User:Jholsenback

Projects

I'm currently working on migrating the current documentation set to this Wiki.

LaTex Markup

These Tex markup segments appear in the reference section. When they are wrapped in the <math></math> tags they ...

Blob Density

% FILE: blobdens
% --------
\begin{displaymath}
  \mathit{density} =
  \mathit{strength}\cdot
  \left(1-\left(\frac{\mathit{distance}}{\mathit{radius}}\right)^2\right)^2
\end{displaymath}

render as:

${\textit {density}}={\textit {strength}}\cdot \left(1-\left({\frac {\min({\textit {distance}},{\textit {radius}})}{\textit {radius}}}\right)^{2}\right)^{2}$

Curve Math

  % FILE: curvmath
% --------
\begin{displaymath}
  \begin{array}{l}
    b = M \cdot x, \mathrm{with:}
    \\ \\
    b = \left[
      \begin{array}{c}
        r(j)^2 \\
        r(j+1)^2 \\
        2 \cdot r(j) \cdot (r(j+1)-r(j-1)) \\
        \hline
        h(j+1)-h(j-1) \\
        2 \cdot r(j+1) \cdot (r(j+2)-r(j)) \\
        \hline
        h(j+2)-h(j)
      \end{array}
    \right]
    \\ \\
    M = \left[
      \begin{array}{c c c c}
        h(j)^3 & h(j)^2 & h(j) & 1 \\
        h(j+1)^3 & h(j+1)^2 & h(j+1) & 1 \\
        3\cdot h(j)^2 & 2\cdot h(j) & 1 & 0 \\
        3\cdot h(j+1)^2 & 2\cdot h(j+1) & 1 & 0
      \end{array}
    \right]
    \\ \\
    x = \left[
      \begin{array}{c}
        A(j)\\ B(j)\\ C(j)\\ D(j)
      \end{array}
    \right]
  \end{array}
\end{displaymath}

render as:

${\begin{array}{l}b=M\cdot x,\mathrm {with:} \\\\b=\left[{\begin{array}{c}r(j)^{2}\\r(j+1)^{2}\\2\cdot r(j)\cdot (r(j+1)-r(j-1))\\\hline h(j+1)-h(j-1)\\2\cdot r(j+1)\cdot (r(j+2)-r(j))\\\hline h(j+2)-h(j)\end{array}}\right]\\\\M=\left[{\begin{array}{c c c c}h(j)^{3}&h(j)^{2}&h(j)&1\\h(j+1)^{3}&h(j+1)^{2}&h(j+1)&1\\3\cdot h(j)^{2}&2\cdot h(j)&1&0\\3\cdot h(j+1)^{2}&2\cdot h(j+1)&1&0\end{array}}\right]\\\\x=\left[{\begin{array}{c}A(j)\\B(j)\\C(j)\\D(j)\end{array}}\right]\end{array}}$

Light Fading

% FILE: lattenua
% --------
\begin{displaymath}
  \mathit{attenuation} =
  \frac{2}
  {1+\left(\frac{d}{\mathit{fade\_distance}}\right)^\mathit{fade\_power}}
\end{displaymath}

render as: ${attenuation}={\frac {2}{1+({\frac {d}{fade\_distance}})^{fade\_power}}}$

Attenuation

% FILE: medatten
% --------
\begin{displaymath}
  \mathit{attenuation} =
  \frac{1}
  {1+\left(\frac{d}{\mathit{fade\_distance}}\right)^\mathit{fade\_power}}
\end{displaymath}

render as: ${attenuation}={\frac {1}{1+({\frac {d}{fade\_distance}})^{fade\_power}}}$

Product

% FILE: prod
% ----
\begin{displaymath}
  \prod_{i=b}^n a
\end{displaymath}

render as: $\prod _{i=b}^{n}a$

Surface of Revolution

% sormath
% -------
\begin{displaymath}
  r^2 = f(h) = A\cdot h^3 + B\cdot h^2 + C\cdot h + D
\end{displaymath}

render as: $r^{2}=f(h)=A\cdot h^{3}+B\cdot h^{2}+C\cdot h+D$

Superquadric Ellipsoid

% FILE: sqemath
% -------
\begin{displaymath}
  f(x,y,z) =
  \left(|x|^{\left(\frac{2}{e}\right)} + |y|^{\left(\frac{2}{e}\right)}
  \right)^{\left(\frac{e}{n}\right)} + |z|^{\left(\frac{2}{n}\right)} - 1 = 0
\end{displaymath}

render as: $f(x,y,z)=(|x|^{({\frac {2}{e}})}+|y|^{({\frac {2}{e}})})^{({\frac {e}{n}})}+|z|^{({\frac {2}{n}})}-1=0$

Sum

% FILE: sum
% ---
\begin{displaymath}
  \sum_{i=b}^n a
\end{displaymath}

render as: $\sum _{i=b}^{n}a$

These Tex segments in the tutorial section. When they are wrapped in the <math></math> tags they ...

Creating the polynomial function

% FILE: polyfunc1
% ---------
\begin{displaymath}
  \sqrt{x^2+y^2+z^2} = r
\end{displaymath}

render as: ${\sqrt {x^{2}+y^{2}+z^{2}}}=r$

% FILE: polyfunc2
% ---------
\begin{displaymath}
  x^2+y^2+z^2-r = 0
\end{displaymath}

render as: $x^{2}+y^{2}+z^{2}-r=0$

% FILE: polyfunc3
% ---------
\begin{displaymath}
  z = \frac{2xy^2}{x^2+y^4}
\end{displaymath}

render as: $z={\frac {2xy^{2}}{x^{2}+y^{4}}}$

% FILE: polyfunc4
% ---------
\begin{displaymath}
  x^2z + y^4z - 2xy^2 = 0
\end{displaymath}

render as: $x^{2}z+y^{4}z-2xy^{2}=0$

% FILE: polyfunc5
% ---------
\begin{displaymath}
  \sqrt{\left(\sqrt{x^2+z^2}-r_1\right)^2+y^2} = r_2
\end{displaymath}

render as: ${\sqrt {({\sqrt {x^{2}+z^{2}}}-r_{1})^{2}+y^{2}}}=r_{2}$

% FILE: polyfunc6
% ---------
\begin{displaymath}
  x^4+2x^2y^2+2x^2z^2-2(r_1^2+r_2^2)x^2+y^4+2y^2z^2+2(r_1^2-r_2^2)y^2+
  z^4-2(r_1^2+r_2^2)z^2+(r_1^2-r_2^2)^2 = 0
\end{displaymath}

render as: $x^{4}+2x^{2}y^{2}+2x^{2}z^{2}-2(r_{1}^{2}+r_{2}^{2})x^{2}+y^{4}+2y^{2}z^{2}+2(r_{1}^{2}-r_{2}^{2})y^{2}+z^{4}-2(r_{1}^{2}+r_{2}^{2})z^{2}+(r_{1}^{2}-r_{2}^{2})^{2}=0$