Chapter V. Lifting Line Theory
Part IV. Elliptically Loaded Wings
In the previous handout titled “Chapter V. Part III: Forces on the Airfoil Section at Point P on the
Lifting Line” we derived the following equation for the distribution of circulation (y) along the
span of a wing.

1
  y0    L  0  y0  

V c( y0 ) 4V
y b
 y
y  b
1
d
dy
0  y  dy
(1)
We also derived an expression for the downwash velocity:
w 
1
4
y b
1
d
dy
0  y  dy
 y
y  b
(2)
In this section, we study the solution of equation (1) for a special case – an elliptically loaded
wing. For this case, the loading is given by the equation of an ellipse:
 y2

1
0 b 2
(3)
Note that we call ‘b’ the semi-span (i.e. half of the wing span) in our notes. Anderson calls ‘b’
the full span. These discrepancies will, hopefully, add to your general confusion about the course
material. The quantity 0 is the maximum circulation that will occur at mid-span, at y=0. This
distribution may be visualized for one-half of a wing as follows:
(y)
0
y/b
We will use this assumed solution in equation (2) and see what happens. To facilitate
this, let us do a simple transformation:
y  b cos
(4)
Equation (3) then yields:
 y   0 sin 
d
dy  d  0 cos d
dy
(5)
Also, in equations (1) and (2)
y  y0  bcos  cos 0 
(6)
Plugging (5) and (6) into equation (2), and using the following identity from the integral tables:
(7)

cos
 cos  cos
0
d  
0
Equation (2) becomes:
w  
0
4b
(8)
In other words, for an elliptically loaded wing, the downwash velocity is a constant along the
span! The negative sign simply indicates that this velocity will be directed downwards.
Before proceeding further, we need to tie the maximum circulation 0 to the total wing lift, L.
From Kutta-Joukowski theorem, lift per unit span is given by:
Lift per unit span  V 
(9)
The total lift L is obtained by integrating the lift per unit span along the span. That is,
b
L  V   y dy
b
(10)
Using the assumed elliptical distribution for circulation (given in equation 3) in (10), and
performing the integration, we can show that:
L  V 0
b
2
Or ,
0 
2L
Vb
(11)
Using the definition of the lift coefficient CL for a wing
L
CL 
1
V2 S
2
(13)
in equation (11), we get:
V SC L
V C  S 
V C
 4b  L  2   4b  L
b
  4b 
AR
where,
0 
AR  Aspect ratio 
2b 
span 2

Wing _ area
S
2
(14)
Plugging equation (14) into equation (8), we get the expression for the downwash w’ in terms of
the lift coefficient CL for the wing as:
w'  
V C L
AR
(15)
The induced angle of attack i is given by (see previous hand-outs):
C
w'
i  
 L
V AR
(16)
Lift acts
normal to
altered
flow
direction
Altered flow
direction due to
downwash w’
Original flow
direction
As discussed in the previous hand-out and shown in the figure above, at each span station the lift
vector is titled slightly to the aft by an angle i. This aft-tilt gives a drag force (per unit span),
approximately equal to sectional lift force per unit span times i. The total drag is obtained by
integrating the drag force over the entire wing:
b
D
 Lift per unit span    ( y)dy
i
b
(17)
In the present elliptically loaded wing case, the induced angle of attack i is a constant.
Thus, the induced rag for an elliptically loaded wing is given by:
DInduced  Di  L i
C Di  C L i
(18)
Using equation (16) into equation (18), we get the induced drag coefficient for an
elliptically loaded wing:
C2
C D ,i  L
AR
(19)
This component of drag is just the inviscid component of drag, to which viscous
contributions (due to skin friction, separated flow, stall effects) must be added to get the total
wing drag. Notice that the smaller the aspect ratio, the higher the induced drag.
Why did we study elliptically loaded wings?
It turns out that an elliptically loaded wing has the lowest possible induced drag for a given wing
aspect ratio, and lift coefficient. Aerodynamic designers strive to achieve an elliptic lift
distribution to achieve this desired effect. They can do this by adjusting the chord distribution,
the twist distribution, and/or the angle of zero lift distribution.
An academic example is an untwisted wing with same airfoils along the span, which has
an elliptical plan form- i.e. c(y) varies with y/b as an ellipse. For such a wing, since the angle of
attack is constant along the span because there is no twist;the angle of zero lift 0 is constant
along the span since the same airfoils are used from root to tip; the induced angle of attack i is
constant due to the elliptical loading. The lift per unit span is given by:
1
Lift per Unit Span  V2 a 0    0   i c y 
2
(20)
Equating this with the circulation (y) we can show that (y) is behaves like an ellipse since c(y)
is elliptic. In equation (20), the quantity a0 is the lift curve slope. According to thin airfoil theory,
a0 is 2. Engineers use an experimentally arrived at lift curve slope (which will be slightly lower
than 2) to account for viscous effects in the analysis.
The total lift is given by integrating the above expression along the span, with respect to y. the
end result is
b
1
1
Lift L  V2 a0    0   i   c( y)dy  V2 a0    0   i S
2
2
b
(21)
The lift coefficient for an elliptic plan-form-untwisted wing with identical airfoil sections all
along the span is, therefore, given by:
L
CL 
 a 0    i   0 
1
2
V S
2
(22)
Using equation (16) in (22) we get, after some minor algebra:
a0
   0 
CL 
a0
1
AR
(23)
The lift curve slope for a 3-D wing with untwisted identical airfoil sections everywhere and an
elliptical plan form is given by:
a0
C L

a

1 0
AR
(24)
Notice that the finite wing has a lower lift curve slope compared to 2-D airfoil (which has a slope
of a0). The smaller the aspect ratio, the lower the lift curve slope.
For more general wings, the drag coefficient will be higher than that given by (19), and the lift
curve slope will be lower than that given by equation (24). General wings are handled by
numerically solving equation (1), using a procedure similar to that in section 5.3.2 in the text
(Equation 5.51 is solved; CL and CD are then found from 5.53 and 5.61, respectively.) See our
web site for a downloadable program from the U. of Sydney that does this.