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)