4
$\begingroup$

If I have a straight line in 3D space and I use a camera matrix to do a perspective projection into a 2D plane/screen will the resulting line always be straight too? If I project the start point and end point and join the dots is that equivalent to having projected the line?

I suspect this is true, but have no basis for this assumption.

Is this still true for off-centre perspective projection or a cylindrical projection? I assume this isn't true for real cameras as they have all kinds of non-ideal optical effects right?

$\endgroup$
2
$\begingroup$

"A not so simple approach". I may have messed a little bit too much with grouping the terms, do forgive my elementary math skills, it's a side effect of using tools like wolfram and mathematica too much.

Without loss of generality we can assume that the near plane is given as $z=1$ (that is having normal $(0,0,1)$ and a point on it $(0,0,1)$) and the camera center (the reflected pinhole camera aperture point) is at $(0,0,0)$. Let $p,q \in \mathbb{R}^3$ be two points on a line (non-degenerate, meaning $p\ne q$, otherwise it's just a point and the projection is also a point - unless it's at $(0,0,0)$ then it's undefined). If the line defined by $p,q$ passes through $(0,0,0)$ the projection is a point and the requirement is trivially satisfied. Now let this not be the case. A perspective projection onto the plane through $(0,0,0)$ is given by $p' = \frac{p}{p_z}$, and $q' = \frac{q}{q_z}$. The line defined by $p,q$ is given as $r(\lambda) = p + \lambda(q-p), \lambda \in \mathbb{R}$. Its projection is given by $r'(\lambda) = \frac{r(\lambda)}{r_z(\lambda)}$. Since $z=1$ for all projected points, then we need only study the $x,y$ coordinates. We want to show that any projected point $r'(\lambda)$ can be written as an affine combination of $p',q'$ (this defines a line). Taking that into account: $$p'+\mu(q'-p') = \frac{r(\lambda)}{r_z(\lambda)}$$ $$(1-\mu)\frac{p}{p_z} + \mu\frac{q}{q_z} = \frac{p + \lambda(q-p)}{p_z + \lambda(q_z-p_z)}$$ $$(1-\mu)q_z(p_z+\lambda(q_z-p_z))p + \mu p_z(p_z+\lambda(q_z-p_z))q = q_zp_z(p + \lambda(q-p))$$ $$q_zp_z(p - \lambda p) + q^2_z\lambda p - \mu q_z(p_z+\lambda(q_z-p_z))p + \mu p_z(p_z+\lambda(q_z-p_z))q - q_zp_z(p - \lambda p) - q_zp_z\lambda q = 0$$ $$(1-\mu)\lambda q^2_z p -\mu(1-\lambda)q_zp_z p + \mu(1-\lambda)p^2_zq - (1-\mu)\lambda p_zq_z q = 0$$ $$(1-\mu)\lambda q_z(q_zp - p_zq) - \mu(1-\lambda)p_z(q_zp-p_zq) = 0$$ $$((1-\mu)\lambda q_z - \mu(1-\lambda)p_z)(q_zp - p_zq) = 0$$

For this to be satisfied one of the terms must be $0$. The first case concerns the second term: $p' = q'$, which we already considered as the trivial case. Then we must show that we can always pick a $\mu$ such that $(1-\mu)\lambda q_z - \mu(1-\lambda)p_z = 0$. Solve for $\mu$ and you get: $\mu = \frac{\lambda q_z}{(1-\lambda)p_z + \lambda q_z}$. Thus I have not only found that it is a straight line, but also the relationship between the parameterizations.

In general a projection onto a different 2-manifold (not a plane) will not result in the lines remaining straight (think of a sphere or a cylinder).

$\endgroup$
  • $\begingroup$ Great thank you! $\endgroup$ – Owen Apr 10 at 12:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.