bundle block adjustment with 3d natural cubic splines

bundle block adjustment with 3d natural cubic splines bundle block adjustment with 3d natural cubic splines

13.07.2015 Views

Substituting (3.2) into (3.3 [1]) leads toand substituting (3.2) into (3.3) leads to2a (i−1)2 + 6a (i−1)3 = 2a i2 (3.4)a i0 = x ia i1 = D ia i2 = 3(x i+1 − x i ) − 2D i − D i+1 (3.5)where D is the derivative.Substituting (3.5) into (3.4) leads towith 1 ≤ i ≤ n − 1.a i3 = 2(x i − x i+1 ) + D i + D i+1D i−1 + 4D i + D i+1 = 3(x i+1 − x i−1 ) (3.6)Since the total number of D i unknowns is n and the total number of equationsis n − 2, two more equations are required to solve the underdetermined system. Ina natural cubic spline, two boundary conditions are given to complete the system ofn − 2 equations. The second derivatives at the end points are set to zero as (3.7).X (2)0 (0) = 0X (2)n−1(0) = 0 (3.7)Otherwise, the first and the last segment can be set to the 2nd order polynomial toreduce two unknowns, for which two boundary conditions are not required.Substituting (3.5) into (3.7) can be described as followings.2D 0 + D 1 = 3(x 1 − x 0 )D n−1 + 2D n = 3(x n − x n−1 ) (3.8)36

⎡⎢⎣2 11 4 11 4 1· · ·1 4 11 2⎤ ⎡⎥⎦ ⎢⎣D 0D 1D 2.D n−1D n⎤⎡=⎥ ⎢⎦ ⎣3(x 1 − x 0 )3(x 2 − x 0 )3(x 3 − x 1 ).3(x n − x n−2 )3(x n − x n−1 )⎤⎥⎦(3.9)D i is obtained in the case of close curves as⎡⎤ ⎡4 11 4 11 4 1⎥⎢⎢⎣· · ·1 4 11 1 4⎥⎦ ⎢⎣D 0D 1D 2.D n−1D n⎤⎡=⎥ ⎢⎦ ⎣3(x 1 − x 0 )3(x 2 − x 0 )3(x 3 − x 1 ).3(x n − x n−2 )3(x n − x n−1 )⎤⎥⎦(3.10)The normalized spline system is⎡⎢⎣b 11 4 11 4 1· · ·1 4 11 1 b⎤ ⎡⎥⎦ ⎢⎣D 0D 1D 2.D n−1D n⎤⎡= k⎥ ⎢⎦ ⎣(x 1 − x 0 )(x 2 − x 0 )(x 3 − x 1 ).(x n − x n−2 )(x n − x n−1 )⎤⎥⎦(3.11)where the value of b and k depends on the boundary conditions of a spline and kdepends on the type of a spline. Normally the value of b and k in a natural cubicare 2 and 3 in an unclosed curve case and 4 and 3 in a closed curve case respectively.The values of b and k in B-Splines are 5 and 6 respectively [13].In case of two parameters, the corresponding relationship between two parameterscan be calculated without an intermediate t parameter. n + 1 point pairs,(x 0 , y 0 ), (x 1 , y 1 ), · · ·, (x n , y n ), have 4n unknown spline parameters in n segments. 2nequations from 0th continuity condition, n − 1 equations from 1st continuity condition,n − 1 equations from 2nd continuity condition, 2 equations from boundaryconditions that the second derivatives at the end points are set to zero.37

⎡⎢⎣2 11 4 11 4 1· · ·1 4 11 2⎤ ⎡⎥⎦ ⎢⎣D 0D 1D 2.D n−1D n⎤⎡=⎥ ⎢⎦ ⎣3(x 1 − x 0 )3(x 2 − x 0 )3(x 3 − x 1 ).3(x n − x n−2 )3(x n − x n−1 )⎤⎥⎦(3.9)D i is obtained in the case of close curves as⎡⎤ ⎡4 11 4 11 4 1⎥⎢⎢⎣· · ·1 4 11 1 4⎥⎦ ⎢⎣D 0D 1D 2.D n−1D n⎤⎡=⎥ ⎢⎦ ⎣3(x 1 − x 0 )3(x 2 − x 0 )3(x 3 − x 1 ).3(x n − x n−2 )3(x n − x n−1 )⎤⎥⎦(3.10)The normalized spline system is⎡⎢⎣b 11 4 11 4 1· · ·1 4 11 1 b⎤ ⎡⎥⎦ ⎢⎣D 0D 1D 2.D n−1D n⎤⎡= k⎥ ⎢⎦ ⎣(x 1 − x 0 )(x 2 − x 0 )(x 3 − x 1 ).(x n − x n−2 )(x n − x n−1 )⎤⎥⎦(3.11)where the value of b and k depends on the boundary conditions of a spline and kdepends on the type of a spline. Normally the value of b and k in a <strong>natural</strong> <strong>cubic</strong>are 2 and 3 in an unclosed curve case and 4 and 3 in a closed curve case respectively.The values of b and k in B-Splines are 5 and 6 respectively [13].In case of two parameters, the corresponding relationship between two parameterscan be calculated <strong>with</strong>out an intermediate t parameter. n + 1 point pairs,(x 0 , y 0 ), (x 1 , y 1 ), · · ·, (x n , y n ), have 4n unknown spline parameters in n segments. 2nequations from 0th continuity condition, n − 1 equations from 1st continuity condition,n − 1 equations from 2nd continuity condition, 2 equations from boundaryconditions that the second derivatives at the end points are set to zero.37

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