Calculation of solid angle for rectangle in 6DOF [closed]

Question

I am an undergrad trying to understand and use solid angle calculations:

I have a point source in R3 space (x_source, y_source, z_source) and a rectangle with given center (x_center, y_center, z_center), orientation (euler angles ZYX) and dimensions (width w, height h).

My goal is to calculate the solid angle under which the point source is visible. So far I dealt only with a case where the rectangle is facing directly towards the source by solving the integral $$ \int_{0}^{h} \int_{0}^{w} \frac{1}{(x-x_{source})^2+(y-y_{source})^2+(z-z_{source})^2} \,dx dy $$

Is this the right solution? And how to extend this to an arbitrary orientation of the rectangle?

Thanks!

Answer 1

You can do this without integration, by performing the following steps.

1. Calculate the coordinates of the vertices of your rectangle. Let the vertices be $v_1,v_2,v_3,v_4$.

2. Let your point "source" be $O$. Normalize the vectors $Ov_j$ by dividing each vector on its length. This gives you 4 points on the unit sphere centered at $O$.

3. Connect those 4 points by great circles in the correct order. You obtain a spherical quadrilateral. Find its interior angles, by breaking the quadrilateral into two triangles and applying the spherical rule of cosines. (They are the same as dihedral angles of the cone built on $Ov_j$. Cosines of the sides and diagonals are simply the dot products of corresponding normalized vectors.)

4. Your solid angle is the area of this spherical quadrilateral, and once you know the interior angles it is equal to the sum of these angles minus $2\pi$.