CN104634248A - Revolving shaft calibration method under binocular vision - Google Patents

Abstract

The invention discloses a revolving shaft calibration method under binocular vision. The method comprises the following steps: placing a rotating platform into the visual fields of two cameras, and placing a calibration board on the rotating platform; detecting the angular points of the calibration board by using two cameras simultaneously, and marking an angular point which is farthest away from the center of the rotating platform as a mark point Q1; rotating the rotating platform by two angles in sequence to obtain a mark point Q2 and a mark point Q3; solving the three-dimensional coordinates of the three mark points in a world coordinate system respectively; converting the geometrical relation between the revolving shaft and a rotating platform plane into an operation relation among vectors, and solving the direction of a revolving shaft and the circle center coordinate of the rotating platform according the three-dimensional coordinates of any two mark points obtained in the step 3 in order to finish the calibration of the revolving shaft. By adopting the revolving shaft calibration method under binocular vision, the problem of low calibration efficiency due to complexity of a calibration device and the requirement of acquisition of the mark points by means of a large amount of calculation in the prior art is solved.

Description

Rotating axis calibration method under a kind of binocular vision

Technical field

The invention belongs to technical field of computer vision, relate to the rotating axis calibration method under a kind of binocular vision.

Background technology

Binocular vision is based on principle of parallax and utilize imaging device from two width images of different position acquisition testees, by the position deviation between computed image corresponding point, obtains the method for object dimensional geological information.Binocular measuring technique based on binocular vision is a kind of important form of machine vision, utilize the structure of two video camera simulation human eyes to measure space characteristics point, it is the image that obtains of fusion two eyes and the difference of observing between them, obvious depth perception can be obtained, set up the corresponding relation between feature, the photosites in space a bit in different images is mapped.In binocular measuring technique, testee can be placed on universal stage by usual people, testee is rotated under the rotation of universal stage thereupon, thus realize the measurement of testee 360 degree, and in order to ensure that testee is with the accuracy measured after universal stage rotation, just need to demarcate universal stage, find out center and the direction of universal stage rotation.

At present, turning axle is demarcated generally two kinds of methods: a kind of method is the method for demarcating for turning axle in computer vision measurement, its primarily of standard flat or high precision and the known standard ball of radius realize, first obtained the surface three dimension data of standard flat or standard ball from multiple position of rotation by the visual measuring equipment demarcated, then according to the surface equation of three-dimensional point data fitting out-of-plane or ball, the position of the rotation of universal stage is finally sought out according to the geometric properties of plane or standard ball; But this method needs standard flat or the known standard ball of high precision radius to be used as scaling board, and need the surface three dimension data in multiple position acquisition plane or standard ball, make the requirement for scaling board just higher, and calibration cost can be caused higher, demarcate efficiency comparison low.

Another method utilizes spherical target, sought out roughly the edge of target again by the searching of gray-scale value saltus step pixel after rotating several positions, then screen according to the Curvature varying edge on target edge between neighbor pixel, obtain spherical target marginal point accurately, and then use least square method to obtain the center of circle of spherical target, finally carry out fit Plane by each center of circle three-dimensional point data again and ask for turning axle.The levels of precision that threshold value is chosen in the process finding target edge directly affects the determination at edge, and choosing of threshold value is normally determined according to gray-scale value saltus step and empirical value, there is very large uncertainty; Secondly, when calculating, data calculated amount is also larger.

To sum up, in the existing method to turning axle demarcation, existence needs extra making marked circle or spherical displacer, makes caliberating device more complicated, and needs obtain gauge point by certain computing method, thus causes demarcation efficiency low.

Summary of the invention

The object of this invention is to provide the rotating axis calibration method under a kind of binocular vision, solve the caliberating device existed in prior art complicated, and need obtain gauge point by a large amount of calculating and cause demarcating inefficient problem.

The technical solution adopted in the present invention is, the rotating axis calibration method under a kind of binocular vision, comprises the following steps:

Step 1, is placed in universal stage in the visual field of two video cameras, and puts on universal stage by scaling board, makes universal stage scaling board entirety before and after rotation all be in the common field range of two video cameras; Two video cameras are designated as left video camera and right video camera respectively;

Step 2, utilizes left video camera and right video camera to detect the angle point of scaling board simultaneously, and a selected distance universal stage center angle point is farthest designated as gauge point Q ₁(x ₁, y ₁, z ₁); Universal stage is rotated at least two angles successively, and selected distance universal stage center two angle points farthest are also designated as gauge point Q respectively ₂(x ₂, y ₂, z ₂) and gauge point Q ₃(x ₃, y ₃, z ₃), above-mentioned 3 gauge points are all positioned in same level; Wherein the coordinate of gauge point Q1 in left video camera and right video camera is designated as Q respectively _l1(u ₁, v ₁) and Q _r1(u ₂, v ₂), the coordinate of gauge point Q2 in left video camera and right video camera is designated as Q respectively _l2(u ₁', v ₁') and Q _r2(u ₂', v ₂'), the coordinate of gauge point Q3 in left video camera and right video camera is designated as Q respectively _l3(u ₁", v ₁") and Q _r3(u ₂", v ₂");

Step 3, asks for the three-dimensional coordinate of 3 gauge points under world coordinate system respectively;

Step 4, according to the three-dimensional coordinate of three gauge points obtained in step 3, asks for the direction of turning axle and the central coordinate of circle of universal stage, thus completes the demarcation of countershaft.

Feature of the present invention is also,

In step 1, scaling board is gridiron pattern.

Step 3 is specially:

Step 3.1, demarcates left video camera and right video camera respectively, obtains the parameter matrix of left video camera and the parameter matrix of right video camera;

Step 3.2, obtains the projection matrix M 1 of left video camera, the projection matrix M 2 of right video camera according to the parameter matrix of left video camera and right video camera respectively;

Wherein, the projection matrix M 1 of left video camera is designated as:

$M_1=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]---\left(10\right)$

In formula, be respectively the i-th row and the jth column element of left video camera projection matrix M 1;

The projection matrix M 2 of right video camera is designated as:

$M_2=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]---\left(11\right)$

In formula, be respectively the i-th row and i-th column element of right video camera projection matrix M 2;

Step 3.3, asks for gauge point Q1, Q2, Q3 three-dimensional coordinate under world coordinate system according to the projection matrix M 2 of left video camera projection matrix M 1 and right video camera respectively;

The concrete grammar asking for the three-dimensional coordinate of gauge point Q1 under world coordinate system is as follows:

$Z_{c1}\left[\begin{array}{c}{u}_1\\ v_1\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]---\left(12\right)$

$Z_{c2}\left[\begin{array}{c}{u}_2\\ v_2\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]---\left(13\right)$

In formula, (u ₁, v ₁, 1) and (u ₂, v ₂, 1) and be respectively Q _l1with Q _r1homogeneous coordinates; (x ₁, y ₁, z ₁, 1) and be gauge point Q ₁homogeneous coordinates, Z _c1and Z _c2scale-up factor respectively;

The concrete grammar asking for the three-dimensional coordinate of gauge point Q2 under world coordinate system is as follows:

$Z_{c1}\left[\begin{array}{c}{u_1}^{\′}\\ {v_1}^{\′}\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]---\left(14\right)$

$Z_{c2}\left[\begin{array}{c}{u_2}^{\′}\\ {v_2}^{\′}\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]---\left(15\right)$

The concrete grammar asking for the three-dimensional coordinate of gauge point Q3 under world coordinate system is as follows:

$Z_{c1}\left[\begin{array}{c}{u_1}^{\′\′}\\ {v_1}^{\′\′}\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]---\left(16\right)$

$Z_{c2}\left[\begin{array}{c}{u_2}^{\′\′}\\ {v_2}^{\′\′}\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]---\left(17\right)$

Step 3.4, by solving the expression formula obtained in step 3.3, obtaining the expression formula of 3 gauge points, being specially:

By to formula 12 and formula 13 simultaneous solution, the expression formula obtaining gauge point Q1 is:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(18\right)$

By to formula 14 and formula 15 simultaneous solution, obtain the expression formula of gauge point Q2:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(19\right)$

By to formula 16 and formula 17 simultaneous solution, the expression formula obtaining gauge point Q3 is:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(20\right)$

Step 3.5, utilizes the formula 9, formula 10 and the formula 11 that obtain in least square method solution procedure 3.4 respectively, obtains gauge point Q1, gauge point Q2 and the three-dimensional coordinate of gauge point Q3 under world coordinate system.

Ask for turning axle direction in step 4 to be specially:

Steps A, chooses two gauge point composition of vector Q on scaling board ₁q ₂, according to turning axle vector R (a', b', c') and the vector normal in universal stage plane, the first expression formula obtaining turning axle vector is:

R＝Q ₁Q ₂×Q ₂Q ₃(21)

Or wherein two the gauge point composition of vector Q chosen on universal stage ₂q ₃, the second expression formula obtaining turning axle vector is:

R＝Q ₂Q ₁×Q ₁Q ₃(22)

Or wherein two the gauge point composition of vector Q chosen on universal stage ₁q ₃, the 3rd expression formula obtaining turning axle vector is:

R＝Q ₁Q ₃×Q ₃Q ₂(23)

Step B, launches the first expression formula of turning axle vector:

R＝(x ₂-x ₁,y ₂-y ₁,z ₂-z ₁)×(x ₃-x ₂,y ₃-y ₂,z ₃-z ₂) (24)

Or the second expression formula of turning axle vector is launched:

R＝(x ₁-x ₂,y ₁-y ₂,z ₁-z ₂)×(x ₃-x ₁,y ₃-y ₁,z ₃-z ₁) (25)

Or the 3rd expression formula of turning axle vector is launched:

R＝(x ₃-x ₁,y ₃-y ₁,z ₃-z ₁)×(x ₂-x ₃,y ₂-y ₃,z ₂-z ₃) (26)

Step C, can obtain three coordinate figure a' about turning axle vector according to the first expression formula of the turning axle vector launched in step B, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_2-y_1& z_2-z_1\\ y_3-y_2& z_3-z_2\end{array}\right|---\left(27\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_2-z_1& x_2-x_1\\ z_3-z_2& x_3-x_2\end{array}\right|---\left(28\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_2-x_1& y_2-y_1\\ x_3-x_2& y_3-y_2\end{array}\right|---\left(29\right)$

Or the coordinate a' about turning axle vector can be obtained according to the second expression formula of the turning axle vector launched in step B, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_1-y_2& z_1-z_2\\ y_3-y_1& z_3-z_1\end{array}\right|---\left(30\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_1-z_2& x_1-x_2\\ z_3-z_1& x_3-x_1\end{array}\right|---\left(31\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_1-x_2& y_1-y_2\\ x_3-x_1& y_3-y_1\end{array}\right|---\left(32\right)$

Or the coordinate a' about turning axle vector can be obtained according to the 3rd expression formula of the turning axle vector launched in step B, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_3-y_1& z_3-z_1\\ y_2-y_3& z_2-z_3\end{array}\right|---\left(33\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_3-z_1& x_3-x_1\\ z_2-z_3& x_2-x_3\end{array}\right|---\left(34\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_3-x_1& y_3-y_1\\ x_2-x_3& y_2-y_3\end{array}\right|---\left(35\right)$

Asking for universal stage central coordinate of circle in step 4 is optional two gauge point composition of vector in universal stage plane, and the principle that the perpendicular bisector going up the line of any two points according to circle must cross the center of circle asks for universal stage central coordinate of circle.

The invention has the beneficial effects as follows, by utilizing the scaling board of camera calibration and then demarcating universal stage, do not need additionally to make spherical displacer or marked circle, and the putting position of scaling board is unrestricted, achieve the simplification of caliberating device, also achieve the fusion that camera calibration and turning axle are demarcated; By identifying scaling board angle point and then demarcating turning axle as gauge point, without the need to asking the centre of sphere to mark ball or marked circle or ask the center of circle again, achieve the simplification of turning axle calibration process, improve demarcation efficiency.

Accompanying drawing explanation

Fig. 1 is the schematic flow sheet of the rotating axis calibration method under a kind of binocular vision of the present invention;

Fig. 2 is the part-structure schematic diagram of the binocular vision Shaft calibration system in the present invention;

Fig. 3 is the schematic diagram that in the present invention, under binocular vision, spatial point is rebuild.

In Fig. 2, OXYZ coordinate system is world coordinate system; O in Fig. 3 _lx _ly _lz _lwith O _rx _ry _rz _rbe respectively the coordinate system in left and right cameras, coordinate system Q ₁(x ₁, y ₁, z ₁) be coordinate in space a bit under world coordinate system, Q _l1(u ₁, v ₁) be Q ₁pixel coordinate under left camera coordinate system, Q _r1(u ₂, v ₂) be Q ₁pixel coordinate under right camera coordinate system.

Embodiment

Below in conjunction with the drawings and specific embodiments, the present invention is described in detail.

Rotating axis calibration method under a kind of binocular vision of the present invention, as shown in Figure 1, specifically implement according to following steps:

Step 1, is placed in universal stage in the visual field of two video cameras, and puts on universal stage by scaling board; Two video cameras are designated as left video camera and right video camera respectively; Wherein, the putting position of scaling board depends on the relative position of two video cameras and universal stage, scaling board is wanted when putting to ensure that universal stage scaling board entirety before and after rotation is all in the common field range of two video cameras, and tessellated angle point can be detected, when camera lens is higher than universal stage plane, scaling board and universal stage plane are that acute angle is arranged; Here, scaling board can be gridiron pattern; For the structural representation of the calibration system of gridiron pattern as shown in Figure 2;

Step 2, utilizes left video camera and right video camera to detect the angle point of scaling board simultaneously, and a selected distance universal stage center angle point is farthest designated as gauge point Q ₁(x ₁, y ₁, z ₁); Universal stage is rotated at least two angles successively, and selected distance universal stage center two angle points are farthest designated as gauge point Q respectively ₂(x ₂, y ₂, z ₂) and gauge point Q ₃(x ₃, y ₃, z ₃), above-mentioned 3 gauge points are all positioned in same level, and all near rotation platform edge, can reduce the error of calculation like this; Wherein the coordinate of gauge point Q1 in left video camera and right video camera is designated as Q respectively _l1(u ₁, v ₁) and Q _r1(u ₂, v ₂), the coordinate of gauge point Q2 in left video camera and right video camera is designated as Q respectively _l2(u ₁', v ₁') and Q _r2(u ₂', v ₂'), the coordinate of gauge point Q3 in left video camera and right video camera is designated as Q respectively _l3(u ₁", v ₁") and Q _r3(u ₂", v ₂");

Step 3, asks for the three-dimensional coordinate of 3 gauge points under world coordinate system respectively, as shown in Figure 3, and Q _l1with Q _r1spatial point Q ₁at left video camera and point respectively corresponding in right video camera, the transformation relation of foundation triangle geometry can uniquely determine space mid point Q by corresponding 2 of left and right cameras ₁, be specially:

Step 3.1, demarcates left video camera and right video camera respectively, obtains the parameter matrix of left video camera and the parameter matrix of right video camera;

The above-mentioned demarcation for video camera is actually and solves inside and outside parameter in video camera, according to the relation in the unique point on scaling board and video camera between picture plane corresponding point, namely homography matrix can calculate the inside and outside parameter of video camera, all applicable in left video camera and right video camera to the scaling method of video camera below, left video camera and right video camera are referred to as video camera, are specially:

Step a, utilizes video camera to carry out Corner Detection to scaling board;

Step b, according to pin-hole imaging model, calculates homography matrix as the corresponding point in plane to all angle points detected in step a and its, can obtain expression formula in the middle of video camera:

$s\left[\begin{array}{c}u\\ v\\ 1\end{array}\right]=A\left[\begin{array}{cc}R& t\end{array}\right]\left[\begin{array}{c}{X}_w\\ Y_x\\ Z_w\\ 1\end{array}\right]=\left[\begin{array}{ccc}\mathrm{\α}& \mathrm{\γ}& u_0\\ 0& \mathrm{\β}& v_0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{cccc}{r}_1& r_2& r_3& t\end{array}\right]\left[\begin{array}{c}{X}_w\\ Y_w\\ Z_w\\ 1\end{array}\right]---\left(1\right)$

In formula, s is scale-up factor, and (u, v, 1) is the homogeneous coordinates of unique point in two-dimensional pixel coordinate system; (X _w, Y _w, Z _w, 1) and be the homogeneous coordinates of this unique point under world coordinate system, A is Intrinsic Matrix, α and β is the scale factor of U axle under image coordinate system and V axle, and γ is the parameter of the deflection of description two image axles, u ₀and v ₀be figure principal point coordinate, [R t] is outer parameter matrix, is rotation and the translation relation of camera coordinate system and world coordinate system;

Step c, is placed in scaling board place plane by world coordinate system plane, therefore Z-direction is 0, then formula (1) can be write as:

$s\left[\begin{array}{c}u\\ v\\ 1\end{array}\right]=A\left[\begin{array}{cc}R& t\end{array}\right]\left[\begin{array}{c}{X}_w\\ Y_x\\ Z_w\\ 1\end{array}\right]=\left[\begin{array}{ccc}\mathrm{\α}& \mathrm{\γ}& u_0\\ 0& \mathrm{\β}& v_0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}{r}_1& r_2& t\end{array}\right]\left[\begin{array}{c}{X}_w\\ Y_w\\ 1\end{array}\right]---\left(2\right)$

Steps d, makes H=λ A [r ₁r ₂t]=[h ₁h ₂h ₃], H is homography matrix, and λ is scale factor; Due to r ₁and r ₂orthogonal, therefore obtain two restriction relations:

$h_1^T{A}^{-T}{A}^{-1}{h}_2=0---\left(3\right)$

$h_1^T{A}^{-T}{A}^{-1}{h}_1=h_2^T{A}^{-T}{A}^{-1}{h}_2---\left(4\right)$

Step e, order

$B=A^{-T}{A}^{-1}=\left[\begin{array}{ccc}{B}_{11}& B_{12}& B_{13}\\ B_{21}& B_{22}& B_{23}\\ B_{31}& B_{32}& B_{33}\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{{\mathrm{\α}}^2}& \frac{\mathrm{\γ}}{{\mathrm{\α}}^2\mathrm{\β}}& \frac{v_0\mathrm{\γ}-u_0\mathrm{\β}}{{\mathrm{\α}}^2\mathrm{\β}}\\ -\frac{\mathrm{\γ}}{{\mathrm{\α}}^2\mathrm{\β}}& \frac{\mathrm{\γ}}{{\mathrm{\α}}^2\mathrm{\β}}+\frac{1}{{\mathrm{\β}}^2}& -\frac{v_0\mathrm{\γ}-u_0\mathrm{\β}}{{\mathrm{\α}}^2}-\frac{v_0}{{\mathrm{\β}}^2}\\ \frac{v_0\mathrm{\γ}-u_0\mathrm{\β}}{{\mathrm{\α}}^2\mathrm{\β}}& -\frac{v_0\mathrm{\γ}-u_0\mathrm{\β}}{{\mathrm{\α}}^2\mathrm{\β}}-\frac{v_0}{{\mathrm{\β}}^2}& \frac{{(v_0\mathrm{\γ}-u_0\mathrm{\β})}^2}{{\mathrm{\α}}^2\mathrm{\β}}+\frac{v_0}{{\mathrm{\β}}^2}+1\end{array}\right]---\left(5\right)$

Step f, defines symmetric matrix B with six-vector b, as follows:

b＝[B ₁₁B ₁₂B ₂₂B ₁₃B ₂₃B ₃₃] ^T(6)

I-th column vector of homography matrix H is expressed as h _i=[h _i1h _i2h _i3] ^t, namely obtain following expression formula:

$h_i^{T}B{h}_j=v_{\mathrm{ij}}^{T}b---\left(7\right)$

In formula,

v _ij＝[h _i1h _j1h _i1h _j2+h _i2h _j1h _i2h _j2h _i3h _j1+h _i1h _j3h _i3h _j2+h _i2h _j3h _i3h _j3] ^T

Step g, turns to the form about b by (3) (4) formula, as follows:

$\left[\begin{array}{c}{v}_{12}^T\\ v_{11}^T-v_{22}^T\end{array}\right]b=0---\left(8\right)$

Then for the image of N width template, obtain expression formula as follows:

vb＝0 (9)

In formula, v is the matrix of 2N × 6, and when N >=3, b just can be solved, thus can solve the inside and outside parameter of video camera.

Step 3.2, obtains the projection matrix M 1 of left video camera, the projection matrix M 2 of right video camera according to the parameter matrix of left video camera and right video camera respectively;

Wherein, the projection matrix M 1 of left video camera is:

$M_1=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]---\left(10\right)$

In formula, be respectively the i-th row and the j column element of left video camera projection matrix M 1;

The projection matrix M 2 of right video camera is:

$M_2=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]---\left(11\right)$

In formula, be respectively the i-th row and the j column element of right video camera projection matrix M 2;

Step 3.3, ask for the three-dimensional coordinate of gauge point Q1 under world coordinate system according to the projection matrix M 2 of left video camera projection matrix M 1 and right video camera respectively, concrete grammar is as follows:

$Z_{c1}\left[\begin{array}{c}{u}_1\\ v_1\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]---\left(12\right)$

$Z_{c2}\left[\begin{array}{c}{u}_2\\ v_2\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]---\left(13\right)$

In formula, (u ₁, v ₁, 1) and (u ₂, v ₂, 1) and be respectively Q _l1with Q _r1homogeneous coordinates; (x ₁, y ₁, z ₁, 1) and be gauge point Q ₁homogeneous coordinates, Z _c1and Z _c2scale-up factor respectively;

Ask for the three-dimensional coordinate of gauge point Q2 under world coordinate system according to the projection matrix M 2 of left video camera projection matrix M 1 and right video camera respectively, concrete grammar is as follows:

$Z_{c1}\left[\begin{array}{c}{u_1}^{\′}\\ {v_1}^{\′}\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]---\left(14\right)$

$Z_{c2}\left[\begin{array}{c}{u_2}^{\′}\\ {v_2}^{\′}\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]---\left(15\right)$

The projection matrix M 2 of left video camera projection matrix M 1 and right video camera asks for the three-dimensional coordinate of gauge point Q3 under world coordinate system, and concrete grammar is as follows:

$Z_{c1}\left[\begin{array}{c}{u_1}^{\′\′}\\ {v_1}^{\′\′}\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]---\left(16\right)$

$Z_{c2}\left[\begin{array}{c}{u_2}^{\′\′}\\ {v_2}^{\′\′}\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]---\left(17\right)$

Step 3.4, by the formula 12 in step 3.3 and formula 13 simultaneous solution, thus cancellation scale-up factor Z _c1with scale-up factor Z _c2cancellation, obtains gauge point Q ₁expression formula:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(18\right)$

By to the formula 14 in step 3.3 and formula 15 simultaneous solution, thus cancellation scale-up factor Z _c1with scale-up factor Z _c2, obtain gauge point Q ₂expression formula:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(19\right)$

By to the formula 16 in step 3.3 and formula 17 simultaneous solution, thus cancellation scale-up factor Z _c1with scale-up factor Z _c2, obtain gauge point Q ₃expression formula:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(20\right)$

Step 3.5, utilizes the formula 9, formula 10 and the formula 11 that obtain in least square method solution procedure 3.4 respectively, obtains gauge point Q1, gauge point Q2 and the three-dimensional coordinate of gauge point Q3 under world coordinate system;

Step 4, according to the coordinate of any two gauge points obtained in step 3, is converted to the operation relation between vector by the geometric relationship of turning axle and universal stage plane, thus asks for the direction of turning axle and the central coordinate of circle of universal stage, turning axle is designated as R;

Asking for the axial method of rotation is: any two gauge points chosen on scaling board can form three vectors, due to turning axle and universal stage plane orthogonal, so can think that turning axle is the normal direction of universal stage plane, namely turning axle vector is all vertical with the vector of three in universal stage plane, can obtain turning axle vector; If turning axle vector is R (a', b', c');

Such as: choose wherein two the gauge point composition of vector Q on universal stage ₁q ₂with vectorial Q ₂q ₃, then the first expression formula obtaining turning axle vector is:

R＝Q ₁Q ₂×Q ₂Q ₃(21)

First expression formula of turning axle vector is launched:

R＝(x ₂-x ₁,y ₂-y ₁,z ₂-z ₁)×(x ₃-x ₂,y ₃-y ₂,z ₃-z ₂) (24)

Can obtain three coordinate figure a' about turning axle vector according to the first expression formula of the turning axle vector launched, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_2-y_1& z_2-z_1\\ y_3-y_2& z_3-z_2\end{array}\right|---\left(27\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_2-z_1& x_2-x_1\\ z_3-z_2& x_3-x_2\end{array}\right|---\left(28\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_2-x_1& y_2-y_1\\ x_3-x_2& y_3-y_2\end{array}\right|---\left(29\right)$

Such as: choose wherein two the gauge point composition of vector Q on universal stage ₂q ₁with vectorial Q ₁q ₃, then the second expression formula obtaining turning axle vector is:

R＝Q ₂Q ₁×Q ₁Q ₃(22)

Second expression formula of turning axle vector is launched:

R＝(x ₁-x ₂,y ₁-y ₂,z ₁-z ₂)×(x ₃-x ₁,y ₃-y ₁,z ₃-z ₁) (25)

Can obtain three coordinate figure a' about turning axle vector according to the second expression formula of the turning axle vector launched, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_1-y_2& z_1-z_2\\ y_3-y_1& z_3-z_1\end{array}\right|---\left(30\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_1-z_2& x_1-x_2\\ z_3-z_1& x_3-x_1\end{array}\right|---\left(31\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_1-x_2& y_1-y_2\\ x_3-x_1& y_3-y_1\end{array}\right|---\left(32\right)$

Such as: choose wherein two the gauge point composition of vector Q on universal stage ₁q ₃with vectorial Q ₃q ₂, then the 3rd expression formula obtaining turning axle vector is:

R＝Q ₁Q ₃×Q ₃Q ₂(23)

3rd expression formula of turning axle vector is launched:

R＝(x ₃-x ₁,y ₃-y ₁,z ₃-z ₁)×(x ₂-x ₃,y ₂-y ₃,z ₂-z ₃) (26)

Can obtain three coordinate figure a' about turning axle vector according to the 3rd expression formula of the turning axle vector launched, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_3-y_1& z_3-z_1\\ y_2-y_3& z_2-z_3\end{array}\right|---\left(33\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_3-z_1& x_3-x_1\\ z_2-z_3& x_2-x_3\end{array}\right|---\left(34\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_3-x_1& y_3-y_1\\ x_2-x_3& y_2-y_3\end{array}\right|---\left(35\right)$

Three expression formulas about a', b', c' obtained above be all normalized, the normalized vector R (a, b, c) obtaining turning axle R is:

$a=\frac{a^{\′}}{\stackrel{\‾}{a^{\′2}+b^{\′2}+c^{\′2}}}---\left(36\right)$

$b=\frac{b^{\′}}{\stackrel{\‾}{a^{\′2}+b^{\′2}+c^{\′2}}}---\left(37\right)$

$c=\frac{c^{\′}}{\stackrel{\‾}{a^{\′2}+b^{\′2}+c^{\′2}}}---\left(38\right)$

Note universal stage central coordinate of circle is O (x, y, z), must cross this character of the center of circle and ask for the circular coordinate of universal stage, choose wherein two the gauge point composition of vector Q on universal stage below according to the perpendicular bisector of the line of the upper any two points of circle ₁q ₂with vectorial Q ₂q ₃be described in detail to the solution procedure of circular coordinate, the gauge point in note universal stage plane and the line in the center of circle are Q ₁o, Q ₂o and Q ₃o, then obtain expression:

Q ₁Q ₂·m ₁O＝0 (39)

Q ₂Q ₃·m ₂O＝0 (40)

In formula, m ₁o is the perpendicular bisector of the line of gauge point Q1 and gauge point Q2 in universal stage plane, m ₂o is the perpendicular bisector of the line of gauge point Q2 and gauge point Q3 in universal stage plane;

Again according to line and its planar process vector normal of justifying a bit upper and the center of circle, obtain expression formula:

Q ₁O·R＝0 (41)

Step b, obtains expression formula by arranging 3 expression formulas obtained in step a:

$\left[\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_2& y_3-y_2& z_3-z_2\\ a& b& c\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}[x_2^2+y_2^2+z_2^2-(x_1^2+y_1^2+z_1^2)]\\ \frac{1}{2}[x_3^2+y_3^2+z_3^2-(x_2^2+y_2^2+z_2^2)]\\ {\mathrm{ax}}_1+{\mathrm{by}}_1+{\mathrm{cz}}_1\end{array}\right]---\left(42\right)$

Step c, by solving the expression formula obtained in step b, then obtains universal stage central coordinate of circle O (x, y, z).

Below by experiment, the rotating axis calibration method under binocular vision of the present invention is verified.

For " the commodity excessive packaging intelligent checking system based on three-dimensional modeling is studied ", first, select gridiron pattern as scaling board, be placed on turntable, binocular camera is fixed, it can be measured in a very large corner, measure without dead angle apart from 360 degree can be carried out time near, needing when testing to select multiple angle point for demarcating, this is done to the noise effect eliminating data.For one group of data, if standard deviation or its data fluctuations of the less explanation of variance less, also just mean that the precision of demarcation is higher.Table 1 is turning axle nominal data, and wherein, a, b, c represent three directions of rotating shaft R (a, b, c), and x, y, z represents three coordinates of center of circle O (x, y, z).The data of table 1 have carried out six times to test the result drawn, can see that standard deviation is close to 0, illustrative experiment data are more stable, and degree of accuracy is higher.

Table 1 turning axle nominal data

Testing data	Mean value	Standard deviation
			a	-0.852722	0.002533
b	-0.085351	0.001829
			c	0.512872	0.008229
x	-15.234195	0.019164
			y	31.519875	0.068249
z	16.150922	0.043606

Claims (5)

1. the rotating axis calibration method under binocular vision, is characterized in that, comprises the following steps:

Step 1, is placed in universal stage in the visual field of two video cameras, and puts on universal stage by scaling board, makes universal stage scaling board entirety before and after rotation all be in the common field range of two video cameras; Two video cameras are designated as left video camera and right video camera respectively;

Step 2, utilizes left video camera and right video camera to detect the angle point of scaling board simultaneously, and a selected distance universal stage center angle point is farthest designated as gauge point Q ₁(x ₁, y ₁, z ₁); Universal stage is at least rotated two angles successively, and selected distance universal stage center two angle points farthest are also designated as gauge point Q respectively ₂(x ₂, y ₂, z ₂) and gauge point Q ₃(x ₃, y ₃, z ₃), above-mentioned 3 gauge points are all positioned in same level; Wherein the coordinate of gauge point Q1 in left video camera and right video camera is designated as Q respectively _l1(u ₁, v ₁) and Q _r1(u ₂, v ₂), the coordinate of gauge point Q2 in left video camera and right video camera is designated as Q respectively _l2(u ₁', v ₁') and Q _r2(u ₂', v ₂'), the coordinate of gauge point Q3 in left video camera and right video camera is designated as Q respectively _l3(u ₁", v ₁") and Q _r3(u ₂", v ₂");

Step 3, asks for the three-dimensional coordinate of 3 gauge points under world coordinate system respectively;

Step 4, according to the three-dimensional coordinate of three gauge points obtained in step 3, asks for the direction of turning axle and the central coordinate of circle of universal stage, thus completes the demarcation of countershaft.

2. the rotating axis calibration method under a kind of binocular vision according to claim 1, is characterized in that, in described step 1, scaling board is gridiron pattern.

3. the rotating axis calibration method under a kind of binocular vision according to claim 1, it is characterized in that, described step 3 is specially:

Step 3.1, demarcates left video camera and right video camera respectively, obtains the parameter matrix of left video camera and the parameter matrix of right video camera;

Step 3.2, obtains the projection matrix M 1 of left video camera, the projection matrix M 2 of right video camera according to the parameter matrix of left video camera and right video camera respectively;

Wherein, the projection matrix M 1 of left video camera is designated as:

$M_1=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]---\left(10\right)$

In formula, be respectively the i-th row and the jth column element of left video camera projection matrix M 1;

The projection matrix M 2 of right video camera is designated as:

$M_2=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]---\left(11\right)$

In formula, be respectively the i-th row and the jth column element of right video camera projection matrix M 2;

Step 3.3, asks for gauge point Q1, Q2, Q3 three-dimensional coordinate under world coordinate system according to the projection matrix M 2 of left video camera projection matrix M 1 and right video camera respectively;

The described concrete grammar asking for the three-dimensional coordinate of gauge point Q1 under world coordinate system is as follows:

$Z_{c1}\left[\begin{array}{c}{u}_1\\ v_1\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]---\left(12\right)$

$Z_{c2}\left[\begin{array}{c}{u}_2\\ v_2\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\\ 1\end{array}\right]---\left(13\right)$

In formula, (u ₁, v ₁, 1) and (u ₂, v ₂, 1) and be respectively Q _l1with Q _r1homogeneous coordinates; (x ₁, y ₁, z ₁, 1) and be gauge point Q ₁homogeneous coordinates, Z _c1and Z _c2scale-up factor respectively;

The described concrete grammar asking for the three-dimensional coordinate of gauge point Q2 under world coordinate system is as follows:

$Z_{c1}\left[\begin{array}{c}{u_1}^{\′}\\ {v_1}^{\′}\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]---\left(14\right)$

$Z_{c2}\left[\begin{array}{c}{u_2}^{\′}\\ {v_2}^{\′}\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\\ 1\end{array}\right]---\left(15\right)$

The described concrete grammar asking for the three-dimensional coordinate of gauge point Q3 under world coordinate system is as follows:

$Z_{c1}\left[\begin{array}{c}{u_1}^{\′\′}\\ {v_1}^{\′\′}\\ 1\end{array}\right]=M_1\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^1& m_{12}^1& m_{13}^1& m_{14}^1\\ m_{21}^1& m_{22}^1& m_{23}^1& m_{24}^1\\ m_{31}^1& m_{32}^1& m_{33}^1& m_{34}^1\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]---\left(16\right)$

$Z_{c2}\left[\begin{array}{c}{u_2}^{\′\′}\\ {v_2}^{\′\′}\\ 1\end{array}\right]=M_2\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]=\left[\begin{array}{cccc}{m}_{11}^2& m_{12}^2& m_{13}^2& m_{14}^2\\ m_{21}^2& m_{22}^2& m_{23}^2& m_{24}^2\\ m_{31}^2& m_{32}^2& m_{33}^2& m_{34}^2\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\\ 1\end{array}\right]---\left(17\right)$

Step 3.4, by solving the expression formula obtained in step 3.3, obtaining the expression formula of 3 gauge points, being specially:

By to formula 12 and formula 13 simultaneous solution, the expression formula obtaining gauge point Q1 is:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(18\right)$

By to formula 14 and formula 15 simultaneous solution, obtain the expression formula of gauge point Q2:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(19\right)$

By to formula 16 and formula 17 simultaneous solution, the expression formula obtaining gauge point Q3 is:

$\left[\begin{array}{ccc}{u}_1{m}_{31}^1-m_{11}^1& u_1{m}_{32}^1-m_{12}^1& u_1{m}_{33}^1-m_{13}^1\\ v_1{m}_{31}^1-m_{21}^1& v_1{m}_{31}^1-m_{22}^1& v_1{m}_{31}^1-m_{23}^1\\ u_2{m}_{31}^2-m_{11}^2& u_2{m}_{32}^2-m_{12}^2& u_2{m}_{33}^2-m_{13}^2\\ v_2{m}_{31}^2-m_{21}^2& v_2{m}_{32}^2-m_{22}^2& v_1{m}_{33}^3-m_{23}^2\end{array}\right]\left[\begin{array}{c}{x}_3\\ y_3\\ z_3\end{array}\right]=\left[\begin{array}{c}{m}_{14}^1-u_1{m}_{34}^1\\ m_{24}^1-v_1{m}_{34}^1\\ m_{14}^2-u_2{m}_{34}^2\\ m_{24}^2-v_2{m}_{34}^2\end{array}\right]---\left(20\right)$

Step 3.5, utilizes the formula 9, formula 10 and the formula 11 that obtain in least square method solution procedure 3.4 respectively, obtains gauge point Q1, gauge point Q2 and the three-dimensional coordinate of gauge point Q3 under world coordinate system.

4. the rotating axis calibration method under a kind of binocular vision according to claim 1, is characterized in that, asks for turning axle direction and be specially in described step 4:

Steps A, chooses two gauge point composition of vector Q on scaling board ₁q ₂, according to turning axle vector R (a', b', c') and the vector normal in universal stage plane, the first expression formula obtaining turning axle vector is:

R＝Q ₁Q ₂×Q ₂Q ₃(21)

Or wherein two the gauge point composition of vector Q chosen on universal stage ₂q ₃, the second expression formula obtaining turning axle vector is:

R＝Q ₂Q ₁×Q ₁Q ₃(22)

Or wherein two the gauge point composition of vector Q chosen on universal stage ₁q ₃, the 3rd expression formula obtaining turning axle vector is:

R＝Q ₁Q ₃×Q ₃Q ₂(23)

Step B, launches the first expression formula of turning axle vector:

R＝(x ₂-x ₁,y ₂-y ₁,z ₂-z ₁)×(x ₃-x ₂,y ₃-y ₂,z ₃-z ₂) (24)

Or the second expression formula of turning axle vector is launched:

R＝(x ₁-x ₂,y ₁-y ₂,z ₁-z ₂)×(x ₃-x ₁,y ₃-y ₁,z ₃-z ₁) (25)

Or the 3rd expression formula of turning axle vector is launched:

R＝(x ₃-x ₁,y ₃-y ₁,z ₃-z ₁)×(x ₂-x ₃,y ₂-y ₃,z ₂-z ₃) (26)

Step C, can obtain three coordinate figure a' about turning axle vector according to the first expression formula of the turning axle vector launched in step B, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_2-y_1& z_2-z_1\\ y_3-y_2& z_3-z_2\end{array}\right|---\left(27\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_2-z_1& x_2-x_1\\ z_3-z_2& x_3-x_2\end{array}\right|---\left(28\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_2-x_1& y_2-y_1\\ x_3-x_2& y_3-y_2\end{array}\right|---\left(29\right)$

Or the coordinate a' about turning axle vector can be obtained according to the second expression formula of the turning axle vector launched in step B, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_1-y_2& z_1-z_2\\ y_3-y_1& z_3-z_1\end{array}\right|---\left(30\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_1-z_2& x_1-x_2\\ z_3-z_1& x_3-x_1\end{array}\right|---\left(31\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_1-x_2& y_1-y_2\\ x_3-x_1& y_3-y_1\end{array}\right|---\left(32\right)$

Or the coordinate a' about turning axle vector can be obtained according to the 3rd expression formula of the turning axle vector launched in step B, the expression formula of b', c' is:

$a^{\′}=\left|\begin{array}{cc}{y}_3-y_1& z_3-z_1\\ y_2-y_3& z_2-z_3\end{array}\right|---\left(33\right)$

$b^{\′}=\left|\begin{array}{cc}{z}_3-z_1& x_3-x_1\\ z_2-z_3& x_2-x_3\end{array}\right|---\left(34\right)$

$c^{\′}=\left|\begin{array}{cc}{x}_3-x_1& y_3-y_1\\ x_2-x_3& y_2-y_3\end{array}\right|.---\left(35\right)$

5. the rotating axis calibration method under a kind of binocular vision as claimed in any of claims 1 to 4, it is characterized in that, asking for the circular coordinate of universal stage in described step 4 is optional two gauge point composition of vector in universal stage plane, and asks for the circular coordinate of universal stage according to the principle that the perpendicular bisector of the line of the upper any two points of circle must cross the center of circle.