Mathematical description of a Perspective Projection

A perspective transformation is found by prescribing a center of projection and a viewing plane. Let here assume that P(x,y,z) be any object point in 3-D and center of projection is at E(0,0,-d). The problem is to find out the image point coordinates P'(x',y',z') on the Z=0 plane as in Figure 18.

The parametric equation of a l EP, beginning from E and passing via P is:

E+ t(P-E)  0

=(0,0,-d)+t[(x,y,z)-(0,0,-d)]

=(0,0,-d)+t(x,y,z+d)

=[t.x, t.y, -d+t.(z+d)]

Point P' is acquired, when t=t*

There is, P'=(x',y',z') =[t*.x, t*.y, -d+t*.(z+d)]

Because P' lies on Z=0 plane implies -d+t*.(z+d)=0 must be true, there is t*=d/(z+d) is actual.

Therefore x'=t*.x=x.d/(z+d)

                  y'=t*.y=y.d/(z+d)

                  z'=-d+t*(z+d)=0,

 

Hence P'=( x.d/(z+d), y.d/(z+d), 0)

                  =(x/((z/d)+1),y/((z/d)+1),0)

In terms of Homogeneous coordinate system here P'=(x,y,0,(z/d)+1).  ---------(5)

 

The equation 5 in above can be written in matrix form as:

-------(1)

There is, P'_h = P_h.P_per,z   ------    (2)

Here P_per,z in equation (4) represents the single point perspective transformation on z-axis.

The Ordinary coordinates are as:

[x',y',z',1]=[x/(r.z+1),y/(r.z+1),0,1]  where r=1/d                             ------ (3)

Posted Date: 4/4/2013 2:50:14 AM | Location : United States

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