Veltmann’s device for

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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

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Abstract

In 1892 the ‘Deutsche Mathematiker Vereinigung’ (German mathematician society) published 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 device 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 restrictions that need to be posed on amounts of fluid that are added in order to make the device work properly.

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Chapter 1 Introduction

In 1892 the ‘Deutsche Mathematiker Vereinigung’ (German mathematician society) published 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 algebraic 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 development 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.

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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¨uren where 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 professor 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).

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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¨osung linearer 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 errors 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 geometrical 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.

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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 exposition of 1892 he writes that there has been a great development in the use and usefulness of mathematical models. He says:

“So erschien es naturgemäss, die Gelegenheit der diesjährigen Versammlung der Deutschen Mathematiker-Vereinigung, die gleichzeitig mit der Versammlung der Gesellschaft deut- scher Naturforscher und Aerzte in Nürnberg tagen sollte, zu benützen, um ein zusam- menhängendes Bild dieser Entwickelung vorzuführen.” [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 exposition 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 instruments 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¨osung von 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.

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The catalogue also contains a ‘Nachtrag’ (supplement), that was published at the occasion 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¨ahlung der 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¨ur die hiebei n¨otigen Betrachtungen 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

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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 “Résolution hydrostatique de l’équation du troisième degré”

[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 `a r´esoudre les

´

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.

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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]

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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.

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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

I²· f = O²· w ⇒ O² I² = f

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 I² = 1²

0.9² = 1

0.81 ≈ 1.23

The density, w, of water, at 20^◦C, is 0.998 g/cm³ [wat]. The density, f , of the fluid has to satisfy _0.998^f = 1.23. This gives f = 0.998 · 1.23 = 1.23 g/cm³. 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.

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3.3 Solving the equation

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

a_1.0+ a_1.1· 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 water 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 a_1.0 = −15 and a1.1= 5.

Figure 3.2: The box filled with water and the cylinders filled with (red) fluid for the case a_1.0= −15, a_1.1 = 5

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

π · I²· t_i cm³ = π · I²· t_i 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 by ˜c0 and ˜c1 respectively.

The following values are calculated:

˜

u₀ = ˜c₀− ˜b, ˜u₁ = ˜c₁− ˜b.

These values are used to calculate

y0 = ˜u0− u₀ = ˜c0− ˜b − (c₀− b) (3.2) y1 = ˜u1− u₁ = ˜c1− ˜b − (c₁− b). (3.3) The y₀ and y₁ satisfy the equation a_1.0 · y₀ + a_1.1· y₁ = 0. The solution for x in the equation a1.0+ a1.1· x = 0 is thus given by x = ^y_y¹

0.

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.

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Chapter 4 Why the device works

Why do y₀ and y₁ from the preceding chapter satisfy a_1.0· y₀+ a_1.1· 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

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All distances are measured in centimeters. The meaning of the indicated variables is as follows (for i = 0, 1):

a_1.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.

f_1.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, e_1.i = f_1.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 and with radius O is π · O²· f_1.0. The mass of this column is

π · O²· f_1.0· w = m_1.0

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

Here n_1.0 is given by

n_1.0 = π · O²· e_1.0· w

The net downward force F1.0 working on cylinder 1.0 is now given by F1.0 = Fg1.0− F_A1.0

= π · O²· f_1.0· w · g − π · O²· e_1.0· w · g

= π · O²· w · g · (f_1.0− e_1.0). (4.1)

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

F1.1 = π · O²· w · g · (f_1.1− e_1.1). (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 F_1.0· p_1.0= F_1.1· p_1.1 where p_1.i is the perpendicular distance from cylinder 1.i to the fulcrum on the line AB. F1.i· p_1.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 p_1.i’s positive. (4.3) can thus be written as:

X

cylinder 1.i on the left of the fulcrum

(−1)F_1.i· p_1.i = X

cylinder 1.j on the right of the fulcrum

F_1.j· p_1.j

This can be rewritten as

n

X

i=1

F1.i· p_1.i= 0.

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.

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It can be seen in figure 4.2 that cos(θ) = _a^p^1.0

1.0, such that p_1.0 = a_1.0· cos(θ) and likewise

p1.1 = a1.1· cos(θ).

The thus obtained expression for the law of the lever is:

X

i

F_1.i· a_1.i· cos(θ) = 0.

Substituting (4.1) and (4.2) gives X

i

π · O²· w · g · (f_1.i− e_1.i) · a_1.i· cos(θ) = 0.

This expression is divided by π · O²· w · g · cos(θ) to obtain X

i

(f_1.i− e_1.i) · a_1.i = 0.

This also can be written as 0 =X

i

(f1.i+ d1.i− (e_1.i+ d1.i)) · a1.i.

From figure 4.1 we see that this is equal to 0 =X

i

(ci− b) · a_1.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) + a_1.1· (˜c1− ˜b) = 0 (4.4) (tildes are added to the variables to indicate that fluid is added).

(y0, y1) = (˜u0− u₀, ˜u1− u₁) = (˜c0− ˜b, ˜c1− ˜b) (4.5) is thus a solution of the equation a1.0· y₀+ a1.1· y₁= 0.

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Dividing this equation on both sides by y0 gives a1.0+ a1.1·^y_y¹

0 = 0. The solution for x in the equation a1.0+ a1.1· x = 0 is thus given by x = ^y_y¹

0.

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) + a_1.1· (c₁− b) − k = 0

where k represents the additional force. Therefore, the additional force k is given by k = a1.0· (c₀− b) + a_1.1· (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 = a_1.0· (˜c₀− ˜b) + a_1.1· (˜c₁− ˜b) − k

= a_1.0· (˜c₀− ˜b) + a_1.1· (˜c₁− ˜b) − (a_1.0· u₀+ a_1.1· u₁)

= a_1.0· (˜c₀− ˜b − u₀) + a_1.1· (˜c₁− ˜b − u₁)

= a_1.0· (˜u₀− u₀) + a_1.1· (˜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, ˜u1− u₁).

The solution for x in the equation a1.0+ a1.1· x = 0 is still given by x = ^y_y¹

0.

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

sin(˜θ) = −(¹₂(a_1.0· t₀+ a_1.1· t₁) − a_1.0· u₀− a_1.1· u₁)(1500 − 2 · π · O²)

750(a²_1.0+ a²_1.1) + 2 · a1.0· a_1.1· π · O² (4.8) 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.

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We also calculated that

˜b = 20 +(a1.0+ a1.1) · (¹₂(a1.0· t₀+ a1.1· t₁) − a1.0· u₀− a_1.1· u₁) · π · O²

750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O² (4.9) and

˜

ci= 20 + 1

2ti− a_1.i·(¹₂(a1.0· t₀+ a1.1· t₁) − a1.0· u₀− a_1.1· u₁)(1500 − 2 · π · O²) 1500(a²_1.0+ a²_1.1) + 4 · a_1.0· a_1.1· π · O² .

(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 (a_1.0, a_1.1 ∈ 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 a_1.0 = 3 and a_1.1 = 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

sin(˜θ) = −(¹₂(3 · 1 + 4 · 1) − 3 · 0 − 4 · 0)(1500 − 2 · π · 1²) 750(3²+ 4²) + 2 · 3 · 4 · π · 1²

= −

7

2(1500 − 2 · π)

18750 + 24 · π = 7 · π − 5250 24 · π + 18750

≈ −0.277710398967654, which leads to

θ ≈ sin˜ ⁻¹(−0.277710398967654) = −0.281409934832440 rad

= −0.281409934832440 · 360 2 · π

◦

≈ −16.1236015789504^◦.

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

˜b = 20 +(a_1.0+ a_1.1) · (¹₂(a_1.0· t₀+ a_1.1· t₁) − a_1.0· u₀− a_1.1· u₁) · π · O² 750(a²_1.0+ a²_1.1) + 2 · a1.0· a_1.1· π · O²

= 20 + (3 + 4) · (¹₂(3 · 1 + 4 · 1) − 3 · 0 − 4 · 0) · π · 1² 750(3²+ 4²) + 2 · 3 · 4 · π · 1²

= 20 +

49 2 · π

18750 + 24 · π ≈ 20.0040885732720 cm

The water level in the box was 20 cm in starting position, so it has risen about 40.8 µm.

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

˜

c0 = 20 +1

2t0− a_1.0·(¹₂(a1.0· t₀+ a1.1· t₁) − a1.0· u₀− a_1.1· u₁)(1500 − 2 · π · O²) 1500(a²_1.0+ a²_1.1) + 4 · a_1.0· a_1.1· π · O²

= 20 +1

2 · 1 − 3 ·(¹₂(3 · 1 + 4 · 1) − 3 · 0 − 4 · 0)(1500 − 2 · π · 1²) 1500(3²+ 4²) + 4 · 3 · 4 · π · 1²

= 41 2 − 3 ·

7

2(1500 − 2 · π)

37500 + 48 · π ≈ 20.0834344015485 cm.

The fluid level in cylinder chain 0 was 20 cm in starting position, so it has risen about 0.08 cm.

˜

c₁ = 20 +1

2t₁− a_1.1·(¹₂(a_1.0· t₀+ a_1.1· t₁) − a_1.0· u₀− a_1.1· u₁)(1500 − 2 · π · O²) 1500(a²_1.0+ a²_1.1) − 4 · a1.0· a_1.1· π · O²

= 20 +1

2 · 1 − 4 ·(¹₂(3 · 1 + 4 · 1) − 3 · 0 − 4 · 0)(1500 − 2 · π · 1²) 1500(3²+ 4²) + 4 · 3 · 4 · π · 1²

= 41 2 − 4 ·

7

2(1500 − 2 · π)

37500 + 48 · π ≈ 19.9445792020647 cm.

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 = ˜ui− u₁, where ˜ui = ˜ci− ˜b. We thus have y₀ = ˜c₀− ˜b − 0 ≈ 20.0834344015485 − 20.0040885732720 ≈ 0.0793458282764729 y₁ = ˜c₁− ˜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 x = y₁

y0

= −0.0595093712073547

0.0793458282764729 = −0.75. (4.11)

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.

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4.3 The device in the general case

A system of more than one linear equation can be written as a_1.0+ a_1.1· x₁+ a_1.2· x₂+ ... + a_1.n· x_n= 0 a_2.0+ a_2.1· x₁+ a_2.2· x₂+ ... + a_2.n· x_n= 0 ...

a_m.0+ a_m.1· x₁+ a_m.2· x₂+ ... + a_m.n· x_n= 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· y_n= 0 a2.0· y₀+ a2.1· y₁+ a2.2· y₂+ ... + a2.n· y_n= 0 ...

a_m.0· y₀+ a_m.1· y₁+ a_m.2· y₂+ ... + a_m.n· y_n= 0, (4.13) which can be written as

n

X

i=0

aj.i· a_j.i= 0, j = 1, ..., m. (4.14) 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 L_j, 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 communicating 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 u_i.

Now t_i 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 L_j 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 called ˜ui.

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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, ..., y_n) = (˜u₀− u₀, ˜u₁− u₁, ..., ˜u_n− u_n).

Dividing by y0 gives the solution to system (4.12): (x1, ..., xn) = (^y_y¹

0, ...,^y_yⁿ

0).

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

We use the subscript ∗_j.i for a variable belonging to cylinder j.i on lever L_j.

As we have seen in section 4.1 the net downward force action on cylinder j.i in the starting position is given by

F_j.i= π · O²· w · g · (f_j.i− e_j.i).

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

n

X

i=0

F_j.i· p_j.i= 0,

where pj.i is the perpendicular distance from cylinder j.i to the fulcrum of lever Lj. This distance is given by p_j.i= a_j.i· cos(˜θ_j). So we have that

n

X

i=0

π · O²· w · g · (f_j.i− e_j.i) · a_j.i· cos(˜θ_j) = 0.

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Dividing by π · O²· w · g · cos(˜θj) gives

n

X

i=0

(fj.i− e_j.i) · aj.i= 0 (4.15)

for each lever L_j. This is equal to

n

X

i=0

(f_j.i+ d_j.i− (e_j.i+ d_j.i) · a_j.i= 0.

From figure B.1 in appendix B we see that this equals

n

X

i=0

(ci− b) · a_j.i= 0.

The difference c_i− b is named u_i. In the ideal case u_i = 0 for every i. If lever L_j is not in exact horizontal position before adding fluid, we take in account an additional force kj

working on lever Lj such that

(

n

X

i=0

(ci− b) · a_j.i) − kj = (

n

X

i=0

ui· a_j.i) − kj = 0 ⇒ kj =

n

X

i=0

ui· a_j.i.

After adding fluid we have that

n

X

i=0

˜

ui· a_j.i− k_j = 0

n

X

i=0

(˜u_i− u_i) · a_j.i = 0, (4.16)

which means that (˜u0− u₀, ˜ui− u₁, ..., ˜un− u_n) = (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(˜θ1), sin(˜θ2),

˜

c₀, ˜c₁, ˜c₂ and ˜b are way too large to express at this point.

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Chapter 5 Restrictions

In this chapter the restrictions that need to be put on t_i 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, a_j.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 t

_i

First we look at the case of one equation.

From equation (B.4) we know that

d˜_1.i = 10 + a_1.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

0 ≤ 10 + a1.i· −(¹₂(a1.0· t₀+ a1.1· t₁) − a1.0· u₀− a_1.1· u₁)(1500 − 2 · π · O²) 750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O² ≤ 20 Which leads to

−10 ≤a1.i·¹₂(a1.0· t₀+ a1.1· t₁)(1500 − 2 · π · O²) 750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O²

−a1.i· (a_1.0· u₀+ a1.1· u₁)(1500 − 2 · π · O²) 750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O² ≤ 10

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This is equal to

−10 + a_1.i· (a_1.0· u₀+ a_1.1· u₁)(1500 − 2 · π · O²) 750(a²_1.0+ a²_1.1) + 2 · a1.0· a_1.1· π · O²

≤ a1.i·¹₂(a1.0· t₀+ a1.1· t₁)(1500 − 2 · π · O²) 750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O² ≤ 10 +a_1.i· (a_1.0· u₀+ a_1.1· u₁)(1500 − 2 · π · O²)

750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O² Multiplying by ^750(a²^1.0^+a1 ²^1.1^)+2·a^1.0^·a^1.1^·π·O²

2(1500−2·π·O²) , which is in general a positive term because 750(a²_1.0+ a²_1.1) >

2 · a_1.0· a_1.1· π · O²

and 1500 > 2 · π · O², results in

−7500(a²_1.0+ a²_1.1) + 20 · a_1.0· a_1.1· π · O²

(750 − π · O²) + 2 · a_1.i (a_1.0· u₀+ a_1.1· u₁) ≤ a1.i(a1.0· t₀+ a1.1· t₁) ≤ 7500(a²_1.0+ a²_1.1) + 20 · a1.0· a_1.1· π · O²

(750 − π · O²) + 2 · a1.i (a1.0· u₀+ a1.1· u₁) (5.1) 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 t_i.

For our numerical example where a_1.0 = 3, a_1.1= 4, u₀ = u₁= 0 and O = 1 we thus have i = 0 : −7500(3²+ 4²) + 20 · 3 · 4 · π · 1²

750 − π · 1² ≤ 3(3 · t₀+ 4 · t1) ≤ 7500(3²+ 4²) + 20 · 3 · 4 · π · 1² 750 − π · 1²

and

i = 1 : −7500(3²+ 4²) + 20 · 3 · 4 · π · 1²

750 − π · 1² ≤ 4(3 · t₀+ 4 · t1) ≤ 7500(3²+ 4²) + 20 · 3 · 4 · π · 1² 750 − 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

3(750 − π · 1²) ≤ 3 · t₀+ 4 · t₁ ≤ 7500(3²+ 4²) + 20 · 3 · 4 · π · 1² 3(750 − π · 1²)

−84.0203802957015 ≤ 3 · t₀+ 4 · t1 ≤ 84.0203802957015 and

−7500(3²+ 4²) + 20 · 3 · 4 · π · 1²

4(750 − π · 1²) ≤ 3 · t₀+ 4 · t1 ≤ 7500(3²+ 4²) + 20 · 3 · 4 · π · 1² 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.

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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, f_j.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:

0 ≤ ˜c_i= 20 + 1 2t_i+ 1

2a_1.i· sin(˜θ) ≤ 30 0 ≤ ˜f_1.i= 10 + 1

2(t_i− a_1.i· sin(˜θ)) ≤ 20.

Rewriting gives

−20 − 1 2t_i≤ 1

2 · a_1.i· sin(˜θ) ≤ 10 −1 2t_i

−10 − 1

2t_i ≤ −1

2 · a_1.i· sin(˜θ) ≤ 10 −1 2t_i or

−20 − 1 2ti≤ 1

2 · a_1.i· sin(˜θ) ≤ 10 −1 2ti

−10 + 1 2ti ≤ 1

2· a_1.i· sin(˜θ) ≤ 10 + 1 2ti. Because t0 and t1 are nonnegative this can be summarized by

−10 + 1 2ti ≤ 1

2 · a_1.i· sin(˜θ) ≤ 10 −1 2ti

or

1

2· a_1.i· sin(˜θ)

≤ 10 −1 2t_i. Substituting (4.8) gives

1

2 · a_1.i· −(¹₂(a1.0· t₀+ a1.1· t₁) − a1.0· u₀− a_1.1· u₁)(1500 − 2 · π · O²) 750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O²

≤ 10 −1 2ti

1

2 · a_1.i· (¹₂(a_1.0· t₀+ a_1.1· t₁) − a_1.0· u₀− a_1.1· u₁) 10 −¹₂t_i

≤ 750(a²_1.0+ a²_1.1) + 2 · a_1.0· a_1.1· π · O² 1500 − 2 · π · O² ,

(5.2) assuming that

750(a²_1.0+ a²_1.1) ≥ 2 · a_1.0· a_1.1· π · O² O² < 1500 2 · π and

10 −1 2ti 6= 0.

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

1

2 · 3 · (¹₂(3 · t₀+ 4 · t₁)) 10 − ¹₂t0

≤ 750(3²+ 4²) + 2 · 3 · 4 · π · 1² 1500 − 2 · π · 1²

9

4t₀+ 3t₁ 10 −¹₂t0

≤ 18750 + 24 · π

1500 − 2π ≈ 12.6030570443552 and

1

2 · 4 · (¹₂(3 · t0+ 4 · t1)) 10 −¹₂t₁

≤ 750(3²+ 4²) + 2 · 3 · 4 · π · 1² 1500 − 2 · π · 1²

3t0+ 4t1

10 − ¹₂t₁

≤ 18750 + 24 · π

1500 − 2π ≈ 12.6030570443552.

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 ≤ ˜fj.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, ..., y_n) where y_i = ˜u_i− u_i= ˜c_i− ˜b − (c_i− b).

The solution to the inhomogeneous system (4.12) is then given by (x₁, ..., x_n) where xi = _y^yⁱ

0.

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 y_i is at most four millimeter (one millimeter for each measured level). The worst final answers we can get are thus x_i = y_i+ 0, 4

y₀− 0, 4 or x_i = y_i− 0, 4 y₀+ 0, 4.

Veltmann’s device for

faculty of mathematics and natural sciences