Best fitting circle to points in 3D
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:
- Find the centroid C (x0, y0, z0) of the points.
- 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
- Orthogonal project all points to the plane
- Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.
- Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.
- 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
- 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 3d svd least-squares
edited Apr 13 '17 at 12:20
Community♦
1
asked Nov 28 '14 at 12:36
borgerodsjo
64
add a comment |
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:
- Find the centroid C (x0, y0, z0) of the points.
- 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
- Orthogonal project all points to the plane
- Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.
- Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.
- 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
- 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 3d svd least-squares
edited Apr 13 '17 at 12:20
Community♦
1
asked Nov 28 '14 at 12:36
borgerodsjo
64
1
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/…
– JJacquelin
Feb 23 '16 at 22:09
add a comment |
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:
- Find the centroid C (x0, y0, z0) of the points.
- 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
- Orthogonal project all points to the plane
- Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.
- Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.
- 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
- 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 3d svd least-squares
edited Apr 13 '17 at 12:20
Community♦
1
asked Nov 28 '14 at 12:36
borgerodsjo
64
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:
- Find the centroid C (x0, y0, z0) of the points.
- 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
- Orthogonal project all points to the plane
- Subtract C from all points, such that C becomes the new origin O for the points. I now have a plane that crosses origin.
- Translate the projected points into a 2D coordinate system that lies in the plane, using this algorithm.
- 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
- 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 3d svd least-squares
circle 3d svd least-squares
edited Apr 13 '17 at 12:20
Community♦
1
asked Nov 28 '14 at 12:36
borgerodsjo
64
edited Apr 13 '17 at 12:20
Community♦
1
asked Nov 28 '14 at 12:36
borgerodsjo
64
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Nov 28 '14 at 12:36
borgerodsjo
64
asked Nov 28 '14 at 12:36
borgerodsjo
64
asked Nov 28 '14 at 12:36
borgerodsjo
64
64
1
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/…
– JJacquelin
Feb 23 '16 at 22:09
add a comment |
1
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/…
– JJacquelin
Feb 23 '16 at 22:09
1
1
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/…
– JJacquelin
Feb 23 '16 at 22:09
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/…
– JJacquelin
Feb 23 '16 at 22:09
add a comment |
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.
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 17 '15 at 14:08
borgerodsjo
64
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
add a comment |
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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.
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 17 '15 at 14:08
borgerodsjo
64
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
add a comment |
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.
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 17 '15 at 14:08
borgerodsjo
64
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
add a comment |
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.
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 17 '15 at 14:08
borgerodsjo
64
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.
edited Apr 13 '17 at 12:19
Community♦
1
answered Feb 17 '15 at 14:08
borgerodsjo
64
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
answered Feb 17 '15 at 14:08
borgerodsjo
64
answered Feb 17 '15 at 14:08
borgerodsjo
64
answered Feb 17 '15 at 14:08
borgerodsjo
64
64
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
add a comment |
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
Here's a MathJax tutorial :)
– Shaun
Feb 17 '15 at 14:27
add a comment |
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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/…
– JJacquelin
Feb 23 '16 at 22:09