diff --git a/Manuals/FDS_User_Guide/FDS_User_Guide.tex b/Manuals/FDS_User_Guide/FDS_User_Guide.tex
index 73ed691c496..33cb624fe2f 100644
--- a/Manuals/FDS_User_Guide/FDS_User_Guide.tex
+++ b/Manuals/FDS_User_Guide/FDS_User_Guide.tex
@@ -2874,7 +2874,7 @@ \subsection{Solid Phase Gas Transport}
 
 The solid phase conduction/reaction algorithm does not have an explicit transport mechanism for pyrolyzed gases. Rather, the pyrolyzates are assumed to appear instantaneously at the surface. Which surface is controlled by the \ct{SURF} line parameter \ct{LAYER_DIVIDE}, a real number that specifies the number of layers whose gaseous pyrolyzates are to be applied to the surface with the given \ct{SURF} label. For example, if \ct{LAYER_DIVIDE=1.5}, the gases generated by the first layer and half of the second layer shall be applied at the given surface. This same partitioning holds whether or not the solid is shrinking or swelling. Be careful to ensure that the specified values of \ct{LAYER_DIVIDE} are consistent for the \ct{SURF} lines that control opposing surfaces. If two different \ct{SURF} lines govern the front and back of a solid obstruction, the specified values of \ct{LAYER_DIVIDE} should sum to the total number of layers. Otherwise, the mass of evolved gases may not be correct.
 
-If \ct{LAYER_DIVIDE} is not set, it is assumed that for a solid with \ct{BACKING='EXPOSED'}, the gases generated within a depth of half the total thickness are to be applied at the front surface and the other half at the back. If the \ct{BACKING} is not \ct{'EXPOSED'}, all of the gases will be applied at the front surface.  
+If \ct{LAYER_DIVIDE} is not set, it is assumed that for a solid with \ct{BACKING='EXPOSED'}, the gases generated within a depth of half the total thickness are to be applied at the front surface and the other half at the back. If the \ct{BACKING} is not \ct{'EXPOSED'}, all of the gases will be applied at the front surface.
 
 Suppose, for example, that the solid is composed of a layer of insulation on top of a layer of plastic on top of a layer of steel. The steel is impermeable. By setting \ct{LAYER_DIVIDE=2.0} to the \ct{SURF} line whose first layer is the insulation directs the vapors generated by the insulation and plastic to be driven out of the exterior surface of the insulation. Similarly, for the \ct{SURF} line that is applied to the steel, specifying \ct{LAYER_DIVIDE=0.0} would indicate that no fuel vapors are to escape the steel surface. Note that in this instance, the sum of the values of \ct{LAYER_DIVIDE} is not equal to the number of layers, but this is not a problem because the layer of steel does not generate any gases.
 
@@ -7239,9 +7239,9 @@ \subsection{Gas Phase}
    \underbrace{\mathrm{C_{x'}O_{z'}A \; (s)}}_{\rm Char} \; + \; \underbrace{\mathrm{\nu_{\rm O_2} O_2 \; (g)}}_{\rm Oxygen} &\rightarrow& \underbrace{\mathrm{\nu_{\rm CO_2} \, CO_2 \; (g)}}_{\rm Carbon\ Dioxide} \; + \; \underbrace{\mathrm{A \; (s)}}_{\rm Ash} \label{char_chemistry}
 \end{eqnarray}
 \be
-   \nu_{\rm O_2} = \frac{\nu_{\rm O_2,char} \, \nu_{\rm char} \, W_{\rm fuel}}{W_{\rm O_2}}   \quad ; \quad
-   \nu_{\rm CO_2} =  \frac{(1+ \nu_{\rm O_2,char} - \nu_{\rm ash}) \, \nu_{\rm char} \, W_{\rm fuel}}{W_{\rm CO_2}}  \quad ; \quad
-   W_{\rm fuel} = \frac{12 {\rm x} + {\rm y} + 16 {\rm z}}{1- \nu_{\rm ash} \, \nu_{\rm char}}
+   \nu_{\rm O_2} = \frac{\nu_{\rm O_2,char} \, \nu_{\rm char} \, W_{\rm veg}}{W_{\rm O_2}}   \quad ; \quad
+   \nu_{\rm CO_2} =  \frac{(1+ \nu_{\rm O_2,char} - \nu_{\rm ash}) \, \nu_{\rm char} \, W_{\rm veg}}{W_{\rm CO_2}}  \quad ; \quad
+   W_{\rm veg} = \frac{12 {\rm x} + {\rm y} + 16 {\rm z}}{1- \nu_{\rm ash} \, \nu_{\rm char}}
 \ee
 \be
    {\rm x}' = \nu_{\rm CO_2}  \quad ; \quad