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+%% $Id: pst-3d-doc.tex 289 2010-02-13 14:35:35Z herbert $
+\documentclass[11pt,english,BCOR10mm,DIV12,bibliography=totoc,parskip=false,smallheadings
+ headexclude,footexclude,oneside,dvipsnames,svgnames]{pst-doc}
+\listfiles
+
+\usepackage[utf8]{inputenc}
+\usepackage{pst-3d}
+\SpecialCoor
+\let\pstFV\fileversion
+\let\belowcaptionskip\abovecaptionskip
+%
+\makeatletter
+\renewcommand*\l@subsection{\bprot@dottedtocline{2}{1.5em}{3.6em}}
+\renewcommand*\l@subsubsection{\bprot@dottedtocline{3}{3.8em}{4.5em}}
+\makeatother
+\def\bgImage{}
+\lstset{explpreset={pos=l,width=-99pt,overhang=0pt,hsep=\columnsep,vsep=\bigskipamount,rframe={}},
+ escapechar=?}
+\def\textat{\char064}%
+\usepackage{shortvrb}
+\MakeShortVerb{|}
+\def\la{<}
+\def\ra{>}
+\def\arc{\mathrm{arc}}
+\def\sign{\mathrm{sign}}
+\def\PiCTeX{\texttt{PiCTeX}}
+\def\endmacro{}
+
+\begin{document}
+\title{\texttt{pst-3d}\\basic three dimension functions \\\small v.\pstFV}
+\docauthor{Herbert Vo\ss}
+\author{Timothy Van Zandt\\Herbert Vo\ss}
+\date{\today}
+
+\maketitle
+
+\begin{abstract}
+This version of \LPack{pst-3d} uses the extended keyval handling
+of \LPack{pst-xkey}.
+
+\vfill
+\noindent
+Thanks to:
+\end{abstract}
+
+\clearpage
+\tableofcontents
+
+\clearpage
+
+\section[PostScript]{PostScript functions \nxLps{SetMatrixThreeD},\nxLps{ProjThreeD}, and \nxLps{SetMatrixEmbed}}
+ \xLps{SetMatrixThreeD}\xLps{ProjThreeD}\xLps{SetMatrixEmbed}
+The \Index{viewpoint} for 3D coordinates is given by three angles: $\alpha$, $\beta$ and
+ $\gamma$. $\alpha$ and $\beta$ determine the direction from which one is
+ looking. $\gamma$ then determines the orientation of the observing.
+
+ When $\alpha$, $\beta$ and $\gamma$ are all zero, the observer is looking
+ from the negative part of the $y$-axis, and sees the $xz$-plane the way in
+ 2D one sees the $xy$ plan. Hence, to convert the 3D coordinates to their 2D
+ project, $\la x, y, z\ra$ map to $\la x, z\ra$.
+
+ When the orientation is different, we rotate the coordinates, and then
+ perform the same projection.
+
+ We move up to latitude $\beta$, over to longitude $\alpha$, and then rotate
+ by $\gamma$. This means that we first rotate around $y$-axis by $\gamma$,
+ then around $x$-axis by $\beta$, and the around $z$-axis by $\alpha$.
+
+ Here are the matrices:
+ \begin{eqnarray*}
+ R_z(\alpha) & = & \left[
+ \begin{array}{ccc}
+ \cos \alpha & -\sin \alpha & 0 \\
+ \sin \alpha & cos \alpha & 0 \\
+ 0 & 0 & 1
+ \end{array} \right] \\
+ R_x(\beta) & = & \left[
+ \begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & \cos \beta & \sin \beta \\
+ 0 & -\sin \beta & \cos \beta
+ \end{array} \right] \\
+ R_y(\gamma) & = & \left[
+ \begin{array}{ccc}
+ \cos \gamma & 0 & -\sin \gamma \\
+ 0 & 1 & 0 \\
+ \sin \gamma & 0 & \cos \gamma
+ \end{array} \right]
+ \end{eqnarray*}
+
+ The rotation of a coordinate is then performed by the matrix $R_z(\alpha)
+ R_x(\beta) R_y(\gamma)$. The first and third columns of the matrix are the
+ basis vectors of the plan upon which the 3D coordinates are project (the old
+ basis vectors were $\la 1, 0, 0\ra$ and $\la 0, 0, 1$\ra; rotating these
+ gives the first and third columns of the matrix).
+
+ These new basis vectors are:
+ \begin{eqnarray*}
+ \tilde{x} & = & \left[
+ \begin{array}{c}
+ \cos\alpha \cos\gamma - \sin\beta \sin\alpha \sin\gamma \\
+ \sin\alpha \cos\gamma + \sin\beta \cos\alpha \sin\gamma \\
+ \cos\beta \sin\gamma
+ \end{array} \right] \\
+ \tilde{z} & = & \left[
+ \begin{array}{c}
+ -\cos\alpha \sin\gamma - \sin\beta \sin\alpha \cos\gamma \\
+ -\sin\alpha \sin\gamma + \sin\beta \cos\alpha \cos\gamma \\
+ \cos\beta \cos\gamma
+ \end{array} \right]
+ \end{eqnarray*}
+
+ Rather than specifying the angles $\alpha$ and $\beta$, the user gives a
+ vector indicating where the viewpoint is. This new viewpoint is the rotation
+ o the old viewpoint. The old viewpoint is $\la 0, -1, 0\ra$, and so the new
+ viewpoint is
+ \[
+ R_z(\alpha) R_x(\beta) \left[ \begin{array}{c} 0\\-1\\0 \end{array} \right]
+ \, = \,
+ \left[ \begin{array}{c}
+ \cos\beta \sin\alpha \\
+ -\cos\beta \cos\alpha \\
+ \sin\beta
+ \end{array} \right]
+ \, = \,
+ \left[ \begin{array}{c} v_1 \\ v_2 \\ v_3 \end{array} \right]
+ \]
+ Therefore,
+ \begin{eqnarray*}
+ \alpha & = & \arc\tan (v_1 / -v_2) \\
+ \beta & = & \arc\tan (v_3 \sin\alpha / v_1)
+ \end{eqnarray*}
+ Unless $p_1=p_2=0$, in which case $\alpha=0$ and $\beta=\sign(p_3)90$, or
+ $p_1=p_3=0$, in which case $\beta=0$.
+
+
+
+The syntax of \Lps{SetMatrixThreeD} is
+ $v_1$ $v_2$ $v_3$ $\gamma$ SetMatrixThreeD
+
+\Lps{SetMatrixThreeD} first computes
+ \[
+ \begin{array}{ll}
+ a=\sin\alpha & b=\cos\alpha\\
+ c=\sin\beta & d=\cos\beta\\
+ e=\sin\gamma & f=\cos\gamma
+ \end{array}
+ \]
+ and then sets \Lps{Matrix3D} to |[|$\tilde{x}$ $\tilde{z}$|]|.
+
+
+\begin{lstlisting}
+/SetMatrixThreeD {
+ dup sin /e ED cos /f ED
+ /p3 ED /p2 ED /p1 ED
+ p1 0 eq
+ { /a 0 def /b p2 0 le { 1 } { -1 } ifelse def
+ p3 p2 abs
+ }
+ { p2 0 eq
+ { /a p1 0 lt { -1 } { 1 } ifelse def /b 0 def
+ p3 p1 abs
+ }
+ { p1 dup mul p2 dup mul add sqrt dup
+ p1 exch div /a ED
+ p2 exch div neg /b ED
+ p3 p1 a div
+ }
+ ifelse
+ }
+ ifelse
+ atan dup sin /c ED cos /d ED
+ /Matrix3D
+ [
+ b f mul c a mul e mul sub
+ a f mul c b mul e mul add
+ d e mul
+ b e mul neg c a mul f mul sub
+ a e mul neg c b mul f mul add
+ d f mul
+ ] def
+} def
+\end{lstlisting}
+
+The syntax of \Lps{ProjThreeD} is $x$ $y$ $z$ \Lps{ProjThreeD} $x'$ $y'$
+where $x'=\la x, y, z\ra \cdot \tilde{x}$ and $y'=\la x, y, z\ra \cdot
+\tilde{z}$.
+
+\begin{lstlisting}
+/ProjThreeD {
+ /z ED /y ED /x ED
+ Matrix3D aload pop
+ z mul exch y mul add exch x mul add
+ 4 1 roll
+ z mul exch y mul add exch x mul add
+ exch
+} def
+\end{lstlisting}
+
+ To embed 2D $\la x, y\ra$ coordinates in 3D, the user specifies the normal
+ vector and an angle. If we decompose this normal vector into an angle, as
+ when converting 3D coordinates to 2D coordinates, and let $\hat\alpha$,
+ $\hat\beta$ and $\hat\gamma$ be the three angles, then when these angles are
+ all zero the coordinate $\la x, y\ra$ gets mapped to $\la x, 0, y\ra$, and
+ otherwise $\la x, y\ra$ gets mapped to
+ \[
+ R_z(\hat\alpha) R_x(\hat\beta) R_y(\hat\gamma)
+ \left[ \begin{array}{c} x \\ 0 \\ y \end{array} \right]
+ \, = \,
+ \left[ \begin{array}{c}
+ \hat{x}_1 x + \hat{z}_1 y\\
+ \hat{x}_2 x + \hat{z}_2 y\\
+ \hat{x}_3 x + \hat{z}_3 y
+ \end{array} \right]
+ \]
+ where $\hat{x}$ and $\hat{z}$ are the first and third columns of
+ $R_z(\hat\alpha) R_x(\hat\beta) R_y(\hat\gamma)$.
+
+ Now add on a 3D-origin:
+ \[
+ \left[ \begin{array}{c}
+ \hat{x}_1 x + \hat{z}_1 y + x_0\\
+ \hat{x}_2 x + \hat{z}_2 y + y_0\\
+ \hat{x}_3 x + \hat{z}_3 y + z_0
+ \end{array} \right]
+ \]
+
+ Now when we project back onto 2D coordinates, we get
+ \begin{eqnarray*}
+ x' & = & \tilde{x}_1(\hat{x}_1 x + \hat{z}_1 y + x_0) +
+ \tilde{x}_2(\hat{x}_2 x + \hat{z}_2 y + y_0) +
+ \tilde{x}_3(\hat{x}_3 x + \hat{z}_3 y + z_0)\\
+ & = &
+ (\tilde{x}_1\hat{x}_1 + \tilde{x}_2\hat{x}_2 + \tilde{x}_3\hat{x}_3) x\\
+ + (\tilde{x}_1\hat{z}_1 + \tilde{x}_2\hat{z}_2 + \tilde{x}_3\hat{z}_3) y\\
+ + \tilde{x}_1 x_0 + \tilde{x}_2 y_0 + \tilde{z}_3 z_0
+ y' & = & \tilde{z}_1(\hat{x}_1 x + \hat{z}_1 y + x_0) +
+ \tilde{z}_2(\hat{x}_2 x + \hat{z}_2 y + y_0) +
+ \tilde{z}_3(\hat{x}_3 x + \hat{z}_3 y + z_0)\\
+ & = &
+ (\tilde{z}_1\hat{x}_1 + \tilde{z}_2\hat{x}_2 + \tilde{z}_3\hat{x}_3) x\\
+ + (\tilde{z}_1\hat{z}_1 + \tilde{z}_2\hat{z}_2 + \tilde{z}_3\hat{z}_3) y\\
+ + \tilde{z}_1 x_0 + \tilde{z}_2 y_0 + \tilde{z}_3 z_0
+ \end{eqnarray*}
+ Hence, the transformation matrix is:
+ \[
+ \left[ \begin{array}{c}
+ \tilde{x}_1\hat{x}_1 + \tilde{x}_2\hat{x}_2 + \tilde{x}_3\hat{x}_3) \\
+ \tilde{z}_1\hat{x}_1 + \tilde{z}_2\hat{x}_2 + \tilde{z}_3\hat{x}_3) \\
+ \tilde{x}_1\hat{z}_1 + \tilde{x}_2\hat{z}_2 + \tilde{x}_3\hat{z}_3) \\
+ \tilde{z}_1\hat{z}_1 + \tilde{z}_2\hat{z}_2 + \tilde{z}_3\hat{z}_3) \\
+ \tilde{x}_1 x_0 + \tilde{x}_2 y_0 + \tilde{z}_3 z_0 \\
+ \tilde{z}_1 x_0 + \tilde{z}_2 y_0 + \tilde{z}_3 z_0
+ \end{array} \right]
+ \]
+
+The syntax of \Lps{SetMatrixEmbed} is
+ $x_0$ $y_0$ $z_0$ $\hat{v_1}$ $\hat{v_2}$ $\hat{v_3}$ $\hat{\gamma}$
+ $v_1$ $v_2$ $v_3$ $\gamma$ \Lps{SetMatrixEmbed}
+
+\Lps{SetMatrixEmbed} first sets |<x1 x2 x3 y1 y2 y3>| to the basis vectors for
+ the viewpoint projection (the tilde stuff above). Then it sets |Matrix3D| to
+ the basis vectors for the embedded plane. Finally, it sets the
+ transformation matrix to the matrix given above.
+
+\begin{lstlisting}
+/SetMatrixEmbed {
+ SetMatrixThreeD
+ Matrix3D aload pop
+ /z3 ED /z2 ED /z1 ED /x3 ED /x2 ED /x1 ED
+ SetMatrixThreeD
+ [
+ Matrix3D aload pop
+ z3 mul exch z2 mul add exch z1 mul add 4 1 roll
+ z3 mul exch z2 mul add exch z1 mul add
+ Matrix3D aload pop
+ x3 mul exch x2 mul add exch x1 mul add 4 1 roll
+ x3 mul exch x2 mul add exch x1 mul add
+ 3 -1 roll 3 -1 roll 4 -1 roll 8 -3 roll 3 copy
+ x3 mul exch x2 mul add exch x1 mul add 4 1 roll
+ z3 mul exch z2 mul add exch z1 mul add
+ ]
+ concat
+} def
+\end{lstlisting}
+
+
+\section{Keywords}
+\subsection{\nxLkeyword{viewpoint}}
+
+\begin{lstlisting}
+\let\pssetzlength\pssetylength
+\define@key[psset]{pst-3d}{viewpoint}{%
+ \pst@expandafter\psset@@viewpoint#1 {} {} {} \@nil
+ \let\psk@viewpoint\pst@tempg}
+\def\psset@@viewpoint#1 #2 #3 #4\@nil{%
+ \begingroup
+ \pssetxlength\pst@dima{#1}%
+ \pssetylength\pst@dimb{#2}%
+ \pssetzlength\pst@dimc{#3}%
+ \xdef\pst@tempg{%
+ \pst@number\pst@dima \pst@number\pst@dimb \pst@number\pst@dimc}%
+ \endgroup}
+\psset[pst-3d]{viewpoint=1 -1 1}
+\end{lstlisting}
+
+\subsection{\nxLkeyword{viewangle}}
+
+\begin{lstlisting}
+\define@key[psset]{pst-3d}{viewangle}{\pst@getangle{#1}\psk@viewangle}
+\psset[pst-3d]{viewangle=0}
+\end{lstlisting}
+
+\subsection{\nxLkeyword{normal}}
+
+\begin{lstlisting}
+\define@key[psset]{pst-3d}{normal}{%
+ \pst@expandafter\psset@@viewpoint#1 {} {} {} \@nil
+ \let\psk@normal\pst@tempg}
+\psset[pst-3d]{normal=0 0 1}
+\end{lstlisting}
+
+
+\subsection{\nxLkeyword{embedangle}}
+\begin{lstlisting}
+\define@key[psset]{pst-3d}{embedangle}{\pst@getangle{#1}\psk@embedangle}
+\psset[pst-3d]{embedangle=0}
+\end{lstlisting}
+
+
+\section{Transformation matrix}
+
+\begin{lstlisting}
+/TMSave {
+ tx@Dict /TMatrix known not { /TMatrix { } def /RAngle { 0 } def } if end
+ /TMatrix [ TMatrix CM ] cvx def
+} def
+/TMRestore { CP /TMatrix [ TMatrix setmatrix ] cvx def moveto } def
+/TMChange {
+ TMSave
+ /cp [ currentpoint ] cvx def % ??? Check this later.
+ CM
+} def
+\end{lstlisting}
+ Set standard coor. system , with |pt| units and origin at \Index{currentpoint}.
+ This let's us rotate, or whatever, around \TeX's current point, without
+ having to worry about strange coordinate systems that the dvi-to-ps
+ driver might be using.
+\begin{lstlisting}
+CP T STV
+\end{lstlisting}
+
+ Let M = old matrix (on stack), and M' equal current matrix. Then
+ go from M' to M by applying M Inv(M').
+\begin{lstlisting}
+CM matrix invertmatrix % Inv(M')
+matrix concatmatrix % M Inv(M')
+\end{lstlisting}
+ Now modify transformation matrix:
+\begin{lstlisting}
+exch exec
+\end{lstlisting}
+Now apply M Inv(M')
+\begin{lstlisting}
+concat cp moveto
+\end{lstlisting}
+
+
+\section{Macros}
+\subsection{\nxLcs{ThreeDput}}
+
+\begin{lstlisting}
+\def\ThreeDput{\pst@object{ThreeDput}}
+\def\ThreeDput@i{\@ifnextchar({\ThreeDput@ii}{\ThreeDput@ii(\z@,\z@,\z@)}}
+\def\ThreeDput@ii(#1,#2,#3){%
+ \pst@killglue\pst@makebox{\ThreeDput@iii(#1,#2,#3)}}
+\def\ThreeDput@iii(#1,#2,#3){%
+ \begingroup
+ \use@par
+ \if@star\pst@starbox\fi
+ \pst@makesmall\pst@hbox
+ \pssetxlength\pst@dima{#1}%
+ \pssetylength\pst@dimb{#2}%
+ \pssetzlength\pst@dimc{#3}%
+ \leavevmode
+ \hbox{%
+ \pst@Verb{%
+ { \pst@number\pst@dima
+ \pst@number\pst@dimb
+ \pst@number\pst@dimc
+ \psk@normal
+ \psk@embedangle
+ \psk@viewpoint
+ \psk@viewangle
+ \tx@SetMatrixEmbed
+ } \tx@TMChange}%
+ \box\pst@hbox
+ \pst@Verb{\tx@TMRestore}}%
+ \endgroup
+ \ignorespaces}
+\end{lstlisting}
+
+\section{Arithmetic}\label{Arithmetic}
+
+ {\verb+\pst@divide+}
+ This is adapted from Donald Arseneau's |shapepar.sty|.
+ Syntax:
+ \begin{verbatim}
+ \pst@divide{<numerator>}{<denominator>}{<command>}
+ \pst@@divide{<numerator>}{<denominator>}
+ \end{verbatim}
+ <numerator> and <denominator> should be dimensions. |\pst@divide| sets
+ <command> to <num>/<den> (in points). |\pst@@divide| sets |\pst@dimg| to
+ <num>/<den>.
+ \begin{lstlisting}
+ \def\pst@divide#1#2#3{%
+ \pst@@divide{#1}{#2}%
+ \pst@dimtonum\pst@dimg{#3}}
+ \def\pst@@divide#1#2{%
+ \pst@dimg=#1\relax
+ \pst@dimh=#2\relax
+ \pst@cntg=\pst@dimh
+ \pst@cnth=67108863
+ \pst@@@divide\pst@@@divide\pst@@@divide\pst@@@divide
+ \divide\pst@dimg\pst@cntg}
+ \end{lstlisting}
+ The number 16 is the level of uncertainty. Use a lower power of 2 for more
+ accuracy (2 is most precise). But if you change it, you must change the
+ repetions of |\pst@@@divide| in |\pst@@divide| above:
+ \[
+ \mbox{precision}^{\mbox{repetitions}} = 65536
+ \]
+ (E.g., $16^4 = 65536$).
+\begin{lstlisting}
+ \def\pst@@@divide{%
+ \ifnum
+ \ifnum\pst@dimg<\z@-\fi\pst@dimg<\pst@cnth
+ \multiply\pst@dimg\sixt@@n
+ \else
+ \divide\pst@cntg\sixt@@n
+ \fi}
+\end{lstlisting}
+
+ {\verb+\pst@pyth+}
+ Syntax:
+ \begin{verbatim}
+ \pst@pyth{<dim1>}{<dim2>}{<dimen register>}
+ \end{verbatim}
+ <dimen register> is set to $((dim1)^2+(dim2)^2)^{1/2}$.
+
+ The algorithm is copied from \PiCTeX, by Michael Wichura (with permission).
+ Here is his description:
+ \begin{quote}
+ Suppose $x>0$, $y>0$. Put $s = x+y$. Let $z = (x^2+y^2)^{1/2}$. Then $z =
+ s\times f$, where
+ \[
+ f = (t^2 + (1-t)^2)^{1/2} = ((1+\tau^2)/2)^{1/2}
+ \]
+ and $t = x/s$ and $\tau = 2(t-1/2)$.
+ \end{quote}
+ \begin{lstlisting}
+\def\pst@pyth#1#2#3{%
+ \begingroup
+ \pst@dima=#1\relax
+ \ifnum\pst@dima<\z@\pst@dima=-\pst@dima\fi % dima=abs(x)
+ \pst@dimb=#2\relax
+ \ifnum\pst@dimb<\z@\pst@dimb=-\pst@dimb\fi % dimb=abs(y)
+ \advance\pst@dimb\pst@dima % dimb=s=abs(x)+abs(y)
+ \ifnum\pst@dimb=\z@
+ \global\pst@dimg=\z@ % dimg=z=sqrt(x^2+y^2)
+ \else
+ \multiply\pst@dima 8\relax % dima= 8abs(x)
+ \pst@@divide\pst@dima\pst@dimb % dimg =8t=8abs(x)/s
+ \advance\pst@dimg -4pt % dimg = 4tau = (8t-4)
+ \multiply\pst@dimg 2
+ \pst@dimtonum\pst@dimg\pst@tempa
+ \pst@dima=\pst@tempa\pst@dimg % dima=(8tau)^2
+ \advance\pst@dima 64pt % dima=u=[64+(8tau)^2]/2
+ \divide\pst@dima 2\relax % =(8f)^2
+ \pst@dimd=7pt % initial guess at sqrt(u)
+ \pst@@pyth\pst@@pyth\pst@@pyth % dimd=sqrt(u)
+ \pst@dimtonum\pst@dimd\pst@tempa
+ \pst@dimg=\pst@tempa\pst@dimb
+ \global\divide\pst@dimg 8 % dimg=z=(8f)*s/8
+ \fi
+ \endgroup
+ #3=\pst@dimg}
+\def\pst@@pyth{% dimd = g <-- (g + u/g)/2
+ \pst@@divide\pst@dima\pst@dimd
+ \advance\pst@dimd\pst@dimg
+ \divide\pst@dimd 2\relax}
+ \end{lstlisting}
+
+
+ {\verb+\pst@sinandcos+}
+ Syntax:
+ \begin{verbatim}
+ \pst@sinandcos{<dim>}{<int>}
+ \end{verbatim}
+ <dim>, in |sp| units, should equal 100,000 times the angle, in degrees
+ between 0 and 90. <int> should equal the angle's quadrant (0, 1, 2 or 3).
+ |\pst@dimg| is set to $\sin(\theta)$ and |\pst@dimh| is set to
+ $\cos(\theta)$ (in pt's).
+
+ The algorithms uses the usual McLaurin expansion.
+ \begin{lstlisting}
+\def\pst@sinandcos#1{%
+ \begingroup
+ \pst@dima=#1\relax
+ \pst@dima=.366022\pst@dima %Now 1pt=1/32rad
+ \pst@dimb=\pst@dima % dimb->32sin(angle) in pts
+ \pst@dimc=32\p@ % dimc->32cos(angle) in pts
+ \pst@dimtonum\pst@dima\pst@tempa
+ \pst@cntb=\tw@
+ \pst@cntc=-\@ne
+ \pst@cntg=32
+ \loop
+ \ifnum\pst@dima>\@cclvi % 256
+ \pst@dima=\pst@tempa\pst@dima
+ \divide\pst@dima\pst@cntg
+ \divide\pst@dima\pst@cntb
+ \ifodd\pst@cntb
+ \advance\pst@dimb \pst@cntc\pst@dima
+ \pst@cntc=-\pst@cntc
+ \else
+ \advance\pst@dimc by \pst@cntc\pst@dima
+ \fi
+ \advance\pst@cntb\@ne
+ \repeat
+ \divide\pst@dimb\pst@cntg
+ \divide\pst@dimc\pst@cntg
+ \global\pst@dimg\pst@dimb
+ \global\pst@dimh\pst@dimc
+ \endgroup}
+ \end{lstlisting}
+
+
+ {\verb+\pst@getsinandcos+}
+ |\pst@getsinandcos| normalizes the angle to be in the first quadrant, sets
+ |\pst@quadrant| to 0 for the first quadrant, 1 for the second, 2 for the
+ third, and 3 for the fourth, invokes |\pst@sinandcos|, and sets |\pst@sin|
+ to the sine and |\pst@cos| to the cosine.
+ \begin{lstlisting}
+\def\pst@getsinandcos#1{%
+ \pst@dimg=100000sp
+ \pst@dimg=#1\pst@dimg
+ \pst@dimh=36000000sp
+ \pst@cntg=0
+ \loop
+ \ifnum\pst@dimg<\z@
+ \advance\pst@dimg\pst@dimh
+ \repeat
+ \loop
+ \ifnum\pst@dimg>\pst@dimh
+ \advance\pst@dimg-\pst@dimh
+ \repeat
+ \pst@dimh=9000000sp
+ \def\pst@tempg{%
+ \ifnum\pst@dimg<\pst@dimh\else
+ \advance\pst@dimg-\pst@dimh
+ \advance\pst@cntg\@ne
+ \ifnum\pst@cntg>\thr@@ \advance\pst@cntg-4 \fi
+ \expandafter\pst@tempg
+ \fi}%
+ \pst@tempg
+ \chardef\pst@quadrant\pst@cntg
+ \ifdim\pst@dimg=\z@
+ \def\pst@sin{0}%
+ \def\pst@cos{1}%
+ \else
+ \pst@sinandcos\pst@dimg
+ \pst@dimtonum\pst@dimg\pst@sin
+ \pst@dimtonum\pst@dimh\pst@cos
+ \fi}
+ \end{lstlisting}
+
+
+ \section{Tilting}
+
+ {\verb+\pstilt+}
+ \begin{lstlisting}
+\def\pstilt#1{\pst@makebox{\pstilt@{#1}}}
+\def\pstilt@#1{%
+ \begingroup
+ \leavevmode
+ \pst@getsinandcos{#1}%
+ \hbox{%
+ \ifcase\pst@quadrant
+ \kern\pst@cos\dp\pst@hbox
+ \pst@dima=\pst@cos\ht\pst@hbox
+ \ht\pst@hbox=\pst@sin\ht\pst@hbox
+ \dp\pst@hbox=\pst@sin\dp\pst@hbox
+ \or
+ \kern\pst@sin\ht\pst@hbox
+ \pst@dima=\pst@sin\dp\pst@hbox
+ \ht\pst@hbox=\pst@cos\ht\pst@hbox
+ \dp\pst@hbox=\pst@cos\dp\pst@hbox
+ \or
+ \kern\pst@cos\ht\pst@hbox
+ \pst@dima=\pst@sin\dp\pst@hbox
+ \pst@dimg=\pst@sin\ht\pst@hbox
+ \ht\pst@hbox=\pst@sin\dp\pst@hbox
+ \dp\pst@hbox=\pst@dimg
+ \or
+ \kern\pst@sin\dp\pst@hbox
+ \pst@dima=\pst@sin\ht\pst@hbox
+ \pst@dimg=\pst@cos\ht\pst@hbox
+ \ht\pst@hbox=\pst@cos\dp\pst@hbox
+ \dp\pst@hbox=\pst@dimg
+ \fi
+ \pst@Verb{%
+ { [ 1 0
+ \pst@cos\space \ifnum\pst@quadrant>\@ne neg \fi
+ \pst@sin\space
+ \ifnum\pst@quadrant>\z@\ifnum\pst@quadrant<\thr@@ neg \fi\fi
+ \ifodd\pst@quadrant exch \fi
+ 0 0
+ ] concat
+ } \tx@TMChange}%
+ \box\pst@hbox
+ \pst@Verb{\tx@TMRestore}%
+ \kern\pst@dima}%
+ \endgroup}
+ \end{lstlisting}
+
+
+ {\verb+\psTilt+}
+ \begin{lstlisting}
+\def\psTilt#1{\pst@makebox{\psTilt@{#1}}}
+\def\psTilt@#1{%
+ \begingroup
+ \leavevmode
+ \pst@getsinandcos{#1}%
+ \hbox{%
+ \ifodd\pst@quadrant
+ \pst@@divide{\dp\pst@hbox}{\pst@cos\p@}%
+ \ifnum\pst@quadrant=\thr@@\kern\else\pst@dima=\fi\pst@sin\pst@dimg
+ \pst@@divide{\ht\pst@hbox}{\pst@cos\p@}%
+ \ifnum\pst@quadrant=\@ne\kern\else\pst@dima=\fi\pst@sin\pst@dimg
+ \else
+ \ifdim\pst@sin\p@=\z@
+ \@pstrickserr{\string\psTilt\space angle cannot be 0 or 180}\@ehpa
+ \def\pst@sin{.7071}%
+ \def\pst@cos{.7071}%
+ \fi
+ \pst@@divide{\dp\pst@hbox}{\pst@sin\p@}%
+ \ifnum\pst@quadrant=\z@\kern\else\pst@dima=\fi\pst@cos\pst@dimg
+ \pst@@divide{\ht\pst@hbox}{\pst@sin\p@}%
+ \ifnum\pst@quadrant=\tw@\kern\else\pst@dima=\fi\pst@cos\pst@dimg
+ \fi
+ \ifnum\pst@quadrant>\@ne
+ \pst@dimg=\ht\pst@hbox
+ \ht\pst@hbox=\dp\pst@hbox
+ \dp\pst@hbox=\pst@dimg
+ \fi
+ \pst@Verb{%
+ { [ 1 0
+ \pst@cos\space \pst@sin\space
+ \ifodd\pst@quadrant exch \fi
+ \tx@Div
+ \ifnum\pst@quadrant>\z@\ifnum\pst@quadrant<\thr@@ neg \fi\fi
+ \ifnum\pst@quadrant>\@ne -1 \else 1 \fi
+ 0 0
+ ] concat
+ } \tx@TMChange}%
+ \box\pst@hbox
+ \pst@Verb{\tx@TMRestore}%
+ \kern\pst@dima}%
+ \endgroup}
+ \end{lstlisting}
+
+
+ {\verb+\psset@Tshadowsize,\psTshadowsize+}
+\begin{lstlisting}
+\define@key[psset]{pst-3d}{Tshadowsize}{%
+ \pst@checknum{#1}\psTshadowsize}
+\psset[pst-3d]{Tshadowsize=1}
+\end{lstlisting}
+
+
+{\verb+\psset@Tshadowangle,\psk@Tshadowangle+}
+\begin{lstlisting}
+\define@key[psset]{pst-3d}{Tshadowangle}{%
+ \pst@getangle{#1}\psk@Tshadowangle}
+\psset[pst-3d]{Tshadowangle=60}
+\end{lstlisting}
+
+
+ {\verb+\psset@Tshadowcolor,\psTshadowcolor+}
+\begin{lstlisting}
+\define@key[psset]{pst-3d}{Tshadowcolor}{%
+ \pst@getcolor{#1}\psTshadowcolor}
+\psset[pst-3d]{Tshadowcolor=lightgray}
+\end{lstlisting}
+
+
+ {\verb+\psshadow+}
+\begin{lstlisting}
+\def\psshadow{\def\pst@par{}\pst@object{psshadow}}
+\def\psshadow@i{\pst@makebox{\psshadow@ii}}
+\def\psshadow@ii{%
+ \begingroup
+ \use@par
+ \leavevmode
+ \pst@getsinandcos{\psk@Tshadowangle}%
+ \hbox{%
+ \lower\dp\pst@hbox\hbox{%
+ \pst@Verb{%
+ { [ 1 0
+ \pst@cos\space \psTshadowsize mul
+ \ifnum\pst@quadrant>\@ne neg \fi
+ \pst@sin\space \psTshadowsize mul
+ \ifnum\pst@quadrant>\z@\ifnum\pst@quadrant<\thr@@ neg \fi\fi
+ \ifodd\pst@quadrant exch \fi
+ 0 0
+ ] concat
+ } \tx@TMChange}}%
+ \hbox to\z@{{\@nameuse{\psTshadowcolor}\copy\pst@hbox\hss}}%
+ \pst@Verb{\tx@TMRestore}%
+ \box\pst@hbox}%
+ \endgroup}
+ \end{lstlisting}
+
+\section{Affin Transformations}
+
+\begin{BDef}
+\Lcs{psAffinTransform}\OptArgs\Largb{transformation matrix}\Largb{object}
+\end{BDef}
+
+\begin{LTXexample}[width=3cm]
+\pspicture(3,6)\psset{linewidth=4pt,arrows=->}
+\psline(0,0)(1.5,0)(3,3)\rput*(2.25,1.5){foo}
+\psAffinTransform{0.5 0 0 2 0 0}{\color{red}%
+ \psline[linecolor=red](0,0)(1.5,0)(3,3)\rput*(2.25,1.5){foo}}%
+\endpspicture
+\end{LTXexample}
+
+The transformation matrix must be a list of 6 values divided by a space.
+For a translation modify the last two values of $1 0 0 1 dx dy$. The values for
+$dx$ and $dy$ must be of the unit pt! For a rotation
+we have the transformation matrix
+
+\begin{align}
+\left[\begin{aligned} \cos(\alpha) & \sin(\alpha) & 0 \\
+-\sin(\alpha) & \cos(\alpha) & 0 \\
+0 & 0 & 1\end{aligned}\right]
+\end{align}
+
+For \Lcs{psAffinTransform} the four values have to be modifies \texttt{a cos a sin a sin neg a cos 0 0}.
+Tilting can be done with $sx 0 0 sy 0 0$. All effects can be combined.
+
+\begin{LTXexample}[width=3cm]
+\pspicture(3,6)\psset{linewidth=4pt,arrows=->}
+\psline(0,0)(1.5,0)(3,3)\rput*(2.25,1.5){foo}
+\psAffinTransform{0.5 0.8 0.3 2 20 -20}{\color{red}%
+ \psline[linecolor=red](0,0)(1.5,0)(3,3)\rput*(2.25,1.5){foo}}%
+\endpspicture
+\end{LTXexample}
+
+
+\clearpage
+\section{List of all optional arguments for \texttt{pst-3d}}
+
+\xkvview{family=pst-3d,columns={key,type,default}}
+
+
+
+
+
+\nocite{*}
+\bgroup
+\RaggedRight
+\bibliographystyle{plain}
+\bibliography{pst-3d-doc}
+\egroup
+
+\printindex
+
+
+
+
+
+\end{document}