\section{The prism} A prism is determined by two parameters: \begin{itemize} \item The base of the prism can be defined by the coordinates of the vertices in the $xy$-plane. Note that it is necessary that the four vertices be given in counterclockwise order with respect to the barycenter of the base; \item the direction of the prism axis (the components of the shearing vector). \end{itemize} \subsubsection{Example 1: right and oblique prisms with polygonal section} \begin{center} \psset{unit=0.5} \psset{lightsrc=10 5 50,viewpoint=50 20 30 rtp2xyz,,Decran=50} \begin{minipage}{5cm} \begin{pspicture*}(-4,-4)(6,9) \psframe(-4,-4)(6,9) \psSolid[object=grille,base=-4 4 -4 4,action=draw]% \psSolid[object=prisme,h=6,base=0 1 -1 0 0 -2 1 -1 0 0]% \axesIIID(4,4,6)(4.5,4.5,8) \end{pspicture*} \small\texttt{[base=\psframebox[fillstyle=solid,fillcolor=black]{\textcolor{white}{0 1 -1 0 0 -2 1 -1 0 0}},h=6]} \\ \end{minipage} \hspace{2cm} \begin{minipage}{5cm} \begin{pspicture*}(-4,-4)(6,9) \psframe(-4,-4)(6,9) \psSolid[object=grille,base=-4 4 -4 4,action=draw]% \psSolid[object=prisme,axe=0 1 2,h=8,base=0 -2 1 -1 0 0 0 1 -1 0]% \axesIIID(4,4,4)(4.5,4.5,8) \psPoint(0,4,8){V} \psPoint(0,4,0){Vy} \psPoint(0,0,8){Vz} \uput[l](Vz){8} \uput[ur](Vy){4} \psline[linecolor=blue]{->}(O)(V) \psline[linestyle=dashed](Vz)(V)(Vy) \end{pspicture*} \small\texttt{[base=\psframebox[fillstyle=solid,fillcolor=black]{\textcolor{white}{0 -2 1 -1 0 0 0 1 -1 0}},}% \\ \texttt{ axe=\psframebox[fillstyle=solid,fillcolor=black]{\textcolor{white}{0 4 8}},h=8]} \end{minipage} \end{center} \subsubsection{Example 2: a right prism with cross-section a rounded square} \psset{lightsrc=10 -20 50,viewpoint=50 -20 30 rtp2xyz,,Decran=50} \begin{LTXexample}[width=6.5cm] \psset{unit=0.5} \begin{pspicture}(-5,-4)(3,9) \psSolid[object=grille,base=-4 4 -4 4,action=draw]% \psSolid[object=prisme,h=6,fillcolor=yellow,% base=% 0 10 90 {/i exch def i cos 1 add i sin 1 add } for % 90 10 180 {/i exch def i cos 1 sub i sin 1 add} for % 180 10 270 {/i exch def i cos 1 sub i sin 1 sub} for % 270 10 360 {/i exch def i cos 1 add i sin 1 sub} for ]% \axesIIID(4,4,6)(6,6,8) \end{pspicture} \end{LTXexample} \subsubsection{Example 3: a right prism with a star-shaped section]} \begin{LTXexample}[width=6.5cm] \psset{unit=0.5} \psset{lightsrc=10 -20 50,viewpoint=50 -20 30 rtp2xyz,,Decran=50} \begin{pspicture*}(-5,-4)(6,9) \defFunction{F}(t){3 t cos 3 exp mul}{3 t sin 3 exp mul}{} \psSolid[object=grille,base=-4 4 -4 4,action=draw]% \psSolid[object=prismecreux,h=8,fillcolor=red!50, resolution=36, base=0 350 {F} CourbeR2+ ]% \end{pspicture*} \end{LTXexample} \subsubsection{Example 4: a prism with an elliptic section} \begin{LTXexample}[width=6.5cm] \psset{unit=0.5} \begin{pspicture}(-6,-3)(4,10) \psSolid[object=grille,base=-6 6 -4 4,action=draw] \psset{lightsrc=10 20 30,viewpoint=50 20 25 rtp2xyz,Decran=50} \defFunction{F1}(t){t cos 4 mul}{t sin 2 mul}{} \psSolid[object=prisme,h=8,fillcolor=green!20, base=0 350 {F1} CourbeR2+]% \defFunction{F2}(t){t cos 4 mul}{t sin 2 mul}{8} \psSolid[object=courbe, r=0, function=F2,range=0 360, linewidth=2\pslinewidth, linecolor=green] \axesIIID(6,4,8)(8,6,10) \end{pspicture} \end{LTXexample} \subsubsection{Example 5: a roof gutter with a semi-circular section} \begin{LTXexample}[width=7cm] \psset{unit=0.5} \psset{lightsrc=10 20 30,viewpoint=50 30 25 rtp2xyz,Decran=50} \begin{pspicture}(-8,-5)(6,10) \defFunction[algebraic]{F}(t) {3*cos(t)}{3*sin(t)}{} \defFunction[algebraic]{G}(t) {2.5*cos(t)}{2.5*sin(t)}{} \psSolid[object=grille, base=-6 6 -6 6,action=draw]% \psSolid[object=prisme,h=12, fillcolor=blue!30,RotX=-90, resolution=19, base=0 pi {F} CourbeR2+ pi 0 {G} CourbeR2+ ](0,-6,3) \axesIIID(6,6,2)(8,8,8) \end{pspicture} \end{LTXexample} We draw the exterior face (semicircle of radius 3~cm) in counterclockwise order: \verb!0 pi {F} CourbeR2+! Then the interior face (semicircle of radius 2{,}5~cm), is drawn in clockwise order: \verb!pi 0 {G} CourbeR2+! We can turn the solid $-90^{\mathrm{o}}$ and place it at the point $(0,-6,3)$. If we use the \verb+algebraic+ option to define the functions $F$ and $G$, the functions $\sin$ and $\cos$ are in radians. \subsubsection{The parameter \texttt{decal}} We wrote above that the four first vertices must be given in counterclockwise order with respect to the barycenter of the vertices of the base. In fact, this is the default version of the following rule: If the base has $n+1$ vertices, and if $G$ is their barycenter, then $(s_0,s_1)$ on one hand and $s_{n-1},s_n)$ on the other, should be in counterclockwise order with respect to $G$. This rule puts constraints on the coding of the base of a prism which sometimes rendering the last[?] unaesthetic. For this reason we have introduced the argument \verb+[decal]+ (default value$=-2$) which allows us to consider the list of vertices of the base as a circular file which one [d\'{e}calera] if needed. An example: default behavior with $decal=-2$:\par \psset{lightsrc=10 20 30,viewpoint=50 80 35 rtp2xyz,Decran=50} \begin{LTXexample}[width=6cm] \psset{unit=0.5} \begin{pspicture}(-6,-4)(6,7) \defFunction{F}(t){t cos 3 mul}{t sin 3 mul}{} \psSolid[object=prisme,h=8, fillcolor=yellow,RotX=-90, num=0 1 2 3 4 5 6, show=0 1 2 3 4 5 6, resolution=7, base=0 180 {F} CourbeR2+ ](0,-10,0) \end{pspicture} \end{LTXexample} We see that the vertex with index~$0$ is not where we expect to find it. We start again, but this time [en supprimant le d\'{e}calage]: \par % \psset{lightsrc=10 20 30,viewpoint=50 80 35 rtp2xyz,Decran=50} \begin{LTXexample}[width=6cm] \psset{unit=0.5} \begin{pspicture}(-6,-4)(6,7) \defFunction{F}(t){t cos 3 mul}{t sin 3 mul}{} \psSolid[object=prisme,h=8, fillcolor=yellow,RotX=-90, decal=0, num=0 1 2 3 4 5 6, show=0 1 2 3 4 5 6, resolution=7, base=0 180 {F} CourbeR2+ ](0,-10,0) \end{pspicture} \end{LTXexample}