MAE 3241: AERODYNAMICS AND FLIGHT MECHANICS

Thin Airfoil Theory

Mechanical and Aerospace Engineering DepartmentFlorida Institute of Technology

D. R. Kirk

OVERVIEW: THIN AIRFOIL THEORY• In words: Camber line is a streamline• Written at a given point x on the chord line• dz/dx is evaluated at that point x• Variable is a dummy variable of integration

which varies from 0 to c along the chord line• Vortex strength = () is a variable along the

chord line and is in units of • In transformed coordinates, equation is written

at a point, 0. is the dummy variable of integration– At leading edge, x = 0, = 0– At trailed edge, x = c, =

• The central problem of thin airfoil theory is to solve the fundamental equation for () subject to the Kutta condition, (c)=0

• The central problem of thin airfoil theory is to solve the fundamental equation for () subject to the Kutta condition, ()=0

dxdzVd

cx

dd

c

dxdzV

xdc

0 0

0

0

coscossin

21

Equation dTransforme

cos12

sin

cos12

tionTransforma Coordinate

21

:Theory AirfoilThin ofEquation lFundamenta

SUMMARY: SYMMETRIC AIRFOILS

Vd

cx

dd

c

dxdz

dxdzV

xdc

0 0

0

0

coscossin

21

Equation dTransforme

cos12

sin

cos12

tionTransforma Coordinate

0

:airfoils Symmetric

21

:Theory AirfoilThin ofEquation lFundamenta

SUMMARY: SYMMETRIC AIRFOILS

0cossin2

002

sincos12

coscossin

21 2

0 0

V

V

V

Vd• Fundamental equation of thin airfoil theory for

a symmetric airfoil (dz/dx=0) written in transformed coordinates

• Solution– “A rigorous solution for () can be

obtained from the mathematical theory of integral equations, which is beyond the scope of this book.” (page 324, Anderson)

• Solution must satisfy Kutta condition ()=0 at trailing edge to be consistent with experimental results

• Direct evaluation gives an indeterminant form, but can use L’Hospital’s rule to show that Kutta condition does hold.

SUMMARY: SYMMETRIC AIRFOILS• Total circulation, , around the airfoil (around the

vortex sheet described by ())

• Transform coordinates and integrate

• Simple expression for total circulation

• Apply Kutta-Joukowski theorem (see §3.16), “although the result [L’=∞V ∞

2] was derived for a circular cylinder, it applies in general to cylindrical bodies of arbitrary cross section.”

• Lift coefficient is linearly proportional to angle of attack

• Lift slope is 2/rad or 0.11/deg

2

2

sin2

2

0

0

ddcc

VcVL

cV

dc

d

l

l

c

EXAMPLE: NACA 65-006 SYMMETRIC AIRFOIL

• Bell X-1 used NACA 65-006 (6% thickness) as horizontal tail

• Thin airfoil theory lift slope:dcl/d = 2 rad-1 = 0.11 deg-1

• Compare with data– At = -4º: cl ~ -0.45– At = 6º: cl ~ 0.65– dcl/d = 0.11 deg-1

dcl/d = 2

SUMMARY: SYMMETRIC AIRFOILS

0

4

4

221

221

4,

,4,

,

2,

22

00

cm

llemcm

llem

LElem

LE

cc

LE

c

ccc

cc

ScV

Mc

cVM

dVdLM

• Total moment about the leading edge (per

unit span) due to entire vortex sheet

• Total moment equation is then transformed to new coordinate system based on

• After performing integration (see hand out, or Problem 4.4), resulting moment coefficient about leading edge is –/2

• Can be re-written in terms of the lift coefficient

• Moment coefficient about the leading edge can be related to the moment coefficient about the quarter-chord point

• Center of pressure is at the quarter-chord point for a symmetric airfoil

EXAMPLE: NACA 65-006 SYMMETRIC AIRFOIL

• Bell X-1 used NACA 65-006 (6% thickness) as horizontal tail

• Thin airfoil theory lift slope:dcl/d = 2 rad-1 = 0.11 deg-1

• Compare with data– At = -4º: cl ~ -0.45– At = 6º: cl ~ 0.65– dcl/d = 0.11 deg-1

• Thin airfoil theory:cm,c/4 = 0

• Compare with data

cm,c/4 = 0

CENTER OF PRESSURE AND AERODYNAMIC CENTER• Center of Pressure: Point on an airfoil (or body) about which aerodynamic

moment is zero– Thin Airfoil Theory:

• Symmetric Airfoil:

• Aerodynamic Center: Point on an airfoil (or body) about which aerodynamic moment is independent of angle of attack– Thin Airfoil Theory:

• Symmetric Airfoil:

4cxcp

4..cx CA

CAMBERED AIRFOILS: THEORY• In words: Camber line is a streamline• Written at a given point x on the chord line• dz/dx is evaluated at that point x• Variable is a dummy variable of integration

which varies from 0 to c along the chord line• Vortex strength = () is a variable along the

chord line and is in units of • In transformed coordinates, equation is written

at a point, 0. is the dummy variable of integration– At leading edge, x = 0, = 0– At trailed edge, x = c, =

• The central problem of thin airfoil theory is to solve the fundamental equation for () subject to the Kutta condition, (c)=0

• The central problem of thin airfoil theory is to solve the fundamental equation for () subject to the Kutta condition, ()=0

dxdzVd

cx

dd

c

dxdzV

xdc

0 0

0

0

coscossin

21

Equation dTransforme

cos12

sin

cos12

tionTransforma Coordinate

21

:Theory AirfoilThin ofEquation lFundamenta

CAMBERED AIRFOILS• Fundamental Equation of

Thin Airfoil Theory• Camber line is a streamline

• Solution– “a rigorous solution for

() is beyond the scope of this book.”

• Leading term is very similar to the solution result for the symmetric airfoil

• Second term is a Fourier sine series with coefficients An. The values of An depend on the shape of the camber line (dz/dx) and

sincos12

:Compare

sinsin

cos12

:Solution

coscossin

21

10

0 0

V

nAAV

dxdzVd

nn

EVALUATION PROCEDURE

dxdzdnAdA

nAAV

dxdzVd

n

n

nn

1 0 00 0

0

10

0 0

coscossinsin1

coscoscos11

sinsin

cos12

coscossin

21

PRINCIPLES OF IDEAL FLUID AERODYNAMICSBY K. KARAMCHETI, JOHN WILEY & SONS, INC., NEW YORK, 1966. APPENDIX E

PRINCIPLES OF IDEAL FLUID AERODYNAMICSBY K. KARAMCHETI, JOHN WILEY & SONS, INC., NEW YORK, 1966. APPENDIX E

CAMBERED AIRFOILS

0

00

10

100

100

cos2

1

cos

cos

cos

dnfB

dfB

nBBf

nAAdxdz

dxdznAA

n

nn

nn

nn

• After making substitutions of standard forms available in advanced math textbooks

• We can solve this expression for dz/dx which is a Fourier cosine series expansion for the function dz/dx, which describes the camber of the airfoil

• Examine a general Fourier cosine series representation of a function f() over an interval 0 ≤ ≤

• The Fourier coefficients are given by B0 and Bn

ADVANCED CALCULUS FOR APPLICATIONS, 2nd EDITIONBY F. B. HILDEBRAND, PRENTICE-HALL, INC., ENGLEWOOD CLIFFS, N.J., 1976

ADVANCED CALCULUS FOR APPLICATIONS, 2nd EDITIONBY F. B. HILDEBRAND, PRENTICE-HALL, INC., ENGLEWOOD CLIFFS, N.J., 1976

ADVANCED CALCULUS FOR APPLICATIONS, 2nd EDITIONBY F. B. HILDEBRAND, PRENTICE-HALL, INC., ENGLEWOOD CLIFFS, N.J., 1976

CAMBERED AIRFOILS

000

000

000

cos2

1

1

dndxdzA

ddxdzA

ddxdzA

n

• Compare Fourier expansion of dz/dx with general Fourier cosine series expansion

• Analogous to the B0 term in the general expansion

• Analogous to the Bn term in the general expansion

CAMBERED AIRFOILS

10

0 1 00

10

0

0

2

sinsincos1

sinsin

cos12

:for solution general Recall

sin2

AAcV

dnAdAcV

nAAV

dc

d

nn

nn

c

• We can now calculate the overall circulation around the cambered airfoil

• Integration can be done quickly with symbolic math package, or by making use of standard table of integrals (certain terms are identically zero)

• End result after careful integration only involves coefficients A0 and A1

CAMBERED AIRFOILS

2

1cos12

2

21

2

2

000

102

102

10

ddc

ddxdzc

AASV

Lc

AAcVL

AAcV

VL

l

l

l

• Calculation of lift per unit span

• Lift per unit span only involves coefficients A0 and A1

• Lift coefficient only involves coefficients A0 and A1

• The theoretical lift slope for a cambered airfoil is 2 , which is a general result from thin airfoil theory

• However, note that the expression for cl differs from a symmetric airfoil

CAMBERED AIRFOILS

0000

000

0

0

1cos1

1cos12

2

ddxdz

ddxdzc

c

ddcc

L

l

Ll

Ll

l

• From any cl vs. data plot for a cambered airfoil

• Substitution of lift slope = 2

• Compare with expression for lift coefficient for a cambered airfoil

• Let L=0 denote the zero lift angle of attack– Value will be negative for

an airfoil with positive (dz/dx > 0) camber

• Thin airfoil theory provides a means to predict the angle of zero lift– If airfoil is symmetric

dz/dx = 0 and L=0=0

SAMPLE DATA: SYMMETRIC AIRFOIL

Lift

Coe

ffic

ient

Angle of Attack,

A symmetric airfoil generates zero lift at zero

SAMPLE DATA: CAMBERED AIRFOIL

Lift

Coe

ffic

ient

Angle of Attack,

A cambered airfoil generates positive lift at zero

SAMPLE DATA• Lift coefficient (or lift) linear

variation with angle of attack, a– Cambered airfoils have

positive lift when = 0– Symmetric airfoils have

zero lift when = 0• At high enough angle of attack,

the performance of the airfoil rapidly degrades → stall

Lift

(for

now

)

Cambered airfoil haslift at =0At negative airfoilwill have zero lift

AERODYNAMIC MOMENT ANALYSIS

22

sin2

cos12

1

221

21

sinsin

cos12

210,

0,

02,

222,

10

00

AAAc

dcV

c

dcV

c

cV

M

ScV

Mc

nAAV

dVdLM

lem

lem

c

lem

LELElem

nn

cc

LE

• Total moment about the leading edge (per unit span) due to entire vortex sheet

• Total moment equation is then transformed to new coordinate system based on

• Expression for moment coefficient about the leading edge

• Perform integration, “The details are left for Problem 4.9”, see hand out

• Result of integration gives moment coefficient about the leading edge, cm,le, in terms of A0, A1, and A2

AERODYNAMIC MOMENT SUMMARY

21

124,

21,

210,

14

4

44

22

AAc

cx

AAc

AAcc

AAAc

lcp

cm

llem

lem

• Aerodynamic moment coefficient about leading

edge of cambered airfoil

• Can re-writte in terms of the lift coefficient, cl– For symmetric airfoil

• dz/dx=0• A1=A2=0• cm,le=-cl/4

• Moment coefficient about quarter-chord point– Finite for a cambered airfoil

• For symmetric cm,c/4=0– Quarter chord point is not center of

pressure for a cambered airfoil– A1 and A2 do not depend on

• cm,c/4 is independent of – Quarter-chord point is theoretical location

of aerodynamic center for cambered airfoils

CENTER OF PRESSURE AND AERODYNAMIC CENTER• Center of Pressure: Point on an airfoil (or body) about which aerodynamic

moment is zero– Thin Airfoil Theory:

• Symmetric Airfoil:• Cambered Airfoil:

• Aerodynamic Center: Point on an airfoil (or body) about which aerodynamic moment is independent of angle of attack– Thin Airfoil Theory:

• Symmetric Airfoil:• Cambered Airfoil:

2114

4

AAc

cx

cx

lcp

cp

4

4

..

..

cx

cx

CA

CA

ACTUAL LOCATION OF AERODYNAMIC CENTER

NACA 23012xA.C. < 0.25c

NACA 64212xA.C. > 0.25 c

x/c=0.25

x/c=0.25

IMPLICATIONS FOR STALL

• Flat Plate Stall

• Leading Edge Stall

• Trailing Edge Stall

Increasing airfoilthickness

LEADING EDGE STALL• NACA 4412 (12% thickness)

• Linear increase in cl until stall

• At just below 15º streamlines are highly curved (large lift) and still attached to upper surface of airfoil

• At just above 15º massive flow-field separation occurs over top surface of airfoil → significant loss of lift

• Called Leading Edge Stall• Characteristic of relatively thin

airfoils with thickness between about 10 and 16 percent chord

TRAILING EDGE STALL

• NACA 4421 (21% thickness)• Progressive and gradual movement of separation from trailing edge toward leading

edge as is increased• Called Trailing Edge Stall

THIN AIRFOIL STALL• Example: Flat Plate with 2% thickness (like a NACA 0002)• Flow separates off leading edge even at low ( ~ 3º)• Initially small regions of separated flow called separation bubble• As a increased reattachment point moves further downstream until total separation

NACA 4412 vs. NACA 4421• NACA 4412 and NACA 4421 have

same shape of mean camber line• Theory predicts that linear lift slope

and L=0 same for both

• Leading edge stall shows rapid drop of lift curve near maximum lift

• Trailing edge stall shows gradual bending-over of lift curve at maximum lift, “soft stall”

• High cl,max for airfoils with leading edge stall

• Flat plate stall exhibits poorest behavior, early stalling

• Thickness has major effect on cl,max

AIRFOIL THICKNESS

AIRFOIL THICKNESS: WWI AIRPLANES

English Sopwith Camel

German Fokker Dr-1

Higher maximum CLInternal wing structureHigher rates of climbImproved maneuverability

Thin wing, lower maximum CLBracing wires required – high drag

OPTIMUM AIRFOIL THICKNESS• Some thickness vital to achieving high maximum lift coefficient• Amount of thickness influences type of stall• Expect an optimum• Example: NACA 63-2XX, NACA 63-212 looks about optimum

cl,max

NACA 63-212

MODERN LOW-SPEED AIRFOILSNACA 2412 (1933)Leading edge radius = 0.02c

NASA LS(1)-0417 (1970)Whitcomb [GA(w)-1] (Supercritical Airfoil)Leading edge radius = 0.08cLarger leading edge radius to flatten cpBottom surface is cusped near trailing edgeDiscourages flow separation over topHigher maximum lift coefficientAt cl~1 L/D > 50% than NACA 2412

MODERN AIRFOIL SHAPES

http://www.nasg.com/afdb/list-airfoil-e.phtml

Root Mid-Span Tip

Boeing 737