Advanced Computer Graphics
• Exercise 3

Due: March 27

• We are given the equation $L = \frac{d\Phi}{cos(\theta) ~ dA d\Omega}$ where L is the radiance eminating from the source. This equation can be arranged such that $d\Phi = L ~ cos(\theta) ~ dA d\Omega$ and can then be integrated such that $\Phi = \bigg\int L ~ cos(\theta) ~ dA d\Omega$. The only thing left to do is integrate. Since L is constant it can drop out completely and since we are given that the area of interest is A, the dA also drops leaving $\Phi = L ~ A \bigg\int cos(\theta) ~ d\Omega$. $d\Omega$ is the differential solid angle, which in spherical coordinates translates to $sin(\theta) ~ d\theta d\phi$. Plugging all of this in we are left with $\Phi = L ~ A \bigg\int_0^{2\pi}\bigg\int_0^{\pi/2} cos(\theta) ~ sin(\theta) ~ d\theta d\phi = 2\pi ~ L ~ A \bigg\int_0^{\pi/2} cos(\theta) ~ sin(\theta) ~ d\theta = \pi ~ L ~ A ~ sin^2(\theta) \bigg\lbrack_0^{\pi/2} = \mathbf{\pi ~ L ~ A}$.
  

2. A point light source having an intensity of 1 watt per steradian (emitting uniformly in all directions) is 1 meter away from the center of a square patch of surface, of edge length 1 meter. Find the irradiance at the center of the square, and at the corner.

• For both cases we can use the simple equation derived in class $E = \frac{I(\hat\mathbf{n} \cdot \hat\mathbf{l})}{r^2}$. For the case of the center of the square the dot product is simply unity and since the light also happens to be of unit length away from the center $E = I$.
• For the second one we need only abit of trigonometry. Any corner of the square will work so the general case will be solved. First, notice that at any of the corners, the distance to the center of the square is $\frac{\sqrt{2}}{2}$. This means that if we look perpendicular to a diagonal, we will see the center and a corner separated by that distance. What we want, is the angle in this view that that normal at the corner makes with the vector pointing from the corner to the light:
  
  
  
  
  
• From the diagonal length we calculated before, we use trigonometry to find the angle between the two to be $\frac{\sqrt6}{2}$. We could have also used vectors and arrived at the same answer (I did to check). Now we only need to distance from the corner to the center which is easily extracted via Pythagoras to be $\frac{\sqrt{6}}{2}$ again. Plugging this into the equation, we find $E = \frac{\sqrt{6}}{3}I$.
  

3. Following are three different versions of the rendering equation. For each one, provide a picture and short phrase or equation describing each term (e.g., I(x' -> x'') is the intensity of light traveling from point x' to point x''). Then, show that the equations are the same by deriving two of them from one of the others.

• I will start from the second one and derive the others from it. First Let me enumerate all of the terms of all three equations.
• 1st Equation: The equation reads: "The radiance traveling from point x' to point x'' is given by the radiance eminating from x' towards x'' plus the sum of the product of the BRDF, the radiance, the visibility, and the geometry from every point in the scene to x' (and then to x'' for the BRDF)".
• 2nd Equation: The equation reads: "The intensity seen at x'' coming from x' is given by the intensity emitted from x' towards x'' plus the geometry between x' and x'' times the sum of the product of the BRDF, the two point transport intensity, and the visibility from every point in the scene to x' (and then to x'' for the BRDF)".
• 3rd Equation: The equation reads: "The outgoing radiance from a point x' in the direction of w' is given by the emitted radiance from x' in the direction w' plus the sum over the hemisphere centered at x' of the BRDF from an incoming direction of w to an outgoing direction of w' at point x times the radiance arriving at point x' coming from the direction w times the dot product between the normal at x' and the incoming direction w for every incoming direction w".
• I will now derive the other two equations from the second one. To go from the second to the first, notice that $I(x\tox') = L(x,\omega)G(x,x')$. Now the only thing left to do is to convert the intensities into radiances. We can use the fact that radiance is intensity times the solid angle. If we make these substitutions, we find the extra $G(x',x'')$ term disappears due to the change. We have therefore turned equation 2 into equation 1. Now we can go to equation 3 pretty easily. First we can change to the solid angle notation by noting that any two points in space are actually connected by a solid angle. Therefore we can easily change from two points to a point and a direction. This takes care of the left-hand side, the emitted radiance, the BRDF, and the incoming radiance. Now, we can convert the dA, the V(x,x'), and the G(x,x') to dw and (w•n) because $d\omega = \frac{dA}{r^2}$ and $(\vec\omega \cdot \vecn) = cos(\theta)$. The other $cos(\theta)$ comes from the transition from $L(x\tox')$ to $L_i(x',\vec\omega)$.
• The pictures of the equations are pictured next in order:



• Notice that the first two pictures are identical in the abstract sense because the only thing that is changing really is the measured quantity.

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  © 2008 Sean M. Arietta
  University of Virginia