How to demonstrate that an elliptical circulation distribution can be induced by an elliptical chord distribution?

Question

I intend to demonstrate that a wing with an elliptical chord distribution, without torsion ($\alpha(z)=\alpha=\text{const}$) and constant section ($\beta(z)=\beta=\text{const}$, and $a_{2\text{D}}(z)=a_{2\text{D}}=\text{const}$), can induce an elliptical circulation distribution.

Well, for an elliptical chord distribution, the chord is defined by:

$$c(z)=c_C\sqrt{1-\left(\frac{z}{s}\right)^2}$$

where $c_C$ is a constant.

On the other hand, the circulation distribution is defined by:

$$\Gamma(z)=\frac{1}{2}a_{2\text{D}}V_\infty c(z)\left[\alpha+\beta+\alpha_i(z)\right]$$

So, by substituting $c(z)$ on $\Gamma(z)$, we get:

$$\Gamma(z)=\frac{1}{2}a_{2\text{D}}V_\infty \left[c_C\sqrt{1-\left(\frac{z}{s}\right)^2}\right]\left[\alpha+\beta+\alpha_i(z)\right]$$

which can be written as: $$\Gamma(z)=\Gamma_C(z)\sqrt{1-\left(\frac{z}{s}\right)^2}$$

where $\Gamma_C(z)=\frac{1}{2}a_{2\text{D}}V_\infty c_C\left[\alpha+\beta+\alpha_i(z)\right]$

So, to demonstrate that an elliptical chord distribution can induce an elliptical circulation distribution, we would also need to demonstrate that $\alpha_i(z)=\alpha_i=\text{const}$, to make $\Gamma_C$ a constant.

I don't know how to do this last step. How would you do it?

Note that:

$s:=$ semispan of the wing,

$z:=$ transversal coordinate of the wing, it goes from $-s$ to $s$,

$\alpha:=$ geometric angle of attack,

$\alpha_i:=$ induced angle of attack,

$\beta:=$ symmetric of the null lift (geometric) angle of attack,

$a_{2\text{D}}:=$ airfoil lift coefficient slope,

$V_\infty:=$ freestream airspeed.

Answer 1

First, let's transform the coordinates to polar coordinates for simplicity by introducing a variable $\theta$ such that:

$$z=-s\cos{\theta}$$

In this coordinate, the chord distribution is then:

$$c(\theta)=c_C\sin{\theta}$$

Let's write the unknown lift distribution of the bound vortex as a Fourier sine series, again in the $\theta$ coordinate:

$$\Gamma(\theta) = \sum_{n=1}^{\infty}A_n\sin{n\theta}$$

Through Biot-Savart, the induced AOA can be computed from the trailing vortices, such that:

$$\alpha_i(z) = \frac{1}{4\pi V_\infty}\int_{-s}^{s}{\frac{(d\Gamma/d\zeta)}{z-\zeta}d\zeta} = \frac{1}{4\pi V_\infty}\int_{0}^{\pi}{\frac{\sum_{n=1}^{\infty}{nA_n\cos{n\theta_0}}}{\cos{\theta_0}-\cos{\theta}}d\theta_0} = \frac{1}{4V_\infty}\sum_{n=1}^{\infty}{nA_n\frac{\sin{n\theta}}{\sin{\theta}}}$$

Now collecting everything together and put into your original equation, we have:

$$\Gamma(\theta) = \frac{1}{2}a_{2D}V_{\infty}c_C\sin{\theta} \left[ \alpha - \beta - \frac{1}{4V_\infty}\sum_{n=1}^{\infty}{nA_n\frac{\sin{n\theta}}{\sin{\theta}}} \right] = \sum_{n=1}^{\infty}A_n\sin{n\theta}$$

The key, at this point, is to realize that each $\frac{\sin{n\theta}}{\sin{\theta}}$ is linearly independent to the others, so in order for the other terms to satisfy the above relation for all $\theta$, they must be simultaneously zero. That is,

$$A_n = \begin{cases} A_1 & n=1\\ 0 & n>1\\ \end{cases}$$

So now the circulation and chord distributions have the same form, which are elliptical.