lifting line theory revisited

8/9/2019 lifting line theory revisited

1/19

Lifting Line Theory


2/19

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

1875-19531868-1946

Γ h

V

π 4

Γ=

h

Biot Savart Law:Velocity produced by asemi-infinite segment of

a vortex filament

Thin-airfoil theoryC l=2 π(α-αo)


3/19

Large Aspect Ratio? Unswept?


4/19

O u t w

a s h

Physics of an Unswept Wingl, Γ

y

pu


5/19

Simplest Possible ModelSection A-A

α ε

V ∞ -w

l

ε

d i

c

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

Induced dragb

Total drag coeff

S V L

C L 221

∞

= ρ

S V

DC i

D i 221 ∞

= ρ

Total lift coeff

A

A

Wake model Section model


6/19

LLT – The Wake ModelΓ

y1 y

dy1

Strength ofvortex shed at y1=

• Assume role up of wake unimportant• Assume wake remains in a planeparallel to the free stream• Model wake using single vortex sheetstarting at the quarter chord

Downwash at y dueto vortex shed at y1 )(4

)(1

11

y y

dydyd

ydw y−

Γ−=−

π

Downwash at ydue to entire wake ∫

− −

Γ=

s

s

y

y y

dydyd

yw)(4

)(1

11

π

s-s


7/19

LLT – The Section Model

α ε

V ∞ -w

l

ε

d i

=Γ==∞

∞

∞ cV V

cV l

C l 2212

21 ρ

ρ ρ

wccV π α α π +−=Γ ∞ )( 0

Sectional lift coefficient

• Assume flow over each section 2Dand determined by downwash at ¼

chord, and thin airfoil theory

So

Sectional forces Γ≈ ∞V l ρ Γ−≈Γ≈ ∞ wV d i ρ ε ρ

∫−

∞ Γ≈s

s

dyV L ρ ∫−

Γ−≈s

si dyw D ρ

∫−∞∞

Γ≈=s

s

L dy

S V S V

LC

22

21

ρ ∫

−∞∞

Γ≈=s

s

i D dyw

S V S V

DC

i 22

2

1

2

ρ

Total Forcesintegrated over span

Total Coefficients


8/19

The Monoplane EquationWake model

Section model∫− −

Γ=

s

s

y

y ydydy

d

yw)(4

)(1

11

π

wccV π α α π +−=Γ ∞ )( 0

l, Γ

ys-s

∫−

∞ −

Γ+−=Γ

s

s

y

y y

dydyd

ccV

1

1

01

4)( α α π

0 π θ

θ cos/ −=s y

Substitute for θ , and expressΓ as a sine series in θ

∑∞

=∞=Γ

odd nn n AsU

,1

)sin(4 θ

⎥⎦

⎤

⎢⎣

⎡ +=− ∑∞

=θ

π θ θ α α

π sin

4)sin(sin)(

4 ,10 s

cnn A

sc

odd nn The Monoplane Eqn.


9/19

Results

∫− −

Γ=

s

s

y

y y

dydyd

yw)(4

)(1

11

π ∫−∞Γ=

s

s L dyS V

C 2 ∫

−∞

Γ=s

s D dywS V

C i 2

2

∑∞

=∞=Γ

odd nn n AsU

,1

)sin(4 θ Substituting

into

• Lift increases with aspect ratio• For planar wings at least lift goes linearly with angle of attack and liftcurve slope increases with aspect ratio (to 2 π 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 1 A ARC L π =)1(

2

δ π

+= ARC

C L D i

∑∞

=

=odd n

n A An,3

21 )/(δ θ

θ

sin

)sin(,1∑

∞

=

∞

−= odd nn nnA

V wα

⎥⎦

⎤

⎢⎣

⎡ +=− ∑∞

=θ

π θ θ α α

π sin

4)sin(sin)(

4 ,10

s

cnn A

s

c

odd nn


10/19

Solution of monoplane equation⎥⎦

⎤

⎢⎣

⎡ +=− ∑∞

=θ

π θ θ α α

π sin

4)sin(sin)(

4 ,10 s

cnn A

sc

odd nn

ys-s0 π θ

θ cos/ −=s y

1. Decide on the number of terms N needed for the sine series for Γ 2. Select N points across the half span, evenly spaced in θ 3. At each point evaluate c, α , α 0 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).


11/19

s=2. 8; %Hal f span ( di st ances nor mal i zed on r oot chor d)al pha=5*pi / 180; %5 degr ees angl e of at t ackal pha0=- 5. 4*pi / 180; %Zer o l i f t AoA=- 5. 4 deg. f or Cl ar k YN=20; %N=20 poi nt s acr oss hal f spant h=[ 1: N] ' / N*pi / 2; %Col umn vect or of t het a' sy=- cos( t h) *s; %Spanwi se posi t i onc=ones( si ze( t h) ) ; %Rect angul ar wi ng, so c = c_r ever ywher en=1: 2: 2*N- 1; %Row vect or of odd i ndi ces

r es=pi *c/ 4/ s . *( al pha- al pha0) . *s i n( t h) ; %N by 1 r esul t vect orcoef =si n( t h*n) . *( pi *c*n/ 4/ s+r epmat ( si n( t h) , 1, N) ) ; %N by N coef f i ci ent mat r i xa=coef \ r es; %N by 1 sol ut i on vect or

gamma=4*si n( t h*n) *a; %Nor mal i zed on ui nf and sw=- ( s i n( t h* n) * ( a. * n' ) ) . / si n( t h) ;AR=2*s/ mean( c) ;CL=AR*pi *a( 1) ;CDi =CL 2̂/ pi / AR*( 1+n( 2: end) *( a( 2: end) . 2̂/ a( 1) . 2̂) ) ;

1. Decide on the number of terms N needed for the sine series for Γ 2. Select N points across the half span, evenly spaced in θ 3. At each point evaluate c, α , α 0 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).

⎥⎦

⎤

⎢⎣

⎡ +=− ∑∞

=θ

π θ θ α α

π sin

4)sin(sin)(

4 ,10 s

cnn A

sc

odd nn

1 A ARC L π =)1(

2

δ π

+= ARC

C L D i

∑∞

=∞=Γ

odd nn n AsU

,1

)sin(4 θ

llt.m


12/19

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, C Di=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≈-5.4 o

Determine aerodynamic characteristics

of our rectangular Clark Y wing

Our AR=5.6 Rectangular Clark Y Wing


13/19

Drag Polar ARC

C L D π

2

= Curve for minimumdrag (elliptical wing)

Note that friction drag coefficient of 0.01 added to CDi


14/19

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, C Di=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

==

=+

−=

ARC

C

AR AR

C

L D

L

i π

α α π

AR=5.6, α =5 o, α 0=-5.6 o

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


15/19

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

( )s y /cos −=θ

14

22

1

=⎟ ⎠ ⎞

⎜⎝ ⎛ +

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ Γ⇒

∞ s y

s AV

)sin(4)sin(4 1,1

θ θ sAU n AsU odd n

n ∞

∞

=∞ ==Γ ∑

1,1

sin

)sin( A

nnA

V w odd n

n

−=−=∑∞

=

∞ θ

θ

wccV π α α π +−=Γ ∞ )( 0

π α α π 10 )( AV V c

∞∞ −−Γ=⇒

• If the wing is untwisted, the chordlength isproportional to circulation and thus also hasan elliptical form

• Lift distribution has an elliptical shape.

• Downwash velocity is constant across span

1.

2.

3.


16/19

Spitfire

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


17/19

Further results1 A ARC L π = AR

C C L D i π

2

= 1 AV w −=∞

But what is A1?

s AV ss AV r

r 1

22

1

410

4 ∞

∞

=Γ⇒=⎟ ⎠ ⎞

⎜⎝ ⎛ +

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ Γ

π α α π 10 )( AV V c r r

∞∞ −−Γ=

Planform area of elliptic wing is r scS π 21=

Now

and

Substituting and solving for A1 gives )2/()(2 01 +−= AR A α α

And thus2

)(22

)(2 00+

−−=+

−=∞ ARV

w AR

ARC L

α α α α π

…confirming our earlier presumption about aspect ratio effects on CL


18/19


19/19

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

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=−

B A

L B A AR AR

C 11π α α ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=−

B A

L D D AR AR

C C C iBiA

112

π

Prandtl’s ClassicRectangular WingData for Different

Aspect Ratios

Prandtl’srescaling usingLLT result to

AR=5

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lifting line theory revisited