First of all the result of Nusselt is proven using figure 1.3, which shows that the inner form factor integral can be calculated by a double projection of , first onto the unit hemisphere centered above , then to the base circle of this hemisphere in the plane of , and finally by calculating the ratio of the double projected area and the area of the unit circle ( ). By geometric arguments, or by the definition of solid angles, the projected area of a differential area on the surface of the hemisphere is . This area is orthographically projected onto the plane of , multiplying the area by factor . The ratio of the double projected area and the area of the base circle is:
Since the double projection is a one-to-one mapping, if surface is above the plane of , the portion, taking the whole surface into account, is:
This is exactly the formula of an inner form factor integral.
Now we turn to the problem of the hemispherical projection of a planar polygon. To simplify the problem, consider only one edge line of the polygon first, and two vertices, and , on it (figure 1.4). The hemispherical projection of this line is a half great circle. Since the radius of this great circle is 1, the area of the sector formed by the projections of and and the center of the hemisphere is simply half the angle of and . Projecting this sector orthographically onto the equatorial plane, an ellipse sector is generated, having the area of the great circle sector multiplied by the cosine of the angle of the surface normal and the normal of the segment ( ).
The area of the doubly projected polygon can be obtained by adding and subtracting the areas of the ellipse sectors of the different edges, as is demonstrated in figure 1.4, depending on whether the projections of vectors and follow each other clockwise. This sign value can also be represented by a signed angle of the two vectors, expressing the area of the double projected polygon as a summation:
Having divided this by to calculate the ratio of the area of the double projected polygon and the area of the equatorial circle, equation 1.26 can be generated.
These methods have supposed that surface is above the plane of and is totally visible. Surfaces below the equatorial plane do not pose any problems, since we can get rid of them by the application of a clipping algorithm. Total visibility, that is when visibility term is everywhere 1, however, is only an extreme case in the possible arrangements. The other extreme case is when the visibility term is everywhere 0, and thus the form factor will obviously be zero.
When partial occlusion occurs, the computation can make use of these two extreme cases according to the following approaches: