In a field of 3x3 solar dishes that can track the position of the sun, how to calculate the area of shadow cast by one dish on its adjacent? Consider 9 dishes placed in a matrix form 3x3. The center to center spacing between them is x in east-west direction and y in north-south direction. Area of each dish is A $m^2$. For a given time, we know the angles the sun is going to make, namely elevation angle and azimuth angle. As the dish will be normal to the sun's direction, we know the orientation of the dish. If the shadow is being cast, out of total area of 9A, I'd like to know how much area is under the shadow?
Please refer these images for reference: 

9 Dishes, sun's POV.

9 Dishes from another POV, to see the shadow area. Blue is under sunlight and Pink is under shadow. 6.11% is under shadow according to the software. Hence 0.0611*9A = 0.55A is under shadow.

$y$ spacing and Gamma Angle.

$x$ spacing and Alpha Angle.
$x$, $y$, and angles of the sun are known. The area under shadow is to be found.
Please refer the images for a better understanding of the problem.
I have a software (Creo) that tells me how much the shadow in percentage is (the pictures are taken in Creo). But I have to find that percentage for $3000$ similar orientations of sun at different values of $x$ and $y$. Hence I want a mathematical expression of shadow in terms of $x$, $y$ and the angles for my Solar Field Design.
Not sure if this is possible. If it is, I'm not sure how to go on about finding it. Any idea how I could find an expression for this? Also, this is a simple case if the dishes are circular. What will change if the dishes are square, hexagon, etc?
 A: Hints:
If you look at the dishes from the sun, what you see are ellipses. The coordinates of the centers can be found by means of the projection matrix defined by the azimuth and elevation. In that projection, the axis of the ellipses are horizontal and vertical and their lengths are drawn from the projections of the corresponding radii.
Then you need to intersect the ellipses in pairs, solving the system of quadratic equations
$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1,\\
\frac{(x-x_1)^2}{a^2}+\frac{(y-y_1)^2}{b^2}=1.$$
Fortunately, as the axis are the same, by subtracting the equations you get a linear one, so that the problem remains quadratic. You will find zero or two intersections.
If you project them back to the 3D dishes, the shadow area is twice the area of the circular segment defined by the two points.
The whole computation is quite tractable, but you need some familiarity with 3D geometry.
A: The interesting things happen in a "virtual" plane orthogonal to the light rays. This plane has a horizontal $u$-axis and a second $v$-axis orthogonal to the $u$-axis. The direction of the $u$-axis with respect to the already present $x$- and $y$-axes is determined by the azimuth angle $\gamma$, and the angle between the $v$-axis and the vertical $z$-axis is the elevation angle $\alpha$.
Project the lattice of the dishes centers along the light rays orthogonally to this $(u,v)$-plane. You will then obtain a parallelogram lattice $\Lambda$  in this plane. The given angles and other data define the $(u,v)$-measurements of these parallelograms in a simple way (this can be done through vector algebra). At the lattice points of $\Lambda$ draw circular discs of the original radius. Then you can see how much these discs overlap. Hopefully the situation is not too complicated, so that any disc overlaps at most one other disc.
You shall need to build up the following auxiliary function: If the centers of two unit discs have a distance $h\geq0$ then the area overlap is $f(h)$ unit discs. One has $f(0)=1$ and $f(h)=0$ when $h>2$.
