Aerodynamics notes

1. Solution
• 0
=
⋅ n
u at control pt #1:
The velocity at control pt #1 is the sum of the freestream + 3 point vortices’
velocities at that point:
1 2 3
1
2 2 2
2 2 2
u V i i i j
π π π
∞
Γ Γ Γ
= + − +
     
     
     
The normal at control pt #1 is:
1
1 2
1 1 0
2 2
2 2
n i
u n V
π π
∞
= −
Γ Γ
⇒ ⋅ = − − + =
   
   
   
Rearranging:
1 2
V
π π ∞
Γ Γ
− = − (1)
• 0
=
⋅ n
u at control pt #2:
Now, following the same procedure for control pt #2:
=10 miles
V = 100 mph
8
Γ1
Γ2
Γ3
x
y

2. Solution
16.100 2002 2
2
2 3
2 2
1 3
2 2
0
2
n i j
V
u n
π π
∞
= +
Γ Γ
⋅ = − + =
2 3
2
V
π π
∞
Γ Γ
− = (2)
• 0
=
⋅ n
u at control pt #3:
3
1 3
3 3
1 3
2 2
0
2
n i j
V
u n
π π
∞
= −
Γ Γ
⋅ = + − =
1 3
2
V
π π
∞
Γ Γ
⇒ − + = (3)
Final System of Equations
Combine the numbered equations:
1
2
3
vortex
strengths
(unknowns)
Influence matrix
1 1
0
1 1
0
2
1 1
0
2
i
V n
V
V
V
π π
π π
π π
∞
∞
∞
∞
   
−
  −
 
Γ
 
   
 
   
− Γ =
 
   
Γ
 
     
 
−  
 
 
i
The problem with these equations is that they have infinitely many solutions.
One clue is that the determinant of the matrix is zero. In particular we can add a
constant strength to any solution because:
0
0
0
influence
0
matrix
 
Γ
 
 
 
  Γ =
 
 
Γ
 
   
 

3. Solution
16.100 2002 3
⇒ Given a solution
0
0
0
Γ
 
 
Γ
 
Γ
 
 
, then
1 0
2 0
3 0
Γ Γ
   
   
Γ + Γ
   
Γ Γ
   
   
is also a solution where 0
Γ is arbitrary.
So, how do we resolve this?
Answer: the Kutta condition!
2 2
. . . .
1 1
2 2
t e upper t e lower
p V p V
ρ ρ
+ = +
⇒ 0
upper lower
V V
= ≠
What’s the Kutta condition for the windy city problem:
Kutta: 3 0
Γ = ⇒ no flow around node 3!
So, we can now solve our system of equations starting with 0
3 =
Γ
1
2
3
2
2
0
V
V
π
π
∞
∞
Γ = −
⇒ Γ =
Γ =
V
8
Γ1
Γ2
Γ3