CN105069274A - Method for screening and determining effective shielding ground objects based on sunshine space-time characteristics - Google Patents

Abstract

The invention provides a method for screening and determining effective shielding ground objects based on sunshine space-time characteristics. Firstly, the shadow movement track formula of the shielding boundary of a height-known shielding object to a test point is determined, furthermore the shadow movement track of a critical point from the starting time to the stopping time is determined, the shadow movement track of the critical point from the starting time to the stopping time and the test point form two vertical planes which are intersected and perpendicular to the ground horizontal plane, during the forming process of the shadow movement track, the plane, on the top of the shielding object, being parallel to the ground horizontal plane and the sunshine conical surface form an arc-shaped movement track, the arc-shaped movement track forms an arc-shaped vertical plane which is perpendicular to the ground horizontal plane, the two vertical planes and the arc-shaped vertical plane from the effective shielding scope of the test point, redundant ground objects in the effective shielding scope are removed, and effective ground objects in the effective shielding scope are determined. On the premise that the calculating precision is not affected, the ground object data involved in the calculation is reduced as much as possible, and the overall operational efficiency is improved.

Description

Based on sunshine, the atural object that effectively blocks of space-time characteristic screens defining method

Technical field

The present invention relates to field at a kind of sunshine, particularly relate to a kind of based on sunshine space-time characteristic effectively block atural object reconstruct.

Background technology

When carrying out Sunlight Analysis to complicated atural object, earth object not only comprises the building of rule, also comprises the ground surface modifier etc. of landform that height rises and falls, the vegetation with certain altitude and body sign and city and residential district.The difference of research object causes in method and efficiency, presenting problems towards the Sunlight Analysis model of complicated geographic scenes, is mainly reflected in the following aspects:

First, when calculating the sunshine time of certain test point, people cannot block sunshine for judging in the complicated atural object of periphery which atural object will produce test point, but unrestrictedly can not calculate all atural object of periphery.

Secondly, the atural object data of blocking sunshine may be produced to test point even if only calculate, still there is a large amount of redundant computation.

Summary of the invention

The technical problem to be solved in the present invention be to provide a kind of based on sunshine space-time characteristic effectively block atural object reconstruct, to solve prior art Problems existing.

For solving the problems of the technologies described above, the present invention by the following technical solutions:

Based on sunshine, the atural object that effectively blocks of space-time characteristic screens defining method, the method is first by determining the shadows cast by the sun motion track formula of the shelter of known altitude to test point Ouluding boundary, and then determine rising, the only shadows cast by the sun motion track of the critical point of time, then rise, only the shadows cast by the sun motion track of the critical point of time and test point form two and to intersect and perpendicular to two facades of the earth surface level, simultaneously in shadows cast by the sun motion track forming process, shelter top be parallel to the earth surface level plane and sunshine circular conical surface form the movement locus of an arc, the movement locus of arc forms an arc facade perpendicular to the earth surface level, described two facades and arc facade composition test point effectively block scope, then the redundancy atural object effectively blocked in scope is removed, determine the effective atural object effectively blocked in scope,

The method of the redundancy atural object that described removal is effectively blocked in scope is: the data structure [tin first setting up atural object TIN model ₁, tin ₂..., tin _m], in described data structure, the internal storage structure of a single point comprises: the earth coordinates be made up of longitude L, latitude B, elevation H, the topocentric coordinate system be made up of station longitude centroid L ', the heart latitude B ' and elevation H ' that stands, X, Y, Y rectangular coordinate system in space X centered by the earth's core, Y and Z, also comprises the transition flag Flag between them; Then the memory mapping table of timeslice is built in the periphery of atural object TIN data structure, the node of each timeslice in memory mapping table corresponds to the memory address of TIN data set, when calculating some moment are between timeslice, address date pointed by this timeslice is exactly the data that actual needs participates in the atural object TIN model of computing, and the data of other atural object TIN model are the atural object data that redundancy need be deleted;

The shadows cast by the sun motion track formula of described shelter to the maximum Ouluding boundary of test point is:

$\left\{\begin{array}{c}{f}_1^{\′}=f_1(B_x,L_x,A_x)\⇒f_1^{\′}(B_x,L_x,F\left(B_x,L_x,T\right))\⇒f_1^{\′\′}(B_x,L_x)=0\\ f_2^{\′}=f_2(B_x,L_x,A_x)\⇒f_2^{\′}(B_x,L_x,F\left(B_x,L_x,T\right))\⇒f_2^{\′\′}(B_x,L_x)=0\end{array}\right.$ Formula (1)

Wherein, moment T produces the initial time of shade or produces the stop timing of shade, (L _x, B _x) be the latitude and longitude coordinates in T moment;

And in above-mentioned formula (1),

$\begin{array}{c}{f}_1(B_x,L_x,A_x)=H^{\′}-(e^2{H}_0+\frac{N_x^{\′}}{S})\sin {\mathrm{BsinB}}_x+{\tanh }_T{N}_x^{\′}{\mathrm{cosB}}_x\[{\mathrm{cosA}}_x{\mathrm{cosL}}_x^{\′}\sin B\\ -\sin A_x{\mathrm{sinL}}_x^{\′}\]+\cos B\{-N_x^{\′}{\mathrm{cosB}}_x{\mathrm{cosL}}_x^{\′}-{\mathrm{cosA}}_x{\tanh }_T\[e^{2}N\sin B+(e^2{H}_0+\frac{N_x^{\′}}{S}){\mathrm{sinB}}_x\]\}\end{array}$ Formula (2),

$\begin{array}{c}{f}_2(B_x,L_x,A_x)=N(-\frac{2a^2}{N}+2N_x{\mathrm{cosBcosB}}_x\cos (L-L_x)+\frac{2N_x{\sin }^{2}B}{S})\\ +{(-\frac{a^2}{N}+N_x{\mathrm{cosBcosB}}_x\cos (L-L_x)+\frac{N_x{\sin }^{2}B}{S})}^2(1+e^{\′2}{\sin }^{2}B)\\ +e^{\′2}{\mathrm{cosA}}_x\cos B\sin B(-\frac{2a^2}{N}+2N_x{\mathrm{cosBcosB}}_x\cos (L-L_x)+\frac{2N_x{\sin }^{2}B}{S})\\ \{{\mathrm{cosA}}_x\sin B\[(e^{2}N+\frac{N_x}{S})\cos B-N_x{\mathrm{cosB}}_x\cos (L-L_x)\]+N_x{\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin (L-L_x)\}\\ +(1+e^{\′2}{\cos }^2{A}_x{\cos }^{2}B)\{{\mathrm{cosA}}_x\sin B\[(e^{2}N+\frac{N_x}{S})\cos B-N_x{\mathrm{cosB}}_x\cos (L-Lx)\]\\ +N_x{\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin (L-L_x){\}}^2\end{array}$ Formula (3),

Wherein, B is latitude; A _xfor the solar azimuth that T moment test point observes, a is semimajor axis of ellipsoid, and b is minor semi-axis, L _x, B _xfor the earth latitude coordinates of test point, N _xthe fourth of the twelve Earthly Branches chief of a tribe corresponding for test point point encloses radius-of-curvature, and e ' is ellipsoid second excentricity, and S is the position of the test point point observation sun in celestial sphere.

When the shadows cast by the sun motion track of shelter to test point Ouluding boundary of described known altitude refers to that solar irradiation is mapped to the summit of shelter, the movement locus that when the summit projection of shelter overlaps with test point all the time, shelter is formed on the ground.

Beneficial effect of the present invention: the present invention by block height, analysis time section and shutter body geographic range between " space-time " funtcional relationship, by effective sunlight " time " and sunshine light " space " feature, screen participating in the atural object data of computing; And based on the atural object data after screening, the secondary data realized based on " timeslice " index reconstructs, and is not affecting under computational accuracy prerequisite, decrease the atural object data participating in computing as much as possible, improve integral operation efficiency.

Accompanying drawing explanation

Fig. 1 is the memory data structure of atural object TIN model.

Fig. 2 be topocentric the earth's core rectangular space coordinate and sunshine two dimensional surface coordinate corresponding relation.

Fig. 3 is shadows cast by the sun running orbit schematic diagram.

Fig. 4 is for effectively to block three-dimensional spatial area schematic diagram.

Fig. 5 is that the simplification of effectively blocking three-dimensional spatial area is expressed.

Fig. 6 is the position relationship of shelter and test point on earth surface.

Fig. 7 be sunshine two-dimensional coordinate system test point and shelter between relation.

Fig. 8 is for effectively to block scope sphere 3 area schematic.

Fig. 9 atural object is being subject to perspective view on shadow face, and (A) is earth surface shelter and test point relation, (B)

For atural object under geocentric coordinate system projects by under shadow face.

Figure 10 is based on the corresponding memory data structure of atural object TIN of timeslice index.

Figure 11 based on timeslice starting point with block the space four rib centrum that height threshold constructs.

Embodiment

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

Because when calculating the sunshine time of certain test point, people cannot block sunshine for judging in the complicated atural object of periphery which atural object will produce test point, but unrestrictedly can not calculate all atural object of periphery.Therefore need to select in advance and may produce the regional extent of blocking to test point, then judge that in selected scope, atural object blocks the sunshine to test point.Secondly, the atural object data of blocking sunshine may be produced to test point even if only calculate, still there is a large amount of redundant computation.

The present invention use based on sunshine space-time feature carry out the determination of effective occlusion area, adopt the index relative setting up atural object and time based on " timeslice " simultaneously, the basis of effectively blocking atural object screening realizes the reconstruct of atural object data secondary, when making the sunshine calculating certain moment, only calculate the atural object data relevant to timeslice, significantly reduce the atural object data volume participating in computing, significantly improve operation efficiency.Meet " precisely efficient " demand of Sunlight Analysis model.The method is first by determining the shadows cast by the sun motion track formula of the shelter of known altitude to test point Ouluding boundary, determine again rising, the only shadows cast by the sun motion track of the critical point of time, then rise, only the shadows cast by the sun motion track of the critical point of time and test point form two and to intersect and perpendicular to two facades of the earth surface level, simultaneously in shadows cast by the sun motion track forming process, shelter top be parallel to the earth surface level plane and sunshine circular conical surface form the movement locus of an arc, the movement locus of arc forms an arc facade perpendicular to the earth surface level, described two facades and arc facade composition test point effectively block scope, then the redundancy atural object effectively blocked in scope is removed, determine the effective atural object effectively blocked in scope.

The present invention uses the astronomical algorithm position of sun computing method that namely " AstronomicalAlgorithms " (Meeus, 1998) are announced to be that the representative algorithm of TEP-SPA (i.e. theoretical expansion method) is as benchmark SPA algorithm.

The method removing the redundancy atural object effectively blocked in scope is set up the data structure [tin1 of atural object TIN model, tin2, tinm], then the memory mapping table of timeslice is built in the periphery of atural object TIN data structure, the node of each timeslice in memory mapping table corresponds to the memory address of TIN data set, when calculating some moment are between timeslice, address date pointed by this timeslice is exactly the data that actual needs participates in the atural object TIN model of computing, and the data of other atural object TIN model are the atural object data that redundancy need be deleted;

Therefore first the present invention carries out the definition of the data structure of atural object TIN model, because scene of the present invention relates to building, landform, the atural object key element such as vegetation and mark in complicated geographic scenes, select TIN data model expression atural object data effectively can not only realize fusion and the conversion of multi-source data, and simplify the deterministic process of atural object projection and test point topological relation.

As shown in Figure 1, in atural object TIN process of establishing, building, landform and other types of ground objects are unified adopts TIN structure representation, and single TIN is then made up of unordered three points (_ 3DPoint self-defined structure).Indivisible for avoiding ignoring in Floating-point Computation process, improve computational accuracy, need to be converted to topocentric coordinate system to carry out correlation computations to the earth's core rectangular space coordinate is unified.Therefore, the internal storage structure of a single point needs to comprise: earth coordinates (longitude L, latitude B and elevation H), topocentric coordinate system (the longitude centroid L ' that stands, stand heart latitude B ' and elevation H '), the earth's core rectangular coordinate system in space (X, Y and Z) and transition flag Flag each other.

Then, in order to describe and convenience of calculation, change between need expressing the different coordinates of same point.Adopt WGS84 the earth (barycenter) coordinate system as coordinate system reference in the present invention, namely following middle longitude and latitude refers to longitude and latitude under WGS84 coordinate system.Simultaneously for distinguishing astronomical coordinate system and earth coordinates, following best the earth (barycenter) coordinate system that " earth-based coordinate system " expression will be adopted to be applicable to Sunlight Analysis research, namely refers to WGS84 the earth geocentric coordinate system in the present invention.Set by the planar two dimensional coordinate system (hereinafter referred to as " coordinate system at sunshine ") of blocking point, ground point and position of sun structure of blocking atural object simultaneously.

As shown in Figure 2, first build the corresponding relation of ground point P (X, Y, Z) and (x, y, z) under sunshine coordinate system under the rectangular coordinate system in space of the earth's core, performing step is as follows:

(1) space coordinates and P point topocentric coordinate system (x, y, z direction is constant) relational expression is built

The earth rectangular coordinate system in space O-XYZ is moved to P point, is transformed into the coordinate system P-XYZ centered by P point, because P coordinate under O-XYZ coordinate system is: $\left\{\begin{array}{c}X=(N+H_p)\cos B\cos L\\ Y=(N+H_p)\cos B\sin L\\ Z=\left(N\right(1-e^2)+H_p)\sin B\end{array}\right.,$ N is that the P chief of a tribe that calls the roll in the morning encloses radius-of-curvature, and e is ellipsoid first excentricity, and L, B are P point longitude and latitude, H _pfor P point geodetic height (if P point is on sphere, H _p=0).

Now, space any point (do not overlap with P, be assumed to be M) coordinate (X in P-XYZ coordinate system _p-M, Y _p-M, Z _p-M) with O-XYZ coordinate system under coordinate (X _m, Y _m, Z _m) corresponding relation be:

$\left\{\begin{array}{c}{X}_{p-M}=X_M-(N+H_p)\cos B\cos L\\ Y_{p-M}=Y_M-(N+H_p)\cos B\sin L\\ Z_{p-M}=Z_M-\left(N\right(1-e^2)+H_p)\sin B\end{array}\right.---\left(1\right)$

(2) P point topocentric coordinate system is converted to P station center right-angle coordinate in horizon system (, to Horizon north, z is to zenith for x)

For coordinate system P-XYZ, first to turn clockwise longitude 180 ° of-L (L is P point longitude) around Z axis, then Y-axis reciprocal transformation is Y ', to turn clockwise 90 ° of-B (B is P point latitude) around Y ' axle again, then formed centered by P point, longitude and latitude tangential direction is respectively the new rectangular coordinate system P-X ' Y ' Z ' of X ', Y '.Then put M coordinate in P-X ' Y ' Z ' coordinate system to become:

Wherein, $R_Y^T\left(\mathrm{\θ}\right)=\left(\begin{array}{ccc}cos\mathrm{\θ}& 0& sin\mathrm{\θ}\\ 0& 1& 0\\ -\sin \mathrm{\θ}& 0& cos\mathrm{\θ}\end{array}\right),R_z^T\left(\mathrm{\θ}\right)=\left(\begin{array}{ccc}cos\mathrm{\θ}& -sin\mathrm{\θ}& 0\\ sin\mathrm{\θ}& \cos \mathrm{\θ}& 0\\ 0& 0& 1\end{array}\right).$ Therefore can obtain:

$\left(\begin{array}{c}{X_{p-M}}^{\′}\\ {Y_{p-M}}^{\′}\\ {Z_{p-M}}^{\′}\end{array}\right)=\left(\begin{array}{ccc}-\cos L\sin B& -\sin B\sin L& \cos B\\ -\sin L& \cos L& 0\\ \cos B\cos L& \cos B\sin L& \sin B\end{array}\right)\left(\begin{array}{c}{X}_{p-M}\\ Y_{p-M}\\ Z_{p-M}\end{array}\right)---\left(2\right)$

(3) P station center right-angle coordinate in horizon system is converted to two-dimensional coordinate system at sunshine (x is projecting direction at sunshine)

Suppose a certain moment, solar azimuth is A, if coordinate system P-X ' Y ' Z ' to turn clockwise A (position angle be negative in the morning, be just in the afternoon) around Z ' axle, forms new coordinate system P-xyz, in P-xyz coordinate system, sunshine light and shelter exist

In one plane.And this time point M coordinate in P-xyz coordinate system becomes:

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=R_Z^T\left(A\right)\left(\begin{array}{c}{X_{p-M}}^{\′}\\ {Y_{p-M}}^{\′}\\ {Z_{p-M}}^{\′}\end{array}\right)$

P-X ' Y ' Z ' and O-XYZ relation can be obtained by formula (1) and formula (2).Again because R is rotation matrix (being also orthogonal matrix), therefore exist: R ^-1(θ)=R ^t(θ) and

Can show that P-xyz and O-XYZ closes thus is:

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}\sin A(X\sin L-Y\cos L)+\cos A(\cos B\left({\mathrm{NsinBe}}^2+Z\right)-\sin B\left(X\cos L+Y\sin L\right))\\ \cos A(Y\cos L-X\sin L)+\sin A(\cos B\left({\mathrm{NsinBe}}^2+Z\right)-\sin B\left(X\cos L+Y\sin L\right))\\ -N{\cos }^{2}B+(X\cos L+Y\sin L)\cos B+\sin B(Z+\left(e^2-1\right)N\sin B)\end{array}\right)---\left(3\right)$

$\left(\begin{array}{c}X\\ Y\\ Z\end{array}\right)=\left(\begin{array}{c}(N+z)\cos B\cos L+\sin A(xsinL-y\cos LsinB)-\cos A(x\cos LsinB+y\sin L)\\ -x\cos L\sin A+(\left(N+z\right)\cos B-y\sin A\sin B)\sin L+\cos A(y\cos L-x\sin B\sin L)\\ x\cos A\cos B+y\sin A\cos B+(-N{e}^2+N+z)\sin B\end{array}\right)---\left(4\right)$

Because the straight line that sunshine, light formed with test point and ground point is in same plane, therefore, relatively a certain moment y ≡ 0, that is:

$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}\sin A(X\sin L-Y\cos L)+\cos A\left(\cos B\right({\mathrm{NsinBe}}^2+Z)-\sin B(X\cos L+Y\sin L\left)\right)\\ 0\\ -{\mathrm{Ncos}}^{2}B+(X\cos L+Y\sin L)csoB+\sin B(Z+(e^2-1\left)N\sin B\right)\end{array}\right)---\left(5\right)$

$\left(\begin{array}{c}X\\ Y\\ Z\end{array}\right)=\left(\begin{array}{c}(N+z)\cos B\cos L+x\sin L\sin A-x\cos L\sin B\cos A\\ -x\cos L\sin A+(N+z)\cos B\sin L-x\sin B\sin L\cos A\\ x\cos A\cos B+(-N{e}^2+N+z)\sin B\end{array}\right)---\left(6\right)$

(4) determine sunshine coordinate system with the earth longitude-latitude coordinates relation

Relational expression between the earth's core rectangular space coordinate and latitude coordinates:

$\left\{\begin{array}{c}X=(N+H_p)\cos B\cos L\\ Y=(N+H_p)\cos B\sin L\\ Z=\left(N\right(1-e^2)+H_p)\sin B\end{array}\right.---\left(7\right)$

Adopt formula (7) that O-XYZ, P-xyz and (L can be constructed, B) corresponding relation between, that is: be transformed to P-xyz by O-XYZ and adopt formula (5), be transformed to O-XYZ by P-xyz and adopt formula (6), calculate O-XYZ according to longitude and latitude and position angle and adopt formula (7).In above-mentioned formula, L, B are P ₀and P ₁latitude and longitude, N is that the fourth of the twelve Earthly Branches chief of a tribe corresponding to (L, B) position encloses radius-of-curvature, and e is ellipsoid first excentricity; H _pfor calculation level geodetic height (if P is on ground, H _p=0).

Next, asking for of projection formula at sunshine is carried out.

As shown in Fig. 9 (A), under rectangular coordinate system in space, M ₀for shutter body, M ₂and M ₃for the thing that is blocked, M ₀geodetic height is H ₀, M ₃geodetic height is H ₁.P ₀for M ₀ground point, P ₁for M ₀certain a bit (being illustrated as peak); P ₂for the body M that is blocked ₃ground point, P ₃for M ₃summit.Suppose P ₃for test point, block P ₃atural object be M ₀, so M ₀do not block P completely ₃the sufficient and necessary condition of some light at sunshine is: with M ₀upper any point P _iwith position of sun constructed by the light at sunshine that goes out, be H apart from ground height ₁the intersection point of shadow surface, be not equal to P ₂point longitude and latitude.So the solving and be converted into of occlusion issue at sunshine: calculate a certain moment P _ithe light at sunshine of point, with the shadow surface intersection point problem of certain geodetic height.

As shown in Fig. 9 (B), P ₀for blocking atural object M ₀ground point, P ₁for block atural object produces shade certain a bit, P ₁apart from P ₀distance is H ₀.At a time, P ₁place's sun altitude is h, G2 was P ₁the light at sunshine of point, P ₃for test point, also namely G2 with by shadow face intersection point.P ₂with P ₃there is identical terrestrial coordinate but be highly 0 ground point.Suppose P ₀place's longitude and latitude is respectively L and B, P ₂place's longitude and latitude is longitude and latitude Lx and Bx to be asked, then P ₁and P ₃the earth latitude coordinates is respectively (L, B, H ₀) and (L _x, B _x, H ₁).

P is built under P-xyz coordinate system ₀, P ₁, P ₂with P ₃geometric relationship, as shown in Fig. 9 (B), G1 be sunshine facade and the tangent method formed in earth ellipsoid face cut arc.If allow the Z axis of P-xyz coordinate system and P ₀p ₁overlap, point to zenith, X-axis is then defined as P ₀the tangent line of point on ellipsoid, points to solar rays G2 in floor projection direction; P ₄for G2 and x-axis intersection point, a certain moment P ₁place's sun altitude is h is known quantity, only needs according to B, L, H now ₀, H ₁and the time variable in a certain moment can calculate P ₂and P ₃latitude coordinates L _x, B _x, then by judging L _x, B _xp can be judged with practical test points position relationship ₀p ₁whether block P ₃.

Ask for L _x, B _xprocess as follows:

1. the G1 method built under P-xyz coordinate system cuts arc curvilinear equation

G ₁(x,z)＝K ₁x ²+K ₂z ²+K ₃z+K ₄xz

Wherein: $\left\{\begin{array}{c}{K}_1=1+(S-1)co{s}^{2}Aco{s}^{2}B\\ K_2=1+(S-1)si{n}^{2}B\\ K_3=2N,\\ K_4=2(S-1)\cos A\cos B\sin B\\ S=a^2/b^2\end{array}\right.,$ A is semimajor axis of ellipsoid, and b is minor semi-axis; A is P ₁the solar azimuth that the point a certain moment is corresponding.

2. build and solve P ₂point coordinate z under P-xyz coordinate system ₂, x ₂system of equations

Suppose P ₂, P ₃under P-xyz coordinate system, coordinate is respectively (z ₂, x ₂) and (z ₃, x ₃), can P be determined according to geometric relationship simultaneously ₀(0,0), P ₁(H ₀, 0), P ₃(0, H ₀coth), wherein h is sun altitude corresponding to a certain moment.

Condition one: P ₃meet straight-line equation G2, wherein G2 is: z+xtanh-H ₀=0, that is:

f ₁＝z ₃+x ₃tanh-H ₀＝0(8)

Condition two: P ₂meet curvilinear equation G1, that is:

$f_2=K_1{x_2}^2+K_2{z}_2^2+K_3{z}_2+K_4{x}_2{z}_2=0---\left(9\right)$

Under condition three: O-XYZ coordinate system, P ₂, P ₃at XOY plane subpoint P ₂', P ₃' and initial point O conllinear.

Due to P ₂and P ₃there is same latitude and longitude, therefore under O-XYZ coordinate system, P ₂, P ₃at XOY plane subpoint P ₂', P ₃' and initial point O conllinear.Draw according to formula (6):

f ₃＝(C ₁x ₃+C ₂z ₃+C ₃)(D ₁x ₂+D ₂z ₂+D ₃)-(D ₁x ₃+D ₂z ₃+D ₃)(C ₁x ₂+C ₂z ₂+C ₃)＝0(10)

Wherein: $\left\{\begin{array}{c}{C}_1=-\cos L\sin A-\cos A\sin B\sin L\\ C_2=\cos B\sin L\\ C_3=N\cos B\sin L\\ D_1=-\cos A\cos L\sin B+\sin A\sin L\\ D_2=\cos L\cos B\\ D_3=N\cos B\cos L\end{array}\right.$

Condition four: P ₂to P ₃air line distance is H ₁, that is:

$f_4={(z_2-z_3)}^2+{(x_2-x_3)}^2-H_1^2=0---\left(11\right)$

The f drawn by above four conditions ₁, f ₃and f ₄can cancellation z ₃and x ₃, obtain z ₂, x ₂equation with two unknowns group:

$\left\{\begin{array}{c}{f_1}^{\′}=K_1{x_2}^2+K_2{z}_2^2+K_3{z}_2+K_4{x}_2{z}_2=0\\ {f_2}^{\′}=J_0+J_1{z}_2+J_2{z}_2^2+J_3{z}_2^3+J_4{z}_2^4+J_5{x}_2^4+x_2^3(J_6+J_7{z}_2)\\ +x_2(J_8+J_9{z}_2^3+J_{10}{z}_2^2+J_{11}{z}_2)+x_2^2(J_{12}+J_{13}{z}_2^2+J_{14}{z}_0)=0\end{array}\right.---\left(12\right)$

Wherein: $\left\{\begin{array}{cc}{J}_1=2N(H_0^2-H_1^2-H_{0}N)& J_8=-2{\mathrm{NH}}_1^2\tanh -2H_0{N}^2\tanh \\ J_2=H_0^2-H_1^2-4H_{0}N+N^2& J_9=2\tanh \\ J_3-2H_0+2N& J_{10}=-2H_0\tanh +4N\tanh \\ J_4=1& J_{11}=-2H_1^2\tanh -4H_{0}N\tanh +2N^2\tanh \\ J_5={\tan }^{2}h& J_{12}=H_0^2-H_1^2{\tan }^{2}h+N^2{\tan }^{2}h\\ J_6=-2H_0\tanh & J_{13}=1+{\tan }^{2}h\\ J_7=2\tanh & J_{14}=-2H_0+2{\mathrm{Ntan}}^{2}h\end{array}\right.$

If adopt newton's method of approximation to solve x ₂and z ₂, restraining 8-12 time can obtain 10 ^-12data precision.

3. P is solved ₁subpoint P ₃the earth longitude and latitude (L _x, B _x)

P ₂coordinate figure (x under some P-xyz coordinate system ₂, 0, z ₂) bring formula (6) into and draw P ₂point X under O-XYZ coordinate system ₂, Y ₂and Z ₂value, then according to formula (7) backwards calculation (H _pwhen=0) go out L _x, B _x, be P ₂and P ₃point longitude and latitude (L _x, B _x).

In the present invention, adopt the effectively scope of blocking based on sunshine " space-time " feature to choose to screen with atural object.Because under geographic position, analysis time section known case, solar motion track can adopt nonlinear equation to express.And under given shutter body altitudes, according to the scope that analysis may exist the initial time determination shutter body that test point produces the time period of blocking.Wherein, shutter body height is called as and blocks height threshold, analysis time section and shutter body geographic range between function representation be sunshine " space-time " relation.

Under the large ground of hypothesis is surface level condition, can clearly draw out shadows cast by the sun running orbit.As shown in Figure 3, X represents direction, due east, Y represent Due South to, O is test point, and on XOY the earth surface level, Due South is to existence one limit S, and existing between S and test point O is highly the mark post L of M.If move L in T to the T ' time period, the summit of L is projected and overlaps with test point O all the time, then L will form path curves CC ' in XOY plane.

After CC ' determines, can think, within T to the T ' time period, in southern S point and CC ' scope, the atural object being highly less than or equal to M can not block O point.Therefore, can draw to draw a conclusion: when CC ' determines, M value is the critical altitude threshold value of test point O; Otherwise when M value is determined, CC ' is the critical boundary of test point O.

Based on above-mentioned principle, running track from sun can be derived at X, in Y-direction with height threshold M and analysis time t funtcional relationship, then determine to block height threshold M, effective occlusion area O-CC ' can be determined.

First, represent test point O to sun distance with Parameter N, derive containing N, analyze the running track from sun of moment t in X (east), the equation of locus group G on Y (south) and Z (zenith) three directions _{x, y, z}=F _{x, y, z}(t, N); Utilize G _zvalue eliminates G _xand G _yin solar distance Parameter N, construct G _x,y=F (Z, t) two dimensional surface functional equation, wherein, Z represents track of sun a certain height on zenith direction.Now, problem has been converted to the sun at X, the track that Y-direction is run and the funtcional relationship of Z height.

According to the two dimensional surface equation G relevant to Z _x,y=F (Z, t), supposes around test point O, there is the atural object that a height is Z value, this atural object is assumed to be a transportable mark post, within T to the T ' time period, if this atural object can block the generation of O point, its position that may exist be on curve C C ' certain a bit; And for the atural object of any height < Z value, within T to the T ' time period, if this atural object can block the generation of O point, its position that may exist must be in (being exactly scope between curve C C ' and O point) within Δ COC' Delta Region.Otherwise block test point O within T to the T ' time period if southern S point and CC ' scope have atural object, its height must be greater than Z value.Therefore, two dimensional surface equation G can be thought _x,y=F (Z, t) is CC ' curvilinear equation, and wherein, namely Z value is equivalent to blocks height threshold M.

Block after height threshold M determines, C and C ' point coordinate can be calculated according to sunrise, sunset moment, and then determine CC ' curve, can definite threshold height be M corresponding effectively block range delta COC'.

But above-mentioned effective occlusion area is only the two dimensional surface scope on the earth surface level, only rely on two dimensional surface scope, other redundancy atural object cannot be screened out according to " elevation constraint ", best screening effect can not be reached, therefore, need effective occlusion area to be converted to three-dimensional spatial area.Therefore, after the planar range of effective occlusion area is determined, the three-dimensional spatial area effectively blocked residing for atural object that need determine further can be made up of the critical zone of left and right and upper and lower both direction.Right boundary is mainly based on effective sunshine " time " feature (i.e. effective time section), the reverse calculating of projection formula at sunshine is adopted to determine, and upper and lower two borders mainly based on sunshine light " " feature (the maximum significant height that namely effective planar range is corresponding) still adopts the reverse calculating of projection formula at sunshine to determine in space.

Specifically determine that the method for effectively blocking 3D region is as follows:

(1) according to the right boundary of sunshine " time " feature determination effective range

Suppose that test point O is blocked all the time, block after height threshold determines, the position of upper two-end-point C and C ' of its critical boundary CC ' can be determined (as shown in Figure 3) according to T and the T ' moment.As shown in Figure 4, the corresponding P of C ₁point, the corresponding P of C ' ₀point, O is test point, once P ₀, P ₁determine with O point, then the left and right facade effectively blocking three-dimensional spatial area can be determined, that is: cross straight line and perpendicular to two space facades of the earth surface level.Obviously, between two facades, the corresponding solar azimuth of angle is from analyzing the variable quantity of initial time to end time, and the intersection of two facades was the straight line of test point O perpendicular to the earth surface level.

Curve P ₀p ₁(i.e. CC '), for blocking critical boundary corresponding to height threshold (M), the shutter body of any height < M, block if produced test point O, its planimetric position must between O-P ₀p ₁between, also must between crossing straight line and within facade border, left and right perpendicular to the earth surface level.

(2) according to the up-and-down boundary of sunshine " space " feature determination effective range

As shown in Figure 4, suppose that blocking height threshold is M, diagonal line hatches filling part is the circular conical surface at sunshine relative to test point O in analysis time.If be the space level face that M place structure one is parallel to the earth surface level in the earth surface level height, and make this space level face crossing with circular conical surface at sunshine, by sunshine circular conical surface surface form a curve arc P ₆p ₇.

According to circular conical surface principle at sunshine, sunshine, circular conical surface was " class circular cone " curved surface be formed by connecting by the ray of test point O and position of sun.Principle is chosen, the curve P on the earth surface level according to blocking height threshold ₀p ₁(i.e. CC ') is " highly for the mark post of M (blocking height threshold) is under guaranteeing that test point O situation can be blocked in its summit all the time, travel path on the ground ".Therefore, relative to same test point O, in the same analysis moment, the mark post summit of M (blocking height threshold) " highly for " should be on the ray that test point O and position of sun form all the time, that is: " mark post summit " should be on circular conical surface at this all the time at sunshine.In addition, P ₆p ₇form by the space level face being highly M is crossing with circular conical surface at sunshine, P ₆p ₇upper any point is M apart from the height of the earth surface level.Therefore, P ₆p ₇be " being highly the mark post summit of M (blocking height threshold) " travel path spatially, P ₀p ₁(CC ') i.e. P ₆p ₇track projection on the earth surface level.

In sum, by P ₆p ₇the curved surface formed with test point O and P ₆p ₇the plane at place is the coboundary of effectively blocking three-dimensional spatial area, and the earth surface level is its lower boundary.In conjunction with the right boundary of effective range, both can determine effectively to block three-dimensional spatial area, this three-dimensional spatial area is by curve P ₀p ₁, P ₆p ₇with space curved surface, the P of the vertical surface level greatly that height M determines ₆p ₇circular conical surface at the sunshine subregion formed with test point, excessively straight line and form with the earth surface level perpendicular to two, the left and right facade of the earth surface level.

(3) simplification of effectively blocking three-dimensional spatial area is expressed

Effectively block three-dimensional spatial area and comprise two space curved surfaces, in actual applications, the expression of space curved surface is very complicated, consuming time process with calculating.If can adopt the thought of uniformly subdivision, the three-dimensional enveloping solid building curved surface replaces expression and the calculating of curved surface, neither affects the attributive function of effectively blocking three-dimensional spatial area, will simplify the expression of three-dimensional spatial area again, improves counting yield.

As shown in Figure 5, P' _mfor positive noon 12 time (true solar time) sunray and curve P ₆p ₇intersection point, P _mfor P' _msubpoint on the earth surface level.Cross P _mdo vertical line respectively with intersect at P ₂and P ₃point, P ₄, P ₅for distance P above the earth horizontal plane ₃, P ₂for space two point of M, obviously, P _mwith P ₂, P ₃conllinear, P' _mwith P ₄, P ₅conllinear.Connect P ₆with P' _m, P ₇with P' _m, P' _mwith P _m, P ₀with P _m, P ₁with P _m, then by P ₁, P ₂, P ₅, P ₆, P' _mwith P _mcomposition wedge S ₁, by P ₀, P ₃, P ₄, P ₇, P' _mwith P _mcomposition wedge S ₂, by P ₃, P ₂, P ₅, P ₄rectangular pyramid S is formed with test point O ₃.

According to S ₁, S ₂with S ₃formation basic theory known, effectively block three-dimensional spatial area and will be completely contained in S ₁, S ₂with S ₃the area of space of common composition.Therefore, wedge S ₁, S ₂with rectangular pyramid S ₃can replace in actual applications above-mentionedly effectively blocking three-dimensional spatial area.

The complicated three-dimensional spatial area that effectively blocks is reduced to two wedges and a rectangular pyramid, and the unification of atural object data have employed atural object TIN model tormulation, therefore, on the topological relation that namely key issue of effective atural object screening is converted into atural object TIN and two wedge and a rectangular pyramid judges.

Wedge or rectangular pyramid are all regular geometric bodies, and the topological relation that conventional geometric figure algorithm can be adopted to carry out between itself and atural object TIN judges.For rectangular pyramid, if atural object TIN is crossing with rectangular pyramid or comprise, then can thinks that this atural object TIN belongs to and effectively block atural object data.

The three-dimensional spatial area that scope is effectively blocked in above-mentioned calculating carries out premised on " earth surface is surface level ".But in reality, blocking atural object projection line is on the ground not ideal line, but relevant to earth ellipsoid parameter curve.Therefore, if take the body characteristics of the earth into account, on the position concentrating on C and C ' is determined by the key issue of determining effectively to block three-dimensional spatial area, in order to improve precision and accurately fixed, need the position calculation formula based on geoid parameter again derivation C and C '.

By test point position with block the reverse computation process that process that height threshold calculates critical key point can think to project sunshine, wherein, " critical key point " is the terminal in critical edge boundary line, i.e. C and C ' point in Fig. 3.

As shown in Figure 6, for the Northern Hemisphere, have a test point P on the ground, existing around it is highly the shelter L of H (blocking height threshold).Choose principle according to blocking height threshold, within the Sunlight Analysis time period, if L can block P point all the time, then the position of L analyzing, only residing for the time is " the critical key point " in critical edge boundary line.

Adopt above-mentioned coordinate conversion principle, under " coordinate system at sunshine ", shelter, test point and sunshine light be in facade at same sunshine.As shown in Figure 7, at the P of y=0 ₀under-xz coordinate system (" coordinate system at sunshine " that namely certain moment T is corresponding), G1 be sunshine facade and the method for the tangent formation in earth ellipsoid face cut arc, P ₁for test point, be highly H ₁; P ₀for the ground point that test point is corresponding, its longitude and latitude (L, B) is known, meanwhile, and P ₀also be P ₀-xz coordinate origin; P ₂for threshold value H ₀the position that exists in the T moment of shelter, its terrestrial coordinate is coordinate (L to be asked _x, B _x); P ₃for shelter summit, h _xfor P ₃the sun altitude that position calculated in the T moment, G2 is that the T moment is through P ₃the light at sunshine of point; P ₄for G2 and x-axis intersection point; h _tfor G2 and x-axis angle, known according to sun altitude definition: h _t=h _x-(B-B _x); Due to h _tp ₁point T moment sun altitude, if consider that parallax and astronomical refraction are to P ₁and P ₃in the impact of the difference of T moment elevation angle, then: h _t=h _x-(B-B _x)+Δ _x-T, wherein Δ _x-Tfor parallax and astronomical refraction are to the correcting value of elevation angle difference.Think Δ _x-T≈ 0.

Therefore, consult shown in Fig. 7, the formula proving process chosen based on earth valid parameter value is as follows:

(1) P is built ₀under-xz coordinate system, G1 method cuts arc equation

Formula (6) is brought into ellipsoid equation :

G1(x,z,A _x)＝k1x ²+k2z ²+k3z+k4xz(13)

Wherein: $\left\{\begin{array}{c}k1=1+(S-1)co{s}^2{A}_x{\cos }^{2}B\\ k2=1+(S-1)si{n}^{2}B\\ k3=2N\\ k4=2(S-1)\cos A_x\cos B\sin B\\ S=a^2/b^2\end{array}\right.,$

A is semimajor axis of ellipsoid, and b is minor semi-axis; A _xp is engraved in for during T ₃the solar azimuth observed; B is P ₀-xz true origin P ₀latitude; N is P ₀the chief of a tribe that calls the roll in the morning encloses radius-of-curvature.

(2) P is built ₀g2 straight-line equation under-xz coordinate system

As shown in Figure 7, at P ₀p under-xz coordinate system ₀, P ₁and P ₄z, x coordinate be respectively: P ₀(z, x)=(0,0), P ₁(z, x)=(H ₁, 0) and P ₄(z, x)=(0, H ₁coth _t).According to P ₁and P ₄determine that straight-line equation is:

$\frac{1}{{\coth }_T}=\frac{H_1-z}{x}$

That is:

G2(z,x)＝(H ₁-z)coth _T-x(14)

(3) structure solves L _x, B _xsystem of equations

Condition one: P ₃(z ₃, x ₃) meet G2 equation (14)

According to formula (6) and rectangular space coordinate and longitude and latitude relation:

$P_3=\left(\begin{array}{c}{x}_3\\ z_3\end{array}\right)=\left(\begin{array}{c}{\mathrm{sinA}}_x(X_3\sin L-Y_3\cos L)+{\mathrm{cosA}}_x\left(\cos B\right({\mathrm{NsinBe}}^2+Z_3)-\sin B(X_3\cos L+Y_3\sin L\left)\right)\\ -{\mathrm{Ncos}}^{2}B+(X_3\cos L+Y_3\sin L)\cos B+\sin B(Z_3+(e^2-1\left)N\sin B\right)\end{array}\right)$

And

$\left\{\begin{array}{c}{X}_3=(N_x+H_0){\mathrm{cosB}}_x{\mathrm{cosL}}_x\\ Y_3=(N_x+H_0){\mathrm{cosB}}_x{\mathrm{sinL}}_x\\ Z_3=\left(N_x\right(1-e^2)+H_0){\mathrm{sinB}}_x\end{array}\right.,$

L _x, B _xfor test point P ₂the earth latitude coordinates, N _xfor P ₂the fourth of the twelve Earthly Branches chief of a tribe that point is corresponding encloses radius-of-curvature.

That is:

$\begin{array}{c}\left(\begin{array}{c}{z}_3\\ x_3\end{array}\right)=\left(\begin{array}{c}1/2(-e^2{\mathrm{Ncos}}^{2}B+2(H_0+N_x\left){\mathrm{cosBcosB}}_x\cos \right(L-Lx\left)\right)\\ {\mathrm{cosA}}_x(-(H_0+N_x\left){\mathrm{cosB}}_x\cos \right(L-Lx)\sin B+\cos B(e^{2}N\sin B+\left(H_0+N_x-e^2{N}_x\right){\mathrm{sinB}}_x\left)\right)\end{array}\right)\\ +\left(\begin{array}{c}0.5\left(N\right(-2+e^2+e^2{\sin }^{2}B)+(H_0+N_x-e^2{N}_x\left){\mathrm{sinBsinB}}_x\right)\\ (H_0+N_x){\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin (L-Lx)\end{array}\right)\end{array}$

Due to P ₃(z ₃, x ₃) meet G2 equation (14), therefore:

$\begin{array}{c}{f}_1(B_x,L_x,A_x)=H^{\&prime;}-(e^2{H}_0+\frac{N_x^{\&prime;}}{S}){\mathrm{sinBsinB}}_x+{\tanh }_T{N}_x^{\&prime;}{\mathrm{cosB}}_x\&lsqb;{\mathrm{cosA}}_x{\mathrm{cosL}}_x^{\&prime;}\sin B\\ -{\mathrm{sinA}}_x{\mathrm{sinL}}_x^{\&prime;}\&rsqb;+\cos B\{-N_x^{\&prime;}{\mathrm{cosB}}_x{\mathrm{cosL}}_x^{\&prime;}-{\mathrm{cosA}}_x{\tanh }_T\&lsqb;e^{2}N\sin B+(e^2{H}_0+\frac{N_x^{\&prime;}}{S}){\mathrm{sinB}}_x\&rsqb;\}\end{array}---\left(15\right)$

Wherein $\left\{\begin{array}{c}{L}_x^{\&prime;}=L-L_x\\ N_x=\frac{a}{\sqrt{1-e^2{\sin }^{2}B}}\\ N_x^{\&prime;}=H_0+N_x\\ H^{\&prime;}=H_1+N-\frac{3e^{2}N}{4}+\frac{1}{2}{e}^2{\mathrm{Ncos}}^{2}B+\frac{1}{4}{e}^{2}Ncos\left(2B\right)\end{array}\right.$

Condition two: P ₂(z ₂, x ₂) meet G1 equation (13)

$\left(\begin{array}{c}{z}_2\\ x_2\end{array}\right)=\left(\begin{array}{c}{N}_x\cos B\cos B_{x}cos(L-L_x)+\frac{N_x{\sin }^{2}B}{S}-\frac{a^2}{N}\\ \cos A_x\sin B(\cos B\left(e^{2}N+\frac{N_x}{S}\right)-N_x\cos B_{x}cos\left(L-L_x\right))+N_x\cos B_x\sin A_{x}sin(L-L_x)\end{array}\right)$

P simultaneously ₂meet G1 equation (13), therefore have

$\begin{array}{c}{f}_2(B_x,L_x,A_x)=N(-\frac{2a^2}{N}+2N_x{\mathrm{cosBcosB}}_x\cos (L-L_x)+\frac{2N_x{\sin }^{2}B}{S})\\ +{(-\frac{a^2}{N}+N_x{\mathrm{cosBcosB}}_x\cos (L-L_x)+\frac{N_x{\sin }^{2}B}{S})}^2(1+e^{\&prime;2}{\sin }^{2}B)\\ +e^{\&prime;2}{\mathrm{cosA}}_x\cos B\sin B(-\frac{2a^2}{N}+2N_x{\mathrm{cosBcosB}}_x\cos (L-L_x)+\frac{2N_x{\sin }^{2}B}{S})\\ \{{\mathrm{cosA}}_x\sin B\&lsqb;(e^{2}N+\frac{N_x}{S})\cos B-N_x{\mathrm{cosB}}_x\cos (L-L_x)\&rsqb;+N_x{\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin (L-L_x)\}\\ +(1+e^{\&prime;2}{\cos }^2{A}_x{\cos }^{2}B)\{{\mathrm{cosA}}_x\sin B\&lsqb;(e^{2}N+\frac{N_x}{S})\cos B-N_x{\mathrm{cosB}}_x\cos (L-Lx)\&rsqb;\\ +N_x{\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin (L-L_x){\}}^2\end{array}$ (16)

Wherein e ' is ellipsoid second excentricity.

Condition three: L _x, B _xwith A _xbetween there is conversion relation

According to SPA formula:

A _x＝F(B _x,L _x,T)(16)

Employing formula (16) can cancellation formula (15), the middle P of formula (16) ₂point position angle A to be asked _xparameter, finally draws containing L _x, B _xthe equation with two unknowns group of parameter:

$\left\{\begin{array}{c}{f}_1^{\&prime;}=f_1(B_x,L_x,A_x)\&DoubleRightArrow;f_1^{\&prime;}(B_x,L_x,F\left(B_x,L_x,T\right))\&DoubleRightArrow;f_1^{\&prime;\&prime;}(B_x,L_x)=0\\ f_2^{\&prime;}=f_2(B_x,L_x,A_x)\&DoubleRightArrow;f_2^{\&prime;}(B_x,L_x,F\left(B_x,L_x,T\right))\&DoubleRightArrow;f_2^{\&prime;\&prime;}(B_x,L_x)=0\end{array}\right.$ (17)

Based on formula (17), suppose that certain moment T is known, adopt newton's method of approximation to solve L _x, B _xparameter, restraining 8-12 time can obtain 10 ^-12data precision.

(4) by determining critical key point analysis time

Due to height for the shelter that blocks height threshold analyzing, the position of stopping residing for the time is " critical key point " in critical edge boundary line.Therefore, blocking under height threshold and analysis time section known case, the Analysis on Selecting beginning and ending time, as calculating parameter, calculates " critical key point ".Relative to test point P ₀(L, B), supposes P _s(L _s, B _s) and P _e(L _e, B _e) be respectively that Sunlight Analysis rises, only Time Calculation " the critical key point " that go out, then by P ₀, P _sand P _e3 can be constructed earth surface effectively blocks scope.

Screening effectively blocking atural object based on sunshine " space-time " feature, bulk redundancy atural object data can be filtered out, but when the sunshine of carrying out discrete instants when circulating calculates, all effectively atural object data still need to participate among each calculating process.Therefore, only screening effectively blocking atural object, inherently can not improve the counting yield of model.

What the present invention is based on timeslice index effectively blocks atural object data reconstruction, is to reconstruct in the secondary data that atural object screening basis is carried out of effectively blocking based on sunshine " space-time " feature.It, when the calculating some time sets a date photograph, makes the atural object data volume of participation computing be contracted on data set corresponding to timeslice, and then inherently improves the counting yield of model.

The atural object data reconstruction that effectively blocks based on timeslice index reconstructs the secondary of original atural object TIN data, is expand the secondary of original atural object TIN data structure, do not redefine original TIN structure.

As shown in Figure 10, build the memory mapping table of timeslice in original TIN data structure periphery, the node Nodes of each timeslice is the memory address of corresponding TIN data set, when calculating moment T is between timeslice T _ibetween time, [T _i] address date that points to of .Nodes is the atural object data set of actual participation computing.

Atural object TIN data reconstruction based on timeslice Indexing Mechanism realizes index construct and data reconstruction by memory pointer, does not need the internal memory that expense is new; In addition, participate in all effectively blocking atural object data set in calculating process by [T _i] .Nodes point to address date replace, make to the Time & Space Complexity spent by the shadowing of test point be contracted to timeslice limit scope, counting yield can be improved in essence.

The realization of effectively blocking atural object data reconstruction based on timeslice index can be summarized as two large steps:

First, with given " time step ", differential is carried out to the bulk analysis time, form uniform time slice sequence.This step needs fixed time step parameter TStep and border to swing threshold value (T-SWayTolerance).Under threshold level and test point known case, can be constructed by the start/stop time of arbitrary timeslice with test point is the rectangular pyramid on summit.Wherein, T-SwayTolerance is mainly used in: when scope face is in critical conditions, eliminates polyhedron and TIN topological relation error in judgement.The timeslice initial time adopted is: " computing time, sheet initial time deducted T-SWayTolerance ", end time is: " computing time, sheet end time added T-SWayTolerance ".

Secondly, with the rising of timeslice, only moment with block height threshold and build space rectangular pyramid, judge the topological relation of atural object TIN data and rectangular pyramid, so decision whether this TIN belongs to Nodes node corresponding to its timeslice.If dotted portion in Figure 11 is rectangular pyramid corresponding to timeslice, this rectangular pyramid and atural object TIN topological relation exist crossing, from comprise three kinds of situations, only intersect and relation of inclusion exists time, atural object TIN is just associated with timeslice T.Nodes.

Claims (2)

1. based on sunshine space-time characteristic effectively block atural object screening defining method, it is characterized in that:

The method is first by determining the shadows cast by the sun motion track formula of the shelter of known altitude to test point Ouluding boundary, and then determine rising, the only shadows cast by the sun motion track of the critical point of time, then rise, only the shadows cast by the sun motion track of the critical point of time and test point form two and to intersect and perpendicular to two facades of the earth surface level, simultaneously in shadows cast by the sun motion track forming process, shelter top be parallel to the earth surface level plane and sunshine circular conical surface form the movement locus of an arc, the movement locus of arc forms an arc facade perpendicular to the earth surface level, described two facades and arc facade composition test point effectively block scope, then the redundancy atural object effectively blocked in scope is removed, determine the effective atural object effectively blocked in scope,

The method of the redundancy atural object that described removal is effectively blocked in scope is: the data structure [tin first setting up atural object TIN model ₁, tin ₂..., tin _m], in described data structure, the internal storage structure of a single point comprises: the earth coordinates be made up of longitude L, latitude B, elevation H, the topocentric coordinate system be made up of station longitude centroid L ', the heart latitude B ' and elevation H ' that stands, X, Y, Y rectangular coordinate system in space X centered by the earth's core, Y and Z, also comprises the transition flag Flag between them; Then the memory mapping table of timeslice is built in the periphery of atural object TIN data structure, the node of each timeslice in memory mapping table corresponds to the memory address of TIN data set, when calculating some moment are between timeslice, address date pointed by this timeslice is exactly the data that actual needs participates in the atural object TIN model of computing, and the data of other atural object TIN model are the atural object data that redundancy need be deleted;

The shadows cast by the sun motion track formula of described shelter to the maximum Ouluding boundary of test point is:

$\left\{\begin{array}{c}{f}_1^{\&prime;}=f_1\left(B_x,L_x,A_x\right)\&DoubleRightArrow;f_1^{\&prime;}\left(B_x,L_x,F\left(B_x,L_x,T\right)\right)\&DoubleRightArrow;f_1^{\&prime;\&prime;}\left(B_x,L_x\right)=0\\ f_2^{\&prime;}=f_2\left(B_x,L_x,A_x\right)\&DoubleRightArrow;f_2^{\&prime;}\left(B_x,L_x,F\left(B_x,L_x,T\right)\right)\&DoubleRightArrow;f_2^{\&prime;\&prime;}\left(B_x,L_x\right)=0\end{array}\right.$ Formula (1)

Wherein, moment T produces the initial time of shade or produces the stop timing of shade, (L _x, B _x) be the latitude and longitude coordinates in T moment;

And in above-mentioned formula (1),

$\begin{array}{c}{f}_1\left(B_x,L_x,A_x\right)=H^{\&prime;}-(e^2{H}_0+\frac{N_x^{\&prime;}}{S})\sin B\sin B_x+\tan h_T{N}_x^{\&prime;}\cos B_x[\cos A_x\cos L_x^{\&prime;}\sin B\\ -{\mathrm{sinA}}_x{\sin L}_x^{\&prime;}\&rsqb;+\cos B\left\{-N_x^{\&prime;}{\mathrm{cosB}}_x\cos L_x^{\&prime;}-{\mathrm{cosA}}_x{\tanh }_T\&lsqb;e^{2}N\sin B+\left(e^2{H}_0+\frac{N_x^{\&prime;}}{S}\right){\mathrm{sinB}}_x\&rsqb;\right\}\end{array}$ Formula (2),

$\begin{array}{c}{f}_2\left(B_x,L_x,A_x\right)=N\left(-\frac{2a^2}{N}+2N_x{\mathrm{cosBcosB}}_x\cos \left(L-L_x\right)+\frac{2N_x{\sin }^{2}B}{S}\right)\\ +{\left(-\frac{a^2}{N}+N_x{\mathrm{cosBcosB}}_x\cos \left(L-L_x\right)+\frac{N_x{\sin }^{2}B}{S}\right)}^2\left(1+e^{\&prime;2}{\sin }^{2}B\right)\\ +e^{\&prime;2}{\mathrm{cosA}}_x\cos B\sin B\left(-\frac{2a^2}{N}+2N_x{\mathrm{cosBcosB}}_x\cos \left(L-L_x\right)+\frac{2N_x{\sin }^{2}B}{S}\right)\\ \left\{{\mathrm{cosA}}_x\sin B\&lsqb;\left(e^{2}N+\frac{N_x}{S}\right)\cos B-N_x{\mathrm{cosB}}_x\cos \left(L-L_x\right)\&rsqb;+N_x{\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin \left(L-L_x\right)\right\}\\ +\left(1+e^{\&prime;2}{\cos }^2{A}_x{\cos }^{2}B\right)\{{\mathrm{cosA}}_x\sin B\&lsqb;\left(e^{2}N+\frac{N_x}{S}\right)\cos B-N_x{\mathrm{cosB}}_x\cos \left(L-L_x\right)\&rsqb;\\ +N_x{\mathrm{cosB}}_x{\mathrm{sinA}}_x\sin \left(L-L_x\right){\}}^2\end{array}$ Formula (3),

Wherein, B is latitude; A _xfor the solar azimuth that T moment test point observes, a is semimajor axis of ellipsoid, and b is minor semi-axis, L _x, B _xfor the earth latitude coordinates of test point, N _xthe fourth of the twelve Earthly Branches chief of a tribe corresponding for test point point encloses radius-of-curvature, and e ' is ellipsoid second excentricity, and S is the position of the test point point observation sun in celestial sphere.

2. according to claim 1 based on sunshine space-time characteristic effectively block atural object screening defining method, it is characterized in that: when the shadows cast by the sun motion track of shelter to test point Ouluding boundary of described known altitude refers to that solar irradiation is mapped to the summit of shelter, the movement locus that when the summit projection of shelter overlaps with test point all the time, shelter is formed on the ground.