Best fitting circle to points in 3D












1












$begingroup$


I have a set of n ≥ 3 points in 3D that are measurements of a possible circle. The measured points are "noisy" so best-fitting algorithms are involved. I'm programming in C# and have put together some algorithms to do my procedure to find the best fitting circle to the points.



My procedure is the following:




  1. Find the centroid C (x0, y0, z0) of the points.

  2. Find the best fitting plane for the points using SVD. I now have a normalvector N [a, b, c] and a point C, fully describing the plane P

  3. Orthogonal project all points to the plane

  4. Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.

  5. Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.

  6. With the points now described in a 2D coordinate system, i can use Levenberg-Marquardt to find the best fitting cirlce. This gives me a circle center A (a, b) and a radius R

  7. The found circle center is described in the 2D coordinate system of the plane P


Now; my question:




How do I translate the found circle center A to the original 3D coordinate system?




I've tried to do A + C but this produces the following, which places the circle center in the centroid of the points:



Circle center is placed at the centroid of the points










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  • 1




    $begingroup$
    I agree with the process that you use. It is mainely the same as in paper : fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D , pages 28-32 , applied to ellipse, parabola and hyperbola in 3D. Circle is a particular case. The main difference is that the final fitting is not carried out with a Levenberg-Marquardt algorithm. It uses a direct method (no itteration, no guessed initial values). In case of circle, the direct fitting is explained pages 11-13 in the paper : fr.scribd.com/doc/14819165/…
    $endgroup$
    – JJacquelin
    Feb 23 '16 at 22:09
















1












$begingroup$


I have a set of n ≥ 3 points in 3D that are measurements of a possible circle. The measured points are "noisy" so best-fitting algorithms are involved. I'm programming in C# and have put together some algorithms to do my procedure to find the best fitting circle to the points.



My procedure is the following:




  1. Find the centroid C (x0, y0, z0) of the points.

  2. Find the best fitting plane for the points using SVD. I now have a normalvector N [a, b, c] and a point C, fully describing the plane P

  3. Orthogonal project all points to the plane

  4. Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.

  5. Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.

  6. With the points now described in a 2D coordinate system, i can use Levenberg-Marquardt to find the best fitting cirlce. This gives me a circle center A (a, b) and a radius R

  7. The found circle center is described in the 2D coordinate system of the plane P


Now; my question:




How do I translate the found circle center A to the original 3D coordinate system?




I've tried to do A + C but this produces the following, which places the circle center in the centroid of the points:



Circle center is placed at the centroid of the points










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I agree with the process that you use. It is mainely the same as in paper : fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D , pages 28-32 , applied to ellipse, parabola and hyperbola in 3D. Circle is a particular case. The main difference is that the final fitting is not carried out with a Levenberg-Marquardt algorithm. It uses a direct method (no itteration, no guessed initial values). In case of circle, the direct fitting is explained pages 11-13 in the paper : fr.scribd.com/doc/14819165/…
    $endgroup$
    – JJacquelin
    Feb 23 '16 at 22:09














1












1








1


1



$begingroup$


I have a set of n ≥ 3 points in 3D that are measurements of a possible circle. The measured points are "noisy" so best-fitting algorithms are involved. I'm programming in C# and have put together some algorithms to do my procedure to find the best fitting circle to the points.



My procedure is the following:




  1. Find the centroid C (x0, y0, z0) of the points.

  2. Find the best fitting plane for the points using SVD. I now have a normalvector N [a, b, c] and a point C, fully describing the plane P

  3. Orthogonal project all points to the plane

  4. Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.

  5. Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.

  6. With the points now described in a 2D coordinate system, i can use Levenberg-Marquardt to find the best fitting cirlce. This gives me a circle center A (a, b) and a radius R

  7. The found circle center is described in the 2D coordinate system of the plane P


Now; my question:




How do I translate the found circle center A to the original 3D coordinate system?




I've tried to do A + C but this produces the following, which places the circle center in the centroid of the points:



Circle center is placed at the centroid of the points










share|cite|improve this question











$endgroup$




I have a set of n ≥ 3 points in 3D that are measurements of a possible circle. The measured points are "noisy" so best-fitting algorithms are involved. I'm programming in C# and have put together some algorithms to do my procedure to find the best fitting circle to the points.



My procedure is the following:




  1. Find the centroid C (x0, y0, z0) of the points.

  2. Find the best fitting plane for the points using SVD. I now have a normalvector N [a, b, c] and a point C, fully describing the plane P

  3. Orthogonal project all points to the plane

  4. Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.

  5. Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.

  6. With the points now described in a 2D coordinate system, i can use Levenberg-Marquardt to find the best fitting cirlce. This gives me a circle center A (a, b) and a radius R

  7. The found circle center is described in the 2D coordinate system of the plane P


Now; my question:




How do I translate the found circle center A to the original 3D coordinate system?




I've tried to do A + C but this produces the following, which places the circle center in the centroid of the points:



Circle center is placed at the centroid of the points







circle 3d svd least-squares






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share|cite|improve this question













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edited Apr 13 '17 at 12:20









Community

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asked Nov 28 '14 at 12:36









borgerodsjoborgerodsjo

64




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  • 1




    $begingroup$
    I agree with the process that you use. It is mainely the same as in paper : fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D , pages 28-32 , applied to ellipse, parabola and hyperbola in 3D. Circle is a particular case. The main difference is that the final fitting is not carried out with a Levenberg-Marquardt algorithm. It uses a direct method (no itteration, no guessed initial values). In case of circle, the direct fitting is explained pages 11-13 in the paper : fr.scribd.com/doc/14819165/…
    $endgroup$
    – JJacquelin
    Feb 23 '16 at 22:09














  • 1




    $begingroup$
    I agree with the process that you use. It is mainely the same as in paper : fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D , pages 28-32 , applied to ellipse, parabola and hyperbola in 3D. Circle is a particular case. The main difference is that the final fitting is not carried out with a Levenberg-Marquardt algorithm. It uses a direct method (no itteration, no guessed initial values). In case of circle, the direct fitting is explained pages 11-13 in the paper : fr.scribd.com/doc/14819165/…
    $endgroup$
    – JJacquelin
    Feb 23 '16 at 22:09








1




1




$begingroup$
I agree with the process that you use. It is mainely the same as in paper : fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D , pages 28-32 , applied to ellipse, parabola and hyperbola in 3D. Circle is a particular case. The main difference is that the final fitting is not carried out with a Levenberg-Marquardt algorithm. It uses a direct method (no itteration, no guessed initial values). In case of circle, the direct fitting is explained pages 11-13 in the paper : fr.scribd.com/doc/14819165/…
$endgroup$
– JJacquelin
Feb 23 '16 at 22:09




$begingroup$
I agree with the process that you use. It is mainely the same as in paper : fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D , pages 28-32 , applied to ellipse, parabola and hyperbola in 3D. Circle is a particular case. The main difference is that the final fitting is not carried out with a Levenberg-Marquardt algorithm. It uses a direct method (no itteration, no guessed initial values). In case of circle, the direct fitting is explained pages 11-13 in the paper : fr.scribd.com/doc/14819165/…
$endgroup$
– JJacquelin
Feb 23 '16 at 22:09










1 Answer
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Found the answer my self.



Using notation from this question: 2D Coordinates of Projection of 3D Vector onto 2D Plane



Find the matrix M that converts coordinates from the X, Y, Z-system to the coordinate system of the plane: N, e'1, e'2. This is a 3 x 3 matrix with e'1 as row 1, e'2 as row 2 and N as row 3. By taking the inverse (transpose) of this matrix you can convert coordinates the other way around.






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    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Feb 17 '15 at 14:27











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Found the answer my self.



Using notation from this question: 2D Coordinates of Projection of 3D Vector onto 2D Plane



Find the matrix M that converts coordinates from the X, Y, Z-system to the coordinate system of the plane: N, e'1, e'2. This is a 3 x 3 matrix with e'1 as row 1, e'2 as row 2 and N as row 3. By taking the inverse (transpose) of this matrix you can convert coordinates the other way around.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Feb 17 '15 at 14:27
















0












$begingroup$

Found the answer my self.



Using notation from this question: 2D Coordinates of Projection of 3D Vector onto 2D Plane



Find the matrix M that converts coordinates from the X, Y, Z-system to the coordinate system of the plane: N, e'1, e'2. This is a 3 x 3 matrix with e'1 as row 1, e'2 as row 2 and N as row 3. By taking the inverse (transpose) of this matrix you can convert coordinates the other way around.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Feb 17 '15 at 14:27














0












0








0





$begingroup$

Found the answer my self.



Using notation from this question: 2D Coordinates of Projection of 3D Vector onto 2D Plane



Find the matrix M that converts coordinates from the X, Y, Z-system to the coordinate system of the plane: N, e'1, e'2. This is a 3 x 3 matrix with e'1 as row 1, e'2 as row 2 and N as row 3. By taking the inverse (transpose) of this matrix you can convert coordinates the other way around.






share|cite|improve this answer











$endgroup$



Found the answer my self.



Using notation from this question: 2D Coordinates of Projection of 3D Vector onto 2D Plane



Find the matrix M that converts coordinates from the X, Y, Z-system to the coordinate system of the plane: N, e'1, e'2. This is a 3 x 3 matrix with e'1 as row 1, e'2 as row 2 and N as row 3. By taking the inverse (transpose) of this matrix you can convert coordinates the other way around.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 13 '17 at 12:19









Community

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answered Feb 17 '15 at 14:08









borgerodsjoborgerodsjo

64




64












  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Feb 17 '15 at 14:27


















  • $begingroup$
    Here's a MathJax tutorial :)
    $endgroup$
    – Shaun
    Feb 17 '15 at 14:27
















$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Feb 17 '15 at 14:27




$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Feb 17 '15 at 14:27


















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