faculty of mathematics and natural sciences

Veltmann’s device for solving a system of linear equations

Bachelor Thesis Mathematics July 2012 Student: H.J. Veenstra First Supervisor: Prof. dr. J. Top Second Supervisor: dr. M.K. Camlibel

Abstract

In 1892 the ‘Deutsche Mathematiker Vereinigung’ (German mathematician society) pub- lished a catalogue of (physico-)mathematical models, devices and instruments. In this catalogue we find the description of a device invented by professor W. Veltmann. His device could be used to solve a system of linear equations by using hydromechanics. In this thesis we tried to discover Veltmann’s motives to build such a device. Hereafter we looked at the device itself. We found that Veltmann might have built this device for educational purposes. The de- vice works because of an equilibrium according to the law of the lever and an equilibrium between the buoyant force and the gravitational force. Veltmann’s device works properly for inconsistent systems and systems without a unique solution. However, there are re- strictions that need to be posed on amounts of fluid that are added in order to make the device work properly.

Contents

1 Introduction 4

2 Historical Background 5 2.1 Biography of Wilhelm Veltmann ...... 5 2.2 Walther Dyck’s catalogue ...... 6 2.3 Why did Veltmann build his device? ...... 9 2.4 Similar devices ...... 9

3 The device for one equation 10 3.1 The description by Veltmann ...... 10 3.2 The appearance of the device ...... 14 3.3 Solving the equation ...... 16 3.4 Practicability ...... 17

4 Why the device works 18 4.1 Physical background ...... 18 4.2 Computational background ...... 22 4.3 The device in the general case ...... 25

5 Restrictions 28

5.1 Restrictions on ti ...... 28 5.2 Precision ...... 31 5.3 Error reduction ...... 32

6 Properties 33 6.1 Systems without a unique solution ...... 33 6.2 Inconsistent systems ...... 35

7 Conclusion 38

A A list of Veltmann’s publications 44

2 B Computations for one equation 47 B.1 Properties of the box ...... 49

B.2 Expressing sin(θ˜) in terms of a1.0 and a1.1 ...... 50

B.3 Expressing the other variables in terms of a1.0 and a1.1 ...... 52

B.4 Calculating y0 and y1 and the solution of the linear equation ...... 54

C Computations for a system with more than one equation 57 C.1 Some equations for the general case ...... 58 C.2 Results for two equations ...... 60

3 Chapter 1

Introduction

In 1892 the ‘Deutsche Mathematiker Vereinigung’ (German mathematician society) pub- lished a catalogue of (physico-)mathematical models, devices and instruments [vD92]. This catalogue contains descriptions of all kinds of devices and geometric models, some of which still can be found in museums or in the archives of several universities. However, some other devices seem to have completely disappeared. In this catalogue we find the description of a device invented by professor W. Veltmann, hereafter called ‘Veltmann’s device’. This device could be used to solve a system of linear equations, not in an alge- braic way but by using hydromechanics.

In mathematics there are more prevailing methods to solve such systems, which raises the question: “Why would somebody invent and build such a device?”. In order to answer this question we first look at the life of professor Veltmann. Who was he and what where the main subjects of his research? We will also look at the develop- ment of the previously mentioned catalogue. Once this historical background is explored, we will look at the device itself: “How is it used and why does it work?” To fully understand this, some physical principles are recalled. Subsequently a way to reduce the error in our solution is proposed. In the end the behaviour of this device in some special cases is investigated. What happens if we have a system without a unique solution? Or a system without a solution at all?

This thesis would not have been possible without the guidance and support of many people. I want to thank my first supervisor, prof. dr. J. Top, who has managed to make me enthusiastic about an old, strange device. I also want to thank dr. M.K. Camlibel, for being my second supervisor and for reading this thesis during his holiday. Furthermore I want to thank my study mate, Lianne. Day after day we were working on our theses, motivating and helping each other through. I also want to thank my parents and my boyfriend, who had to listen to long stories about old devices every time I spoke to them.

4 Chapter 2

Historical Background

In this chapter some information will be given about Wilhelm Veltmann, the inventor of the device at issue. Furthermore, background information is given about Walther Dyck and his catalogue. In the end we will look at devices that are similar to Veltmann’s device.

2.1 Biography of Wilhelm Veltmann

Wilhelm Veltmann was born on the 29th of December 1832 in Hagen-Bathey, Germany [HAN]. We have no information about his youth. The first record that we found was that of his first article, which appeared in ‘Astrono- mische Nachrichten’ in 1870 [Vel70]. This article suggests that he then lived in Bonn. Based on the addresses in his publications we know where he lived in the subsequent years. In 1871 he lived in Wiedenbr¨uck [Vel71]. In 1873 he had moved to Holzminden, where he worked at the Baugewerkschule (a school for builders)[Vel73]. In 1875 he lived in D¨urenwhere he was a teacher at a secondary school [Vel75]. In 1877 he became a rector in Remagen [Vel77]. After 1877 his whereabouts are unknown for 5 years. In 1882 he published some articles while he was living in Frankenthal(Pfalz) [Vel82]. In 1883 or 1884 he became a teacher at the landwirtschaftlichen Akademie (Agricultural Academy) in Poppelsdorf-Bonn, which later became part of the ‘Rheinischen Friedrich- Wilhelms-Universit¨at’in Bonn (University of Bonn)[Vel84b]. In 1892 he became a profes- sor at this academy.[Vel92] On the 6th of March 1902 he died in Hagen-Bathey, his birthplace [HAN].

Wilhelm Veltmann wrote many papers. A list of these is found in the Appendix (A).

5 From this list it can be seen that in the first years of his scientific career Veltmann was mainly interested in astronomy and physics. Later he treats a lot of isolated subjects. One subject that keeps coming back however, is the theory of ‘Beobachtungsfehler’(observation errors) in inter alia [Vel92]. This cited article was published in the same year as Dyck’s catalogue, in which the description of Veltmann’s device is found. Six years earlier Velt- mann wrote an article about “Aufl¨osunglinearer Gleichungen” (the solution of linear equations) [Vel86]. It is a possibility that Veltmann wanted to tell his students something about iterative approximations of the solution of a linear system and the observation er- rors that thereby emerge. As an example of an iterative process, his device can be used. This theory is supported by the fact that Veltmann himself poses, in the description of his device, a way to reduce the observation errors by using an iterative process. There is however no real source for this theory, so it is just a speculation.

2.2 Walther Dyck’s catalogue

In 1892 the Deutsche Mathematiker Vereinigung decided to organize an exposition of mathematical and physico-mathematical, instruments and devices on the occasion of their annual meeting. This meeting was to be held in September 1892 in N¨urnberg, Germany [vD92, p. III]. The task of organizing the exposition was assigned to Walther Dyck, a professor at the ‘Technische Hochschule’ (Technical University) in M¨unchen.

Walther Franz Anton Dyck was born on the 6th of December 1856 in M¨unchen [zSW59, p. 210]. Here he went to high school, where Oskar Miller, who later founded the German Museum, was one of his friends. He studied engineering at the Technical University in M¨unchen. Dyck was much influenced by his contact with Felix Klein, a famous mathematician at this university [Has99, p. 7]. Klein was convinced of the usefulness of looking at geometri- cal models for educational purposes. These models were produced by his own students. Walther Dyck also produced some of these models. In 1879 he received his PhD as a student of Klein with a dissertation about Riemannian surfaces. After his graduation he became one of Klein’s assistants. He worked on the theory of groups and was the first mathematician to define the abstract definition of a group. In 1884 Dyck became a professor in M¨unchen. He used a lot of the educational principles that he learned from Klein. He also made his students produce mathematical models and graphic representations of geometrical objects.

6 Dyck played an important role in the foundation of the Deutsche Mathematiker Vereini- gung in 1889. At their assembly of 1891 he suggested to organize an exposition in 1892 [Has99, p. 8]. He might have obtained this idea from a similar exposition in London in 1876. In the introduction to the catalogue which appeared on the occasion of the exposi- tion of 1892 he writes that there has been a great development in the use and usefulness of mathematical models. He says: “So erschien es naturgem¨ass,die Gelegenheit der diesj¨ahrigenVersammlung der Deutschen Mathematiker-Vereinigung, die gleichzeitig mit der Versammlung der Gesellschaft deut- scher Naturforscher und Aerzte in N¨urnberg tagen sollte, zu ben¨utzen,um ein zusam- menh¨angendesBild dieser Entwickelung vorzuf¨uhren.”[vD92, p. IV] (So it came naturally, to use the occasion of this years meeting of the German Mathe- matical Society, which would take place simultaneously with the meeting of the Society of German Scientists and Physicians in N¨urnberg, to demonstrate a coherent picture of this development)

A lot of institutes made devices and models available. Submissions came from Ger- many, the USA, France, Italy, the Netherlands, Norway, Austria-Hungary, Russia and Switzerland. The cooperating institutes decided to produce a catalogue with accurate descriptions and figures. This catalogue had to be composed in the last five weeks before the exposition. But, when the moment the show would take place was there, a disaster occurred. There was a cholera-breakout in Germany, which caused the organizers to cancel the exposition. The members of the Deutsche Mathematiker Vereinigung decided to postpone the exposi- tion to their meeting in 1893 in M¨unchen. Nevertheless the catalogue was still published. Not anymore to give a complete overview of all models and devices, but to show the intention to organize a new exposition and to show the gaps in the collection that still had to be filled.

The catalogue starts with 8 articles from renowned mathematicians including F.Klein and L. Boltzmann. Subsequently there are about 90 pages with devices and instru- ments that can be used for arithmetic, algebra, function theory and integral calculus. Veltmann’s device is described in the section about algebra and function theory, in the subsection “Apparate zur Aufl¨osungvon Gleichungen und zur Construction functioneller Abh¨angigkeiten” (Devices for solving equations and constructing functional dependen- cies). The second part of the catalogue consists of descriptions of geometrical models (80 pages) and in the third part instruments and models for applied mathematics are described. This leads to a total of 270 pages with descriptions of models, instruments and devices.

7 The catalogue also contains a ‘Nachtrag’ (supplement), that was published at the occa- sion of the exposition in 1893 in M¨unchen. The introduction to this supplement tells that this supplement was meant to fill in some gaps and correct some errors in the original catalogue. But the supplement also contains some newly discovered components. Klein says at the end of his article about “Geometrisches zur Abz¨ahlungder reellen Wurzeln algebraischer Gleichungen” (geometry for counting the number of real roots of algebraic equations) in the original catalogue: “Hier ist offenbar ohne geeignete Modelle nicht durchzukommen. Es wird sehr dankenswert sein, wenn jemand die Herstellung solcher Modelle in die Hand nehmen wollte”[vD92, p. 15]. (You obviously can’t get through this without suitable models. It would be commendable when someone would undertake the manufacturing of such models) In the supplement we find a model of Karl Doehlemann, also a docent in M¨unchen, called “zur Discussion der Gleichung 3. Grades” (to discuss the third degree equation) [vD92, Nachtrag p. 28]. In the description of this model Doehlemann refers to Klein’s article: “F¨urdie hiebei n¨otigenBetrachtungen kann das folgende Modell von Vorteil sein”[vD92, Nachtrag p. 28]. (For the considerations necessary in this case, the following model can be beneficial) Apparently a new model was manufactured in the year between the cancelled exposition and the exposition of 1893. Once the supplement was released, as far as we know no new additions or editions appeared.

After the exposition in 1893 in M¨unchen, which really took place, Dyck also organized an exposition at one of the famous world fairs, ‘The World’s Columbian Exposition’, which was held in Chicago in 1893. Dyck’s exposition was mainly used to show Germany’s superiority in mathematics. The exposition consisted of a collection of mathematical models, mathematical writings and some statues of famous German mathematicians.

When he was back in Germany, he still didn’t continue his own mathematical work. He worked together with Klein on some anthologies that gave an overview of the develop- ments in mathematics at that time. In M¨unchen he was committed to the educational work there. He changed the Technical University into a scientific University, where mathematics and physics were applied to technical problems [Has99]. In 1900 he became rector at this university. In 1901 he was awarded the ‘Verdienstordens der Bayerischen Krone (Ritter)’ (knight- hood in the Order of Merit of the Bavarian Crown) [zSW59, p. 210], [Isi]. From then on his name became ‘Walther von Dyck’. Von Dyck was closely involved in the foundation of ‘Das Deutsche Museum’ (the German museum) in M¨unchen in June 1903. The founding of this museum was commissioned

8 by the chairman of the Bavarian District Association of German Engineers and Royal architect Oskar von Miller, von Dycks old high school friend. For more information about the influence of von Dyck on the museum, see [Has99]. Walther von Dyck died on the 5th of November 1934 in M¨unchen, his place of birth [zSW59, p. 210].

2.3 Why did Veltmann build his device?

As we saw before, we are not aware of any source in which Veltmann’s motives for building his device are found. In his days there were more prevailing methods to solve a linear system, so it doesn’t seem logical that he actually used his device to solve systems. It is possible that he used it to explain some concepts to his students. This is in line with the motives of Walther Dyck and Felix Klein to make models and devices. They were convinced of the usefulness of looking at geometrical models for educational purposes.

2.4 Similar devices

We are also interested in devices similar to Veltmann’s device, because these devices might have influenced Veltmann in designing his device. J. Frame gives a short description of devices used to solve (systems of) equations, which are not necessarily linear [Fra45]: C.V. Boys’ “Machine for solving equations” [Boy86] is a device that uses levers to represent each coefficient in a non-linear equation, but no water or other fluid. This device was introduced in 1886 mainly to show some interesting principles, and had no practical use. M. Demanet writes about “R´esolutionhydrostatique de l’´equationdu troisi`emedegr´e” [Dem98]. He uses a device to solve cubic equations. This device uses the principle of communicating vessels, just like Veltmann’s device but was introduced later, in 1898. M.G. Meslin uses levers and communicating vessels for “Une machine `ar´esoudreles ´equations”[Mes00], that was introduced in 1900. The constant term is represented by weights on a scale. Another device is a device that is designed by sir William Thomson (Lord Kelvin) [Tho78] in 1878. This device can be used to solve systems of linear equations, just like Veltmann’s device. However, this device works by a different principle; it uses pulleys to transfer distances from one equation to the other. Thomson’s device is built for 9 equations with 9 unknowns by John Wilbur, a professor at MIT in 1936 [Wil36]. This original device has disappeared (photo’s can however still be found), but in 1945 a replica was made in Japan, which still is exposed in the ‘Japan National Museum of Nature and Science’[CEED]. Veltmann’s device was one of the first devices using a combination of levers and the principle of communicating vessels. It seems logical that he knew some of the older devices and combined them.

9 Chapter 3

The device for one equation

Little is known about Veltmann’s device. Below the entire text is displayed, as it can be found in [vD92]. This describes the general case. It doesn’t really make clear what the device looks like and why it works. Another description is given in [Vel84a]. This gives an a little bit more detailed description of the device, but the essential parts of this source have become unreadable in time. Furthermore, we have found no indication that Veltmann’s device still exists. In order to understand the device, Veltmann’s device is described in this chapter in the special case of solving one linear equation. In section 4.3 we will discuss the general case.

3.1 The description by Veltmann

The description of the device as it is given by Wilhelm Veltmann in [vD92, p. 155-158]

10 11 12 13 3.2 The appearance of the device

The device consists of a box, which we choose to have size 50 cm×30 cm×30 cm, because this seems to be a manageable size for a device. Other dimensions can be chosen, but this will change the expressions that are obtained later. A top view of this box is given in figure 3.1. The corners of the box are labeled C,D,E and F . Over this box there is a lever, L1, with its fulcrum on the line AB. This axis of rotation is with the top of the box on one level.

Figure 3.1: A top view of the device

The box is filled with water. On one side of the box there is a scale on which the water level inside the box can be measured. There are two cylinders attached to lever L1 with radius O and height 20 cm, closed at the bottom and open at the top. These cylinders are named 1.0 and 1.1 and are indicated by circles in figure 3.1. On the outside of the box there are two cylinders with radius O and height 30 cm (the same height as the box). These cylinders are named 0 and 1. Cylinder 0 is connected with cylinder 1.0 by a little tube. Cylinder 1 is connected with cylinder 1.1 in the same way. Cylinders 0 and 1.0 together are called cylinder chain 0.

14 Likewise cylinders 1 and 1.1 together are called cylinder chain 1. The cylinders 0 and 1 can be filled with a fluid, and are provided with a scale on which the fluid level in centimeters inside these cylinders can be read. Because these cylinders are connected to the cylinders on the lever by a tube, the fluid level in all the cylinders of a cylinder chain will be equal. This is called the principle of communicating vessels [PT34, p. 20].

For the process it is necessary that a fluid column inside a cylinder, with radius I (the inner radius of a cylinder) has the same weight as a water column of the same height and with radius O (the outer radius of a cylinder). That is

π · I2 · h · f = π · O2 · h · w, where h is the height of the water or fluid column, f is the density of the fluid and w is the density of water. This can be reduced to

O2 f I2 · f = O2 · w ⇒ = . I2 w This means that the ratio of the density of the fluid and the density of water has to be the same as the ratio of O2 and I2.

3.2.1 Numerical example

Assume the outer radius of a cylinder, O, is 1 cm. If the thickness of the sides of the cylinder is 0.1 cm, the inner radius, I, is 1 − 0.1 = 0.9 cm. The ratio of O2 and I2 is

O2 12 1 = = ≈ 1.23 I2 0.92 0.81

The density, w, of water, at 20◦C, is 0.998 g/cm3 [wat]. The density, f, of the fluid has to f 3 satisfy 0.998 = 1.23. This gives f = 0.998 · 1.23 = 1.23 g/cm . For instance glycerol could be used, which has a density of 1.26 g/cm3. We could also use a saline with a proper concentration. Now we have that a water column of 1 cm has a volume of π · O2 · 1 = π · 12 · 1 = π cm3, this has a weight of π · 0.998 ≈ 3.14 grams. A fluid column of 1 cm has a volume of π · I2 · 1 = π · 0.92 · 1 = 0.81π cm3, this has a weight of 0.81π · 1.23 ≈ 3.14 grams, what was necessary for the process.

Now a fluid column in a cylinder can be indicated by the height in cm of a water column with a radius of O cm.

15 3.3 Solving the equation

The aim is to solve a linear equation. This is an equation of the form

a1.0 + a1.1 · x = 0 (3.1)

To solve this equation, cylinder 1.0 is attached to the lever at a distance a1.0 from the fulcrum of the lever. Here the positive direction is taken to be pointing towards side CF , as indicated in figure 3.1. Cylinder 1.1 is attached to the lever at a distance a1.1 from the fulcrum. While the lever is kept in horizontal position, the box is filled with water until the wa- ter level is 20 cm. Still keeping the lever in horizontal position, cylinders 0 and 1 are filled with fluid until the fluid levels of both cylinder chains are equal to the water level. Releasing the lever can cause a little perturbation in the equilibrium. The lever can be balanced again by adding a little amount of, for instance, sand on the balance. This is called ‘taring the equilibrium’. In figure 3.2 this situation is shown for a1.0 = −15 and a1.1 = 5.

Figure 3.2: The box filled with water and the cylinders filled with (red) fluid for the case a1.0 = −15, a1.1 = 5

16 The lever still might not be in an exact horizontal position, which will cause the water level in the box and the fluid levels in the cylinders to rise or to fall a little. The water level, b, in the box is measured. Likewise the fluid levels in cylinder 0 and 1 are measured, these levels are called c0 and c1 respectively. The following values are calculated for later use:

u0 = c0 − b, u1 = c1 − b.

In the ideal case the water and fluid levels are equal, so u0 = u1 = 0.

Now ti cm of fluid is added to cylinder i, for at least one i.‘ti cm’ indicates a fluid column of ti cm in cylinder i. This is equal to

2 3 2 π · I · ti cm = π · I · ti ml.

This will cause the lever to tilt over to one side. The water level in the box and the fluid levels in the cylinders will change, because the cylinders attached to the lever will be submerged further or less far into the water than they were in the starting position. The new water level in the box is measured and is denoted by ˜b. Also the fluid levels in cylinder 0 and 1 are measured, and these levels are denoted byc ˜0 andc ˜1 respectively. The following values are calculated:

u˜0 =c ˜0 − ˜b, u˜1 =c ˜1 − ˜b.

These values are used to calculate

y0 =u ˜0 − u0 =c ˜0 − ˜b − (c0 − b) (3.2)

y1 =u ˜1 − u1 =c ˜1 − ˜b − (c1 − b). (3.3)

The y0 and y1 satisfy the equation a1.0 · y0 + a1.1 · y1 = 0. The solution for x in the equation a + a · x = 0 is thus given by x = y1 . 1.0 1.1 y0

3.4 Practicability

We had plans to build this device, but concluded that this was technically not feasible. The little tubes connecting the cylinders and the attaching of the cylinders to the levers would be almost impossible to realize. Veltmann himself already states some difficulties in the technical realization in [Vel84a]. We can’t use much of his instructions, because that part of the article is unreadable in our source. The original device was constructed by ‘Mechaniker Wolz’ in Bonn, who must have been a true craftsman. Although we don’t have a real working model, we are still able to explain in the next chapter why the device works.

17 Chapter 4

Why the device works

Why do y0 and y1 from the preceding chapter satisfy a1.0 · y0 + a1.1 · y1 = 0? This will be shown in this chapter. Also a numerical example is given, in which the equation 3x = −4 is solved. Furthermore the use of Veltmann’s device for larger linear systems is explained.

4.1 Physical background

This description starts with the starting position as described in section 3.3. A detailed image of this position is given in figure 4.1. The lever is tilting over a little to the left, this could of course also be to the right.

Figure 4.1: A detailed image of the starting position

18 All distances are measured in centimeters. The meaning of the indicated variables is as follows (for i = 0, 1): a1.i is the distance from cylinder 1.i to the fulcrum, corresponding to the coefficients of a1.0 + a1.1x = 0. b is the water level inside the box. d1.i is the distance from the bottom of the box to the bottom of cylinder 1.i. e1.i is the distance from the bottom of cylinder 1.i to the water surface. f1.i is the height of the fluid column in cylinder 1.i. In the ideal case, where the weight of the little tubes, the cylinders and the lever can be neglected, e1.i = f1.i; the fluid level in cylinder chain i is exactly equal to the water level in the box.

There are two different forces acting on cylinder 1.0. Firstly, there is the gravitational force. The mass of this cylinder is assumed to be equal to the mass of the fluid column inside this cylinder (radius = I). The fluid was chosen in such a way that this mass is equal to the mass of a water column of the same height and with radius O. The height of the fluid column in the cylinder is f1.0, so the volume of a water column of this height 2 and with radius O is π · O · f1.0. The mass of this column is 2 π · O · f1.0 · w = m1.0 where w is the density of water. The gravitational force on the cylinder is given by

Fg1.0 = m1.0 · g where g is the standard gravity, g = 9.81 m/s2.

The second force is the buoyant force, the magnitude of this force is given by Archimedes’ principle: “The vertical force of buoyancy on a submerged object is equal to the weight of fluid the object displaces” [Pic08, p. 41].

Cylinder 1.0 displaces a water column of radius O and height e1.0. The buoyant force is then given by

FA1.0 = n1.0 · g where n1.0 is the mass of a water column of height e1.0 and g is again the standard gravity.

Here n1.0 is given by 2 n1.0 = π · O · e1.0 · w

The net downward force F1.0 working on cylinder 1.0 is now given by

F1.0 = Fg1.0 − FA1.0 2 2 = π · O · f1.0 · w · g − π · O · e1.0 · w · g 2 = π · O · w · g · (f1.0 − e1.0). (4.1)

19 Likewise the downward force F1.1 working on cylinder 1.1 can be found. This force is given by

2 F1.1 = π · O · w · g · (f1.1 − e1.1). (4.2)

Now Archimedes’ ‘Law of the lever’ [Pic08, p. 497] can be used. This law states that, when the lever has reached an equilibrium, it holds that F1.0 · p1.0 = F1.1 · p1.1 where p1.i is the perpendicular distance from cylinder 1.i to the fulcrum on the line AB. F1.i · p1.i is called the moment of cylinder 1.i. This law can also be expressed as:

sum of the moments of the cylinders on the left of the fulcrum = sum of the moments of the cylinders on the right of the fulcrum. (4.3)

In the law of the lever all distances are assumed to be positive. In the case of Veltmann’s device, a1.i, and thus p1.i is negative if cylinder 1.i is on the left of the fulcrum. All terms belonging to cylinders on th left of the fulcrum have to be multiplied by −1 to make these p1.i’s positive. (4.3) can thus be written as: X X (−1)F1.i · p1.i = F1.j · p1.j cylinder 1.i on the left of the fulcrum cylinder 1.j on the right of the fulcrum This can be rewritten as n X F1.i · p1.i = 0. i=1

Figure 4.2: The distance from a cylinder to the fulcrum

The distance p1.i can be expressed in terms of a1.i and the angle θ between lever L1 and the horizontal line, measured in radians. Here θ is taken to be positive in the counter- clockwise direction.

20 It can be seen in figure 4.2 that cos(θ) = p1.0 , such that a1.0

p1.0 = a1.0 · cos(θ) and likewise

p1.1 = a1.1 · cos(θ).

The thus obtained expression for the law of the lever is: X F1.i · a1.i · cos(θ) = 0. i Substituting (4.1) and (4.2) gives

X 2 π · O · w · g · (f1.i − e1.i) · a1.i · cos(θ) = 0. i

This expression is divided by π · O2 · w · g · cos(θ) to obtain X (f1.i − e1.i) · a1.i = 0. i This also can be written as X 0 = (f1.i + d1.i − (e1.i + d1.i)) · a1.i. i From figure 4.1 we see that this is equal to X 0 = (ci − b) · a1.i. i So the forces acting on the cylinders are proportionate to the difference between the fluid level in the cylinder and the water level in the box.

In the ideal situation the lever is in exact horizontal position before adding any fluid.

In horizontal position f1.i = e1.i, or ci = b, thus all moments are equal to zero. If the notation u0 = c0 −b and u1 = c1 −b is used, this is equal to u0 = u1 = 0. After a specified amount of fluid is added, the lever moves to a new equilibrium. The law of the lever now states, for 2 cylinders, that

a1.0 · (˜c0 − ˜b) + a1.1 · (˜c1 − ˜b) = 0 (4.4)

(tildes are added to the variables to indicate that fluid is added).

(y0, y1) = (˜u0 − u0, u˜1 − u1) = (˜c0 − ˜b, c˜1 − ˜b) (4.5) is thus a solution of the equation a1.0 · y0 + a1.1 · y1 = 0.

21 Dividing this equation on both sides by y gives a + a · y1 = 0. The solution for x in 0 1.0 1.1 y0 the equation a + a · x = 0 is thus given by x = y1 . 1.0 1.1 y0

In general the lever will tilt over a little to one side in the starting position. In this position an additional force, caused by for instance the weight of the little tubes, is acting on the lever. Before fluid is added it holds that

a1.0 · (c0 − b) + a1.1 · (c1 − b) − k = 0 where k represents the additional force. Therefore, the additional force k is given by

k = a1.0 · (c0 − b) + a1.1 · (c1 − b)

= a1.0 · u0 + a1.1 · u1. (4.6)

After fluid is added the lever moves to a new equilibrium, but the additional force k stays the same (the device is not changed). For this equilibrium the law of the lever states

0 = a1.0 · (˜c0 − ˜b) + a1.1 · (˜c1 − ˜b) − k

= a1.0 · (˜c0 − ˜b) + a1.1 · (˜c1 − ˜b) − (a1.0 · u0 + a1.1 · u1)

= a1.0 · (˜c0 − ˜b − u0) + a1.1 · (˜c1 − ˜b − u1)

= a1.0 · (˜u0 − u0) + a1.1 · (˜u1 − u1), (4.7) which is the same as equation(4.4) where we had u0 = u1 = 0. So the solution of a1.0 · y0 + a1.1 · y1 = 0 is again given by

(y0, y1) = (˜u0 − u0, u˜1 − u1).

The solution for x in the equation a + a · x = 0 is still given by x = y1 . 1.0 1.1 y0

4.2 Computational background

Now the solution is known. But to further investigate the device, it would be nice to be able to predict the final water and fluid levels and the position of the lever. Computing these outcomes takes a lot of effort. Below we only give the resulting expressions, the exact derivation can be found in appendix B.

We found that ( 1 (a · t + a · t ) − a · u − a · u )(1500 − 2 · π · O2) ˜ 2 1.0 0 1.1 1 1.0 0 1.1 1 sin(θ) = − 2 2 2 (4.8) 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O where θ˜ is the angle between the lever and the horizontal line after fluid is added, measured in the counter-clockwise direction. ti is, as before, the amount of fluid that is added to cylinder i, expressed in centimeters.

22 We also calculated that (a + a ) · ( 1 (a · t + a · t ) − a · u − a · u ) · π · O2 ˜ 1.0 1.1 2 1.0 0 1.1 1 1.0 0 1.1 1 b = 20 + 2 2 2 (4.9) 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O and

1 2 1 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) c˜i = 20 + ti − a1.i · 2 2 2 . 2 1500(a1.0 + a1.1) + 4 · a1.0 · a1.1 · π · O (4.10)

4.2.1 Numerical example for one equation

In practice we won’t work with a1.0 and a1.1 but with real numbers as coefficients in the linear equation (a1.0, a1.1 ∈ R). Assume that we wanted to solve the equation 4x = −3.

Writing it in the form a1.0 + a1.1 · x = 0 gives 3 + 4x = 0

Take a1.0 = 3 and a1.1 = 4. Assume the outer radius of the cylinders to be 1 cm, so

O = 1 cm. Add an equal amount of fluid to both cylinder chains, say t0 = t1 = 1 cm.

We assume the lever to be in exact horizontal starting position, so u0 = u1 = 0. Substituting all of these values in (4.8) and calculating in Maple gives

( 1 (3 · 1 + 4 · 1) − 3 · 0 − 4 · 0)(1500 − 2 · π · 12) sin(θ˜) = − 2 750(32 + 42) + 2 · 3 · 4 · π · 12 7 (1500 − 2 · π) 7 · π − 5250 = − 2 = 18750 + 24 · π 24 · π + 18750 ≈ −0.277710398967654, which leads to

θ˜ ≈ sin−1(−0.277710398967654) = −0.281409934832440 rad −0.281409934832440 · 360◦ = ≈ −16.1236015789504◦. 2 · π

Calculating the fluid levels in the cylinders and the water level in the box gives:

(a + a ) · ( 1 (a · t + a · t ) − a · u − a · u ) · π · O2 ˜ 1.0 1.1 2 1.0 0 1.1 1 1.0 0 1.1 1 b = 20 + 2 2 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O (3 + 4) · ( 1 (3 · 1 + 4 · 1) − 3 · 0 − 4 · 0) · π · 12 = 20 + 2 750(32 + 42) + 2 · 3 · 4 · π · 12 49 · π = 20 + 2 ≈ 20.0040885732720 cm 18750 + 24 · π The water level in the box was 20 cm in starting position, so it has risen about 40.8 µm.

23 Now the fluid levels in cylinder chain 0 and 1 are calculated:

1 2 1 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) c˜0 = 20 + t0 − a1.0 · 2 2 2 2 1500(a1.0 + a1.1) + 4 · a1.0 · a1.1 · π · O 1 ( 1 (3 · 1 + 4 · 1) − 3 · 0 − 4 · 0)(1500 − 2 · π · 12) = 20 + · 1 − 3 · 2 2 1500(32 + 42) + 4 · 3 · 4 · π · 12 41 7 (1500 − 2 · π) = − 3 · 2 ≈ 20.0834344015485 cm. 2 37500 + 48 · π The fluid level in cylinder chain 0 was 20 cm in starting position, so it has risen about 0.08 cm.

1 2 1 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) c˜1 = 20 + t1 − a1.1 · 2 2 2 2 1500(a1.0 + a1.1) − 4 · a1.0 · a1.1 · π · O 1 ( 1 (3 · 1 + 4 · 1) − 3 · 0 − 4 · 0)(1500 − 2 · π · 12) = 20 + · 1 − 4 · 2 2 1500(32 + 42) + 4 · 3 · 4 · π · 12 41 7 (1500 − 2 · π) = − 4 · 2 ≈ 19.9445792020647 cm. 2 37500 + 48 · π The fluid level in cylinder chain 1 was 20 cm in starting position, so it has fallen about 0.05 cm.

With these values we calculate y0 and y1.

From equation (4.7) it is know that yi =u ˜i − u1, whereu ˜i =c ˜i − ˜b. We thus have

y0 =c ˜0 − ˜b − 0 ≈ 20.0834344015485 − 20.0040885732720 ≈ 0.0793458282764729

y1 =c ˜1 − ˜b − 0 ≈ 19.9445792020647 − 20.0040885732720 ≈ −0.0595093712073547.

So a solution to the equation 3y0 + 4y1 = 0 should be given by

(y0, y1) = (0.0793458282764729; −0.0595093712073547).

Let’s check this: 3 · 0.0793458282764729 + 4 · −0.0595093712073547 = 0.

The solution to the original equation 3 + 4 · x = 0 is now given by y −0.0595093712073547 x = 1 = = −0.75. (4.11) y0 0.0793458282764729 As we see throughout this derivation the change in water and fluid levels is very small. It might not even be noticeable in practice. The restrictions that need to be put on the amount of fluid that is added, ti, are discussed in chapter 5.

24 4.3 The device in the general case

A system of more than one linear equation can be written as

a1.0 + a1.1 · x1 + a1.2 · x2 + ... + a1.n · xn = 0

a2.0 + a2.1 · x1 + a2.2 · x2 + ... + a2.n · xn = 0 . .

am.0 + am.1 · x1 + am.2 · x2 + ... + am.n · xn = 0. (4.12)

This system can be transformed to a homogeneous system by taking

a1.0 · y0 + a1.1 · y1 + a1.2 · y2 + ... + a1.n · yn = 0

a2.0 · y0 + a2.1 · y1 + a2.2 · y2 + ... + a2.n · yn = 0 . .

am.0 · y0 + am.1 · y1 + am.2 · y2 + ... + am.n · yn = 0, (4.13) which can be written as n X aj.i · aj.i = 0, j = 1, ..., m. (4.14) i=0 Now the device can be put into starting position.

In the case of system (4.13) the device consists of m levers L1, ..., Lm that all have their fulcrum on the line AB. To each lever Lj, n + 1 cylinders are attached at distances aj.0, ..., aj.n from the fulcrum. Now cylinder 1.i is connected to cylinder i on the outside of the box, but also to cylinder 2.i. Cylinder 2.i is connected to cylinder 3.i and so on. This means that the fluid levels in cylinder chain i (the cylinders i, 1.i, 2.i, ..., m.i) are all equal by the principle of com- municating vessels. A top view for the case n = m = 3 is given in figure 4.3.

The levers are held in horizontal position and the box is filled with water until b = 20 cm. Then all cylinders are filled with fluid until the fluid level in each chain is equal to the water level in the box. This means that (20 + m · 10) cm of fluid has to be added (20 cm in cylinder i and 10 cm in cylinder 1.i, 2.i, ..., m.i). The levers are released and the initial difference between the water level and the fluid level in cylinder chain i, ci − b, is measured. This difference is called ui.

Now ti centimeter of fluid is added to cylinder chain i (fluid can be added to a selection of cylinder chains or to all cylinder chains). The positions of the levers will change until an equilibrium is reached. Lever Lj is then making an angle of θ˜j with the horizontal line.

The difference between the fluid level,c ˜i in cylinder i and the water level ˜b in the box is measured. This difference is calledu ˜i.

25 Figure 4.3: A top view of the device for a system of three equations with three unknowns

The solution to system (4.13) is given by (y0, y1, ..., yn) = (˜u0 − u0, u˜1 − u1, ..., u˜n − un). Dividing by y gives the solution to system (4.12): (x , ..., x ) = ( y1 , ..., yn ). 0 1 n y0 y0

Just like in the case of one linear equation it can easily be seen why (y0, ..., yn) gives a solution to system (4.13).

We use the subscript ∗j.i for a variable belonging to cylinder j.i on lever Lj. As we have seen in section 4.1 the net downward force action on cylinder j.i in the starting position is given by 2 Fj.i = π · O · w · g · (fj.i − ej.i).

The law of the lever states that for each lever Lj it holds that

n X Fj.i · pj.i = 0, i=0 where pj.i is the perpendicular distance from cylinder j.i to the fulcrum of lever Lj. This distance is given by pj.i = aj.i · cos(θ˜j). So we have that

n X 2 π · O · w · g · (fj.i − ej.i) · aj.i · cos(θ˜j) = 0. i=0

26 2 Dividing by π · O · w · g · cos(θ˜j) gives

n X (fj.i − ej.i) · aj.i = 0 (4.15) i=0 for each lever Lj. This is equal to n X (fj.i + dj.i − (ej.i + dj.i) · aj.i = 0. i=0 From figure B.1 in appendix B we see that this equals

n X (ci − b) · aj.i = 0. i=0

The difference ci − b is named ui. In the ideal case ui = 0 for every i. If lever Lj is not in exact horizontal position before adding fluid, we take in account an additional force kj working on lever Lj such that

n n n X X X ( (ci − b) · aj.i) − kj = ( ui · aj.i) − kj = 0 ⇒ kj = ui · aj.i. i=0 i=0 i=0 After adding fluid we have that

n X u˜i · aj.i − kj = 0 i=0 n X (˜ui − ui) · aj.i = 0, (4.16) i=0 which means that (˜u0 − u0, u˜i − u1, ..., u˜n − un) = (y0, y1, ..., yn) is a solution of (4.13).

The computational background for the case m = n = 2 and the relevant equations for the general case can be found in appendix C. The expressions obtained for sin(θ˜1), sin(θ˜2), c˜0, c˜1, c˜2 and ˜b are way too large to express at this point.

27 Chapter 5

Restrictions

In this chapter the restrictions that need to be put on ti are discussed, as announced in section 4.2.1. We also take a look at the precision of Veltmann’s device. Furthermore a method to reduce the error in the final answer, introduced by Veltmann himself, is given.

First of all, notice that each lever Lj has a length up to a maximum of 50 cm. This means that the maximum distance from the fulcrum to each cylinder, aj.i can be at most

25 cm. If one of the coefficients aj.i is bigger than 25, try to divide all coefficients by a common factor or by for example 10. This won’t change the solution of the system.

5.1 Restrictions on ti

First we look at the case of one equation. From equation (B.4) we know that

d˜1.i = 10 + a1.i · sin(θ˜).

This number is not allowed to become smaller than 0, because it represents the distance between the bottom of the box and the bottom of cylinder 1.i. It is also not allowed to exceed 20 cm because then the cylinder would rise out of the water. Substituting the expression for sin(θ˜) (B.10) gives

1 2 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) 0 ≤ 10 + a1.i · − 2 2 2 ≤ 20 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O Which leads to 1 2 a1.i · 2 (a1.0 · t0 + a1.1 · t1)(1500 − 2 · π · O ) −10 ≤ 2 2 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O 2 a1.i · (a1.0 · u0 + a1.1 · u1)(1500 − 2 · π · O ) − 2 2 2 ≤ 10 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

28 This is equal to 2 a1.i · (a1.0 · u0 + a1.1 · u1)(1500 − 2 · π · O ) −10 + 2 2 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O 1 2 a1.i · 2 (a1.0 · t0 + a1.1 · t1)(1500 − 2 · π · O ) ≤ 2 2 2 ≤ 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O 2 a1.i · (a1.0 · u0 + a1.1 · u1)(1500 − 2 · π · O ) 10 + 2 2 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

2 2 2 750(a1.0+a1.1)+2·a1.0·a1.1·π·O Multiplying by 1 2 , which is in general a positive term because 2 (1500−2·π·O ) 2 2 2 2 750(a1.0 + a1.1) > 2 · a1.0 · a1.1 · π · O and 1500 > 2 · π · O , results in 7500(a2 + a2 ) + 20 · a · a · π · O2 − 1.0 1.1 1.0 1.1 + 2 · a (a · u + a · u ) ≤ (750 − π · O2) 1.i 1.0 0 1.1 1

a1.i(a1.0 · t0 + a1.1 · t1) ≤ 7500(a2 + a2 ) + 20 · a · a · π · O2 1.0 1.1 1.0 1.1 + 2 · a (a · u + a · u ) (5.1) (750 − π · O2) 1.i 1.0 0 1.1 1 The right and left hand terms of this expression can be calculated for i ∈ {0, 1} which gives a restriction on the amount of fluid added in cylinder chain i, given by ti.

5.1.1 Numerical example

For our numerical example where a1.0 = 3, a1.1 = 4, u0 = u1 = 0 and O = 1 we thus have 7500(32 + 42) + 20 · 3 · 4 · π · 12 7500(32 + 42) + 20 · 3 · 4 · π · 12 i = 0 : − ≤ 3(3 · t + 4 · t ) ≤ 750 − π · 12 0 1 750 − π · 12 and 7500(32 + 42) + 20 · 3 · 4 · π · 12 7500(32 + 42) + 20 · 3 · 4 · π · 12 i = 1 : − ≤ 4(3 · t + 4 · t ) ≤ . 750 − π · 12 0 1 750 − π · 12

Dividing by 3 in the first expression and by 4 in the second (which is allowed, because these are positive numbers) this reduces to 7500(32 + 42) + 20 · 3 · 4 · π · 12 7500(32 + 42) + 20 · 3 · 4 · π · 12 − ≤ 3 · t + 4 · t ≤ 3(750 − π · 12) 0 1 3(750 − π · 12)

−84.0203802957015 ≤ 3 · t0 + 4 · t1 ≤ 84.0203802957015 and 7500(32 + 42) + 20 · 3 · 4 · π · 12 7500(32 + 42) + 20 · 3 · 4 · π · 12 − ≤ 3 · t + 4 · t ≤ 4(750 − π · 12) 0 1 4(750 − π · 12)

−63.0152852217761 ≤ 3 · t0 + 4 · t1 ≤ 63.0152852217761.

Because t0 and t1 can’t be negative by definition, this can be replaced by

3 · t0 + 4 · t1 ≤ 63.0152852217761.

29 Furthermore we know that the amount of fluid in cylinder i, ci, has to stay smaller than

30 cm, to prevent the fluid from overflowing. Besides that, ci has to be positive, a fluid level can not be negative. Also the amount of fluid in each cylinder j.i, fj.i, has to stay smaller than 20 cm and has to stay positive. From (B.5) and (B.16) these conditions can be expressed as: 1 1 0 ≤ c˜ = 20 + t + a · sin(θ˜) ≤ 30 i 2 i 2 1.i 1 0 ≤ f˜ = 10 + (t − a · sin(θ˜)) ≤ 20. 1.i 2 i 1.i Rewriting gives 1 1 1 −20 − t ≤ · a · sin(θ˜) ≤ 10 − t 2 i 2 1.i 2 i 1 1 1 −10 − t ≤ − · a · sin(θ˜) ≤ 10 − t 2 i 2 1.i 2 i or 1 1 1 −20 − t ≤ · a · sin(θ˜) ≤ 10 − t 2 i 2 1.i 2 i 1 1 1 −10 + t ≤ · a · sin(θ˜) ≤ 10 + t . 2 i 2 1.i 2 i

Because t0 and t1 are nonnegative this can be summarized by 1 1 1 −10 + t ≤ · a · sin(θ˜) ≤ 10 − t 2 i 2 1.i 2 i or

1 ˜ 1 · a1.i · sin(θ) ≤ 10 − ti. 2 2 Substituting (4.8) gives

1 2 1 ( (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) 1 · a · − 2 ≤ 10 − t 1.i 2 2 2 i 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O 2

1 1 2 2 2 · a1.i · ( (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) 750(a + a ) + 2 · a · a · π · O 2 2 ≤ 1.0 1.1 1.0 1.1 , 1 1500 − 2 · π · O2 10 − 2 ti (5.2) assuming that

2 2 2 750(a1.0 + a1.1) ≥ 2 · a1.0 · a1.1 · π · O 1500 O2 < 2 · π and 1 10 − t 6= 0. 2 i

30 5.1.2 Numerical example

For our numerical example where a1.0 = 3, a1.1 = 4, u0 = u1 = 0 and O = 1 we now have

1 1 2 2 2 · 3 · ( (3 · t0 + 4 · t1)) 750(3 + 4 ) + 2 · 3 · 4 · π · 1 2 2 ≤ 1 1500 − 2 · π · 12 10 − 2 t0 9 t0 + 3t1 18750 + 24 · π 4 ≤ ≈ 12.6030570443552 1 1500 − 2π 10 − 2 t0 and

1 1 2 2 2 · 4 · ( (3 · t0 + 4 · t1)) 750(3 + 4 ) + 2 · 3 · 4 · π · 1 2 2 ≤ 1 1500 − 2 · π · 12 10 − 2 t1

3t + 4t 18750 + 24 · π 0 1 ≤ ≈ 12.6030570443552. 1 1500 − 2π 10 − 2 t1

For systems with more equations the restrictions can’t be found so easily in detail, because the expression for sin(θ˜j) is much more complicated. It still has to hold ∀i, j that

0 ≤ d˜j.i ≤ 20,

0 ≤ c˜i ≤ 30,

0 ≤ f˜j.i ≤ 20. (5.3)

It might be wise to start by adding small amounts of fluid (for example ti = 1 cm). If the cylinders are not overflowing or reaching the bottom of the box, add some more fluid. Repeat this until a significant change in the water and fluid levels in the box and cylinders can be seen.

5.2 Precision

The final solution to the homogeneous system (4.13) is given by (y0, ..., yn) where

yi =u ˜i − ui =c ˜i − ˜b − (ci − b).

The solution to the inhomogeneous system (4.12) is then given by (x1, ..., xn) where x = yi . i y0 There could be inaccuracies in reading the water and fluid levels. If the scale is given in millimeters, we can measure the fluid levels in centimeters up to one decimal. Suppose that the error for each measurement is at most one millimeter. Then the error made in each yi is at most four millimeter (one millimeter for each measured level). The worst yi + 0, 4 yi − 0, 4 final answers we can get are thus xi = or xi = . y0 − 0, 4 y0 + 0, 4

31 The smaller the exact values of yi, the more influence a small perturbation has. There- fore, in order to make the precision as high as possible, we prefer to add as much fluid as possible. In practice, if the cylinders are not overflowing or reaching the bottom of the box, keep adding fluid until this almost is the case.

5.3 Error reduction

Veltmann himself already stated that the obtained values for the xi might not be exact. He proposes the following error reduction method. [vD92, p. 158]

Say the obtained values are (x1, ..., xn) while the true solutions are (x1 + f1, ..., xn + fn).

So fi is the error in the ith solution xi. Now consider the following system

(a1.0 + a1.1 · x1 + a1.2 · x2 + ... + a1.n · xn) + a1.1 · f1 + a1.2 · f2 + ... + a1.n · fn = 0

(a2.0 + a2.1 · x1 + a2.2 · x2 + ... + a2.n · xn) + a2.1 · f1 + a2.2 · f2 + ... + a2.n · fn = 0 . .

(am.0 + am.1 · x1 + am.2 · x2 + ... + am.n · xn) + am.1 · f1 + am.2 · f2 + ... + am.n · fn = 0. (5.4)

The xi are known, so this is a system in the variables f1, ..., fn. The parts between brackets are just constants, that might be very close to 0 but which are in general not 0

(if this is the case, (x1, ..., xn) already is an exact solution). Transforming this to an homogeneous system gives

(a1.0 + a1.1 · x1 + a1.2 · x2 + ... + a1.n · xn)z0 + a1.1 · z1 + a1.2 · z2 + ... + a1.n · zn = 0

(a2.0 + a2.1 · x1 + a2.2 · x2 + ... + a2.n · xn)z0 + a2.1 · z1 + a2.2 · z2 + ... + a2.n · zn = 0 . .

(am.0 + am.1 · x1 + am.2 · x2 + ... + am.n · xn)z0 + am.1 · z1 + am.2 · z2 + ... + am.n · zn = 0. (5.5)

This system can be solved by Veltmann’s device. Since we have just solved (4.12), the cylinders can stay at the same positions on the levers, only the cylinders of cylinder chain 0 have to be moved.

If the values of the coefficients of z0 are close to zero, they can be multiplied by a constant c. After solving this system using Veltmann’s device we obtain new solutions 1 zi (z ¯0, z1, ..., zn). The solution to system (5.4) is then given by (f1, ..., fn) with fi = c . z¯0 This solution is an estimate of the error incorporated in our initial solution (x1, ..., xn). 0 Add fi to xi to get a better estimate xi of the true solution. Repeat this process to improve the solution.

32 Chapter 6

Properties

When solving systems of linear equations it is possible to have systems without an unique solution or systems with no solutions at all (inconsistent systems). These cases are dis- cussed in this chapter.

6.1 Systems without a unique solution

Systems without unique solutions are mostly systems with m < n (the number of equa- tions is smaller than the number of variables). We can also have systems where some pairs of equations are linearly dependent, but these systems can be reduced to systems with m < n by dropping one of these equations. The simplest system without a unique solutions is the homogeneous system

b1.1x1 + b1.2x2 = 0. (6.1)

It can be easily seen that this has a solution (x1, x2) = (−b1.2, b1.1). All multiples of this solution are also solutions to (6.1), including the trivial solution (x1, x2) = (0, 0). To solve this system using Veltmann’s device, we can just use the device for one equation stating y0 = x1, y1 = x2 and a1.0 = b1.1, a1.1 = b1.2. From equations (B.21) and (B.23) we know that the particular solution given by the device is 1 y = a ·[375t · a + t · a · π · O2 − 375t · a + 750 · a · u ... 0 1.1 0 1.1 2 0 1.0 1 1.0 1.0 1 1 2 2 2 2 t1 · a1.1 · π · O − a1.0 · u0 · π · O + a1.1 · u1 · π · O − 750 · a1.1 · u0] 2 2 2 (6.2) 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

A similar expression for y1 can be found in appendix B.4

Which solution is found depends on the amount of fluid added (ti), the radius of the cylinders (O) and the initial error (ui).

If we don’t know the expressions for yi in advance (which would be logical, otherwise

33 we wouldn’t use the device), we can’t predict which solution the device will give. Thus it is possible that the device gives a very small or an otherwise not nice solution. To find the, in most cases, nicest solution for a system with integer coefficients we can use the following strategy: We know that the equation b + b · z = 0 has a unique solution, given by z = y1 . 1.1 1.2 y0 Thus y1 has to be a constant, independent of t , O and u . Expressing the number z as y0 i i a fraction gives a solution (x1, x2) = (denominator z, numerator z) where the solution is given in integers.

6.1.1 Numerical example

Suppose we want to solve the equation 3x1 + 4x2 = 0.

We take a1.0 = 3, a1.1 = 4 and solve the equation 3y0 + 4y1 = 0 using Veltmann’s device. From section 4.2.1 we know that this has a solution

(y0, y1) = (0.0793458282764729; −0.0595093712073547) for t0 = t1 = 1 cm, O = 1 cm and u0 = u1 = 0. These solutions for y0 and y1, and thus for x and x , aren’t nice numbers. The equation 3+4z = 0 has the solution z = y1 = −0.75, 1 2 y0 from (4.11). −3 −0.75 can be written as 4 so a nicer solution is given by (x1, x2) = (−3, 4)

For the general case, suppose we want to solve

a1.0 + a1.1 · x1 + a1.2 · x2 + ... + a1.n · xn = 0

a2.0 + a2.1 · x1 + a2.2 · x2 + ... + a2.n · xn = 0 . .

am.0 + am.1 · x1 + am.2 · x2 + ... + am.n · xn = 0 with m < n. Using Veltmann’s device we find a solution (x1, ..., xn). First keep adding fluid until the difference between the water level and the fluid level of cylinder chain 0 has changed , such that y0 6= 0, otherwise that would force us to divide by 0. Now one non-trivial solution is found. The other solutions can be found by adding different amounts of fluid, but it seems pretty random which solution will be obtained in this way.

34 For systems with integer coefficients there is another way to obtain a ’nice’ solution, although it is time-consuming. Write all obtained values for the xi for one solution as fractions. All these fractions are converted to like quantities in order to make them all have the same denominator. Now this denominator is used as the value for y0 and the numerator of xi can be used to be the nicer solutions for xi.

Suppose, for example, that the following solution is obtained: x1 = 0.4736842105, x2 =

0.3333333333, x3 = 0.5333333333. 9 4 8 We write these as fractions: x1 = 19 , x2 = 12 , x3 = 15 . The lowest common multiple of 19, 12 and 15 is 1140. 540 380 608 Writing the xi with this denominator gives x1 = 1140 , x2 = 1140 , x3 = 1140 . So a nicer 0 0 0 solution to the considered system might be x1 = 540, x2 = 380, x3 = 608.

6.2 Inconsistent systems

Another special case of linear systems are systems without a solution. Suppose we want to solve the following inhomogeneous system of two equations:

a1.0 + a1.1 · x1 + a1.2 · x2 = 0

a2.0 + a2.1 · x1 + a2.2 · x2 = 0.

From linear algebra we know that this system is inconsistent if the equations are linearly independent and its determinant is 0:

a1.1 · a2.2 − a1.2 · a2.1 = 0. (6.3)

There is nothing that prevents us from solving this system with Veltmann’s device. But what would happen if we do so? First this system is transformed into an homogeneous system

a1.0 · y0 + a1.1 · y1 + a1.2 · y2 = 0

a2.0 · y0 + a2.1 · y1 + a2.2 · y2 = 0.

In appendix C.2 it is found that the solution (C.23), (C.24) to the system is given by

a1.2 · a2.0 − a1.0 · a2.2 x1 = , a1.1 · a2.2 − a1.2 · a2.1 and

a1.0 · a2.1 − a1.1 · a2.0 x2 = . a1.1 · a2.2 − a1.2 · a2.1 For the inconsistent system the denominators of these solutions are 0 from (6.3), thereby we have to divide by 0 and thus have no solutions at all.

35 But it still isn’t clear how this can be seen from the device. (6.3) can also be written as a1.2 · a2.1 a1.1 = . (6.4) a2.2 Substituting this in the obtained expressions for y1 and y2, using Maple, still gives us large expressions. But substituting this in y0 by using the following command subs({a11 = a12*a21/a22}, y0)

gives y0 = 0. This means that the difference between the fluid level in cylinder chain 0 and the water level in the box hasn’t changed after adding fluid.

The just found values for y1 and y2 are a solution to the system

a1.1 · y1 + a1.2 · y2 = −a1.0 · 0 = 0

a2.1 · y1 + a2.2 · y2 = −a2.0 · 0 = 0.

6.2.1 Numerical example

Suppose we want to solve

x1 + 2 · x2 = 4

2 · x1 + 4 · x2 = 9 The equations are linearly independent, but the determinant of this system is 1·4−2·2 = 0, so this system is inconsistent. Transforming it into a homogeneous system gives

−4 · y0 + y1 + 2 · y2 = 0

−9 · y0 + 2 · y1 + 4 · y2 = 0

Using maple to find the solutions for y0, y1 and y2 that the device would give, we obtain

> evalf(subs({O = 1, a10 = -4, a11 = 1, a12 = 2, a20 = -9, a21 = 2, a22 = 4, t0 = 1, t1 = 1, t2 = 1, u0 = 0, u1 = 0, u2 = 0}, y0)); 0. > evalf(subs({O = 1, a10 = -4, a11 = 1, a12 = 2, a20 = -9, a21 = 2, a22 = 4, t0 = 1, t1 = 1, t2 = 1, u0 = 0, u1 = 0, u2 = 0}, y1)); 0.1309993428 > evalf(subs({O = 1, a10 = -4, a11 = 1, a12 = 2, a20 = -9, a21 = 2, a22 = 4, t0 = 1, t1 = 1, t2 = 1, u0 = 0, u1 = 0, u2 = 0}, y2)); -0.06549967140

36 So y0 = 0, y1 = 0.1309993428 and y2 = −0.06549967140, which is what we expected from the theory above. (y1, y2) = (0.1309993428; −0.06549967140) is now a solution to the system

y1 + 2 · y2 = 0

2 · y1 + 4 · y2 = 0

This is what the device does for each inconsistent system

a1.0 + a1.1 · x1 + a1.2 · x2 + ... + a1.n · xn = 0

a2.0 + a2.1 · x1 + a2.2 · x2 + ... + a2.n · xn = 0 . .

am.0 + am.1 · x1 + am.2 · x2 + ... + am.n · xn = 0.

It gives y0 = 0 and therefore a solution to the system

a1.1 · y1 + a1.2 · y2 + ... + a1.n · yn = 0

a2.1 · y1 + a2.2 · y2 + ... + a2.n · yn = 0 . .

am.1 · y1 + am.2 · y2 + ... + am.n · yn = 0.

37 Chapter 7

Conclusion

In this chapter we will answer the questions that were posed in the introduction.

The first question was “Why would somebody invent and build such a device?”. We found that it seems most likely that Veltmann built his device for educational purposes. Using models to clarify mathematical principles was also the main goal of Walther Dyck. Veltmann might have used it as an example of an iterative error reducing process.

Based on the description in Dyck’s catalogue, that was published on the occasion of an exposition, we conclude that there must have been a real exemplar of Veltmann’s de- vice. However it seems improbable that he really used his device to solve linear systems. It is time-consuming and in practice the errors might be pretty large. In theory, however, we can exactly predict what the device would do in the ideal case.

We also found that there are a lot of devices similar to Veltmann’s device. Some of these devices can even solve equations of higher order. Veltmann might have been in- spired by these devices and combined some of the principles that they were using.

The next subject of our research was “How is Veltmann’s device used and why does it work?”. We found that the device works because of some physical principles. The levers represent the equations in the linear system, the distance from cylinders to the fulcrum of these levers represent the coefficients of the unknowns. This lever reaches an equilibrium according to the law of the lever. The force acting on each cylinder is a com- bination of the buoyant force and the gravitational force. Using computational software like Maple, explicit expressions for the water and fluid levels can be obtained.

38 Hereafter we looked at restrictions on the amount of fluid that was added. These restric- tions depend on the fact that fluid levels can’t be negative and that cylinders are not allowed to overflow. The found restrictions are given by

2 2 2 7500(a1.0 + a1.1) + 20 · a1.0 · a1.1 · π · O − 2 + 2 (a1.0 · u0 + a1.1 · u1) ≤ ... maxi a1.i(750 − π · O )

a1.0 · t0 + a1.1 · t1 ≤ ... 2 2 2 7500(a1.0 + a1.1) + 20 · a1.0 · a1.1 · π · O 2 + 2 (a1.0 · u0 + a1.1 · u1) maxi a1.i(750 − π · O ) and

1 1 2 2 2 · a1.i · ( (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) 750(a + a ) + 2 · a · a · π · O 2 2 ≤ 1.0 1.1 1.0 1.1 1 1500 − 2 · π · O2 10 − 2 ti Veltmann himself states a method to reduce the error in the final answer. Unfortunately, this method is very time-consuming.

Lastly we looked at some special cases. For a system without a unique solution, the obtained solution depends on the amount of fluid that is added. There are some methods to change an obtained solution to a nice solution for a system with integer coefficients, but they are not very elegant. When solving inconsistent systems, Veltmann’s device is able to solve the system by giving the constant term coefficient 0. It is therefore impossible to solve the inhomogeneous sys- tem, because we are then forced to divide by 0.

39 Bibliography

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[CEED] Civil and Environmental Engineering Department. Original still missing, but copy of Wilbur mechanical calculator reappears in Tokyo. Accessed through MIT http://web.mit.edu/civenv/html/people/alumni_newsletters/winter_01/art7. htm.

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[Vel70] Wilhelm Veltmann. Ueber die Fortpflanzung des Lichts in bewegten Medien. Astronomische Nachrichten, 76(No. 1809), 1870.

41 Accessed through SAO/NASA Astrophysics Data System (ADS) http://articles.adsabs.harvard.edu/full/seri/AN.../0076//0000069.000. html.

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[Vel73] Wilhelm Veltmann. Ueber die Fortpflanzung des Lichts in bewegten Medien. Annalen der Physik und Chemie, 150(No. 12):497–535, 1873. Accessed through Gallica biblioth`equenum´erique http://gallica.bnf.fr/ark:/12148/bpt6k152347/f517.image.r=Annalen\%20de\ %20Physik.langFR.

[Vel75] Wilhelm Veltmann. Bewegung in Kegelschnitten von mehr als zwei K¨orpern, welche sich nach dem Newton’schen Gesetz anziehen. Astronomische Nachrichten, 86(No. 2042):17–30, 1875. Accessed through SAO/NASA Astrophysics Data System (ADS) http://articles.adsabs.harvard.edu//full/seri/AN.../0086//0000014.000. html.

[Vel76] Wilhelm Veltmann. Ueber die Bewegung einer Glocke. Polytechnischen Journals, 220:481–495, 1876. Accessed through Digitalisierung des Polytechnischen Journals http://dingler.culture.hu-berlin.de/article/pj220/ar220120.

[Vel77] Wilhelm Veltmann. Beitrag zu den Grundlagen der Invariantentheorie. Zeitschrift f¨urMathematik und Physik, XXII:277–298, 1877. Accessed through G¨ottingerDigitalisierungszentrum http://resolver.sub.uni-goettingen.de/purl?PPN599415665_0022.

[Vel82] Wilhelm Veltmann. Die Fourier’sche Reihe. Zeitschrift f¨urMathematik und Physik, XXVII:193–235, 1882. Accessed through G¨ottingerDigitalisierungszentrum http://resolver.sub.uni-goettingen.de/purl?PPN599415665_0027.

[Vel84a] Wilhelm Veltmann. Apparat zur Aufl¨osunglinearer Gleichungen. Zeitschrift f¨urInstrumentenkunde, 4:338–342, 1884. Accessed through Internet Archive http://archive.org/details/zeitschriftfrin04gergoog.

42 [Vel84b] Wilhelm Veltmann. Die algebraische Transformation der doppeltperiodischen Functionen. Zeitschrift f¨ur Mathematik und Physik, XXIX:73–85, 1884. Accessed through G¨ottingerDigitalisierungszentrum http://resolver.sub.uni-goettingen.de/purl?PPN599415665_0029.

[Vel86] Wilhelm Veltmann. Aufl¨osunglinearer Gleichungen. Zeitschrift f¨urMathematik und Physik, XXXI:257–272, 1886. Accessed through G¨ottingerDigitalisierungszentrum http://resolver.sub.uni-goettingen.de/purl?PPN599415665_0031.

[Vel92] Wilhelm Veltmann. Zur Theorie der Beobachtungsfehler. Astronomische Nachrichten, 131(No. 3121):1–15, 1892. Accessed through SAO/NASA Astrophysics Data System (ADS) http://articles.adsabs.harvard.edu//full/seri/AN.../0131//0000005.000. html.

[wat] http://antoine.frostburg.edu/chem/senese/javascript/water-density.html.

[Wil36] John B. Wilbur. The mechanical solution of simultaneous equations. Journal of the Franklin Institute, 222(6):715–724, 1936. Accessed through http://www.cs.princeton.edu/courses/archive/fall07/ cos323/papers/wilbur36.pdf.

[zSW59] Otto Graf zu Stolberg-Wernigerode. Neue Deutsche Biographie, vierter Band. Duncker & Humblot, Berlin, 1959. Accessed through Deutsche Forschungsgemeinschaft (DFG) http://daten.digitale-sammlungen.de/0001/bsb00016320/images/index.html? seite=226.

43 Appendix A

A list of Veltmann’s publications

A list of the articles that Veltmann has written is found in the “Jahrbuch f¨urMathematik database” [JfM]. In addition to these articles one more article is found about the bells of the tower of the cathedral of Cologne [Vel76]. A list of his publications that are known to us is given below, sorted by Veltmann’s residence at the time of appearance of these articles.

Residence Year Title of publication Reference 1870 “Fresnel’s Hypothese zur Erkl¨arungder Aberrationserschein- ungen” (Fresnel’s Hypothesis to explain the phenomena of abbera- tions) Bonn 1870 “Ueber die Fortpflanzung des Lichts in bewegten Medien” [Vel70] (On the propagation of light in moving media) 1870 “Die Helmholtz’sche Theorie der Fl¨ussigkeitswirbel” (The Helmholtz theory of the fluid vortex) Wieden- 1871 “Beitr¨agezur Theorie der Determinanten” [Vel71] br¨uck (Contribution to the theory of determinants) Holz- 1873 “Ueber die Fortpflanzung des Lichts in bewegten Medien” [Vel73] minden (On the propagation of light in moving media) 1874 “Theorie der Influenzmaschine” (Theory of the influence machine) D¨uren 1875 “Bewegung in Kegelschnitten von mehr als zwei K¨orpern, [Vel75] welche sich nach dem Newton’schen Gesetz anziehen” (Motion in conic sections of more than two bodies which at- tract according to Newton’s law) 1876 “Ueber die Bewegung einer Glocke” [Vel76] (On the movement of a bell)

44 Remagen 1877 “Beitrag zu den Grundlagen der Invariantentheorie” [Vel77] (Contribution to the foundations of the theory of invariants) 1882 “Die Fourier’sche Reihe” [Vel82] (The Fourier series) Franken- 1882 “Zur Theorie der Punktmengen” thal (On the theory of point sets) (Pfalz) 1883 “Bemerkung ¨uber den Ausdruck, Theilung einer Strecke in unendlich kleine Theile” (Remark on the expression of the division of a line into in- finitely small parts) 1884 “Apparat zur Aufl¨osunglinearer Gleichungen” [Vel84a] (Device for solving linear equations) 1884 “Die algebraische Transformation der doppeltperiodischen [Vel84b] Functionen” (The algebraic transformation of double periodic functions) 1886 “Formeln der niederen und h¨oherenMathematik sowie der Theorie der Beobachtungsfehler und der Ausgleichung dersel- ben nach der Methode der kleinsten Quadrate” (Formulas of the lower and higher mathematics and the theory of errors and its adjustment to the method of least squares) 1886 “Ausgleichung der Beobachtungsfehler nach dem princip sym- metrisch berechneter Mittelgr¨ossen” (Adjustment of the observational error on the principle of sym- metric calculated averages) Poppels- 1886 “Aufl¨osunglinearer Gleichungen” [Vel86] dorf- (Solving linear equations) Bonn 1886 “Bestimmung der Unbekannten einer Ausgleichungsaufgabe mittels der Gauss’schen Transformation der Summe der Fehlerquadrate” (Determination of the unknowns of an adjustment task using the Gaussian transformation of the sum of squares) 1887 “Berechnung des Inhalts eines Vielecks aus den Coordinaten der Eckpunkte” (Calculation of the content of a polygon from the coordinates of the vertices) 1887 “Ueber Kettenbr¨uche” (On continued fractions)

45 1892 “Zur Theorie der Beobachtungsfehler” [Vel92] (On the theory of observation errors) 1892 “Apparat zur Aufl¨osunglinearer Gleichungen” [vD92] (Device for solving linear equations) 189* “Formeln der niederen und h¨oherenMathematik sowie f¨urdie Teilung der Grundst¨ucke und fur Tracirungsarbeiten. Zum Gebrauch beim geod¨atischen Studium und in der geod¨atischen Praxis” (Formulas of the lower and higher mathematics as well as for the division of the lands and for tracing work. For use in geodetic studies and in geodetic practice) Poppels- 1897 “Der mittlere Beobachtungsfehler” dorf- (The average observation error) Bonn 1899 “Die Interpolation” (The interpolation) 1899 “Geometrische S¨atze¨uber die Fl¨ache und die Winkelsumme des Dreiecks und des Vierecks” (Geometrical theorem about the area and the sum of the an- gles of the triangle and the quadrilateral) 1900 “Nachtrag zu meiner Herleitung der Interpolationsformeln” (Addendum to my derivation of the interpolation formulas) 1876 “Kriterien der singul¨arenIntegrale der Differentialgleichungen erster Ordnung” (Criteria for singular integrals of first order differential equations) 1876 “Ueber eine besondere Art von successiven linearen Substitu- tionen” (On a special kind of successive linear substitutions) Unknown 1879 “Die dreiaxigen Coordinaten in den Gleichungen ersten und zweiten Grades” (The three-axis coordinates in the equations of first and second degree) 1881 “Die Bestimmung einer Function auf einer Kreisfl¨ache aus gegebenen Randbedingungen” (The determination of a function on a circular area from given boundary conditions) 1882 “Ueber die Anordnung unendlich vieler Singularit¨ateeiner Function” (On the arrangement of infinitely many singular- ities of a function)

Table A.1: Veltmann’s publications that are known to us

46 Appendix B

Computations for one equation

In this appendix the computational background is given for solving a1.0 + a1.1 · x = 0 with Veltmann’s device. We assume that the reader has read section (4.1). The notation that is used there won’t be introduced again here.

We start by looking at the device in the starting position: It can be seen from figure 4.1 that

d1.i + e1.i = b.

The fluid level in cylinder 1.i and cylinder i are equal because of the principle of commu- nicating vessels. This gives that

d1.i + f1.i = ci. In the ideal starting position, the water level was set to be 20 cm, so b = 20. If the lever is in horizontal position, the distance from the lever to the water surface is 10 cm. Therefore the cylinders that are attached to the lever are submerged 10 cm into the water: e1.i = 10. In this starting position e1.i = f1.i ⇒ ci = b (the water level and the fluid level are equal). From these statements the following is found for the ideal case:

b = 20

c0 = c1 = 20

d1.0 = d1.1 = 10

e1.0 = e1.1 = 10

f1.0 = f1.1 = 10.

The total amount of fluid in cylinder chain i is given by f1.i + ci = 10 + 20 = 30, where fluid in the little tubes is neglected.

47 As said before, in reality the lever might tilt over a little. The amount of fluid in a cylinder chain stays the same, but the other values will change a little. Also an additional force k is added. This force is given by equation (4.6). For the real case the following equalities hold:

d1.i + e1.i = b

d1.i + f1.i = ci

f1.i + ci = 30

a1.0 · u0 + a1.1 · u1 = k.

Now ti cm of fluid is added in cylinder i. This will cause the angle θ between the lever and the horizontal line to change to θ˜. The variables after adding fluid are denoted with a ∼ over it. After adding fluid the following equalities hold:

d˜1.i +e ˜1.i = ˜b (B.1)

d˜1.i + f˜1.i =c ˜i (B.2)

f˜1.i +c ˜i = 30 + ti. (B.3)

This situation for cylinder 1.0 is shown in figure B.1.

Figure B.1: The device after adding ti cm of fluid

It can be seen that sin(θ˜) = horizontal shift of cylinder1.i , now the horizontal shift of cylinder a1.i 1.i can be written as sin(θ˜) · a1.i.

In the ideal starting position, where the lever is in horizontal position, d1.i = 10.

48 From figure B.1 it can be seen that after fluid is added it holds that

d˜1.i = 10 + a1.i · sin(θ˜). (B.4)

From equations (B.2), (B.3) and (B.4) it follows that

d˜1.i + f˜1.i =c ˜i

10 + a1.i · sin(θ˜) + f˜1.i = 30 + ti − f˜1.i, from which it follows that 1 f˜ = 10 + (t − a · sin(θ˜)). (B.5) 1.i 2 i 1.i Now equation (B.1) is used in which equation (B.4) is substituted

d˜1.0 +e ˜1.0 = d˜1.1 +e ˜1,1

10 + a1.0 · sin(θ˜) +e ˜1.0 = 10 + a1.1 · sin(θ˜) +e ˜1.1

e˜1.0 = (a1.1 − a1.0) · sin(θ˜) +e ˜1.1. (B.6)

B.1 Properties of the box

Before this calculation is continued some properties of the box are used. First the amount of water in the box is calculated. In the ideal starting position there is 20 cm of water in the box, but cylinders 1.0 and 1.1 are both 10 cm submerged in the water. They both occupy a volume of 10 · π · O2 cm3 = 10 · π · O2 ml. There is 50 · 30 · 20 − 2 · 10 · π · O2 cm3 = 30000 − 20 · π · O2ml water in the box. From now on all water volumes are expressed in ‘cylinder centimeters’, 1 cylinder cen- timeter (1 ccm) is equal to the volume occupied by one cm of a cylinder, which is equal 2 to π · O . The amount of water in the box Wb of the box is thus 30000 − 20 · π · O2 30000 W = = − 20 ccm. (B.7) b π · O2 π · O2

After fluid is added the amount of water in the box, Wb stays the same. But the water level b changes to ˜b. From equation (B.1) and equations (B.4) and (B.6) we have

˜b = d˜1.0 +e ˜1.0

= 10 + a1.0 · sin(θ˜) + (a1.1 − a1.0) · sin(θ˜) +e ˜1.1

= 10 + a1.1 · sin(θ˜) +e ˜1.1 = d˜1.1 +e ˜1.1. (B.8)

49 After fluid is added to the cylinders, we have that the water level times the area of the box has to be equal to Wb plus the volume occupied by the submerged cylinders. Cylinders 1.0 and 1.1 are submergede ˜1.0 cm ande ˜1.1 cm respectively. This gives rise to the following equation: 30 · 50 · ˜b = W +e ˜ +e ˜ . π · O2 b 1.0 1.1 Substituting equations (B.6), (B.7) and (B.8) gives 1500 · (10 + a · sin(θ˜) +e ˜ ) π · O2 1.1 1.1 30000 = − 20 + (a − a ) · sin(θ˜) +e ˜ +e ˜ . π · O2 1.1 1.0 1.1 1.1 Multiplying both sides by π · O2 gives

15000+1500 · a1.1 · sin(θ˜) + 1500 · e˜1.1 2 2 2 =30000 − 20 · π · O + (a1.1 − a1.0) · π · O · sin(θ˜) + 2 · π · O · e˜1.1,

which is equal to

2 2 2 15000−20 · π · O − (a1.0 · π · O + (1500 − π · O ) · a1.1) · sin(θ˜) 2 =(1500 − 2 · π · O ) · e˜1.1.

Both sides are divided by (1500 − 2 · π · O2)

15000 − 20 · π · O2 − (a · π · O2 + (1500 − π · O2) · a ) · sin(θ˜) e˜ = 1.0 1.1 1.1 1500 − 2 · π · O2 (a + a ) · π · O2 = 10 − a · sin(θ˜) − 1.0 1.1 · sin(θ˜). (B.9) 1.1 1500 − 2 · π · O2 ˜ B.2 Expressing sin(θ) in terms of a1.0 and a1.1

From the derivation in chapter 4 we know that the following holds as a consequence of the law of the lever (from equation(4.7))

0 = a1.0 · (˜u0 − u0) + a1.1 · (˜u1 − u1)

= a1.0 · (˜c0 − ˜b − u0) + a1.1 · (˜c1 − ˜b − u1).

Substituting (B.1) and (B.2) gives

0 = a1.0 · [d˜1.0 + f˜1.0 − (d˜1.0 +e ˜1.0) − u0] + a1.1 · [d˜1.1 + f˜1.1 − (d˜1.1 − e˜1.1) − u1]

= a1.0 · [f˜1.0 − e˜1.0 − u0] + a1.1 · [f˜1.1 − e˜1.1 − u1].

50 Substituting (B.5 and (B.6) gives

 1  0 = a · 10 + (t − a · sin(θ˜)) − ((a − a ) · sin(θ˜) +e ˜ ) − u 1.0 2 0 1.0 1.1 1.0 1.1 0  1  + a · 10 + (t − a · sin(θ˜)) − e˜ − u 1.1 2 1 1.1 1.1 1  1 1  = a · 10 + t + ( a − a ) · sin(θ˜) − e˜ − u 1.0 2 0 2 1.0 1.1 1.1 0  1 1  + a · 10 + t − a · sin(θ˜) − e˜ − u . 1.1 2 1 2 1.1 1.1 1

The just found expression fore ˜1.1 (equation(B.9)) is substituted in the last equation

 1 1 0 = a · 10 + t + ( a − a ) · sin(θ˜) − 10 + a · sin(θ˜) 1.0 2 0 2 1.0 1.1 1.1 (a + a ) · π · O2  + 1.0 1.1 · sin(θ˜) − u 1500 − 2 · π · O2 0  1 1 +a · 10 + t − a · sin(θ˜) − 10 + a · sin(θ˜) 1.1 2 1 2 1.1 1.1 (a + a ) · π · O2  + 1.0 1.1 · sin(θ˜) − u . 1500 − 2 · π · O2 1

Simplifying this expression and eliminating some brackets gives

1 1 (a + a ) · π · O2   0 = a · t + a + 1.0 1.1 · sin(θ˜) − u 1.0 2 0 2 1.0 1500 − 2 · π · O2 0 1 1 (a + a ) · π · O2   + a · t + a + 1.0 1.1 · sin(θ˜) − u 1.1 2 1 2 1.1 1500 − 2 · π · O2 1 1 1 (a2 + a · a ) · π · O2  = a · t + a2 + 1.0 1.0 1.1 · sin(θ˜) − a · u 1.0 2 0 2 1.0 1500 − 2 · π · O2 1.0 0 1 1 (a · a + a2 ) · π · O2  + a · t + a2 + 1.0 1.1 1.1 · sin(θ˜) − a · u 1.1 2 1 2 1.1 1500 − 2 · π · O2 1.1 1 1 1 = (a · t + a · t ) − a · u − a · u + (a2 + a2 ) · sin(θ˜) 2 1.0 0 1.1 1 1.0 0 1.1 1 2 1.0 1.1 (a2 + 2 · a · a + a2 ) · π · O2 + 1.0 1.0 1.1 1.1 sin(θ˜) 1500 − 2 · π · O2 or 1 (a · t + a · t ) − a · u − a · u 2 1.0 0 1.1 1 1.0 0 1.1 1 1 (a2 + 2 · a · a + a2 ) · π · O2  = − (a2 + a2 ) + 1.0 1.0 1.1 1.1 sin(θ˜) 2 1.0 1.1 1500 − 2 · π · O2

51 1 750−π·O2 2 can be written as 1500−2·π·O2 Substituting this gives 1 (a · t + a · t ) − a · u − a · u 2 1.0 0 1.1 1 1.0 0 1.1 1 (750 − π · O2)(a2 + a2 ) + (a2 + 2 · a · a + a2 ) · π · O2 = − 1.0 1.1 1.0 1.0 1.1 1.1 sin(θ˜) 1500 − 2 · π · O2 750(a2 + a2 ) + 2 · a · a · π · O2 = − 1.0 1.1 1.0 1.1 sin(θ˜). 1500 − 2 · π · O2 Now sin(θ˜) can be isolated to obtain ( 1 (a · t + a · t ) − a · u − a · u )(1500 − 2 · π · O2) ˜ 2 1.0 0 1.1 1 1.0 0 1.1 1 sin(θ) = − 2 2 2 . (B.10) 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

B.3 Expressing the other variables in terms of a1.0 and a1.1

This expression can be substituted in the expressions we obtained fore ˜1.1, ˜b, f˜1.i,c ˜i and u˜0. From equation (B.9) it is known that  (a + a ) · π · O2  e˜ = 10 − a + 1.0 1.1 · sin(θ˜). 1.1 1.1 1500 − 2 · π · O2 Substituting (B.10) gives:  (a + a ) · π · O2  e˜ =10 − a + 1.0 1.1 1.1 1.1 1500 − 2 · π · O2 1 2 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) · − 2 2 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O =10 1 2 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1 + (1500 · a1.1 + (a1.0 − a1.1) · π · O ) · 2 2 2 . 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O (B.11) From equation (B.6) we know that

e˜1.0 = (a1.1 − a1.0) · sin(θ˜) +e ˜1.1  (a + a ) · π · O2  = (a − a ) · sin(θ˜) + 10 − a + 1.0 1.1 · sin(θ˜) 1.1 1.0 1.1 1500 − 2 · π · O2  (a + a ) · π · O2  = 10 − a + 1.0 1.1 · sin(θ˜) 1.0 1500 − 2 · π · O2 Like in equation (B.11) this becomes

e˜1.0 = 10 1 2 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1 + (1500 · a1.0 + (a1.1 − a1.0) · π · O ) · 2 2 2 . 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O (B.12)

52 From equation (B.8) it is known that

˜b = 10 + a1.1 · sin(θ˜) +e ˜1.1.

Substituting (B.9) gives

(a + a ) · π · O2 ˜b = 10 + a · sin(θ˜) + 10 − a · sin(θ˜) − 1.0 1.1 · sin(θ˜) 1.1 1.1 1500 − 2 · π · O2 (a + a ) · π · O2 = 20 − 1.0 1.1 · sin(θ˜). (B.13) 1500 − 2 · π · O2 Now the expression for sin(θ˜) (B.10) can be substituted

˜b = 20 2 1 2 (a1.0 + a1.1) · π · O ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) . − 2 · − 2 2 2 1500 − 2 · π · O 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O 1 2 (a1.0 + a1.1) · ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) · π · O = 20 + 2 2 2 . (B.14) 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

From equation (B.5) it is known that 1 f˜ = 10 + (t − a · sin(θ˜)). 1.i 2 i 1.i Substituting the expression for sin(θ˜) (B.10) gives

( 1 (a · t + a · t ) − a · u − a · u )(1500 − 2 · π · O2) ˜ 1 2 1.0 0 1.1 1 1.0 0 1.1 1 f1.i = 10 + ti + a1.i · 2 2 2 . 2 1500(a1.0 + a1.1) + 4 · a1.0 · a1.1 · π · O (B.15)

From equation (B.3) it is known that

c˜i = 30 + ti − f˜1.i.

Substituting equation (B.5) gives 1 1 1 c˜ = 30 + t − 10 − (t − a · sin(θ˜)) = 20 + t + a · sin(θ˜). (B.16) i i 2 i 1.i 2 i 2 1.i Substituting the expression for sin(θ˜) (B.10) gives

1 2 1 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1)(1500 − 2 · π · O ) c˜i = 20 + ti + a1.i · − 2 2 2 . 2 1500(a1.0 + a1.1) + 4 · a1.0 · a1.1 · π · O (B.17)

As before the following notation is used

u˜i =c ˜i − ˜b.

53 Substituting equations (B.16) and (B.13) gives

1 1 (a + a ) · π · O2 u˜ = 20 + t + a · sin(θ˜) − 20 + 1.0 1.1 · sin(θ˜) i 2 i 2 1.i 1500 − 2 · π · O2 1 1 (a + a ) · π · O2  = t + a + 1.0 1.1 · sin(θ˜). 2 i 2 1.i 1500 − 2 · π · O2

1 750−π·O2 ˜ Substituting 2 = 1500−2·π·O2 and the expression for sin(θ) (B.10) gives

1 (750 − π · O2) · a + (a + a ) · π · O2 u˜ = t + 1.i 1.0 1.1 · sin(θ˜) i 2 i 1500 − 2 · π · O2 1 = t − ((750 − π · O2) · a + (a + a ) · π · O2) 2 i 1.i 1.0 1.1 1 ( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) · 2 2 2 (B.18) 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O ) Rewriting this gives

2 1 1 (750 · a1.0 + π · O · a1.1)( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) u˜0 = t0 − 2 2 2 (B.19) 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O and

2 1 1 (750 · a1.1 + π · O · a1.0)( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) u˜1 = t1 − 2 2 2 . (B.20) 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

B.4 Calculating y0 and y1 and the solution of the linear equation

Now y0 and y1 can be calculated

y0 =u ˜0 − u0 2 1 1 ((750 · a1.0 + π · O · a1.1)( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) = t0 − 2 2 2 − u0. 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O

1 2 ti can be written as

2 2 2 1 375 · ti · (a1.0 + a1.1) + ti · a1.0 · a1.1 · π · O ti = 2 2 2 . 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O u0 can be written as

2 2 2 750 · (a1.0 + a1.1) · u0 + 2 · a1.0 · a1.1 · u0 · π · O u0 = 2 2 2 . 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O Eliminating brackets in 1 (750 · a + π · O2 · a )( (a · t + a · t ) − a · u − a · u ) 1.0 1.1 2 1.0 0 1.1 1 1.0 0 1.1 1

54 gives

2 2 375 · t0 · a1.0 + 375 · t1 · a1.0 · a1.1 − 750 · a1.0 · u0 − 750 · a1.0 · a1.1 · u1 1 1 + t · a · a · π · O2 + t · a2 · π · O2 2 0 1.0 1.1 2 1 1.1 2 2 2 −a1.0 · a1.1 · u0 · π · O − a1.1 · u1 · π · O .

The numerator of y0 is given by

2 2 2 375 · t0 · (a1.0 + a1.1) + t0 · a1.0 · a1.1 · π · O  2 2 − 375 · t0 · a1.0 + 375 · t1 · a1.0 · a1.1 − 750 · a1.0 · u0 − 750 · a1.0 · a1.1 · u1+ 1 1 + t · a · a · π · O2 + t · a2 · π · O2 2 0 1.0 1.1 2 1 1.1 2 2 2 − a1.0 · a1.1 · u0 · π · O − a1.1 · u1 · π · O 2 2 2 −(750 · (a1.0 + a1.1) · u0 + 2 · a1.0 · a1.1 · u0 · π · O ) 1 = 375 · t · a2 + t · a · a · π · O2 − 375 · t · a · a + 750 · a · a · u 0 1.1 2 0 1.0 1.1 1 1.0 1.1 1.0 1.1 1 1 − t · a2 · π · O2 − a · a · u · π · O2 + a2 · u · π · O2 2 1 1.1 1.0 1.1 0 1.1 1 2 −750 · a1.1 · u0  1 = a · 375t · a + t · a · π · O2 − 375t · a + 750 · a · u 1.1 0 1.1 2 0 1.0 1 1.0 1.0 1 1  − t · a · π · O2 − a · u · π · O2 + a · u · π · O2 − 750 · a · u . 2 1 1.1 1.0 0 1.1 1 1.1 0 (B.21)

Likewise

y1 =u ˜1 − u1 2 1 1 ((750 · a1.1 + π · O · a1.0)( 2 (a1.0 · t0 + a1.1 · t1) − a1.0 · u0 − a1.1 · u1) = t1 − 2 2 2 − u1. 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O u1 can be written as 2 2 2 750 · (a1.0 + a1.1) · u1 + 2 · a1.0 · a1.1 · u1 · π · O u1 = 2 2 2 . 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O Eliminating brackets in 1 (750 · a + π · O2 · a )( (a · t + a · t ) − a · u − a · u ) 1.1 1.0 2 1.0 0 1.1 1 1.0 0 1.1 1 gives

2 2 375 · t0 · a1.0 · a1.1+375 · t1 · a1.1 − 750 · a1.0 · a1.1 · u0 − 750 · a1.1 · u1 1 1 + t · a2 · π · O2 + t · a · a · π · O2 2 0 1.0 2 1 1.0 1.1 2 2 2 −a1.0 · u0 · π · O − a1.0 · a1.1 · u1 · π · O .

55 The numerator of y1 is given by

2 2 2 375 · t1 · (a1.0 + a1.1) + t1 · a1.0 · a1.1 · π · O  2 2 − 375 · t0 · a1.0 · a1.1 + 375 · t1 · a1.1 − 750 · a1.0 · a1.1 · u0 − 750 · a1.1 · u1 1 1 + t · a2 · π · O2 + t · a · a · π · O2 2 0 1.0 2 1 1.0 1.1 2 2 2 − a1.0 · u0 · π · O − a1.0 · a1.1 · u1 · π · O 2 2 2 −(750 · (a1.0 + a1.1) · u1 + 2 · a1.0 · a1.1 · u1 · π · O ) 1 = 375 · t · a2 + t · a · a · π · O2 − 375 · t · a · a + 750 · a · a · u 1 1.0 2 1 1.0 1.1 0 1.0 1.1 1.0 1.1 0 1 − t · a2 · π · O2 + a2 · u · π · O2 − a · a · u · π · O2 2 0 1.0 1.0 0 1.0 1.1 1 2 −750 · a1.0 · u1  1 = a · 375 · t · a + t · a · π · O2 − 375t · a + 750 · a · u 1.0 1 1.0 2 1 1.1 0 1.1 1.1 0 1  − t · a · π · O2 + a · u · π · O2 − a · u · π · O2 − 750 · a · u . 2 0 1.0 1.0 0 1.1 1 1.0 1 (B.22)

The claim was that a solution of a + a x = 0 is given by x = y1 1.0 1.1 y0

numerator y1 y 750(a2 + a2 ) + 2 · a · a · π · O2) numerator y 1 = 1.0 1.1 1.0 1.1 = 1 (B.23) y0 numerator y0 numerator y0 2 2 2 750(a1.0 + a1.1) + 2 · a1.0 · a1.1 · π · O ) 1 2 a1.0 · [375t1 · a1.0 + t1 · a1.1 · π · O − 375t0 · a1.1 + 750 · a1.1 · u0 = 2 1 2 a1.1 · [−375t1 · a1.0 − 2 t1 · a1.1 · π · O + 375t0 · a1.1 − 750 · a1.1 · u0 1 2 2 2 − 2 t0 · a1.0 · π · O + a1.0 · u0 · π · O − a1.1 · u1 · π · O − 750 · a1.0 · u1] 1 2 2 2 + 2 t0 · a1.0 · π · O − a1.0 · u0 · π · O + a1.1 · u1 · π · O + 750 · a1.0 · u1] a = − 1.0 (B.24) a1.1

And this is indeed a solution of a1.0 + a1.1x = 0:

a1.0 a1.0 + a1.1 · − = a1.0 − a1.0 = 0 a1.1

56 Appendix C

Computations for a system with more than one equation

In this appendix the computational background is given for solving a linear system with m equations and n unknowns, as described in section 4.3. The expressions in terms of the aj.i are derived for a system with two linear equations and two unknown variables. The relevant equalities will be given for the general case.

Figure C.1: Cylinder j.i before fluid is added

57 C.1 Some equations for the general case

We take j ∈ {1, ..., m}, i ∈ {0, ..., n} For each cylinder j.i we have in the starting position that

dj.i + ej.i = b

dj.i + fj.i = ci.

The box and the cylinders are filled until a water or fluid level of 20 cm is reached.

b = 20

ci = 20

ej.i = 10

fj.i = 10.

The total amount of fluid in cylinder chain i is equal to the sum of the amount of fluid in cylinder i and the amount of fluid in the cylinders j.i.

m X total amount = ci + fj.i = 20 + 10 · m. (C.1) j=1

If the lever is not in exact horizontal position we take in account an initial error kj

n X kj = aj.i · ui (C.2) i=0 where ui = ci − b.

After adding ti ccm of fluid to cylinder chain i we have

d˜j.i +e ˜j.i = ˜b (C.3)

d˜j.i + f˜j.i =c ˜i. (C.4)

And from (C.1) it follows that

m X c˜i + f˜j.i = 20 + 10 · m + ti. (C.5) j=1 Just like in the case of one equation we have for each cylinder j.i that

d˜j.i = 10 + aj.i · sin(θ˜j). (C.6)

Combining equations (C.4),(C.5) and (C.6) gives

m X 10 + aj.i · sin(θ˜j) + f˜j.i + f˜j.i = 20 + 10 · m + ti. j=1

58 This is equal to m X f˜j.i = 10 + 10 · m + ti − aj.i · sin(θ˜j) − f˜j.i, j=1 which for fixed k ∈ {1, ..., m} leads to m ˜ ˜ ˜ X ˜ fk.i = 10 + 10 · m + ti − ak.i · sin(θk) − fk.i − fj.i, j=1,j6=k  m  1 X f˜ = 5 + 5 · m + t − a · sin(θ˜ ) − f˜ . (C.7) k.i 2  i k.i k j.i j=1,j6=k ˜ In this way we obtain m equations with m unknowns, expressions for the fk.i in terms of ˜ ak.i, sin(θk) and ti can be calculated from these equations. Later we will do this for the case of a system of two equations.

From equation (C.3) we know that it holds that

d˜j.i +e ˜j.i = ˜b = d˜j.n +e ˜j.n for i ∈ {0, ..., n}.

From these equations we can solvee ˜j.i in terms of ej.n for i ∈ {0, ..., n − 1}:

e˜j.i = d˜j.n − d˜j.i +e ˜j.n.

Substituting (C.6) gives   e˜j.i = 10 + aj.n · sin(θ˜j) − 10 + aj.i · sin(θ˜j) +e ˜j.n

= (aj.n − aj.i) sin(θ˜j) +e ˜j.n. (C.8)

Now we look at the properties of the box. The amount of water in the box is equal to 30000 W = − 10 · number of cylinders i.j b π · O2 30000 = − 10 · m · (n + 1). (C.9) π · O2 After fluid is added to the cylinders, we have that the water level times the area of the box has to be equal to Wb plus the volume occupied by the submerged cylinders: 1500 · ˜b X = W + e˜ . π · O2 b j.i i,j Again use (C.3) to obtain ˜ 1500 · (dj.n +e ˜j.n) X = W + e˜ (C.10) π · O2 b j.i i,j for j ∈ {1, ..., m}.

59 From these m equations (C.10) we can find expressions fore ˜j.n, which gives us expressions for alle ˜j.i in terms of aj.i, sin(θ˜j) and O using (C.8).

Now for every j ∈ {1, ..., m} the law of the lever states (see (4.16)

n n X X (˜ui − ui) · aj.i = (f˜j.i − e˜j.i − ui) · aj.i = 0. i=0 i=0

Substituting the just found equations fore ˜j.i and f˜j.i gives us m equations in terms of aj.i, sin(θ˜j), O, ui and ti. From these equations we can obtain expressions for sin(θ˜j). In the next section we will do this for the case of a system of two equations, but we won’t give the explicit expressions for sin(θ˜j), because they are way too big.

Substituting the just found expression for sin(θ˜j) in (C.6) and in our just found ex- pressions fore ˜j.i, f˜j.i and d˜j.i gives us the possibility to calculate ˜b andc ˜i from (C.3) and (C.4).

Now we havec ˜i and ˜b in terms of aj.i, ti, ui and O only!

Calculatingu ˜i =c ˜i − ˜b and yi =u ˜i − ui gives a solution to the homogeneous system.

Dividing by y0 gives a solution to the inhomogeneous system in terms of its coefficients aj.i.

C.2 Results for two equations

Suppose we want to solve the system

a1.0 + a1.1 · x1 + a1.2 · x2 = 0

a2.0 + a2.1 · x1 + a2.2 · x2 = 0. (C.11)

This system can be transformed to a homogeneous system by taking

a1.0 · y0 + a1.1 · y1 + a1.2 · y2 = 0

a2.0 · y0 + a2.1 · y1 + a2.2 · y2 = 0. (C.12)

Thus we have m = 2, n = 2, such that i ∈ {0, 1, 2} and j ∈ {1, 2}

The total amount of fluid in a cylinder chain before adding extra fluid is (from (C.1))

20 + 10 · m = 20 + 10 · 2 = 40 ccm.

After the fluid is added we have (C.5)

c˜i + f˜1.i + f˜2.i = 40 + ti.

60 From (C.7) this leads to the following equations 1   f˜ = 15 + t − a · sin(θ˜ ) − f˜ , (C.13) 1.i 2 i 1.i 1 2.i 1   f˜ = 15 + t − a · sin(θ˜ ) − f˜ . (C.14) 2.i 2 i 2.i 2 1.i Substituting (C.14) in (C.13) gives 1   1   f˜ = 15 + t − a · sin(θ˜ ) − 15 + t − a · sin(θ˜ ) − f˜ 1.i 2 i 1.i 1 2 i 2.i 2 1.i 1 1 1 1  = 15 + t − a · sin(θ˜ ) + a · sin(θ˜ ) − 15 + f˜ 2 2 i 1.i 1 2 2.i 2 2 1.i 15 1 1 1 1 = + t − a · sin(θ˜ ) + a · sin(θ˜ ) + f˜ . 2 4 i 2 1.i 1 4 2.i 2 4 1.i 1 ˜ 3 Subtracting 4 f1.i on both sides and dividing by 4 leads to 1 2 1 f˜ = 10 + t − a · sin(θ˜ ) + a · sin(θ˜ ). (C.15) 1.i 3 i 3 1.i 1 3 2.i 2 Substituting (C.15) in (C.14) gives 1   1 2 1  f˜ = 15 + t − a · sin(θ˜ ) − 10 + t − a · sin(θ˜ ) + a · sin(θ˜ ) 2.i 2 i 2.i 2 3 i 3 1.i 1 3 2.i 2 1 2 4 2  = 15 + t − a · sin(θ˜ ) − 10 + a · sin(θ˜ ) 2 3 i 3 2.i 2 3 1.i 1 1 2 1 = 10 + t − a · sin(θ˜ ) + a · sin(θ˜ ). (C.16) 3 i 3 2.i 2 3 1.i 1 In the end we will use Maple to solve some equations. We now already know the following input:

> d10 := 10+a10*sinth1; > d20 := 10+a20*sinth2; > d11 := 10+a11*sinth1; > d21 := 10+a21*sinth2; > d12 := 10+a12*sinth1; > d22 := 10+a22*sinth2;

> f10 := 10+(1/3)*t0-(2/3)*a10*sinth1+(1/3)*a20*sinth2; > f11 := 10+(1/3)*t1-(2/3)*a11*sinth1+(1/3)*a21*sinth2; > f12 := 10+(1/3)*t2-(2/3)*a12*sinth1+(1/3)*a22*sinth2; > f20 := 10+(1/3)*t0-(2/3)*a20*sinth2+(1/3)*a10*sinth1; > f21 := 10+(1/3)*t1-(2/3)*a21*sinth2+(1/3)*a11*sinth1; > f22 := 10+(1/3)*t2-(2/3)*a22*sinth2+(1/3)*a12*sinth1;

61 Expressinge ˜j.0 ande ˜j.1 in terms ofe ˜j.2 gives (from (C.8))

e˜j.0 = (aj.2 − aj.0) sin(θ˜j) +e ˜j.2 (C.17)

e˜j.1 = (aj.2 − aj.1) sin(θ˜j) +e ˜j.2. (C.18)

Looking at the amount of water in the box we have (from (C.9)) 30000 W = − 10 · 2 · (2 + 1) b π · O2 30000 = − 60 ccm. (C.19) π · O2 After adding fluid we have (from (C.10))

1500 · (d˜ +e ˜ ) j.2 j.2 = W +e ˜ +e ˜ +e ˜ +e ˜ +e ˜ +e ˜ . π · O2 b 1.0 1.1 1.2 2.0 2.1 2.2 Substituting (C.6), (C.19),(C.17) and (C.18) gives

1500 · (10 + a · sin(θ˜ ) +e ˜ ) 30000 j.2 j j.2 = − 60 + (a − a ) sin(θ˜ ) +e ˜ π · O2 π · O2 1.2 1.0 1 1.2

+(a1.2 − a1.1) sin(θ˜1) +e ˜1.2 +e ˜1.2

+(a2.2 − a2.0) sin(θ˜2) +e ˜2.2 + (a2.2 − a2.1) sin(θ˜2) +e ˜2.2 +e ˜2.2 30000 = − 60 + (2a − a − a ) sin(θ˜ ) π · O2 1.2 1.0 1.1 1

+(2a2.2 − a2.0 − a2.1) sin(θ˜2) + 3˜e1.2 + 3˜e2.2. (C.20)

From now on we use Maple to solve these expression fore ˜1.2 ande ˜2.2 . Using the following input

> e10 := (a12-a10)*sinth1+e12; > e11 := (a12-a11)*sinth1+e12; > e20 := (a22-a20)*sinth2+e22; > e21 := (a22-a21)*sinth2+e22;

> Wb := 30000/(pi*O^2)-60; > eq1 := (1500*(d12+e12))/(pi*O^2) = Wb+e10+e11+e12+e20+e21+e22; > se12 := solve(eq1, e12); > e12 := se12;

> eq2 := (1500*(d22+e22))/(pi*O^2) = Wb+e10+e11+e12+e20+e21+e22; > se22 := solve(eq2, e22); > e22 := se22

62 We obtain e˜1.2 = 10 h i 2 (a1.0 + a1.1 − 5 · a1.2) sin(θ˜1) + (a2.0 + a2.1 + a2.2) sin(θ˜2) π · O + 1500 · a1.2 · sin(θ˜1) + 6 · π · O2 − 1500 (C.21) e˜2.2 = 10 h i 2 (a1.0 + a1.1 + a1.2) sin(θ˜1) + (a2.0 + a2.1 − 5 · a2.2) sin(θ˜2) π · O + 1500 · a2.2 · sin(θ˜2) + . 6 · π · O2 − 1500 (C.22)

Now we use the law of the lever to obtain expressions for sin(θ˜1) and sin(θ˜2)

(f˜1.0 − e˜1.0 − u0) · a1.0 + (f˜1.1 − e˜1.1 − u1) · a1.1 + (f˜1.2 − e˜1.2 − u2) · a1.2 = 0,

(f˜2.0 − e˜2.0 − u0) · a2.0 + (f˜2.1 − e˜2.1 − u1) · a2.1 + (f˜2.2 − e˜2.2 − u2) · a2.2 = 0.

> eq3 := 0 = (f10-e10-u0)*a10+(f11-e11-u1)*a11+(f12-e12-u2)*a12e; > ssinth1 := solve(eq3, sinth1) > sinth1 := ssinth1 > eq4 := 0 = (f20-e20-u0)*a20+(f21-e21-u1)*a21+(f22-e22-u2)*a22; > ssinth2 := solve(eq41, sinth2); > sinth2 := ssinth2

These expressions are too large to put them here. sin(θ˜1) and sin(θ˜2) can be substituted in

˜b = d˜1.0 +e ˜1.0

c˜0 = d˜1.0 + f˜1.0

c˜1 = d˜1.1 + f˜1.1

c˜2 = d˜1.2 + f˜1.2.

Now we calculate yi =u ˜i − ui =c ˜i − ˜b − ui and finally

y1 x1 = y0 y2 x2 = . y0

63 > b := d10+e10; > c0 := d10+f10; > c1 := d11+f11; > c2 := d12+f12; > tu0 := c0-b; > tu1 := c1-b; > tu2 := c2-b; > y0 := tu0-u0; > y1 := tu1-u1; > y2 := tu2-u2; > sx1 := y1/y0; > x1 := simplify(sx1); > sx2 := y2/y0; > x2 := simplify(sx2);

This leads to our final expressions:

a1.2 · a2.0 − a1.0 · a2.2 x1 = (C.23) a1.1 · a2.2 − a1.2 · a2.1 and

a1.0 · a2.1 − a1.1 · a2.0 x2 = (C.24) a1.1 · a2.2 − a1.2 · a2.1

We can now just check if this indeed is a solution to (C.11).

Substituting (C.23) and (C.24) in a1.0 + a1.1 · x1 + a1.2 · x2:     a1.2 · a2.0 − a1.0 · a2.2 a1.0 · a2.1 − a1.1 · a2.0 a1.0 + a1.1 · + a1.2 · a1.1 · a2.2 − a1.2 · a2.1 a1.1 · a2.2 − a1.2 · a2.1 a1.1 · a1.2 · a2.0 − a1.1 · a1.0 · a2.2 + a1.2 · a1.0 · a2.1 − a1.2 · a1.1 · a2.0 = a1.0 + a1.1 · a2.2 − a1.2 · a2.1 −a1.1 · a1.0 · a2.2 + a1.2 · a1.0 · a2.1 = a1.0 + a1.1 · a2.2 − a1.2 · a2.1 a1.1 · a2.2 − a1.2 · a2.1 = a1.0 − a1.0 = a1.0 − a1.0 = 0. a1.1 · a2.2 − a1.2 · a2.1

64 And substituting (C.23) and (C.24) in a2.0 + a2.1 · x1 + a2.2 · x2:     a1.2 · a2.0 − a1.0 · a2.2 a1.0 · a2.1 − a1.1 · a2.0 a2.0 + a2.1 · + a2.2 · a1.1 · a2.2 − a1.2 · a2.1 a1.1 · a2.2 − a1.2 · a2.1 a2.1 · a1.2 · a2.0 − a2.1 · a1.0 · a2.2 + a2.2 · a1.0 · a2.1 − a2.2 · a1.1 · a2.0 = a2.0 + a1.1 · a2.2 − a1.2 · a2.1 a2.1 · a1.2 · a2.0 − a2.2 · a1.1 · a2.0 = a2.0 + a1.1 · a2.2 − a1.2 · a2.1 a1.1 · a2.2 − a1.2 · a2.1 = a2.0 − a2.0 = a2.0 − a2.0 = 0. a1.1 · a2.2 − a1.2 · a2.1

65