6. Rectification 2 hours

Size: px

Start display at page:

Download "6. Rectification 2 hours"

Helen Kelly
5 years ago
Views:

1 Lecture /2/2003 Concept Hell/Pfeiffer February ectification 2 hours Aim: Principles for rectification Theory: Indirect rectification in digital image processing Methods of rectification and integration in programs 6.1. The deformations of metric photograph A line map is unsatisfactory for some regions (undeveloped areas). Photo-map is a map that shows the content of the aerial photographs and is better solution. Photo maps are significantly less expensive and quicker for producing in comparison to conventional topographic line maps. The deformations of images, including photographs are described in differential geometry by means of the Tissot indicatrix. This theory is developed for analytical surfaces. Ground surface is represented in photogrammetry by digitizing a large set of individual points on the surface Another method is based on the closely spaced discrete points, defined by their XYZ-coordinates. The deformation between photograph and terrain are described by co-linearity equations r11.( Xi X0) + r21.( Yi Y0 ) + r31.( Zi Z0) x xp = c. r.( X X ) + r.( Y Y ) + r.( Z Z ) 13 i 0 23 i 0 33 i 0 r12.( Xi X0) + r22.( Yi Y0 ) + r32.( Zi Z0) y yp = c. r.( X X ) + r.( Y Y ) + r.( Z Z ) 13 i 0 23 i 0 33 i 0 Geometrically this deformation is presented in the next figure (6.1)

2 Lecture /2/2003 Figure 6.1. elations between grids in object and photo The estimation of deformation can be obtain by the deformation of square grid, shown on the next figure

given by expressions δ12 δ13 λx =, λy = (6.2) X Y where 2 ik = ( i k ) + ( i k 2 δ ξ ξ η η ) (6.

3 Lecture /2/2003 η Y ξ X Figure 6.2. Deformations of square grid Deformations in length in X and Y directions are given by expressions δ12 δ13 λx =, λy = (6.2) X Y where 2 ik = ( i k ) + ( i k 2 δ ξ ξ η η ) (6.3) Deformations of grid and photos due to the projection and terrain are presented in the next figure Figure 6.3. Deformed grid of aerial photograph For plane object surface the deformation can be see very clearly in comparison with XY-grid of the object

4 Lecture /2/2003 η Y ξ X Figure 6.4. Projective deformation of grid The line of main vertical (of the large slope) passes from upper-left to the lower-right corner of the grid. The simple presentation of deformations is shown on the following figure

5 Lecture /2/2003 ν τ 0 τ a n n p 0 0 ρ p a 0 ρ 0 N A Figure 6.5. eduction of length in tilted photograph. The radial displacement of point position with radius ρ for small angles of tilt ν is presented by expression 2 ρ ρ =.v (6.4) c 2 2 where ρ = ξ + η and 2 2 ρ = ξ + η The increase or reduction of small lengths dρ is obtained by differentiating the previous expression ρ d( ρ) = 2.. ν. dρ (6.5) c The photograph is similar to the orthophoto (without geometrical distortions) if the object plane and image plane are made parallel. The images that does not lye in this plane are displaced from there actual position depending on the difference of distance from the projection center (that means different height for aerial photos).

6 Lecture /2/2003 ρ b ρ a b b p a a ρ b ρ a P 2 c Z 0 A Z B B b Z A A A B Figure 6.6. adial displacement due to the height of the objects From figure 6.6 is possible to derive the value of error in displacement that is due to heights the objects c ρ ρ ρ = = Z = Z (6.6) Z Z c m p For image rectification now are used two procedures. rectification with photographic rectifier and rectification procedure called differential rectification. In the photographic rectifier the new photograph is a central projection of the original but with geometric properties of an orthophoto. This method of rectification can be used only for tilted photographs of plane objects.

7 Lecture /2/ ectification by central perspective transformation There are two possible methods for perspective rectification with and without establishment of inner orientation parameters in the rectifier Perspective rectification with establishment of inner orientation The inner and outer orientation of original photograph must reestablished for the solution of rectification task. Figure 6.7. ectifier with reestablished inner and outer orientation The projector has fiducial marks for reestablishements of the inner orientation. Three elements of outer orientation can be set: z 0 of the projector, rotations ω and φ. z 0 Z0 = (6.7) m 0 where m 0 is the scale factor of the orthophoto

8 Lecture /2/2003 The focal length of the projector lens must be different from the focal length of the original camera lens. This rectifier requires three control points from which the six elements of outer orientation can be determined (from them only three are important for rectification) Perspective rectification with reestablishment of inner orientation. It is necessary to satisfy two optical condition to achieve a sharp projected image over the full format. ptical conditions The plane of the photograph in the rectifier must be tilted relative to the optical axis. This condition however is not satisfied in metric cameras where the plane of photograph is perpendicular to the optical axis of the camera. First optical condition is called Scheimpflug condition, because T.Sheimpflug first developed it. importance. It says that plane of the photograph IP, the principle plane of the objective H e and the plane of the projection stage PS must intersect in the same straight line S. This is shown at the figure 6.8.

9 Lecture /2/2003 Figure 6.7. ptical conditions for perspective rectification Second optical condition is Newton s condition 1 1 1, s + s = f (6.8) It could be presented in Newtonian form x. x' = f 2 where x = s f (6.9) x' = s' f If this condition called also lens equation is satisfied for points on the optical axes the Sheimpflug condition ensures that it also be satisfied for all homologous points in the plane of the photograph and projection stage.

10 Lecture /2/2003 The lens equation and Scheimpflug condition are satisfied by mechanical devices, so-called inversors and by electronic control systems in newer instruments. Geometric conditions The Scheimpflug condition is not satisfied when photograph is taken. It is seen in the next figure. Figure 6.8. Geometric relations when taking photograph Geometric relations of taking the image and in rectifier differ. The relations are pointed below. 1. The image plane of the rectifier is identical with the image plane of the taking camera, which is tilted by nadir distance ν. The intersection of the planes IP and PS is made coincident with the intersection S of the object plane P and the image plane IP. Line S must be imaged in itself. 2.The image plane is perpendicular to the camera axis. (The image points 1,2,3 are central projection of object points. 3.The image plane IP intersects the projection plane PS in any arbitrary tilt. 4.bject points 1, 2, 3 must be transferred on the projection stage by circular arcs with center S. 5.Connection of points 1, 2, 3 on the projection stage to the points 1, 2, 3 gives new projection center e. This center belongs to the optical axis tilted relatively to the image plane. We speak for the rotation of projection center. Two planes IP and PS are projective to one another. 6. All projection centers corresponding to arbitrary tilts of projection stage PS, also satisfy the Scheimpflug condition and lens equation.

11 Lecture /2/2003 Figure 6.9.ectification with not congruent bundle of rays The relations between geometrical characteristics are shown in Appendix 1. This rectifiers are with bundle of rays that is not congruent to that of the original photograph ectification with known instrument settings When we calculate the outer orientation parameters of photo we obtain coordinates of the projection center X0, Y0, Z0 and three rotation angles ω, φ, κ. From this parameters we must derive the direction α, the nadir distance ϖ and swing κ.

12 Lecture /2/2003 We construct new spatial rotation matrix ανκ a ν κ Figure Definition of rectification angles = (6.10) After expression in analytical form is possible to obtain angle elements from the coefficients of rotation matrix r r tan α = cos = tanκ = (6.11) ν r33 r23 r32 It is possible to make rectification with control points and with knowledge of the inner orientation. For such orientation sufface three control points. However usually four control points are used. The additional point is for check of the orientation process. Two step procedure is used. 1. Change the projection distance 2.Tilt of projection stage about two axis (for some rectifiers only one tilt and rotation are given). In the second case orientation procedure is more complicated. There is possible to arrange the orientation procedure with control points, but without knowledge of the inner orientation. The construction of rectifiers allow two group of possibilities: 1. Group

13 Lecture /2/ Group Tilts of the projection stage in two directions Displacement of the photograph in two directions Tilt of the projection stage in one direction Displacement of the photograph in two directions and rotation of the photograph Errors of perspective rectification Errors of perspective rectification depend on two types of sources. The first one is the error in establishing the settings in the rectifier that are computed or derived from control points. The second source of error is due to the diversity of terrain surface from assumed reference plane for which the rectification is processed.. The errors accordingly to the errors of object heights may be presented for small angles by equation. The derivation of errors is shown in Appendix 2 For error in the orthophoto plane we obtain the equation r r = Z (6.12) cm. p The error are in radial direction from the nadir point of orthophoto. As a conclusion it must be pointed that area of application of perspective rectification is limited to plane areas with small diversity of height. The maximum allowed different in height from assumed plane can be calculated by defining the permissible error (usually 1mm) at the corners of orthophoto by solving the above equation. rcm.. p Z = (6.13) r where Z is on the terrain, and r is the displacement in orthophoto.

14 Lecture /2/2003 ρ p n ρ c ν z = Z 0 0 m 0 Z 0 r b r b N r a r a Z 0 A B b Z A A Z B A B Figure adial image displacements of points outside the reference plane The same is influence in the accuracy of the rectified photographs of buildings facades for elements in front of or behind the plane of the facade.

Lecture 6-15 - 11/2/2003 6.3. ectifiers 6.3.1. Classification of rectifiers ectifiers with congruent bundle of rays are not produced because of the lack of sharpness in the projected image.

15 Lecture /2/ ectifiers Classification of rectifiers ectifiers with congruent bundle of rays are not produced because of the lack of sharpness in the projected image. More over they need different objectives depending of required enlargement ratio that is practically inconvenient. There exist some photographic enlargers working with small angles of tilt Examples of optical-mechanical rectifiers ectifiers with rigorously satisfied geometric and optical conditions are produced by firms Zeiss Jena, Leica Heerbrugg and Zeiss berkochen. Wild E4 has enlargement ratios x. Projection stage can be tilted about two perpendicular axes up to ±15gon. Picture carrier can be displaced in two perpendicular directions up to ±4cm. The focal length of the objective is 15cm. The SEG6 ectifier of Zeiss berkochen has approximately the same ranges as Wild E4. The ectimat C of Zeiss Jena was equipped with objective 15cm for enlargement ratios 0.85 to 8.0x and with objective 7cm for enlargement ratios 3.0 to 18.0x. These rectifiers can operate with preset elements of outer orientation. Figure ectifier ZEG 6 Accuracy of perspective transformation depends on the height error of terrain or the objects 6.4. Digital rectification Th main principal of digital transformation is similar to the general principal of image restoration.

16 Lecture /2/2003 F I (i,j) F (i,j) F (i,j) f (k,l) B g (k,l) Figure Main scheme for image restoration The possible sources for geometrical deformation are following: But in digital rectification the object of restoration are not the radiometric properties but the geometric properties of the images. 1. Projective transformations 2. Influence of the height of the objects 3. Lens distortions 4. Earth curvature 5. Map projection Neverthless of source of geometrical deformation main goal of digital rectification is to reverse the pixel position in this place, that it must have in the undisturb image of the corresponding scale. If we have some source of geometric distortion that is described by formulas x y o o = = f f B, x B, y ( i I ( i I, j I, j I ) ) (6.14) ur goal is to find such reverse transformation x y = g = g, x, y ( i ( i, j, j ) ) (6.15) In the case when the method of nearest neighbour is used the transformations take the form i j = Ent[ g = Ent[ g, x, y ( i ( i, j, j ) + 0.5] ) + 0.5] (6.16) By reverse solution we can reach to the output coordinates (without corrections). Comparing the values for this condition for perfect restoration i j = Ent[ g = Ent[ g 1, x ( i, j ) + 0.5] = Ent[ gi, x 1, y ( i, j ) + 0.5] = Ent[ gi, y These are conditions for geometrical restoration g g 1, x ( y, x) = gi, x 1, y ( y, x) = gi, y ( y, x) ( y, x) ( i ( i, j, j ) + 0.5] ) + 0.5] (6.17) (6.18)

17 Lecture /2/2003 There are possible two schemes for digital rectification - direct method or forward transformation; - indirect method or reverse transformation. In direct method the corrected coordinates of input image are calculated. Every pixel obtain its real value coordinates in the matrix of restored image. It is possible to interpolate this value to calculate the grey values in raster matrix. Thi smethod is shown graphically on the next figure. n o r g (k,l) m o Figure Direct transformation Direct transformation method requires more memory because at the intermidate stage it is necessary to keep not only the greg value but also the coordinates of transformed pixel. The procedure for finding from which transformed pixel to calculate the interpolated values is relatively complex. In the indirect method is used calculation from the pixel coordinates of transformed image. For this purposes the reverse formula is necessary to be applied. The graphic presentation of this method is shown on the next figure

18 Lecture /2/2003 n o r -1 g (k,l) m o Figure Indirect method The calculated values in the input image are not integer. To calculate the grey value at the desired point it is necessary to use some method of grey value interpolation. There are used defferent methods: - nearest neighbour; (square interpolation) -bilinear interpolation; - triangle; -bell-shape curve; -Gauss curve; - sinc function (sinx/x) for different sizes of interpolation window. The Graphical presentation and formulas for interpolation methods are shown in the Appendix 3. Appendixes Appendix 1 From the figure 6.9 can be derived several relations Tilt of the photograph β: Z0 f e SF = = (6.19) m 0.sin ν sin β fe. m0 sin β =.sinν (6.20) Z m0 scale number of orthophoto. Tilt of the stage γ: 0

19 Lecture /2/2003 F a = c fe Fe = sinν = sinγ (6.21) f e sin γ =.sinν (6.22) c Image distance s - s' = A' e s' = f + F.cos γ. tanβ (6.23) Substituting the value of e e F e we find tan β s' = f e. 1+ tanγ (6.24) Projection distance s - s = A Substituting s in lens equation we obtain e tanγ s= f e.1 + tan β (6.25) Enlargement ratio (along the optical axis) for straight lines in image and object perpendicular to the optical axis s tanγ v = = (6.26) s' tanβ Displacement e of the principal point p from optical axis AA. f.cot '.cot e γ e= FP FA = c ν (6.27) cos β Substituting the value of sinγ e c.sin f γ cotγ e= fe (6.28) sinγ cos β Appendix 2.. The errors accordingly to the errors of object heights may be presented for small angles by equation. ρ r r = ρ. v = Z. v= Z (6.29) Z Z 0 0 The value of Z0 depends on photo scale and photo scale. Z = cm (6.30) 0. p

20 Lecture /2/2003 For error in the orthophoto plane we obtain the equation r r = Z (6.31) cm. p The error are in radial direction from the nadir point of orthophoto. Appendix 3 The mathematical formulation of interpolation Presentation of restoration function J I F( x, y) = FI( j x, i y) ( x j x, y i y) j= J i= I (6.32) The restoration process is described in spatial area by relation J I ED( x, y) = FI( j x, i y).[ ( x j x, y i y) I( x j x, y i y)] j= J i= I (6.33) The specral presentation of interpolation is obtained by Furier transformation 2 F 4π ( x, y) ( x, y) I( x xs, y ys). ω ω = ω ω x y F ω j ω ω i ω (6.34) j= i= This presentation allows easier to estimate the errors of interpolation The presentation of different types of iterpolation function is shown in the next table

21 Lecture /2/2003 Function Graphic presentation Definition in spatial area Spectral Presentation sinc 4 sin(2πx/ T ) sin(2 / (, ) x πy Ty xy =. 1, ωx ωxs, ωy ωys TT x y (2 πx/ Tx) (2 πy/ Ty) ( ωx, ω y) = 0 - otherwise Tx = 2 π/ ω xs, Ty = 2 π/ ωys ectangular 1 sin( ω, x Tx /2, y Ty /2 xtx / 2).sin( ωyty / 2) 0 ( x, y) = TT 0( ωx, ω y) = x y ( ωxtx /2).( ωyty /2) 0 - otherwise Triangle 1( x, y) = 0( x, y) 0( x, y) 2 1( ωx, ω y) = 0( ωx, ω y) Bell-shape type 2( x, y) = 0( x, y) 1( x, y) 3 2( ωx, ω y) = 0( ωx, ω y) Cubic B-spline 3( x, y) = 0( x, y) 2( x, y) 4 3( ωx, ω y) = 0( ωx, ω y) Gauss curve 2 2 x + y 1 2 2σ xy (, ) =. e ω 2 2 πσω ( ωx, ω y) = e σω( ω x +ωy ) 2

23 Lecture /2/2003 The derivation of relation for bilinear interpolation Z Z 11 Z 00 Z 0d Z 01 Y Z1d Dy Z 10 Dx X δx δx zxy = z0δ.(1 ) + z1δ. = x x δy δy δx δy δy δx = z 00.(1 ) + z01..(1 ) z 10.(1 ) z11.. y y + x + y y = x δx δy δx δy δx δy δx δy = z z 10..(1 ) + z01.(1 ). + z 00.(1 ).(1 ) x y x y x y x y (6.35)

Training i Course Remote Sensing Basic Theory & Image Processing Methods September 2011

Training i Course Remote Sensing Basic Theory & Image Processing Methods 19 23 September 2011 Geometric Operations Michiel Damen (September 2011) damen@itc.nl ITC FACULTY OF GEO-INFORMATION SCIENCE AND