Coulomb Damping - Fay

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Coulomb dampingTemple H. Fay

a b

a

Department of Mathematical Technology , Tshwane University of Technology , Pretoria 0001 , South Africab Department of Mathematics , University of Southern Mississippi ,

P.O. Box 5045, Hattiesburg , MS 39406 , USA

Published online: 23 Nov 2011.

To cite this article: Temple H. Fay (2012) Coulomb damping, International Journal of Mathematical

Education in Science and Technology, 43:7, 923-936, DOI: 10.1080/0020739X.2011.633624

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International Journal of Mathematical Education in

Science and Technology, Vol. 43, No. 7, 15 October 2012, 923–982

CLASSROOM NOTES

Coulomb damping

Temple H. Fayab*

aDepartment of Mathematical Technology, Tshwane University of Technology,Pretoria 0001, South Africa; bDepartment of Mathematics, University of Southern

Mississippi, P.O. Box 5045, Hattiesburg, MS 39406, USA

(Received 29 April 2011)

Viscous damping is commonly discussed in beginning differential equationsand physics texts but dry friction or Coulomb friction is not despite dry

friction being encountered in many physical applications. One reason foravoiding this topic is that the equations involve a jump discontinuity in thedamping term. In this article, we adopt an energy approach which permits ageneral discussion on how to investigate trajectories for second-orderdifferential equations representing mechanical vibration models having dryfriction. This approach is suitable for classroom discussion and computerlaboratory investigation in beginning courses, hence introduction of dryfriction need not be delayed for more advanced courses in mechanics ormodelling. Our method is applied to a harmonic oscillator example and apendulum model. One advantage of this method is that the values of themaximum deflections of a solution can be calculated without solving thedifferential equation either analytically or numerically, a technique that

depends on only the initial conditions.Keywords: dry friction; dissipation; Coulomb-damped harmonic oscillator;Coulomb-damped pendulum

1. Introduction

In beginning differential equations, linear models are developed for the explanation

of simple mechanical and electrical vibrations which include viscous damping.

In particular, linear spring models are fully developed using Newton’s second law

and Hooke’s law: forces resulting from the weight and stiffness are the first two terms

usually accounted for. However, damping does not arise from a single physical

phenomenon; damping is a measure of energy dissipation in a vibrational model andplays an important role in the modelling dynamical systems. There are as many types

of damping as there are ways to convert mechanical energy into heat. These include

material damping, structural damping, interfacial damping, aerodynamic and

hydrological drag; the mathematical description of these are quite different. Many

mechanical systems are complicated and efforts to lump together many facets of the

system into a simpler linear single degree of freedom model are employed in the early

design process as such linear models have known exact solutions.

*Email: [email protected]

ISSN 0020–739X print/ISSN 1464–5211 online

2012 Taylor & Francis

http://dx.doi.org/10.1080/0020739X.2011.633624

http://www.tandfonline.com


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Studying Coulomb damping using our elementary methods can achieve two ends:

the first being the introduction of new, deeper and perhaps more relevant syllabus

material and second, illustrates and provides practice for the use of a computer

algebra system, numerical algorithms and graphical interpretation. Furthermore, we

demonstrate an integration between theory, numerics and graphics that shows the

student that one goes hand-in-hand with the other.There is no question that dissipation plays an important role in modelling of

damped nonlinear dynamical systems. Despite this importance, only linear dissipa-

tive forces are discussed in beginning texts. Thus many frictional forces which

influence the motion of an object through a fluid or a gas are ignored as they are

usually considered complex.

Material damping arises from complex molecular interactions and thus is

highly dependent upon the type of material and its fabrication. Structural damping

is a more macroscopic effect and is important as, for example, modern high-rise

buildings may oscillate excessively in high winds. Interfacial damping arises from

dry friction in a structural system. Generally aerodynamic damping in structures issmall compared to mechanical damping and is interesting as, at low velocities,

it is positive but can be negative at certain high-wind conditions. Hydrological

drag is relatively large, especially for ocean structures. For more details on

these items see [1]

The attention given to viscous damping in beginning differential equations

courses is reasonable as it fits into a linear differential equation of motion. If one

assumes that the damping force is a function D(v) dependent upon velocity v,

then D(0)¼ 0 as with no motion there is no damping. Thus, assuming D(v) has a

Taylor approximation about v¼ 0, a first-order approximation of D(v) is simply cv

where c is a constant. But there is another type of damping of major import,Coulomb damping, the result of two dry (or sometimes lubricated) surfaces rubbing

together.

In this article, we illustrate how one might introduce in the classroom or

computer laboratory setting an investigation of this important nonlinear friction

through two fundamental examples, one with a linear restoring force (harmonic

oscillator with dry friction) and the other with a sinusoidal restoring force

(Coulomb-damped pendulum). These examples provide a glimpse into dealing

with discontinuous models and can serve as an introduction to discontinuous models

such as dissipative oscillators with quadratic damping, see [2]. All numerical and

graphical investigations were carried out using Mathematica version 7 [3], but any

computer algebra system or ode solver package would suffice.

In Section 2, we give a brief description of Coulomb friction. Section 3 is devoted

to an energy approach and an ‘energy function’. A harmonic oscillator with dry

friction example is discussed in Section 4, where the function will be used to calculate

the relative maxima and minima for the oscillations. The Coulomb-damped

pendulum model is examined in Section 5 where again we use the energy approach

to calculate relative extrema and to determine basins of attraction for the attracting

spiral points (stable equilibria) for this model. For the harmonic oscillator equation

with Coulomb damping, the decrease in amplitude of the oscillations is linear, but

for the Coulomb-damped pendulum this is not the case. This observation leads to a

Student Project, culminating Section 5, which suggests an investigation seeking, forCoulomb damping, a generalization of the notion of logarithmic decrement used for

viscous damping of the harmonic oscillator.

924 Classroom Notes


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2. Coulomb damping

A common type of mechanical damping arises from dry friction. Charles-Augustin

de Coulomb (1736–1806) won the 1781 prize from the Acade ´ mie des Sciences for

Thé orie de Machines Simples, Mé moires de Savants é tranges, tomb X (published

separately in Paris in 1809) and therein pointed out the difference between

static and dynamical friction . Due to this study, sliding friction is called Coulomb

friction [4]. Of course, Coulomb is much better known for his foundational work in

electricity and magnetism and in particular his inverse square law of electrostatic

force.

A nice introduction to Coulomb damping can be found in Andronov et al. [5], see

also [6]. However, in elementary textbooks for beginning courses, Coulomb damping

is generally not mentioned or developed. Two interesting examples of Coulomb

friction are given in Jordan and Smith [7] The first example involves a continuous

belt revolving at a constant rate with a surmounted sliding block attached to a spring

which is fixed to a stationary support (stick–slip oscillation). The second

example is a model of a simple brake shoe applied to a the hub of a revolving

wheel (non-oscillatory damped motion). An ‘energy’ approach is used to build phase

plane portraits for each example. Our energy approach is a somewhat deeper

analysis.

Coulomb friction is exhibited in complex structures with non-welded parts, such

as airplane wings. This type of friction between two moving or sliding surfaces (dry

or lubricated) is generally independent of the velocity and the frequency of the

motion, and thus under Coulomb damping, the frictional force is constant, or very

nearly constant. Static friction, also called stiction, is larger than sliding friction

between two surfaces and occurs only when the velocity is instantaneously zero.

Hence the effect is to give the magnitude of the sliding frictional force a spike whenthe velocity is very near zero, Norton [8]. Stiction only affects the model behaviour as

a stopping condition at the turning points of the motion [9]. We assume the stiction is

small enough that it can be ignored and will assume that the Coulomb frictional

force is constant in absolute magnitude.

Herein we only consider strictly dissipative oscillators. A frictional force usually

opposes the direction of the motion (sometimes Coulomb friction can add energy

into a system as for a violin string and thus need not be strictly dissipative, the Van

der Pol equation that models a relaxation oscillator is a non-strictly dissipative

oscillator), and consequently always has the opposite sign of the velocity. To

interpret this force for an equation of motion, with the correct sign to adjust for thedirection of motion, the equation of motion usually becomes

x::þ c Sgnðx

:Þ þ f ðxÞ ¼ 0: ð1Þ

The function f (x) will be called the restoring force (actually f (x) is the restoring

force but it is traditional to transpose all the forces to the same side to the equation

as the acceleration x::

); the constant c is the coefficient of friction and is material-

dependent; the signum function Sgnðx:Þ is defined by

Sgnðx:Þ ¼

1 for x:40,

0 for x:¼ 0,

1 for x: 50:

8


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Something else to note, the critical values for the system

x:¼ y,

y:¼ c Sgnð yÞ f ðxÞ

ð3Þ

depend only on the roots of f (x) and not at all on the coefficient of friction c.

3. Energy

The so-called energy approach for conservative systems is very useful and easy to

implement, for example see [10,11]. We can produce a useful energy surface here as

well despite the jump discontinuity in the equation/system. Let us return to

Equation (1) and multiply through by x:

d and integrate from ¼ 0 to ¼ t to obtain

Z t

0

x::

x:d þ Z

t

0

c Sgnðx:Þx:

d þ Z t

0

f ðxÞx:

d ¼ 0: ð4Þ

The first integral represents kinetic energy 12 mv

2:

1

2x:ðtÞ2

1

2x:ð0Þ2: ð5Þ

The last integral represents potential energy since it depends solely upon displace-

ment x(t). The middle integral represents the loss of mechanical energy that is

converted into heat and we call this term the dissipation. Setting y ¼ x:

and

integrating, we have

y2

2

y202 þ c Sgnð yÞðx x0Þ þ

Z xx0

f ð Þd ¼ 0, ð6Þ

where the initial values are x(0)¼x0 x:ð0Þ ¼ y0:

Note that the right-hand side appears to be a constant so that kinetic plus

potential plus dissipation is constant in the surface determined by the energy

function and thus contours in the energy surface represent trajectories for the system

with the initial values determining the contour constant. But the factor Sgn( y) means

that as the trajectory crosses the x-axis ( y¼ 0), the value of the contour constant

changes. Note that again we have a discontinuity at x:¼ y ¼ 0:

We will see that this energy function is more than the usual ‘kinetic plus potential’

since the dissipation term is present. From physical considerations, we know the sumof the kinetic and potential energies is not constant but decreases and these combined

energies reduce as friction causes a ‘heat sink’ and stable critical values are thus

attracting spiral points for these oscillating systems.

4. The Coulomb-damped harmonic oscillator

Let us assume the restoring force is linear of the form f (x)¼x so that the equation of

motion is that of a harmonic oscillator with dry friction damping

x::þ c Sgnðx

:Þ þ x ¼ 0: ð7Þ

There is no loss of generality to assume the coefficient of x is 1 since it is at most a

dilation of the time variable t to obtain this. Of course, Equation (7) is really two

926 Classroom Notes


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equations depending upon the sign of the velocity x:

and note that there is a jump

discontinuity when x:¼ 0: This means that the motion is divided into intervals

bounded by the conditions that the velocity is zero and between endpoints is of

constant sign.

If we assume x:40, and we have initial conditions ðxð0Þ, x

:ð0Þ Þ ¼ ðx0, y0Þ, then the

solution to

x::þ c þ x ¼ 0 ð8Þ

is

xðtÞ ¼ ðx0 þ cÞ cosðtÞ þ y0 sinðtÞ c: ð9Þ

If x:50, and we have initial conditions ðxð0Þ, x

:ð0Þ Þ ¼ ðx0, y0Þ, then the solution to

x:: c þ x ¼ 0 ð10Þ

is

xðtÞ ¼ ðx0 cÞ cosðtÞ þ y0 sinðtÞ þ c: ð11Þ

Thus a solution to Equation (7) consists of basically an oscillating function whose

increasing segments from a relative minimum to a relative maximum (when x:40Þ is

determined by the solution (9) and decreasing segment from the relative maximum to

the next relative minimum (when x:50Þ comes from the solution (11). Note that the

frequency of the oscillation is the same as the natural frequency.

We can predict the decrease in amplitudes from one cycle to the next. Suppose we

choose the initial conditions (x(0), y(0))¼ (x0, 0) with x0> 0 so that we are beginning

the oscillations at a relative maximum. The solution for the next half-cycle is given by

xðtÞ ¼ ðx0 cÞ cosðtÞ þ c, ð12Þ

since over this portion of the cycle y< 0 and accordingly x is decreasing. This

half-cycle ends at t¼, and x()¼x0þ 2c. Thus the reduction in amplitude is 2c

and consequently from one relative maximum (minimum) to the next relative

maximum (minimum) is 4c.

An equivalent form of Equation (7) is the system

x:¼ y,

y: ¼ c Sgnð yÞ x, ð13Þ

for which (0, 0) is the unique critical value. Intuitively, this critical value should be

classified as an attracting spiral point, but due to the jump discontinuity at x:¼ 0,

eigenvalue classification fails since the derivative of c Sgn( y) fails to exist at y¼ 0.

In Figure 1, we show a vector field plot for the system (13) with c¼ 1/9 indicating

that (0, 0) is an attracting spiral point as expected. The value c¼ 1/9 was chosen

solely for scaling and other visual effects, and the figure is intended to be

representative of the model behaviour.

The relative maxima and minima for the solution to (7) with c ¼ 1/9 are given in

Table 1, where the first maxima is at (t1, x1) and t1¼ 0.732815 since 10 sin(t1)/9þcos(t1)¼ 0. It readily follows that t2¼ t1þ, t3¼ t2þ, t4¼ t3þ, and so on.

Evaluating the numerical solution at these times provides the values listed in Table 1.

International Journal of Mathematical Education in Science and Technology 927


7/15

The change from one maximum to the next is 4c¼ 4/9 and similarly from one

minimum to the next. This means that all the peaks lie on a pair of straight lines

x¼4c(t t1)þx1. This is indicated in Figure 2.

One need not solve the system numerically in order to determine the relative

extrema of the solution to the initial value problem. Knowing a formula for the

energy of the system which depends upon both displacement and velocity is one way

to determine these extrema as they occur when the velocity is zero. The only problem

is that the energy is determined by two formulae, one when the velocity is positive

and other when the velocity is negative, thus requiring an iterative method.

The relative extrema for this solution are easy to find from Equations (9) and

(11), but we will use an energy approach to determine these maximum amplitudes of

the oscillation. To that end, multiply Equation (7) by x:

d and integrate from ¼ 0 to

¼ t, to obtain (setting y ¼ x:Þ

y2

2 y

20

2 þ x

2

2 x

20

2 ¼ c Sgnð yÞðx x0Þ: ð14Þ

Figure 1. Vector field plot for system (13), c¼ 1/9.

Table 1. Relative extrema (Equation (7)).

x1¼ 1.38374 x4¼0.717069x2¼1.16151 x5¼ 0.494847x3¼ 0.939292 x6¼0.272625

928 Classroom Notes


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We interpret the kinetic energy as y2/2 and the potential energy as x2/2 and set the

energy as

E ðx, yÞ ¼ y2

2 þ

x2

2 ¼

y202 þ

x202 c Sgnð yÞðx x0Þ: ð15Þ

The initial energy is

E 0 ¼ E ðx0, y0Þ ¼ y20

2 þx202 , ð16Þ

and we interpret the right-hand side of Equation (15) as dissipation.

Returning to our example above, since y0> 0, the first increasing segment of the

solution peaks at (t1, x1) (see Figure 2) which can be found from (9). The energy

along this segment is given by (15)

y2

2 ¼

y202

x2

2

x202

1

9ðx x0Þ

: ð17Þ

Since (1, 1) are the initial conditions, Equation (17) becomes

y2

2 ¼

10

9

x2

2

1

9x: ð18Þ

Now x1 is found using Newton–Raphson method as y¼ 0 when x¼ x1. Recall that

Newton–Raphson method requires a starting point which we choose to be x0¼ 1.

This yields x1¼ 1.38374.

The next extrema is a relative minimum at (t2, x2) and x2 is that value of x which

yields y¼ 0 for the energy equation

y2

2 ¼

x2

2

x212

þ

1

9ðx x1Þ

: ð19Þ

Here we use the starting point 1. This produces x2¼1.16151. Alternating beween

the two forms of Equations (15), the relative extrema are readily calculted without

Figure 2. The solution x(t) for (x0, y0)¼ (1, 1).



9/15

ever having to the solve the differential equation. These first six extrema are listed

in Table 2.

5. The Coulomb-damped pendulum

The pendulum equation is found in almost every beginning text, but usually, because

of its nonlinearity and exact solution involving an arcsine of an elliptic sine, the

equation is given a linear approximation. For a derivation of the analytic solution

see [12]. The undamped pendulum equation is investigated numerically together with

various approximations in [13]. The viscous damped pendulum equation is seldom

discussed in any detail. The Coulomb-damped pendulum is a bit easier to investigate.

In Figure 3, we show a vector field plot for the system

x:¼ y,

y: ¼ c Sgnðx: Þ sinðxÞ, ð20Þ

where again we have set c¼ 1/9 for visual convenience.

Table 2. Relative extrema (Equation (15)).

x1¼ 1.38374 x2¼1.16151x3¼ 0.93929 x40.71707x5¼ 0.494845 x6¼0.27262

Figure 3. Vector field plot for the system (20), c¼ 1/9.

930 Classroom Notes


10/15

The critical values are (n, 0), with attracting spiral points for n even and saddle

points for n odd.

The energy function is determined by the energy equation

y2

2

y20

2 þZ

x

x0 sinð Þd þ c Sgnð yÞðx x0Þ ¼ 0, ð21Þ

and we set

E ðx, yÞ ¼ y2

2 cosðxÞ þ c Sgnð yÞx, ð22Þ

and call this the energy function. Normally, for any given initial condition

(x(0), y(0))¼ (x0, y0), the contour

E ðx, yÞ ¼ y20

2

cosðx0Þ þ c Sgnð yÞx0 ð23Þ

would be the trajectory in the phase plane for the solution to the initial value

problem, except due to the discontinuity. We have to adjust for when the trajectory

crosses the x-axis. The trajectory consists of segments pieced together from the upper

half-plane and the lower half-plane and the energy surface is constant on each of

these segments; the value of the contour constant will change from the upper half-

plane to the lower half-plane. We illustrate this with the initial conditions

(x0, y0)¼ (1, 1). These initial conditions were chosen as the trajectory produced can

be viewed as representative.

In Figure 4, we plot the phase plane trajectory for

x::þ

1

9 Sgnðx

:Þ þ sinðxÞ ¼ 0, ð24Þ

with (x0, y0)¼ (1, 1) shown solved numerically in both forward and backward time.

Again c ¼ 1/9 was chosen as representative and to keep matters of scale appropriate.

This trajectory consists of a number of segments above the x-axis and a number

below. On each trajectory segment, the energy is constant. Even though the

trajectory is continuous as it crosses the x-axis, the contour constants above and

below are different. In Figure 4, we plot the trajectory indicating that (0, 0) is an

attracting spiral point and eight other points on the x-axis where the trajectory

crosses.With the exception of the furthest most point to the right (2.42045, 0), and the

furthest point to the left (1.85091,0), six of these points are listed in Table 3

alternating positive and negative. Starting at the initial point (1, 1) travelling in a

clockwise direction to the first crossing of the x-axis at (x1, 0) and then continuing

travelling below to the point (x2, 0), then travelling above to (x3,0) and so on.

These points are listed in Table 3, with the time values that generate them from

the numerical algorithm. For the last point closest to the origin, (x7, 0), the

numerical algorithm stalls at t¼ 20.0989, which produces the value of

x(20.0989)¼0.0403, but as we will see, the energy approach predicts the value to

be x7¼0.0263356.A plot of the solution x(t) with these six local maxima and minima indicated is

given in Figure 5.



11/15

In this figure, the upper dashed line is the line passing through the points ( t1, x1)

and (t3, x3); its reflection is the lower dashed line. We see that the decay in amplitude

no longer is linear.

But to find these points in Figure 5, we need not resort to a numerical trajectory

solution. We need only use the energy surface (25) determined by Equation (24) and

the initial condition (x0, y0)¼ (1, 1):

y2

2 cosðxÞ þ

1

9 Sgnð yÞx ¼

y202 cosðx0Þ þ

1

9 Sgnð yÞx0: ð25Þ

Starting at (1, 1) and moving clockwise, we first reach (x1, 0). The value x1 can be

calculated from the equation

cosðx1Þ þ1

9x1 ¼

1

2 cosð1Þ þ

1

9: ð26Þ

Figure 4. The full trajectory for (x0, y0)¼ (1, 1) and crossing points.

Table 3. Relative extrema with times values.

(t1, x1)¼ (0.9351325, 1.47732) (t4, x4)¼ (11.148161, 0.656845)(t2, x2)¼ (4.485212, 1.17258) (t5, x5)¼ (14.3592246, 0.422916)(t3, x3)¼ (7.869422, 0.903965) (t6, x6)¼ (17.52999, 0.196617)

932 Classroom Notes


12/15

This solution is x1¼ 1.47732. Newton–Raphson’ method is a good technique for this

calculation. Continuing, the point (x2, 0) is reached by travelling clockwise from

(x1, 0) below the x-axis, and the value x2¼1.17258 is found from the equation

cosðx2Þ 1

9x2 ¼ cosðx1Þ

1

9x1: ð27Þ

In general, if i > 1 is odd, then xi þ1 is calculated by solving

cosðxi þ1Þ 19

xi þ1 ¼ cosðxi Þ 19

xi , ð28Þ

and if i is even then xi þ1 is calculated by solving

cosðxi þ1Þ þ1

9xi þ1 ¼ cosðxi Þ þ

1

9xi ð29Þ

Of course this does not give the corresponding ti value, but in practical applications,

one is generally interested in finding the maximum amplitudes of the oscillations and

not when they occur since this depends upon the choice of initial values.

Using numerical methods, it is not difficult to produce a phase plane portrait for

this situation and we do not have to piece together many parts of the trajectory but

only solve an initial value problem in both forward and backward time. Since there

are saddle points involved, one can expect separatrices dividing the plane into

distinct regions of trajectory behaviour and which provide boundaries for basins of

attraction for the spiral points. A partial phase plane portrait including separatrices

and three initial starting points off the axes is given in Figure 6.

The wavy bands shown that are slightly thicker form the boundaries of the basins

of attraction for the attracting spiral points (2n, 0). These band comprise a

separatrix, passing through the saddle points ((2nþ 1), 0). Of course, many of these

wavy lines are omitted in the figure in order to keep things from becoming too

crowded and obscuring the representative local behaviour near the critical points.Note that some points are indicated by dots on the y-axis. These points are of

the form (0, vn) where the separatrices emanating from the six saddle points (, 0),

Figure 5. The solution x(t) and relative extrema.



13/15

(3, 0) and (5, 0) cross the y -axis. The energy equations (23) and (25) permit the

calculation of these points easily and numerical solutions are not necessary. Suppose

we wish to find where the separatrix from (5, 0) crosses the y-axis. Then the contour

in the energy surface is

E ðx, yÞ ¼ y2

2 cosðxÞ þ

x

9¼ 1 þ 5

9 , ð30Þ

and the value v5¼ 2.73691 is obtained by setting x ¼ 0 and solving E [0, y]¼ 1þ 5/9

for y¼ v5. It is clear that the points (0, vn) are symmetrically placed. The first sixpoints are tabulated below.

v1 ¼ 2:16752,

v3 ¼ 2:46868,

v5 ¼ 2:73691:

Imagine the pendulum hanging at rest so that x0¼ 0 and at time t¼ 0 the

pendulum is struck with an impulse giving it an initial velocity y0. For small values of

y0 the maximum displacement is less than and the pendulum begins its damped

oscillatory motion tending towards the rest position again. But if y0 is sufficientlylarge (greater in absolute value than v1), the effect is of giving the pendulum a

number of revolutions, and the trajectory will eventually spiral down to one of the

Figure 6. An abbreviated phase portrait.

934 Classroom Notes


14/15

attracting spiral points. Determining the range of values of y0 so that the trajectory

will spiral down to a specified attracting spiral point, (2n, 0) is determined by the

upper (for y0> 0) and lower ( y0


15/15

[12] H.T. Davis, Nonlinear Differential and Integral Equations, Dover Publications,

New York, 1962.

[13] T.H. Fay, The pendulum equation, Int. J. Math. Educ. Sci. Technol. 33 (2002),

pp. 505–519.

[14] W. Boyce and R. DiPrima, Elementary Differential Equations, 7th ed., Wiley & Sons,

New York, 2001.

Modelling the landing of a plane in a calculus lab

Antonio Morante and Jose ´ A. Vallejo*

Facultad de Ciencias, Universidad Autó noma de San Luis Potosı́ ,Lat. Av. Salvador Nava s/n, CP 78290, San Luis Potosı́ (SLP), Mé xico

(Received 4 June 2011)

We exhibit a simple model of a plane landing that involves only basicconcepts of differential calculus, so it is suitable for a first-year calculus lab.We use the computer algebra system Maxima and the interactive geometrysoftware GeoGebra to do the computations and graphics.

Keywords: mathematical modelling; calculus labs; GeoGebra; CASMaxima

1. Introduction

Usually, the problems posed in freshman calculus courses are just formal in character

and, therefore, not quite exciting. Students are not satisfied when we say that

mathematics can be used in different fields of knowledge, they want to see math

applied to real-world situations!

Obviously, the problem is that most real mathematical applications need the

study of more advanced subjects than the ones offered at freshman level, but with a

little bit of imagination it is possible to adapt complex real problems to tractable case

studies requiring a minimum of mathematical formalism. An example of this

approach is [1], of which we present an adaptation using Maxima and GeoGebra.

The choice of this couple is dictated by two main reasons: their power and easiness of

use, and the fact that they are free software, so the students can work at home withthe same programs with which they learn at the classroom, at no cost. We include a

brief Appendix on the basic usage of both programs, to make the lab self-contained.

Before June 1965, when autolanding was first used in a commercial flight,

there were several accidents with landing aircrafts, like that of Paradise Airlines

Flight 901A, where a Constellation crashed with Genoa peak on approach to Lake

Tahoe airport (see, e.g. http://aviation-safety.net/database/record.php?id=19640301-1).

Nowadays, even with good visual conditions, most of the medium/large range

airliners are able to land by means of the so-called instrument landing systems (or

ILS for short). There is a well-defined protocol to decide when to use these ILS,

indeed, landings are classified taking into account several conditions related to

*Corresponding author. Email: [email protected]

936 Classroom Notes

http://dx.doi.org/10.1080/0020739X.2011.633626

Coulomb Damping - Fay

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