By Andreas Nüchter

The monograph written by means of Andreas Nüchter is concentrated on buying spatial versions of actual environments via cellular robots. The robot mapping challenge is usually often called SLAM (simultaneous localization and mapping). 3D maps are essential to stay away from collisions with complicated stumbling blocks and to self-localize in six levels of freedom

(*x*-, *y*-, *z*-position, roll, yaw and pitch angle). New suggestions to the 6D SLAM challenge for 3D laser scans are proposed and a large choice of functions are presented.

**Read or Download 3D Robotic Mapping: The Simultaneous Localization and Mapping Problem with Six Degrees of Freedom PDF**

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**Extra resources for 3D Robotic Mapping: The Simultaneous Localization and Mapping Problem with Six Degrees of Freedom**

**Example text**

The Simultaneous Local. & Map. , STAR 52, pp. 35–75. com © Springer-Verlag Berlin Heidelberg 2009 36 4 3D Range Image Registration Nm Nd ˆ i − (Rdˆj + t) wi,j m E(R, t) = 2 . 1) i=1 j=1 ˆ describes the same point in space as wi,j is assigned 1 if the i-th point of M ˆ Otherwise wi,j is 0. Two things have to be calculated: the j-th point of D. First, the corresponding points, and second, the transformation (R, t) that minimizes E(R, t) on the base of the corresponding points. The ICP algorithm calculates iteratively the point correspondences.

Simpliﬁcation of the Error Function The double summation in Eq. 2) All corresponding points can be represented in a tuple (mi , di ). Note 1. Implementations of the ICP algorithm use Eq. 2). Instead of storing a matrix W for the weights a vector of point correspondences is built. There is no inﬂuence to following computations; the amount of memory is usually reduced from O(n2 ) to O(n). Closed Form Solution Four algorithms are currently known that solve the error function of the ICP algorithm in closed form [76].

STAR 52, pp. 35–75. com © Springer-Verlag Berlin Heidelberg 2009 36 4 3D Range Image Registration Nm Nd ˆ i − (Rdˆj + t) wi,j m E(R, t) = 2 . 1) i=1 j=1 ˆ describes the same point in space as wi,j is assigned 1 if the i-th point of M ˆ Otherwise wi,j is 0. Two things have to be calculated: the j-th point of D. First, the corresponding points, and second, the transformation (R, t) that minimizes E(R, t) on the base of the corresponding points. The ICP algorithm calculates iteratively the point correspondences.