Shadow of an ellipsoid at various angles

Question

I have an ellipsoid which represents an elliptical wing. It has a chord ($2\times r_A$), a semi-span ($r_C$), and a maximum thickness ($r_B$). Here, $r$ is the radius, and $A$, $B$, and $C$ are the coordinate system based at the base of the ellipsoid.

A global coordinate system is defined $x$, $y$, and $z$. $C$ is always $z$, but $A$ & $B$ are rotations of $x$ and $y$ by some angle $\alpha$. In other words, the test domain is fixed by $x$, $y$ and $z$, but the wing is at some angle-of-attack $\alpha$ to the flow (coming in the positive $x$ direction).

What is the projection (shadow) of the ellipsoid on the $yz$ plane for any $\alpha$?

I know that for $\alpha=0$, the area will be $\pi \times r_C \times r_B$, and should increase as the angle increases as long as $r_A$ is larger than $r_B$. For $\alpha=\pi/2$, the area should be $\pi \times r_C \times r_A$.

Given that $r_C$ will always appear in the equation, I feel the problem may just be working out the projection of the ellipse that is formed at the base of the wing (with radii $r_A$ and $r_B$) on to the $y$ axis.

If this is correct, then the problem is to find the ? in the schematic below.

Answer 1

Your hypothesis is right: the projection on the $yz$ plane is an ellipse and its area is $\pi r_C h$, where $h$ is the distance marked "?" in your diagram.

It is easier to work with an ellipse having $x$ and $y$ axes as axes of symmetry, in which case your $x$ axis is line $OH$ in the diagram below and distance $h$ is $DH$ (I'm supposing $0\le\alpha\le\pi/2$).

The ellipse has an equation $x^2/r_A^2+y^2/r_B^2=1$ and at $D$ you must have $y'=-(r_B^2/r_A^2)(x/y)=-\tan\alpha$. From these equations you can find the coordinates of $D$: $$ x={r_A^2\sin\alpha\over\sqrt{r_A^2\sin^2\alpha+r_B^2\cos^2\alpha}}, \quad y={r_B^2\cos\alpha\over\sqrt{r_A^2\sin^2\alpha+r_B^2\cos^2\alpha}}. $$ You then easily get: $$ DH=DF\cos\alpha=(DX+XF)\cos\alpha=y\cos\alpha+x\sin\alpha= \sqrt{r_A^2\sin^2\alpha+r_B^2\cos^2\alpha}. $$