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In perspective projection, group of parallel lines have the same vanishing point. I am interesting about the reverse calculation: Getting the group of parallel lines equations that their vanishing point specific point.

Say I know that the camera is perspective camera at $(0,0,0)$ and it's direction $(0,0,1)$, the view plane is $z = 1$ and I am interesting about the lines in plane $y= y_0$ that their vanishing point is $P = (p_x,p_y,p_z)$.

I have tried to calculate the projection point of some point $(x,y_0,z)$ and get the equations:
(i) $p_x = x(\frac1z)$
(ii) $p_y = y_0(\frac1z)$
(iii) $p_z = 1$

But it seems wrong because if the vanishing point is like $(x_0,0,*)$ then form (ii) we will get $z\rightarrow \infty$ but then (i) is wrong because $x(\frac1z)\rightarrow0$ but it need to be equals to $x_0$.

So how can I get the group of parallel lines have the same vanishing point in these conditions?

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    $\begingroup$ The lines in plane $y = y_0$ are not all parallel. $\endgroup$ – ratchet freak May 8 '16 at 13:06
  • $\begingroup$ Of course, I need a group of parallel lines in plane $y=y_0$ that their vanishing point is some point, say $(x_0,0,1)$ (the camera position and direction and the view plane are defined in the question) $\endgroup$ – nrofis May 8 '16 at 13:07
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Here's a hint to get you started: Parallel lines include the line through the camera.

So really all you need is the direction from the camera to the vanishing point on the view plane. Then create lines parallel to that line.

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  • $\begingroup$ I have tough about that, so if my camera in $(0,0,0)$ and it's looking to $(0,0,1)$, the view plane is $z=1$ and lets say that the vanishing point is $(10,0,1)$. Then the line from the camera to the vanishing point is $(0,0,0)+t(10,0,1)$. So all the parallel lines that are going to that vanishing point will be $(a_1,a_2,a_3) + t(10,0,1)$? And then all the lines in plane $y=y_0$ that have that vanishing point will be $(a_1,y_0,a_3) + t(10,0,1)$? $\endgroup$ – nrofis May 8 '16 at 13:58
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    $\begingroup$ stop focusing on the $y = y_0$ plane, just having the vanishing point on the view plane and the camera position is enough. $\endgroup$ – ratchet freak May 8 '16 at 14:08
  • $\begingroup$ OK, so $(a_1,a_2,a_3)+t(10,0,1)$ are all the lines that their vanishing point is $(10,0,1)$? $\endgroup$ – nrofis May 8 '16 at 14:09

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