Pin-hole camera revisited

There are three coordinate systems involved - camera, image and the world:

1. Camera: perspective projection.
   
   
   $$\left[ \begin{array}{c} x_c \\ y_c \\ f \end{array}\right] = k \left[ \begin{array}{c} X_c \\ Y_c \\ Z_c \end{array}\right]$$
   where $k = f/Z_c$ . This can be written as
   $$\left[ \begin{array}{c}x_c \\ y_c \\ f \end{array} \right] = \lef... ...ray} \right] \left[ \begin{array}{c}X_c \\ Y_c \\ Z_c \\ 1 \end{array} \right]$$

2. Image: (intrinsic/internal camera parameters)
   
   
   $$\begin{array}{ccc} k_u x_c & = & u - u_0 \\ k_v y_c & = & v_0 - v \\ \end{array}$$
   where the unit of $k$ 's are pixel/length. This can be expressed as
   $$\left[ \begin{array}{c}u \\ v \\ 1 \end{array} \right] = \left[ \... ...} \right] = {\bf C} \left[ \begin{array}{c}x_c \\ y_c \\ f \end{array} \right]$$
   ${\bf C}$ is called the camera calibration matrix and it provides the transformation between an image point and a ray in Euclidean 3-space.
3. World: (extrinsic/external camera parameters)
   
   

   The Euclidean transformation between the camera and world coordinates is:

   $${\bf X_c} = {\bf R}{\bf X_w} + {\bf T}$$
   and is expressed projectively as:
   $$\left[ \begin{array}{c}X_c \\ Y_c \\ Z_c \\ 1 \end{array} \right]... ...rray}\right] \left[ \begin{array}{c}X_w \\ Y_w \\ Z_w \\ 1 \end{array} \right]$$

   Finally, concatenating the three matrices, we have

   $$\left[ \begin{array}{c}u \\ v \\ 1 \end{array} \right] = {\bf C}\... ...f T} \right] \left[ \begin{array}{c}X_w \\ Y_w \\ Z_w \\ 1 \end{array} \right]$$
   which defines the $3 \times 4$ projection from Euclidean 3-space to an image:
   $${\bf x} = {\bf P_E}{\bf X} \;\;\;\;\;\; {\bf P_E} = {\bf C}\left[ {\bf R}\mid {\bf T} \right]$$