how to generate barcode in c# asp.net Transformation: ECEF to Tangent Plane in Software

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Transformation: ECEF to Tangent Plane
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e = [x, y, z]e [xo , yo , zo ]e x (228)
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where [xo , yo , zo ]e are the ECEF coordinates of the origin of the local tangent plane Then e is a vector from the local tangent plane origin to x an arbitrary location P e = [x, y, z]e , with the vector and point coordinates each expressed relative to the ECEF axis The transformation of vectors from ECEF to tangent plane (TP) can be constructed by two plane rotations, as depicted in Figure 212 First, a plane rotation about the ECEF z-axis to align the rotated y-axis (denoted y ) with the tangent plane east axis; second, a plane rotation about the new y -axis to align the new z-axis (denoted z ) with tangent plane inward pointing normal vector The rst plane rotation is de ned by cos( ) sin( ) 0 R1 = [ ]3 = sin( ) cos( ) 0 (229) e 0 0 1
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25 SPECIFIC VECTOR TRANSFORMATIONS
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Figure 212: Variables for derivation of Rt e where is the longitude of the point [xo , yo , zo ]e The second plane rotation is de ned by 0 sin + cos + 2 2 0 1 0 (230) = Rt = + 1 2 2 0 cos + 2 sin + 2 sin( ) 0 cos( ) 0 1 0 , = (231) cos( ) 0 sin( ) where is the latitude of the point [xo , yo , zo ]e The overall transformation for vectors from ECEF to tangent plane representation is then vt = Rt ve where e sin( ) cos( ) sin( ) Rt = Rt R1 = e 1 e cos( ) cos( ) sin( )sin( ) cos( ) cos( )sin( ) cos( ) 0 sin( ) (232)
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The inverse transformation for vectors from tangent plane to ECEF is ve = Re vt where Re = (Rt ) t t e Example 23 The angular rate of the ECEF frame with respect to the inertial frame represented in the ECEF frame is e = [0, 0, 1] ie Thereie fore, the angular rate of the ECEF frame with respect to the inertial frame represented in the tangent frame is ie cos( ) 0 t = Rt e = ie e ie ie sin( )
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CHAPTER 2 REFERENCE FRAMES
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Let P t = [n, e, d] denote the coordinates of the point P represented in the tangent plane reference system, then t n e = Re e x t d Using eqn (228), the transformation of the coordinates of a point from the tangent plane system to the ECEF system is e t e n xo x y = yo + Re e (233) t zo d z Example 24 For the ECEF position given in the Example 21 on page 33, the matrix 02569 04967 08290 00000 Rt = 08882 04594 e 03808 07364 05592 transforms vectors from the ECEF coordinate system to tangent plane coordinates The point transform is de ned as e t 2430601 x x y = Rt y 4702442 106 e 3546587 z z where the origin location [xo , yo , zo ] is de ned in eqn (212) The inverse transformations are easily derived from the preceding text For example, the local unit gravity vector which is assumed to be (0, 0, 1) in tangent plane coordinates, transforms to (03808, 07364, 05592) in ECEF coordinates
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Transformation: ECEF to Geographic
The geographic frame has a few points that distinguish it from the other frames First, because the origin of the geographic frame moves with the vehicle and is the projection of vehicle frame origin onto the reference ellipsoid, the position of the vehicle in the geographic frame is xg = [0, 0, h] The latitude and longitude de ne the position of the geographic frame d origin (vehicle frame projection) on the reference ellipsoid Second, dt xg = ; which is not the velocity vector for the vehicle The Earth rel[0, 0, h] d e ative velocity vector represented in the ECEF frame is ve = dt xe This vector can be represented in the geographic frame as
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