projective camera: back-projecting a point on the image plane into 3-space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:43:41Z http://mathoverflow.net/feeds/question/86730 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86730/projective-camera-back-projecting-a-point-on-the-image-plane-into-3-space projective camera: back-projecting a point on the image plane into 3-space tdomhan 2012-01-26T15:58:45Z 2012-02-14T20:22:12Z <p>suppose I got a projective camera model. for this model I would like to back-project a ray through a point in the image plane. I know that the equation for this is the following: $$y(\lambda) = P^+_0 x_0 + \lambda c_0$$ where $P^+_0$ denotes the pseudoinverse of the camera matrix. $x_0$ the point on the image plane and $c_0$ the center of the camera.</p> <p>Now I don't fully get this equation. I get that $P^+_0 x_0$ results in a point on the line we are looking for. Hence we have two points that we can use for constructing a line. However I don't get the parametrization using $\lambda$. Why is the equation not in the form like: $$y(\lambda) = (1-\lambda) a + \lambda b$$</p> <p>Any help in understanding the original equation of the resulting ray would be appreciated! :D</p> http://mathoverflow.net/questions/86730/projective-camera-back-projecting-a-point-on-the-image-plane-into-3-space/86751#86751 Answer by John S. Adair for projective camera: back-projecting a point on the image plane into 3-space John S. Adair 2012-01-26T19:38:11Z 2012-01-26T19:57:53Z <p>The equation is in your form, with <code>$a = P^+_0 x_0$</code> and <code>$b = P^+_0 x_0 + c_0$</code>.</p> <p>Since <code>$P c_0 = 0$</code>, <code>$P \lambda c_0 = \lambda P c_0 = 0$</code>. I <em>think</em> that adding a linear multiple of <code>$c_0$</code> corresponds to sliding along a line normal to the image plane (<em>i.e.</em> parallel to the principal axis), which doesn't change the projected point in the image plane.</p> <p>By the way, I found the original poster's specific equation on p.162 of Multiple View Geometry in Computer Vision by Richard Hartley and Andrew Zisserman via a web preview.</p> http://mathoverflow.net/questions/86730/projective-camera-back-projecting-a-point-on-the-image-plane-into-3-space/87170#87170 Answer by tdomhan for projective camera: back-projecting a point on the image plane into 3-space tdomhan 2012-01-31T20:20:46Z 2012-01-31T20:20:46Z <p>I hope it's ok if I provide an answer myself.</p> <p>Algebraic explanation: we're trying to solve the equation $$PX=x$$ This is a linear system which can be solved using the pseudo-inverse(<a href="http://en.wikipedia.org/wiki/Moore%25E2%2580%2593Penrose_pseudoinverse" rel="nofollow">see</a>): $$X(\lambda)=P^+x+(I-P^+P)\lambda$$ We now $PC=0$, hence $I-P^+P$ is exactly our $C$. </p>