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\newcommand{\lecturetitle}[4]{
  \noindent
  CS 280, Computer Vision (Prof. Jitendra Malik) \hfill #3\\
  Lecture \##1 (#2) \hfill {\bf DRAFT} Notes by #4\\[-.05in]
  \mbox{}\hrulefill \mbox{}\\}

\begin{document}

% MODIFY THE FOLLOWING TO INCLUDE THE LECTURE #, TOPIC, DATE, and YOUR NAME
\lecturetitle{15}{Occlusion and Radiometry/Photometry}{March 15, 1999}{Jeng Lung}

% INCLUDE AN OUTLINE OF THE CLASS TOPICS
\begin{itemize} 
  \item Occlusion
  \subitem Stereo
  \subitem Motion

  \item Radiometry and Photometry
  \subitem Events
  \subitem Irradiance
  \subitem Radiance
  \subitem Reflectance BRDF
  
\end{itemize}

% THE NOTES WILL GO HERE

\section{Occlusion}

In the previous lecture, we talked about cues to identify occlusion.  The cues include
texture, junction, stereo, and motion.  Today we will look at identifying occlusion 
with stereo and motion.

\subsection{Stereo}

>From the last lecture, we saw that some points does not have correspondence between
the left eye's image plane and the right eye's image plane.  We called this
{\bf half occlusion} and it is best demonstrated with the DaVinci Stereopsis model.

In the case where the length of the wall in front is greater than the interocular distance
between the left and right eyes, objects behind the wall can not been seen as shown in Figure 1.

\begin{figure}[htb]
\centerline{\psfig{file=fig1.eps,height=2in}}
\caption{Length of the wall is longer than interocular distance}
\end{figure}

In the case where the length of the wall in front is less than the interocular distance,
however, objects behind the wall can be seen as shown in Figure 2.

\begin{figure}[htb]
\centerline{\psfig{file=fig2.eps,height=2in}}
\caption{Length of the wall is shorter than interocular distance}
\end{figure}

\subsection{Motion}

When you have images from time t and time t+1, there are points that are visible at time t
but disappear at time t+1, known as {\bf accretion/deletion of texture}.  There are also
points that are not visible at time t but visible at time t+1, known as
{\bf occlusion/disocclusion}.  These regions occurs when you have occluding surfaces along
with motion as shown in Figure 3.

\begin{figure}[htb]
\centerline{\psfig{file=fig3.eps,height=2in}}
\caption{Occlusion from motion}
\end{figure}

\section{Radiometry and Photometry}

So far we have been studying the statics and kinematics of projection where we were
concerned about coordinates and directions.  We asked question like, "where are the points
imaged?"  Now with radiometry and photometry we ask questions such as, "what is the
brightness and intensity of points imaged?"  Radiometry deals with light of all wavelengths
while photometry deals only with visible lights.

\subsection{Events}

Events in photometry include {\bf self shadow/attached shadow} and {\bf cast shadow}, as
illustrated in Figure 4.  Note that the shadow is not completely dark since light bounces
and causes {\bf mutual illumination}.  For example, the sun is a primary light source and
all secondary light sources are surfaces on which the sun's ray bounces off of.  Also, the
more alignment between the light ray and the surface normal, the brighter the patch.

\begin{figure}[htb]
\centerline{\psfig{file=fig4.eps,height=2in}}
\caption{self shadow/attached shadow and cast shadow}
\end{figure}

\subsection{Irradiance}

We measure the irradiance using pixel values on the image.  The pixel values represent the
energy received in that area in a given amount of time.  These values are proportional to
power per unit area (W/$m^2$).  The image brightness can be thought of as a intensity 
function at all the points on the image plane I(x, y).

\subsection{Radiance}

Unlike irradiance, radiance is a directional concept and it is measured as power per unit
area per unit solid angle.  In a 2-D circle, the solid angle $\theta_s$ is merely the arclength over the radius s/r as shown in Figure 5.  For a 2-D square, the solid angle would be
$\theta_s$ = scos$\theta$/r.  In 2-D, the maximum value that the solid angle can be is 2$\pi$.

\begin{figure}[htb]
\centerline{\psfig{file=fig5.eps,height=2in}}
\caption{Solid angle of a circle}
\end{figure}

For a 3-D sphere, the solid angle subtended at the center of sphere is equal to 
Acos$\theta$/$r^2$, where A is the area.  In 3-D, the maximum value that the 
solid angle can be is 4$\pi$.  In the example shown in Figure 6, the area of the 
element is $r^2$sin($\theta$)d$\theta$d$\phi$ so the solid angle would be
sin($\theta$)d$\theta$d$\phi$.

\begin{figure}[htb]
\centerline{\psfig{file=fig6.eps,height=2in}}
\caption{Solid angle of a sphere}
\end{figure}

The solid angle subtended by the northern hemisphere in this case would be
\[\int_0^{\pi/2}\int_0^{2\pi}sin\theta\,d\theta\,d\phi = 2\pi\int_0^{\pi/2}sin\theta\,d\theta = 2\pi\]

\begin{figure}[htb]
\centerline{\psfig{file=fig7.eps,height=2in}}
\caption{Power transfer between patch dx and patch dy}
\end{figure}

In Figure 7, the power transferred from x to y would be:
\[L(x, \theta_x, \phi_x)\,dxcos\theta_x(\theta_s)\]
\[=L(x, \theta_x, \phi_x)\,dx\,dy\frac{cos\theta_xcos\theta_y}{r^2}\]
where $\theta_x$ is the angle between the normal of dx and the ray and $\theta_y$ is
the angle between the normal of dy and the ray.  $\theta_s$ is the solid angle subtended
by dy at x.

The power received by y from x would be:
\[L(y, \theta_y, \phi_y)\,dycos\theta_y(\theta_s')\]
\[=L(x, \theta_y, \phi_y)\,dx\,dy\frac{cos\theta_xcos\theta_y}{r^2}\]
where $\theta_s'$ is the solid angle subtended by x as seen from y.

Hence, L(x, $\theta_x$, $\phi_x$) = L(y, $\theta_y$, $\phi_y$) and this equation states
that the radiance is conserved along a ray (but this is only true in vacuum where there is
no absorption).

Inverse Square Law results from the fact that the radiance remains the same with distance
along the ray but the solid angle changes by a factor of 1/$r^2$.

\subsection{Reflectance BRDF}

To be continued in the next lecture ...

\end{document}

