Iterative solution of Helmert transformation based on a unit dual quaternion

Abstract

The rigid motion involving both rotation and translation in the 3D space can be simultaneously described by a unit dual quaternion. Considering this excellent property, the paper constructs the Helmert transformation (seven-parameter similarity transformation) model based on a unit dual quaternion and then presents a rigid iterative algorithm of Helmert transformation using a unit dual quaternion. Because of the singularity of the coefficient matrix of the normal equation, the nine parameter (including one scale factor and eight parameters of a dual quaternion) Helmert transformation model is reduced into five parameter (including one scale factor and four parameters of a unit quaternion which can represent the rotation matrix) Helmert transformation one. Besides, a good start estimate of parameter is required for the iterative algorithm, hence another algorithm employed to compute the initial value of parameter is put forward. The numerical experiments involving a case of small rotation angles i.e. geodetic coordinate transformation and a case of big rotation angles i.e. the registration of LIDAR points are studied. The results show the presented algorithms in this paper are correct and valid for the two cases, disregarding the rotation angles are big or small. And the accuracy of computed parameter is comparable to the classic Procrustes algorithm from Grafarend and Awange (J Geod 77:66–76, 2003), the orthonormal matrix algorithm from Zeng (Earth Planets Space 67:105, 2015), and the algorithm from Wang et al. (J Photogramm Remote Sens 94:63–69, 2014).

Appendices

Appendix 1: The relationship between ${\mathbf{R}}$ and three rotation angles

Assume ${\mathbf{R}}$ is depicted by rotating angles (Eulerian angles) $\theta_{x}$, $\theta_{y}$, $\theta_{z}$ counterclockwise about the X, Y and Z axes respectively, then ${\mathbf{R}}$ is expressed by rotation angles as

$${\mathbf{R}} = \left[ {\begin{array}{*{20}c} {\cos \theta_{z} \cos \theta_{y} } & {\sin \theta_{z} \cos \theta_{x} + \cos \theta_{z} \sin \theta_{y} \sin \theta_{x} } & {\sin \theta_{z} \sin \theta_{x} - \cos \theta_{z} \sin \theta_{y} \cos \theta_{x} } \\ { - \sin \theta_{z} \cos \theta_{y} } & {\cos \theta_{z} \cos \theta_{x} - \sin \theta_{z} \sin \theta_{y} \sin \theta_{x} } & {\cos \theta_{z} \sin \theta_{x} + \sin \theta_{z} \sin \theta_{y} \cos \theta_{x} } \\ {\sin \theta_{y} } & { - \cos \theta_{y} \sin \theta_{x} } & {\cos \theta_{y} \cos \theta_{x} } \\ \end{array} } \right].$$

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Reversely if the rotation matrix ${\mathbf{R}}$ is given, the rotation angles $\theta_{x}$, $\theta_{y}$, $\theta_{z}$ can be computed by Eq. (3) as

$$\theta_{x} = - \tan^{ - 1} \frac{{R_{32} }}{{R_{33} }},\;\theta_{y} = \sin^{ - 1} (R_{31} ),\;\theta_{z} = - \tan^{ - 1} \frac{{R_{21} }}{{R_{11} }} .$$

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where $R_{ij}$ is the element of ${\mathbf{R}}$ in the i-th row and j-th column.

Appendix 2: Partial derivatives of ${\mathbf{W}}(\varvec{r})^{T}$, $Q(\varvec{r})$, $\varvec{r}$ with respect to $r_{1}$, $r_{2}$, $r_{3}$, $r_{4}$ respectively

$$\frac{{\partial {\mathbf{W}}(\varvec{r})^{T} }}{{\partial r_{1} }} = \left[ {\begin{array}{*{20}c} 0 & 0 & 0 & { - 1} \\ 0 & 0 & { - 1} & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} } \right],\quad \frac{{\partial {\mathbf{W}}(\varvec{r})^{T} }}{{\partial r_{2} }} = \left[ {\begin{array}{*{20}c} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & { - 1} \\ { - 1} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} } \right]$$

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$$\frac{{\partial {\mathbf{W}}(\varvec{r})^{T} }}{{\partial r_{3} }} = \left[ {\begin{array}{*{20}c} 0 & { - 1} & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & { - 1} \\ 0 & 0 & 1 & 0 \\ \end{array} } \right],\quad \frac{{\partial {\mathbf{W}}(\varvec{r})^{T} }}{{\partial r_{4} }} = \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right]$$

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$$\frac{{\partial Q(\varvec{r})}}{{\partial r_{1} }} = \left[ {\begin{array}{*{20}c} 0 & 0 & 0 & 1 \\ 0 & 0 & { - 1} & 0 \\ 0 & 1 & 0 & 0 \\ { - 1} & 0 & 0 & 0 \\ \end{array} } \right],\quad \frac{{\partial Q(\varvec{r})}}{{\partial r_{2} }} = \left[ {\begin{array}{*{20}c} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ { - 1} & 0 & 0 & 0 \\ 0 & { - 1} & 0 & 0 \\ \end{array} } \right]$$

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$$\frac{{\partial Q(\varvec{r})}}{{\partial r_{3} }} = \left[ {\begin{array}{*{20}c} 0 & { - 1} & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & { - 1} & 0 \\ \end{array} } \right],\quad \frac{{\partial Q(\varvec{r})}}{{\partial r_{4} }} = \left[ {\begin{array}{*{20}c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} } \right]$$

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$$\frac{{\partial \varvec{r}}}{{\partial r_{1} }} = \left[ {\begin{array}{*{20}c} 1 \\ 0 \\ 0 \\ 0 \\ \end{array} } \right],\quad \frac{{\partial \varvec{r}}}{{\partial r_{2} }} = \left[ {\begin{array}{*{20}c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array} } \right],\quad \frac{{\partial \varvec{r}}}{{\partial r_{3} }} = \left[ {\begin{array}{*{20}c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array} } \right],\quad \frac{{\partial \varvec{r}}}{{\partial r_{4} }} = \left[ {\begin{array}{*{20}c} 0 \\ 0 \\ 0 \\ 1 \\ \end{array} } \right]$$

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