CN101639355A - Three dimensional plane extraction method - Google Patents

Abstract

The invention provides a three dimensional plane extraction method. The input of the technical scheme is that any method is used for obtaining a three dimensional line feature set and a sparse digitalelevation model data in a target scene, obtaining stereo image of a target scene from two different visual angles and obtaining all the inner parameters and the outer parameters of the stereoscopic camera as well as the line extracting result of the stereo image; the output of the technical scheme is the three dimensional plane with high positioning precision and favorable boundary integrity andaccuracy. In the treating process of the technical scheme, the positioning of the three dimensional plane and the extraction and separation of the plane zone boundary are divided into two stages whichare connected for independently finishing two tasks and reducing error impact between the two tasks, which causes the results of the two stages to be respectively optimal, so that the three dimensional plane extraction is optimal.

Description

A kind of three dimensional plane extraction method

Technical field

The present invention relates to Photogrammetry and Remote Sensing, computer vision field, relate in particular to a kind of extracting method of three-dimensional planar.

Background technology

The man-made target three-dimensional reconstruction has in fields such as city planning, disaster monitoring, communications facility construction very widely to be used, and for a long time, is the important topic that Photogrammetry and Remote Sensing, computer vision field study for a long period of time always.In the target three-dimensional reconstruction scheme of various polyhedral structures, it is one of them committed step that three-dimensional planar extracts.

At present, the method for extracting three-dimensional planar mainly contains four kinds: first kind, utilize intensive digital elevation model (Digital Elevation Model, the vehicle economy M) data that directly obtain, and obtain three-dimensional planar by match; Second kind, by obtaining the target area image under the different illumination conditions, the reflection characteristic model on combining target surface is set up a plurality of reflectograms, utilizes the photometry stereo vision method to calculate the surface direction of each object table millet cake, realizes the reconstruction to target surface; The third, utilize the method for linear feature perception marshalling, suppose the shape in known target surface plane zone, the linear feature that has extracted is carried out the marshalling of ad hoc rules, as profile marshalling at the rectangle plane zone, generate a plurality of hypothesises of plane mechanism, and then carry out the checking of the hypothesis of plane mechanism according to image or other priori; The 4th kind, the linear feature that extracts is mated, calculate corresponding 3 d-line, 3 d-line is carried out the coplane analysis by forward intersection, generate all possible hypothesis of plane mechanism and formation level border to form closed plane domain, utilize various evidences that these enclosed region are confirmed.Comparatively speaking, the shortcoming of first method is the increase along with the spot elevation packing density, and procurement cost can increase simultaneously, and for baroque target, if the general type that lacks plane quantity or object construction is as priori, all planes of extracting target surface are very difficult.The shortcoming of second method is to need accurately known a plurality of illumination conditions, thereby this method is more common in the resurfacing of the small size target under the simple photoenvironment, for the illumination condition complexity, background is various or the target surface reflection characteristic such as there are differences at various non-cooperation situations, this method is difficult to realize satisfactory results.The shortcoming of the third method is that the geometric configuration that can extract the plane is restricted.Than first three methods, the 4th kind of method applicability is stronger, and little to the dependence of altitude figures, even can not need the support of altitude figures, simultaneously, this method adopts a kind of bottom-up leaching process, and the three-dimensional planar that can solve various polygonal shapes extracts.

In the three-dimensional planar leaching process, the three-dimensional planar information that need obtain comprises two aspects: the one, and the three-dimensional planar locating information shows as the plane equation that the plane is opened at its place; The 2nd, the three-dimensional planar boundary information, the complete sum that shows as closed plane is the boundary line accurately.In existing the 4th kind of three-dimensional planar extractive technique, to extract these two kinds of information finishes simultaneously as a task, the bearing accuracy on plane can be subjected to the influence of boundary alignment error, and can be reduced because of the positional parameter error on plane by the boundary alignment precision that Plane intersects obtains.

Summary of the invention

The technical problem to be solved in the present invention is in the three-dimensional planar leaching process, the location of three-dimensional planar and the extraction on plane domain border are separated into two successive stages, independently finish this two tasks respectively, reduce the error effect between them, make the result in two stages reach optimum respectively, thereby realize the optimum that three-dimensional planar extracts.

The stereo-picture of the object scene that the input of technical solution of the present invention is 3 d-line feature set in the object scene that obtains with any means and sparse digital elevation model data, obtain from two different visual angles and obtain all inside and outside parameters of their stereoscopic camera, the straight line of stereoscopic image extracts the result, the output of technical scheme is bearing accuracy height, border integrality and the good three-dimensional planar of accuracy, and the purpose of technical application scheme is that the resurfacing work for follow-up polyhedral structure target lays the first stone.

Thinking of the present invention is: the three-dimensional planar leaching process is divided into two successive optimization step, first step is to obtain the hi-Fix information of the open three-dimensional planar in target surface place, implementation method is a kind of hypothesis substantive approach of cutting apart, idiographic flow is that the member's linear feature that utilizes the open three-dimensional planar of 3 d-line feature generation to suppose → utilize that three-dimensional planar is supposed is cut apart the flat blocks hypothesis that optimal planar piece hypothesis → merging optimum is supposed → chosen to the open a plurality of flat blocks of three-dimensional planar hypothesis formation, the open three-dimensional planar of generation high position precision; Second step is the border of formation level, based on the open three-dimensional planar that generates, the plane intersection line that obtains with Plane intersects, linear feature that extracts in the image and the heuristic knowledge relevant with object construction are as the foundation that generates planar boundary, be the complete border of open planar configuration, idiographic flow is the linear feature that extracts in the stereo-picture of the border that utilizes the member's linear feature and the interplanar intersection formation level on open plane → utilize and the heuristic rule planar boundary that replenishes disappearance → generate the closed hypothesis of plane mechanism by planar boundary, utilizes optimization method to select the optimum closed hypothesis of plane mechanism.

Technical scheme of the present invention is a kind of three dimensional plane extraction method, specifically comprises the steps:

Two width of cloth images of the different visual angles of known a certain scene are called left and right image, and all inside and outside parameters of absorbing the left and right camera of this two width of cloth image, and the coordinate of wherein left and right camera shooting center in world coordinate system is designated as (X respectively _O1, Y _O1, Z _O1) and (X _O2, Y _O2, Z _O2); Linear feature collection by extracting in this two width of cloth image is designated as H respectively ₁And H ₂N in known this scene _SBar 3 d-line feature is designated as S={L _i| i=1,2 ..., N _S, the arbitrary element L in this set _iRepresent a three-dimensional straight line segment, its two end points are respectively (X _I1, Y _I1, Z _I1) and (X _I2, Y _I2, Z _I2); Sparse dem data in known this scene is designated as the spot elevation set

The average error of element is λ in this set.

The first step is found the solution the open three-dimensional planar at target surface place

In (one) step, the three-dimensional planar hypothesis generates

The set angle threshold phi is approximate consistent if the angle of two 3 d-lines, is then judged their direction vector less than φ; Setpoint distance thresholding ε, if three dimensions o'clock to the distance of a 3 d-line or a three-dimensional planar less than ε, judge that then this three dimensions point is on this 3 d-line or on this three-dimensional planar.The value of two thresholdings is decided according to known 3 d-line features location precision.

The angular range of two adjacent boundary place straight lines on any one plane on target setting surface is

Minimum angle between any two adjacent planes of target surface is

Their value is relevant with the design feature of target surface.

In the 1st step, calculate the three-dimensional planar hypothesis

3 d-line feature among the pair set S makes up in twos, generates the three-dimensional planar hypothesis, and concrete steps are:

Optional two 3 d-line feature L _u∈ S and L _v∈ S, L _uEnd points be (X _U1, Y _U1, Z _U1) and (X _U2, Y _U2, Z _U2), L _vEnd points be (X _V1, Y _V1, Z _V1) and (X _V2, Y _V2, Z _V2), utilize their objective definition function f (a, b, c, d):

f(a，b，c，d)＝(a·X _u1+b·Y _u1+c·Z _u1+d) ²+(a·X _u2+b·Y _u2+c·Z _u2+d) ²+(a·X _v1+b·Y _v1+c·Z _v1+d) ²+(a·X _v2+b·Y _v2+c·Z _v2+d) ²

Adopt the gradient descent method calculate f (a, b, c, when d) getting minimum value (c d), obtains by L for a, b _uAnd L _vThe three-dimensional planar hypothesis equation of appointment is aX+bY+cZ+d=0.

Suppose set omega={ P according to all three-dimensional planars that above-mentioned steps obtains _i| i=1,2 ..., N} is for any one three-dimensional planar hypothesis P wherein _i, plane equation is a _iX+b _iY+c _iZ+d _i=0, two 3 d-line features that are used to generate it are called P _iMember's linear feature.

In the 2nd step, the three-dimensional planar that merges coplane is supposed

For any one three-dimensional planar hypothesis P _i∈ Ω, the mid point of two member's linear feature mid point lines that calculates it is in plane P _iOn projection, be designated as (X _M ⁱ, Y _M ⁱ, Z _M ⁱ).

The 1st) step, judge the coplanar relation that three-dimensional planar is supposed

For any two three-dimensional planars hypothesis P _i∈ Ω and P _j∈ Ω is if the inequality group below satisfying is then judged P _iAnd P _jCoplane:

$\left\{\begin{array}{c}\frac{|a_i\&CenterDot;a_j+b_i\&CenterDot;b_j+c_i\&CenterDot;c_j|}{\sqrt{a_i^2+b_i^2+c_i^2}\&CenterDot;\sqrt{a_{\mathrm{<figure-callout\; id="2"\; label="j"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">j</figure-callout>}}^2+{\mathrm{<figure-callout\; id="2"\; label="b\; j"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">b</figure-callout>}}_j^{}+{\mathrm{<figure-callout\; id="2"\; label="c\; j"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">c</figure-callout>}}_j^{}}}>\cos \mathrm{\&phi;}\\ \frac{|a_i\&CenterDot;X_M^j+b_i\&CenterDot;Y_M^j+c_i\&CenterDot;Z_M^j+d_i|}{\sqrt{a_i^2+b_i^2+c_i^2}}<\mathrm{\&epsiv;}\\ \frac{|a_j\&CenterDot;X_M^i+b_j\&CenterDot;Y_M^i+c_j\&CenterDot;Z_M^i+d_j|}{\sqrt{a_{\mathrm{<figure-callout\; id="2"\; label="j"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">j</figure-callout>}}^2+{\mathrm{<figure-callout\; id="2"\; label="b\; j"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">b</figure-callout>}}_j^{}+{\mathrm{<figure-callout\; id="2"\; label="c\; j"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">c</figure-callout>}}_j^{}}}<\mathrm{\&epsiv;}\end{array}\right.$

The 2nd) step, the three-dimensional planar hypothesis of search coplane

Generate a non-directed graph G, with each three-dimensional planar hypothesis P _i∈ Ω, then keeps between the node of their correspondences connecting, otherwise does not connect if any two three-dimensional planars hypothesis is a coplane as a node.

Calculate all very big groups of non-directed graph G, be designated as { Q _W| w=1,2 ..., N _Q, any one greatly rolls into a ball Q _wBe the subclass of Ω, be designated as $\left\{P_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;m_w\},$ Any two three-dimensional planars hypothesis all is a coplane in the subclass.

The 3rd) step, the three-dimensional planar that merges coplane is supposed

Greatly roll into a ball Q for any one _w, merge all three-dimensional planar hypothesis that it comprises, obtain a new three-dimensional planar hypothesis P _w, its plane equation is designated as a _wX+b _wY+c _wZ+d _w=0, wherein:

${\stackrel{\&OverBar;}{a}}_w=\frac{1}{m_w}\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}\frac{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}{\sqrt{{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{b}}_w=\frac{1}{m_w}\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}\frac{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}{\sqrt{{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{c}}_w=\frac{1}{m_w}\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}\frac{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}{\sqrt{{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{d}}_w=-\frac{1}{m_w}({\stackrel{\&OverBar;}{a}}_w\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}{X}_M^{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}+{\stackrel{\&OverBar;}{b}}_w\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}{Y}_M^{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}+{\stackrel{\&OverBar;}{c}}_w\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}{Z}_M^{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)})$

Will $\left\{P_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;m_w\}$ In all member's linear features of each three-dimensional planar hypothesis project to P _wOn, all projection straight line sections that obtain become P _wMember's linear feature.

Greatly roll into a ball the three-dimensional planar hypothesis set that obtains by all and be designated as Ω={ P _w| w=1,2 ..., N}, wherein, N=N _Q, three-dimensional planar is supposed P arbitrarily _wMember's linear feature set be designated as ${\stackrel{\&OverBar;}{M}}_w=\left\{{\stackrel{\&OverBar;}{L}}_j^w\right|j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,{\stackrel{\&OverBar;}{n}}_w\},$ Any one member's linear feature L _j ^wEnd points be designated as (X _{J, 1} ^w, Y _{J, 1} ^w, Z _{J, 1} ^w) and (Z _{J, 2} ^w, Y _{J, 2} ^w, Z _{J, 2} ^w).

In (two) step, the three-dimensional planar hypothesis confirms

In the 1st step, cut apart three-dimensional planar and be assumed to be the flat blocks hypothesis

For any one three-dimensional planar hypothesis P _i∈ Ω utilizes its member's linear feature M _iIt is cut apart, obtain some flat blocks hypothesis.Concrete grammar may further comprise the steps:

The 1st) step, member's linear feature of merging conllinear

For P _iAny two member's linear feature L _j ⁱ, ${\stackrel{\&OverBar;}{L}}_k^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ Calculate the angle of their place straight lines, be designated as μ, and L _j ⁱMid point to L _k ⁱThe distance of place straight line and L _k ⁱMid point to L _j ⁱThe distance of place straight line is designated as η respectively ₁And η ₂

If μ＜φ and η ₁＜ε and μ ₂＜ε then judges L _j ⁱAnd L _k ⁱConllinear generates a new 3 d-line feature with them, and two end points of this 3 d-line feature are L _j ⁱAnd L _k ⁱAt a distance of farthest two end points, add new 3 d-line feature to set M _iIn, and from set M _iMiddle deletion L _j ⁱAnd L _k ⁱ

The 2nd) step, calculate P _iThe intersection point of all member's linear features

For P _iAny two member's linear feature L _α ⁱ, ${\stackrel{\&OverBar;}{L}}_{\mathrm{\&beta;}}^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ If the angle theta of their place straight lines satisfies Then calculate L _α ⁱAnd L _β ⁱIntersection point or the intersection point of their extended line.

The set that the intersection point of all calculating is formed is designated as R _i

The 3rd) step, generate the non-directed graph that is used for the segmentation plane hypothesis

Generate a non-directed graph, be designated as G _i, its node is by two class dot generation: the one, and intersection point set R _iIn each element, the 2nd, for M _iIn any one member's linear feature, if do not have and M on the extended line outside its end points _iIn the intersection point of other member's linear feature, and this end points is not set R _iIn element, then this end points becomes figure G _iA node.

Non-directed graph G _iIn any two internodal annexations comprise two classes: the one, if the point of their correspondences on same member's linear feature or its extended line, and on the connection line segment of these two points not and figure G _iIn the corresponding point of other node, then their keep to connect; The 2nd, the annexation of structure, building method is as follows:

For any one member's linear feature L _α ⁱIf its one or two end points is figure G _iNode, then handle in two kinds of situation:

First kind of situation, L _α ⁱWith M _iIn any other member's linear feature all non-intersect

Select any member's linear feature ${\stackrel{\&OverBar;}{L}}_{\mathrm{\&beta;}}^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ And β ≠ α specifies on its place straight line the G with figure _iIn two corresponding two points of node, these two points are positioned at L _β ⁱThe mid point both sides and the connection line segment between them on not and figure G _iThe middle corresponding point of node is designated as (X with them ₁, Y ₁, Z ₁) and (X ₂, Y ₂, Z ₂), and calculate

$e_1=({\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i$

$e_2=({\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},2}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},2}^i$

Wherein, (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and (X _{α, 2} ⁱ, Y _{α, 2} ⁱ, Z _{α, 2} ⁱ) be L _α ⁱEnd points.For arbitrarily ${\stackrel{\&OverBar;}{L}}_l^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ And l ≠ α, l ≠ β calculate

${\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">f</figure-callout>}}_1=({\stackrel{\&OverBar;}{X}}_{l,1}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{l,1}^i$

${\mathrm{<figure-callout\; id="2"\; label="f"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">f</figure-callout>}}_2=({\stackrel{\&OverBar;}{X}}_{l,2}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{l,2}^i$

$f_3=({\stackrel{\&OverBar;}{X}}_{l,1}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i}{{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i}+{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i-{\stackrel{\&OverBar;}{Y}}_{l,1}^i$

$f_4=({\stackrel{\&OverBar;}{X}}_{l,2}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i}{{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i}+{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i-{\stackrel{\&OverBar;}{Y}}_{l,2}^i$

Wherein, (X _{L, 1} ⁱ, Y _{L, 1} ⁱ, Z _{L, 1} ⁱ) and (X _{L, 2} ⁱ, Y _{L, 2} ⁱ, Z _{L, 2} ⁱ) be L _l ⁱEnd points.If satisfy

$\left\{\begin{array}{c}{e}_1\&CenterDot;e_2>0\\ {\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">f</figure-callout>}}_1\&CenterDot;f_3>0\\ {\mathrm{<figure-callout\; id="2"\; label="f"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">f</figure-callout>}}_2\&CenterDot;f_4>0\\ {\mathrm{<figure-callout\; id="1"\; label="f"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">f</figure-callout>}}_1\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="f"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">f</figure-callout>}}_2>0\end{array}\right.$

Then calculate

$f_5=({\stackrel{\&OverBar;}{X}}_{l,1}^i-X_1)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{l,2}^i-Y_1}{{\stackrel{\&OverBar;}{X}}_{l,2}^i-{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{l,1}^i$

$f_6=(X_2-X_1)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{l,2}^i-Y_1}{{\stackrel{\&OverBar;}{X}}_{l,2}^i-{\mathrm{<figure-callout\; id="1"\; label="X"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_1}+Y_1-{\mathrm{<figure-callout\; id="2"\; label="Y"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">Y</figure-callout>}}_2$

If f ₅F ₆＞0, tie point (X then _{α, i} ¹, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and (X ₂, Y ₂, Z ₂) at figure G _iMiddle corresponding node, and point (X _{α, 2} ⁱ, Y _{α, 2} ⁱ, Z _{α, 2} ⁱ) and (X ₁, Y ₁, Z ₁) at figure G _iMiddle corresponding node.

Second kind of situation, L _α ⁱWith M _iIn at least one other member's linear feature intersect

If (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and figure G _iA node corresponding, select member's linear feature L _α ⁱOr on its extended line with (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) nearest intersection point, be designated as (X ₃, Y ₃, Z ₃), suppose that this intersection point is L _α ⁱWith member's linear feature L _β ⁱIntersection point.Select L _β ⁱOr on its extended line with figure G _iIn the corresponding point of node, if at it and point (X ₃, Y ₃, Z ₃) between line on not and figure G _iThe middle corresponding point of node then is designated as (X with it ₄, Y ₄, Z ₄), if such point has two, then respectively they are designated as (X ₄, Y ₄, Z ₄) and (X ₅, Y ₅, Z ₅).

If (X ₄, Y ₄, Z ₄) be L _β ⁱWith the intersection point of other member's linear feature, suppose that this member's linear feature is L _l ⁱ, and L _l ⁱOr exist on its extended line and figure G _iThe middle corresponding point of node, it satisfies and point (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) drop on L _β ⁱThe same side, and with point (X ₄, Y ₄, Z ₄) between line on not and figure G _iIn the corresponding point of node, then connect it and point (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) at figure G _iMiddle corresponding node; Otherwise, tie point (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and point (X ₄, Y ₄, Z ₄) at figure G _iMiddle corresponding node.If (X ₅, Y ₅, Z ₅) exist, then according to (X ₄, Y ₄, Z ₄) same method handles.

If (X _{α, 2} ⁱ, Y _{α, 2} ⁱ, Z _{α, 2} ⁱ) be figure G _iNode, method of then constructing annexation and (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) identical.

The 4th) step, cut apart generation flat blocks hypothesis

Search non-directed graph G _iIn all rings, for any one ring wherein, if the node number that belongs to same member's linear feature or its extended line that it comprises is no more than 2, then generate a flat blocks hypothesis with it, all nodes that it comprises are formed an orderly point set according to its annexation, represent the vertex set of this flat blocks hypothesis.

The set of being cut apart the flat blocks hypothesis that obtains by all planes among the Ω is designated as $\widehat{\mathrm{\&Omega;}}=\left\{{\hat{P}}_w\right|w=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,\hat{N}\},$ Wherein, any one flat blocks hypothesis

Vertex set by it is represented, is designated as $\left\{({\hat{X}}_{\mathrm{wj}},{\hat{Y}}_{\mathrm{wj}},{\hat{Z}}_{\mathrm{wj}})\right|j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,{\hat{n}}_w\},$ Cut apart among the Ω and obtain The equation of three-dimensional planar hypothesis be designated as ${\hat{a}}_w\&CenterDot;X+{\hat{b}}_w\&CenterDot;Y+{\hat{c}}_w\&CenterDot;Z+{\hat{d}}_w=0.$

In the 2nd step, calculate the reliability measure that each flat blocks is supposed

Spot elevation set J and left and right image that the basis of flat blocks hypothesis reliability measure is known calculate each some projection in left and right image among the J, obtain two plane point sets, are designated as respectively

With

Wherein,

With

It is respectively the spot elevation among the J Subpoint in left and right image.

The arbitrary plane piece is supposed ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ Calculate its reliability measure, comprise the steps:

The 1st) step, the search quilt

The known spot elevation that covers

Projection plane piece hypothesis

Vertex set to left and right image, remember any summit

Projection in left and right image is respectively

With

The order projection on connection summit in left and right image respectively according to each fixed point in the vertex set forms

Two polygon view fields.

For any spot elevation

If the projection in its left image Be positioned at

In the view field in left image, and the projection in its right image

Be positioned at

In the view field in right image, then judge spot elevation Quilt

Cover.

Quilt

The set that all spot elevations that cover are formed is designated as ${\mathrm{\&Theta;}}_w\&Subset;J,$ The quantity of element is α in the set _w

The 2nd) step, calculate the elevation consistance and estimate

If

The spot elevation number α that covers _w＞0, then calculate

The elevation consistance estimate for

(formula one)

Wherein,

If α _w=0, then $E_w^D=0.$

The 3rd) step, calculate the gray scale similarity measure

All images point is formed a point set in the view field in left image, is designated as K _w ^L, the gradation of image value of being had a few in them is formed an array, is designated as { h _Wt| t=1,2 ..., o _w.

Foundation Three pairs of same places in left and right image

With

With

And

With

Set of computations K _w ^LThe middle same place of all elements point in right image, the set that these same places are formed is designated as K _w ^R, the gradation of image value of being had a few in them is obtained by interpolation algorithm, and these gray-scale values are formed another array, are designated as { h _Wt' | t=1,2 ..., o _w.Calculate

Gray scale similarity measure between the view field in left and right image is

$E_w^G=\frac{\underset{t=1}{\overset{O_w}{\mathrm{\&Sigma;}}}(h_{\mathrm{wt}}-{\stackrel{\&OverBar;}{h}}_w)({h_{\mathrm{wt}}}^{\&prime;}-{{\stackrel{\&OverBar;}{h}}_w}^{\&prime;})}{\sqrt{\left[\underset{t=1}{\overset{O_w}{\mathrm{\&Sigma;}}}{(h_{\mathrm{wt}}-{\stackrel{\&OverBar;}{h}}_w)}^2\right]\&CenterDot;\left[\underset{t=1}{\overset{O_w}{\mathrm{\&Sigma;}}}{({h_{\mathrm{wt}}}^{\&prime;}-{{\stackrel{\&OverBar;}{h}}_w}^{\&prime;})}^2\right]}}$ (formula two)

Wherein, ${\stackrel{\&OverBar;}{h}}_w=\frac{1}{O_w}\underset{t=1}{\overset{O_w}{\mathrm{\&Sigma;}}}{h}_{\mathrm{wt}},$ ${{\stackrel{\&OverBar;}{h}}_w}^{\&prime;}=\frac{1}{O_w}\underset{t=1}{\overset{O_w}{\mathrm{\&Sigma;}}}{h_{\mathrm{wt}}}^{\&prime;}.$

The 4th) step, calculate Reliability measure

Reliability measure be

$E_w=\left\{\begin{array}{cc}{E}_w^D\&CenterDot;(E_w^G+1)/2& E_w^D>0\\ (E_w^G+1)/2& E_w^D=0\end{array}\right.$ (formula three)

In the 3rd step, search for reliable flat blocks hypothesis

Divide two steps to finish:

The 1st) step, the reliable flat blocks hypothesis that search has known spot elevation to cover

Set up a non-directed graph, for any one flat blocks hypothesis ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ If $E_w^D>0,$ Then generate a node of this non-directed graph with it, the property value of node is E _wFor any two nodes,, then connect them if all there is not lap in the view field of the flat blocks of their correspondences hypothesis in left and right image.

Calculate all very big groups of this non-directed graph, therefrom choose the set of all flat blocks hypothesis that the very big group of node attribute values sum maximum comprises, be designated as K ₁, it represents that chlamydate spot elevation is as the reliable flat blocks hypothesis of weighing foundation.

The 2nd) step, the reliable flat blocks hypothesis that search does not have known spot elevation to cover

Set up a non-directed graph, for any one flat blocks hypothesis ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ If $E_w^D=0$ And it and K ₁In any one flat blocks hypothesis view field in left and right image all do not have lap, then use

Generate a node of this non-directed graph, the property value of node is E _wFor any two nodes,, then connect them if all there is not lap in the view field of the flat blocks of their correspondences hypothesis in left and right image.

Calculate all very big groups of this non-directed graph, therefrom choose the set of all flat blocks hypothesis that the very big group of node attribute values sum maximum comprises, be designated as K ₂, the spot elevation that its expression does not have to cover is as weighing foundation and and K ₁In all flat blocks suppose coexisting reliable flat blocks hypothesis.

With K ₁∪ K ₂All flat blocks hypothesis that comprise are as reliable flat blocks hypothesis.

In the 4th step, the reliable flat blocks that merges coplane is supposed

The 1st) step, judge the coplanar relation that flat blocks is supposed

For any two reliable flat blocks hypothesis

${\hat{P}}_v\&Element;{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">K</figure-callout>}}_1\&cup;{\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">K</figure-callout>}}_2,$ If following inequality is formed upright, then judge

With

Coplane:

$\left\{\begin{array}{c}\frac{|{\hat{a}}_u\&CenterDot;{\hat{a}}_v+{\hat{b}}_u\&CenterDot;{\hat{b}}_v+{\hat{c}}_u+{\hat{c}}_v|}{\sqrt{{{\hat{a}}_{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}}}^2+{{\hat{b}}_{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}}}^2+{{\hat{c}}_{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}}}^2}\&CenterDot;\sqrt{{{\hat{a}}_{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}}}^2+{{\hat{b}}_{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}}}^2+{{\hat{c}}_{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}}}^2}}>\cos (\frac{3}{2}\&CenterDot;\mathrm{\&phi;})\\ \frac{|{\hat{a}}_u\&CenterDot;{\hat{X}}_{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}1}+{\hat{b}}_u\&CenterDot;{\hat{Y}}_{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}1}+{\hat{c}}_u\&CenterDot;{\hat{Z}}_{\mathrm{<figure-callout\; id="1"\; label="v"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}1}+{\hat{d}}_u|}{\sqrt{{{\hat{a}}_{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}}}^2+{{\hat{b}}_{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}}}^2+{{\hat{c}}_{\mathrm{<figure-callout\; id="2"\; label="u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}}}^2}}<\frac{3}{2}\&CenterDot;\mathrm{\&epsiv;}\\ \frac{|{\hat{a}}_v\&CenterDot;{\hat{X}}_{\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}1}+{\hat{b}}_v\&CenterDot;{\hat{Y}}_{\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}1}+{\hat{c}}_v\&CenterDot;{\hat{Z}}_{\mathrm{<figure-callout\; id="1"\; label="u"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">u</figure-callout>}1}+{\hat{d}}_v|}{\sqrt{{{\hat{a}}_{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}}}^2+{{\hat{b}}_{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}}}^2+{{\hat{c}}_{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">v</figure-callout>}}}^2}}<\frac{3}{2}\&CenterDot;\mathrm{\&epsiv;}\end{array}\right.$

The 2nd) step, the reliable flat blocks hypothesis of search coplane

Generate a non-directed graph, be designated as

Use K ₁∪ K ₂In each reliable flat blocks hypothesis generate a node, if any two reliable flat blocks hypothesis are coplanes, then connect the node of their correspondences.

Calculating chart

All very big groups, be designated as $\left\{{\hat{Q}}_s\right|s=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,{\hat{N}}_Q\},$ Any one is greatly rolled into a ball

Be figure A subclass, be designated as $\left\{{\hat{P}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;{\hat{m}}_s\}\&Subset;{\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">K</figure-callout>}}_1\&cup;{\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">K</figure-callout>}}_2,$ Any two reliable flat blocks hypothesis all are coplanes in this subclass.

The 3rd) step, the reliable flat blocks that merges coplane is supposed

For any one very big group

Merge all reliable flat blocks hypothesis that it comprises, obtain an open plane Plane equation is designated as ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s\&CenterDot;X+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s\&CenterDot;Y+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s\&CenterDot;Z+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{d}}}_s=0,$ Wherein:

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s=\frac{1}{{\hat{m}}_s}\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}\frac{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}{\sqrt{{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s=\frac{1}{{\hat{m}}_s}\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}\frac{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}{\sqrt{{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s=\frac{1}{{\hat{m}}_s}\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}\frac{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}{\sqrt{{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{d}}}_s=-\frac{1}{{\hat{m}}_s}({\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}{\hat{X}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right),1}+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}{\hat{Y}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right),1}+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}{\hat{Z}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right),1})$

Will $\left\{{\hat{P}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;{\hat{m}}_s\}$ In all summits of each reliable flat blocks hypothesis project to

On, connecting all subpoints in turn according to the order between the summit in the flat blocks hypothesis, all projection straight line sections that obtain become

Member's linear feature.

The open plane of all that obtain after the merging is designated as $\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}}=\left\{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_s\right|s=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{N}}\},$ Wherein, $\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{N}}={\hat{N}}_Q,$ The three-dimensional planar real border of their correspondences is unknown, open arbitrarily plane

Plane equation be ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s\&CenterDot;X+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s\&CenterDot;Y+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s\&CenterDot;Z+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{d}}}_s=0,$ The set of its all member's linear features is designated as ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_s=\left\{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{L}}}_l^s\right|l=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{n}}}_s\},$ Any one member's linear feature

End points be designated as

With

Second step, the border of formation level

In (one) step, utilize member's linear feature on open plane and plane intersection line formation level border to suppose

In the 1st step, generate the plane intersection line set

Choose two open planes arbitrarily

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_j\&Element;\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}},$ If

Judge that they can intersect, calculate their intersection equation, be designated as $\frac{X-{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{X}}}_0}{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{p}}}_{\mathrm{ij}}}=\frac{Y-{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{Y}}}_0}{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{q}}}_{\mathrm{ij}}}=\frac{Z-{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{Z}}}_0}{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{r}}}_{\mathrm{ij}}}.$

The set of being made up of all plane intersection lines is designated as U.

In the 2nd step, judge the collinear relationship between plane intersection line and plane member's linear feature and revise member's linear feature set on open plane

For any open plane

In U, select it with

The intersection on any open plane, if its any member's linear feature ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{L}}}_l^i\&Element;{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_i$ The angle of place straight line and this plane intersection line is less than φ, and

Mid point to the distance of plane intersection line less than ε, then judge

With this plane intersection line conllinear, and from open plane

Member's linear feature set Middle deletion

By all with

The set formed of the plane intersection line of certain member's linear feature conllinear, be designated as

In the 3rd step, replenish the open plane that lacks

For any one open plane Any one member's linear feature ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{L}}}_l^i\&Element;{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_i,$ Use a kind of method, generate a new open plane by it by the wall scroll 3 d-line generation hypothesis of plane mechanism,

Become its member's linear feature, add this plane to set In.

The 4th step is at the open plane computations plane intersection line that replenishes

For any one open plane ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_u\&Element;\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}},$ If it is generated by the wall scroll 3 d-line, then utilize and identical method of the 1st step, calculate it and any other open plane ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_v\&Element;\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}}$ Intersection, and with this plane intersection line add to the set U in; If this plane intersection line with The angle of any member's linear feature place straight line less than φ, and

Mid point to the distance of plane intersection line less than ε, then this plane intersection line is added to the straight line set

In, and from set

Member's linear feature of middle this conllinear of deletion.

In the 5th step, add the member linear feature of plane intersection line for open plane

For any plane intersection line among the set U, suppose that it is open plane With

Intersection, then this plane intersection line is added to

With Member's linear feature set

With

In go.

The end points of the member's linear feature that increases newly is uncertain at this moment.

In the 6th step, whether arbitrary plane intersection and other member's linear feature intersect in member's linear feature set on the open plane of judgement

For any open plane

Any one member's linear feature, if it belong to the set U, and it and

The angle of any other member's linear feature belong to scope

Judge that then these two member's linear features intersect.

The 7th step is by member's linear feature formation level border on open plane

Method may further comprise the steps:

The 1st) go on foot, split all member's linear feature to two subclass of the face of setting level

For any one open plane Its all member's linear features are split into two subclass, and a subclass is ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_u\&cap;{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_u^{\&prime;},$ Be designated as U ₁, another subclass is

Be designated as U ₂

The 2nd) step, generate the non-directed graph set

Generate U ₂All subclass.

Choose U ₂Any one subclass and U ₁Form new member's linear feature collection, be designated as U ₀, generate a non-directed graph.This non-directed graph is with U ₀In each member's linear feature as node, if member's linear feature of any two node correspondences intersects, then they are coupled together.

The 3rd) step, generate planar boundary

Its all Hamilton circles of search in each non-directed graph that previous step obtains, for each unduplicated circle,, calculate the intersection point of member's linear feature of adjacent node correspondence according to the order of connection between the node that wherein comprises, they as the summit, are generated a closed hypothesis of plane mechanism; If there is no the Hamilton circle is then searched for all Hamilton paths, for each unduplicated path, according to the order of connection between the node that wherein comprises, calculate the intersection point of member's linear feature of adjacent node correspondence, they as the summit, are generated a semi-open hypothesis of plane mechanism.

The set that all the closed hypothesises of plane mechanism that generate and the semi-open hypothesis of plane mechanism are formed be designated as Ω '=P ' _k| k=1,2 ..., N ' }, any one element P ' wherein _kOrderly vertex set with it is represented, is designated as $M_k^{\&prime;}=\left\{T_k^j\right|j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,n_k^{\&prime;}\},$ If ${\mathrm{<figure-callout\; id="1"\; label="T\; k"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">T</figure-callout>}}_k^{}=T_k^{n_k^{\&prime;}},$ Expression P ' _kBe a closed hypothesis of plane mechanism, otherwise, expression P ' _kIt is a semi-open hypothesis of plane mechanism.Any one closed or semi-open hypothesis of plane mechanism P ' _kPlane equation be identical with the open plane that generates it, for corresponding with Ω ', this plane equation is rewritten as a ' _kX+b ' _kY+c ' _kZ+d ' _k=0.

In (two) step, utilize the disappearance border of extracting the linear feature and the additional semi-open hypothesis of plane mechanism of heuristic rule

For arbitrary plane hypothesis P ' _k∈ Ω ' if it is semi-open, then replenishes the border of disappearance for it, at first utilize and extract linear feature H ₁And H ₂If, can search the linear feature that can be used for replenishing the disappearance border, then this step finishes, and generates one or more closed hypothesises of plane mechanism; Otherwise, utilize heuristic rule to replenish the border of disappearance, generate a closed hypothesis of plane mechanism.Method is as follows:

In the 1st step, utilize the linear feature that extracts in the image to replenish the disappearance border of the semi-open hypothesis of plane mechanism

The 1st) step, calculate P ' _kThe region of disappearance border projection in left and right image

Projection P ' _kTo left and right image, obtain point set respectively $\left\{{\mathrm{Tl}}_k^j\right|j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,n_k^{\&prime;}\}$ With $\left\{{\mathrm{Tr}}_k^j\right|j=\mathrm{1,2},\&CenterDot;\&CenterDot;\&CenterDot;,n_k^{\&prime;}\}.$

Set the distance threshold h of an expression linear feature integrality _d, its expression, in piece image, if linear feature is not extract completely, the distance between the end points that is extracted to it of its real end points is less than h so _d

Specify P ' _kThe zone at projection place in left image, disappearance border be a quadrilateral, be designated as R _k ^L, its summit is designated as g respectively ₁, g ₂, g ₃And g ₄, wherein, ${\mathrm{<figure-callout\; id="1"\; label="g"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">g</figure-callout>}}_1={\mathrm{<figure-callout\; id="1"\; label="Tl\; k"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">Tl</figure-callout>}}_k^{},$ ${\mathrm{<figure-callout\; id="2"\; label="g"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">g</figure-callout>}}_2={\mathrm{Tl}}_k^{n_k^{\&prime;}},$ g ₃It is line segment

Extended line on to the some g ₂Distance be h _dPoint, g ₄Be line segment Tl _k ²Tl _k ¹Extended line on to the some g ₁Distance be h _dPoint, connecting them becomes R _k ^LFour edges g ₁g ₂, g ₂g ₃, g ₃g ₄And g ₄g ₁

Utilize similar method, calculate P ' _kThe region R at projection place in right image, disappearance border _k ^R, its four summits are designated as g ' respectively ₁, g ' ₂, g ' ₃And g ' ₄

The 2nd) in the step, utilize region R _k ^LAnd R _k ^RInterior H ₁And H ₂In linear feature replenish the disappearance border of the semi-open hypothesis of plane mechanism

For H ₁In any one linear feature, if its two end points all drop on R _k ^LIn, and its extended line and line segment g ₂g ₃And g ₄g ₁Intersect, then connect two intersection points and obtain the straight line section, according to P ' _kPlane equation and the collinearity equation in the photogrammetric Theory, calculate the three dimensions point of two end points correspondences of this straight-line segment, with P ' _kVertex set in remove T _k ¹With

Outside all summits together, generate a closed hypothesis of plane mechanism.For H ₂In any one linear feature, if its two end points all drop on R _k ^RIn, and its extended line and line segment g ' ₂G ' ₃And g ' ₄G ' ₁Intersect, calculate its same place of end points in left image, in the computing method of same place and the first step of the present invention the 3rd of the 2nd step in (two) step the) the same place computing method in step are identical, connecting two same places is the straight line section, if any disappearance border that searches in this straight-line segment and the left image is conllinear not, then calculate the extended line and the line segment g ' of this linear feature in right image ₂G ' ₃And g ' ₄G ' ₁Intersection point, connect these two intersection points and obtain the straight line section, according to P ' _kPlane equation and the collinearity equation in the photogrammetric Theory, calculate the three dimensions point of two end points correspondences of this straight-line segment, with P ' _kVertex set in remove T _k ¹With

Outside all summits together, generate a closed hypothesis of plane mechanism.

The 2nd step is according to the disappearance border of the additional semi-open hypothesis of plane mechanism of heuristic rule

If respectively with T _k ²T _k ¹With

Be shared border and P _k' two adjacent hypothesises of plane mechanism have all obtained complete border, according to the isometric rule of adjacent polygon plane common edge, at T _k ²T _k ¹Or 1 Φ of appointment on its extended line ₁, make line segment T _k ²Φ ₁With with T _k ²T _k ¹For the shared border of the adjacent plane on shared border hypothesis isometric,

Or 1 Φ of appointment on its extended line ₂, make line segment

With with

For the shared border of the adjacent plane on shared border hypothesis isometric, line segment Φ ₁Φ ₂Become additional disappearance border.

If respectively with T _k ²T _k ¹With

Be shared border and P ' _kHave only a hypothesis of plane mechanism to obtain complete border in two adjacent hypothesises of plane mechanism, suppose that this adjacent plane is assumed to be P ' _l, and T _k ²T _k ¹Be P ' _lWith P ' _kShared border, then according to the shared border of adjacent polygon plane isometric rule, at T _k ²T _k ¹Or specify on its extended line a bit, make line segment T _k ²Φ ₁With P ' _lShared border isometric,

Or 1 Φ of appointment on its extended line ₂, make line segment With T _k ²Φ ₁Isometric, line segment Φ ₁Φ ₂Become additional disappearance border.

According to P ' _kPlane equation and the collinearity equation in the photogrammetric Theory, calculate Φ ₁And Φ ₂Corresponding three dimensions point is with P ' _kVertex set in remove T _k ¹With

Outside all summits together, generate a closed hypothesis of plane mechanism.

The shared border here is meant adjacent two boundary lines that polygon plane is owned together.

The closed hypothesis of plane mechanism among the Ω ' and the closed hypothesis of plane mechanism that is generated by the semi-open hypothesis of plane mechanism among the Ω ' merged becomes closed hypothesis of plane mechanism set, is designated as Ω "={ P " _i| i=1,2 ..., N " }, wherein, any closed hypothesis of plane mechanism P " _iBy n " _iThe bar boundary line is formed.

(3) goes on foot, and chooses the closed hypothesis of plane mechanism of global optimum

The 1st goes on foot, and calculates the reliability measure of the closed hypothesis of plane mechanism

For any one closed hypothesis of plane mechanism P " _i∈ Ω ", utilize formula one to calculate P " _iThe elevation consistance estimate, if P " _iIt is 0 that the known elevation that covers is counted, and then specifying its elevation consistance to estimate is 1/ (2 λ), utilizes formula three, calculates its reliability measure, is designated as E " _iAt P " _iOn the plane, place, specify next-door neighbour outside its every boundary line, with the rectangle plane zone of this boundary line as a long limit, the long length that equals the boundary line, wide half that equals to grow, the elevation consistance of utilizing formula one to calculate this rectangle plane zone is estimated, if it is 0 that the known elevation that it covers is counted, then specifying its elevation consistance to estimate is 1/ (2 λ), utilize formula three, calculate its reliability measure, note is E by the reliability measure in the rectangle plane zone of j bar boundary line correspondence " _{I, j}, calculate closed hypothesis of plane mechanism P according to following formula " _iThe reliability measure on corresponding true planar border, border be

${\mathrm{\&gamma;}}_i=\frac{E_i^{\&prime;\&prime;}}{\underset{j=1}{\overset{n_i^{\&prime;\&prime;}}{\mathrm{\&Sigma;}}}{E_{i,j}}^{\&prime;\&prime;}}$ (formula four)

In the 2nd step, find the solution the optimum closed hypothesis of plane mechanism

Generate a non-directed graph, each closed hypothesis of plane mechanism is a node to be designated as G in "; with set omega ", utilize formula four computation's reliability estimate as with the attribute of the corresponding node of the closed hypothesis of plane mechanism, if all there is not lap in the view field of the closed hypothesis of plane mechanism of any two node correspondences in left image and right image, then connect this two nodes, otherwise do not connect.Calculating chart G " all very big groups, therefrom choose all closed hypothesises of plane mechanism that the very big group of node attribute values sum maximum comprises, as the closed hypothesis of plane mechanism of optimum, promptly three-dimensional planar extracts the result.

Description of drawings

Fig. 1 is the synoptic diagram of three dimensional plane extraction method flow process of the present invention;

Fig. 2 is the synoptic diagram that the elevation consistance of calculating flat blocks hypothesis is estimated;

Fig. 3 is the synoptic diagram that replenishes the disappearance border of the semi-open hypothesis of plane mechanism according to heuristic rule.

Embodiment

Below in conjunction with accompanying drawing the present invention is further explained.

Fig. 1 is the synoptic diagram of the described three dimensional plane extraction method flow process of the inventive method; Technical scheme comprises: the first step, find the solution the open three-dimensional planar at target surface place, and this step was divided into for two steps again, (1) step, the three-dimensional planar hypothesis generates, and this step is divided into two sub-steps again, the 1st step, calculating three-dimensional planar hypothesis, in the 2nd step, the three-dimensional planar that merges coplane is supposed, (two) step, the three-dimensional planar hypothesis confirms, this step is divided into four sub-steps again, in the 1st step, cuts apart three-dimensional planar and is assumed to be the flat blocks hypothesis, the 2nd step, calculate the reliability measure of each flat blocks hypothesis, in the 3rd step, search for the most reliable flat blocks hypothesis, in the 4th step, the reliable flat blocks that merges coplane is supposed; Second step, the border of formation level, this step was divided into for three steps again, (1) step, utilize the member's linear feature and the plane intersection line formation level border hypothesis on open plane, this step is divided into seven sub-steps again, the 1st step, the set of generation plane intersection line, the 2nd step, judge the collinear relationship between plane intersection line and plane member's linear feature and revise member's linear feature set on open plane, the 3rd step, replenish the open plane of disappearance, the 4th step, at the open plane computations plane intersection line that replenishes, the 5th step, add the member linear feature of plane intersection line for open plane, in the 6th step, whether arbitrary plane intersection and other member's linear feature intersect in member's linear feature set on the open plane of judgement, the 7th step, by member's linear feature formation level border on open plane, in (two) step, utilize the disappearance border of extracting the linear feature and the additional semi-open hypothesis of plane mechanism of heuristic rule, this step is divided into two sub-steps again, the 1st step, utilize the linear feature that extracts in the image to replenish the disappearance border of the semi-open hypothesis of plane mechanism, the 2nd goes on foot, replenish the disappearance border of the semi-open hypothesis of plane mechanism according to heuristic rule, (3) goes on foot, and chooses the closed hypothesis of plane mechanism of global optimum, and this step is divided into two sub-steps again, the 1st step, calculate the reliability measure of the closed hypothesis of plane mechanism, in the 2nd step, find the solution the optimum closed hypothesis of plane mechanism.

Fig. 2 is the synoptic diagram that the elevation consistance of the 2nd step calculating flat blocks hypothesis during (two) of the described first step of the inventive method goes on foot is estimated: O ₁And O ₂Be camera center, the left and right sides, spot elevation

With O ₁And O ₂Line hand over flat blocks in a F respectively ₁And F ₂This method is calculated spot elevation respectively

With a F ₁, F ₂Distance, if these two distances are more little, represent this flat blocks hypothesis and spot elevation

Consistance good more.

Fig. 3 is the synoptic diagram that the 2nd step was utilized the disappearance border of the additional semi-open hypothesis of plane mechanism of heuristic rule in going on foot in (two) of described second step of the inventive method: suppose that the semi-open hypothesis of plane mechanism has two known borders, Fig. 3 a) in, A and B are the closed hypothesises of plane mechanism with integral edge, C is a semi-open hypothesis of plane mechanism, T _k ²T _k ¹Be the common edge boundary line of A and C, T _k ²T _k ³Be the common edge boundary line of B and C, dotted line is the disappearance boundary line Φ that replenishes for plane C according to the isometric enlightening rule in common edge boundary line ₁Φ ₂Fig. 3 b) in, with Fig. 3 in a) different place be that B also is a semi-open hypothesis of plane mechanism, at first, utilizes the isometric rule in common edge boundary line of A and C, specifies an end points Φ of the disappearance boundary line of C ₁,, specify another end points Φ of disappearance boundary line again according to the common edge boundary line of B and C and the isometric rule in common edge boundary line of A and C ₂, connect them, generate the disappearance boundary line of semi-open hypothesis of plane mechanism C.

Describe the relevant details among the present invention below in detail.

First point calculates the projection of three dimensions point on three-dimensional planar

Projection three-dimensional spatial point described in the present invention is all adopted in the following method to the step of three-dimensional planar, and particular content is:

Suppose that three dimensions point is (X ₀, Y ₀, Z ₀), the three-dimensional planar equation is aX+bY+cZ+d=0, then projection (the X of this three dimensions point on this three-dimensional planar ₀', Y ₀', Z ₀'), wherein

$\left\{\begin{array}{c}{X}_0^{\&prime;}=X_0-\frac{a\&CenterDot;(a\&CenterDot;X_0+b\&CenterDot;Y_0+c\&CenterDot;Z_0+d)}{a^2+{\mathrm{<figure-callout\; id="2"\; label="b"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">b</figure-callout>}}^2+c^2}\\ Y_0^{\&prime;}=Y_0-\frac{b\&CenterDot;(a\&CenterDot;X_0+b\&CenterDot;Y_0+c\&CenterDot;Z_0+d)}{a^2+{\mathrm{<figure-callout\; id="2"\; label="b"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">b</figure-callout>}}^2+c^2}\\ Z_0^{\&prime;}=Z_0-\frac{c\&CenterDot;(a\&CenterDot;X_0+b\&CenterDot;Y_0+c\&CenterDot;Z_0+d)}{a^2+{\mathrm{<figure-callout\; id="2"\; label="b"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">b</figure-callout>}}^2+{\mathrm{<figure-callout\; id="2"\; label="c"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">c</figure-callout>}}^2}\end{array}\right.$

Second point calculates the projection of three dimensions point on left and right image

Projection three-dimensional spatial point described in the present invention is all adopted in the following method to the step of known left and right image, and particular content is:

According to photogrammetric Theory, utilize known left and right image imaging parameter, can obtain the relevant collinearity equation group of every width of cloth image, can calculate the projection of three dimensions point in left and right image according to collinearity equation.

Obtain the method for collinearity equation group specifically referring to Wang Zhizhuo by the image imaging parameter, Principlesof photogrammetry with remote sensing, Press of Wuhan Technical University ofSurveying and Mapping, 1990,18-26.

Thirdly, calculate the projection of three-dimensional straight line segment on three-dimensional planar

Projection three-dimensional straight-line segment described in the present invention is all adopted in the following method to the step of three-dimensional planar, and particular content is:

According to the method for projection three-dimensional spatial point to three-dimensional planar, two end points of projection straight line section are to three-dimensional planar respectively, and connecting two subpoints that obtain is the straight line section, and this straight-line segment is exactly the projection of being asked.

The 4th point calculates the projection of three-dimensional straight line segment on left and right image

Projection three-dimensional straight-line segment described in the present invention is all adopted in the following method to the step of known left and right image, and particular content is:

According to the method for projection three-dimensional straight-line segment to left and right image, two end points of projection straight line section are to left and right image respectively, and connecting two subpoints that obtain in every width of cloth image is the straight line section, and these two straight-line segments are exactly the projection of being asked.

The 5th point, the gradient descent method is calculated plane equation

In (one) step of the first step of the present invention, in the 2nd step, adopt the gradient descent method to calculate by L _uAnd L _vThe method of the plane equation of appointment, particular content is:

Become known for generating two 3 d-line feature L on plane _u∈ S and L _v∈ S, end points is respectively (X _U1, Y _U1, Z _U1) and (X _U2, Y _U2, Z _U2) and (X _V1, Y _V1, Z _V1) and (X _V2, Y _V2, Z _V2), calculate by them and determine that the method for plane equation is made up of following three steps:

The 1st) go on foot, find the solution the initial value of plane equation coefficient

Calculation level (X ₀, Y ₀, Z ₀) and two value D _u, D _v, wherein

$\left\{\begin{array}{c}{X}_0=({\mathrm{<figure-callout\; id="1"\; label="X\; u"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_u+{\mathrm{<figure-callout\; id="2"\; label="X\; u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_u+{\mathrm{<figure-callout\; id="1"\; label="X\; v"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">X</figure-callout>}}_v+X_{v2})/4\\ Y_0=({\mathrm{<figure-callout\; id="1"\; label="Y\; u"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">Y</figure-callout>}}_u+{\mathrm{<figure-callout\; id="2"\; label="Y\; u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">Y</figure-callout>}}_u+{\mathrm{<figure-callout\; id="1"\; label="Y\; v"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">Y</figure-callout>}}_v+Y_{v2})/4\\ Z_0=({\mathrm{<figure-callout\; id="1"\; label="Z\; u"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">Z</figure-callout>}}_u+{\mathrm{<figure-callout\; id="2"\; label="Z\; u"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">Z</figure-callout>}}_u+{\mathrm{<figure-callout\; id="1"\; label="Z\; v"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">Z</figure-callout>}}_v+Z_{v2})/4\end{array}\right.$

$D_u=\sqrt{{(X_{u1}-X_{u2})}^2+{(Y_{u1}-Y_{u2})}^2+{(Z_{u1}-Z_{u2})}^2}$

$D_v=\sqrt{{(X_{v1}-X_{v2})}^2+{(Y_{v1}-Y_{v2})}^2+{(Z_{v1}-Z_{v2})}^2}$

Obtain L _uAnd L _vThe normalization direction vector be $v_u=\left(\begin{array}{ccc}\frac{X_{u1}-X_{u2}}{D_u}& \frac{Y_{u1}-Y_{u2}}{D_u}& \frac{Z_{u1}-Z_{u2}}{D_u}\end{array}\right)$ With $v_v=\left(\begin{array}{ccc}\frac{X_{v1}-X_{v2}}{D_v}& \frac{Y_{v1}-Y_{v2}}{D_v}& \frac{Z_{v1}-Z_{v2}}{D_v}\end{array}\right),$ And their cross product v _u* v _v=(α β γ).

Calculating χ=-(α X ₀+ β Y ₀+ λ Z ₀), obtain a four-vector (α β γ χ); Calculate trivector (α ' β ' γ ')=(v _u+ v _v) * (α β γ), further calculate χ '=-(α ' X ₀+ β ' Y ₀+ γ ' Z ₀) obtain another four-vector (α ' β ' γ ' χ ').

According to objective function

f(a，b，c，d)＝(a·X _u1+b·Y _u1+c·Z _u1+d) ²+(a·X _u2+b·Y _u2+c·Z _u2+d) ²+(a·X _v1+b·Y _v1+c·Z _v1+d) ²+(a·X _v2+b·Y _v2+c·Z _v2+d) ²

If f (α β γ χ)＞f (α ' β ' γ ' χ '), the initial value of plane equation coefficient Be (α ' β ' γ ' χ '), otherwise be (α β γ χ).

The 2nd) step, utilize the gradient descent method to calculate plane equation coefficient

Objective function be f (a, b, c, d), the plane equation coefficient initial value is

The iteration convergence condition is

Utilize the gradient descent method to calculate optimum plane equation coefficient.

The computation process of gradient descent method is referring to " principle of optimality, the method and find the solution software " of Yang Mingsheng, Luo Changtong work, Beijing: Science Press,, 23-31 page or leaf in 2006.

The 3rd) in the step, obtain the hypothesis of plane mechanism

Calculate I=f (a ^(k), b ^(k), c ^(k), d ^(k)), if I＞4 ε ², then do not generate the hypothesis of plane mechanism; Otherwise obtain hypothesis of plane mechanism equation is a ^(k)X+b ^(k)Y+c ^(k)Z+d ^(k)=0.

The 6th point is greatly rolled into a ball the rapid solving method

All very big groups find the solution and all adopt a kind of method for solving of greatly rolling into a ball fast among the present invention, specifically referring to Tomita E, Tanaka A, Takahashia H.The worst-case time complexity for generating allmaximal cliques and computational experiments.Theoretical Computer Science, 2006,363:28-42.

The 7th point, foundation

Three pairs of same places in left and right image

With

With And

With

Set of computations K _w ^LIn any some same place in right image

The 2nd step in (two) step of the first step of the present invention the 3rd) in the step, utilize

Three pairs of same places in left and right image

With

With

And

With

Calculate the same place of arbitrfary point in right image in the left projection polygon, particular content is:

Generating vector V and matrix M is respectively

$V=\left[\begin{array}{c}\hat{x}{r}_{w1}\\ \hat{x}{r}_{w2}\\ \hat{x}{r}_{w3}\\ \hat{y}{r}_{w1}\\ \hat{y}{r}_{w2}\\ \hat{y}{r}_{w3}\end{array}\right],$ $M=\left[\begin{array}{cccccc}\hat{x}{l}_{w1}& \hat{y}{l}_{\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">w</figure-callout>}1}& 1& 0& 0& 0\\ \hat{x}{l}_{w2}& \hat{y}{l}_{\mathrm{<figure-callout\; id="2"\; label="w"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">w</figure-callout>}2}& 1& 0& 0& 0\\ \hat{x}{l}_{w2}& \hat{y}{l}_{w3}& 1& 0& 0& 0\\ 0& 0& 0& \hat{x}{l}_{w1}& \hat{y}{l}_{\mathrm{<figure-callout\; id="1"\; label="w"\; filenames="A2009100441920045H2.png,A2009100441920046H1.png"\; state="\{\{state\}\}">w</figure-callout>}1}& 1\\ 0& 0& 0& \hat{x}{l}_{w2}& \hat{y}{l}_{\mathrm{<figure-callout\; id="2"\; label="w"\; filenames="A2009100441920046H1.png"\; state="\{\{state\}\}">w</figure-callout>}2}& 1\\ 0& 0& 0& \hat{x}{l}_{w2}& \hat{y}{l}_{w3}& 1\end{array}\right]$

Calculate

Transform coefficient on the plane between the same place

[a ₁₁?a ₁₂?a ₁₃?a ₂₁?a ₂₂?a ₂₃] ^T＝M ^-1V

Then

Arbitrfary point in the view field in left image

Same place in right image is $(a_{11}\&CenterDot;{\hat{x}}_1+a_{12}\&CenterDot;{\hat{y}}_1+a_{13},a_{21}\&CenterDot;{\hat{x}}_1+a_{22}\&CenterDot;{\hat{y}}_1+a_{23}).$

The 8th point, the first step of the present invention (two) step the 2nd the step the 3rd) step in calculating K _w ^RThe interpolation algorithm of middle arbitrary image point gray-scale value adopts the bilinear interpolation algorithm.

The 9th point is by the method for the wall scroll 3 d-line generation hypothesis of plane mechanism

In the 3rd step in (one) step in second step of the present invention, utilize wall scroll 3 d-line feature to generate the method for three-dimensional planar hypothesis referring to a kind of method of utilizing many views picture to extract three-dimensional planar and rebuild buildings, be Baillard C, Zisserman A.A plane-sweep strategy for the 3D reconstruction ofbuildings from multiple images.ISPRS Journal of Photogrammetry and RemoteSensing, 2000,33 (B2): 56-62.

The 10th point, search Hamilton circle and Hamilton route method in non-directed graph

In the 7th step in (one) step in second step of the present invention, search Hamilton circle and Hamilton route method are with reference to the political affairs of model benefit in non-directed graph, Wang Yi, Gong's generation just waits " the graph theory guiding " of translating, Beijing: People's Telecon Publishing House, Gary Chartrand, Ping Zhang work .2007,122-136 page or leaf.

Claims (11)

1. three dimensional plane extraction method, the stereo-picture of 3 d-line feature set in the known target scene and sparse digital elevation model data, the object scene obtained from two different visual angles and obtain all inside and outside parameters of their stereoscopic camera, the straight line of stereoscopic image extracts the result, it is characterized in that, specifically comprise the steps:

The first step is obtained the hi-Fix information that three-dimensional planar is opened at the target surface place

Utilize the 3 d-line feature to generate open three-dimensional planar hypothesis, utilize member's linear feature of three-dimensional planar hypothesis to cut apart the open a plurality of flat blocks hypothesis of three-dimensional planar hypothesis formation, utilize a plurality of flat blocks hypothesis to choose optimal planar piece hypothesis, merge optimum flat blocks hypothesis at last, generate open three-dimensional planar;

Second step, the border of formation level

Based on the open three-dimensional planar that generates, utilize the member's linear feature and the interplanar intersection formation level border of open three-dimensional planar, utilize linear feature and the heuristic rule extracted in the stereo-picture to replenish the planar boundary of disappearance again, generate the closed hypothesis of plane mechanism by planar boundary, utilize optimization method to select the optimum closed hypothesis of plane mechanism.

2. three dimensional plane extraction method according to claim 1 is characterized in that:

Two width of cloth images of the different visual angles of known a certain scene are called left and right image, and all inside and outside parameters of absorbing the left and right camera of this two width of cloth image, and the coordinate of wherein left and right camera shooting center in world coordinate system is designated as (X respectively _O1, Y _O1, Z _O1) and (X _O2, Y _O2, Z _O2); Linear feature collection by extracting in this two width of cloth image is designated as H respectively ₁And H ₂N in known this scene _SBar 3 d-line feature is designated as S={L _i| i=1,2 ..., N _S, the arbitrary element L in this set _iRepresent a three-dimensional straight line segment, its two end points are respectively (X _I1, Y _I1, Z _I1) and (X _I2, Y _I2, Z _I2); Sparse dem data in known this scene is designated as the spot elevation set

The average error of element is λ in this set;

The first step is found the solution the open three-dimensional planar at target surface place

In (one) step, the three-dimensional planar hypothesis generates the set angle threshold phi; Setpoint distance thresholding ε;

The angular range of two adjacent boundary place straight lines on any one plane on target setting surface is

Minimum angle between any two adjacent planes of target surface is

In the 1st step, calculate the three-dimensional planar hypothesis

3 d-line feature among the pair set S makes up in twos, generates three-dimensional planar hypothesis set omega={ P _i| i=1,2 ..., N} is for any one three-dimensional planar hypothesis P wherein _i, plane equation is a _iX+b _iY+c _iZ+d _i=0, two 3 d-line features that are used to generate it are called P _iMember's linear feature;

In the 2nd step, the three-dimensional planar that merges coplane is supposed

The three-dimensional planar of all coplanes hypothesis merges among the pair set Ω, and the three-dimensional planar hypothesis set that obtains is designated as Ω={ P _w| w=1,2 ..., N}, three-dimensional planar hypothesis P arbitrarily _wMember's linear feature set be designated as ${\stackrel{\&OverBar;}{M}}_w=\left\{{\stackrel{\&OverBar;}{L}}_j^w\right|j=\mathrm{1,2},...,{\stackrel{\&OverBar;}{n}}_w\},$ Any one member's linear feature L _j ^wEnd points be designated as (X _{J, 1} ^w, Y _{J, 1} ^w, Z _{J, 1} ^w) and (X _{J, 2} ^w, Y _{J, 2} ^w, Z _{J, 2} ^w);

In (two) step, the three-dimensional planar hypothesis confirms

In the 1st step, cut apart three-dimensional planar and be assumed to be the flat blocks hypothesis

For any one three-dimensional planar hypothesis P _i∈ Ω utilizes its member's linear feature M _iIt is cut apart, obtain some flat blocks hypothesis;

The set of being cut apart the flat blocks hypothesis that obtains by all planes among the Ω is designated as $\widehat{\mathrm{\&Omega;}}=\left\{{\hat{P}}_w\right|w=\mathrm{1,2},...,\hat{N}\},$ Wherein, any one flat blocks hypothesis

Vertex set by it is represented, is designated as $\left\{({\hat{X}}_{\mathrm{wj}},{\hat{Y}}_{\mathrm{wj}},{\hat{Z}}_{\mathrm{wj}})\right|j=\mathrm{1,2},...,{\hat{n}}_w\},$ Cut apart among the Ω and obtain The equation of three-dimensional planar hypothesis be designated as ${\hat{a}}_w\&CenterDot;X+{\hat{b}}_w\&CenterDot;Y+{\hat{c}}_w\&CenterDot;Z+{\hat{d}}_w=0;$

In the 2nd step, calculate the reliability measure that each flat blocks is supposed

Spot elevation set J and left and right image that the basis of flat blocks hypothesis reliability measure is known calculate each some projection in left and right image among the J, obtain two plane point sets, are designated as respectively

With

Wherein,

With

It is respectively the spot elevation among the J

Subpoint in left and right image;

The arbitrary plane piece is supposed ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;},}$ Calculate its reliability measure;

In the 3rd step, search for reliable flat blocks hypothesis;

In the 4th step, the reliable flat blocks that merges coplane is supposed;

The open plane of all that obtain after the merging is designated as $\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}}=\left\{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_s\right|s=\mathrm{1,2},...,\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{N}}\},$ The three-dimensional planar real border of their correspondences is unknown, open arbitrarily plane

Plane equation be ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s\&CenterDot;X+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s\&CenterDot;Y+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s\&CenterDot;Z+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{d}}}_s=0,$ The set of its all member's linear features is designated as ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_s=\left\{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{L}}}_l^s\right|l=\mathrm{1,2},...,{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{n}}}_s\},$ Any one member's linear feature

End points be designated as With

Second step, the border of formation level

In (one) step, utilize member's linear feature on open plane and plane intersection line formation level border to suppose

In the 1st step, generate the plane intersection line set

Choose two open planes arbitrarily ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_i,{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_j\&Element;\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}},$ If

Judge that they can intersect, calculate their intersection equation, be designated as $\frac{X-{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{X}}}_0}{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{p}}}_{\mathrm{ij}}}=\frac{Y-{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{Y}}}_0}{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{q}}}_{\mathrm{ij}}}=\frac{Z-{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{Z}}}_0}{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{r}}}_{\mathrm{ij}}};$

The set of being made up of all plane intersection lines is designated as U;

In the 2nd step, judge the collinear relationship between plane intersection line and plane member's linear feature and revise member's linear feature set on open plane

For any open plane

In U, select it with

The intersection on any open plane, if its any member's linear feature ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{L}}}_l^i\&Element;{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_i$ The angle of place straight line and this plane intersection line is less than φ, and

Mid point to the distance of plane intersection line less than ε, then judge

With this plane intersection line conllinear, and from open plane

Member's linear feature set

Middle deletion

By all with

The set formed of the plane intersection line of certain member's linear feature conllinear, be designated as

In the 3rd step, replenish the open plane that lacks

For any one open plane

Any one member's linear feature ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{L}}}_l^i\&Element;{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_i,$ Use a kind of method, generate a new open plane by it by the wall scroll 3 d-line generation hypothesis of plane mechanism,

Become its member's linear feature, add this plane to set

In;

The 4th step is at the open plane computations plane intersection line that replenishes

For any one open plane ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_u\&Element;\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}},$ If it is generated by the wall scroll 3 d-line, then calculate it and any other open plane ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{P}}}_v\&Element;\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{\mathrm{\&Omega;}}}$ Intersection, and with this plane intersection line add to the set U in; If this plane intersection line with

The angle of any member's linear feature place straight line less than φ, and

Mid point to the distance of plane intersection line less than ε, then this plane intersection line is added to the straight line set

In, and from set

Member's linear feature of middle this conllinear of deletion;

In the 5th step, add the member linear feature of plane intersection line for open plane

For any plane intersection line among the set U, suppose that it is open plane

With

Intersection, then this plane intersection line is added to

With

Member's linear feature set

With In go;

In the 6th step, whether arbitrary plane intersection and other member's linear feature intersect in member's linear feature set on the open plane of judgement

For any open plane

Any one member's linear feature, if it belong to the set U, and it and

The angle of any other member's linear feature belong to scope

Judge that then these two member's linear features intersect;

The 7th step is by member's linear feature formation level border on open plane

The set that all the closed hypothesises of plane mechanism that generate and the semi-open hypothesis of plane mechanism are formed be designated as Ω '=P ' _k| k=1,2 ..., N ' }, any one element P ' wherein _kOrderly vertex set with it is represented, is designated as ${M_k}^{\&prime;}=\left\{T_k^j\right|j=\mathrm{1,2},...,{n_k}^{\&prime;}\}{P_k}^{\&prime;}$ Plane equation be a ' _kX+b ' _kY+c ' _kZ+d ' _k=0;

In (two) step, utilize the disappearance border of extracting the linear feature and the additional semi-open hypothesis of plane mechanism of heuristic rule

For arbitrary plane hypothesis P ' _k∈ Ω ' if it is semi-open, then replenishes the border of disappearance for it, at first utilize and extract linear feature H ₁And H ₂If, can search the linear feature that can be used for replenishing the disappearance border, then this step finishes, and generates one or more closed hypothesises of plane mechanism; Otherwise, utilize heuristic rule to replenish the border of disappearance, generate a closed hypothesis of plane mechanism;

(3) goes on foot, and chooses the closed hypothesis of plane mechanism of global optimum

The 1st goes on foot, and calculates the reliability measure of the closed hypothesis of plane mechanism;

In the 2nd step, find the solution the optimum closed hypothesis of plane mechanism

Generate a non-directed graph, each closed hypothesis of plane mechanism is a node to be designated as G in "; with set omega ", its reliability measure as with the attribute of the corresponding node of the closed hypothesis of plane mechanism, if all there is not lap in the view field of the closed hypothesis of plane mechanism of any two node correspondences in left image and right image, then connect this two nodes, otherwise do not connect; Calculating chart G " all very big groups, therefrom choose all closed hypothesises of plane mechanism that the very big group of node attribute values sum maximum comprises, extract the result as three-dimensional planar.

3. three dimensional plane extraction method according to claim 2 is characterized in that, the 3 d-line feature among the pair set S makes up in twos, generates the three-dimensional planar hypothesis, and concrete steps are:

Optional two 3 d-line feature L _u∈ S and L _v∈ S, L _uEnd points be (X _U1, Y _U1, Z _U1) and (X _U2, Y _U2, Z _U2), L _vEnd points be (X _V1, Y _V1, Z _V1) and (X _V2, Y _V2, Z _V2), utilize their objective definition function f (a, b, c, d):

f(a，b，c，d)＝(a·X _u1+b·Y _u1+c·Z _u1+d) ²+(a·X _u2+b·Y _u2+c·Z _u2+d) ²

+(a·X _v1+b·Y _v1+c·Z _v1+d) ²+(a·X _v2+b·Y _v2+c·Z _v2+d) ²

Adopt the gradient descent method calculate f (a, b, c, when d) getting minimum value (c d), obtains by L for a, b _uAnd L _vThe three-dimensional planar hypothesis equation of appointment is aX+bY+cZ+d=0.

4. three dimensional plane extraction method according to claim 3 is characterized in that, the method that merges the three-dimensional planar hypothesis of coplane is:

For any one three-dimensional planar hypothesis P _i∈ Ω, the mid point of two member's linear feature mid point lines that calculates it is in plane P _iOn projection, be designated as (X _M ⁱ, Y _M ⁱ, Z _M ⁱ);

The 1st) step, judge the coplanar relation that three-dimensional planar is supposed

For any two three-dimensional planars hypothesis P _i∈ Ω and P _j∈ Ω is if the inequality group below satisfying is then judged P _iAnd P _jCoplane:

$\left\{\begin{array}{c}\frac{|a_i\&CenterDot;a_j+b_i\&CenterDot;b_j+c_i\&CenterDot;c_j|}{\sqrt{{a_i}^2+{b_i}^2+{c_i}^2}\&CenterDot;\sqrt{{a_j}^2+{b_j}^2+{c_j}^2}}>\cos \mathrm{\&phi;}\\ \frac{|a_i\&CenterDot;X_M^j+b_i\&CenterDot;Y_M^j+c_i\&CenterDot;Z_M^j+d_i|}{\sqrt{{a_i}^2+{b_i}^2+{c_i}^2}}<\mathrm{\&epsiv;}\\ \frac{|a_j\&CenterDot;X_M^i+b_j\&CenterDot;Y_M^i+c_j\&CenterDot;Z_M^i+d_j|}{\sqrt{{a_j}^2+{b_j}^2+{c_j}^2}}<\mathrm{\&epsiv;}\end{array}\right.$

The 2nd) step, the three-dimensional planar hypothesis of search coplane

Generate a non-directed graph G, with each three-dimensional planar hypothesis P _i∈ Ω, then keeps between the node of their correspondences connecting, otherwise does not connect if any two three-dimensional planars hypothesis is a coplane as a node;

Calculate all very big groups of non-directed graph G, be designated as { Q _w| w=1,2 ..., N _Q, any one greatly rolls into a ball Q _wBe the subclass of Ω, be designated as $\left\{P_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},...m_w\};$

The 3rd) step, the three-dimensional planar that merges coplane is supposed

Greatly roll into a ball Q for any one _w, merge all three-dimensional planar hypothesis that it comprises, obtain a new three-dimensional planar hypothesis P _w, its plane equation is designated as a _wX+b _wY+c _wZ+d _w=0, wherein:

${\stackrel{\&OverBar;}{a}}_w=\frac{1}{m_w}\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}\frac{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}{\sqrt{{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{b}}_w=\frac{1}{m_w}\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}\frac{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}{\sqrt{{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{c}}_w=\frac{1}{m_w}\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}\frac{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}{\sqrt{{a_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{b_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2+{c_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{d}}_w=-\frac{1}{m_w}({\stackrel{\&OverBar;}{a}}_w\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}{X}_M^{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}+{\stackrel{\&OverBar;}{b}}_w\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}{Y}_M^{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}+{\stackrel{\&OverBar;}{c}}_w\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{m_w}{\mathrm{\&Sigma;}}}{Z}_M^{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)})$

Will $\left\{P_{{\mathrm{\&Gamma;}}_w\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},...,m_w\}$ In all member's linear features of each three-dimensional planar hypothesis project to P _wOn, all projection straight line sections that obtain become P _wMember's linear feature.

5. three dimensional plane extraction method according to claim 4 is characterized in that, for any one three-dimensional planar hypothesis P _i∈ Ω utilizes its member's linear feature M _iIt is cut apart, obtain some flat blocks hypothesis, concrete grammar may further comprise the steps:

The 1st) step, member's linear feature of merging conllinear

For P _iAny two member's linear feature L _j ⁱ, ${\stackrel{\&OverBar;}{L}}_k^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ Calculate the angle of their place straight lines, be designated as μ, and L _j ⁱMid point to L _k ⁱThe distance of place straight line and L _k ⁱMid point to L _j ⁱThe distance of place straight line is designated as η respectively ₁And η ₂

If μ＜φ and η ₁＜ε and μ ₂＜ε then judges L _j ⁱAnd L _k ⁱConllinear generates a new 3 d-line feature with them, and two end points of this 3 d-line feature are L _j ⁱAnd L _k ⁱAt a distance of farthest two end points, add new 3 d-line feature to set M _iIn, and from set M _iMiddle deletion L _j ⁱAnd L _k ⁱ

The 2nd) step, calculate P _iThe intersection point of all member's linear features

For P _iAny two member's linear feature L _α ⁱ, ${\stackrel{\&OverBar;}{L}}_{\mathrm{\&beta;}}^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ If the angle theta of their place straight lines satisfies

Then calculate L _α ⁱAnd L _β ⁱIntersection point or the intersection point of their extended line;

The set that the intersection point of all calculating is formed is designated as R _i

The 3rd) step, generate the non-directed graph that is used for the segmentation plane hypothesis

Generate a non-directed graph, be designated as G _i, its node is by two class dot generation: the one, and intersection point set R _iIn each element, the 2nd, for M _iIn any one member's linear feature, if do not have and M on the extended line outside its end points _iIn the intersection point of other member's linear feature, and this end points is not set R _iIn element, then this end points becomes figure G _iA node;

Non-directed graph G _iIn any two internodal annexations comprise two classes: the one, if the point of their correspondences on same member's linear feature or its extended line, and on the connection line segment of these two points not and figure G _iIn the corresponding point of other node, then their keep to connect; The 2nd, the annexation of structure, building method is as follows:

For any one member's linear feature L _α ⁱIf its one or two end points is figure G _iNode, then handle in two kinds of situation:

First kind of situation, L _α ⁱWith M _iIn any other member's linear feature all non-intersect

Select any member's linear feature ${\stackrel{\&OverBar;}{L}}_{\mathrm{\&beta;}}^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ And β ≠ α specifies on its place straight line the G with figure _iIn two corresponding two points of node, these two points are positioned at L _β ⁱThe mid point both sides and the connection line segment between them on not and figure G _iThe middle corresponding point of node is designated as (X with them ₁, Y ₁, Z ₁) and (X ₂, Y ₂, Z ₂), and calculate

$e_1=({\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-X_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i$

$e_2=({\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},2}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-X_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},2}^i$

Wherein, (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and (X _{α, 2} ⁱ, Y _{α, 2} ⁱ, Z _{α, 2} ⁱ) be L _α ⁱEnd points; For arbitrarily ${\stackrel{\&OverBar;}{L}}_l^i\&Element;{\stackrel{\&OverBar;}{M}}_i,$ And l ≠ α, l ≠ β calculate

$f_1=({\stackrel{\&OverBar;}{X}}_{l,1}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-X_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{l,1}^i$

$f_2=({\stackrel{\&OverBar;}{X}}_{l,2}^i-X_1)\&CenterDot;\frac{Y_2-Y_1}{X_2-X_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{l,2}^i$

$f_3=({\stackrel{\&OverBar;}{X}}_{l,1}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i}{{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i}+{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i-{\stackrel{\&OverBar;}{Y}}_{l,1}^i$

$f_4=({\stackrel{\&OverBar;}{X}}_{l,2}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i}{{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},2}^i-{\stackrel{\&OverBar;}{X}}_{\mathrm{\&alpha;},1}^i}+{\stackrel{\&OverBar;}{Y}}_{\mathrm{\&alpha;},1}^i-{\stackrel{\&OverBar;}{Y}}_{l,2}^i$

Wherein, (X _{L, 1} ⁱ, Y _{L, 1} ⁱ, Z _{L, 1} ⁱ) and (X _{L, 2} ⁱ, Y _{L, 2} ⁱ, Z _{L, 2} ⁱ) be L _l ⁱEnd points; If satisfy

$\left\{\begin{array}{c}{e}_1\&CenterDot;e_2>0\\ f_1\&CenterDot;f_3>0\\ f_2\&CenterDot;f_4>0\\ f_1\&CenterDot;f_2>0\end{array}\right.$

Then calculate

$f_5=({\stackrel{\&OverBar;}{X}}_{l,1}^i-X_1)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{l,2}^i-Y_1}{{\stackrel{\&OverBar;}{X}}_{l,2}^i-X_1}+Y_1-{\stackrel{\&OverBar;}{Y}}_{l,1}^i$

$f_6=(X_2-X_1)\&CenterDot;\frac{{\stackrel{\&OverBar;}{Y}}_{l,2}^i-Y_1}{{\stackrel{\&OverBar;}{X}}_{l,2}^i-X_1}+Y_1-Y_2$

If f ₅F ₆＞0, tie point (X then _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and (X ₂, Y ₂, Z ₂) at figure G _iMiddle corresponding node, and point (X _{α, 2} ⁱ, Y _{α, 2} ⁱ, X _{α, 2} ⁱ) and (X ₁, Y ₁, Z ₁) at figure G _iMiddle corresponding node;

Second kind of situation, L _α ⁱWith M _iIn at least one other member's linear feature intersect

If (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and figure G _iA node corresponding, select member's linear feature L _α ⁱOr on its extended line with (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) nearest intersection point, be designated as (X ₃, Y ₃, Z ₃), suppose that this intersection point is L _α ⁱWith member's linear feature L _β ⁱIntersection point; Select L _β ⁱOr on its extended line with figure G _iIn the corresponding point of node, if at it and point (X ₃, Y ₃, Z ₃) between line on not and figure G _iThe middle corresponding point of node then is designated as (X with it ₄, Y ₄, Z ₄), if such point has two, then respectively they are designated as (X ₄, Y ₄, Z ₄) and (X ₅, Y ₅, Z ₅);

If (X ₄, Y ₄, Z ₄) be L _β ⁱWith the intersection point of other member's linear feature, suppose that this member's linear feature is L _l ⁱ, and L _l ⁱOr exist on its extended line and figure G _iThe middle corresponding point of node, it satisfies and point (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) drop on L _β ⁱThe same side, and with point (X ₄, Y ₄, Z ₄) between line on not and figure G _iIn the corresponding point of node, then connect it and point (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) at figure G _iMiddle corresponding node; Otherwise, tie point (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) and point (X ₄, Y ₄, Z ₄) at figure G _iMiddle corresponding node; If (X ₅, Y ₅, Z ₅) exist, then according to (X ₄, Y ₄, Z ₄) same method handles;

If (X _{α, 2} ⁱ, Y _{α, 2} ⁱ, Z _{α, 2} ⁱ) be figure G _iNode, method of then constructing annexation and (X _{α, 1} ⁱ, Y _{α, 1} ⁱ, Z _{α, 1} ⁱ) identical;

The 4th) step, cut apart generation flat blocks hypothesis

Search non-directed graph G _iIn all rings, for any one ring wherein, if the node number that belongs to same member's linear feature or its extended line that it comprises is no more than 2, then generate a flat blocks hypothesis with it, all nodes that it comprises are formed an orderly point set according to its annexation, represent the vertex set of this flat blocks hypothesis.

6. three dimensional plane extraction method according to claim 5 is characterized in that, the arbitrary plane piece is supposed ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ Calculate its reliability measure, comprise the steps:

The 1st) step, the search quilt

The known spot elevation that covers

Projection plane piece hypothesis Vertex set to left and right image, remember any summit

Projection in left and right image is respectively

With

The order projection on connection summit in left and right image respectively according to each fixed point in the vertex set forms

Two polygon view fields;

For any spot elevation

If the projection in its left image

Be positioned at

In the view field in left image, and the projection in its right image

Be positioned at

In the view field in right image, then judge spot elevation

Quilt

Cover;

Quilt

The set that all spot elevations that cover are formed is designated as ${\mathrm{\&Theta;}}_w\&Subset;J,$ The quantity of element is α in the set _w

The 2nd) step, calculate the elevation consistance and estimate

If

Then calculate spot elevation number α w＞0 that covers

The elevation consistance estimate for

(formula one)

Wherein,

If α _w=0, then $E_w^D=0;$

The 3rd) step, calculate the gray scale similarity measure

All images point is formed a point set in the view field in left image, is designated as K _w ^L, the gradation of image value of being had a few in them is formed an array, is designated as { h _Wt| t=1,2 ..., o _w;

Foundation

Three pairs of same places in left and right image

With

With

And

With

Set of computations K _w ^LThe middle same place of all elements point in right image, the set that these same places are formed is designated as K _w ^R, the gradation of image value of being had a few in them is obtained by interpolation algorithm, and these gray-scale values are formed another array, are designated as { h _Wt' | t=1,2 ..., o _w; Calculate

Gray scale similarity measure between the view field in left and right image is

$E_w^G=\frac{\underset{t=1}{\overset{o_w}{\mathrm{\&Sigma;}}}(h_{\mathrm{wt}}-{\stackrel{\&OverBar;}{h}}_w)({h_{\mathrm{wt}}}^{\&prime;}-{{\stackrel{\&OverBar;}{h}}_w}^{\&prime;})}{\sqrt{\left[\underset{t=1}{\overset{o_w}{\mathrm{\&Sigma;}}}{(h_{\mathrm{wt}}-{\stackrel{\&OverBar;}{h}}_w)}^2\right]\&CenterDot;\left[\underset{t=1}{\overset{o_w}{\mathrm{\&Sigma;}}}{({h_{\mathrm{wt}}}^{\&prime;}-{{\stackrel{\&OverBar;}{h}}_w}^{\&prime;})}^2\right]}}$ (formula two)

Wherein, ${\stackrel{\&OverBar;}{h}}_w=\frac{1}{o_w}\underset{t=1}{\overset{o_w}{\mathrm{\&Sigma;}}}{h}_{\mathrm{wt}},$ ${{\stackrel{\&OverBar;}{h}}_w}^{\&prime;}=\frac{1}{o_w}\underset{t=1}{\overset{o_w}{\mathrm{\&Sigma;}}}{h_{\mathrm{wt}}}^{\&prime;};$

The 4th) step, calculate

Reliability measure

Reliability measure be

$E_w=\left\{\begin{array}{cc}{E}_w^D\&CenterDot;(E_w^G+1)/2& E_w^D>0\\ (E_w^G+1)/2& E_w^D=0\end{array}\right.$ (formula three).

7. three dimensional plane extraction method according to claim 6 is characterized in that, the arbitrary plane piece is supposed ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ The step of searching for the most reliable flat blocks hypothesis is:

The 1st) step, the reliable flat blocks hypothesis that search has known spot elevation to cover

Set up a non-directed graph, for any one flat blocks hypothesis ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ If $E_w^D>0,$ Then generate a node of this non-directed graph with it, the property value of node is E _wFor any two nodes,, then connect them if all there is not lap in the view field of the flat blocks of their correspondences hypothesis in left and right image;

Calculate all very big groups of this non-directed graph, therefrom choose the set of all flat blocks hypothesis that the very big group of node attribute values sum maximum comprises, be designated as K ₁, it represents that chlamydate spot elevation is as the reliable flat blocks hypothesis of weighing foundation;

The 2nd) step, the reliable flat blocks hypothesis that search does not have known spot elevation to cover

Set up a non-directed graph, for any one flat blocks hypothesis ${\hat{P}}_w\&Element;\widehat{\mathrm{\&Omega;}},$ If $E_w^D>0$ And it and K ₁In any one flat blocks hypothesis view field in left and right image all do not have lap, then use

Generate a node of this non-directed graph, the property value of node is E _wFor any two nodes,, then connect them if all there is not lap in the view field of the flat blocks of their correspondences hypothesis in left and right image;

Calculate all very big groups of this non-directed graph, therefrom choose the set of all flat blocks hypothesis that the very big group of node attribute values sum maximum comprises, be designated as K ₂, the spot elevation that its expression does not have to cover is as weighing foundation and and K ₁In all flat blocks suppose coexisting reliable flat blocks hypothesis;

With K ₁∪ K ₂All flat blocks hypothesis that comprise are as reliable flat blocks hypothesis.

8. three dimensional plane extraction method according to claim 7 is characterized in that, the method that merges the reliable flat blocks hypothesis of coplane is;

The 1st) step, judge the coplanar relation that flat blocks is supposed

For any two reliable flat blocks hypothesis

${\hat{P}}_v\&Element;K_1\&cup;K_2,$ If following inequality is formed upright, then judge

With

Coplane:

$\left\{\begin{array}{c}\frac{|{\hat{a}}_u\&CenterDot;{\hat{a}}_v+{\hat{b}}_u\&CenterDot;{\hat{b}}_v+{\hat{c}}_u\&CenterDot;{\hat{c}}_v|}{\sqrt{{{\hat{a}}_u}^2+{{\hat{b}}_u}^2+{{\hat{c}}_u}^2}\&CenterDot;\sqrt{{{\hat{a}}_v}^2+{{\hat{b}}_v}^2+{{\hat{c}}_v}^2}}>\cos (\frac{3}{2}\&CenterDot;\mathrm{\&phi;})\\ \frac{|{\hat{a}}_u\&CenterDot;{\hat{X}}_{v1}+{\hat{b}}_u\&CenterDot;{\hat{Y}}_{v1}+{\hat{c}}_u\&CenterDot;{\hat{Z}}_{v1}+{\hat{d}}_u|}{\sqrt{{{\hat{a}}_u}^2+{{\hat{b}}_u}^2+{{\hat{c}}_u}^2}}<\frac{3}{2}\&CenterDot;\mathrm{\&epsiv;}\\ \frac{|{\hat{a}}_v\&CenterDot;{\hat{X}}_{u1}+{\hat{b}}_v\&CenterDot;{\hat{Y}}_{u1}+{\hat{c}}_v\&CenterDot;{\hat{Z}}_{u1}+{\hat{d}}_v|}{\sqrt{{{\hat{a}}_v}^2+{{\hat{b}}_v}^2+{{\hat{c}}_v}^2}}<\frac{3}{2}\&CenterDot;\mathrm{\&epsiv;}\end{array}\right.$

The 2nd) step, the reliable flat blocks hypothesis of search coplane generates a non-directed graph, is designated as

Use K ₁∪ K ₂In each reliable flat blocks hypothesis generate a node, if any two reliable flat blocks hypothesis are coplanes, then connect the node of their correspondences;

Calculating chart

All very big groups, be designated as $\left\{{\hat{Q}}_s\right|s=\mathrm{1,2},...,{\hat{N}}_Q\},$ Any one is greatly rolled into a ball

Be figure

A subclass, be designated as $\left\{{\hat{P}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},...{\hat{m}}_s\}\&Subset;K_1\&cup;K_2;$

The 3rd) step, the reliable flat blocks that merges coplane is supposed

For any one very big group

Merge all reliable flat blocks hypothesis that it comprises, obtain an open plane Plane equation is designated as ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s\&CenterDot;X+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s\&CenterDot;Y+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s\&CenterDot;Z+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{d}}}_s=0,$ Wherein:

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s=\frac{1}{{\hat{m}}_s}\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}\frac{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}{\sqrt{{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s=\frac{1}{{\hat{m}}_s}\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}\frac{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}{\sqrt{{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s=\frac{1}{{\hat{m}}_s}\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}\frac{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}{\sqrt{{{\hat{a}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{b}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2+{{\hat{c}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}}^2}}$

${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{d}}}_s=-\frac{1}{{\hat{m}}_s}({\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{a}}}_s\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}{\hat{X}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right),1}+{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{b}}}_s\&CenterDot;\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}{\hat{Y}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right),1}{+\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{c}}}_s\underset{\mathrm{\&kappa;}=1}{\overset{{\hat{m}}_s}{\mathrm{\&Sigma;}}}{\hat{Z}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right),1})$

Will $\left\{{\hat{P}}_{{\widehat{\mathrm{\&Gamma;}}}_s\left(\mathrm{\&kappa;}\right)}\right|\mathrm{\&kappa;}=\mathrm{1,2},...{\hat{m}}_s\}$ In all summits of each reliable flat blocks hypothesis project to

On, connecting all subpoints in turn according to the order between the summit in the flat blocks hypothesis, all projection straight line sections that obtain become

Member's linear feature.

9. three dimensional plane extraction method according to claim 8 is characterized in that, be may further comprise the steps by the method on member's linear feature formation level border on open plane:

The 1st) go on foot, split all member's linear feature to two subclass of the face of setting level

For any one open plane

Its all member's linear features are split into two subclass, and a subclass is ${\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_u\&cap;{{\stackrel{\&OverBar;}{\stackrel{\&OverBar;}{M}}}_u}^{\&prime;},$ Be designated as U ₁, another subclass is

Be designated as U ₂

The 2nd) step, generate the non-directed graph set

Generate U ₂All subclass;

Choose U ₂Any one subclass and U ₁Form new member's linear feature collection, be designated as U ₀, generate a non-directed graph; This non-directed graph is with U ₀In each member's linear feature as node, if member's linear feature of any two node correspondences intersects, then they are coupled together;

The 3rd) step, generate planar boundary

Its all Hamilton circles of search in each non-directed graph that previous step obtains, for each unduplicated circle,, calculate the intersection point of member's linear feature of adjacent node correspondence according to the order of connection between the node that wherein comprises, they as the summit, are generated a closed hypothesis of plane mechanism; If there is no the Hamilton circle is then searched for all Hamilton paths, for each unduplicated path, according to the order of connection between the node that wherein comprises, calculate the intersection point of member's linear feature of adjacent node correspondence, they as the summit, are generated a semi-open hypothesis of plane mechanism.

10. three dimensional plane extraction method according to claim 9 is characterized in that, utilizes the disappearance border of extracting the linear feature and the additional semi-open hypothesis of plane mechanism of heuristic rule, and method is as follows:

In the 1st step, utilize the linear feature that extracts in the image to replenish the disappearance border of the semi-open hypothesis of plane mechanism

The 1st) step, calculate P ' _kThe projection in left and right image of disappearance border the region projection P ' _kTo left and right image, obtain point set respectively $\left\{{\mathrm{Tl}}_k^j\right|j=\mathrm{1,2},...,{n_k}^{\&prime;}\}$ With $\left\{{\mathrm{Tr}}_k^j\right|j=\mathrm{1,2},...,{n_k}^{\&prime;}\};$

Set the distance threshold h of an expression linear feature integrality _d

Specify P ' _kThe zone at projection place in left image, disappearance border be a quadrilateral, be designated as P _k ^L, its summit is designated as g respectively ₁, g ₂, g ₃And g ₄, wherein, $g_1={\mathrm{Tl}}_k^1,$ $g_2={\mathrm{Tl}}_k^{{n_k}^{\&prime;}},$ g ₃It is line segment

Extended line on to the some g ₂Distance be h _dPoint, g ₄Be line segment Tl _k ²Tl _k ¹Extended line on to the some g ₁Distance be h _dPoint, connecting them becomes R _k ^LFour edges g ₁g ₂, g ₂g ₃, g ₃g ₄And g ₄g ₁

Utilize similar method, calculate P ' _kThe region R at projection place in right image, disappearance border _k ^R, its four summits are designated as g respectively ₁', g ₂', g ₃' and g ₄';

The 2nd) in the step, utilize region R _k ^LAnd R _k ^RInterior H ₁And H ₂In linear feature replenish the disappearance border of the semi-open hypothesis of plane mechanism

For H ₁In any one linear feature, if its two end points all drop on R _k ^LIn, and its extended line and line segment g ₂g ₃And g ₄g ₁Intersect, then connect two intersection points and obtain the straight line section, according to P ' _kPlane equation and the collinearity equation in the photogrammetric Theory, calculate the three dimensions point of two end points correspondences of this straight-line segment, with P ' _kVertex set in remove T _k ¹With

Outside all summits together, generate a closed hypothesis of plane mechanism; For H ₂In any one linear feature, if its two end points all drop on R _k ^RIn, and its extended line and line segment g ₂' g ₃' and g ₄' g ₁' intersect, calculate its same place of end points in left image, connecting two same places is the straight line section, if any disappearance border that searches in this straight-line segment and left image conllinear not then calculates the extended line and the line segment g of this linear feature in right image ₂' g ₃' and g ₄' g ₁' intersection point, connect these two intersection points and obtain the straight line section, according to P ' _kPlane equation and the collinearity equation in the photogrammetric Theory, calculate the three dimensions point of two end points correspondences of this straight-line segment, with P ' _kVertex set in remove T _k ¹With

Outside all summits together, generate a closed hypothesis of plane mechanism;

The 2nd step is according to the disappearance border of the additional semi-open hypothesis of plane mechanism of heuristic rule

If respectively with T _k ²T _k ¹With

For shared border, with P ' _kTwo adjacent hypothesises of plane mechanism have all obtained complete border, according to the isometric rule of adjacent polygon plane common edge, at T _k ²T _k ¹Or 1 Φ of appointment on its extended line ₁, make line segment T _k ²Φ ₁With with T _k ²T _k ¹For the shared border of the adjacent plane on shared border hypothesis isometric,

Or 1 Φ of appointment on its extended line ₂, make line segment

With with

For the shared border of the adjacent plane on shared border hypothesis isometric, line segment Φ ₁Φ ₂Become additional disappearance border;

If respectively with T _k ²T _k ¹With

Be shared border and P ' _kHave only a hypothesis of plane mechanism to obtain complete border in two adjacent hypothesises of plane mechanism, suppose that this adjacent plane is assumed to be P ' _l, and T _k ²T _k ¹Be P ' _lWith P ' _kShared border, then according to the shared border of adjacent polygon plane isometric rule, at T _k ²T _k ¹Or specify on its extended line a bit, make line segment T _k ²Φ ₁With P ' _lShared border isometric,

Or 1 Φ of appointment on its extended line ₂, make line segment

With T _k ²Φ ₁Isometric, line segment Φ ₁Φ ₂Become additional disappearance border;

According to P ' _kPlane equation and the collinearity equation in the photogrammetric Theory, calculate Φ ₁And Φ ₂Corresponding three dimensions point is with P ' _kVertex set in remove T _k ¹With

Outside all summits together, generate a closed hypothesis of plane mechanism;

The closed hypothesis of plane mechanism among the Ω ' and the closed hypothesis of plane mechanism merging that is generated by the semi-open hypothesis of plane mechanism among the Ω ' are become closed hypothesis of plane mechanism set, are designated as Ω "=P ' _i| i=1,2 ..., N " }, wherein, any closed hypothesis of plane mechanism P " _iBy n " _iThe bar boundary line is formed.

11. three dimensional plane extraction method according to claim 10 is characterized in that, the method for calculating the reliability measure of the closed hypothesis of plane mechanism is:

For any one closed hypothesis of plane mechanism P " _i∈ Ω ", utilize formula one to calculate P " _iThe elevation consistance estimate, if P " _iIt is 0 that the known elevation that covers is counted, and then specifying its elevation consistance to estimate is 1/ (2 λ), utilizes formula three, calculates its reliability measure, is designated as E " _iAt P " _iOn the plane, place, specify next-door neighbour outside its every boundary line, with the rectangle plane zone of this boundary line as a long limit, the long length that equals the boundary line, wide half that equals to grow, the elevation consistance of utilizing formula one to calculate this rectangle plane zone is estimated, if it is 0 that the known elevation that it covers is counted, then specifying its elevation consistance to estimate is 1/ (2 λ), utilize formula three, calculate its reliability measure, note is E by the reliability measure in the rectangle plane zone of j bar boundary line correspondence _{I, j}", calculate closed hypothesis of plane mechanism P according to following formula " _iThe reliability measure on corresponding true planar border, border be

${\mathrm{\&gamma;}}_i=\frac{{E_i}^{\&prime;\&prime;}}{\underset{j=1}{\overset{n_i^{\&prime;\&prime;}}{\mathrm{\&Sigma;}}}{E_{i,j}}^{\&prime;\&prime;}}$ (formula four).

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