Talk:PlanetPhysics/Direction Cosine Matrix

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: direction cosine matrix %%% Primary Category Code: 45.40.-f %%% Filename: DirectionCosineMatrix.tex %%% Version: 15 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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A direction cosine matrix (DCM) is a transformation \htmladdnormallink{matrix}{http://planetphysics.us/encyclopedia/Matrix.html} that transforms one coordinate \htmladdnormallink{reference frame}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} to another. If we extend the \htmladdnormallink{concept}{http://planetphysics.us/encyclopedia/PreciseIdea.html} of how the three dimensional \htmladdnormallink{direction cosines}{http://planetphysics.us/encyclopedia/DirectionCosines.html} locate a \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html}, then the DCM locates three \htmladdnormallink{unit vectors}{http://planetphysics.us/encyclopedia/PureState.html} 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 \htmladdnormallink{works}{http://planetphysics.us/encyclopedia/Work.html} as a transformation matrix through simple \htmladdnormallink{matrix multiplication}{http://planetphysics.us/encyclopedia/Matrix.html}.

\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 orthonormal 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 the 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

$$ \left[ \begin{array}{ccc} v_1 & v_2 & v_3 \end{array} \right] = (\left[ \begin{array}{ccc} v_1\prime & v_2\prime & v_3\prime  \end{array} \right] R_{1-2} )^T $$

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

$$ \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right] = R_{1-2}^T \left[ \begin{array}{c} v_1\prime \\ v_2\prime \\ v_3\prime  \end{array} \right] $$

Finally giving us the ability to transform a vector from the second (prime) frame to the first frame.

$$ \vec{v} = R_{2-1} \vec{v \prime} $$

Much much more can be found in the general entry about the Transformation matrix.

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