CN105157725A - Hand-eye calibration method employing two-dimension laser vision sensor and robot - Google Patents

Abstract

A disclosed hand-eye calibration method employing a two-dimension laser vision sensor and a robot comprises the following steps: step A, establishing a mathematic module of a calibration algorithm; step B, making enforcement operation step for hand-eye calibration. The step of establishing the mathematic module of the calibration algorithm realizes derivation of the mathematic module of the calibration algorithm; and the step of making the enforcement operation step for hand-eye calibration solves the concrete enforcement operation flow in the h and-eye calibration process by employing the two-dimension laser vision sensor and the robot. The method possesses the advantages of being simple, practical, agile, good in precision and the like.

Description

The hand and eye calibrating method of a kind of two-dimensional laser vision sensor and robot

Technical field

The present invention relates to the hand and eye calibrating technology of a kind of laser sensor and robot, in particular to the hand and eye calibrating method of a kind of two-dimensional laser vision sensor and robot, this hand and eye calibrating method becomes appearance Motor execution device with the artificial machinery of industrial machine, with industrial robot self and two-dimensional laser sensor for measurement mechanism, robot becomes appearance and drives laser sensor to measure some spaces point of fixity, the attitude of difference recorder people and the coordinate figure of this spatial point in basis coordinates system of robot and laser sensor surving coordinate system, by the transformation relation between certain Algorithm for Solving sensor measurement coordinate system and robot end's flange coordinate system, namely the hand and eye calibrating of two-dimensional laser sensor is carried out.

Background technology

Because vision system has good detection perform and positioning performance, so robotic vision system develops focus and emphasis that oneself becomes robot research field.The informative of vision-sensing method because obtaining, and high sensitivity, high precision, with the advantage such as workpiece is contactless, and to be more and more subject to people's attention.

At present, the image of visual sensing collection has based on the image of natural light, artificial common light and take laser as the structure light image of active light source.In the industrial environment that some is special, such as there is the bad disturbing factors such as strong arc light, dust, smog at welding scene, the performance of traditional C CD camera receives comparatively serious interference, and CCD camera traditional in such a case just can not be finished the work well, poor practicability.By contrast, described two-dimensional laser sensor is based on principle of triangulation, object section profile measurement is carried out by linear beam laser, adopt and filter all parasitic lights comprising arc light with laser with the optical filter of equiwavelength, the integrated optical receiver assembly of sensor internal, CMOS area detector receive only and form the image of laser stripe.The advantage of this two-dimensional laser sensor does not adopt any portable parts, sturdy and durable, not by interference such as arclight, flue dust, splashings.

The advantages such as laser has high-energy, high brightness as active light source, monochromaticity is good, antijamming capability is strong, therefore two-dimensional laser vision sensor has very large development prospect.CCD camera is face battle array vision sensor, and described two-dimensional laser sensor is linear array vision sensor.Machine vision, undoubtedly can the flexibility efficiency of significantly hoisting machine people operation as one of the core technology of detection field and artificial intelligence field.Wherein, mapping relations between visual coordinate system and robot end's joint coordinate system must be known by hand and eye calibrating, the precision of demarcating determines the homework precision of robot to a great extent, for this demarcates vision system, the vision system demarcated can obtain higher precision becomes the technical issues that need to address.

Summary of the invention

The object of the invention is to overcome the shortcoming of prior art and deficiency, the hand and eye calibrating method of a kind of two-dimensional laser vision sensor and robot is provided, this hand and eye calibrating method comprises the mathematical model setting up calibration algorithm, the implementation and operation step formulating hand and eye calibrating, has the features such as easy, practical, flexible, precision is good.

Object of the present invention is achieved through the following technical solutions: the hand and eye calibrating method of a kind of two-dimensional laser vision sensor and robot, this hand and eye calibrating method adopts an inside to be integrated with optical receiver assembly, the two-dimensional laser sensor of CMOS area detector and robot (containing robot controller, teach box), wherein two-dimensional laser sensor is fixedly mounted on robot end's flange by support, forms Eye-in-hand hand-eye system; Note robot basis coordinates be Base}, robot the 6th joint end flange coordinate be that { End}, two-dimensional laser sensor measurement coordinate are that { M}, the object of hand and eye calibrating is and solves coordinate system { M} is relative to the coordinate system { transformation matrix of End}

The linear beam that the two-dimensional laser sensor that the method adopts sends project measured object on the surface time, laser beam can form the image consistent with measured object surface profile, this laser beam has a series of continuous, uniform P laser sampling point, and then sensor returns this P sampled point relative to the sensor measurement coordinate system { Z of M} _maxle and X _maxial coordinate value;

The method also adopts computing machine to obtain the algorithm computing of the measurement data of two-dimensional laser sensor, the typing completing nominal data, execution demarcation;

The method also needs to adopt some miscellaneous parts, such as: scaling board.

Described two-dimensional laser vision sensor and the hand and eye calibrating method of robot, comprise the following steps:

Steps A, set up the mathematical model of calibration algorithm;

The implementation and operation step of step B, formulation hand and eye calibrating;

Described steps A comprises the following steps:

A1) 1, space P is obtained ₁in basis coordinates system of robot, { coordinate in Base} is ^bp ₁, ^bp ₁=[ ¹x ₁, ¹y ₁, ¹z ₁, 1] ^t, acquisition point P ₁at two-dimensional laser sensor measurement coordinate system, { coordinate in M} is ^mp ₁, ^mp ₁=[ ^mx ₁, ^my ₁, ^mz ₁, 1] ^t.Note trick matrix, namely { relative to robot end's flange coordinate system, { transformation matrix of End} is M} coordinate system $T_M^{End}=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{32}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right],$ { relative to coordinate system, { transformation relation of Base} is End} note coordinate system

A2) according to the transformation relation between coordinate system, known by equation two ends premultiplication matrix inverse, obtain:

${\left(T_{End}^{Base}\right)}^{-1}{P_B}_1=T_M^{End}{P_M}_1,---\left(1\right)$

In formula, for coordinate system End} relative to coordinate system the transformation matrix of Base} inverse, ^bp is 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Note $T={\left(T_{End}^{Base}\right)}^{-1}{P_B}_1={\[c_1,d_1,e_1,1\]}^T,$ Obtain:

$T{P_B}_1=T_M^{End}{P_M}_1,---\left(2\right)$

In formula, T be column matrix, ^bp ₁for 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Formula (2) is launched:

$\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right],---\left(3\right)$

In formula, $\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]$ For column matrix, $\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]$ For coordinate system M} relative to robot end's flange coordinate system the concrete form of the transformation matrix of End}, $\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right]$ For column matrix, represent some P ₁at two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Formula (3) is launched further obtain:

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{21}{x_M}_1+r_{23}{z_M}_1+\mathrm{\Δ}y=d_1\\ r_{31}{x_M}_1+r_{33}{z_M}_1+\mathrm{\Δ}z=e_1\end{array}\right.,---\left(4\right)$

In formula, the implication of variable is see formula (3);

A3) space point P is similar to ₁, in like manner, for space point P ₂, P ₃, have:

$\left\{\begin{array}{c}{r}_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{21}{x_M}_2+r_{23}{z_M}_2+\mathrm{\Δ}y=d_2\\ r_{31}{x_M}_2+r_{33}{z_M}_2+\mathrm{\Δ}z=e_2\end{array}\right.,---\left(5\right)$

$\left\{\begin{array}{c}{r}_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\\ r_{21}{x_M}_3+r_{23}{z_M}_3+\mathrm{\Δ}y=d_3\\ r_{31}{x_M}_3+r_{33}{z_M}_3+\mathrm{\Δ}z=e_3\end{array}\right.,---\left(6\right)$

Formula (5) and symbol implication in formula (6) are see formula (4);

Can be obtained by formula (4), formula (5), formula (6):

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\end{array}\right.,---\left(7\right)$

In formula, symbol implication is see formula (3)---formula (6);

Formula (7) can be write as following form:

$\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(8\right)$

To convert further:

$\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(9\right)$

In formula (8), formula (9), symbol implication is see formula (7);

So far, r has been solved ₁₁, r ₁₃, Δ x;

In like manner, can be obtained by formula (4), formula (5), formula (6):

$\left[\begin{array}{c}{r}_{21}\\ r_{23}\\ \mathrm{\Δ}y\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right],---\left(10\right)$

$\left[\begin{array}{c}{r}_{31}\\ r_{33}\\ \mathrm{\Δ}z\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{e}_1\\ e_2\\ e_3\end{array}\right],---\left(11\right)$

In formula, ${\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}$ For $\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]$ Inverse.

So far, r has been solved ₁₁, r ₁₃, Δ x and r ₂₁, r ₂₃, Δ y and r ₃₁, r ₃₃, Δ z;

A4) to determine trick relational matrix completely also need to solve r ₁₂, r ₂₂, r ₃₂, the character according to attitude matrix:

$\begin{array}{c}{r}_{11}^2+r_{21}^2+r_{31}^2=1\\ r_{12}^2+r_{22}^2+r_{32}^2=1\\ r_{13}^2+r_{23}^2+r_{33}^2=1\end{array},---\left(12\right)$

And it is vectorial be vector of unit length and pairwise orthogonal.

In formula, symbol is coordinate system, and { M} is relative to robot end's flange coordinate system { transformation matrix expression amount of End}.

So have:

R can have been solved by formula (13) ₁₂, r ₂₂, r ₃₂, by the vector solved carry out unitization process, obtain normalized vector, so far solves trick relational matrix

A5) obtained by coordinate pose transformation relation:

${P_B}^{\′}=T_{End}^{Base}T_M^{End}P_M,---\left(14\right)$

In formula, ¹p' is according to calibration result derive the space point P that draws coordinate system the theoretical coordinate value in Base}, ^mp is that { coordinate in M}, will at coordinate system for this point ^bp' with ^bp carries out contrasting the precision just checking hand and eye calibrating, only utilizes spatial point P ₁, P ₂, P ₃the possibility of result carrying out hand and eye calibrating can also exist larger error, in order to improve accuracy and the serious forgiveness of calibration algorithm, choosing N (N>3) individual spatial point, therefrom choosing 3 points, total plant combination, choose a minimum combination of its medial error as calibration result.

Described steps A 2) in, according to the transformation relation between coordinate system, obtain by equation two ends premultiplication matrix inverse, obtain:

${\left(T_{End}^{Base}\right)}^{-1}{P_B}_1=T_M^{End}{P_M}_1,---\left(1\right)$

Note $T={\left(T_{End}^{Base}\right)}^{-1}{P_B}_1={\[c_1,d_1,e_1,1\]}^T,$ Obtain:

$T{P_B}_1=T_M^{End}{P_M}_1,---\left(2\right)$

Formula (2) is launched:

$\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right],---\left(3\right)$

Formula (3) is launched further obtain:

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{21}{x_M}_1+r_{23}{z_M}_1+\mathrm{\Δ}y=d_1\\ r_{31}{x_M}_1+r_{33}{z_M}_1+\mathrm{\Δ}z=e_1\end{array}\right.,---\left(4\right)$

In formula, $\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]$ For column matrix, $\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]$ For coordinate system M} relative to robot end's flange coordinate system the concrete form of the transformation matrix of End}, $\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right]$ For column matrix, represent some P ₁at two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Described steps A 3) in, be similar to space point P ₁, get other space point P ₂, P ₃, in like manner have:

$\left\{\begin{array}{c}{r}_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{21}{x_M}_2+r_{23}{z_M}_2+\mathrm{\Δ}y=d_2\\ r_{31}{x_M}_2+r_{33}{z_M}_2+\mathrm{\Δ}z=e_2\end{array}\right.,---\left(5\right)$

$\left\{\begin{array}{c}{r}_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\\ r_{21}{x_M}_3+r_{23}{z_M}_3+\mathrm{\Δ}y=d_3\\ r_{31}{x_M}_3+r_{33}{z_M}_3+\mathrm{\Δ}z=e_3\end{array}\right.,---\left(6\right)$

Can be obtained by formula (4), formula (5) and formula (6):

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\end{array}\right.,---\left(7\right)$

Write formula (7) as following form:

$\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(8\right)$

To convert further:

$\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(9\right)$

In formula (8), formula (9), symbol implication is see formula (7);

So far, r has been solved ₁₁, r ₁₃, Δ x;

In like manner, can be obtained by formula (4), formula (5), formula (6):

$\left[\begin{array}{c}{r}_{21}\\ r_{23}\\ \mathrm{\Δ}y\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right],---\left(10\right)$

$\left[\begin{array}{c}{r}_{31}\\ r_{33}\\ \mathrm{\Δ}z\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{e}_1\\ e_2\\ e_3\end{array}\right],---\left(11\right)$

In formula, ${\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}$ For $\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]$ Inverse;

So far, r has been solved ₁₁, r ₁₃, Δ x, r ₂₁, r ₂₃, Δ y, r ₃₁, r ₃₃with Δ z.

Described step B comprises the following steps:

B1) scaling board is put in space a certain suitable position, manipulation robot makes to be arranged on laser rays that the two-dimensional laser sensor emission on its end flange goes out and to be incident upon on scaling board on certain point, be designated as a P, according to scaling board and geometric relationship between laser light photosensor read now P point at the laser sensor surving coordinate system { coordinate figure in M} ^mp; Keep the motionless reading of robot and write down now robot end's flanged tool coordinate system { End} is relative to basis coordinates system of the robot { pose of Base}

B2) TCP of manipulation robot tool coordinates system that another one has been established arrives described P point, reads P point at basis coordinates system of the robot { coordinate in Base} ^bp.By what obtain ^mp, ^bp is as one group of nominal data;

B3) repeat step B1), B2), obtain the nominal data that N group is different;

B4) by step B3) in obtain N group nominal data input calibrating procedure, draw hand and eye calibrating result by computer calculate

The present invention has following advantage and effect relative to prior art:

1, of the present invention practical, use flexible, easy, hand and eye calibrating precision is high.

2, the hand-eye system being applicable to two-dimensional laser vision sensor and robot and forming of the present invention, utilizes laser vision sensor to replace traditional C CD vision well, meets the application demand of robot under special occasions.

Accompanying drawing explanation

Fig. 1 a obtains space calibration point at sensor measurement coordinate system internal coordinate and robot current pose schematic diagram in the present invention.

Fig. 1 b is the partial enlarged drawing of laser sensor.

Fig. 2 obtains space calibration point at basis coordinates system of robot internal coordinate schematic diagram in the present invention.

Embodiment

Below in conjunction with embodiment and accompanying drawing, the present invention is described in further detail, but embodiments of the present invention are not limited thereto.

Embodiment

Install last two-dimensional laser sensor by mounting bracket at robot (containing robot controller, teach box) end flange, this sensor communicates with computing machine, the measurement data that computing machine receiving sensor returns.

The laser beam projects that two-dimensional laser sensor sends to measured object on the surface time, laser beam can form the image consistent with measured object surface profile, this laser beam has a series of continuous, uniform P laser sampling point, and then sensor returns P sampled point in this Shu Jiguang relative to the Z axis in sensor measurement coordinate system and X-axis coordinate figure.

In conjunction with scaling board, utilize laser sensor and robot to obtain hand and eye calibrating desired data.The method also adopts computing machine to obtain the algorithm computing of the measurement data of two-dimensional laser sensor, the typing completing nominal data, execution demarcation, utilizes these data to calculate and solves trick relation.

This annex of scaling board is additionally used in the present embodiment.

As illustrated in figs. ia and ib, described steps A (setting up the mathematical model of calibration algorithm) comprises the following steps:

A1) 1, space P is obtained ₁in basis coordinates system of robot, { coordinate in Base}1 is ^bp ₁, ^bp ₁=[ ¹x ₁, ¹y ₁, ¹z ₁, 1] ^t, acquisition point P ₁in the surving coordinate system of two-dimensional laser sensor, { coordinate in M}2 is ^mp ₁, ^mp ₁=[ ^mx ₁, ^my ₁, ^mz ₁, 1] ^t.Note trick matrix, namely { relative to robot end's flange coordinate system, { transformation matrix of End}3 is M} coordinate system { relative to coordinate system, { transformation relation of Base} is End} note coordinate system

In formula, for coordinate system End} relative to coordinate system the transformation matrix of Base} inverse, ^bp is 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Wherein:

$T_M^{End}=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{32}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]$ In 12 variablees be all unknown; also be a 4x4 matrix.

Its value is known, directly can read from robot teach box 4 and know.

A2) according to the transformation relation between coordinate system, known by equation two ends premultiplication matrix inverse, obtain:

${\left(T_{End}^{Base}\right)}^{-1}{P_B}_1=T_M^{End}{P_M}_1,---\left(1\right)$

In formula, for coordinate system End} relative to coordinate system the transformation matrix of Base} inverse, ^bp is 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Note $T={\left(T_{End}^{Base}\right)}^{-1}{P_B}_1={\[c_1,d_1,e_1,1\]}^T,$ Obtain:

$T{P_B}_1=T_M^{End}{P_M}_1,---\left(2\right)$

In formula, T be column matrix, ^bp ₁for 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Formula (2) is launched:

$\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right],---\left(3\right)$

In formula, $\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]$ For column matrix, $\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]$ For coordinate system M} relative to robot end's flange coordinate system the concrete form of the transformation matrix of End}, $\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right]$ For column matrix, represent some P ₁at two-dimensional laser sensor measurement coordinate system { coordinate in M}.

Formula (3) is launched further obtain:

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{21}{x_M}_1+r_{23}{z_M}_1+\mathrm{\Δ}y=d_1\\ r_{31}{x_M}_1+r_{33}{z_M}_1+\mathrm{\Δ}z=e_1\end{array}\right.,---\left(4\right)$

In formula, the implication of variable is see formula (3);

A3) space point P is similar to ₁, in like manner, for space point P ₂, P ₃, have:

$\left\{\begin{array}{c}{r}_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{21}{x_M}_2+r_{23}{z_M}_2+\mathrm{\Δ}y=d_2\\ r_{31}{x_M}_2+r_{33}{z_M}_2+\mathrm{\Δ}z=e_2\end{array}\right.,---\left(5\right)$

$\left\{\begin{array}{c}{r}_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\\ r_{21}{x_M}_3+r_{23}{z_M}_3+\mathrm{\Δ}y=d_3\\ r_{31}{x_M}_3+r_{33}{z_M}_3+\mathrm{\Δ}z=e_3\end{array}\right.,---\left(6\right)$

In formula (5), (6), symbol implication is see formula (4);

Can be obtained by formula (4), formula (5), formula (6):

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\end{array}\right.,---\left(7\right)$

In formula, symbol implication is see formula (3)---formula (6);

Formula (7) can be write as following form:

$\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(8\right)$

Inverse of a matrix is multiplied by further by same for equation two ends:

$\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(9\right)$

In formula (8), formula (9), symbol implication is see formula (7);

So far, r has been solved ₁₁, r ₁₃, Δ x;

In like manner, can be obtained by formula (4), formula (5), formula (6):

$\left[\begin{array}{c}{r}_{21}\\ r_{23}\\ \mathrm{\Δ}y\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right],---\left(10\right)$

$\left[\begin{array}{c}{r}_{31}\\ r_{33}\\ \mathrm{\Δ}z\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{e}_1\\ e_2\\ e_3\end{array}\right],---\left(11\right)$

In formula, ${\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}$ For $\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]$ Inverse.

So far, r has been solved ₁₁, r ₁₃, Δ x and r ₂₁, r ₂₃, Δ y and r ₃₁, r ₃₃, Δ z;

A4) to determine trick relational matrix completely also need to solve r ₁₂, r ₂₂, r ₃₂, the character according to attitude matrix:

$\begin{array}{c}{r}_{11}^2+r_{21}^2+r_{31}^2=1\\ r_{12}^2+r_{22}^2+r_{32}^2=1\\ r_{13}^2+r_{23}^2+r_{33}^2=1\end{array},---\left(12\right)$

Vector be vector of unit length and pairwise orthogonal.

In formula, symbol is coordinate system, and { M} is relative to robot end's flange coordinate system { transformation matrix expression amount of End}.

So have:

R can have been solved by formula (13) ₁₂, r ₂₂, r ₃₂, by the vector solved according to with carry out unitization process, obtain unitization vector, so far solves trick relational matrix

A5) obtained by coordinate pose transformation relation:

${P_B}^{\′}=T_{End}^{Base}T_M^{End}P_M,---\left(14\right)$

In formula, ^bp' be according to calibration result derive the space point P that draws coordinate system the theoretical coordinate value in Base}, ^mp is that { coordinate in M}, will at coordinate system for this point ^bp' with ^bp carries out contrasting the precision just checking hand and eye calibrating, only utilizes spatial point P ₁, P ₂, P ₃the possibility of result carrying out hand and eye calibrating can also exist larger error, in order to improve accuracy and the serious forgiveness of calibration algorithm, choosing 5 spatial point, therefrom choosing 3 points, total plant combination, choose a minimum combination of its medial error as calibration result.

Described step B (formulating the implementation and operation step of hand and eye calibrating) comprises the following steps:

B1) as shown in Figure 1, scaling board 5 is put a certain suitable position in space, make to be arranged on by robot teach box 4 manipulation robot laser rays that the two-dimensional laser sensor 2 on its end flange 3 launches and be incident upon a certain P point place on scaling board 5, scaling board 5 is in the range ability of sensor, as can be seen from the geometric relationship in Fig. 1, find out Z in the present sample data that sensor 2 returns _maxle minimum value Z _min{ the Z axis coordinate in M}2, with Z at sensor measurement coordinate system to be P point _mincorresponding X _maxial coordinate value is just for P point is at the coordinate system { X in M} _maxial coordinate, so just have read P point at the laser sensor surving coordinate system { coordinate figure in M}2 ^mp; Keep robot motionless, from robot teach box 4 read and write down represent now robot end's flanged tool coordinate system { End}3 is relative to basis coordinates system of robot { Base}1 pose eulerian angle α, β, γ and offset Δ f _x, Δ f _y, Δ f _z. can be obtained by following formula:

$T_M^{End}=\left[\begin{array}{cccc}c\mathrm{\α}c\mathrm{\β}& c\mathrm{\α}s\mathrm{\β}s\mathrm{\γ}-s\mathrm{\α}c\mathrm{\γ}& c\mathrm{\α}s\mathrm{\β}c\mathrm{\γ}+s\mathrm{\α}s\mathrm{\γ}& {\mathrm{\Δf}}_x\\ s\mathrm{\α}c\mathrm{\β}& s\mathrm{\α}s\mathrm{\β}s\mathrm{\γ}+c\mathrm{\α}c\mathrm{\γ}& s\mathrm{\α}s\mathrm{\β}c\mathrm{\γ}-c\mathrm{\α}s\mathrm{\γ}& {\mathrm{\Δf}}_y\\ -s\mathrm{\β}& c\mathrm{\β}s\mathrm{\γ}& c\mathrm{\β}c\mathrm{\γ}& {\mathrm{\Δf}}_z\\ 0& 0& 0& 1\end{array}\right],$

In formula, for tool coordinates system, { End} is relative to basis coordinates system of robot { Base} position auto―control;

Wherein: s α, c α represent sin α, cos α respectively, all the other symbols by that analogy;

B2) as shown in Figure 2, { TCP of Tool}6 arrives described P point, reads P point at basis coordinates system of the robot { coordinate in Base}1 in the tool coordinates system that manipulation robot makes another one establish ^bp.By what obtain ^mp, ^bp is as one group of nominal data;

B3) repeat step B1), B2), obtain the nominal data that N group is different;

B4) by step B3) in obtain N group nominal data input calibrating procedure, draw hand and eye calibrating result by computing machine 7 according to the computing of calibration algorithm mathematical model

Above-described embodiment is the present invention's preferably embodiment; but embodiments of the present invention are not restricted to the described embodiments; change, the modification done under other any does not deviate from Spirit Essence of the present invention and principle, substitute, combine, simplify; all should be the substitute mode of equivalence, be included within protection scope of the present invention.

Claims (6)

1. a hand and eye calibrating method for two-dimensional laser vision sensor and robot, is characterized in that, comprises the following steps:

Steps A, set up the mathematical model of calibration algorithm;

The implementation and operation step of step B, formulation hand and eye calibrating;

Described step B comprises the following steps:

B1) scaling board is put in space a certain suitable position, manipulation robot makes to be arranged on the laser rays that the two-dimensional laser sensor emission on its end flange goes out and is incident upon P point on scaling board, according to scaling board and geometric relationship between laser light photosensor read now P point at the laser sensor surving coordinate system { coordinate figure in M} ^mp; Keep the motionless reading of robot and write down now robot end's flanged tool coordinate system { End} is relative to basis coordinates system of the robot { pose of Base}

B2) TCP of manipulation robot tool coordinates system that another one has been established arrives described P point, reads P point at basis coordinates system of the robot { coordinate in Base} ^bp, by what obtain ^mp, ^bp is as one group of nominal data;

B3) repeat step B1), B2), until obtain the different nominal data of N group set;

B4) by step B3) in obtain N group nominal data input calibrating procedure, draw hand and eye calibrating result by computer calculate

2. the hand and eye calibrating method of two-dimensional laser vision sensor as claimed in claim 1 and robot, it is characterized in that, described steps A comprises the following steps:

A1) 1, space P is obtained ₁at basis coordinates system of the robot { coordinate in Base} ^bp ₁, ¹p ₁=[ ¹x ₁, ¹y ₁, ¹z ₁, 1] ^t, acquisition point P ₁at two-dimensional laser sensor measurement coordinate system, { coordinate in M} is ^mp ₁, ^mp ₁=[ ^mx ₁, ^my ₁, ^mz ₁, 1] ^t, note trick matrix, namely { relative to robot end's flange coordinate system, { transformation matrix of End} is M} coordinate system $T_M^{End}=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{32}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right],$ { relative to coordinate system, { transformation relation of Base} is End} note coordinate system

A2) according to the transformation relation between coordinate system, obtain:

by equation two ends premultiplication matrix inverse, obtain:

${\left(T_{End}^{Base}\right)}^{-1}{P_B}_1=T_M^{End}{P_M}_1,---\left(1\right)$

In formula, for coordinate system End} relative to coordinate system the transformation matrix of Base} inverse, ^bp is 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M};

Note $T={\left(T_{End}^{Base}\right)}^{-1}{P_B}_1={\[c_1,d_1,e_1,1\]}^T,$ Obtain:

$T{P_B}_1=T_M^{End}{P_M}_1,---\left(2\right)$

In formula, T be column matrix, ^bp ₁for 1, space P ₁basis coordinates system of robot the coordinate in Base}, for coordinate system M} relative to robot end's flange coordinate system the transformation matrix of End}, ^mp ₁for a P ₁at the two-dimensional laser sensor measurement coordinate system { coordinate in M};

Formula (2) is launched to obtain:

$\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]=\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{31}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right],---\left(3\right)$

In formula, $\left[\begin{array}{c}{c}_1\\ d_1\\ e_1\\ 1\end{array}\right]$ For column matrix, $\left[\begin{array}{cccc}{r}_{11}& r_{12}& r_{13}& \mathrm{\Δ}x\\ r_{21}& r_{22}& r_{23}& \mathrm{\Δ}y\\ r_{31}& r_{32}& r_{33}& \mathrm{\Δ}z\\ 0& 0& 0& 1\end{array}\right]$ For coordinate system M} relative to robot end's flange coordinate system the concrete form of the transformation matrix of End}, $\left[\begin{array}{c}{x_M}_1\\ {y_M}_1\\ {z_M}_1\\ 1\end{array}\right]$ For column matrix, represent some P ₁at two-dimensional laser sensor measurement coordinate system { coordinate in M};

Formula (3) is launched further obtain:

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{21}{x_M}_1+r_{23}{z_M}_1+\mathrm{\Δ}y=d_1\\ r_{31}{x_M}_1+r_{33}{z_M}_1+\mathrm{\Δ}z=e_1\end{array}\right.,---\left(4\right)$

A3) for space point P ₂and P ₃, have:

$\left\{\begin{array}{c}{r}_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{21}{x_M}_2+r_{23}{z_M}_2+\mathrm{\Δ}y=d_2\\ r_{31}{x_M}_2+r_{33}{z_M}_2+\mathrm{\Δ}z=e_2\end{array}\right.,---\left(5\right)$

$\left\{\begin{array}{c}{r}_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\\ r_{21}{x_M}_3+r_{23}{z_M}_3+\mathrm{\Δ}y=d_3\\ r_{31}{x_M}_3+r_{33}{z_M}_3+\mathrm{\Δ}z=e_3\end{array}\right.,---\left(6\right)$

Obtained by formula (4), formula (5) and formula (6):

$\left\{\begin{array}{c}{r}_{11}{x_M}_1+r_{13}{z_M}_1+\mathrm{\Δ}x=c_1\\ r_{11}{x_M}_2+r_{13}{z_M}_2+\mathrm{\Δ}x=c_2\\ r_{11}{x_M}_3+r_{13}{z_M}_3+\mathrm{\Δ}x=c_3\end{array}\right.,---\left(7\right)$

Write formula (7) as following form:

$\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]=\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(8\right)$

Formula (7) is converted further:

$\left[\begin{array}{c}{r}_{11}\\ r_{13}\\ \mathrm{\Δ}x\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right],---\left(9\right)$

So far, r has been solved ₁₁, r ₁₃with Δ x;

Obtained by formula (4), formula (5) and formula (6):

$\left[\begin{array}{c}{r}_{21}\\ r_{23}\\ \mathrm{\Δ}y\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{d}_1\\ d_2\\ d_3\end{array}\right],---\left(10\right)$

$\left[\begin{array}{c}{r}_{31}\\ r_{33}\\ \mathrm{\Δ}z\end{array}\right]={\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}\left[\begin{array}{c}{e}_1\\ e_2\\ e_3\end{array}\right],---\left(11\right)$

So far, r has been solved ₁₁, r ₁₃, Δ x, r ₂₁, r ₂₃, Δ y, r ₃₁, r ₃₃with Δ z;

In formula, ${\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]}^{-1}$ For $\left[\begin{array}{ccc}{x_M}_1& {z_M}_1& 1\\ {x_M}_2& {z_M}_2& 1\\ {x_M}_3& {z_M}_3& 1\end{array}\right]$ Inverse;

A4) in order to determine trick relational matrix character according to attitude matrix solves r ₁₂, r ₂₂and r ₃₂:

$r_{11}^2+r_{21}^2+r_{31}^2=1$

$r_{12}^2+r_{22}^2+r_{32}^2=1,---\left(12\right)$

$r_{13}^2+r_{23}^2+r_{33}^2=1$

Vector with be vector of unit length and pairwise orthogonal;

In formula, with be coordinate system M} relative to robot end's flange coordinate system the transformation matrix expression amount of End}, therefore obtains:

R is solved by formula (13) ₁₂, r ₂₂, r ₃₂, by the vector solved with carry out unitization process, obtain normalized with vector, so far, solves trick relational matrix

A5) obtained by coordinate pose transformation relation:

${P_B}^{\′}=T_{End}^{Base}T_M^{End}P_M,---\left(14\right)$

Wherein, ^bp' be according to calibration result derive the above-mentioned space point P that draws coordinate system the theoretical coordinate value in Base}, ^mp is that { coordinate in M}, will at coordinate system for this point ^bp' with ^bp carries out contrasting the precision just checking hand and eye calibrating, only utilizes spatial point P ₁, P ₂, P ₃the possibility of result carrying out hand and eye calibrating can also exist larger error, in order to improve accuracy and the serious forgiveness of calibration algorithm, choosing N (N>3) individual spatial point, therefrom choosing 3 points, total plant combination, choose plant the minimum one combination of combination medial error as calibration result.

3. the hand and eye calibrating method of two-dimensional laser vision sensor as claimed in claim 2 and robot, is characterized in that, described steps A 1) in, obtain 1, space P ₁in basis coordinates system of robot, { coordinate in Base} is ^bp ₁, ^bp ₁=[ ¹x ₁, ¹y ₁, ¹z ₁, 1] ^t, acquisition point P ₁at two-dimensional laser sensor measurement coordinate system, { coordinate in M} is ^mp ₁, ^mp ₁=[ ^mx ₁, ^my ₁, ^mz ₁, 1] ^t.

4. the hand and eye calibrating method of two-dimensional laser vision sensor as claimed in claim 2 and robot, is characterized in that, described steps A 4) in, utilize vector with be vector of unit length and the characteristic of pairwise orthogonal, so have solve r ₁₂, r ₂₂and r ₃₂, in order to reduce calibrated error, by the vector solved with carry out unitization process, obtain normalized with vector, so far, has solved trick relational matrix

5. the hand and eye calibrating method of two-dimensional laser vision sensor as claimed in claim 2 and robot, is characterized in that, described steps A 5) in, obtained by coordinate pose transformation relation:

${P_B}^{\′}=T_{End}^{Base}T_M^{End}P_M,$

Wherein, ^bp' be according to hand and eye calibrating result derive the space point P that draws basis coordinates system of robot the theoretical coordinate value in Base}, ^mp is that { coordinate in M}, will at coordinate system for this point ^bp' with ^bp carries out contrasting the precision just checking hand and eye calibrating, in order to improve accuracy and the serious forgiveness of calibration algorithm, chooses N (N>3) individual spatial point, therefrom chooses at 3 o'clock as one group and substitutes into calibration algorithm mathematical model, total plant combination, choose a minimum combination of its medial error as calibration result.

6. the hand and eye calibrating method of two-dimensional laser vision sensor as claimed in claim 1 and robot, it is characterized in that, described step B1) in, scaling board is put in space a certain suitable position, manipulation robot makes to be arranged on the laser rays that the two-dimensional laser sensor emission on its end flange goes out and is incident upon P point on scaling board, according to scaling board and geometric relationship between laser light photosensor read now P point at the laser sensor surving coordinate system { coordinate figure in M} ^mp.