Solution
Γ2
y
V = 100 mph
8
Γ3
=10 miles
Γ1
•
u ⋅ n = 0 at control pt #1:
The velocity at control pt #1 is the sum of the freestream + 3 point vortices’
velocities at that point:
u1 = V∞i +
Γ1
Γ2
Γ3
i−
i+
j
 
 
 
2π  
2π  
2π  
2
2
2
The normal at control pt #1 is:
n1 = −i
⇒ u1 ⋅ n1 = −V∞ −
Γ1
Γ2
+
=0
 
 
2π   2π  
2
2
Rearranging:
Γ1
π
•
−
Γ2
π
= −V∞
(1)
u ⋅ n = 0 at control pt #2:
Now, following the same procedure for control pt #2:
x
Solution
1
3
i+
j
2
2
Γ
Γ
V
u2 ⋅ n2 = ∞ − 2 + 3 = 0
2 π
π
Γ 2 Γ 3 V∞
−
=
π
π
2
n2 =
•
(2)
u ⋅ n = 0 at control pt #3:
1
3
i−
j
2
2
Γ
Γ
V
u3 ⋅ n3 = ∞ + 1 − 3 = 0
π
2 π
n3 =
⇒ −
Γ1
π
+
Γ3
π
=
V∞
2
(3)
Final System of Equations
Combine the numbered equations:
1
 1



−
0 
 π
−
V

π
∞

  Γ1  

V
1
1
 0
−   Γ2  =  ∞ 

π
π     2 
Γ 
 1
1   3   V∞ 
−
 vortex
0


 2 
π  strengths
 π
(unknowns)
Influence matrix
V∞ i ni
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:






influence
matrix
16.100 2002

  Γ0 
  Γ0  = 0
 
  Γ0 

2
Solution
 Γ0 
⇒ Given a solution  Γ 0  , then
 
 Γ 0 
 Γ1   Γ0 
 Γ  +  Γ  is also a solution where Γ is arbitrary.
0
 2  0
 Γ3   Γ0 
So, how do we resolve this?
Answer: the Kutta condition!
pt .e. +
⇒
Vupper
1
1
2
2
ρVupper
= pt .e. + ρVlower
2
2
= Vlower ≠ 0
What’s the Kutta condition for the windy city problem:
Γ2
V
8
Γ3
Γ1
Kutta: Γ3 = 0 ⇒ no flow around node 3!
So, we can now solve our system of equations starting with Γ3 = 0
π
Γ1 = − V∞
2
⇒ Γ2 =
π
2
Γ3 = 0
V∞
16.100 2002
3