3d wings

Aero 301: Spring 2011 III.5 3D Vortices & Biot–Savart

III.5 Vortices in 3D & The Biot–Savart Law

• We are finally ready to start thinking about the aerodynamics of 3D objects: wings that

do not extend to ±∞ into and out of the page. To do this we start by thinking aboutwhat would happen if we were to impulsively start a motionless airfoil in a 2D world.

• We know 2 (seemingly) contradictory things

1. There will be vorticity/circulation associated with

the lift the moving airfoil generates but,

2. In the initial motionless state ω = 0 everywhere andDω/Dt = 0 so we should have ω = 0 everywhere for all time.

What gives?

• Dω/Dt doesn’t hold at the trailing edge (theKutta condition arises because of viscosity) so

we do generate the vorticity and circulation

required to produce the lift.

Γ = 0

Γ = 0

Γ < 0

Γ > 0

t < 0, No Motion

starting vortex

t > 0, Airfoil moving right

bound vortices

• But, besides that point, the flow remains inviscid

so no torque is applied to a large control volume

enclosing the wing and lots of space around it so the

circulation about that volume’s perimeter remains zero.

• So, for some region close to the wing to have positive circulation, some

negative circulation path must exist around a vortex not bound to the airfoil.

This other negative-Γ vortex is called a starting vortex. It has equal butopposite strength to the net vortex strength that’s bound to the moving airfoil.


• Does this really happen? Yes!

Set aside this 2D picture for a few moments

and let’s move gingerly into 3D. . .

• Imagine that we make the simplest extension

from our 2D airfoil picture into 3D. This

would mean that the point vortices would

become lines that extend to ±∞ inthe y direction, into and out of the page.

• Next, imagine that instead of just vortex lines,

the 3D world can have vortex filaments, twisty

vortex strings that produce infinite vθ -type velocitiesas distance from the filament goes to zero.

(You know these twisty vortices as tornados.)

• Working from the 3D incompressible Euler equations, we

could prove three vortex theorems developed by Helmholtz.

(But we won’t.)

H1 A vortex filament has constant strength, Γ along it’s length.

H2 A vortex filament cannot begin or end in a fluid. It must

end at a boundary, form a closed loop or extend to infinity.

H3 An inviscid fluid that is initially irrotational will remain

irrotational for all time.

• H1 and H2 do not apply in 2D

• H3 is nothing more than D~ω/Dt = 0.


• The ”closed loop” option of H2 explains the starting vortex that’s

observed in 2D flows. The two vorticies — one bound to the airfoil, the

other behind the airfoil — are just two bits of the same vortex loop.

• What about the rest of the loop?

As a finite wing (a 3D shape that

doesn’t extend to ±∞ in y) begins tomove, the Kutta condition generates a

vortex loop with one part of the vortex

bound to the wing, a starting vortex that

remains more or less at its starting

position and two legs called wingtip

vortices that connect the bound vortex

to the starting vortex. Γ = constantaround loop

Start

ing v

ortex

b = w

ingsp

an

• We can observe the wingtip vortices and the starting

vortex. We know that the bound vortex exists

because we can measure the lift on the wing: L = ρ U∞Γ b(although this ignores some nasty details). So, the

theoretical picture is in good agreement with observation.

• So, when a finite wing starts to move, it generates a starting vortex

and this remains pretty much where it originated (unless it’s near the

ground). As the wing moves, the bound vortex moves with it (so the

Kutta condition is satisfied) and the wingtip vortices get longer and

longer.


Top View Rear Views

Close to

trailing

edge

Far from

trailing

edge

Side View

Aero 301: Spring 2011 III.6 Prandtl’s Lifting Line

• How do we analyze the effect of 3D vortex filaments?

• The 3D extension of vθ =−Γ /2πr is known as the

Biot–Savart Law: ~v =−

∫ Γ4π

d~s× r|~r|3

– d~s points along the vortex filament indicating the senseof rotation (right-hand rule);

– ~r is the vector pointing from the point of interest (i.e.,where the velocity induced by the vortex filament is

evaluated) to the point s along the vortex; and

– the integral is evaluated along the length of the vortex

which is usually ±∞ or a closed loop

• The velocity potential of this field satisfies Laplace’s

Equation in 3D as it must.

• When the Biot–Savart Law is applied to a straight vortex

filament that extends from ±∞, the velocity reduces to thecorrect 2D behavior.

Use the setup to the right to verify this.

(Hints: r = h/cosβ and s = h tanβ .) h

ds

du

s

β

r

θ

vortexfilament

• As another example, what is the velocity at the

center of a circular vortex ring?


III.6 Prandtl’s Lifting Line Theory, Downwash and Induced Drag

• What are the implications of H1, H2 and the Biot–Savart

Law on finite wings? How do we model finite wings?

• Imagine that all the little γds vortices arrayed along thecamber line of a 2D airfoil are collected into a single Γthat’s placed at the quarter-chord point.

• In 3D, this vortex cannot end so wingtip vortices

connect the bound vortex back to the starting

vortex that is still sitting on the runway, 1000

miles behind the wing.

Γ = constantaround loop

b = w

ingsp

an

w(y) < 0

x

y

z

• The bound vortex generates lift.

• The starting vortex is so far aft, it does not induce any

velocity near the wing.

• The wingtip vortices generate downwash: w(y)< 0between themselves, including at the location of the

bound vortex.

• What is the downwash velocity induced by the pair of

wingtip vorices along the bound vortex? Calculate using

Biot–Savart. . .


w(y) =−Γ

πb [1− (2y/b)2]

• The preceding expression is a bit scary. It implies we have infinite

downwash velocities at the wing tips. Even disregarding the

impossibility of infinite velocities, we expect our small-angle

approximations do not work and that the wingtips are probably

stalled.

What do we do about this?

• The next level of approximation allows Γ to vary

along the wingspan, despite the fact that Γ must

be constant along a vortex filament. If we let

Γ → 0 at the wing tips, maybe we can avoid infinitedownwash.

How do we have our cake and eat it too?

• Because a single vortex filament must have

constant Γ , we simply stack a number of boundvortices along the lifting line but allow them to

turn back into trailing vortices at different points

along the span. This allows for a varying Γ (y).

Γ1

Γ1

Γ2

Γ2

Γ3

Γ3

Lifting Line


• Why might Γ vary along the span? All the reasons that L′ can

vary for an airfoil: chord, angle of attack and zero-lift angle

can all vary along the span.

– Chord variations are called taper: c = c(y)

– Angle of attack variations are called twist: α = α(y)

– Zero-lift angle variations are called

aerodynamic twist: αL=0 = αL=0(y)

– And, as we will see, there is an induced angle of attack,

αi (y) that decreases the effective angle of attack to lessthan the geometrical angle of attack.

• Now we need to consider what multiple vortices mean for the

downwash velocity w(y) along the lifting line.

• Consider the diagram shown to the right that shows

the lifting line from directly upstream. The

downwash at the point y0 due to each of the

half-infinite vortices is

wn(y0) =−12

Γn

2π(y0− yn)

where n is an index for each of the half vortices andyn is the y location of each of these. Each of the ‘A’vortices is positive, each of the ‘B’ vortices is the

negative of the corresponding ‘A’.

Γ1,A

Γ2,A

Γ3,A

Γ3,B

Γ2,B

Γ1,B

−b/2b/2y

0

y

z


• In the picture the vortices 1A, 2A, 3B, 2B, and 1B induce

velocities down; 3A induces a velocity up. The different

directions is given by the signs of y0 − yn and the signs of Γn

• If we have a large number of very weak vortices we can do a

little calculus and say

dw(y0,y) =−dΓ (y)

4π(y0− y)=−

14π(y0− y)

dΓdy

∣

∣

∣

∣

ydy

and, with this,

w(y0) =−1

4π

∫ b/2

−b/2

1y0− y

dΓdy

∣

∣

∣

∣

ydy.

This is the net downwash at y due to all the vortices.

• The net effect of this downwash is to tilt the

incoming flow vector down the tilt angle is called

an induced angle of attack, αi and, like w(y), thisangle depends on y.

U∞

w(y0)

αi(y

0)

• We define the induced angle of attack to be positive if

w < 0 (as it usually is) so

αi(y0)≈−w(y0)

U∞if w(y)≪U∞


• What are the effects of this tilt?

1. The effective angle of attack is less than the geometrical angle

of attack (the angle between the chord line and U∞ and this

leads to less lift at each station along the blade that you would

expect based on the geometrical angle of attack.

2. The aerodynamic force perpendicular to U∞ at any y0 is L′ cos(αi)(i.e., it is decreased slightly because it is tilted back)

3. There is now an aerodynamic force in the direction of U∞called induced drag: D′

i = L′ sinαi .

• To sort out the implications of all this, we need a way to determine

Γ (y). This function determines L′ and w at each station along the spanand needs to correctly reflect all the geometrical features of the wing.

• The strategy for finding Γ (y) is to equate two separate expressions forL′ for each 2D airfoil section along the wing. First, the

Kutta–Joukowski Theorem gives at y0

L′(y) = ρ U∞Γ (y0).

Second, the 2D airfoil characteristics give

L′(y) = 2π [α(y0)−αi(y0)−αL=0(y0)]×12

ρ U2∞c(y0)

• So, we have two different expressions for L′ at any y0. If we set these

equal to each other we can solve for the one thing we do not know for

a wing we have built, the Γ (y) distribution.

Aero 301: Spring 2011 III.7 Elliptical Wings

• Setting the two L′ espressions equal results in the

Fundamental Equation of Finite Wing Theory

2Γ (y0)

U∞ c(y0)= 2π

[

α(y0)−αL=0(y0)−1

4πU∞

∫ b/2

−b/2

1y0− y

dΓdy

∣

∣

∣

∣

ydy

]

• If we solve this equation for Γ (y0) we know the the lift at eachsection and, from this, the lift on the wing.

• This whole process is very similar to the development of

thin-airfoil theory. Equating two expressions for L′ gives Γ (y0)and integrating Γ (y0) over the wingspan gives the overall lift.

• It is difficult to solve this equation (it is another integral

equation) so our solution procedure will again include a sine series

(this is nice because we would like to have Γ = 0 at the wingtipsto avoid infinite downwash.

• What is unfortunate about all this is that for a given wing at a

given U∞ more lift requires more Γ (via an increased geometricalangle of attack). We see that increasing Γ also increases thedrag. However, increasing Γ also increases αi so, actually, Diincreases like L.


III.7 The Elliptical Lift Distribution

Solving the Fundamental Equation of Finite Wing Theory requires us to guess at a Γ (y) distribution and showit’s a correct guess. (The same approach we used for the γ(x) distribution for thin airfoils.)

As a first guess we consider an elliptic distribution:

Γ (y) = Γ0

[

1−

(

2yb

)2]1/2

This distribution has circulation Γ0 at the root (y = 0) and Γ = 0 at the wingtips (which avoids the infinitedownwash problem).

First, let’s compute the downwash by taking the derivative dΓ /dy and performing the variable transformations:2y/b = cosθ . With this we obtain

w =−Γ0

2b

An elliptic Γ distribution produces uniform downwash.


• Is such a Γ distribution possible? Put it into thefundamental equation of finite wing theory to verify. . .

• Yep! It works for an elliptic c(y) distribution if there is notwist and no aerodynamic twist (i.e., α and αL=0 are

constants along the span):

c(y) = c0

[

1−

(

2yb

)2]1/2

• We can integrate this chord distribution to find the

planform area, S and the aspect ratio, AR= b/S

S =πc0b

4and AR =

4bπc0

• We prefer to cite results in terms of b and AR rather than band c0 because not all wings are elliptical but all have an

unambiguious wingspan and a (nearly) unambiguous

planform area.


• Because we know Γ (y), then we can integrate across the span tofind the lift:

Γ0 =4U∞b(α −αL=0)

AR+2

• And with w <<U∞ such that cosαi ≈ 1 then αi can be written as:

L =πρU2

∞b2(α −αL=0)

AR+2

CL = 2π (α −αL=0)AR

AR+2


• What about the induced drag? Integrating across the span gives

Di =π8

ρΓ 20

• The induced drag coefficient is

CDi =C2

L

πAR

The induced drag coefficient depends on the lift coefficient squared.

Aero 301: Spring 2011 III.8 Non-Elliptic Lift Distributions

III.8 Non-Elliptic Lift Distributions

In general, any combination of chord distribution, twist distribution, and aerodynamic twist distribution will

produce lift and induced drag.

To accommodate all the possible variations, a Fourier series approach is used that maintains Γ = 0 at y =±b/2:

Γ = 2bU∞∞

∑n=1

An sin(nθ) where cosθ = 2y/b

This series approach is nothing more than a generalization of the elliptic Γ distribution because the ellipticdistribution has

An =

{

Γ0/2bU∞ n = 1

0 n > 1

Using this approach the odd coefficients, A1, A3, . . . , represent the symmetric variations in Γ that one typicallyimagines for a “normal” wing.

The even coefficients represent asymmetric variations (i.e., when the left wing is different than the right

wing). This doesn’t seem to be very common at first but these are the terms that are used to model aileron

displacements that induce rolling moments.


As before, we can think about the results we would get with a given set of An’s without actually computing the

coefficients. The approximate lift and induced drag are given by integrations similar to those used for the

elliptic distribution.

Begin by integrating for the lift (watch for helpful orthogonality!)

CL = πA1 AR

The lift coefficient only depends on the first coefficient in the series. However, unlike thin airfoil theory, this

single-term result becomes a better approximation to the lift as the number of An’s computed is increased.

(See below.)

Similarly, for drag,


CDi =C2

L

πAR(1+δ ) where δ =

N

∑n=2

n

(

An

A1

)2

δ is always positive and, for a reasonably well designed wing, δ ≪ 1so the preceding expression is often written

CDi =C2

L

π eARwhere e = (1+δ )−1 ≈ 1−δ

The symbol e is selected because this number is an efficiency that is never greater than 100%.

Note that an elliptic lift distribution gives the minimum possible induced drag for a particular AR because it isthe only distribution that gives e = 1 because all of it’s An terms equal zero for n > 1.

To find the An’s choose some number, N, of terms to keep in the sine series.


Then, choose N points along the span and evaluate the fundamental equation at those N points using thevalues of c, α, and αL=0 for each point. This gives N equations for the N unknown An’s.

3d wings

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