|
|
@@ -4,6 +4,8 @@
|
|
|
\begin{document}
|
|
|
\section{Barycentric Coordinates}
|
|
|
|
|
|
+Barycentric coordinates enable you to evenly interpolate between three values among the verticies of a triangle.
|
|
|
+
|
|
|
\def\xa{0} \def\ya{0}
|
|
|
\def\xb{1} \def\yb{2}
|
|
|
\def\xc{3} \def\yc{-1}
|
|
|
@@ -67,8 +69,7 @@ Now we can turn the system of equations into matrix form:
|
|
|
x_p - x_3 \\
|
|
|
y_p - y_3 \\
|
|
|
\end{bmatrix} \\
|
|
|
- & D = 1 \\
|
|
|
- & T \cdot U = R \cdot D
|
|
|
+ & T \cdot U = R
|
|
|
\end{align}
|
|
|
|
|
|
So the solution is
|
|
|
@@ -77,7 +78,7 @@ So the solution is
|
|
|
U = T^{-1} \cdot R
|
|
|
\end{align}
|
|
|
|
|
|
-So the main effort goes towards finding $T^{-1}$
|
|
|
+The main effort goes towards finding $T^{-1}$
|
|
|
|
|
|
\begin{align}
|
|
|
& T^{-1} = \frac{adj(T)}{det(T)} \\
|
|
|
@@ -86,17 +87,17 @@ So the main effort goes towards finding $T^{-1}$
|
|
|
y_2 - y_3 & x_3 - x_2 \\
|
|
|
y_3 - y_1 & x_1 - x_3 \\
|
|
|
\end{bmatrix} \\
|
|
|
- & T^{-1} = \frac{D}{det(T)} \cdot \begin{bmatrix}
|
|
|
+ & T^{-1} = \frac{1}{det(T)} \cdot \begin{bmatrix}
|
|
|
y_2 - y_3 & x_3 - x_2 \\
|
|
|
y_3 - y_1 & x_1 - x_3 \\
|
|
|
\end{bmatrix} \\
|
|
|
- & T^{-1}\cdot R = \frac{D}{det(T)} \cdot \begin{bmatrix}
|
|
|
+ & T^{-1}\cdot R = \frac{1}{det(T)} \cdot \begin{bmatrix}
|
|
|
(y_2 - y_3)(x_p - x_3) + (x_3 - x_2)(y_p - y_3) \\
|
|
|
(y_3 - y_1)(x_p - x_3) + (x_1 - x_3)(y_p - y_3) \\
|
|
|
\end{bmatrix}
|
|
|
\end{align}
|
|
|
|
|
|
-So, the final formula you need to find $(u_1, u_2, u_3)$ given points $v_1, v_2, v_3, p$ is
|
|
|
+And the final formula you need to find $(u_1, u_2, u_3)$ given points $v_1, v_2, v_3, p$~is
|
|
|
|
|
|
\begin{align}
|
|
|
u_1 &= \frac{(y_2 - y_3)(x_p - x_3) + (x_3 - x_2)(y_p - y_3)}{(x_1 - x_3)(y_2 - y_3) - (x_2 - x_3)(y_1 - y_3)} \\
|
rexim