Issue#75 sa triangle

.gitignore

@@ -50,6 +50,7 @@ build/
 # Tex files
 *.aux
 *.log
+*.synctex.gz
 *.out
 *.toc
 

Notes_Upgrades/SA_tetra/LM_version.tex

@@ -0,0 +1,216 @@
+\documentclass[10pt,a4paper]{article}
+\usepackage[margin=0.5in]{geometry}
+\usepackage[utf8]{inputenc}
+\usepackage[english]{babel}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+\usepackage{amssymb}
+ \usepackage{cancel}
+\usepackage{graphicx}
+
+\title{Solid angle subtended by a tetrahedron computation}
+\begin{document}
+
+\maketitle
+
+
+\includegraphics[scale=0.4]{tetra.png}
+
+
+\section{Notations}
+
+Let
+
+\begin{itemize}
+	\item $A$ be the vector $\vec{OA}$
+	\item $B$ be the vector $\vec{OB}$
+	\item $C$ be the vector $\vec{OC}$
+	\item $a$ be the vector $\vec{GA}$
+	\item $b$ be the vector $\vec{GB}$
+	\item $c$ be the vector $c$
+	\item $\mathbf{A}$ be the magnitude of the vector $\vec{OA}$
+	\item $\mathbf{B}$ be the magnitude of the vector $\vec{OB}$
+	\item $\mathbf{C}$ be the magnitude of the vector $\vec{OC}$
+\end{itemize}
+
+\section{Computation}
+
+The solid angle $\Omega$ subtended by the triangular surface $ABC$ is:
+
+$$
+{\displaystyle \tan \left({\frac {1}{2}}\Omega \right)
+  = {\frac {\left|{A}\ {B}\ {C}\right|}
+    {\mathbf{A}\mathbf{B}\mathbf{C} + \left({A}\cdot {B}\right)\mathbf{C}
+      + \left({A}\cdot {C}\right)\mathbf{B}
+      + \left({B}\cdot {C}\right)\mathbf{A}}}}
+$$
+
+where $ \left|{A}\ {B}\ {C}\right|={A}\cdot ({B}\times {C}) $
+
+
+\subsection{Numerator}
+
+
+Given that $A = \vec{OA} = G + a$ (and resp.
+with $B$ and $C$), we get
+
+
+\begin{align*}
+\left|{A}\ {B}\ {C}\right|
+	& =  {A} \cdot ({B}\times {C}) \\
+	& =  {A} \cdot ({(G + b)} \times {(G + c)}) \\
+	& =  {(G + a)} \cdot (\cancel{G \times G}
+	                      + G \times c
+	                      + b \times G
+	                      + b \times c) \\
+	& = \cancel{G \cdot (G \times c)}
+	                      + \cancel{G \cdot (b \times G)}\\
+         & + G \cdot (b \times c)
+	  + a \cdot (G \times c)
+	  + a \cdot (b \times G)
+	  + \cancel{a \cdot (b \times c)} \\
+	& = G \cdot (b \times c)
+	+ G \cdot (c \times a)
+	+ G \cdot (a \times b) \\
+	& = G \cdot \left(b \times c
+	+c \times a
+	+ a \times b \right)
+\end{align*}
+
+since $G$ is the centroid of $ABC$, 
+
+\begin{equation}
+a + b + c= 0
+\label{eq1}
+\end{equation}. We obtain
+
+\begin{align}
+\left|{A}\ {B}\ {C}\right|
+& = G \cdot \left(b \times c
+	+c \times a
+	+ a \times b \right) \nonumber \\
+& = G \cdot \left(b \times c
+	+c \times (-b - c)
+	+ (-b - c) \times b \right) \nonumber \\
+& = G \cdot \left(b \times c
+	+c \times (-b) + \cancel{c \times (- c)}
+	+ \cancel{(-b) \times b} + (- c) \times b \right) \nonumber \\
+& = G \cdot \left(3 b \times c \right) \nonumber \\
+& = 3 G \cdot \left( b \times c \right)
+\end{align}
+
+\subsection{Denominator}
+
+
+First let's see the term in $\mathbf{C}$
+\begin{align*}
+ \left({A}\cdot {B}\right)\mathbf{C}
+	=& \left( G + a \right) \cdot \left( G + b \right) \mathbf{C} \\
+	 =& (\mathbf{G}^2 +  G \cdot b	+ a \cdot G + a \cdot b    ) \mathbf{C}
+\end{align*}
+
+Using \eqref{eq1}
+
+\begin{align}
+ \left({A}\cdot {B}\right)\mathbf{C}
+	 =& (\mathbf{G}^2 +  G \cdot b	+ (-b - c) \cdot G + (-b - c) \cdot b    ) \mathbf{C} \nonumber \\
+	 =& (\mathbf{G}^2 - G \cdot c - c \cdot b -\mathbf{b}^2)\mathbf{C}
+\end{align}
+
+
+
+Now, the term in $\mathbf{B}$
+
+
+\begin{align*}
+ \left({A}\cdot {C}\right)\mathbf{B} 
+	=& \left( G + a \right) \cdot \left( G + c \right) \mathbf{B} \\
+	 =& (\mathbf{G}^2 +  G \cdot c	+ a \cdot G + a \cdot c    ) \mathbf{B}
+\end{align*}
+
+Using \eqref{eq1}
+
+\begin{align}
+ \left({A}\cdot {C}\right)\mathbf{B}
+	 =& (\mathbf{G}^2 +  G \cdot c	+ (-b - c) \cdot G + (-b - c) \cdot c    ) \mathbf{B} \nonumber \\
+	 =& (\mathbf{G}^2 - G \cdot b - c \cdot b - \mathbf{c}^2)\mathbf{B}
+\end{align}
+
+
+For the third term there is no simplification
+
+\begin{align}
+ \left({B}\cdot {C}\right)\mathbf{A}
+	=& \left( G + b \right) \cdot \left( G + c \right) \mathbf{A}\\
+	=& (\mathbf{G}^2 + G \cdot b + G \cdot c + c \cdot b)\mathbf{A}
+\end{align}
+
+
+For the computation of the norms, we can use:
+
+\begin{align*}
+\mathbf{A}^2 &= ||OG||^2 + \mathbf{a}^2 + 2 G \cdot a \\
+        &=   (G+a) \cdot (G+a) \\
+        &=   (G-b-c) \cdot (G-b-c) \\
+\end{align*}
+
+and respectively with B and C, we obtain
+
+
+\begin{align}
+(\mathbf{A}\mathbf{B}\mathbf{C}) &= ||G-b-c||^2\,||G+b||^2\,||G+c||^2
+\end{align}
+
+and
+
+
+\begin{align}
+ \left({A}\cdot {B}\right)\mathbf{C}
+  +& \left({A}\cdot {C}\right)\mathbf{B}
+   +  \left({B}\cdot {C}\right)\mathbf{A} \\ \nonumber
+	=& \mathbf{G}^2 (\mathbf{A} + \mathbf{B} + \mathbf{C})
+	+ c \cdot b (\mathbf{A} - \mathbf{B} - \mathbf{C})
+	 + G \cdot b (\mathbf{A} - \mathbf{B})
+	  + G \cdot c (\mathbf{A} - \mathbf{C})
+	  -  ||b||^2\, \mathbf{C} -  ||c||^2\, \mathbf{B}
+\end{align}
+
+
+\section{Final formula}
+
+
+$$
+{\displaystyle \tan \left({\frac {1}{2}}\Omega \right)
+  = {\frac {3 G \cdot \left( b \times c \right)}
+    {\mathbf{A}\mathbf{B}\mathbf{C} + \left({A}\cdot {B}\right)\mathbf{C}
+      + \left({A}\cdot {C}\right)\mathbf{B}
+      + \left({B}\cdot {C}\right)\mathbf{A}}}}
+$$
+
+we get
+
+$$
+{\displaystyle \tan \left({\frac {1}{2}}\Omega \right)
+  = {\frac {3 G \cdot \left( b \times c \right)}
+    {\mathbf{A}\mathbf{B}\mathbf{C} +
+    (\mathbf{G}^2 - G \cdot c - c \cdot b)\mathbf{C}
+      + (\mathbf{G}^2 - G \cdot c - c \cdot b)\mathbf{B}
+      + \left( G + b \right) \cdot \left( G + c \right) \mathbf{A}
+      }}}
+$$
+
+or
+
+$$
+{\displaystyle \tan \left({\frac {1}{2}}\Omega \right)
+  = {\frac {3 G \cdot \left( b \times c \right)}
+    {\mathbf{A}\mathbf{B}\mathbf{C} +
+    \mathbf{G}^2 (\mathbf{A} + \mathbf{B} + \mathbf{C})
+	+ c \cdot b (\mathbf{A} - \mathbf{B} - \mathbf{C})
+	 + G \cdot b (\mathbf{A} - \mathbf{B})
+	  + G \cdot c (\mathbf{A} - \mathbf{C})
+      }}}
+$$
+
+
+\end{document}