How can I fit an ellipse in 3D space to find it's center coordinates in (x,y,z)?

Input data and ellipse graph in attachments

Answer 1

[Matt J]

Basically, the procedure would be to

1. Fit a plane to your data.
2. Project your (x,y,z) data on to the plane and write the results in a 2D coordinate system (xp,yp) within that plane.
3. Fit an ellipse to the (xp,yp) data.

You can do so using the tools in this File Exchange package,

https://www.mathworks.com/matlabcentral/fileexchange/87584-object-oriented-tools-for-fitting-conics-and-quadrics

load XYZ;

XYZ=[x,y,z].';

pfit=planarFit(XYZ);

    %%% Project onto plane
    B=null(pfit.normal).';
    b0=(pfit.normal*pfit.distance).';

    xpyp=B*(XYZ-b0); %2xN matrix of (xp,yp) coordinates


efit=ellipticalFit(xpyp);

center3D=B.'*efit.center(:)+b0;  %3D center coordinates

xcenter=center3D(1),
ycenter=center3D(2),
zcenter=center3D(3),

This results in,

xcenter =

   3.2233e+04


ycenter =

 -188.3309


zcenter =

  573.7545

[Matt J]

However, the plane fit reveals that your data is essentially concentrated in a plane very nearly parallel to the y-z plane,

pfit = 

  planarFit with properties:

      normal: [0.9999 -0.0043 0.0152]
    distance: 3.2239e+04

So, a simpler approach might have been to just ignore the x component of your data, treating the data as 2D from the very beginning. When I do so, I find that your (y,z) data is in fact fit very well by a circle,

>> cfit=circularFit([y,z].')

cfit = 

  circularFit with properties:

    center: [-184.9425 570.7484]
    radius: 3.6945e+04

    plot(cfit)
    xlabel Y , ylabel Z

[Evan Hemingway]

Hi Matt,

Thanks for this contribution. It has helped me greatly. Is it possible to use your tools to fit an ellipse of known shape to 3D data points? For example, I have a recording of some markers tracking the wrist, where I have 8 surface markers surrounding the joint. The problem is that in some frames the markers drop off and I don't have full data. So, I find a frame where all 8 markers are visible and determine a and b of the ellipse using the work you've demonstrated in this post. For the rest of the frames, as long as I have 3 or more markers, I would like to fit an ellipse now of known size, where a and b were determined from the calibration frame. I noticed you have a groundtruth method but I don't think that's exactly applicable.

[Image Analyst]

@Evan Hemingway you need at least 5 points to find an ellipse. Three won't do it. See attached article.

[Evan Hemingway]

@Image Analyst Hi Image Analyst. Thanks for the interesting article. The way I read it, you need 5 points to determine where the ellipse is, and its a/b parameters. In my case, I already have the a/b parameters, so my ellipse is known shape, and I am basically trying to find the rigid body transformation to fit the ellipse at each frame. Furthermore, I can actually identify each marker with a coordinate on the ellipse, and that gives me additonal information. However, I am using markers attached to skin, so they move around relative to one another, hence why I need a fitting method.

I've thought a little about the problem. I can use Matt's planar fit method given at least 3 points to determine the plane of best fit of the ellipse. The problem then boils down to as follows:

Given 3 or more points in a plane, and an ellipse of known shape (a,b), determine (x,y) and theta, the center of the ellipse and its angle, resulting in minimal distance error between the analytical shape and the numerical data.

Edit: I think I have a path forward. I need to use nonlinear least squares. I need to minimize

,

where are my data and are my parameters.

[Matt J]

You might be able to use this to calculate the orientation change:

https://www.mathworks.com/matlabcentral/fileexchange/26186-absolute-orientation-horn-s-method

[Matt J]

I've added some methods to the planarFit class to help with the steps in my original answer. The code can now be shortened to,

load XYZ;

XYZ=[x,y,z].';

pfit=planarFit(XYZ);

xpyp=pfit.project2D(XYZ); %2xN matrix of (xp,yp) coordinates

efit=ellipticalFit(xpyp); %elliptical fit

center3D=pfit.unproject3D(efit.center(:));  %map center back to 3D

Answer 2

[Image Analyst]

I have a demo that you could probably adapt. It's attached.