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APPENDIX A Optical Surface Area The trigonometric formula used to produce Figures 3 and 4 in chapter 7 is not exactly complicated though it is not particularly trivial either. The optical surface area of a given bit of terrain surface is the area of the solid angle determined by the projection of that terrain surface onto a sphere centered at a given observation point—the fraction of total ambient light converging onto the observation point from the given terrain surface. Measured in steradians, a solid angle is a three-dimensional analogue of a two-dimensional angle measured in radians. A full circle subtends an angle of 2π radians, while a sphere altogether subtends a solid angle of 4π steradians. One steradian is subtended by any surface on a sphere whose area is equal to the square of the sphere’s radius (whereas one radian is subtended by any arc on a circle whose length is equal to the circle’s radius). The solid angle subtended by an arbitrary surface D from a given observation point P is defined as the surface area ΩD covered by the projection of D onto a unit sphere centered at P. This can be written as ΩD =  D  n · da r2 where da is the differential area of a patch of surface D,  n is a unit vector from the origin of the sphere toward that patch, and r is the distance from the origin of the sphere to that patch. In spherical coordinates, this becomes ΩD =  D sin ϕ dϕ dθ with coaltitude (polar angle) ϕ and longitude (azimuth) θ. In Cartesian coordinates , we have ΩD =  D cos ϕ dx dy (x2 + y2 + z2) =  D z dx dy (x2 + y2 + z2)3/2 where cos ϕ = z/r and r2 = x2 +y2 +z2 . These surface integrals can be difficult to use, but there are simpler ways to calculate solid angles in the present case. 164 APPENDIX A: OPTICAL SURFACE AREA / 165 First, with respect to spheres other than the unit sphere, the solid angle ΩD subtended by an arbitrary surface D from a given observation point P is defined as the ratio of the surface area covered by the projection of D onto the sphere (with radius h centered at P) to the surface area of the sphere altogether (the latter being equal to 4πh2 )—just as the angle determined by an arc on a circle with radius h is the ratio of the arc length to the total length (circumference) of the circle (2πh). Second, the surface area of a spherical triangle can be calculated using Girard ’s Theorem—named after Albert Girard (1595–1632), a Flemish mathematician who first published it in 1626 though a statement and proof of it was discovered and recorded in 1603 in the unpublished notebooks of the English mathematician Thomas Harriot (c1560–1621). This theorem states how the area of a spherical triangle depends on its interior angles. Namely, the surface area of a spherical triangle ABC with respective interior angles α, β, γ on the surface of a sphere with radius h is given as follows: Area(ABC) = h2 (α + β + γ − π) This simple formula states that the surface area of the triangle is equal to the square of the sphere’s radius times the amount that the sum of the interior angles α, β, γ exceeds π. The formula in Girard’s Theorem is thus sometimes referred to as Girard’s Spherical Excess Formula. It follows that the associated solid angle subtended by ABC on the concentric unit sphere is: ΩABC = Area(ABC) 4πh2 = 1 4π (α + β + γ − π) In particular, the solid angle Ω0,n,0,m subtended by a rectangle with one corner at the origin, with side lengths n and m, and lying in a terrain-surface plane at a depth h from the center P of a sphere of radius h is composed of two spherical triangles sharing the diagonal from 0, 0 to n, m as a common side (Figure 5). We can thus calculate Ω0,n,0,m by calculating and adding the respective solid angles subtended by these two triangles. We will show how this is done for one of the triangles, namely, for the one with vertices 0, 0, n, 0, and n, m. The other solid angle is calculated similarly. To facilitate this calculation, we can employ the law of sines for spherical triangles, namely, sin a sin α = sin b sin β = sin c sin γ where angles α, β, γ and opposite arcs a, b, c are as depicted in Figure 6. In this case...


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