Finding $y_{A,i}$ and $y_{B,i}$ in this geometric relationship problem.

Question

Finding $y_{A,i}$ and $y_{B,i}$ in this geometric relationship problem.

I'm an high-speed aerodynamics student. I am studying a sweptback wing like in the figure below (in green). Notice that I divided the right half-wing in two regions I and II (yellow and orange respectively):

Notice that this wing has an infinite wingspan. Notice also these things:

1. The wing is symmetrical with respect the $y$ axis.

2. $x_{\beta 1}$ is a line that makes an angle of $\arctan{\left(\beta\right)}$ with the $y$ axis. Therefore its equation is: $x_{\beta 1}=\beta |y|$.

3. $x_{\beta 2}$ is parallel to $x_{\beta 1}$ and therefore its equation is given by: $x_{\beta 1}=c_r + \beta |y|$

4. The leading edge ($x_a$) makes an angle of $\arctan{\left(\lambda\right)}$ with the $y$ axis. Therefore its equation is: $x_{a}=\lambda |y|$.

5. The trailing edge ($x_s$) is parallel to the leading edge. Its equation is given by: $x_s(y)=c_r+\lambda |y|$

Now let's consider a generic point $P(x,y)$ inside the region II (in orange), like I show in the image below:

The red dashed line that passes through $y_A$ in the leading edge is parallel to $x_{\beta 1}$, whereas the red, dashed line that passes through the point $y_B$ is the same as the previous one but mirrored with respect the y axis.

I know that (by doing some geometrical calculations):

$$y_A = \dfrac{\beta y - x}{\lambda + \beta}$$ $$y_B = \dfrac{\beta y + x}{\lambda + \beta}$$

However, the harder I try to solve for $y_{A,i}$ and $y_{B,i}$ the more difficulties I find. So the problem is, how do I find those coordinates (notice that I'm interested in the y-coordinate only; not x): $$y_{A,i}=?$$ $$y_{B,i}=?$$

Maybe you can give me some hints because I feel I can't do it on my own. Any ideas?

NOTE: I know the answers are: $$y_{A,i}=y_{A}+\dfrac{c_r}{\lambda+\beta}$$ $$y_{B,i}=y_{A}\left(1-\dfrac{2\lambda}{\lambda-\beta}\right)-\dfrac{c_r}{\lambda+\beta}$$

But I'm more interested in the procedure I should follow.

Answer 1

You're essentially just intersecting lines. My preferred means to deal with this is projective geometry and homogeneous coordinates. Then all of this boils down to cross products. So here is a crash course.

1. Represent a point $(x,y)$ in the plane by a vector $(x,y,1)\in\mathbb R^3$. or any multiple thereof. So $(2x,2y,2)$ is the same point, just written down differently.
2. Represent a line $ax+by+c=0$ as a vector $(a,b,c)\in\mathbb R^3$. Multiples of this vector again represent the same line.
3. The line connecting two points $P$ and $Q$ is $P\times Q$. The point where lines $g$ and $h$ intersect is $g\times h$. Easy to write, and not really hard do compute either.
4. A point lies on a line if the dot product between their vectors is zero. You probably don't need this, but it's useful to know.
5. To turn homogeneous coordinates of a point back into “normal” coordinates, you divide by the last coordinate. So a homogeneous coordinate vector $(x,y,z)$ represents the point $(x/z,y/z)$ in the plane. This is called dehomogenization.
6. Points with a zero in their last coordinate can't be dehomogenized. These lie at infinity, in a given direction. That's where parallel lines meet. All points at infinity together form the line at infinity, with coordinates $l_\infty=(0,0,1)$ or any multiple thereof.

To apply this to your case, the part of $x_{\beta1}$ with $y\ge0$ has equation $x=\beta y$ or $x-\beta y+0=0$ so it has coordinates $(1,-\beta,0)$. It intersects the line at infinity at $F=(\beta,1,0)$ as you can compute using the cross product. So to draw a parallel to that line through $P$, just compute the cross product $P\times F$. To reflect points or lines about an axis, just flip the corresponding sign. You can compute the outlines of your wing in a similar fashion, then compute the points of intersection using cross products and read the desired coordinates off the dehomogenized end result.