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

Question

Alex Baranov 2021-6-5

2
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/848700-how-can-i-fit-an-ellipse-in-3d-space-to-find-it-s-center-coordinates-in-x-y-z

编辑： Matt J 2022-10-25

采纳的回答： Matt J

Input data and ellipse graph in attachments

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Answer 1

Matt J 2021-6-5

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/848700-how-can-i-fit-an-ellipse-in-3d-space-to-find-it-s-center-coordinates-in-x-y-z#answer_717620

编辑：Matt J 2021-6-5

在 MATLAB Online 中打开

Basically, the procedure would be to

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

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Evan Hemingway 2022-6-21

编辑：Evan Hemingway 2022-6-21

@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 2022-6-22

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 2022-10-25

编辑：Matt J 2022-10-25

在 MATLAB Online 中打开

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