karman treffitz transformation

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Complex mapping of aerofoils a

different perspectiveMiccal T. Matthews

a

a School of Engineering, Edith Cowan University, Joondalup, WA6027, Australia

Published online: 24 Jun 2011.

To cite this article: Miccal T. Matthews (2012): Complex mapping of aerofoils a different

perspective, International Journal of Mathematical Education in Science and Technology, 43:1,

43-65

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International Journal of Mathematical Education in

Science and Technology, Vol. 43, No. 1, 15 January 2012, 4365

Complex mapping of aerofoils a different perspectivey

Miccal T. Matthews*

School of Engineering, Edith Cowan University, Joondalup, WA 6027, Australia

(Received 25 October 2010)

In this article an application of conformal mapping to aerofoil theory isstudied from a geometric and calculus point of view. The problem issuitable for undergraduate teaching in terms of a project or extended pieceof work, and brings together the concepts of geometric mapping,

parametric equations, complex numbers and calculus. The Joukowskiand KarmanTrefftz aerofoils are studied, and it is shown that theKarmanTrefftz aerofoil is an improvement over the Joukowski aerofoilfrom a practical point of view. For the most part only a spreadsheetprogram and pen and paper is required, only for the last portion of thestudy of the KarmanTrefftz aerofoils a symbolic computer package isemployed. Ignoring the concept of a conformal mapping and insteadviewing the problem from a parametric point of view, some interestingmappings are obtained. By considering the derivative of the mappedmapping via the chain rule, some new and interesting analytical results areobtained for the Joukowski aerofoil, and numerical results for theKarmanTrefftz aerofoil.

Keywords: Joukowski aerofoil; KarmanTrefftz aerofoil; complexmapping

1. Introduction

Modelling the flow of air past the wing of an aircraft is a highly complex problem,

and as is common in applied mathematics the problem is made tractable by applying

simplifying assumptions. One such assumption reduces the problem to looking at

cross-sections of the aerofoil, thus the theory of two-dimensional fluid flow can be

utilized. For potential flow in two dimensions, there exists a stream function and a

velocity potential such that the flow is irrotational and automatically satisfies the

equation of continuity [1]. These two functions satisfy the CauchyRiemann

equations, which imply that there exists a complex potential describing the stream

function and velocity potential for some fluid flow problem [2].

With the existence of an analytic function (i.e. the complex potential) we can

employ the theory of conformal mapping to transform complicated geometries into

simpler ones that can be handled analytically [3]. One such complex function was

studied by Joukowski [4,5] and is known as the Joukowski transformation, which

under certain conditions will map a circle onto a curve shaped like the cross-section

of an aerofoil. The mapped aerofoil has the correct shape for wing design, with a

*Email: [email protected] work is dedicated to the memory of Dr. Grant M. Cox, R.I.P.

ISSN 0020739X print/ISSN 14645211 online

2012 Taylor & Francis

http://dx.doi.org/10.1080/0020739X.2011.582174

http://www.tandfonline.com


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blunt note and a sharp trailing edge. The study of fluid flow past an aerofoil

therefore becomes the simpler task of flow past a circle [6].

The sharp trailing edge of the Joukowski transformation can be shown to form a

cusp, and in reality is virtually impossible to construct. A simple modification of the

Joukowski aerofoil which makes the trailing edge have distinct tangents was

developed by Theodore von Ka rma n and Erich Trefftz and is known as the

KarmanTrefftz transformation [7,8]. This transformation has the advantage that it

is a more physical aerofoil shape to construct, so much so that KarmanTrefftz

aerofoils are closely related to a catalogue of aerofoils known as NACA aerofoils

(National Advisory Committee for Aeronautics, a branch of NASA) [9].

Within the theory of conformal mapping, the Joukowski and KarmanTrefftz

transformations are well-studied [6]. However, if we ignore the definition of where

the mapping is conformal and consider instead the problem of mapping a circle to an

aerofoil shape via the theory of parametric equations, we obtain some interesting

mappings. Thinking about the problem in this way, the only mathematical expertise

required is calculus, parametric equations and the most basic concepts of complexnumbers such as real and imaginary parts, polar form and De Moivres theorem.

When complex numbers in Cartesian form and polar form are studied, their

graphical representation is illustrated in the Argand plane. The connection between

the polar coordinates in the two-dimensional Cartesian plane and the polar form of a

complex number in the Argand plane is quickly made, and therefore a plot in the

complex plane is just like a plot in the Cartesian plane in particular, we can think of

a plot in the Argand plane as a parametric one, enabling the calculus to be applied.

This is the point of view taken here for the Joukowski and KarmanTrefftz

transformations. One naive question that often arises when one first studies complex

numbers is why the need for two forms of complex numbers (Cartesian form andpolar form) and the importance of obtaining one form from the other. One of the

main reasons for the two forms is that addition and subtraction are easier with

Cartesian form, while multiplication and division and particularly powers are

easier with polar form. The KarmanTrefftz transformation is an excellent example

of this to obtain the mapping both forms are required.

To illustrate the basic idea and method considered here, consider the polar form

of a simple cardioid [10], r() 1 cos, where x rcos and y rsin. This

represents a parametric function with parameter , and a plot of y versus x yields a

heart-like shape with a cusp at . Using the chain rule, the derivative dy/dx is

found to be

dy

dx

1 cos 1 2cos

sin 1 2cos :

Therefore the cardioid has horizontal tangents when /3 and 5/3 and vertical

tangents when 2/3 and 4/3. When the derivative dy/dx has the

indeterminate form 00, and using lHospitals rule we find that as !, dy/dx ! 0,

thus there is a horizontal tangent at the cusp and the derivative is a continuous

function of around .

In the following sections the Joukowski and KarmanTrefftz transformations are

studied. For the Joukowski transformation a spreadsheet program is used to obtainthe mappings. The derivative of the mapping is calculated to show that there is a

cusp at the trailing edge. For the KarmanTrefftz transformation a spreadsheet

44 M.T. Matthews


4/24

program is again utilized, and the derivative of the mapping is calculated using

a symbolic software package to show the existence of distinct tangents at the

trailing edge.

2. Joukowski transformation

The Joukowski transformation is a complex function that maps a circle into a shape

resembling the cross-section of an aerofoil. The conformal nature of the transfor-

mation has been studied in detail [6], and here we simply look at it from a purely

geometric and calculus point of view. That is, we think of the mapping as defining a

parametric curve in two-dimensional space.

First, some conventions: the complex plane which contains the set of

complex numbers z x iy to be mapped is called the z-plane with axes

labelled (x,y) (Re(z), Im(z)). The mapping is written as w u iv f(z), and the

complex plane which contains the set of mapped complex numbers w u iv is

called the w-plane with axes labelled (u, v) (Re(w), Im(w)). We will consider map-

pings of a circle (x a)2 (y b)2 R2 with centre (a, b) and radius R, and the

mapping is

w z 1

z, 2:1

which is known as the Joukowski transformation.

For any complex number z x iy the Joukowski transformation, Equation

(2.1), may be written in Cartesian form as

w u iv x 1 1

x2 y2

iy 1

1

x2 y2

,

so that equating the real and imaginary parts we have

u x 1 1

x2 y2

, v y 1

1

x2 y2

: 2:2

It will be more convenient to use polar coordinates x r()cos, y r()sin to

describe the circle, where

r a cos b sin ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

a cos b sin 2R2 a2 b2q

, 2:3

provided R24a2 b2. Then from Equation (2.2) we have

u r 1

r

cos , v r

1

r

sin : 2:4

To obtain the shape of an aerofoil, we demand that the circle being mapped

passes through the point z 1, which implies that the mapping w will pass through

the point w z 1/z 2. The point z 1 implies x 1, y 0 which implies

from (x a)

2

(y b)

2

R

2

that

1 a 2 0 b 2 R2 ) R

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia 1 2 b2

q,

International Journal of Mathematical Education in Science and Technology 45


5/24

which implies that 2a 140 from the condition R24a2 b2, that is a4 1/2. Then

Equation (2.3) becomes

r a cos b sin

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia cos b sin 22a 1

q: 2:5

It is a relatively simple matter to perform the mapping in a spreadsheet

program such as Microsoft Excel. By specifying values a and b and a column

of degrees, , starting from 0.5 and going to 359.5 in increments of 1 for a total

of 359 data points (the reason for the half degrees will become clear when we look

at the point 180), a column of r() is created, then columns of x rcos

and y r sin, and u r(1 1/r)cos and v r(1 1/r)sin may be calculated.

By plotting y versus x and v versus u, the original circle and the mapped image

may be visualized. Take note of how the regions 0 and 2 of the

circle (shown in black and grey, respectively, in all figures) are mapped to their

respective portions.

For a 0 b the circle has its centre at the origin with radius 1 and passesthrough the points z 1. The mapping is simply the u-axis between 2 and 2,

which is obvious from Equation (2.4) for r 1 where u 2 cos and v 0.

For1 1/25a50 and b 0 we get a symmetric profile with a rounded edge (at the

leading edge) and a cusp (at the trailing edge). The mapping starts out very large and

reduces in size as a gets closer to zero, and the leading edge always passes through the

u-axis beyond u 2. For a40 and b 0 we again get a symmetric profile with a

rounded leading edge and a cusp at the trailing edge, which are thinner than those for

1/25a50 for the same value of jaj. These mappings are useful for profiles of fins,

rudders and struts [6]. As a increases, we obtain larger profiles. Note that for a40

the regions 0 and 2 of the circle are mapped to the upper and lowerportion of the profile, while for 1/25a50 they are mapped to the lower and upper

portion of the profile, respectively. These mappings are shown in Figures 1 and 2 for

a 0, 0.2, 0.4.

For a 0 and and fixed b the mapping produces a curved line spanning

2 u 2 above the u-axis for b40 and symmetrically below for b50. This single

line may be used as the mean camber line (the line drawn midway between the

upper and lower surfaces of the aerofoil). Later we will show how to obtain the

mean camber line for a particular aerofoil. These mappings are shown in Figure 3

for b 0, 0.2

For a 6 0 and b40 the mapping generates an aerofoil shape, which areknown as Joukowski aerofoils. If we concentrate on the point w 2, which

corresponds to the trailing edge of the aerofoil, we see that the upper and

lower portions of the aerofoil seem to be tangent to each other. This of course

is well known from a conformal mapping point of view [6], where the

upper and lower portions of the aerofoil meet at an angle of zero (i.e. a cusp).

One way to visualize this is to consider the case a 0 for a fixed b this makes

the upper and lower portions of the aerofoil collapse onto a single line, which is

often said to approximate the mean camber line [6] and is tangent to the upper

and lower portions for a 6 0. These mappings are shown in Figure 4 for b 0.25 and

a 0, 0.2.Using calculus, we can show that the upper and lower portions of the aerofoil do

indeed meet at an angle of zero, but the case a 0 and b fixed is not tangent to these

46 M.T. Matthews


6/24

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

y

x

(I)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u

v

(II)

Figure 1. The mapped circles (I) and Joukowski transformations (II) for b 0 anda 0.4, 0.2, 0.



7/24

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

y

x

(I)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u

v

(II)

Figure 2. The mapped circles (I) and Joukowski transformations (II) for b 0 anda 0, 0.2, 0.4.

48 M.T. Matthews


8/24

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

y

x

(I)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u

v

(II)

Figure 3. The mapped circles (I) and Joukowski transformations (II) for a 0 andb 0.25, 0, 0.25.



9/24

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

y

x

(I)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u

v

(II)

Figure 4. The mapped circles (I) and Joukowski transformations (II) for b 0.25 anda 0.2, 0, 0.2.

50 M.T. Matthews


10/24

portions. What we require is the derivative dv/du as a function of using the chain

rule we have

dv

du

dvddu

d

,

where u and v are given by Equation (2.4) and r is given by Equation (2.5). It can be

shown that

dr

d

r b cos a sin

r a cos b sin ,

and

du

d

dr

d1

1

r2

cos r

1

r

sin ,

dvd

dr

d1 1

r2

sin r 1

r

cos :

Therefore we have

dv

du

r2 r cos a r cos a cos 2 b sin 2

r2 b r sin r sin a sin 2 b cos 2 : 2:6

We can also plot Equation (2.6) in the spreadsheet, which shows that the

derivative is a continuous function of , which implies that the gradient of the

mapping as ! and ! are equal, which implies that the upper and lower

portions of the aerofoil meet at an angle of zero at the trailing edge.As ! , we have r ! 1 and dv/du has the indeterminate form 0

0, so using

lHospitals we have

d

dTop

dr

d3r2 1

cos 2ra

r sin 1 r2

2a sin 2 2b cos 2 ,

d

d

Bottom dr

d

2rb 3r2 1 sin r cos 1 r2 2a cos 2 2b sin 2 :

Now, as ! we have r ! 1, dr/d! b/(a 1), and

d

dTop ! 4b,

d

dBottom !

2 a 1 22b2

a 1,

therefore

lim!

dv

du

2b a 1

a 1 2b2: 2:7

Notice that the slope at the point is dependent on a, so setting a 0 for a fixed

b yields different slopes. To get the mean camber line corresponding to particular



11/24

values ofa and b, we need to find a value of b such that the cases a 6 0 and a a 0

have the same slope at . From Equation (2.7) we have

2b a 1

a 1 2b2

2b

1 b2) ^b

b

a 140,

where we have taken the positive root since b40. This implies

lim!

dv

dua, b lim

!

dv

du0,

b

a 1

:

Therefore, for an aerofoil with a 6 0 and b40, its mean camber line is that where

a 0 and b b= a 1 .

A plot of the derivatives are shown in Figure 5 for (a, b) (0.2, 0.25), (0, 0.25)

and a, ^b where a 0, ^b b= a 1 . The case a 0 is symmetric (see Figure 3) and

the cases (0.2, 0.25), 0, 0:250:8

and (0.2, 0.25), 0, 0:251:2

intersect at 180, while

the cases (0.2, 0.25) and (0, 0.25) intersect at an angle less than 180

. A plotof the Joukowski aerofoils and their corresponding mean camber lines are shown

in Figure 6.

We have therefore shown that the angle at the trailing edge of the aerofoil is zero

by calculating the derivative dv/du as a function of and showing that it is

continuous, in particularly it is continuous at . From an engineering point of

view, a zero angle at the trailing edge is impossible to construct and in reality there

will be a non-zero angle at the trailing edge. A transformation that accomplishes this

is known as the KarmanTrefftz transformation.

3. KarmanTrefftz transformation

From Equation (2.1) we have

w 2 z 1 2

z, w 2

z 1 2

z)

w 2

w 2

z 1

z 1

2: 3:1

The KarmanTrefftz transformation is a modification of the Joukowski transfor-

mation and is defined as

w n

w n

z 1

z 1

n

, ) w nz 1 n z 1 n

z 1 n

z 1 n ! 3:2

where n is slightly less than 2, and reduces to the Joulowski transformation

when n 2.

Since n is not an integer for the KarmanTrefftz transformation, evaluating

the real and imaginary parts of the transformation is a little tricky looking at

Equation (3.1), if we let

w1 z z 1

z 1, w2 z z

n, w3 z n z 1

z 1,

then we have w(z) w3(w2(w1(z))). (The expression for w3 comes from solving

(w n)/(w n) z for w.) We do this to make the transformations easier, and showswhy we use two forms of complex numbers Cartesian form a ib is easier for

addition and subtraction (like z 1), and polar form rei is easier for powers (like zn).

52 M.T. Matthews


12/24

1

0.5

0

0.5

1

0 50 100 150 200 250 300 350 400dv/du

(I)

1

0.5

0

0.5

1

0 50 100 150 200

q

q

250 300 350 400dv/du

(II)

Figure 5. Derivatives dv/dv of the Joukowski transformations for a 0.2 (I) and a 0.2(II), for b 0.25 and a 0, ^b b= a 1 .



13/24

2.5

2

1.5

1

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u

v

(I)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u

v

(II)

Figure 6. The Joukowski transformations and correct mean camber lines for b 0.25 anda 0.1 (I) and a 0.1 (II).

54 M.T. Matthews


14/24

For a complex number z x iy the Cartesian forms of w1 and w3 are

w1 z u1 iv1 x2 y2 1

x 1 2y2 i

2y

x 1 2y2,

w3 z u3 iv3

n x2 y2 1 x 1 2y2 i

2ny

x 1 2y2 ,

while the real and imaginary parts of w2 u2 iv2 are found using De Moivres

theorem.

Therefore, starting with z x iy we find w1 u1 iv1 where

u1 x2 y2 1

x 1 2y2, v1

2y

x 1 2y2,

then convert w1 to polar form w1 r1(cos1 isin1) where

r1 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

u21 v21

q, 1 tan1 v1

u1

,

then find w2 u2 iv2 where

u2 rn1 cos n1 , v2 r

n1 sin n1 ,

then find w3 u3 iv3 where

u3 n u22 v

22 1

u2 1

2v22, v3

2nv2

u2 1 2v22

:

Like with the Jouowski transformation, it is a relatively simple matter to performthe mapping in a spreadsheet program such as Microsoft Excel (using the ATAN2

function in Mircosoft Excel for the polar form ofw1). By specifying values ofa, b and

n and a column of degrees, , starting from 0.5 and going to 359.5 in increments of

1, a column of r() is created using Equation (2.5), then columns of x r cos and

y rsin, and u1, v1, r1, 1, u2, v2, u3 and v3 may be calculated. By plotting y versus x

and v3 versus u3, the original circle and the mapped image may be seen. It is also

possible to view each intermediate mapping v1 versus u1 and v2 versus u2. By setting

n 2 we obtain the original Joukowski aerofoil.

For a 0 b the circle has its centre at the origin with radius 1 and passes

through the points z 1. For n 2 the mapping is simply the u3-axis between 2and 2, like for the Joukowski aerofoil. For n42 the mapping forms two arcs above

and below the u3-axis outside the interval 25u352, and as n increases, the

mapping approached a circle. For 15n52 the same thing happens but now the arcs

are inside the interval 25u352. When n 1 the mapping is identical to the original

circle, which is obvious from Equation (3.2) which yields w z for n 1. For

05n51 the mapping forms a figure eight shape, and the mappings for n50 are

identical (in shape) as those for n40. These mappings are shown in Figure 7 for

n 0.5, 1.2, 1.9 and 2.5.

For2 1/25a50 and b 0 we get a symmetric profile. For 05n51 we get a

figure eight shape that is not closed for positive u3, and for 15n52 we see that themapping crosses itself forming a loop for small n and does not meet for larger n.

For n42 the mapping intersects for small n and is not closed for large n.



15/24

These mappings are shown in Figure 8 for a 0.2 and n 0.5, 1.2, 1.9 and 2.2.

Obviously, none of these mappings resembles an aerofoil!

For a40 and b 0 we get a symmetric profile. For 05n51 we get a cardioid-like

figure like that in Section 1. For 15n52 we obtain a mapping with a rounded

leading edge and a sharp point at the trailing edge. For n near 2, the mappings are

useful for profiles of fins, rudders and struts [6]. As a increases, we obtain larger

profiles. For n42 the mapping crosses itself forming two loops, the relative sizes of

the loops changing as n increases further. These mappings are shown in Figure 8 for

a 0.2 and n 0.5, 1.2, 1.9 and 2.2.

For a 0 and and fixed b the mapping produces two curved lines. For 05n51

the curves are above and below the u3-axis, like a figure eight. For 15n52 the curve

below the u3-axis moves upward, eventually moving above the u3-axis so the

mapping forms a crescent. When n 2 the curves meet (so we get a single arc, like the

Joukowski aerofoil), and for n42 the crescent returns. Replacing b with b just

reflects the mapping in the u3-axis. These mappings are shown in Figure 9 for a 0,

b 0.25 and n 0.5, 1.2, 1.9 and 2.2.

For a 6 0, b40 and 3/25n52 the mapping generates an aerofoil shape, which

are known as KarmanTrefftz aerofoils. If we concentrate on the trailing edge of the

aerofoil (which no longer corresponds to the point w 2 for n 6 2 from Equation

(3.2), but to the point w n) we see that the upper and lower portions of the

aerofoil seem to be not quite tangent to each other. This of course is well knownfrom a conformal mapping point of view [6], where the upper and lower portions of

the aerofoil meet at an angle of (2 n). For aerofoils the only angle of interest

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5u3

v3

Figure 7. KarmanTrefftz transformations for a 0 b and n 0.5, 1.2, 1.9, 2.5.

56 M.T. Matthews


16/24

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u3

v3

(I)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5u3

v3

(II)

Figure 8. KarmanTrefftz transformations for a 0.2 (I) and a 0.2 (II) for b 0 andn 0.5, 1.2, 1.9, 2.2.



17/24

is when 05(2 n)5/2, or 3/25n52. One way to visualize this non-zero angle is

to consider the case a 0 for a fixed b this makes the upper and lower portions of

the form a crescent, which is often said to approximate the core of the aerofoil3 and

is tangent to the upper and lower portions for a 6 0. These mappings are shown in

Figure 10 for a 0.2, b 0.25 and n 1.55, 1.75 and 1.95.

For completeness, Figure 11 shows original circle, intermediate mappings and

aerofoil for a 0.1, b 0.25, n 1.95 and the approximate core of the aerofoil

corresponding to a 0. We see that the original circles are first mapped to a circle-

like figure and a straight line, then a cardioid-like figure and two straight lines, then

finally to the aerofoil and the core. For n 2 the intermediate mappings are a circle

and a cardioid.

Again using calculus we can show that the upper and lower portions of the

aerofoil do indeed meet at a non-zero angle, but the case a 0 and b fixed is not

tangent to these portions. What we require is the derivative dv3/du3 (dv3/d)/

(du3/d) as a function of using the chain rule this would involve calculating

dX

d

@X

@u2

@u2

@r1

@r1

@u1

@u2

@1

@1

@u1

@X

@v2

@v2

@r1

@r1

@u1

@v2

@1

@1

@u1

!

@u1

@x

dx

d

@u1

@y

dy

d @X

@u2

@u2

@r1

@r1

@v1

@u2

@1

@1

@v1

@X

@v2

@v2

@r1

@r1

@v1

@v2

@1

@1

@v1

!@v1

@x

dx

d

@v1

@y

dy

d

,

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5u3

v3

Figure 9. KarmanTrefftz transformations for a 0, b 0.25 and n 0.5, 1.2, 1.9, 2.2.

58 M.T. Matthews


18/24

where X u3,v3. The numerator and denominator of dv3/du3 would each have

16 terms, and would have to be differentiated with respect to since the limit !

would again be an indeterminate form of type 00. This is a formidable task especially

since n is not an integer!

Instead, we employ the use of a symbolic software package, such as MAPLE.

Setting the values of a, b and n, we then define r() first, then defining x,y, u1, v1, r1,

1, u2, v2, u3 and v3. MAPLE will calculate the derivative dv3/du3 as a function of

(the formula is exceedingly long), and using MAPLEs standard4 plot command a

plot of dv3/du3 versus is produced. The MAPLE plots are shown in Figure 12 for

a 0.1, b 0.25 and n 1.95 and 2, and in Figure 13 for a 0, b 0.25 and n 1.95.

The case n 2 corresponds to what was found previously in the spreadsheet, and for

n 1.95 there is a jump discontinuity at . Therefore the limits ! and

! are no longer equal. Notice that the slopes at the point are dependent

on a, so setting a 0 for a fixed b yields the same gap but different slopes (compare

Figure 12 (II) and Figure 13 (II)).

We may roughly calculate the angle each limit makes from the derivative, since

the derivative in its most basic form is riserun

and the angle the tangent makes with the

positive x-axis (or u3-axis in this case) satisfies tan riserun

. Therefore the angle of the

derivative is tan1(dv3/du3), and we find that

tan1dv3

du3

!

tan1

dv3

du3

!

% 2 n :

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5u3

v3

Figure 10. KarmanTrefftz transformations for a 0.2, b 0.25 and n 1.55, 1.75, 1.95.



19/24

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

y

x

(I)

6

4

2

0

2

4

6

8

2 0 2 4 6 8 10 12

v1

u1

(II)

Figure 11. The mapped circle, intermediate mappings and KarmanTrefftz transformation

for n 1.95, b 0.25 and a 0.1, 0.

60 M.T. Matthews


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75

55

35

15

5

25

45

65

85

105

125

50 30 10 10 30 50 70 90 110 130 150

v2

u2

(III)

2.5

2

1.5

1

0.5

0

0.5

1

1.5

2

2.5

2.5 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

u3

v3

(IV)

Figure 11. Continued.



21/24

To find the core of the aerofoil with the same slopes at the trailing edge is a

matter of trial and error it appears that it is approximately the same as for theJoukowski aerofoil, its mean camber line is that where a 0 and ^b b= a 1 , but

surely it would depend on n also.

Figure 12. Derivatives dv3/du3 of the KarmanTrefftz transformation for 0 2 (I) andnear (II) for a 0.1, b 0.25 and n 1.95 (grey) and n 2 (black).

62 M.T. Matthews


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Figure 13. Derivatives dv3/du3 of the KarmanTrefftz transformation for 0 2 (I) andnear (II) for a 0, b 0.25 and n 1.95.



23/24

We have therefore shown that the angle at the trailing edge of the aerofoil

is non-zero by calculating the derivative dv3/du3 as a function ofand showing that

it has a jump discontinuity at .

4. Modelling the flow past the aerofoils

With the spreadsheets created it is possible to study the flow past an aerofoil (or

many of the other shapes generated) using the known results for the streamfunction

and velocity potential for the flow past a circle [2,6] and the (complex) derivative of

the mappings dw/dz. Of course, the mappings now must be conformal it may be

shown that the mappings for a40 are in fact conformal [3,6].

5. Conclusion

In this article two applications of conformal mapping to aerofoil theory, namely theJoulowski and KarmanTrefftz transformations, are studied from a geometric and

calculus point of view. A spreadsheet program (an often underrated but extremely

powerful tool for mathematics education) and knowledge of parametric equations,

complex numbers and calculus is all that is required for the most part, and is suitable

for undergraduate teaching in terms of a project or extended piece of work. Some

interesting mappings are generated, and by calculating the derivative of the mapped

aerofoil, some interesting analytical results are obtained for the Joulowski aerofoil,

and numerical results for the KarmanTrefftz aerofoil.

Acknowledgements

The author wishes to thank colleague A/Prof. Ute Mueller for providing the original article byKidwell Bolger delivered at the Queensland Association of Mathematics Teachers (QAMT)Conference in 1994 which inspired this work, and undergraduate project student ShanaleighOkely. The authors also acknowledge the helpful suggestions of the referees whose commentsmaterially improved the presentation.

Notes

1. The mapping is actually not conformal for this case [3].2. Again, the mapping is actually not conformal for this case [3].3. The midpoint of the core would then be the mean camber line.4. By standard we mean that no extra commands, arguments or tricks are used to tell

MAPLE there is a jump discontinuity.

References

[1] R.B. Bird, W.E. Stewart, and E.N. Lightfoot, Transport Phenomena, John Wiley & Sons,

New York, 1960.

[2] G.K. Batchelor, An Introduction to Fluid Dynamics, Cambridge University Press,

Cambridge, 2000.[3] R.V. Churchill and J.W. Brown, Complex Variables and Applications, 4th ed.,

McGraw-Hill, Tokyo, 1984.

64 M.T. Matthews


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[4] N. Joukowski, On the profiles of aerofoils, Z. Flugtech. Motorluftschiffahrt 1 (1910),

p. 281.

[5] N. Joukowski, On the profiles of aerofoils, Z. Flugtech. Motorluftschiffahrt 3 (1912),

p. 81.

[6] L.M. Milne-Thomson, Theoretical Aerodynamics, 4th ed., Dover, New York, 1973.

[7] T. von Karman and E. Trefftz, Potential flow round aerofoil profiles, Z. Flugtech.Motorluftschiffahrt 9 (1918), p. 111.

[8] W. Kutta, On plane circulation flows and their applications to aeronautics, Sitz. Ber.

Ak.-Wiss., Math. Phys. Kl. 41 (1911), p. 65.

[9] E.N. Jacobs, K.E. Ward, and R.M. Pinkerton, The characteristics of 78 related airfoil

sections from tests in the variable density wind tunnel, NACA Report #460, 1933.

[10] J. Stewart, Calculus, 6th ed., Thomson Brooks/Cole, Toronto, 2007.


karman treffitz transformation

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