Solid angle of a pixel

In the following the question of how much solid angle a pixel representing a projection of a 3D scene covers will be presented. Applications for this include computing the total solid angle an object in a scene covers, for example, for lighting calculations. While what follows can be extended to any projection space, for simplicity here we will consider 90 degree by 90 degree perspective projections such as generated from cube maps.

Consider one face of the cube centered at the origin as shown below. Each pixel of which may be in one or another state and the aim is to find the solid angle of all pixels in a particular state. This involves counting up the solid angle of each pixel in that state. If the aim is to find the solid angle overall then this can be applied to each of the 6 cube faces.

One pixel on this face of the cube is shown highlighted above, the corners labeled as shown. Consider vectors from the origin to c₀ ... c₃ and normalise, these are now points on a unit sphere. Splitting the 4 vertex pixel rectangle into two halves gives two spherical triangles. The shortest curve between two of these vertices are sections of a great circle (they lie on the plane that also contains the origin). The solid angle of such a triangle is simply

Where the angles Ø₀ + Ø₁ + Ø₂ are the angles of the spherical triangle at each vertex as shown below. The tangents at each c_i lie in the plane to the neighbouring point.

Computing the tangents at each c_i is achieved using the "standard" double cross product trick. For example to find the tangent vector from c₀ to c₁, calculate the vector c₁ - c₀, take the cross product of this with the normal vector at c₀ which is also c₀ (sphere is at the origin). The result is a vector perpendicular to the normal at c₀. The cross product of this vector with the normal at c₀ is the desired tangent vector.

XYZ CalcTangent(XYZ p0,XYZ p1)
{
   XYZ p,r,t;

   p = VectorSub(p1,p0);
   r = CrossProduct(p0,p);
   t = CrossProduct(r,p0);
   Normalise(&t);

   return(t);
}

There are a few simple test cases for a coded implementation. If the cube face is entirely contained in the object whose solid angle is to be measured then the solid angle should be 4π/6.

Consider a disk of radius R one unit from the origin. The solid angle should be 2π(1-cos(θ)) where θ is atan(R).

For a bounded longitude and latitude rectangle the solid angle is given by (sin(φ₂) - sin(φ₁)) (θ₂ - θ₁) Where φ is latitude and θ longitude.