Vector Plot of Airfoil Stream Speeds

Introduction

The stream speeds as a function of position around an airfoil can be modeled as a combination of the simple horizontal wind speed, a transitional air speed that is higher near the top and bottom surfaces of the airfoil, and, when there is an angle of attack, a circulating air vortex that tends to cancel the wind speed on the bottom of the airfoil and enhance the wind speed on the top of he airfoil. The strength of this latter vortex is proportional to the wind speed and the angle of attack with peak strength near the stall angle. The vortex is responsible for the lift and some of the drag (called induced drag) of the airfoil.

In this document, I will handle just the free stream as it distorts around a thin ellipse that may be rotated to some angle of attack.

Math

I chose to first use a thin ellipse as the airfoil. An ellipse has the advantage over a circular arc airfoil in that the stagnation point actually moves as the angle of attack changes. We need to be able to find the tangents and normals at all points on this ellipse. In particular, we need to locate point on the rotated ellipse whose normal is along the horizontal direction since that will be the stagnation point where the drift velocity is cancelled by the conformally mapped airflow.

Let the ellipse be centered at the origin of the (x,y) coordinate system and have semi-major axis a and semi-minor axis b and be rotated clockwise by angle α.

In the un-tilted frame of reference, the normal vector is parallel to the following vector:

$N = b \cos (θ_{p a r a}) x + a \sin (θ_{p a r a}) y$

(1.1)

where q_para is the angle referred to by the parametric description of the ellipse:

$\begin{array}{l} x_{p} = a \cos θ_{p a r a} \\ y_{p} = b \sin θ_{p a r a} \end{array}$

(1.2)

To become a useful unit vector N should be normalized as follows.

$n = \frac{b \cos (θ_{p a r a}) x + a \sin (θ_{p a r a}) y}{\sqrt{b^{2} \cos^{2} (θ_{p a r a}) + a^{2} \sin^{2} (θ_{p a r a})}}$

(1.3)

We want to find the parametric angle for the normal that points along the -x axis when the ellipse is rotated clockwise by angle α. Therefore we rotate the arbitrary normal by angle α

$n_{x y} = (\begin{matrix} n_{x}' \\ n_{y}' \end{matrix}) = (\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}) (\begin{matrix} n_{x} \\ n_{y} \end{matrix})$

(1.4)

We seek the value for θ_para where n has zero y component and a negative x component

$\begin{array}{l} b \cos θ_{p a r a} \sin α + a \sin θ_{p a r a} \cos α = 0 \\ b \cos θ_{p a r a} \cos α - a \sin θ_{p a r a} \sin α = - \sqrt{b^{2} \cos^{2} θ_{p a r a} + a^{2} \sin^{2} θ_{p a r a}} \end{array}$

(1.5)

$\begin{array}{l} \frac{a \sin θ_{p a r a}}{b \cos θ_{p a r a}} = - \frac{\sin α}{\cos α} \\ \frac{\sin θ_{p a r a}}{\cos θ_{p a r a}} = - \frac{b}{a} \frac{\sin α}{\cos α} \end{array}$

(1.6)

which indicates that

$\begin{array}{l} \tan θ_{p a r a} = - \frac{b}{a} \tan α \\ θ_{p a r a} = - \tan^{- 1} (\frac{b}{a} \tan α) \end{array}$

(1.7)

Since the description of the ellipse in the un-rotated frame is

$\begin{array}{l} x_{p} = a \cos θ_{p a r a} \\ y_{p} = b \sin θ_{p a r a} \end{array}$

(1.8)

To locate this point in the xy frame, we must apply the rotation matrix:

$r_{x y} = (\begin{matrix} x_{x y} \\ y_{x y} \end{matrix}) = (\begin{matrix} \cos α & - \sin α \\ \sin α & \cos α \end{matrix}) (\begin{matrix} x_{p} \\ y_{p} \end{matrix})$

(1.9)

This is the result for the position of the stagnation point at which both v_x and v_y are zero.

Computation of an elliptical airfoil's velocity potential that has zero imaginary component on the boundary

Define:

$c_{2} = a^{2} - b^{2}$

(1.10)

Airfoil border is described by

$\begin{array}{l} z_{b} = \frac{a + b}{2} \exp (i θ) + \frac{a^{2} - b^{2}}{2 (a + b)} \exp (- i θ) = \\ [\frac{a + b}{2} \exp (i θ) + \frac{c_{2}}{2 (a + b)} \exp (- i θ)] \end{array}$

(1.11)

or:

$z_{b} = a \cos θ + i b \sin θ$

(1.12)

The velocity potential, $ϕ$ (z), for this airfoil is obtained from the following equations:

$\begin{array}{l} 2 z = Z + \frac{c_{2}}{Z} \\ s o l v i n g f o r Z w e h a v e : \\ Z (z) = z + \sqrt{z^{2} - c_{2}} \\ t h e n : \\ ϕ (z) = Z (z) + \frac{d_{2}}{Z (z)} \end{array}$ (1.13)

where d₂is currently unknown.

We want to find the value of d₂ that makes the imaginary part of $ϕ$ =0

$Im [ϕ (z_{b})] = Im [Z (z_{b}) + \frac{d_{2}}{Z (z_{b})}] = 0$

(1.14)

$Im [ϕ (z_{b})] = Im [z_{b} + \sqrt{z_{b}^{2} - c_{2}} + \frac{d_{2}}{z_{b} + \sqrt{z_{b}^{2} - c_{2}}}] = 0$

(1.15)

The simplest way to make sense of this is the use equation (1.10) in equation (1.13) thus

$\begin{array}{l} z_{b}^{2} - c_{2} = a^{2} \cos^{2} θ + 2 i a b \sin θ \cos θ - b^{2} \sin^{2} θ - a^{2} + b^{2} = \\ b^{2} \cos^{2} θ - a^{2} \sin^{2} θ + 2 i a b \sin θ \cos θ = (^{b \cos θ + i a \sin θ) 2} \end{array}$

(1.16)

Thus:

$\begin{array}{l} z_{b} + \sqrt{z_{b}^{2} - c_{2}} = a \cos θ + i b \sin θ + b \cos θ + i a \sin θ = \\ (a + b) \cos θ + i (a + b) \sin θ \end{array}$

(1.17)

$\begin{array}{l} \frac{z_{b} + \sqrt{z_{b}^{2} - c_{2}}}{2} + \frac{d_{2}}{\frac{z_{b} + \sqrt{z_{b}^{2} - c_{2}}}{2}} = (a + b) (c o s θ + i s i n θ) + \frac{d_{2} (c o s θ - i s i n θ)}{(a + b)} \end{array}$

(1.18)

$Im (\frac{z_{b} + \sqrt{z_{b}^{2} - c_{2}}}{2} + \frac{d_{2}}{\frac{z_{b} + \sqrt{z_{b}^{2} - c_{2}}}{2}}) = (a + b) (s i n θ) - \frac{d_{2} (s i n θ)}{(a + b)} = 0$

(1.19)

This implies that

$d_{2} = {(a + b)}^{2}$

(1.20)

Then the velocity potential is proportional to the following complex function:

$ϕ (z) = \frac{z + \sqrt{z_{}^{2} - c_{2}}}{2} + \frac{{(a + b)}^{2}}{\frac{z_{} + \sqrt{z_{}^{2} - c_{2}}}{2}}$

(1.21)

and the velocity is proportional to the gradient of $ϕ$ (z).