The first step of the interpolation leads to the following M1 matrix :
\f[
- M1=\begin{tabular}{|cc|}
+ M1=\left[\begin{tabular}{cc}
0.125 & 0.75 \\
- \end{tabular}
+ \end{tabular}\right]
\f]
\subsection TableNatureOfFieldExampleConservVol Conservative volumic case
If we apply the formula \ref TableNatureOfField "above" it leads to the following \f$ M_{Conservative Volumic} \f$ matrix :
\f[
- M_{Conservative Volumic}=\begin{tabular}{|cc|}
- $\frac{0.125}{0.125+0.75} & $\frac{0.75}{0.125+0.75} \\
- \end{tabular}=\begin{tabular}{|cc|}
+ M_{Conservative Volumic}=\left[\begin{tabular}{cc}
+ $\displaystyle\frac{0.125}{0.125+0.75}$ & $\displaystyle\frac{0.75}{0.125+0.75}$ \\
+ \end{tabular}\right]=\left[\begin{tabular}{cc}
0.14286 & 0.85714 \\
- \end{tabular}
+ \end{tabular}\right]
\f]
\f[
- FT=\begin{tabular}{|cc|}
- $\frac{0.125}{0.875} & $\frac{0.75}{0.875} \\
- \end{tabular}.\begin{tabular}{|c|}
+ FT=\left[\begin{tabular}{cc}
+ $\displaystyle\frac{0.125}{0.875}$ & $\displaystyle\frac{0.75}{0.875}$ \\
+ \end{tabular}\right].\left[\begin{tabular}{c}
4 \\
100 \\
- \end{tabular}
- =\begin{tabular}{|c|}
+ \end{tabular}\right]
+ =\left[\begin{tabular}{c}
86.28571\\
- \end{tabular}
+ \end{tabular}\right]
\f]
As we can see here the maximum principle is respected.This nature of field is particulary recommended to interpolate an intensive
If we apply the formula \ref TableNatureOfField "above" it leads to the following \f$ M_{Integral} \f$ matrix :
\f[
- M_{Integral}=\begin{tabular}{|cc|}
- $\frac{0.125}{9} & $\frac{0.75}{3} \\
- \end{tabular}=\begin{tabular}{|cc|}
+ M_{Integral}=\left[\begin{tabular}{cc}
+ $\displaystyle\frac{0.125}{9}$ & $\displaystyle\frac{0.75}{3}$ \\
+ \end{tabular}\right]=\left[\begin{tabular}{cc}
0.013888 & 0.25 \\
- \end{tabular}
+ \end{tabular}\right]
\f]
\f[
- FT=\begin{tabular}{|cc|}
- $\frac{0.125}{9} & $\frac{0.75}{3} \\
- \end{tabular}.\begin{tabular}{|c|}
+ FT=\left[\begin{tabular}{cc}
+ $\displaystyle\frac{0.125}{9}$ & $\displaystyle\frac{0.75}{3}$ \\
+ \end{tabular}\right].\left[\begin{tabular}{c}
4 \\
100 \\
- \end{tabular}
- =\begin{tabular}{|c|}
+ \end{tabular}\right]
+ =\left[\begin{tabular}{c}
25.055\\
- \end{tabular}
+ \end{tabular}\right]
\f]
This type of interpolation is typically recommended for the interpolation of \b power (\b NOT \b power \b density !) for
This type of interpolation is equivalent to the computation of \f$ FS_{vol} \f$ followed by a multiplication by \f$ M1 \f$ where \f$ FS_{vol} \f$ is given by :
\f[
- FS_{vol}=\begin{tabular}{|c|}
- $\frac{4}{9} \\
- $\frac{100}{3} \\
- \end{tabular}
+ FS_{vol}=\left[\begin{tabular}{c}
+ $\displaystyle\frac{4}{9}$ \\
+ $\displaystyle\frac{100}{3}$ \\
+ \end{tabular}\right]
\f]
In the particular case treated \ref TableNatureOfFieldEx1 "here", it means that only a power of 25.055 W is intercepted by the target cell !
If we apply the formula \ref TableNatureOfField "above" it leads to the following \f$ M_{IntegralGlobConstraint} \f$ matrix :
\f[
- M_{IntegralGlobConstraint}=\begin{tabular}{|cc|}
- $\frac{0.125}{0.125} & $\frac{0.75}{0.75} \\
- \end{tabular}=\begin{tabular}{|cc|}
+ M_{IntegralGlobConstraint}=\left[\begin{tabular}{cc}
+ \displaystyle\frac{0.125}{0.125} & \displaystyle\frac{0.75}{0.75} \\
+ \end{tabular}\right]=\left[\begin{tabular}{cc}
1 & 1 \\
- \end{tabular}
+ \end{tabular}\right]
\f]
\f[
- FT=\begin{tabular}{|cc|}
+ FT=\left[\begin{tabular}{cc}
1 & 1 \\
- \end{tabular}.\begin{tabular}{|c|}
+ \end{tabular}\right].\left[\begin{tabular}{c}
4 \\
100 \\
- \end{tabular}
- =\begin{tabular}{|c|}
+ \end{tabular}\right]
+ =\left[\begin{tabular}{c}
104\\
- \end{tabular}
+ \end{tabular}\right]
\f]
This type of interpolation is typically recommended for the interpolation of \b power (\b NOT \b power \b density !) for
If we apply the formula \ref TableNatureOfField "above" it leads to the following \f$ M_{RevIntegral} \f$ matrix :
\f[
- M_{RevIntegral}=\begin{tabular}{|cc|}
- $\frac{0.125}{1.5} & $\frac{0.75}{1.5} \\
- \end{tabular}=\begin{tabular}{|cc|}
+ M_{RevIntegral}=\left[\begin{tabular}{cc}
+ $\displaystyle\frac{0.125}{1.5}$ & $\displaystyle\frac{0.75}{1.5}$ \\
+ \end{tabular}\right]=\left[\begin{tabular}{cc}
0.083333 & 0.5 \\
- \end{tabular}
+ \end{tabular}\right]
\f]
\f[
- FT=\begin{tabular}{|cc|}
- $\frac{0.125}{1.5} & $\frac{0.75}{1.5} \\
- \end{tabular}.\begin{tabular}{|c|}
+ FT=\left[\begin{tabular}{cc}
+ $\displaystyle\frac{0.125}{1.5}$ & $\displaystyle\frac{0.75}{1.5}$ \\
+ \end{tabular}\right].\left[\begin{tabular}{c}
4 \\
100 \\
- \end{tabular}
- =\begin{tabular}{|c|}
+ \end{tabular}\right]
+ =\left[\begin{tabular}{c}
50.333\\
- \end{tabular}
+ \end{tabular}\right]
\f]
This type of nature is particulary recommended to interpolate an intensive \b density