CN103116892A - Linear solving of video camera internal parameters by utilizing two intersected same circles and common tangent - Google Patents

Linear solving of video camera internal parameters by utilizing two intersected same circles and common tangent Download PDF

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CN103116892A
CN103116892A CN201310082074XA CN201310082074A CN103116892A CN 103116892 A CN103116892 A CN 103116892A CN 201310082074X A CN201310082074X A CN 201310082074XA CN 201310082074 A CN201310082074 A CN 201310082074A CN 103116892 A CN103116892 A CN 103116892A
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point
circle
straight line
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赵越
周炳
钟良宣
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Yunnan University YNU
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Yunnan University YNU
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Abstract

The invention relates to linear solving of video camera internal parameters by utilizing two intersected same circles and a common tangent, and the method is a drone composed of two intersected same circles and the common tangent and used for calibrating video cameras. The drone is shoot for six images from different directions; elliptic equations and characteristic points are extracted from the images, and intersection point of two ovals are solved; vanishing points on image planes at the orthogonal direction is obtained according to perpendicular bisector of circle center line segment connecting the two circles and the common tangent of the two intersected circles and cross ratio of collinear four points; and the video camera internal parameters are performed with linear solving by utilizing the vanishing points at the orthogonal direction and restraint of images of an absolute conic. By utilizing the drone, full-automatic calibration can be achieved, and error caused by measurement during the calibration process is reduced. Circle is an element which is more concise and global, and accordingly calibration accuracy is improved during the calibration process.

Description

Two intersect identical circle and common tangent is found the solution camera intrinsic parameter
Technical field
The invention belongs to the computer research field, relate to a kind of be used to finding the solution camera intrinsic parameter two and intersect identical circle and common tangent target.Utilize on the plane any two to intersect identical circles and common tangent as calibrating template, the straight line quadrature vanishing point coordinate in the public string straight line by finding the solution the circle on six width images and the center of circle by two circles, the linear intrinsic parameter of determining video camera.
Background technology
One of basic task of computer vision recovers the geological information of object three dimensions from the two-dimensional image information that video camera obtains exactly, thus the geometric configuration of object in identification and reconstruction of three-dimensional space.Must determine the three-dimensional geometry position of space object point and the mutual relationship between the corresponding point in its image in this process, and this relation is by the geometric model decision of video camera imaging, the parameter of these geometric models is exactly camera parameters.Under most of conditions, these parameters all obtain by experiment, Here it is camera calibration.It generally is divided into tradition and demarcates and certainly demarcate two kinds of methods, which kind of scaling method no matter, and demarcating object is all to adopt some special geometric models, for example: plane square, triangle, circle, space cube, cylinder etc.How setting up relation certain the linear relation especially between these geometric models and camera parameters, is the target that present camera calibration is pursued, and is also one of focus of present computer vision field research.
Although traditional camera marking method can obtain higher precision, calibrating block is made difficulty, is not easy to operation.For this problem document " A flexible new technique for camera calibration ", (Zhengyou Zhang, IEEE Transactions on Pattern Analysis and Machine Intelligence, vol. 22, no. 11, pp. 1330-1334,2000.) proposed to replace with plane template the method for traditional calibrating block, this method is simple and convenient, cost is low, and can obtain higher precision, but need the physical coordinates of dot matrix on accurate locating template.Document " Planar conic based camera calibration ", (Changjiang Yang, Fengmei Sun, Zhanyi Hu, In Proceedings of International Conference on Pattern Recognition, vol.1, pp. 555-558,2000.) this method has been done popularization, come calibrating camera with the quafric curve correspondence between image and template, rather than utilize correspondence between points.Because quafric curve is that a kind of more succinct more globalize gets primitive, thereby can further improve the stability of method.So solving problem of calibrating with curve is widely studied.document " A new easy camera calibration technique based on circular points ", (Xiaoqiao Meng, Zhanyi Hu, Pattern Recognition, vol. 36, no. 5, pp. 115-1164, 2003.) calibrating template that consists of with a circle and some straight lines by the center of circle proposed, utilize the annulus point to find the solution the method for camera intrinsic parameter, the method is dissolved into the point of the annulus in projective geometry in camera calibration first, so the annulus point has become theoretical foundation (the Hartley Richard of camera self-calibration method, Zisserman Andrew, " Multiple view geometry in computer vision ", Cambridge University Press, Cambridge, 2000.).
Circle is special quafric curve on the plane, and circles all on the plane are all by the annulus point.Utilize circle as calibrating template, carry out the method for camera calibration in conjunction with the theory of annulus point and promoted gradually.Document (Yihong Wu, Haijiang Zhu, Zhanyi Hu, Fuchao Wu, " Camera calibration from the quasi-affine invariance of two parallel circles ", In Proceedings of the ECCV, pp. 190-202,2004.) method of demarcating with parallel circle proposed, complete demarcation according to the intersection point that the intersection point of parallel circle is the picture of two parallel circles of annulus point direct solution.Document (Yihong Wu, Xinju Li, Fuchao Wu, Zhanyi Hu, " Coplanar circle; quasi-affine invariance and calibration ", Image and Vision Computing, vol. 24, no. 4, pp. 319-326,2006.) position relationship of any two circles on the plane has been discussed, try to achieve the picture of annulus point according to the intersection point of quafric curve in the position relationship computed image of circle.Circle is a very special figure, and this paper is the public string and the vertical geometric properties of straight line that passes through two centers of circle, linear solution camera intrinsic parameter that intersects identical circle at two.
Summary of the invention
The invention provides a kind of making simple, widely applicable, the target that is used for finding the solution camera intrinsic parameter of good stability, this target intersects identical circle by two and common tangent forms.In finding the solution the process of camera intrinsic parameter, only need video camera to take 6 width images from different orientation and just can go out 5 camera intrinsic parameters by linear solution.
The present invention adopts following technical scheme:
The present invention is the target that is used for camera self-calibration that is made of two crossing identical circles and common tangent.Concrete step comprises: the pixel coordinate of extract minutiae simulates round picture elliptic equation from the image, and find the solution the intersection point of two ellipses, according to harmonic conjugates, draw on the plane of delineation the two public strings of circle and by the vanishing point on the straight line orthogonal directions in two centers of circle, utilize the constraint linear solution camera intrinsic parameter of the picture of vanishing point on orthogonal directions and absolute conic.
(1) curvilinear equation in the fitted figure picture
Utilize Edge function in the Matlab program to extract the coordinate of image characteristic point, and simulate ellipse in image with least-squares algorithm, obtain each elliptic equation on image.
(2) on the computed image plane intersection point of two ellipses and oval intersection point respectively about the intersection point of the tangential equation of two ellipses
On a plane of world coordinate system, there are two to intersect identical circle and common tangent (as Fig. 1),
Figure 201310082074X100002DEST_PATH_IMAGE001
Be two and intersect identical circle,
Figure 201310082074X100002DEST_PATH_IMAGE002
Be respectively round
Figure 405508DEST_PATH_IMAGE001
The center of circle,
Figure 201310082074X100002DEST_PATH_IMAGE003
It is the intersection point of two crossing circles.Cross a little
Figure 627542DEST_PATH_IMAGE003
Justify respectively
Figure 571227DEST_PATH_IMAGE001
Tangent line, the intersection point of tangent line is respectively
Figure 201310082074X100002DEST_PATH_IMAGE004
, the point
Figure 201310082074X100002DEST_PATH_IMAGE005
Be respectively the round point of contact of outside common tangent on circle, connect With
Figure 623552DEST_PATH_IMAGE003
Intersect at a little
Figure 201310082074X100002DEST_PATH_IMAGE006
, namely Be
Figure 201310082074X100002DEST_PATH_IMAGE007
With
Figure 201310082074X100002DEST_PATH_IMAGE008
The mid point of two line segments; On imaging plane (as Fig. 2), circle
Figure 588414DEST_PATH_IMAGE001
Similarly be oval
Figure 201310082074X100002DEST_PATH_IMAGE009
, the point
Figure 201310082074X100002DEST_PATH_IMAGE010
Picture be respectively
Figure 201310082074X100002DEST_PATH_IMAGE011
, extract oval marginal point with the Edge function of Matlab, use the least square fitting elliptic curve, oval equation
Figure 856322DEST_PATH_IMAGE009
Be expressed as with matrix of coefficients
Figure 201310082074X100002DEST_PATH_IMAGE012
Utilize
Figure 361252DEST_PATH_IMAGE009
Matrix of coefficients try to achieve intersection point
Figure 201310082074X100002DEST_PATH_IMAGE013
Coordinate, cross point
Figure 862772DEST_PATH_IMAGE013
Obtain respectively about
Figure 781050DEST_PATH_IMAGE009
Tangential equation, utilize tangential equation to calculate the intersecting point coordinate of tangent line
Figure 201310082074X100002DEST_PATH_IMAGE014
(3) calculate two circles the center of circle the picture coordinate and
Figure 201310082074X100002DEST_PATH_IMAGE015
With
Figure 196856DEST_PATH_IMAGE008
The mid point of two line segments
Figure 618610DEST_PATH_IMAGE006
The picture coordinate
Extracting the unique point at the point of contact of outside common tangent and two ellipses as (as Fig. 2) on the plane Coordinate, obtain respectively straight line with the Cross function of Matlab
Figure 201310082074X100002DEST_PATH_IMAGE017
Equation and straight line
Figure 201310082074X100002DEST_PATH_IMAGE018
Equation.Straight line
Figure 201310082074X100002DEST_PATH_IMAGE019
And straight line
Figure 201310082074X100002DEST_PATH_IMAGE020
Intersection point for the circle
Figure 201310082074X100002DEST_PATH_IMAGE021
The picture in the center of circle
Figure 201310082074X100002DEST_PATH_IMAGE022
Coordinate, straight line
Figure 201310082074X100002DEST_PATH_IMAGE023
And straight line
Figure 688370DEST_PATH_IMAGE020
Intersection point for the circle The picture in the center of circle
Figure 201310082074X100002DEST_PATH_IMAGE025
Coordinate, straight line
Figure 641414DEST_PATH_IMAGE020
With straight line
Figure 201310082074X100002DEST_PATH_IMAGE026
Intersection point be
Figure 80485DEST_PATH_IMAGE006
The picture of point
Figure 201310082074X100002DEST_PATH_IMAGE027
Coordinate.
(4) calculate the coordinate of the vanishing point on orthogonal directions
At stencil plane (as Fig. 1),
Figure 527385DEST_PATH_IMAGE006
Two circle intersection point line segments
Figure 167445DEST_PATH_IMAGE008
In, and straight line
Figure 794735DEST_PATH_IMAGE008
With straight line Vertically, establish straight line
Figure 444022DEST_PATH_IMAGE028
Infinity point on direction is , straight line
Figure 450156DEST_PATH_IMAGE008
Infinity point on direction is
Figure 201310082074X100002DEST_PATH_IMAGE030
, have
Figure 201310082074X100002DEST_PATH_IMAGE031
,
Figure 201310082074X100002DEST_PATH_IMAGE032
, had on picture plane (as Fig. 2) by cross ratio invariability
Figure 201310082074X100002DEST_PATH_IMAGE033
, Wherein
Figure 201310082074X100002DEST_PATH_IMAGE035
Be straight line
Figure 201310082074X100002DEST_PATH_IMAGE036
The vanishing point of direction,
Figure 201310082074X100002DEST_PATH_IMAGE037
Be straight line
Figure 838281DEST_PATH_IMAGE026
The vanishing point of direction.
(5) find the solution camera intrinsic parameter
Obtain 6 width images, gone out 5 intrinsic parameters, i.e. matrixes of video camera by the vanishing point linear solution of orthogonal directions
Figure 201310082074X100002DEST_PATH_IMAGE038
In 5 variablees, be respectively:
Figure 201310082074X100002DEST_PATH_IMAGE039
Be the distortion factor of image,
Figure 201310082074X100002DEST_PATH_IMAGE040
In image coordinate system
Figure 201310082074X100002DEST_PATH_IMAGE041
Axle and
Figure 201310082074X100002DEST_PATH_IMAGE042
The scale factor of axle, It is the principal point coordinate.
Advantage of the present invention:
(1) this target is made simply, draws two crossing identical circle and common tangents with compasses and gets final product;
(2) physical size of this target do not required, need not to know round position and the world coordinates of each point;
(3) but only need to take from different orientation with video camera 6 width images just linear solution go out 5 intrinsic parameters of video camera.
Description of drawings
Fig. 1 is be used to the target formwork structure schematic diagram of finding the solution camera intrinsic parameter.
Fig. 2 is the target template imaging plane figure that finds the solution camera intrinsic parameter.
Embodiment
A kind of target be used to finding the solution camera intrinsic parameter, it by on the plane two intersect identical circle and common tangent consists of, as Fig. 1.Completing finding the solution of camera intrinsic parameter with this modulation needs through following steps:
(1) curvilinear equation in the fitted figure picture
Utilize Edge function in the Matlab program to extract the coordinate of image characteristic point, and simulate elliptic equation in image with least-squares algorithm, obtain each elliptic equation on image, oval equation
Figure 733293DEST_PATH_IMAGE009
Matrix of coefficients be expressed as
Figure 514167DEST_PATH_IMAGE012
(2) on the computed image plane intersection point of two ellipses and oval intersection point respectively about the intersection point of the tangential equation of two ellipses
On a plane of world coordinate system, exist two to intersect identical circle and common tangent (as Fig. 1),
Figure 201310082074X100002DEST_PATH_IMAGE044
Be two and intersect identical circle,
Figure 702703DEST_PATH_IMAGE002
Be respectively round
Figure 917522DEST_PATH_IMAGE001
The center of circle,
Figure 519404DEST_PATH_IMAGE003
It is the intersection point of two crossing circles.Cross a little Justify respectively Tangent line, the intersection point of tangent line is respectively
Figure 565355DEST_PATH_IMAGE004
, the point
Figure 264320DEST_PATH_IMAGE005
Be respectively the round point of contact of common tangent on circle, connect
Figure 652576DEST_PATH_IMAGE004
With Intersect at a little
Figure 936107DEST_PATH_IMAGE006
, namely
Figure 152063DEST_PATH_IMAGE006
Be
Figure 78430DEST_PATH_IMAGE007
With
Figure 33748DEST_PATH_IMAGE008
The mid point of two line segments; As (as Fig. 2) on the plane, round
Figure 856211DEST_PATH_IMAGE044
Similarly be oval
Figure 654402DEST_PATH_IMAGE009
, the point
Figure 259827DEST_PATH_IMAGE010
Picture be respectively
Figure 194285DEST_PATH_IMAGE011
Utilize Matrix of coefficients
Figure 145241DEST_PATH_IMAGE012
Try to achieve intersection point
Figure 787312DEST_PATH_IMAGE013
Coordinate, cross point
Figure 107435DEST_PATH_IMAGE013
Obtain respectively
Figure 943804DEST_PATH_IMAGE013
About
Figure 451009DEST_PATH_IMAGE009
Tangential equation, utilize tangential equation to calculate the intersecting point coordinate of tangent line
Figure 726132DEST_PATH_IMAGE014
(3) calculate two circles the center of circle the picture coordinate and With
Figure 908032DEST_PATH_IMAGE008
The mid point of two line segments
Figure 777899DEST_PATH_IMAGE006
The picture coordinate
Looking like (as Fig. 2) extract minutiae on the plane
Figure 653451DEST_PATH_IMAGE016
Coordinate, obtain respectively straight line with the Cross function of Matlab
Figure 56488DEST_PATH_IMAGE019
Equation and straight line
Figure 828135DEST_PATH_IMAGE023
Equation, straight line
Figure 450878DEST_PATH_IMAGE019
And straight line Intersection point for the circle The picture in the center of circle
Figure 362836DEST_PATH_IMAGE022
Coordinate, straight line
Figure 207295DEST_PATH_IMAGE023
And straight line Intersection point for the circle
Figure 366061DEST_PATH_IMAGE024
The picture in the center of circle
Figure 853412DEST_PATH_IMAGE025
Coordinate, straight line
Figure 185167DEST_PATH_IMAGE020
With straight line
Figure 201310082074X100002DEST_PATH_IMAGE045
Intersection point be
Figure 347158DEST_PATH_IMAGE006
The picture of point
Figure 201310082074X100002DEST_PATH_IMAGE046
Coordinate.
(4) calculate the coordinate of the vanishing point on orthogonal directions
At stencil plane (as Fig. 1),
Figure 110453DEST_PATH_IMAGE006
It is line segment
Figure 66908DEST_PATH_IMAGE008
In, and straight line
Figure 10593DEST_PATH_IMAGE008
With straight line
Figure 710696DEST_PATH_IMAGE028
Vertically, establish straight line
Figure 298803DEST_PATH_IMAGE028
Infinity point on direction is
Figure 550793DEST_PATH_IMAGE029
, straight line
Figure 90096DEST_PATH_IMAGE008
Infinity point on direction is
Figure 452944DEST_PATH_IMAGE030
, have
Figure 895558DEST_PATH_IMAGE031
,
Figure 52870DEST_PATH_IMAGE032
, had on picture plane (as Fig. 2) by cross ratio invariability
Figure 846514DEST_PATH_IMAGE033
,
Figure 481894DEST_PATH_IMAGE034
Wherein
Figure 44594DEST_PATH_IMAGE035
Be straight line The vanishing point of direction,
Figure 417861DEST_PATH_IMAGE037
Be straight line
Figure 997878DEST_PATH_IMAGE045
The vanishing point of direction.
(5) find the solution camera intrinsic parameter
Obtain 6 width images, gone out 5 intrinsic parameters, i.e. matrixes of video camera by the vanishing point linear solution of orthogonal directions
Figure 274139DEST_PATH_IMAGE038
,
Figure 117461DEST_PATH_IMAGE039
Be the distortion factor of image, In image coordinate system
Figure 502361DEST_PATH_IMAGE041
Axle and
Figure 429865DEST_PATH_IMAGE042
The scale factor of axle,
Figure 240826DEST_PATH_IMAGE043
It is the principal point coordinate.
Embodiment
The present invention proposes and utilize on the plane any two to intersect identical circles and common tangent as the linear intrinsic parameter of determining video camera of target.The experiment module structural representation that the present invention adopts as shown in Figure 1.The below makes more detailed description with an example to embodiment of the present invention:
The experiment pattern that utilizes two camera marking methods that intersect identical circles and common tangent to adopt be on the plane arbitrarily two intersect identical circle and common tangent, as shown in Figure 1.Utilize the method in the present invention that the video camera that is used for experiment is demarcated, concrete steps are as follows:
(1) curvilinear equation in the fitted figure picture
The image resolution ratio that the present invention adopts is 640 * 480 pixels, take from different directions several experiment pictures with video camera, choose 6 width picture comparatively clearly, read in image, utilize function in Matlab to extract the coordinate of image characteristic point, and with each curve in least-squares algorithm fitted figure picture, obtain curvilinear equation.Calculate the ellipse on 6 width images
Figure 201310082074X100002DEST_PATH_IMAGE047
( i=1,2,3,4,5,6), its matrix of coefficients is
Figure 201310082074X100002DEST_PATH_IMAGE048
, as follows:
On the first width image, the matrix of coefficients of two ellipses is respectively:
Figure 201310082074X100002DEST_PATH_IMAGE049
，
；
On the second width image, the matrix of coefficients of two ellipses is respectively:
，
；
On the 3rd width image, the matrix of coefficients of two ellipses is respectively:
，
Figure 201310082074X100002DEST_PATH_IMAGE054
；
On the 4th width image, the matrix of coefficients of two ellipses is respectively:
，
Figure 201310082074X100002DEST_PATH_IMAGE056
；
On the 5th width image, the matrix of coefficients of two ellipses is respectively:
Figure 201310082074X100002DEST_PATH_IMAGE057
，
Figure 201310082074X100002DEST_PATH_IMAGE058
；
On the 6th width image, the matrix of coefficients of two ellipses is respectively:
Figure 201310082074X100002DEST_PATH_IMAGE059
，
。
(2) on the computed image plane intersection point of two ellipses and oval intersection point respectively about the intersection point of the tangential equation of two ellipses
On a plane of world coordinate system, there are two to intersect identical circle and common tangent (as Fig. 1),
Figure 962270DEST_PATH_IMAGE044
Be two and intersect identical circle, Be respectively round
Figure 884090DEST_PATH_IMAGE044
The center of circle,
Figure 275888DEST_PATH_IMAGE003
Two intersect the intersection point of circle.Cross a little
Figure 54488DEST_PATH_IMAGE003
Justify respectively
Figure 850447DEST_PATH_IMAGE044
Tangent line, the intersection point of tangent line is respectively , the point
Figure 681317DEST_PATH_IMAGE005
Be respectively round common tangent and round point of contact, connect With Intersect at a little , namely
Figure 186882DEST_PATH_IMAGE006
Point is
Figure 786228DEST_PATH_IMAGE007
With
Figure 968948DEST_PATH_IMAGE008
The mid point of two line segments; As (as Fig. 2) on the plane, round Similarly be oval
Figure 284840DEST_PATH_IMAGE009
, the point Picture be respectively a little
Figure 859358DEST_PATH_IMAGE011
Utilize
Figure 31451DEST_PATH_IMAGE009
Matrix of coefficients
Figure 495930DEST_PATH_IMAGE012
Try to achieve intersection point
Figure 571334DEST_PATH_IMAGE013
Coordinate, cross point Obtain respectively
Figure 256710DEST_PATH_IMAGE013
About
Figure 993722DEST_PATH_IMAGE009
Tangential equation, thereby utilize tangential equation to obtain the intersecting point coordinate of tangent line
Figure 48265DEST_PATH_IMAGE014
Looking like simultaneous on the plane
Figure 201310082074X100002DEST_PATH_IMAGE063
Solving Equations solve intersection point
Figure 150214DEST_PATH_IMAGE013
Coordinate be respectively With
Figure 201310082074X100002DEST_PATH_IMAGE065
, wherein With
Figure 201310082074X100002DEST_PATH_IMAGE067
Respectively the expression with
Figure 201310082074X100002DEST_PATH_IMAGE068
Corresponding horizontal ordinate and ordinate, homogeneous coordinates are
Figure 201310082074X100002DEST_PATH_IMAGE069
With
Figure 201310082074X100002DEST_PATH_IMAGE070
, i.e. mistake
Figure 201310082074X100002DEST_PATH_IMAGE071
The tangential equation of point two curves is:
Figure 201310082074X100002DEST_PATH_IMAGE072
， （1）
Figure 201310082074X100002DEST_PATH_IMAGE073
； （2）
Namely cross Two tangential equations of point are respectively:
Figure 201310082074X100002DEST_PATH_IMAGE075
， （3）
Figure 201310082074X100002DEST_PATH_IMAGE076
； （4）
Simultaneous equations (1), (3) get the intersection point of two tangent lines Coordinate
Figure 201310082074X100002DEST_PATH_IMAGE078
, simultaneous equations (2), (4) get the intersection point of two tangent lines
Figure 201310082074X100002DEST_PATH_IMAGE079
Figure 155953DEST_PATH_IMAGE079
Coordinate
Figure 201310082074X100002DEST_PATH_IMAGE080
(3) calculate two circles the centers of circle the picture coordinate and
Figure 132874DEST_PATH_IMAGE015
With
Figure 917291DEST_PATH_IMAGE008
The mid point of two line segments
Figure 49195DEST_PATH_IMAGE006
The picture coordinate
Looking like (as Fig. 2) extract minutiae on the plane
Figure 919062DEST_PATH_IMAGE016
Coordinate, obtain respectively straight line with the Cross function of Matlab
Figure 794614DEST_PATH_IMAGE017
Equation and straight line
Figure 27012DEST_PATH_IMAGE018
Equation, straight line And straight line
Figure 686981DEST_PATH_IMAGE020
Intersection point for the circle
Figure 474546DEST_PATH_IMAGE021
The picture in the center of circle
Figure 358188DEST_PATH_IMAGE022
Coordinate, straight line
Figure 441682DEST_PATH_IMAGE023
And straight line
Figure 410775DEST_PATH_IMAGE020
Intersection point for the circle
Figure 503496DEST_PATH_IMAGE024
The picture in the center of circle
Figure 913749DEST_PATH_IMAGE025
Coordinate, straight line
Figure 558357DEST_PATH_IMAGE020
With straight line
Figure 388647DEST_PATH_IMAGE045
Intersection point be
Figure 878534DEST_PATH_IMAGE006
The picture of point Coordinate;
Process
Figure 201310082074X100002DEST_PATH_IMAGE081
Straight line be:
Figure 201310082074X100002DEST_PATH_IMAGE082
：
Figure 201310082074X100002DEST_PATH_IMAGE083
； （5）
Process
Figure 201310082074X100002DEST_PATH_IMAGE084
Straight line be:
Figure 201310082074X100002DEST_PATH_IMAGE085
：
Figure 201310082074X100002DEST_PATH_IMAGE086
； （6）
Process Straight line be:
Figure 201310082074X100002DEST_PATH_IMAGE088
：
Figure 201310082074X100002DEST_PATH_IMAGE089
； （7）
Simultaneous equations (5), (6) must be round
Figure 201310082074X100002DEST_PATH_IMAGE090
The center of circle
Figure DEST_PATH_IMAGE091
Picture
Figure 201310082074X100002DEST_PATH_IMAGE092
Coordinate
Simultaneous equations (5), (7) must be round The center of circle
Figure DEST_PATH_IMAGE095
Picture
Figure 201310082074X100002DEST_PATH_IMAGE096
Coordinate
Figure DEST_PATH_IMAGE097
Cross The straight-line equation of point is:
：
Figure DEST_PATH_IMAGE099
； （8）
Simultaneous equations (5), (8)
Figure 948493DEST_PATH_IMAGE020
With
Figure 201310082074X100002DEST_PATH_IMAGE100
Get intersection point
Figure 412710DEST_PATH_IMAGE046
Coordinate
(4) calculate the coordinate of the vanishing point on orthogonal directions
At stencil plane (as Fig. 1),
Figure 797555DEST_PATH_IMAGE006
It is line segment
Figure 659332DEST_PATH_IMAGE008
In, and straight line With straight line
Figure 62949DEST_PATH_IMAGE028
Vertically, establish straight line
Figure 833459DEST_PATH_IMAGE028
Infinity point on direction is
Figure 990770DEST_PATH_IMAGE029
, straight line
Figure 282949DEST_PATH_IMAGE008
Infinity point on direction is
Figure 183909DEST_PATH_IMAGE030
, have
Figure 746609DEST_PATH_IMAGE031
,
Figure 74822DEST_PATH_IMAGE032
, had on picture plane (as Fig. 2) by cross ratio invariability
Figure 418078DEST_PATH_IMAGE033
， （9）
Figure 732516DEST_PATH_IMAGE034
， （10）
Wherein
Figure 274356DEST_PATH_IMAGE035
Be straight line
Figure 914416DEST_PATH_IMAGE036
The vanishing point of direction,
Figure 541706DEST_PATH_IMAGE037
Be straight line
Figure 158370DEST_PATH_IMAGE045
The vanishing point of direction; If
Figure 201310082074X100002DEST_PATH_IMAGE102
Coordinate be
Figure DEST_PATH_IMAGE103
,
Figure 201310082074X100002DEST_PATH_IMAGE104
Coordinate be
Figure DEST_PATH_IMAGE105
, will
Figure 201310082074X100002DEST_PATH_IMAGE106
Coordinate substitution equation (9):
Figure DEST_PATH_IMAGE107
，
Figure 201310082074X100002DEST_PATH_IMAGE108
；
Will
Figure DEST_PATH_IMAGE109
Coordinate substitution equation (10):
Figure 69563DEST_PATH_IMAGE107
，
Figure 942841DEST_PATH_IMAGE108
。
The data point of substitution 6 width images is found the solution orthogonal directions
Figure 667215DEST_PATH_IMAGE020
,
Figure 713668DEST_PATH_IMAGE036
Common vanishing point
Figure 201310082074X100002DEST_PATH_IMAGE110
With
Figure 636625DEST_PATH_IMAGE100
Vanishing point
Figure DEST_PATH_IMAGE111
, wherein
Figure 201310082074X100002DEST_PATH_IMAGE112
Represent 1 to 6 width image, as follows:
Vanishing point on the first width image orthogonal directions:
Figure DEST_PATH_IMAGE113
，
Figure 201310082074X100002DEST_PATH_IMAGE114
；
Vanishing point on the second width image orthogonal directions:
Figure DEST_PATH_IMAGE115
，
Figure 201310082074X100002DEST_PATH_IMAGE116
；
Vanishing point on the 3rd width image orthogonal directions:
Figure DEST_PATH_IMAGE117
，
Figure 201310082074X100002DEST_PATH_IMAGE118
；
Vanishing point on the 4th width image orthogonal directions:
，
Figure 201310082074X100002DEST_PATH_IMAGE120
；
Vanishing point on the 5th width image orthogonal directions:
Figure DEST_PATH_IMAGE121
，
Figure 201310082074X100002DEST_PATH_IMAGE122
；
Vanishing point on the 6th width image orthogonal directions:
Figure DEST_PATH_IMAGE123
，
Figure 201310082074X100002DEST_PATH_IMAGE124
。
(5) solve camera intrinsic parameter
The coordinate of the vanishing point on the 6 width images that obtain through above step can solve linearly the camera intrinsic parameter matrix and is:
Figure DEST_PATH_IMAGE125
Wherein five intrinsic parameters (unit: pixel) be respectively:
Figure 201310082074X100002DEST_PATH_IMAGE126
, ,
Figure 201310082074X100002DEST_PATH_IMAGE128
,
Figure DEST_PATH_IMAGE129
,
Figure 201310082074X100002DEST_PATH_IMAGE130

Claims (1)

1. one kind is utilized two to intersect identical circles and common tangent linear solution camera intrinsic parameter, it is characterized in that utilizing the target on a plane, this target be by on the plane two intersect identical circle and common tangent consists of; At first extract elliptic equation and unique point and find the solution the intersection point of two ellipses from image, divide and the character of the double ratio of conllinear four vertical according to two public strings that intersect identical circles and the center of circle line segment that is connected two circles, draw the vanishing point on orthogonal directions on the plane of delineation, utilize at last the constraint linear solution camera intrinsic parameter of the picture of vanishing point on orthogonal directions and absolute conic;
(1) curvilinear equation in the fitted figure picture
Utilize Edge function in the Matlab program to extract the coordinate of image characteristic point, and simulate elliptic equation in image with least-squares algorithm, obtain each elliptic equation on image, oval equation
Figure 201310082074X100001DEST_PATH_IMAGE001
Matrix of coefficients be expressed as
Figure 201310082074X100001DEST_PATH_IMAGE002
(2) on the computed image plane intersection point of two ellipses and oval intersection point respectively about the intersection point of the tangential equation of two ellipses
On a plane of world coordinate system, there are two to intersect identical circle and common tangent,
Figure 201310082074X100001DEST_PATH_IMAGE003
Be two and intersect identical circle,
Figure 201310082074X100001DEST_PATH_IMAGE004
Be respectively round
Figure 777046DEST_PATH_IMAGE003
The center of circle,
Figure 201310082074X100001DEST_PATH_IMAGE005
It is the intersection point of two crossing circles;
Cross a little
Figure 469059DEST_PATH_IMAGE005
Justify respectively
Figure 347016DEST_PATH_IMAGE003
Tangent line, the intersection point of tangent line is respectively ,
Figure 201310082074X100001DEST_PATH_IMAGE007
Be respectively the round point of contact of outside common tangent on circle, connect
Figure 339243DEST_PATH_IMAGE006
With
Figure 34666DEST_PATH_IMAGE005
Intersect at a little
Figure 201310082074X100001DEST_PATH_IMAGE008
, namely Be
Figure 201310082074X100001DEST_PATH_IMAGE009
With
Figure 201310082074X100001DEST_PATH_IMAGE010
The mid point of two line segments; On imaging plane, circle
Figure 331841DEST_PATH_IMAGE003
Similarly be oval, point
Figure 201310082074X100001DEST_PATH_IMAGE011
Picture be respectively
Figure 201310082074X100001DEST_PATH_IMAGE012
, extract oval marginal point with the Edge function of Matlab, use the least square fitting elliptic curve, oval equation
Figure 178575DEST_PATH_IMAGE001
Be expressed as with matrix of coefficients
Figure 418801DEST_PATH_IMAGE002
Utilize
Figure 678881DEST_PATH_IMAGE001
Matrix of coefficients try to achieve intersection point
Figure 201310082074X100001DEST_PATH_IMAGE013
Coordinate, cross point
Figure 695379DEST_PATH_IMAGE013
Obtain respectively about
Figure 599881DEST_PATH_IMAGE001
Tangential equation, utilize tangential equation to calculate the intersecting point coordinate of tangent line
Figure 201310082074X100001DEST_PATH_IMAGE014
(3) calculate two circles the center of circle the picture coordinate and
Figure 201310082074X100001DEST_PATH_IMAGE015
With
Figure 840369DEST_PATH_IMAGE010
The mid point of two line segments
Figure 696067DEST_PATH_IMAGE008
The picture coordinate
In the unique point as the point of contact of extraction outside common tangent and two ellipses on the plane
Figure 201310082074X100001DEST_PATH_IMAGE016
Coordinate, obtain respectively straight line with the Cross function of Matlab Equation and straight line Equation; Straight line
Figure 201310082074X100001DEST_PATH_IMAGE019
And straight line
Figure 201310082074X100001DEST_PATH_IMAGE020
Intersection point for the circle The picture in the center of circle
Figure 201310082074X100001DEST_PATH_IMAGE022
Coordinate, straight line
Figure 201310082074X100001DEST_PATH_IMAGE023
And straight line
Figure 827840DEST_PATH_IMAGE020
Intersection point for the circle
Figure 201310082074X100001DEST_PATH_IMAGE024
The picture in the center of circle
Figure 201310082074X100001DEST_PATH_IMAGE025
Coordinate, straight line
Figure 383587DEST_PATH_IMAGE020
With straight line
Figure 201310082074X100001DEST_PATH_IMAGE026
Intersection point be The picture of point
Figure 201310082074X100001DEST_PATH_IMAGE027
Coordinate;
(4) calculate the coordinate of the vanishing point on orthogonal directions
At stencil plane,
Figure 373856DEST_PATH_IMAGE008
Two circle intersection point line segments
Figure 60053DEST_PATH_IMAGE010
In, and straight line
Figure 703261DEST_PATH_IMAGE010
With straight line
Figure 201310082074X100001DEST_PATH_IMAGE028
Vertically, establish straight line
Figure 754394DEST_PATH_IMAGE028
Infinity point on direction is
Figure 201310082074X100001DEST_PATH_IMAGE029
, straight line
Figure 476362DEST_PATH_IMAGE010
Infinity point on direction is
Figure 201310082074X100001DEST_PATH_IMAGE030
, have
Figure 201310082074X100001DEST_PATH_IMAGE031
, , had as the plane by cross ratio invariability
Figure 201310082074X100001DEST_PATH_IMAGE033
,
Figure 201310082074X100001DEST_PATH_IMAGE034
Wherein
Figure 201310082074X100001DEST_PATH_IMAGE035
Be straight line
Figure 201310082074X100001DEST_PATH_IMAGE036
The vanishing point of direction,
Figure 201310082074X100001DEST_PATH_IMAGE037
Be straight line
Figure 949938DEST_PATH_IMAGE026
The vanishing point of direction;
(5) find the solution camera intrinsic parameter
Obtain 6 width images, gone out 5 intrinsic parameters, i.e. matrixes of video camera by the vanishing point linear solution of orthogonal directions In 5 variablees, be respectively:
Figure 201310082074X100001DEST_PATH_IMAGE039
Be the distortion factor of image,
Figure 201310082074X100001DEST_PATH_IMAGE040
In image coordinate system Axle and
Figure 201310082074X100001DEST_PATH_IMAGE042
The scale factor of axle,
Figure 201310082074X100001DEST_PATH_IMAGE043
It is the principal point coordinate.
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