CN112163309B - A fast method for extracting the spatial center of a single planar circle image - Google Patents

Abstract

The invention discloses a method for rapidly extracting the spatial circle center of a single plane circle image, which comprises a first step, a second step, a third step and a fourth step. The invention relates to the technical field of vision measurement systems, in particular to a method for quickly extracting the space circle center of a single plane circle image, which aims at solving the problem that the space coordinate extraction precision and efficiency of the current circle center target circle center are low.

Description

A fast method for extracting the spatial center of a single planar circle image

technical field

The invention relates to the technical field of visual measurement systems, in particular to a method for quickly extracting the spatial center of a single plane circle image.

Background technique

With the development of computer technology and surveying and mapping technology, optical three-dimensional measurement technology has been more and more widely used. In this process, the accurate and fast extraction of the spatial coordinates of the center of the circular target is an important aspect that affects the measurement accuracy and efficiency.

A variety of research data show that the extraction of space center coordinates mainly relies on the stereo vision method. First, the ellipse center is extracted, and then the binocular imaging principle is used to solve the space center coordinates. Because the perspective projection transformation process is often not the space center of the ellipse center Therefore, the accuracy of the reconstructed space center coordinates is not high, especially when facing large-scale center targets. Many scholars have proposed a variety of solutions to this problem, which are mainly divided into two categories: one is the distortion compensation method, that is, the circle center compensation is given according to the angle between the known marker point and the image plane and the distance between the marker point and the image plane. According to the mathematical model, the extracted ellipse center is compensated. Although this method has been verified in the simulation, it is difficult to obtain the attitude information of the marker point and the image plane in practical applications, so this method is used in practical applications. is greatly restricted in practical applications, and has little significance in practical applications; the second is to use nonlinear optimization methods to optimize poses, that is, to take two perspective images, solve the camera pose transformation according to epipolar matching, and then perform edge Contour reprojection, building a least squares model to minimize the edge reprojection error, this method needs to match the edge under two perspective images, and the effect of ellipse edge matching by epipolar line is not good; the literature proposes a method based on The dual quadratic curve method is used to solve the center of the circle, and the center projection vector between any two circle centers is solved according to the three coplanar marker points, and then the center projection is solved by vector cross multiplication. This method improves the accuracy of the center extraction to a certain extent. However, the marker points need to be coplanar, and at least three non-collinear marker points are required. It is difficult to ensure that the marker points are coplanar in actual measurement, and images from two perspectives are also required when extracting the three-dimensional coordinates of the center of the circle. Generally speaking, the current methods require multiple perspective pictures or multiple marker points to complete the extraction of the three-dimensional coordinates of the circular target center, and the accuracy is not reliable, especially in applications that rely on large circular target positioning, such as robot positioning, circular It is difficult to achieve good results in requirements such as pipeline measurement.

Contents of the invention

In order to solve the above-mentioned existing problems, the present invention provides a method to solve the problem of low accuracy and efficiency in extracting the spatial coordinates of the center of the circle target at present. The spatial conic equation is constructed by using the optical center of the camera and the imaging ellipse, and then the positive base of the cone is constructed. According to the positive The bottom surface solves the parallel plane equation of the space circle, and finally finds the correct space circle equation according to the parallel plane and the actual radius, which avoids the problem of solving the space center projection coordinates, and only needs a single image to solve the space center coordinates, stable and reliable A fast method for extracting the spatial center of a single planar circle image.

The technical scheme that the present invention takes is as follows:

A conic section is a curve obtained by the intersection of a plane and a quadratic cone surface. This curve has the characteristics of mirror symmetry. According to the conic section theory and the above imaging model, there are two conclusions as follows:

i. There is a unique acute angle θ, and the positive base of the cone rotates θ clockwise and counterclockwise around the major axis to obtain two circular sections;

ii. If the intersection of plane A and the cone is a circle, for any plane B parallel to plane A, if there is an intersection, then the intersection is a circle.

According to the above conclusions, the method for rapidly extracting the center of space of a single plane circle image of the present invention comprises the following steps:

Step 1: Construct the positive bottom of the cone;

Step 2: Rotate the plane where the bottom surface is located around the long axis, find the angle θ where the section is a circle, and record the plane as α at this time;

Step 3: Find the plane β parallel to α and the diameter of the cross-sectional circle is D;

Step 4: Solve the cross-sectional circle equation at this time, and solve the coordinates of the center of the circle.

Further, in said step one, constructing the conical bottom surface includes the following steps:

(1) Solve the elliptic cone equation in the coordinate system:

其中Y＝(y₀,y₁,y₂)，/>

为3×3对称矩阵，/>

为3×1向量，/>

为标量，取A＝[a_ij]，B＝[b_i]，c＝1，The ellipse cone equation in the image coordinate system PX _p Y _p Z _p is

where Y=(y ₀ ,y ₁ ,y ₂ ),/>

is a 3×3 symmetric matrix, />

is a 3×1 vector, />

is a scalar, take A=[a _ij ], B=[b _i ], c=1,

Since the ellipse is on the XOY plane, then y ₂ is 0,

At this time, if Q(Y)=0, the following equation can be obtained:

According to the standard equation (2) of the ellipse,

Assume that the major axis radius of the ellipse is d ₀ , and the minor axis radius is d ₁ ,

but

The gradient is 0 at the vertex of the ellipse cone, that is

在椭圆锥上，则有/>

联立式(1)、式(2)之可得and

On an elliptical cone, there is />

Simultaneous formula (1) and formula (2) can be obtained

求解如式(3)所示：at this time

The solution is shown in formula (3):

(2) Solve the ellipse cone equation in the camera coordinate system:

Assume that a set of orthonormal bases of the image coordinate system in the camera coordinate system is Re=[Ue Ve Ne], Re can also be regarded as the rotation transformation matrix between the two coordinate systems, assuming C _e is a point on the disk surface in the camera Coordinates in the coordinate system, X is the ellipse coordinates in the image coordinate system, Y is the ellipse coordinates in the camera coordinate system, as shown in the following formula (4):

Therefore, the general elliptic cone equation is derived as shown in the following formula (5):

in:

Assuming that there is a projection plane whose origin is C _p and unit normal vector is N _p , construct the other two orthogonal unit vectors U _p and V _p , then the rotation matrix is R _p =[U _p V _p N _p ] the plane and the ellipse Cones intersect, the line of intersection is an ellipse,

Then the point on the ellipse can be represented by X _p =C _p +y2U _p +y3, N _p =C _p +J _p Y _p , and the derivation of the ellipse equation is shown in the following formula (7):

椭圆方程可以重构为如下形式：(7) of which

The elliptic equation can be reconstructed as follows:

(Y _p -K) ^T M (Y _p -K) = 1 (8)

为圆心坐标，(8) of which

is the coordinates of the center of the circle,

Perform SVD decomposition on M RDR ^T = M,

Where R is a 2×2 matrix representing an orthogonal basis vector,

λ₀和λ₁分别表示椭圆的长短轴；

λ ₀ and λ ₁ represent the major and minor axes of the ellipse, respectively;

(3) Construct the positive bottom surface:

From the symmetry of the elliptical cone, it is easy to know that the line connecting the apex of the cone and the center of the positive bottom surface must be perpendicular to the positive bottom surface. A point c ₁ is randomly selected on the projection contour, with c ₁ as the origin and the straight line c ₁ E as the normal vector direction to construct a plane π ₁ . Intersect the cone with the ellipse l, calculate the coordinate c ₂ of the center of the circle according to the algorithm of the above steps (1) and (2), and judge whether the angle between the connecting line c ₂ E and the normal vector of the plane is smaller than the threshold value, and the threshold value is generally taken as e ^-6 , if it is not satisfied, continue to use c ₂ E as the normal to construct a plane and intersect the cone until the positive bottom is found.

Further, in the step 2, step 3 and step 4, solving the circular target plane includes the following steps:

S1: The plane α intersects the ellipse cone with the ellipse, and the radius of the major and minor axes of the ellipse can be obtained according to the steps in the process of constructing the positive base of the cone in the first step,

Disturbing the process with the minimum increment Δθ is shown in equation (9):

(λ ₀ ,λ ₁ )=f(θ+ηΔθ)

Among them, η represents the learning rate, which is used to control the disturbance step size, and constructs the least squares problem to find the parameter θ to minimize the difference between the long and short axis radii of the ellipse, as shown in formula (10):

S2: Through S1, the plane equation when the intersecting line is a circle can be obtained, and the plane is recorded as β, and the origin is the coordinate c of the center of the circle. It is easy to know that the center of the circular target must be on the ray Ec, assuming that the distance between the center and point c is l, Then the coordinates are the point coordinates that can be obtained according to the linear transformation, and the plane β' parallel to the plane β is constructed with this point as the origin, and the intersection of β' and the elliptical cone is obtained according to the steps in the process of constructing the positive bottom of the cone in the first step. The circle diameter d is known, and the circular target diameter D is known, then the least squares problem can be constructed as shown in the following equations (11) and (12):

d=f(l+ηΔl) (11)

S3: The correct l value can be obtained through iteration, and then the spatial circular target equation and the coordinates of the center of the circle can be obtained.

The beneficial effects obtained by the present invention by adopting the above-mentioned structure are as follows: the rapid extraction method of the spatial center of a single planar circle image in this scheme aims at the problem of low accuracy and low efficiency in the extraction of the spatial coordinates of the current center of the circle target circle, and first uses the optical center of the camera and the imaging ellipse to construct The spatial conic equation, and then construct the positive base of the cone, solve the spatial circle equation according to the positive base, and finally find the correct spatial circle equation according to the parallel plane and the actual radius, avoiding the problem of solving the spatial center projection coordinates, and only need a single image The coordinates of the space center can be solved, the method is stable and reliable, and it has good practicability in problems such as robot positioning and circular pipe measurement. This scheme can also be applied to the field of photogrammetry, and the observation data of multiple viewing angles is used for beam adjustment to improve measurement accuracy.

Description of drawings

Fig. 1 is the schematic diagram of the target imaging model of the method for quickly extracting the center of space of a single plane circle image of the present invention;

Fig. 2 is the schematic diagram of constructing the positive bottom surface of the spatial circle center rapid extraction method of a single plane circle image of the present invention;

Fig. 3 is the analysis schematic diagram of the learning rate of the method for quickly extracting the center of space of a single plane circle image of the present invention;

Fig. 4 is a schematic structural diagram of a pipe surveying instrument and a pipe detection robot in Embodiment 1 of the method for rapidly extracting the center of space of a single plane circle image of the present invention;

Fig. 5 is a comparison diagram of the results of pipe length measurement in Embodiment 1 of the method for quickly extracting the center of space of a single plane circle image of the present invention;

FIG. 6 is a schematic diagram of a circular target in Embodiment 2 of the method for quickly extracting the spatial center of a single plane circular image according to the present invention.

Fig. 7 is a comparison diagram of the results of pipe length measurement in Embodiment 1 of the method for quickly extracting the center of space of a single plane circle image of the present invention;

Fig. 8 is a schematic diagram of a circular target in Embodiment 2 of the method for quickly extracting the spatial center of a single plane circular image of the present invention.

The accompanying drawings are used to provide a further understanding of the present invention, and constitute a part of the description, and are used together with the embodiments of the present invention to explain the present invention, and do not constitute a limitation to the present invention.

In Figure 1, point E is the viewpoint, EX _e Y _e Z _e is the camera coordinate system, point P is the principal point of the camera, and PX _p Y _p Z _p is the image coordinate system; in Figure 2, c ₁ is the origin, π ₁ is The plane constructed in the direction of the normal vector of the straight line c ₁ E, c ₂ is the coordinates of the center of the circle, and π ₂ is the plane constructed in the direction of the normal vector of the straight line c ₂ E; in Figure 6, (a) is the pipeline surveying instrument, (b) is the pipeline surveying and mapping robot ; In Fig. 7, (a) represents the simulation environment, (b) represents the physical environment, and (c) is a measurement comparison chart between the simulation environment and the physical environment.

Detailed ways

The technical solutions in the embodiments of the present invention will be clearly and completely described below in conjunction with the accompanying drawings in the embodiments of the present invention. Obviously, the described embodiments are only some of the embodiments of the present invention, not all of them; based on The embodiments of the present invention and all other embodiments obtained by persons of ordinary skill in the art without making creative efforts belong to the protection scope of the present invention.

As shown in Figures 1-8, the method for quickly extracting the center of space of a single plane circle image of the present invention includes the following steps:

Step 1: Construct the positive bottom of the cone;

Step 2: Rotate the plane where the bottom surface is located around the long axis, find the angle θ where the section is a circle, and record the plane as α at this time;

Step 3: Find the plane β parallel to α and the diameter of the cross-sectional circle is D;

Step 4: Solve the cross-sectional circle equation at this time, and solve the coordinates of the center of the circle.

In said step one, constructing the conical bottom surface includes the following steps:

(1) Solve the elliptic cone equation in the coordinate system:

其中Y＝(y₀,y₁,y₂)，/>

为3×3对称矩阵，/>

为3×1向量，/>

为标量，取A＝[a_ij]，B＝[b_i]，c＝1，The ellipse cone equation in the image coordinate system PX _p Y _p Z _p is

where Y=(y ₀ ,y ₁ ,y ₂ ),/>

is a 3×3 symmetric matrix, />

is a 3×1 vector, />

is a scalar, take A=[a _ij ], B=[b _i ], c=1,

Since the ellipse is on the XOY plane, then y ₂ is 0,

At this time, if Q(Y)=0, the following equation can be obtained:

According to the standard equation (2) of the ellipse,

Assume that the major axis radius of the ellipse is d ₀ , and the minor axis radius is d ₁ ,

but

The gradient is 0 at the vertex of the ellipse cone, that is

在椭圆锥上，则有/>

联立式(1)、式(2)之可得and

On an elliptical cone, there is />

Simultaneous formula (1) and formula (2) can be obtained

求解如式(3)所示：at this time

The solution is shown in formula (3):

(2) Solve the ellipse cone equation in the camera coordinate system:

Assume that a set of orthonormal bases of the image coordinate system in the camera coordinate system is Re=[Ue Ve Ne], Re can also be regarded as the rotation transformation matrix between the two coordinate systems, assuming C _e is a point on the disk surface in the camera The coordinates in the coordinate system, X is the ellipse coordinates in the image coordinate system, and Y is the ellipse coordinates in the camera coordinate system, as shown in the following formula (4):

Therefore, the general elliptic cone equation is derived as shown in the following formula (5):

in:

Assuming that there is a projection plane whose origin is C _p and unit normal vector is N _p , construct the other two orthogonal unit vectors U _p and V _p , then the rotation matrix is R _p =[U _p V _p N _p ] the plane and the ellipse Cones intersect, the line of intersection is an ellipse,

Then the point on the ellipse can be represented by X _p =C _p +y2U _p +y3, N _p =C _p +J _p Y _p , and the derivation of the ellipse equation is shown in the following formula (7):

椭圆方程可以重构为如下形式：in

The elliptic equation can be reconstructed as follows:

(Y _p -K) ^T M (Y _p -K) = 1 (8)

为圆心坐标，(8) of which

is the coordinates of the center of the circle,

Perform SVD decomposition on M RDR ^T = M,

λ₀和λ₁分别表示椭圆的长短轴；Where R is a 2×2 matrix representing an orthogonal basis vector,

λ ₀ and λ ₁ represent the major and minor axes of the ellipse, respectively;

(3) Construct the positive bottom surface:

From the symmetry of the elliptical cone, it is easy to know that the line connecting the apex of the cone and the center of the positive bottom surface must be perpendicular to the positive bottom surface. A point c ₁ is randomly selected on the projection contour, with c ₁ as the origin and the straight line c ₁ E as the normal vector direction to construct a plane π ₁ . Intersect the cone with the ellipse l, calculate the coordinate c ₂ of the center of the circle according to the algorithm of the above steps (1) and (2), and judge whether the angle between the connecting line c ₂ E and the normal vector of the plane is smaller than the threshold value, and the threshold value is generally taken as e ^-6 , if it is not satisfied, continue to use c ₂ E as the normal to construct a plane and intersect the cone until the positive bottom is found.

In said step 2, step 3 and step 4, solving the circular target plane includes the following steps:

S1: The plane α intersects the ellipse cone with the ellipse, and the radius of the major and minor axes of the ellipse can be obtained according to the steps in the process of constructing the positive base of the cone in the first step,

Disturbing the process with the minimum increment Δθ is shown in equation (9):

(λ ₀ ,λ ₁ )=f(θ+ηΔθ)

Among them, η represents the learning rate, which is used to control the disturbance step size, and constructs the least squares problem to find the parameter θ to minimize the difference between the long and short axis radii of the ellipse, as shown in formula (10):

S2: Through S1, the plane equation when the intersecting line is a circle can be obtained, and the plane is recorded as β, and the origin is the coordinate c of the center of the circle. It is easy to know that the center of the circular target must be on the ray Ec, assuming that the distance between the center and point c is l, Then the coordinates are the point coordinates that can be obtained according to the linear transformation, and the plane β' parallel to the plane β is constructed with this point as the origin, and the intersection of β' and the elliptical cone is obtained according to the steps in the process of constructing the positive bottom of the cone in the first step. The circle diameter d is known, and the circular target diameter D is known, then the least squares problem can be constructed as shown in the following equations (11) and (12):

d=f(l+ηΔl) (11)

S3: The correct l value can be obtained through iteration, and then the spatial circular target equation and the coordinates of the center of the circle can be obtained.

In the above steps, the parameter η represents the learning rate. This parameter is generally called a hyperparameter, and its value affects the optimal solution of the least squares problem. It is easy to know that the least squares problem in the circular target plane is a convex optimization problem, so the gradient descent is used The method must be able to obtain the extreme value. Experiments show that when the value is constant, that is, when the least squares model is perturbed with a constant step size, if the value is small, the convergence will be slow, and if the value is large, there will be oscillation convergence. In the case of , as shown in Figure 3, better results can be achieved by gradually reducing η along with the solution process.

As shown in Figure 3-5, the initial decline speed is faster, and as the learning rate decays, it gradually converges to the minimum value, and the number of iterations is also small.

Embodiment 1, applied to pipeline length measurement:

The equipment shown in Figure 6 is a pipeline surveyor and a pipeline inspection robot. The surveyor is used in static measurement scenarios, while the pipeline inspection robot is used for pipeline measurement in narrow spaces. Both measurement equipment include lighting devices and cameras. The cameras are controlled by Zhang’s calibration method completes the calibration of the internal parameters. This program compares the measurement of the physical pipeline with the 3D simulation environment. The length of the pipeline is 2m, and the diameter of the pipeline is known to be 0.4m. Use the algorithm steps in this program to calculate the center coordinates of the circular ports at both ends. , and then calculate the length of the pipeline, and solve it 20 times from multiple angles. The statistical data are shown in Figure 7. It can be seen that the algorithm in this paper can accurately calculate the length of the pipeline. Measurement variance, because in the actual environment, there are more noises on the edge of the circular target, and the extraction accuracy is lower than that in the simulation environment. In actual measurement, multiple measurements can be averaged to reduce the variance of the measurement value.

Embodiment 2, measurement of extraction accuracy of circle center projection:

In order to measure the accuracy of the extraction of the center of the scheme, the symbol circle shown in Figure 8 is used as the target for testing. The radius of the circle to be measured is 5cm, and the middle corner point is the actual center of the circle. When the image plane and the target plane form an angle of 45°, images are collected at different distances, and the sub-pixel coordinates of the corner points are extracted as a reference. This solution is used to solve the The three-dimensional coordinates of the center of the circle are projected to the imaging plane through the camera model to obtain the pixel coordinates, and then the sub-pixel coordinates of the center of the fitted ellipse are extracted.

The results of the three are shown in Table 1 below.

Table 1 Comparison results of the coordinates of the center of the circle

The difference between the coordinates of the center of the circle extracted by this scheme and the reference coordinates is about 0.1 to 2 pixels, but the difference between the center of the ellipse and the reference is 5 to 12 pixels. Therefore, the coordinates of the center of the circle obtained by the method in this paper are more accurate, and the error with the actual coordinates is very small.

The present invention and its implementations have been described above, and this description is not limiting. What is shown in the drawings is only one of the implementations of the present invention, and the actual structure is not limited thereto. All in all, if a person of ordinary skill in the art is inspired by it, without departing from the inventive concept of the present invention, without creatively designing a structure and an embodiment similar to the technical solution, it shall fall within the scope of protection of the present invention.

Claims (1)

1. The method for quickly extracting the center of space of a single plane circle image is characterized in that, a pipeline surveying instrument and a pipeline detection robot are used to measure the physical pipeline, and the pipeline surveying instrument and the pipeline detection robot all include a lighting device and a camera, a camera The internal reference calibration is completed by Zhang’s calibration method, and the pipe length measurement includes the following steps: S1: First obtain a single image through the camera, use the optical center of the camera and the imaging ellipse to form a spatial conic equation, and then construct the positive base of the cone; S2: Solve the parallel plane equation of the space circle according to the positive base of the cone; S3: Finally, find out the correct space circle equation according to the parallel plane and the actual radius; The establishment of the space circle equation specifically includes the following steps: Step 1: Construct the positive bottom of the cone; Step 2: Rotate the plane where the bottom surface is located around the long axis, find the angle θ where the section is a circle, and record the plane as α at this time; Step 3: Find the plane β parallel to α and the diameter of the cross-sectional circle is D; Step 4: Solve the cross-sectional circle equation at this time, and solve the coordinates of the center of the circle; In said step one, constructing the conical bottom surface includes the following steps: (1) Solve the elliptic cone equation in the coordinate system: The ellipse cone equation in the image coordinate system PX _p Y _p Z _p is

where Y=(y ₀ ,y ₁ ,y ₂ ),/>

is a 3×3 symmetric matrix, />

is a 3×1 vector, />

As a scalar, take A=[a _ij ], B=[b _i ], c=1, since the ellipse is on the XOY plane, then _y2 is 0, at this time, Q(Y)=0 can get the following equation:

In formula (1), Y is the ellipse coordinate in the camera coordinate system, Y=(y ₀ , y ₁ , y ₂ ), y ₂ is 0 in the XOY plane, A=[a _ij ] is simplified to a 2*2 matrix, B=[b _i ] is simplified to a 2*1 matrix; According to the standard equation (2) of the ellipse,

In formula (2), y ₀ and y ₁ are the coordinates of the x-axis and y-axis of Y in the XOY plane; assuming that the radius of the major axis of the ellipse is d ₀ and the radius of the minor axis is d ₁ , but

The gradient is 0 at the vertex of the ellipse cone, that is

and/>

On an elliptical cone, there is />

Simultaneous formula (1), formula (2) can get:

at this time

The solution is shown in formula (3):

(2) Solve the ellipse cone equation in the camera coordinate system: Assume that a set of orthonormal bases of the image coordinate system in the camera coordinate system is Re=[Ue Ve Ne], Re can also be regarded as the rotation transformation matrix between the two coordinate systems, assuming C _e is a point on the disk surface in the camera The coordinates in the coordinate system, X is the ellipse coordinates in the image coordinate system, and Y is the ellipse coordinates in the camera coordinate system, as shown in the following formula (4):

Therefore, the general elliptic cone equation is derived as shown in the following formula (5):

in:

Assuming that there is a projection plane whose origin is C _p and unit normal vector is N _p , construct the other two orthogonal unit vectors U _p and V _p , then the rotation matrix is R _p =[U _p V _p N _p ] the plane and the ellipse Cones intersect, the line of intersection is an ellipse, Then the point on the ellipse can be represented by X _p =C _p +y2U _p +y3, N _p =C _p +J _p Y _p , and the derivation of the ellipse equation is shown in the following formula (7):

in:

The elliptic equation can be reconstructed as follows: (Y _p -K) ^T M (Y _p -K) = 1 (8) in

is the coordinates of the center of the circle, perform SVD decomposition on M to obtain RDR ^T = M, where R is a 2×2 matrix representing an orthogonal basis vector, />

λ ₀ and λ ₁ represent the major and minor axes of the ellipse, respectively; (3) Construct the positive bottom surface: From the symmetry of the elliptical cone, it is easy to know that the line connecting the apex of the cone and the center of the positive bottom surface must be perpendicular to the positive bottom surface. A point c ₁ is randomly selected on the projection contour, with c ₁ as the origin and the straight line c ₁ E as the normal vector direction to construct a plane π ₁ . Intersect the cone with the ellipse l, calculate the coordinate c ₂ of the center of the circle according to the algorithm of the above steps (1) and (2), and judge whether the angle between the connecting line c ₂ E and the normal vector of the plane is smaller than the threshold value, and the threshold value is generally taken as e ^-6 , if it is not satisfied, continue to use c ₂ E as the normal to construct the plane and the cone to intersect until the positive bottom is found; In said step 2, step 3 and step 4, solving the section circle equation includes the following steps: S1: The plane α intersects the ellipse cone with the ellipse, and the radius of the major and minor axes of the ellipse can be obtained according to the steps in the process of constructing the positive base of the cone in the first step, Disturbing the process with the minimum increment Δθ is shown in equation (9): (λ ₀ ,λ ₁ )=f(θ+ηΔθ) (9) Among them, η represents the learning rate, which is used to control the disturbance step size, and constructs a least squares problem to find the parameter θ to minimize the difference between the long and short axis radii of the ellipse, as shown in formula (10):

S2: Through S1, the plane equation when the intersection line is a circle can be obtained, and the plane is recorded as β, and the origin is the coordinate c of the center of the circle. It is easy to know that the space circle center of the target must be on the ray Ec, assuming that the distance between the center and point c is l , then the coordinates are the coordinates of the point that can be obtained according to the linear transformation, and the plane β' parallel to the plane β is constructed with this point as the origin, and the relationship between β' and the elliptical cone is obtained according to the steps in the process of constructing the positive base of the cone in the first step. The diameter of the intersecting circle is d, and the diameter of the cross-section circle to be obtained is known to be D, then the least squares problem can be constructed as shown in the following equations (11) and (12): d=f(l+ηΔl) (11)

S3: The correct l value can be obtained through iteration, and then the section circle equation and the coordinates of the circle center can be obtained.

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