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-rw-r--r--Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.pdfbin1817939 -> 1648513 bytes
-rw-r--r--Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.tex54
2 files changed, 29 insertions, 25 deletions
diff --git a/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.pdf b/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.pdf
index beeda452e89..5abe63eb719 100644
--- a/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.pdf
+++ b/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.pdf
Binary files differ
diff --git a/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.tex b/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.tex
index 862ce41c362..d5233e322d4 100644
--- a/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.tex
+++ b/Master/texmf-dist/doc/generic/pstricks-add/pstricks-add-doc.tex
@@ -3137,7 +3137,7 @@ This mode is in math also called double logarithm. It is a combination of the tw
\begin{LTXexample}[width=7cm]
\begin{pspicture}(-3.5,-3.5)(3.5,3.5)
\psplot[linewidth=2pt,linecolor=red]{0.001}{3}{x log}
- \psaxes[xylogBase=10,Oy=-3]{<->}(-3,-3)(3.5,3.5)
+ \psaxes[xylogBase=10,Oy=-3,Ox=-3]{->}(-3,-3)(3.5,3.5)
\uput[-90](3.5,-3){x}
\uput[180](-3,3.5){y}
\rput(2.5,1){$y=\log x$}
@@ -3149,7 +3149,11 @@ This mode is in math also called double logarithm. It is a combination of the tw
%--------------------------------------------------------------------------------------------
\subsubsection{\texttt{ylogBase}}
%--------------------------------------------------------------------------------------------
-The values for the \texttt{psaxes} y-coordinate are now the exponents to the base $10$ and for the right function to the base $e$: $10^{-3} \ldots 10^1$ which corresponds to the given y-intervall $-3\ldots 1.5$, where only integers as exponents are possible. These logarithm labels have no effect to the internal used units. To draw the logarithm function we have to use the math function
+The values for the \texttt{psaxes} y-coordinate are now the exponents to the base $10$ and for
+the right function to the base $e$: $10^{-3} \ldots 10^1$ which corresponds to the given
+y-intervall $-3\ldots 1.5$, where only integers as exponents are possible. These logarithm
+labels have no effect to the internal used units. To draw the logarithm function we have
+to use the math function
\[y=\log\{\log x\}\]
\[y=\ln\{\ln x\}\]
with an drawing intervall of $1.001\ldots 6$.
@@ -3157,12 +3161,12 @@ with an drawing intervall of $1.001\ldots 6$.
\medskip
\begin{LTXexample}[width=7cm]
\begin{pspicture}(-0.5,-3.5)(6.5,1.5)
- \psaxes[ylogBase=10]{<->}(0,-3)(6.5,1.5)
+ \psaxes[ylogBase=10,Oy=-3]{->}(0,-3)(6.5,1.5)
\uput[-90](6.5,-3){x}
\uput[0](0,1.4){y}
\rput(5,1){$y=\log x$}
\psplot[linewidth=2pt,%
- plotpoints=100,linecolor=red]{1.001}{6}{x log log} % log(x)
+ plotpoints=100,linecolor=red]{1.001}{6}{x log log} % log(log(x))
\end{pspicture}
\end{LTXexample}
@@ -3171,7 +3175,7 @@ with an drawing intervall of $1.001\ldots 6$.
\begin{pspicture}(-0.5,-3.5)(6.5,1.5)
\psplot[linewidth=2pt,plotpoints=100,linecolor=red]%
{1.04}{6}{/ln {log 0.4343 div} def x ln ln} % log(x)
- \psaxes[ylogBase=e]{<->}(0,-3)(6.5,1.5)
+ \psaxes[ylogBase=e,Oy=-3]{->}(0,-3)(6.5,1.5)
\uput[-90](6.5,-3){x}
\uput[0](0,1.5){y}
\rput(5,1){$y=\ln x$}
@@ -3183,7 +3187,7 @@ with an drawing intervall of $1.001\ldots 6$.
\medskip
\begin{LTXexample}[width=7cm]
\begin{pspicture}(-0.5,1.75)(6.5,4.5)
- \psaxes[ylogBase=10,Oy=2]{<->}(0,2)(0,2)(6.5,4.5)
+ \psaxes[ylogBase=10,Oy=2]{->}(0,2)(0,2)(6.5,4.5)
\end{pspicture}
\end{LTXexample}
@@ -3195,7 +3199,7 @@ with an drawing intervall of $1.001\ldots 6$.
\psplot{0}{6}{x x cos add log} % x + cox(x)
\psplot[linecolor=red]{0}{6}{x 3 exp x cos add log} % x^3 + cos(x)
\psplot[linecolor=cyan]{0}{6}{x 5 exp x cos add log} % x^5 + cos(x)
- \psaxes[ylogBase=10]{<->}(6.5,4.5)
+ \psaxes[ylogBase=10]{->}(6.5,4.5)
\end{pspicture}
\end{LTXexample}
@@ -3207,7 +3211,7 @@ with an drawing intervall of $1.001\ldots 6$.
\psplot{0}{6}{x x cos add log} % x + cox(x)
\psplot[linecolor=red]{0}{6}{x 3 exp x cos add log} % x^3 + cos(x)
\psplot[linecolor=cyan]{0}{6}{x 5 exp x cos add log} % x^5 + cos(x)
- \psaxes[ylogBase=10]{<->}(0,-1)(0,-1)(6.5,4.5)
+ \psaxes[ylogBase=10]{->}(0,-1)(0,-1)(6.5,4.5)
\end{pspicture}
\end{LTXexample}
@@ -3235,7 +3239,7 @@ Now we have to use the easy math function $y=x$ because the x axis is still $\lo
\begin{pspicture}(-3.5,-3.5)(3.5,3.5)
\psplot[linewidth=2pt,linecolor=red]{-3}{3}{x} % log(x)
\psplot[linewidth=2pt,linecolor=blue]{-1.3}{1.5}{x 0.4343 div} % ln(x)
- \psaxes[ylogBase=10,Oy=-3,Ox=-3]{->}(-3,-3)(3.5,3.5)
+ \psaxes[xlogBase=10,Oy=-3,Ox=-3]{->}(-3,-3)(3.5,3.5)
\uput[-90](3.5,-3){x}
\uput[180](-3,3.5){y}
\rput(2.5,1){$y=\log x$}
@@ -3261,7 +3265,7 @@ Now we have to use the easy math function $y=x$ because the x axis is still $\lo
\uput[0](-1,1){y}
\rput(0,1){$y=\sin x$}
\psplot[linewidth=2pt,plotpoints=5000,linecolor=red]{-1}{3.5}{10 x exp sin }
- \psaxes[xlogBase=10,Oy=-1]{->}(-1,-1)(4.25,1.25)
+ \psaxes[xlogBase=10,Oy=-1,Ox=-1]{->}(-1,-1)(4.25,1.25)
\end{pspicture}
\end{lstlisting}
@@ -3452,14 +3456,14 @@ This is only a demonstration that the default option \verb|logBase={}| still wor
%--------------------------------------------------------------------------------------
When having big numbers as data records then it makes sense to write the values
as ${<number>\cdot 10^{<exp>}}$. These new options allow to define the additional part
-of the value.
+of the value, but it must be set in math mode when using math operators!
\resetOptions
\begin{LTXexample}
\readdata{\data}{demo1.dat}
\pstScalePoints(1,0.000001){}{}% (x,y){additional x operator}{y op}
\psset{llx=-1cm,lly=-1cm}
-\psgraph[ylabelFactor={\cdot 10^6},Dx=5,Dy=100](0,0)(25,750){8cm}{5cm}
+\psgraph[ylabelFactor={$\cdot 10^6$},Dx=5,Dy=100](0,0)(25,750){8cm}{5cm}
\listplot[linecolor=red, linewidth=2pt, showpoints=true]{\data}
\endpsgraph
\pstScalePoints(1,1){}{}% reset
@@ -4451,7 +4455,7 @@ The y-length maybe given as !, then the macro uses the same unit as for the x-ax
\pstScalePoints(1,0.000001){}{}% (x,y){additional x operator}{y op}
\psset{llx=-1cm,lly=-1cm}
\begin{psgraph}[axesstyle=frame,xticksize=0 759,yticksize=0 25,%
- subticks=0,ylabelFactor={\cdot 10^6},%
+ subticks=0,ylabelFactor={$\cdot 10^6$},%
Dx=5,dy=100\psyunit,Dy=100](0,0)(25,750){10cm}{6cm} % parameters
\listplot[linecolor=red,linewidth=2pt,showpoints=true]{\data}
\end{psgraph}
@@ -4462,7 +4466,7 @@ The y-length maybe given as !, then the macro uses the same unit as for the x-ax
\pstScalePoints(1,0.000001){}{}% (x,y){additional x operator}{y op}
\psset{llx=-1cm,lly=-1cm}
§\ON§\begin{psgraph}§\OFF§[axesstyle=frame,xticksize=0 759,yticksize=0 25,%
- subticks=0,ylabelFactor={\cdot 10^6},%
+ subticks=0,ylabelFactor={$\cdot 10^6$},%
Dx=5,dy=100\psyunit,Dy=100](0,0)(25,750){10cm}{6cm} % parameters
\listplot[linecolor=red,linewidth=2pt,showpoints=true]{\data}
§\ON§\end{psgraph}§\OFF§
@@ -4493,7 +4497,7 @@ In the following example, the y unit gets the same value as the one for the x-ax
xAxisLabelPos={3cm,-1cm},yAxisLabelPos={-1.5cm,2.5cm}}
\pstScalePoints(1,0.00000001){}{}
\begin{psgraph}[axesstyle=frame,xticksize=0 7.5,yticksize=0 25,subticksize=1,
- ylabelFactor={\cdot 10^8},Dx=5,Dy=1,xsubticks=2](0,0)(25,7.5){5.5cm}{5cm}
+ ylabelFactor={$\cdot 10^8$},Dx=5,Dy=1,xsubticks=2](0,0)(25,7.5){5.5cm}{5cm}
\listplot[linecolor=red, linewidth=2pt, showpoints=true]{\data}
\end{psgraph}
\end{center}
@@ -4504,7 +4508,7 @@ In the following example, the y unit gets the same value as the one for the x-ax
§\ON§xAxisLabelPos§\OFF§={3cm,-1cm},§\ON§yAxisLabelPos§\OFF§={-1.5cm,2.5cm}}
\pstScalePoints(1,0.00000001){}{}
\begin{psgraph}[axesstyle=frame,xticksize=0 7.5,yticksize=0 25,subticksize=1,
- §\ON§ylabelFactor§\OFF§={\cdot 10^8},Dx=5,Dy=1,xsubticks=2](0,0)(25,7.5){5.5cm}{5cm}
+ §\ON§ylabelFactor§\OFF§={$\cdot 10^8$},Dx=5,Dy=1,xsubticks=2](0,0)(25,7.5){5.5cm}{5cm}
\listplot[linecolor=red, linewidth=2pt, showpoints=true]{\data}
\end{psgraph}
\end{lstlisting}
@@ -5367,12 +5371,12 @@ Just appreciate the difference between the normal behavior and the plotting with
\psset{xunit=.5}
\begin{pspicture}[showgrid=true](0,0)(12.566,2)
\parametricplot[algebraic,linecolor=red,VarStep, showpoints=true,
- VarStepEpsilon=.01]{0}{12.566}{t+cos(-t-PI/2)|1+sin(-t-PI/2)}
+ VarStepEpsilon=.01]{0}{12.566}{t+cos(-t-Pi/2)|1+sin(-t-Pi/2)}
\end{pspicture}
%
\begin{pspicture}[showgrid=true](0,0)(12.566,2)
\parametricplot[algebraic,linecolor=blue,VarStep, showpoints=false,
- VarStepEpsilon=.001]{0}{12.566}{t+cos(-t-PI/2)|1+sin(-t-PI/2)}
+ VarStepEpsilon=.001]{0}{12.566}{t+cos(-t-Pi/2)|1+sin(-t-Pi/2)}
\end{pspicture}
\egroup
\end{center}
@@ -5381,12 +5385,12 @@ Just appreciate the difference between the normal behavior and the plotting with
\psset{xunit=.5}
\begin{pspicture}[showgrid=true](0,0)(12.566,2)
\parametricplot[algebraic,linecolor=red,VarStep, showpoints=true,
- VarStepEpsilon=.01]{0}{12.566}{t+cos(-t-PI/2)|1+sin(-t-PI/2)}
+ VarStepEpsilon=.01]{0}{12.566}{t+cos(-t-Pi/2)|1+sin(-t-Pi/2)}
\end{pspicture}
%
\begin{pspicture}[showgrid=true](0,0)(12.566,2)
\parametricplot[algebraic,linecolor=blue,VarStep, showpoints=false,
- VarStepEpsilon=.001]{0}{12.566}{t+cos(-t-PI/2)|1+sin(-t-PI/2)}
+ VarStepEpsilon=.001]{0}{12.566}{t+cos(-t-Pi/2)|1+sin(-t-Pi/2)}
\end{pspicture}
\end{lstlisting}
@@ -5768,7 +5772,7 @@ the algortihm.
\rput*(3.3,.5){\psline[linecolor=green](-.75cm,0)}
\rput*[l](3.3,.5){\small solution exacte}
\end{pspicture}
-\captionof{figure}{Equation $y'=-y$ with $y_0=1$.} \label{fig:minusexpvarstep}
+{\captionof{figure}{Equation $y'=-y$ with $y_0=1$.}\label{fig:minusexpvarstep}}
\egroup
\end{center}
@@ -5825,7 +5829,7 @@ the algortihm.
\rput*(2.3,.4){\psline[linecolor=green](-.75cm,0)}
\rput*[l](2.3,.4){\small solution exacte}
\end{pspicture}
-\captionof{figure}{Equation $y''=-y$}\label{fig:trigfunc}
+{\captionof{figure}{Equation $y''=-y$}\label{fig:trigfunc}}
\egroup
\end{center}
@@ -6031,7 +6035,7 @@ equation. In the following example the masses of the stars are 1 and 20.
\end{pspicture}
\end{LTXexample}
\vspace{-2ex}
-\captionof{figure}{Gravitational interaction : fixed landmark, trajectory of the stars}\label{fig:InterGravRepFix}
+{\captionof{figure}{Gravitational interaction: fixed landmark, trajectory of the stars}\label{fig:InterGravRepFix}}
@@ -6045,7 +6049,7 @@ equation. In the following example the masses of the stars are 1 and 20.
\end{pspicture}
\end{LTXexample}
\vspace{-2ex}
-\captionof{figure}{Gravitational interaction : landmark defined by one star}\label{fig:IGnewrep}
+{\captionof{figure}{Gravitational interaction : landmark defined by one star}\label{fig:IGnewrep}}
\begin{center}
@@ -6059,8 +6063,8 @@ equation. In the following example the masses of the stars are 1 and 20.
\psplotDiffEqn[linecolor=blue, method=varrkiv, plotpoints=2, varsteptol=.0001,
plotfuncy=dup 2 get dup mul exch 3 get dup mul add sqrt]{0}{8}{\InitCond}{\Grav}
\end{pspicture}
-\egroup
\captionof{figure}{Gravitational interaction : vitessesspeed of the stars}
+\egroup
\end{center}
\begin{lstlisting}