diff --git a/docs/report.tex b/docs/report.tex
index 54d93c8dcfd449fe70b095113e2a1930b12cf4e7..0f816e1c25a7aa22b7199df27a0cd3d8216f9dc2 100644
--- a/docs/report.tex
+++ b/docs/report.tex
@@ -72,6 +72,9 @@
\newcommand{\dder}[2]{\frac{\deriv{#1}}{\deriv{#2}}}
\newcommand{\vv}[1]{\bm{#1}\xspace}
+\newcommand{\unit}[1]{\left[{#1}\right]}
+\newcommand{\txtunit}[1]{$\left[{#1}\right]$}
+
%----------------------------------------------------------------------------------------
% ARTICLE INFORMATION
%----------------------------------------------------------------------------------------
diff --git a/docs/symbols.tex b/docs/symbols.tex
index 19db4a374eb9a3697bcb8971ca2cca9e0b40877e..e12b55820da048ffef27bb790fca4e0b363ba23e 100644
--- a/docs/symbols.tex
+++ b/docs/symbols.tex
@@ -1,3 +1,113 @@
+%-------------------------------------------------------------------------------
+%\subsubsection{Balancing Units}
+%-------------------------------------------------------------------------------
+Ensuring that units are balanced and correct requires care.
+Take the description of the nonlinear system of ODEs that arises from the finite volume discretisation
+\begin{equation}
+ \label{eq:linsys_FV}
+ V_i^{k+1} + \sum_{j\in\mathcal{N}_i} {\frac{\Delta t \alpha_{ij}}{\sigma_i} (V_i^{k+1}-V_j^{k+1})}
+ = V_i^k - \frac{\Delta t}{c_m}(i_m^{k} - i_e).
+\end{equation}
+The choice of units for a parameter, e.g. $\mu m^2$ or $m^2$ for the area $\sigma_{ij}$, introduces a constant of proportionality wherever it is used ($10^{-12}$ in the case of $\mu m^2 \rightarrow m^2$).
+Wherever terms are added in \eq{eq:linsys_FV} the units must be checked, and constants of proportionality balanced.
+
+First, appropriate units for each of the parameters and variables are chosen in~\tbl{tbl:units}.
+We try to use the same units as NEURON, except for the specific membrane capacitance $c_m$, for which $F\cdot m^{-2}$ is used in place of $nF\cdot mm^{-2}$.
+In \eq{eq:linsys_FV} we choose units of $mV \equiv 10^{-3}V$ for each term because of the $V_i$ terms on either side of the equation.
+
+\begin{table}[hp!]
+\begin{tabular}{lllr}
+ \hline
+ term & units & normalized units & NEURON \\
+ \hline
+ $t$ & $ms$ & $10^{-3} \cdot s$ & yes \\
+ $V$ & $mV$ & $10^{-3} \cdot V$ & yes \\
+ $a,~\Delta x$ & $\mu m$ & $10^{-6} \cdot m$ & yes \\
+ $\sigma_{i},~\sigma_{ij}$ & $\mu m^2$ & $10^{-12} \cdot m^2$ & yes \\
+ $c_m$ & $F\cdot m^{-2}$ & $s\cdot A\cdot V^{-1}\cdot m^{-2}$ & no \\
+ $r_L$ & $\Omega\cdot cm$ & $ 10^{-2} \cdot A^{-1}\cdot V\cdot m$ & yes \\
+ $\overline{g}$ & $S\cdot cm^{-2}$ & $10^{4} \cdot A\cdot V^{-1}\cdot m^{-2}$ & yes \\
+ $I_e$ & $nA$ & $10^{-9} \cdot A$ & yes \\
+ \hline
+\end{tabular}
+\caption{The units chosen for parameters and variables in NEST MC. The NEURON column indicates whether the same units have been used as NEURON.}
+\label{tbl:units}
+\end{table}
+
+%------------------------------------------
+\subsubsection{current terms}
+%------------------------------------------
+Membrane current is calculated as follows $i_m = \overline{g}(E-V)$, with units
+\begin{align}
+ \unit{ i_m } &= \unit{ \overline{g} } \unit{ V } \nonumber \\
+ &= 10^{4} \cdot A\cdot V^{-1}\cdot m^{-2} \cdot 10^{-3} \cdot V \nonumber \\
+ &= 10 \cdot A \cdot m^{-2}. \label{eq:im_unit}
+\end{align}
+The injected current $I_e$ has units $nA$, which has to be expressed in terms of current per unit area $i_e=I_e / \sigma_i$ with units
+\begin{align}
+ \unit{ i_e } &= \unit{ I_e } \unit{ \sigma_i }^{-1} \nonumber \\
+ &= 10^{-9}\cdot A \cdot 10^{12} \cdot m^{-2} \nonumber \\
+ &= 10^{3} \cdot A \cdot m ^{-2}, \label{eq:ie_unit}
+\end{align}
+which must be scaled by $10^2$ to match $i_m$ in \eq{eq:im_unit}.
+
+The units for the flux coefficent can be calculated as follows:
+\begin{align}
+ \unit{ \frac{\Delta t}{c_m} } &= 10^{-3} \cdot s \cdot s^{-1}\cdot A^{-1}\cdot V\cdot m^2 \nonumber \\
+ &= 10^{-3} \cdot A^{-1} \cdot V\cdot m^2. \label{eq:dtcm_unit}
+\end{align}
+From \eq{eq:im_unit} and \eq{eq:dtcm_unit} that the units of the full current term are
+\begin{align}
+ \unit{ \frac{\Delta t}{c_m}\left(i_m - i_e\right) }
+ &= 10^{-3} \cdot A^{-1} \cdot V\cdot m^2 \cdot 10 \cdot A \cdot m^{-2} \nonumber \\
+ &= 10^{-2} \cdot V,
+\end{align}
+which must be scaled by $10$ to match the units of $mV\equiv10^{-3}V$.
+%------------------------------------------
+\subsubsection{flux terms}
+%------------------------------------------
+The coefficients in the linear system have the units
+\begin{equation}
+ \unit{ \frac{\Delta t\alpha_{ij}}{\sigma_i} }
+ =
+ \unit{ \frac{\Delta t \sigma_{ij} } {c_m r_L \Delta x_{ij} \sigma_i} }
+ =
+ \unit{ \frac{\Delta t } {c_m r_L \Delta x_{ij} } },
+\end{equation}
+where we we simplify by noting that $\unit{\sigma_{ij}}=\unit{\sigma_i}$.
+The units of the term $c_m r_L$ on the denominator are calculated as follows
+\begin{align}
+ \unit{c_m r_L}
+ &= s \cdot A \cdot V^{-1} \cdot m^{-2} \cdot 10^{-2} \cdot A^{-1} \cdot V \cdot m \nonumber \\
+ &= 10^{-2} \cdot s \cdot m^{-1},
+\end{align}
+so the units of the denominator are
+\begin{align}
+ \unit{c_m r_L \Delta x_{ij}}
+ &= 10^{-2} \cdot s \cdot m^{-1} \cdot 10^{-6} \cdot m \nonumber \\
+ &= 10^{-8} \cdot s,
+\end{align}
+and hence
+\begin{align}
+ \unit{\frac{\Delta t } {c_m r_L \Delta x_{ij} }}
+ &= 10^{8} \cdot s^{-1} \cdot 10^{-3} \cdot s \nonumber \\
+ &= 10^{5}.
+\end{align}
+
+So, the terms with $\alpha_{ij}$ must be scaled by $10^5$ to match the units of $mV$.
+%------------------------------------------
+\subsubsection{discretization with scaling}
+%------------------------------------------
+Here is something that I wish the NEURON documentation had provided:
+\begin{align}
+& V_i^{k+1} + \sum_{j\in\mathcal{N}_i} {10^5 \cdot \frac{\Delta t \alpha_{ij}}{\sigma_i} (V_i^{k+1}-V_j^{k+1})} \nonumber \\
+& = V_i^k - 10\cdot \frac{\Delta t}{c_m}(i_m^{k} - 10^2\cdot I_e/\sigma_i).
+\end{align}
+%------------------------------------------
+\subsection{Supplementary Unit Information}
+%------------------------------------------
+Here is some information about units scraped from Wikipedia for convenience.
+
\begin{table*}[htp!]
\begin{center}
@@ -15,6 +125,8 @@
\hline
\end{tabular}
+ \vspace{20pt}
+
\begin{tabular}{llll}
\hline
symbol & unit & equivalents & SI base \\
@@ -44,27 +156,4 @@
\end{center}
\caption{Symbols and quantities.}
\end{table*}
-%-------------------------------------------------------------------------------
-\subsubsection{Units}
-%-------------------------------------------------------------------------------
-Reffering to the cable equation first defined in~\eq{eq:cable}
-\begin{equation*}
- c_m \pder{V}{t} = \frac{1}{2\pi a r_{L}} \pder{}{x} \left( a^2 \pder{V}{x} \right) - i_m + i_e,
-\end{equation*}
-
-If the units are taken to be
-\begin{itemize}
- \item $c_m = F\cdot cm^{-2}$
- \item $V = V$
- \item $a = cm$
- \item $r_L = \Omega\cdot cm$
-\end{itemize}
-Then the units of each term in equation are $A\cdot cm^{-2}$.
-In practice, the units above are not used, for example distances are usually measured in $\mu m$ and areas in $cm^2$.
-But if we work term-by-term, scaling for these factors is manageable.
-
-A useful identity to use when performing the dimensional analysis relates capacitance and resistance
-\begin{equation*}
- 1~F = 1~\Omega^{-1} \cdot s
-\end{equation*}