Lifting Line Theory• Applies to large aspect ratio unswept wings at small angle of attack.• Developed by Prandtl and Lanchester during the early 20th century. • Relevance

– Analytic results for simple wings– Basis of much of modern wing theory (e.g. helicopter rotor aerodynamic

analysis, extends to vortex lattice method,)– Basis of much of the qualitative understanding of induced drag and aspect ratio

1875-19531868-1946Γ h

Vπ4Γ

=

h

Biot Savart Law:Velocity produced by a semi-infinite segment of a vortex filament

Thin-airfoil theoryCl=2π(α-αo)

Physics of an Unswept Wingl, Γ

y

pu<pl pu≈ pl

Downwash

s-s

Lift varies across span

Circulation is shed (Helmholz thm)

Vortical wake

Vortical wake induces downwash on wing…

…changing angle of attack just enough to produce variation of lift across span

Simplest Possible Model

Section A-AdiInduced dragb A

A

Wake model Section model

LLT – The Section Model

αε

V∞w

l

ε

di

LLT – The Wake ModelΓ

y1

ydy1

s-s

The Monoplane EquationWake model

Section model∫− −

Γ

=s

s

y

yy

dydyd

yw)(4

)(1

11

π

0( )V c wcπ α α π∞Γ = − −

l, Γ

ys-s

1

1

01

( )4

sy

s

d dydycV cy y

π α α∞−

Γ

Γ = − −−∫

0 π θ

θcos/ −=sy

Substitute for θ, and express Γ as a sine series in θ

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θ

+=− ∑

∞

=

θπθθααπ sin4

)sin(sin)(4 ,1

0 scnnA

sc

oddnn The Monoplane Eqn.

Results

∫− −

Γ

=s

s

y

yy

dydyd

yw)(4

)(1

11

π∫−∞

Γ=s

sL dy

SVC 2

∫−∞

Γ=s

sD dyw

SVC

i 22

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θSubstituting into

gives 1AARCL π= )1(2

δπ

+=ARCC L

Di

∑∞

=

=oddn

n AAn,3

21)/(δ

1,sin( )

sin

nn odd

nA nw

V

θ

θ

∞

=

∞

=∑

+=− ∑

∞

=

θπθθααπ sin4

)sin(sin)(4 ,1

0 scnnA

sc

oddnn

Solution of monoplane equation

+=− ∑

∞

=

θπθθααπ sin4

)sin(sin)(4 ,1

0 scnnA

sc

oddnn

ys-s0 π θ

θcos/ −=sy

s=2.8; %Half span (distances normalized on root chord)alpha=5*pi/180; %5 degrees angle of attackalpha0=-5.4*pi/180; %Zero lift AoA=-5.4 deg. for Clark YN=20; %N=20 points across half spanth=[1:N]'/N*pi/2; %Column vector of theta's y=-cos(th)*s; %Spanwise positionc=ones(size(th)); %Rectangular wing, so c = c_r everywheren=1:2:2*N-1; %Row vector of odd indices

res=pi*c/4/s.*(alpha-alpha0).*sin(th); %N by 1 result vectorcoef=sin(th*n).*(pi*c*n/4/s+repmat(sin(th),1,N)); %N by N coefficient matrixa=coef\res; %N by 1 solution vector

gamma=4*sin(th*n)*a; %Normalized on uinf and sw=(sin(th*n)*(a.*n'))./sin(th);AR=2*s/mean(c);CL=AR*pi*a(1);CDi=CL^2/pi/AR*(1+n(2:end)*(a(2:end).^2/a(1).^2));

+=− ∑

∞

=

θπθθααπ sin4

)sin(sin)(4 ,1

0 scnnA

sc

oddnn

1AARCL π=)1(

2

δπ

+=ARCC L

Di

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θ

llt.m

Example

-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 00

0.05

0.1

0.15

0.2

Γ/V∞

s

CL=0.80783, CDi=0.038738

-1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0-0.2

-0.15

-0.1

-0.05

0

y/s

-w/V

∞

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.050.1

x/c

y/c αo≈-5.4o

Determine aerodynamic characteristics of our rectangular Clark Y wing

Our AR=5.6 Rectangular Clark Y Wing

Drag Polar

Note that friction drag coefficient of 0.01 added to CDi

)(tan 0liftLD CC αα −=Curve assuming wing only generates force normal to chord

ARCC L

D π

2

= Curve for minimum drag (elliptical wing)

The Elliptic WingThe minimum drag occurs for a wing for which An=0 for n≥3. For this wing:

( )sy /cos −=θ

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θ

1,sin( )

sin

nn odd

nA nw

V

θ

θ

∞

=

∞

=∑

0( )V c wcπ α α π∞Γ = − −

1.

2.

3.

Further results1AARCL π=

ARCC L

Di π

2

= 1w A

V∞

=

Spitfire

Note that the chordlengths are all lined up along the quarter chord line so the actual wing shape is not an ellipse

Not done yet…

ARCC L

Di π

2

=2

)(2 0

+−

=AR

ARCLααπ

Geometrically Similar WingsThese results work quite well even for non-elliptical wings:

−=−

BA

LBA ARAR

C 11π

αα

−=−

BA

LDD ARAR

CCCiBiA

112

π

Prandtl’s Classic Rectangular Wing Data for Different Aspect Ratios

Prandtl’s rescaling using LLT result to AR=5