Lifting Line Theory
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•
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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
Thin-airfoil theory
Cl=2π(α-αo)
Biot Savart Law:
Velocity produced by a
semi-infinite segment of
a vortex filament
h
Γ
V=
Γ
4πh
1868-1946
1875-1953
Physics of an Unswept Wing
l, Γ
Lift varies across span
-s
s
y
Circulation is shed (Helmholz thm)
pu<pl
pu≈ pl
Vortical wake
Vortical wake induces
downwash on wing…
Downwash
…changing angle of attack
just enough to produce
variation of lift across span
Simplest Possible Model
Wake model
b
Section model
A
A
Induced drag
di
Section A-A
LLT – The Section Model
di
ε
l
V∞
ε w
α
LLT – The Wake Model
Γ
-s
y1
y
s
dy1
The Monoplane Equation
l, Γ
Wake model
dΓ
s
w( y ) =
dy y1
dy1
∫ 4π ( y − y )
−s
Section model
1
s
c
=
Γ π V∞ (α − α 0 )c − ∫
4 −s
dΓ
dy y1
dy1
y − y1
π θ
y / s = − cos θ
=
Γ π V∞ (α − α 0 )c − π wc
πc
s y
-s
0
Substitute for θ, and express
Γ as a sine series in θ
Γ = 4U ∞ s
 πcn

(α − α 0 ) sin θ = ∑ An sin( nθ ) 
+ sin θ 
4s
 4s

n =1, odd
∞
∑ A sin(nθ )
n =1, odd
n
∞
The Monoplane Eqn.
Substituting Γ = 4U ∞ s
Results
∞
∑ A sin(nθ )
n =1, odd
s
n
into
s
2
CL =
Γdy
V∞ S −∫s
2
C Di = 2 ∫ wΓdy
V∞ S − s
s
w( y ) =
dΓ
gives C L = ARπA1
C Di =
C
(1 + δ )
πAR
w
=
V∞
dy1
∫ 4π ( y − y )
−s
∞
2
L
dy y1
∑
n =1, odd
1
nAn sin(nθ )
sin θ
δ=
∞
∑ n( A
n =3, odd
πc
4s
(α − α 0 ) sin θ =
 πcn

sin(
θ
)
sin
θ
+
A
n
∑
n
 4 s

n =1, odd
∞
n
/ A1 ) 2
Solution of monoplane equation
πc
 πcn

(α − α 0 ) sin θ = ∑ An sin( nθ ) 
+ sin θ 
4s
 4s

n =1, odd
∞
s y
-s
0
π θ
y / s = − cos θ
s=2.8;
alpha=5*pi/180;
alpha0=-5.4*pi/180;
N=20;
th=[1:N]'/N*pi/2;
y=-cos(th)*s;
c=ones(size(th));
n=1:2:2*N-1;
%Half span (distances normalized on root chord)
%5 degrees angle of attack
%Zero lift AoA=-5.4 deg. for Clark Y
%N=20 points across half span
%Column vector of theta's
%Spanwise position
%Rectangular wing, so c = c_r everywhere
%Row vector of odd indices
llt.m
res=pi*c/4/s.*(alpha-alpha0).*sin(th); %N by 1 result vector
coef=sin(th*n).*(pi*c*n/4/s+repmat(sin(th),1,N)); %N by N coefficient matrix
∞
a=coef\res;
%N by 1 solution vector
Γ = 4U ∞ s
gamma=4*sin(th*n)*a; %Normalized on uinf and s
w=(sin(th*n)*(a.*n'))./sin(th);
AR=2*s/mean(c);
CL =
CL=AR*pi*a(1);
CDi=CL^2/pi/AR*(1+n(2:end)*(a(2:end).^2/a(1).^2));
πc
4s
(α − α 0 ) sin θ =
ARπA1
 πcn

sin(
)
sin
θ
θ
+
A
n
∑ n
 4 s

n =1, odd
∞
∑ A sin(nθ )
n =1, odd
n
C L2
C Di =
(1 + δ )
πAR
Example
0.1
y/c
Our AR=5.6 Rectangular Clark Y Wing
0
0
CL=0.80783, CDi=0.038738
0.2
Γ/V∞s
0.15
0.1
0.05
0
-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
-0.9
-0.8
-0.7
-0.6
-0.5
y/s
-0.4
-0.3
-0.2
-0.1
0
0
-w/V
∞
-0.05
-0.1
-0.15
-0.2
-1
Determine aerodynamic characteristics
of our rectangular Clark Y wing
αo≈-5.4o
0.05
0.1
0.2
0.3
0.4
0.5
x/c
0.6
0.7
0.8
0.9
1
Drag Polar
C L2
CD =
πAR
Curve for minimum
drag (elliptical wing)
C D = C L (tan α − α 0lift )
Curve assuming wing only
generates force normal to chord
Note that friction drag coefficient of 0.01 added to CDi
The Elliptic Wing
The minimum drag occurs for a wing for which An=0 for n≥3. For this wing:
(cos θ = − y / s )
1.
Γ = 4U ∞ s
∞
n =1, odd
∞
2.
w
=
V∞
∑
∑ A sin(nθ )
n =1, odd
n
nAn sin(nθ )
sin θ
3.
=
Γ π V∞ (α − α 0 )c − π wc
Further results
C L = ARπA1
C L2
C Di =
πAR
w
= A1
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…
2πAR(α − α 0 )
CL =
AR + 2
C L2
C Di =
πAR
Geometrically Similar Wings
These results work quite well even for non-elliptical wings:
C  1
1 

α A − α B = L 
−
π  ARA ARB 
Prandtl’s Classic
Rectangular Wing
Data for Different
Aspect Ratios
C DiA − C DiB
1 
C L2  1


=
−
π  ARA ARB 
Prandtl’s
rescaling using
LLT result to
AR=5