PlanetPhysics/Direction Cosine Matrix

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.

$$ DCM = \left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{matrix} \right] $$

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.

$$ \hat{y}_1 = A_{11} \hat{x}_1 + A_{12} \hat{x}_2 + A_{13} \hat{x}_3 $$

\medskip \begin{figure} \includegraphics[scale=0.78]{DCM.eps} \end{figure} \medskip

Similarily, the other two unit vectors can be described by

$$ \hat{y}_2 = A_{21} \hat{x}_1 + A_{22} \hat{x}_2 + A_{23} \hat{x}_3 $$ $$ \hat{y}_3 = A_{31} \hat{x}_1 + A_{32} \hat{x}_2 + A_{33} \hat{x}_3 $$

It is easy to see how equation 1 works as a transformation matrix through simple matrix multiplication.

$$ \left[ \begin{matrix} \hat{y}_1 \\ \hat{y}_2 \\ \hat{y}_3 \end{matrix} \right] = \left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{matrix} \right] \left[ \begin{matrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{matrix} \right] $$

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

$$ R_{1-2}^{-1} = R_{1-2}^T = R_{2-1} $$ $$ R_{1-2} R_{1-2}^T = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] $$

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

"$ v = v_1 \hat{x}_1 + v_2 \hat{x}_2 + v_3 \hat{x}_3 $" "$ v = v_1\prime \hat{y}_1 + v_2\prime \hat{y}_2 + v_3\prime \hat{y}_3 $"

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

$$ v = \left[ \begin{matrix} v_1 & v_2 & v_3  \end{matrix} \right] \left[ \begin{matrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3  \end{matrix} \right] $$ $$ v = \left[ \begin{matrix} v_1\prime & v_2\prime & v_3\prime  \end{matrix} \right] \left[ \begin{matrix} \hat{y}_1 \\ \hat{y}_2 \\ \hat{y}_3  \end{matrix} \right] $$

Setting them equal and substituting equation 2 in for the second coordinate frame yields

$$ v = \left[ \begin{matrix} v_1 & v_2 & v_3  \end{matrix} \right] \left[ \begin{matrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3  \end{matrix} \right] = \left[ \begin{matrix} v_1\prime & v_2\prime & v_3\prime  \end{matrix} \right]  \left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{matrix} \right] \left[ \begin{matrix} \hat{x}_1 \\ \hat{x}_2 \\ \hat{x}_3 \end{matrix} \right] $$

Then by inspection (or go through the matrix manipulation to cancel the x frame)

$$ \left[ \begin{matrix} v_1 & v_2 & v_3 \end{matrix} \right] = \left[ \begin{matrix} v_1\prime & v_2\prime & v_3\prime  \end{matrix} \right]  \left[ \begin{matrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{matrix} \right] $$

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{matrix} v_1 & v_2 & v_3 \end{matrix} \right] = (\left[ \begin{matrix} v_1\prime & v_2\prime & v_3\prime \end{matrix} \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{matrix} v_1 \\ v_2 \\ v_3 \end{matrix} \right] = R_{1-2}^T \left[ \begin{matrix} v_1\prime \\ v_2\prime \\ v_3\prime \end{matrix} \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.