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Tuesday, 30 April 2013


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A perspective transformation (also called an imaging transformation) projects 3D points onto a
plane. Perspective transformations play a central role in image processing because they provide
an approximation to the manner in which an image is formed by viewing a 3D world. These
transformations are fundamentally different, because they are nonlinear in that they involve
division by coordinate values.

Figure 10 shows a model of the image formation process. The camera coordinate system (x, y, z)
has the image plane coincident with the xy plane and the optical axis (established by the center
of the lens) along the z axis. Thus the center of the image plane is at the origin, and the centre of
the lens is at coordinates (0.0, λ). If the camera is in focus for distant objects, λ is the focal length
of the lens. Here the assumption is that the camera coordinate system is aligned with the world
coordinate system (X, Y, Z).

Let (X,Y,Z) be the world co-ordinates of any point in a 3-D scene, as shown in the Fig 10. We assume throughout the following discussion that Z>λ ; that is all points of interest lie in front of the lens.The first step is to obtain a relationship that gives the coordinates (x,y) of the projection of the point (X,Y,Z) onto the image plane.This is easily accomplished by the use of similar triangles. With reference to Fig 10,

Fig 10 Basic model of the imaging process The camera coordinate system (x, y, z) is aligned with the world coordinate system (X, Y, Z)


 Where the negative signs in front of X and Y indicate that image points are actually inverted, as the geometry of Fig 10 shows.
The image-plane coordinates of the projected 3-D point follow directly from above equations
These equations are nonlinear because they involve division by the variable Z. Although we could use them directly as shown, it is often convenient to express them in linear matrix form, for rotation, translation and scaling. This is easily accomplished by dividing the first three homogeneous coordinates by the fourth. A point in the cartesian world coordinate system may be expressed in vector form as
and its homogeneous counterpart is
If we define the perspective transformation matrix as
The product PWh yields a vector denoted Ch
                                                                             Ch=PWh


The element of ch is the camera coordinates in homogeneous form. As indicated, these
coordinates can be converted to Cartesian form by dividing each of the first three components of
ch by the fourth. Thus the Cartesian of any point in the camera coordinate system are given in
vector form by

 The first two components of c are the (x, y) coordinates in the image plane of a projected 3-D
point (X, Y, Z). The third component is of no interest in terms of the model in Fig. 10. As shown
next, this component acts as a free variable in the inverse perspective transformation

 The inverse perspective transformation maps an image point back into 3-D.
 wh=P-1Ch
 Where P-1 is  

 Suppose that an image point has coordinates (xo, yo, 0), where the 0 in the z location simply
indicates that the image plane is located at z = 0. This point may be expressed in homogeneous
vector form as
 or, in Cartesian coordinates

 This result obviously is unexpected because it gives Z = 0 for any 3-D point. The problem here is
caused by mapping a 3-D scene onto the image plane, which is a many-to-one transformation.
The image point (x0, y0) corresponds to the set of collinear 3-D points that lie on the line passing
through (xo, yo, 0) and (0, 0, λ). The equation of this line in the world coordinate system; that is,

Equations above show that unless something is known about the 3-D point that generated an
image point (for example, its Z coordinate) it is not possible to completely recover the 3-D point
from its image. This observation, which certainly is not unexpected, can be used to formulate the
inverse perspective transformation by using the z component of ch as a free variable instead of 0.
Thus, by letting
It thus follows


 which upon conversion to Cartesian coordinate gives
 In other words, treating z as a free variable yields the equations

 Solving for z in terms of Z in the last equation and substituting in the first two expressions yields

 which agrees with the observation that revering a 3-D point from its image by means of the
inverse perspective transformation requires knowledge of at least one of the world coordinates of
the point.


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