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I am trying to understand the formalism of the projective transform of 2D image. it has 9 parameters (a-i) which the 9th is redundant since we use houmogenous coordinates. this transformation preserves only straight lines.

$\pmatrix{x'\\ y' \\ w'}$ = $\pmatrix{ a& b& c\\\ d& e& f \\\ g& h& i}$ $\pmatrix{x\\ y \\ w}$

I have three questions according this formalism:

  1. Since the redundancy (from 8 to 9) of degrees of freedom is $i=1$ ?
  2. what is the $w$ and $w'$ ? are they parameters? how do i calculate them?
  3. How can I find the parameters so I can transform the image to its original(from the black background image to the square one).

what I want to do is build the matrix to transform this image:

input image

and find matrix $M$

so I get back this image:

enter image description here

All the images above taken from here

What I want finally do is to find the

M = [a  b c; 
     d  e f;
     g  h i];

t_proj = maketform('projective',T);   
I_projective = imtransform(I,t_proj,'FillValues',.3);
imshow(I_projective)
title('unprojective - rectangular carpet')

taken from here

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You have to understand that projective transformation is not linear. It's $(x,y) -> (X(x, y)/Z(x,y) , Y(x, y)/Z(x,y))$, where X, Y, Z - linear functions.

$w$ and $w'$ are essential here - they help represent this non-linear transformation as linear - matrix multiplication - and that way this linear representation is a homomorphism (preserve "chaining" of operations).$w$ for input image is usually taken as $1$; You get $w'$ from matrix multiplication ($w' = gx + hy + iw$ ) and the end result of transform is $(x'/w', y'/w')$

Generally speaking you can't just assign $i=1$, but due to homogeneity you can divide all the matrix by $i$ and get new matrix, representing the same transform with right-left element $1$

I would advise you to read some textbook on projective geometry for computer vision to better understand this subject. Those questions have some deep implications.

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  • 3
    $\begingroup$ good reference: Multiple View Geometry $\endgroup$ – Maurits Feb 22 '12 at 10:33
  • $\begingroup$ It would be nice you someone can add more details and answers all the questions above thanks. $\endgroup$ – 0x90 Mar 6 '12 at 11:19

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