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Defining the Centre of the Volume

In the above formula it is assumed that the volume of integration is centred about $x = 0$. If this is not the case (when the origin of the world coordinates is not at the centre of the brain) then the matrix $M$ can be simply modified to include a general centre.

That is, let $x_c$ be the desired centre, and define a new coordinate

\tilde{x} = x - x_c.
\end{displaymath} (33)

This new coordinate is now zero when $x = x_c$, so that the above derivation is valid for $\tilde{x}$. Furthermore, this translation can be expressed in matrix form (using homogeneous coordinates) as $\tilde{x} = M_c x$ where

\begin{displaymath}M_c = \left[
\begin{array}{cc} I & -x_c \\ 0 \; 0 \; 0 & 1 \end{array}

Therefore, $\Delta x = M \, M_c^{-1} \tilde{x}$, giving $\tilde{M} = M
\, M_c^{-1}$. This implies that $\tilde{A} = A$ and $\tilde{t} = t +
A \, x_c$ so that

E_{RMS}^2 = \frac{1}{5} R^2 \ensuremath{\mathrm{Trace}}(A^\top A) + (t+A \, x_c)^\top (t+A \, x_c).
\end{displaymath} (34)

Mark Jenkinson 2003-02-11