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I have edited the original post and then the question look like as follows;

I have been given set of points having x,y,z coordinates and they already grouped into different segments. The point relevant to each segment is not very well fitting with planes but were located very closer to the planes. I want to find the intersection lines of some of the plane pairs as each and every plane pairs is not intersecting. some plane pairs are parallel to each other.

So, if I consider only 2 planes (plane1, plane2) which are intersected each other as an example case;

first, I did Least Square plane fitting to get the plane parameters of each individual plane.

After that, I computed the intersection line using computed plane parameters. output of my intersection line contains following informations, (distance of plane1 from origin, distance of plane2 from origin,0), and (x component of cross product of plane normal vectors, y component of cross product of plane normal vectors, z component of cross product of plane normal vectors)

this means, now i know the equation of the intersection line. this is endless line. First of all i want to know can i calculate variance-covariance matrix for that line? if so, how can i do it? from that endless line, if need a specific part of the line and if i know the both ends (i can do that by projecting most far end points, located on both planes, to that line). so, what i finally looking for is the high quality line segment. this should contain 2 points coordinates and variance-covariance matrix of the line. I am not sure how the effect of variance and covariance should come to the intersection line. But i guess, it would be something like; x,y,z (for variance) and xy,yz & zx (for covariance).

If I say something about my least square process; i followed orthogonal distance regression plane fitting method. so it was considered as an eigenvector problem. During this process, Initially algorithm takes all points relevant to a one segment, then calculates the centroid, then creates a matrix. It looks like as follows

matrix

Then, this matrix is solved by determining the eigen values and eigen vectors. From there, select the smallest eigen value out of 3 values and obtain the respective eigen vector as plane parameters.

Please, simply explain me about this as I am not an expert in this area.

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"Variance-covariance" applies to fitting functions to data, among other things; if you're finding the line common to two planes, there is nothing to fit. –  J. M. Aug 10 '11 at 8:11
    
Perhaps your planes have random distributions and you're trying to calculate properties of the distribution of the line of intersection from (properties of) the distributions of the planes? You're going to have to provide more context about what you're doing to get a helpful answer. –  joriki Aug 10 '11 at 10:37
    
@g_niro: a) All this should have been in the question when you posed it. b) You should edit it into the question now (and perhaps delete those comments) -- the question should be self-contained and people shouldn't have to read the comments to understand it. c) Some of this is not quite clear to me -- it would be preferable if you could write down the equations that you used to find the plane parameters and their (co)variances. –  joriki Aug 10 '11 at 10:50
    
@ all >> ok, i have edited my original post and deleted the comments added by me. –  niro Aug 10 '11 at 13:26
    
This part isn't clear to me: "This lines is obtained by considering aditional plane going through the origin and perpendicular to the line direction." –  joriki Aug 10 '11 at 13:49

1 Answer 1

up vote 1 down vote accepted

I'm not an expert on this sort of thing; I'm just answering as best I can because you don't seem to be getting any answer otherwise.

I think it would probably be pretty complicated to get an exact covariance matrix for your line parameters. It should be possible to calculate the covariance matrix in linear approximation using propagation of uncertainty as described below. However, note that such linear estimates are usually biased. You can use this to get an idea of how uncertain your line parameters are, but keep in mind that it's only an approximation to the exact covariance matrix.

To avoid misunderstandings: I'll be talking a lot about the covariance matrix of the plane parameters. This is not the covariance matrix you display in the question, which is the covariance matrix of the point coordinates.

I don't know whether the code you're using already gives you a covariance matrix for the plane parameters. I'll say something below about how you might obtain one if it doesn't, but first let's assume you have two plane fits, given by parameter vectors $(d_i,n_{xi},n_{yi},n_{zi})$ with $i=1,2$, where $(n_{xi},n_{yi},n_{zi})$ is the normal for plane $i$ and $d_i$ is the distance of plane $i$ from the origin, together with associated $4\times4$ covariance matrices $M_i$ describing the uncertainty in these parameters. (Since the planes are derived from separate sets of data points, I'm assuming they can be treated as independent.)

Now your line parameters are $(d_1,d_2,(n_1\times n_2)_x,(n_1\times n_2)_y,(n_1\times n_2)_z)$. In a linear approximation, the errors due to errors in $n_1$ and $n_2$ simply add up,

$$(n_1+\Delta n_1)\times (n_2+\Delta n_2)\approx n_1\times n_2 \Delta n_1\times n_2+n_1\times\Delta n_2\;,$$

and the Jacobian matrix for the errors is given by

$$ \begin{pmatrix} \Delta d_1 \\ \Delta d_2 \\ \Delta (n_1\times n_2)_x \\ \Delta (n_1\times n_2)_y \\ \Delta (n_1\times n_2)_z \end{pmatrix} = \begin{pmatrix} 1 \\ &&&&1 \\ &0&n_{z2}&-n_{y2}&&0&-n_{z1}&n_{y1}\\ &-n_{z2}&0&n_{x2}&&n_{z1}&0&-n_{x1}\\ &n_{y2}&-n_{x2}&0&&-n_{y1}&n_{x1}&0\\ \end{pmatrix} \begin{pmatrix} \Delta d_1 \\ \Delta n_{x1} \\ \Delta n_{y1} \\ \Delta n_{z1} \\ \Delta d_2 \\ \Delta n_{x2} \\ \Delta n_{y2} \\ \Delta n_{z2} \end{pmatrix} \;. $$

Let's denote this by $\Delta l=J\Delta p$. Then you can compute the covariance matrix $M_l$ of the line parameters from those of the plane parameters using propagation of uncertainty:

$$M_l=J\begin{pmatrix}M_1\\&M_2\end{pmatrix}J^{\text T}\;.$$

This is relatively straightforward because the line parameters are given as explicit functions of the plane parameters. Approximating the $M_i$ if your code doesn't already provide them is a bit more complicated, since the plane parameters are only given implicitly as solutions of an eigenvalue problem.

The following is somewhat speculative; I haven't tried it out, but this is how I'd approach the problem if I had to solve it. I'll only sketch the solution since I don't know whether you'll be using it.

Determining the change in an eigenvector to linear order in a change in the matrix is a well-known problem in perturbation theory. I'll only treat the non-degenerate case where the lowest eigenvalue of the covariance matrix of the point coordinates is distinct; if it isn't, your plane parameters are useless anyway. I won't use indices $1,2$ for the planes in this part, since this calculation is completely separate for each plane. I'll denote the covariance matrix of the point coordinates by $P$.

You've obtained the eigenvector corresponding to the lowest eigenvalue for the eigenvalue problem

$$Pn=\lambda n\;.$$

To find the change $\Delta n$ in $n$ corresponding to a change $\Delta P$ in $P$, write

$$(P+\Delta P)(n+\Delta n)=(\lambda+\Delta\lambda)(n+\Delta n)\;.$$

To zeroth order, this just yields the eigenvalue problem again. To linear order, we get

$$P\Delta n+\Delta P n=\lambda\Delta n+\Delta\lambda n\;.\tag1$$

We can choose the eigenvectors to be orthonormal. Then multiplying $(1)$ by $n^{\text T}$ yields

$$n^{\text T}P\Delta n+n^{\text T}\Delta P n=\lambda n^{\text T}\Delta n+\Delta\lambda n^{\text T}n\;.$$

In the first term on the left-hand side, $n^{\text T}P=n^{\text T}\lambda$ (since $P$ is symmetric), so the first terms cancel. With $n^{\text T}n=1$, that leaves

$$\Delta\lambda=n^{\text T}\Delta P n\;.$$

The result for the change in the eigenvector is unfortunately slightly more complicated. We can multiply $(1)$ by the other two eigenvectors $e_1$ and $e_2$ to obtain

$$e_i^{\text T}P\Delta n+e_i^{\text T}\Delta P n=\lambda e_i^{\text T}\Delta n+\Delta\lambda e_i^{\text T}n\;.$$

The second term on the right-hand side vanishes because the eigenvectors are orthogonal. In the first term on the left-hand side, $e_i^{\text T}P=e_i^{\text T}\lambda_i$, where $\lambda_i$ is the eigenvalue corresponding to $e_i$. Thus we get

$$e_i^{\text T}\Delta n=\frac{e_i^{\text T}\Delta P n}{\lambda-\lambda_i}\;.$$

Since the eigenvectors are orthonormal, this is the coefficient of $e_i$ in an expansion of $\Delta n$ in the eigenvectors (the coefficient of $n$ being zero since the change in a unit vector is orthogonal to that vector). So we have

$$\Delta n=\sum_{i=1,2}\frac{e_i^{\text T}\Delta P n}{\lambda-\lambda_i}e_i\;.$$

Thus, if you solve the entire eigenvalue problem for the covariance matrix of the point coordinates (rather than just finding the eigenvector corresponding to the lower eigenvalue), you have all the ingredients to put together the Jacobian that propagates the uncertainty from the errors $\Delta P$ to the errors $\Delta n$.

The distance $d$ of the plane from the origin is just the scalar product of the centre of mass $c$ of the points with the normal vector $n$,

$$d=c^{\text T}n\;,$$

so the error for that is

$$\Delta d=\Delta c^{\text T}n+c^{\text T}\Delta n\;,$$

where $\Delta c$ is just the average of the errors in the point coordinate vectors. The errors $\Delta P$ are also readily obtained to linear order in terms of the errors in the point coordinates. That gives you all the Jacobians all the way from the point coordinates through $\Delta P$ and $\Delta c$ and then through $\Delta d$ and $\Delta n$, which together give $\Delta p$, and then finally to $\Delta l$; you can apply them one after the other to the covariance matrix of the point coordinates, and the end result will be a linear approximation to the covariance matrix of the line parameters.

Note that the result diverges as the lowest eigenvalue of $P$ becomes degenerate; that makes sense since in that case you don't really have any idea about the normal direction of the plane.

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