effectively acts. Take \(\rho=1000\,\mathrm{kg\,m^{-3}}\), \(g=9.8\,\mathrm{m\,s^{-2}}\).

Show solution
\(F=\tfrac12\rho g b h^2=\tfrac12\cdot1000\cdot9.8\cdot6\cdot16=470\,400\) N. The moment about the surface is \(\int_0^h\rho g b y^2\,dy=\tfrac13\rho g b h^3\), so the centre of pressure is at depth \(\tfrac{(1/3)h^3}{(1/2)h^2}=\tfrac23h\approx2.67\) m.

Exercise 9.2A conical flask (wide base, narrow neck) and a cylinder both have base area \(0.01\,\mathrm{m^2}\) and contain water to height 0.2 m. The flask holds 0.5 kg of water, the cylinder 2 kg. Find the force of the water on each base. Why does a scale under the flask still read only the weight of flask plus water?

Show solution
Both bases feel \(\rho g h A=1000\cdot9.8\cdot0.2\cdot0.01=19.6\) N. In the cylinder this equals the water’s weight. In the flask the water weighs only 4.9 N, yet pushes 19.6 N on the base. The inward-sloping walls push down on the water, and the water pushes up on the walls with the difference, 14.7 N. That upward push on the glass cancels the excess on the base, so the scale reads the total weight.
Lesson 10

Equal temperament

For the general reader. A keyboard tuned to pure intervals sounds lovely in one key and sour in others. Divide the octave into twelve exactly equal semitones and every key sounds alike. In an unfinished manuscript, Vande Spiegheling der Singconst, written around 1605 and printed only in 1884, Stevin argued that this equal division is the correct one and computed it using the twelfth root of two.

Goals
  • Express equal temperament with the ratio \(2^{1/12}\).
  • Measure intervals in cents and compare equal and pure intervals.
  • Understand the Pythagorean comma.
Equal temperament

In twelve-tone equal temperament, the frequency ratio of a semitone is \(s=2^{1/12}\approx1.059463\), and an interval of \(k\) semitones has ratio \(2^{k/12}\). On a string of fixed tension, length is inversely proportional to frequency, so the successive string lengths form a geometric progression with ratio \(2^{-1/12}\). The interval between ratios \(r_1\) and \(r_2\) measures \(1200\log_2(r_2/r_1)\) cents.

Why the octave cannot be divided purely

A pure fifth has ratio \(3/2\). Twelve fifths should come back to seven octaves, but \((3/2)^{12}=531441/4096\approx129.75\) while \(2^7=128\). No power of \(3/2\) equals a power of 2, since \(3^a=2^b\) has no solution in positive integers. The excess ratio \(3^{12}/2^{19}\approx1.01364\), about 23.5 cents, is the Pythagorean comma. Equal temperament shrinks each fifth by a twelfth of it, to exactly 700 cents.

Stevin’s claim was polemical. He held that the equal intervals are the true ones and that the simple ratios of Greek theory are an error. Historians, notably H. Floris Cohen, read the treatise as an argument about consonance more than a tuning manual. Zhu Zaiyu in China computed \(2^{1/12}\) to high precision at about the same time; the two were independent.

Exercise 10.1Find the size in cents of the pure fifth \(3/2\) and the pure major third \(5/4\), and compare with the equal-tempered fifth and third.

Show solution
\(1200\log_2 1.5\approx701.96\) cents against 700: the tempered fifth is about 2 cents flat. \(1200\log_2 1.25\approx386.31\) cents against 400: the tempered third is about 13.7 cents sharp, which is why it sounds rough to trained ears.

Exercise 10.2A monochord string is 1000 units long. Give, to the nearest unit, the lengths sounding the tempered major third and fifth above it.

Show solution
Lengths are \(1000\cdot2^{-k/12}\). Third, \(k=4\): \(1000/1.259921\approx794\). Fifth, \(k=7\): \(1000/1.498307\approx667\) (the pure fifth would be \(666\tfrac23\)).
Part III · Life and legacy
Lesson 11

Quartermaster, teacher, Copernican, 1600–1617

In January 1600 Leiden University opened a separate course for surveyors and military engineers, taught in Dutch, at the request of Maurice and Stevin. Stevin wrote its curriculum, an Instruction that combined theory with practice: arithmetic with decimals, a selection of Euclid, trigonometric tables and surveying, measurement of volumes and wine gauging, and then fortification. The first teachers were Ludolph van Ceulen, famous for computing \(\pi\) to 35 places, and Simon Fransz van Merwen. Frans van Schooten the elder later continued the course. It is often called the first engineering school in the Netherlands, and it led to no degree.

From about 1604 Stevin served as quartermaster-general of the army of the States-General, responsible for laying out camps and fortifications. His Castrametatio (1617) describes the layout of a military camp, drawing on Maurice’s camp at the siege of Jülich in 1610. With it appeared a treatise on new kinds of sluices to control water levels in fortress moats.

Between 1605 and 1608 he published the Wisconstighe Ghedachtenissen, a large collection based on his lessons for Maurice. It covers trigonometry, geodesy, perspective, navigation and astronomy. Willebrord Snellius translated much of it into Latin as Hypomnemata mathematica. Its astronomical part, De Hemelloop (1608), defended the Copernican system without hedging. Earlier, in De Havenvinding (1599), Stevin had proposed finding longitude at sea from the deviation of the compass needle from true north. The method was sound in principle but failed in practice, because magnetic declination does not vary with longitude in any simple way.

DiscussionStevin wanted engineers trained in Dutch, in a practical course outside the Latin degree. What did this gain and lose for Dutch science, compared with the Latin culture of the universities?

Lesson 12

Death and legacy

Stevin married late, in 1610 or 1614 according to different sources, and had four children. In 1612 he bought a house on the Raamstraat in The Hague. He died in 1620, probably in February, most likely in The Hague; some sources give Leiden. He left a widow and young children. His son Hendrick Stevin became a scientist in his own right and later edited his father’s unpublished papers.

In 1634 Albert Girard’s French edition, Les Œuvres mathématiques de Simon Stevin, brought his work to a wider European readership. His name survives in many places: a statue by Eugène Simonis stands on the Simon Stevinplein in Bruges, unveiled in 1847; the Dutch research council awards an annual Stevin Prize; a Belgian research vessel bears his name.

Legacy

  • Arithmetic: decimal fractions for everyday use, and with them the habit of approximating rather than computing exactly. Metric units realized his proposal for decimal measures.
  • Number: the view that all magnitudes, rational or not, positive or negative, are numbers, which became the real-number line.
  • Mechanics: the law of the inclined plane, the composition of forces, and the use of the impossibility of perpetual motion as a principle.
  • Fluids: pressure depends on depth alone; the solidification argument; forces on walls by bounding sums.
  • Language and teaching: a Dutch scientific vocabulary and a model of engineering education that joined theory and practice.

Final projectChoose one of Stevin’s arguments from Part II (the wreath, solidification, or decimal search for roots). Write a two-page account for a strong first-year student that presents it in his terms, then in modern notation, and identifies exactly which assumption a modern proof must add.

Reference

Connections

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

Students

  • Maurice of Nassau

Influenced by

Influenced

Colleagues and collaborators

  • Jan Cornets de Groot 1586
  • Ludolph van Ceulen 1600-1610
  • Willebrord Snellius

Family

  • Hendrick Stevin

Also mentioned in this life

Fields

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Simon Stevin, MacTutor History of Mathematics.
  2. Simon Stevin, Wikipedia (with references).
  3. Delft tower experiment, Wikipedia.
  4. The early years of mathematics education at Leiden University, 1575–1645, Leiden Special Collections blog.
  5. E. J. Dijksterhuis, Simon Stevin: Science in the Netherlands around 1600, Martinus Nijhoff, 1970.
  6. J. T. Devreese and G. Vanden Berghe, ‘Magic is No Magic’: The Wonderful World of Simon Stevin, WIT Press, 2008.

Works and mathematics

  1. The Principal Works of Simon Stevin, 5 vols., ed. E. J. Dijksterhuis, D. J. Struik, A. Pannekoek and others, Swets & Zeitlinger, 1955–1966 (Dutch text with English translation).
  2. S. Stevin, De Thiende, Leiden, 1585; English translation by R. Norton, Disme: The Art of Tenths, London, 1608.
  3. S. Stevin, De Beghinselen der Weeghconst and De Beghinselen des Waterwichts, Leiden, 1586.
  4. A. Girard (ed.), Les Œuvres mathématiques de Simon Stevin de Bruges, Leiden, 1634.
  5. E. Mach, The Science of Mechanics, ch. 1 (on the inclined plane and the wreath).
  6. H. F. Cohen, Quantifying Music: The Science of Music at the First Stage of the Scientific Revolution, 1580–1650, Reidel, 1984.
  7. D. J. Struik, The Land of Stevin and Huygens, Reidel, 1981.