Lifting Line Theory

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)

Large Aspect Ratio? Unswept?

Out

was

h

Physics of an Unswept Wingl, Γ

y

pu<pl pu≈ pl

Downwash

Inwash

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-A

αε

V∞

-w

l

ε

di

c

α=α (y) Geometric angle of attackε=ε (y) Downwash angle-w=-w (y) Downwash velocityc=c (y) Chordlengths Half spanl Lift per unit spandi Drag per unit span

Induced dragb

Total drag coeff

SVLCL 2

21

∞

=ρ

SVDC i

Di 221

∞

=ρ

Total lift coeff

A

A

Wake model Section model

LLT – The Wake ModelΓ

y1

ydy1

Strength of vortex shed at y1=

• Assume role up of wake unimportant• Assume wake remains in a plane parallel to the free stream• Model wake using single vortex sheet starting at the quarter chord

Downwash at y due to vortex shed at y1 )(4

)(1

11

yy

dydyd

ydw y

−

Γ−=−

π

Downwash at ydue to entire wake ∫

− −

Γ

=s

s

y

yy

dydyd

yw)(4

)(1

11

π

s-s

LLT – The Section Model

αε

V∞

-w

l

ε

di

=Γ

==∞

∞

∞ cVV

cVlCl 2

212

21 ρ

ρρ

wccV πααπ +−=Γ ∞ )( 0

Sectional lift coefficient

• Assume flow over each section 2D and determined by downwash at ¼chord, and thin airfoil theory

So

Sectional forces Γ≈ ∞Vl ρ Γ−≈Γ≈ ∞ wVdi ρερ

∫−

∞ Γ≈s

s

dyVL ρ ∫−

Γ−≈s

si dywD ρ

∫−∞∞

Γ≈=s

sL dy

SVSVLC 2

221 ρ ∫

−∞∞

Γ≈=s

s

iD dyw

SVSVDC

i 2221

2ρ

Total Forcesintegrated over span

Total Coefficients

The Monoplane EquationWake model

Section model∫− −

Γ

=s

s

y

yy

dydyd

yw)(4

)(1

11

π

wccV πααπ +−=Γ ∞ )( 0

l, Γ

ys-s

∫−

∞ −

Γ

+−=Γs

s

y

yy

dydyd

ccV1

1

01

4)( ααπ

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 2

2

∑∞

=∞=Γ

oddnn nAsU

,1)sin(4 θSubstituting

into

• Lift increases with aspect ratio• For planar wings at least lift goes linearly with angle of attack and lift curve slope increases with aspect ratio (to 2π at ∞) • Drag decreases with aspect ratio and goes as the lift squared?• Downwash tends to be largest at the wing tips ?• Drag is minimum for a wing for which An=0 for n≥3.

So,

gives 1AARCL π=

)1(2

δπ

+=ARCC L

Di

∑∞

=

=oddn

n AAn,3

21)/(δ θ

θ

sin

)sin(,1∑∞

=

∞

−= oddnn nnA

Vwα

⎥⎦⎤

⎢⎣⎡ +=− ∑

∞

=

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

1. Decide on the number of terms N needed for the sine series for Γ2. Select N points across the half span, evenly spaced in θ3. At each point evaluate c, α, α0 and thus the NxN matrix of terms that

multiplies the An’s and the N terms on the left hand side4. Solve for the An’s by matrix division5. Evaluate CL, CDi , w(y), and Γ(y).

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));

1. Decide on the number of terms N needed for the sine series for Γ2. Select N points across the half span, evenly spaced in θ3. At each point evaluate c, α, α0 and thus the NxN matrix of terms that

multiplies the An’s and the N terms on the left hand side4. Solve for the An’s by matrix division5. Evaluate CL, CDi , w(y), and Γ(y).

⎥⎦⎤

⎢⎣⎡ +=− ∑

∞

=

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

0.05

0.1

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 PolarARCC L

D π

2

= Curve for minimum drag (elliptical wing)

Note that friction drag coefficient of 0.01 added to CDi

If we pretend wing is elliptical…

-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

∞

041.0

856.02

)(2

2

0

==

=+−

=

ARCC

ARARC

LD

L

i π

ααπ

AR=5.6, α=5o, α0=-5.6o

Thus, an elliptical lift distribution can often be a good approximation!

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

( )sy /cos −=θ

14

22

1

=⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ Γ⇒

∞ sy

sAV

)sin(4)sin(4 1,1

θθ sAUnAsUoddn

n ∞

∞

=∞ ==Γ ∑

1,1

sin

)sin(A

nnA

Vw oddn

n

−=−=∑∞

=

∞ θ

θ

wccV πααπ +−=Γ ∞ )( 0

πααπ 10 )( AVVc

∞∞ −−Γ

=⇒• If the wing is untwisted, the chordlength is proportional to circulation and thus also has an elliptical form

• Lift distribution has an elliptical shape.

• Downwash velocity is constant across span

1.

2.

3.

Spitfire

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

Further results1AARCL π=

ARCC L

Di π

2

= 1AVw

−=∞

But what is A1?

sAVssAV r

r1

22

1

4104 ∞

∞

=Γ⇒=⎟⎠⎞

⎜⎝⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛ Γ

πααπ 10 )( AVVc r

r∞∞ −−

Γ=

Planform area of elliptic wing is rscS π21=

Now

and

Substituting and solving for A1 gives )2/()(2 01 +−= ARA αα

And thus 2

)(22

)(2 00

+−

−=+−

=∞ ARV

wAR

ARCLααααπ

…confirming our earlier presumption about aspect ratio effects on CL

Not done yet…

ARCC L

Di π

2

=

Consider two elliptical wings with the same section but different AR producing the same lift coefficient:

2)(2 0

+−

=AR

ARCLααπ

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−⇒

+=−

+=−

BA

LBA

B

BLB

A

ALA

ARARC

ARARC

ARARC

11

2)2(

2)2(

00

παα

παα

παα

Similarly, we can show the two drag coefficients are related as:

⎟⎟⎠

⎞⎜⎜⎝

⎛−=−

BA

LDiBDiA ARAR

CCC 112

π

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’srescaling using LLT result to AR=5