2
$\begingroup$

I am reading about homography in images and such. One thing pops up a lot:

$\mathbf{P} = [\mathbf{R}|\mathbf{t}]$

What does this mean?

Does this mean: If $\mathbf{R} = \begin{bmatrix}a & b\\\ c &d\end{bmatrix}$ and $ \mathbf{t} = \begin{bmatrix}x\\\ y\end{bmatrix}$, I get $ \mathbf{P} = \begin{bmatrix}a &b &x\\\ c& d& y\end{bmatrix}?$

$\endgroup$
5
  • $\begingroup$ Could you mention what book are you looking at? $\endgroup$ Jul 21, 2011 at 14:37
  • 3
    $\begingroup$ Yes, I think you got this right. Think of matrices $\mathbf{R}, \mathbf{S}$ as consisting of their columns. Then $[\mathbf{R}|\mathbf{S}]$ usually means the matrix consisting of the columns of $\mathbf{R}$ then $\mathbf{S}$ (assuming that $\mathbf{R},\mathbf{S}$ have the same number of rows. $\endgroup$
    – t.b.
    Jul 21, 2011 at 14:38
  • $\begingroup$ @J.M. I am looking at Hartley & Zisserman, Multiple View Geometry (2000/2003) $\endgroup$
    – Unapiedra
    Jul 21, 2011 at 14:53
  • 1
    $\begingroup$ Yes, if you read through the book carefully, you'll see that it's their notation for a "camera matrix". $\endgroup$ Jul 21, 2011 at 15:01
  • $\begingroup$ Well, I only have access to the chapter posted on Zisserman's website. $\endgroup$
    – Unapiedra
    Jul 22, 2011 at 8:46

2 Answers 2

2
$\begingroup$

It called as the augmented matrix. Quite useful while solving linear equations. Please see:

$\endgroup$
1
$\begingroup$

P denotes an augmented matrix (in this case a projection matrix) and your assumptions are correct about R and t.

$\endgroup$
2
  • $\begingroup$ I don't understand: projection matrix? $\endgroup$
    – t.b.
    Jul 21, 2011 at 14:41
  • 1
    $\begingroup$ Projection matrix is the matrix P such that: x = P X. Where X is your 3D point in homogeneous coordinates and x is your 2D point on the image plane. $\endgroup$
    – Unapiedra
    Jul 21, 2011 at 14:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .