DirectionCosineMatrix 2005-08-25 19:56:33 2005-08-28 20:21:31 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here A direction cosine matrix (DCM) is a transformation matrix that transforms one coordinate reference frame to another. If we extend the concept of how the three dimensional direction cosines locate a vector, then the DCM locates three unit vectors that describe a coordinate reference frame. Using the notation in equation 1, we need to find the matrix elements that correspond to the correct transformation matrix. \begin{equation} DCM = \left[ \begin{array}{ccc} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array} \right] \end{equation} The first unit vector of the second coordinate frame can be located in the first frame by normal vector notation. See figure 1 for relationship. \begin{center} $ \hat{y}_1 = A_{11} \hat{x}_1 + A_{12} \hat{x}_2 + A_{13} \hat{x}_3 $ \end{center} \medskip \begin{figure} \includegraphics[scale=0.78]{DCM.eps} \end{figure} \medskip Similarily, the other two unit vectors can be described by \begin{center} $$ \hat{y}_2 = A_{21} \hat{x}_1 + A_{22} \hat{x}_2 + A_{23} \hat{x}_3 $$ $$ \hat{y}_3 = A_{31} \hat{x}_1 + A_{32} \hat{x}_2 + A_{33} \hat{x}_3 $$ \end{center} It is easy to see how equation 1 works as a transformation matrix through simple matrix multiplication. \begin{equation} \left[ \begin{array}{c} \hat{y}_1 \\ \hat{y}_2 \\ \hat{y}_3 \end{array} \right] = \left[ \begin{array}{ccc} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array} \right] \left[ \begin{array}{c} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{array} \right] \end{equation} Once this transformation matrix is found, it can be used to transform vectors from the second frame to the first frame and vice versa. Equation 2 transforms the x frame to the y frame and can be denoted as $R{1-2}$. In order to get $R{2-1}$, which transforms the y frame to the x frame, we use a property of transformation matrices of orthonomal reference frames (a frame that is described by unit vectors and are perpindicular to each other). See the entry on a transformation matrix for more info on its properties. We use the properties that $$ R_{1-2}^-1 = R_{1-2}^T = R_{2-1} $$ $$ R_{1-2} R_{1-2}^T = \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] $$ so using these properties and rearranging equation 2 $$ \hat{y} = R_{1-2} \hat{x} $$ yields $$ R_{1-2}^-1 \hat{y} = R_{1-2}^-1 R_{1-2} \hat{x} $$ giving transformation of the y frame to the x frame $$ \hat{x} = R_{2-1} \hat{y} So to extend this concept to transform vectors from one frame to another a closer examination of a vector being represented in both frames is needed. If we denote the second frame as the prime ($\prime$) frame, then a vector expressed in each of these is given by \begin{equation} v = v_1 \hat{x}_1 + v_2 \hat{x}_2 + v_3 \hat{x}_3 \end{equation} \begin{equation} v = v_1\prime \hat{y}_1 + v_2\prime \hat{y}_2 + v_3\prime \hat{y}_3 \end{equation} Since both equations describe the same vector, let us set them equal to each other so $$ v_1 \hat{x}_1 + v_2 \hat{x}_2 + v_3 \hat{x}_3 = v_1\prime \hat{y}_1 + v_2\prime \hat{y}_2 + v_3\prime \hat{y}_3 $$ This notation is clumsy so we want to represent it in matrix notation. This is simple enough if you have an understanding of multiplying a column vector by a row vector. This allows us to describe equations 3 and 4 by $$ v = \left[ \begin{array}{ccc} v_1 & v_2 & v_3 \end{array} \right] \left[ \begin{array}{c} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{array} \right] $$ $$ v = \left[ \begin{array}{ccc} v_1\prime & v_2\prime & v_3\prime \end{array} \right] \left[ \begin{array}{c} \hat{y}_1 \\ \hat{y}_2 \\ \hat{y}_3 \end{array} \right] $$ Setting them equal and substituting equation 2 in for the second coordinate frame yields $$ v = \left[ \begin{array}{ccc} v_1 & v_2 & v_3 \end{array} \right] \left[ \begin{array}{c} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{array} \right] = \left[ \begin{array}{ccc} v_1\prime & v_2\prime & v_3\prime \end{array} \right] \left[ \begin{array}{ccc} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array} \right] \left[ \begin{array}{c} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{array} \right] $$ Then by inspection (or go through the matrix manipulation to cancel the x frame) $$ \left[ \begin{array}{ccc} v_1 & v_2 & v_3 \end{array} \right] = \left[ \begin{array}{ccc} v_1\prime & v_2\prime & v_3\prime \end{array} \right] \left[ \begin{array}{ccc} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array} \right] $$ Representing the transformation matrix as $R_{1-2}$ as the transformation from the first frame to the second frame and transposing the previous equation gives $$ \left[ \begin{array}{ccc} v_1 & v_2 & v_3 \end{array} \right] = (\left[ \begin{array}{ccc} v_1\prime & v_2\prime & v_3\prime \end{array} \right] R_{2-1} )^T $$ Performing the transposition and using a transposition property for two matrices A and B such that $$ (AB)^T = B^TA^T $$ leads to the relationship $$ \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = R_{2-1}^T \left[ \begin{array}{c} v_1\prime \\ v_2\prime \\ v_3\prime \end{array} \right] $$