1
$\begingroup$

I have a set of $N$ points in form of their $xyz$ coordinates. I am not using all of them due to high computation time. I am looking to find best plane fit to given set of points.

I used this as reference: Hugh transform on page 3, there's algorithm number 2 that interests me

I am calculating their Hough space coordinates using this formula:

$$p = x_n \cos(\theta)\sin(\phi) + y_n\sin(\phi)\sin (\theta) + z_n\cos(\phi),$$

wher $\phi \in [-90°, 90°]$, $\theta \in [0°, 360°]$.

This is where I am probably wrong:

So for each point I am calculating $180 × 360$ possible combinations of degrees in sin and cos. I defined a matrix that has $360$ rows and $180$ columns for each combination of $\phi$ and $\theta$. I increment cells in this matrix if threshold $p$ is less than $0.03$.

If this is incorrect please tell me what I am doing wrong there.

So in the end I have matrix of votes and highest numbers in matrix represents combination of $\phi$ and $\theta$ that are best solution for algorithm.

But what I need is classic plane formula:

$$Ax + By + Cz + D = 0.$$

How do I get that equation from $\phi$ and $\theta$ that I get from algorithm?

$\endgroup$
  • $\begingroup$ But why? Complexity is abysmal. Why not just solve for plane's eq? $\endgroup$ – mathreadler Apr 22 '18 at 21:10
  • $\begingroup$ I need to implement 3 different algorithms for plane fit. I have already used least squares and RANSAC. My final choice was Hough $\endgroup$ – Macedon Apr 22 '18 at 22:04
2
$\begingroup$

You need three parameters to specify a surface in 3D: two angles $\phi$ and $\theta$, and a radius $r$.

That means you need a 3-dimensional matrix, with values for $\phi$, $\theta$ as you've done, and values for $r \in [0, R]$, where $R$ is the maximal radius possible for your data.

After running the algorithm, you will get values $\phi^\ast$, $\theta^\ast$, and $r^\ast$ which correspond to the higher number of votes in your matrix.

Now you know the points on the optimal plane satisfy $r^\ast = x \cos(\theta^\ast)\sin(\phi^\ast) + y\sin(\theta^\ast)\sin(\phi^\ast) + z\cos(\phi^\ast)$.

In other words, if we set $A = \cos(\theta^\ast)\sin(\phi^\ast)$, $B = \sin(\theta^\ast)\sin(\phi^\ast)$, $C = \cos(\phi^\ast)$, and $D = -r^\ast$, your plane is parametrized by $Ax+By+Cz+D=0$.

$\endgroup$

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.