Remove slide rule
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@ -6,28 +6,8 @@
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]{../../../lib/tex/ormc_handout}
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\usepackage{../../../lib/tex/macros}
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\usepackage{pdfpages}
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\usepackage{sliderule}
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\usepackage{changepage}
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\usepackage{listings}
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% Args:
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% x, top scale y, label
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\newcommand{\slideruleind}[3]{
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\draw[
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line width=1mm,
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draw=black,
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opacity=0.3,
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text opacity=1
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]
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({#1}, {#2 + 1})
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--
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({#1}, {#2 - 1.1})
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node [below] {#3};
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}
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\uptitlel{Advanced 2}
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\uptitler{\smallurl{}}
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\title{Fast Inverse Square Root}
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@ -37,48 +17,8 @@
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\maketitle
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%\begin{center}
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%\begin{minipage}{6cm}
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% Dad says that anyone who can't use
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% a slide rule is a cultural illiterate
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%and should not be allowed to vote.
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%
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% \vspace{1ex}
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%
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% \textit{Have Space Suit --- Will Travel, 1958}
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%\end{minipage}
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%\end{center}
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%\hfill
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%\input{parts/0 logarithms.tex}
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%\input{parts/1 intro.tex}
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%\input{parts/2 multiplication.tex}
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%\input{parts/3 division.tex}
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%\pagebreak
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\input{parts/4 int.tex}
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\input{parts/5 float.tex}
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\input{parts/6 approximate.tex}
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\input{parts/7 quake.tex}
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% Make sure the slide rule is on an odd page,
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% so that double-sided prints won't require
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% students to tear off problems.
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\checkoddpage
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\ifoddpage\else
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\vspace*{\fill}
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\begin{center}
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{
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\Large
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\textbf{This page unintentionally left blank.}
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}
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\end{center}
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\vspace{\fill}
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\pagebreak
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\fi
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%\includepdf[
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% pages=1,
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% fitpaper=true
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%]{resources/rule.pdf}
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\input{parts/1 int.tex}
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\input{parts/2 float.tex}
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\input{parts/3 approximate.tex}
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\input{parts/4 quake.tex}
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\end{document}
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@ -1,72 +0,0 @@
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\section{Logarithms}
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\definition{}<logdef>
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The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \par
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In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?''
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\vspace{2mm}
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In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
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\problem{}
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Evaluate the following by hand:
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\begin{enumerate}
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\item $\log_{10}{(1000)}$
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\vfill
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\item $\log_2{(64)}$
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\vfill
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\item $\log_2{(\frac{1}{4})}$
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\vfill
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\item $\log_x{(x)}$ for any $x$
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\vfill
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\item $log_x{(1)}$ for any $x$
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\vfill
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\end{enumerate}
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\pagebreak
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\problem{}<logids>
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Prove the following identities:
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\begin{enumerate}[itemsep=2mm]
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\item $\log_b{(b^x)} = x$
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\item $b^{\log_b{x}} = x$
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\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
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\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
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\item $\log_b{(x^y)} = y \log_b{(x)}$
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\end{enumerate}
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\vfill
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\begin{instructornote}
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A good intro to the following sections is the linear slide rule:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\linearscale{2}{1}{}
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\linearscale{0}{0}{}
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\slideruleind
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{5}
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{1}
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{2 + 3 = 5}
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\end{tikzpicture}
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\end{center}
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Take two linear rulers, offset one, and you add. \par
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If you do the same with a log scale, you multiply!
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\vspace{2mm}
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Note that the slide rules above start at 0.
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\linehack{}
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After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
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\end{instructornote}
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\pagebreak
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@ -1,44 +0,0 @@
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\section{Slide Rules}
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Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
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\vspace{2mm}
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The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. \par
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Before you continue, tear off the last page of this handout and assemble your slide rule.
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\vspace{2mm}
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There are four scales on your slide rule, each labeled with a letter on the left side:
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\def\sliderulewidth{13}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\tscale{0}{9}{T}
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\kscale{0}{8}{K}
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\abscale{0}{7}{A}
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\abscale{0}{5.5}{B}
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\ciscale{0}{4.5}{CI}
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\cdscale{0}{3.5}{C}
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\cdscale{0}{2}{D}
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\lscale{0}{1}{L}
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\sscale{0}{0}{S}
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\end{tikzpicture}
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\end{center}
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Each scale's ``generating function'' is on the right:
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\begin{itemize}
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\item T: $\tan$
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\item K: $x^3$
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\item A,B: $x^2$
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\item CI: $\frac{1}{x}$
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\item C, D: $x$
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\item L: $\log_{10}(x)$
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\item S: $\sin$
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\end{itemize}
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Once you understand the layout of your slide rule, move on to the next page.
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\pagebreak
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@ -1,278 +0,0 @@
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\section{Multiplication}
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We'll use the C and D scales of your slide rule to multiply. \par
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Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index}
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of the C scale over the smaller number, $2$:
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\def\sliderulewidth{10}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\end{tikzpicture}
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\end{center}
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Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(6)}
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{1}
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{6}
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\end{tikzpicture}
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\end{center}
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Of course, our answer is 6.
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\problem{}
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What is $1.15 \times 2.1$? \par
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Use your slide rule.
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(1.15)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(1.15)}
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{1}
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{1.15}
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\slideruleind
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{\cdscalefn(1.15) + \cdscalefn(2.1)}
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{1}
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{2.415}
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within
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two decimal places is close enough for most practical applications.
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\pagebreak
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Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
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What should we do if we want to calculate $32 \times 210$? \par
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\problem{}
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Using your slide rule, calculate $32 \times 210$.
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2.1)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(2.1)}
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{1}
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{2.1}
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\slideruleind
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{\cdscalefn(2.1) + \cdscalefn(3.2)}
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{1}
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{6.72}
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\end{tikzpicture}
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\end{center}
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Placing the decimal point correctly is your job. \par
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$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
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\end{solution}
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\vfill
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\problem{}
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Compute the following:
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\begin{enumerate}
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\item $1.44 \times 52$
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\item $0.38 \times 1.24$
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\item $\pi \times 2.35$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $1.44 \times 52 = 74.88$
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\item $0.38 \times 1.24 = 0.4712$
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\item $\pi \times 2.35 = 7.382$
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\end{enumerate}
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\end{solution}
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\vfill
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\problem{}<provemult>
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Note that the numbers on your C and D scales are logarithmically spaced.
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\def\sliderulewidth{13}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{0}{1}{C}
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\cdscale{0}{0}{D}
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\end{tikzpicture}
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\end{center}
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Why does our multiplication procedure work?
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\vfill
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\pagebreak
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Now we want to compute $7.2 \times 5.5$:
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\def\sliderulewidth{10}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\cdscale{\cdscalefn(5.5)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(5.5)}
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{1}
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{5.5}
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\slideruleind
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{\cdscalefn(5.5) + \cdscalefn(7.2)}
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{1}
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{???}
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\end{tikzpicture}
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\end{center}
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No matter what order we go in, the answer ends up off the scale. There must be another way.
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\vspace{2mm}
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Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}.
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Move it over the \textit{larger} number, $7.2$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\end{tikzpicture}
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\end{center}
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Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(3.96)}
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{1}
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{3.96}
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\end{tikzpicture}
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\end{center}
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Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$.
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We can do this by hand to verify our answer.
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\vspace{2mm}
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\problem{}
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Why does this work?
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\begin{solution}
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Consider the following picture, where we have two D scales next to each other:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{}
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\cdscale{-10}{0}{}
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\draw[
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draw=black,
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]
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(0, 0)
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--
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(0, -0.3)
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node [below] {D};
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\draw[
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draw=black,
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]
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(-10, 0)
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--
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(-10, -0.3)
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node [below] {D};
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\slideruleind
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{-10 + \cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(3.96)}
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{1}
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{3.96}
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\end{tikzpicture}
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\end{center}
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\vspace{2mm}
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The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$
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smaller than it should be.
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\vspace{2mm}
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\vspace{2mm}
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In other words, the answer we get from reverse multiplication is $\log{a} + \log{b} - \log{10}$. \par
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This reduces to $\log{(\frac{a \times b}{10})}$, and explains the misplaced decimal point in $7.2 \times 5.5$.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Compute the following using your slide rule:
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\begin{enumerate}
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\item $9 \times 8$
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\item $15 \times 35$
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\item $42.1 \times 7.65$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $9 \times 8 = 72$
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\item $15 \times 35 = 525$
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\item $42.1 \times 7.65 = 322.065$
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\end{enumerate}
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\end{solution}
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\vfill
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\pagebreak
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@ -19,15 +19,11 @@ Namely, show that
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\begin{equation*}
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\log_2(x_f) ~=~ \frac{x_i}{2^{23}} - 127 + \varepsilon
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\end{equation*}
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for some correction term term $\varepsilon$
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\vspace{5mm}
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Before you start, make sure you understand what we're trying to do: \par
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We're finding an expression for $\log_2(x_f)$ in terms of $x_i$ and an correction term $\varepsilon$. \par
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That is, we're finding a closed-form expression that connects the two interpretations \par
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(\texttt{float} and \texttt{int}) of the bit string $x$.
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for some correction term term $\varepsilon$ \par
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\note{
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In other words, we're finding an expression for $x$ as a float
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in terms of $x$ as an int.
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}
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\begin{solution}
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Let $E$ and $F$ be the exponent and float bits of $x_f$. \par
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@ -1,91 +0,0 @@
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\section{Division}
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Now that you can multiply, division should be easy. All you need to do is work backwards. \\
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Let's look at our first example again: $3 \times 2 = 6$.
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\medskip
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We can easily see that $6 \div 3 = 2$
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(6)}
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{1}
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{Align here}
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\slideruleind
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{\cdscalefn(2)}
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{1}
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{2}
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\end{tikzpicture}
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\end{center}
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and that $6 \div 2 = 3$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(3)}{-3}{C}
|
||||
\cdscale{0}{-4}{D}
|
||||
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(6)}
|
||||
{-3}
|
||||
{Align here}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(3)}
|
||||
{-3}
|
||||
{3}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
If your left-hand index is off the scale, read the right-hand one. \\
|
||||
Consider $42.25 \div 6.5 = 6.5$:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(4.225)}
|
||||
{1}
|
||||
{Align here}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(6.5)}
|
||||
{1}
|
||||
{6.5}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Place your decimal points carefully.
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Compute the following using your slide rule. \par
|
||||
|
||||
\begin{enumerate}
|
||||
\item $135 \div 15$
|
||||
\item $68.2 \div 0.575$
|
||||
\item $(118 \times 0.51) \div 6.6$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $135 \div 15 = 9$
|
||||
\item $68.2 \div 0.575 = 118.609$
|
||||
\item $(118 \times 0.51) \div 6.6 = 9.118$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
File diff suppressed because one or more lines are too long
|
Before Width: | Height: | Size: 862 KiB |
@ -1,534 +0,0 @@
|
||||
\NeedsTeXFormat{LaTeX2e}
|
||||
\ProvidesPackage{sliderule}[2022/08/22 Slide rule tools]
|
||||
|
||||
\RequirePackage{tikz}
|
||||
\RequirePackage{ifthen}
|
||||
|
||||
|
||||
% Scale functions:
|
||||
% See https://sliderulemuseum.com/SR_Scales.htm
|
||||
%
|
||||
% l: length of the rule
|
||||
% n: the number on the rule
|
||||
%
|
||||
% A/B: (l/2) * log(n)
|
||||
% C/D: l / log(n)
|
||||
% CI: abs(l * log(10 / n) - l)
|
||||
% K: (l/3) * log(n)
|
||||
%
|
||||
% L: n * l
|
||||
% T: l * log(10 * tan(n))
|
||||
% S: l * log(10 * sin(n))
|
||||
|
||||
\def\sliderulewidth{10}
|
||||
|
||||
\def\abscalefn(#1){(\sliderulewidth/2) * log10(#1)}
|
||||
\def\cdscalefn(#1){(\sliderulewidth * log10(#1))}
|
||||
\def\ciscalefn(#1){(\sliderulewidth - \cdscalefn(#1))}
|
||||
\def\kscalefn(#1){(\sliderulewidth/3) * log10(#1)}
|
||||
\def\lscalefn(#1){(\sliderulewidth * #1)}
|
||||
\def\tscalefn(#1){(\sliderulewidth * log10(10 * tan(#1)))}
|
||||
\def\sscalefn(#1){(\sliderulewidth * log10(10 * sin(#1)))}
|
||||
|
||||
|
||||
% Arguments:
|
||||
% Label
|
||||
% x of start
|
||||
% y of start
|
||||
\newcommand{\linearscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks
|
||||
\foreach \i in {0,..., 10}{
|
||||
\draw[black]
|
||||
({#1 + (\sliderulewidth / 10) * \i}, #2) --
|
||||
({#1 + (\sliderulewidth / 10) * \i}, #2 + 0.3)
|
||||
node[above] {\i};
|
||||
}
|
||||
|
||||
% Submarks
|
||||
\foreach \n in {0, ..., 9} {
|
||||
\foreach \i in {1,..., 9} {
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
|
||||
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2) --
|
||||
({#1 + (\sliderulewidth / 10) * (\n + \i / 10)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
|
||||
% Arguments:
|
||||
% Label
|
||||
% x of start
|
||||
% y of start
|
||||
\newcommand{\abscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks 1 - 9
|
||||
\foreach \i in {1,..., 9}{
|
||||
\draw[black]
|
||||
({#1 + \abscalefn(\i)}, #2) --
|
||||
({#1 + \abscalefn(\i)}, #2 + 0.3)
|
||||
node[above] {\i};
|
||||
}
|
||||
% Numbers and marks 10 - 100
|
||||
\foreach \i in {1,..., 10}{
|
||||
\draw[black]
|
||||
({#1 + \abscalefn(10 * \i)}, #2) --
|
||||
({#1 + \abscalefn(10 * \i)}, #2 + 0.3)
|
||||
node[above] {\ifthenelse{\i=10}{1}{\i}};
|
||||
}
|
||||
|
||||
% Submarks 1 - 9
|
||||
\foreach \n in {1, ..., 9} {
|
||||
\ifthenelse{\n<5}{
|
||||
\foreach \i in {1,..., 9}
|
||||
} {
|
||||
\foreach \i in {2,4,6,8}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + \abscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \abscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \abscalefn(\n + \i / 10)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 10 - 100
|
||||
\foreach \n in {10,20,...,90} {
|
||||
\ifthenelse{\n<50}{
|
||||
\foreach \i in {1,..., 9}
|
||||
} {
|
||||
\foreach \i in {2,4,6,8}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + \abscalefn(\n + \i)}, #2) --
|
||||
({#1 + \abscalefn(\n + \i)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \abscalefn(\n + \i)}, #2) --
|
||||
({#1 + \abscalefn(\n + \i)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
\newcommand{\cdscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks 1 - 10
|
||||
\foreach \i in {1,..., 10}{
|
||||
\draw[black]
|
||||
({#1 + \cdscalefn(\i)}, #2) --
|
||||
({#1 + \cdscalefn(\i)}, #2 + 0.3)
|
||||
node[above] {\ifthenelse{\i=10}{1}{\i}};
|
||||
}
|
||||
|
||||
% Submarks 1 - 9
|
||||
\foreach \n in {1, ..., 9} {
|
||||
\ifthenelse{\n<3}{
|
||||
\foreach \i in {5,10,...,95}
|
||||
} {
|
||||
\foreach \i in {10,20,...,90}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=50}{
|
||||
\draw[black]
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2) --
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2);
|
||||
\ifthenelse{\n=1}{
|
||||
\draw
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.2)
|
||||
node [above] {1.5};
|
||||
}{}
|
||||
} {
|
||||
\ifthenelse{
|
||||
\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
|
||||
\i=60 \OR \i=70 \OR \i=80 \OR \i=90
|
||||
}{
|
||||
\draw[black]
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2) --
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.15);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2) --
|
||||
({#1 + \cdscalefn(\n + \i / 100)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
\newcommand{\ciscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks
|
||||
\foreach \i in {1,...,10}{
|
||||
\draw[black]
|
||||
({#1 + \ciscalefn(\i)}, #2) --
|
||||
({#1 + \ciscalefn(\i)}, #2 + 0.3)
|
||||
node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
|
||||
}
|
||||
|
||||
% Submarks 1 - 9
|
||||
\foreach \n in {1, ..., 9} {
|
||||
\ifthenelse{\n<3}{
|
||||
\foreach \i in {5,10,...,95}
|
||||
} {
|
||||
\foreach \i in {10,20,...,90}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=50}{
|
||||
\draw[black]
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2) --
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2);
|
||||
\ifthenelse{\n=1}{
|
||||
\draw
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.2)
|
||||
node [above] {1.5};
|
||||
}{}
|
||||
} {
|
||||
\ifthenelse{
|
||||
\i=10 \OR \i=20 \OR \i=30 \OR \i=40 \OR
|
||||
\i=60 \OR \i=70 \OR \i=80 \OR \i=90
|
||||
}{
|
||||
\draw[black]
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2) --
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.15);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2) --
|
||||
({#1 + \ciscalefn(\n + \i / 100)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
\newcommand{\kscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks 1 - 9
|
||||
\foreach \i in {1,...,9}{
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\i)}, #2) --
|
||||
({#1 + \kscalefn(\i)}, #2 + 0.3)
|
||||
node[above] {\i};
|
||||
}
|
||||
% Numbers and marks 10 - 90
|
||||
\foreach \i in {1,..., 9}{
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(10 * \i)}, #2) --
|
||||
({#1 + \kscalefn(10 * \i)}, #2 + 0.3)
|
||||
node[above] {\ifthenelse{\i=10}{1}{\i}};
|
||||
}
|
||||
% Numbers and marks 100 - 1000
|
||||
\foreach \i in {1,..., 10}{
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(100 * \i)}, #2) --
|
||||
({#1 + \kscalefn(100 * \i)}, #2 + 0.3)
|
||||
node[above] {\ifthenelse{\i=10}{1}{\i}};
|
||||
}
|
||||
|
||||
% Submarks 1 - 9
|
||||
\foreach \n in {1, ..., 9} {
|
||||
\ifthenelse{\n<4}{
|
||||
\foreach \i in {1,..., 9}
|
||||
} {
|
||||
\foreach \i in {2,4,6,8}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \kscalefn(\n + \i / 10)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 10 - 90
|
||||
\foreach \n in {10,20,...,90} {
|
||||
\ifthenelse{\n<40}{
|
||||
\foreach \i in {1,..., 9}
|
||||
} {
|
||||
\foreach \i in {2,4,6,8}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\n + \i)}, #2) --
|
||||
({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\n + \i)}, #2) --
|
||||
({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 100 - 1000
|
||||
\foreach \n in {100,200,...,900} {
|
||||
\ifthenelse{\n<400}{
|
||||
\foreach \i in {10,20,...,90}
|
||||
} {
|
||||
\foreach \i in {20,40,60,80}
|
||||
}
|
||||
{
|
||||
\ifthenelse{\i=50}{
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\n + \i)}, #2) --
|
||||
({#1 + \kscalefn(\n + \i)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \kscalefn(\n + \i)}, #2) --
|
||||
({#1 + \kscalefn(\n + \i)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
\newcommand{\lscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks
|
||||
\foreach \i in {0,..., 10}{
|
||||
\draw[black]
|
||||
({#1 + \lscalefn(\i / 10)}, #2) --
|
||||
({#1 + \lscalefn(\i / 10)}, #2 + 0.3)
|
||||
node[above] {\ifthenelse{\i=10}{1}{\ifthenelse{\i=0}{0}{.\i}}};
|
||||
}
|
||||
|
||||
% Submarks
|
||||
\foreach \n in {0, ..., 9} {
|
||||
\foreach \i in {1,...,19} {
|
||||
\ifthenelse{\i=10}{
|
||||
\draw[black]
|
||||
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
|
||||
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.2);
|
||||
} {
|
||||
\ifthenelse{
|
||||
\i=1 \OR \i=3 \OR \i=5 \OR \i=7 \OR
|
||||
\i=9 \OR \i=11 \OR \i=13 \OR \i=15 \OR
|
||||
\i=17 \OR \i=19
|
||||
}{
|
||||
\draw[black]
|
||||
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
|
||||
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.1);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2) --
|
||||
({#1 + \lscalefn((\n + (\i / 20))/10)}, #2 + 0.15);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
\newcommand{\tscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
% First line
|
||||
\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks 6 - 10
|
||||
\foreach \i in {6,...,9,10,15,...,45}{
|
||||
\draw[black]
|
||||
({#1 + \tscalefn(\i)}, #2) --
|
||||
({#1 + \tscalefn(\i)}, #2 + 0.3)
|
||||
node[above] {\i};
|
||||
}
|
||||
|
||||
% Submarks 6 - 10
|
||||
\foreach \n in {6, ..., 9} {
|
||||
\foreach \i in {1,...,9}{
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + \tscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \tscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \tscalefn(\n + \i / 10)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 15 - 45
|
||||
\foreach \n in {10, 15, ..., 40} {
|
||||
\foreach \i in {1,...,24}{
|
||||
\ifthenelse{
|
||||
\i=5 \OR \i=10 \OR \i=15 \OR \i=20
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \tscalefn(\n + \i / 5)}, #2) --
|
||||
({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \tscalefn(\n + \i / 5)}, #2) --
|
||||
({#1 + \tscalefn(\n + \i / 5)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
\newcommand{\sscale}[3]{
|
||||
\draw[black] ({#1}, #2) -- ({#1 + \sliderulewidth}, #2);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.9);
|
||||
\draw[black] ({#1}, #2 + 0.9) -- ({#1}, #2 + 0.7);
|
||||
\draw[black] ({#1 + \sliderulewidth}, #2 + 0.9) -- ({#1 + \sliderulewidth}, #2 + 0.7);
|
||||
|
||||
% First line
|
||||
\draw[black] ({#1}, #2) -- ({#1}, #2 + 0.2);
|
||||
|
||||
|
||||
\draw ({#1 - 0.1}, #2 + 0.5) node[left] {#3};
|
||||
|
||||
% Numbers and marks
|
||||
\foreach \i in {6,...,9,10,15,...,30,40,50,...,60,90}{
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\i)}, #2) --
|
||||
({#1 + \sscalefn(\i)}, #2 + 0.3)
|
||||
node[above] {\i};
|
||||
}
|
||||
|
||||
% Submarks 6 - 10
|
||||
\foreach \n in {6, ..., 9} {
|
||||
\foreach \i in {1,...,9}{
|
||||
\ifthenelse{\i=5}{
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i / 10)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i / 10)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 15 - 30
|
||||
\foreach \n in {10, 15, ..., 25} {
|
||||
\foreach \i in {1,...,24}{
|
||||
\ifthenelse{
|
||||
\i=5 \OR \i=10 \OR \i=15 \OR \i=20
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i / 5)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i / 5)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i / 5)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 30
|
||||
\foreach \n in {30} {
|
||||
\foreach \i in {1,...,19}{
|
||||
\ifthenelse{
|
||||
\i=2 \OR \i=4 \OR \i=6 \OR \i=8 \OR
|
||||
\i=10 \OR \i=12 \OR \i=14 \OR \i=16 \OR
|
||||
\i=18
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i / 2)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i / 2)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i / 2)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 40 - 50
|
||||
\foreach \n in {40, 50} {
|
||||
\foreach \i in {1,...,9}{
|
||||
\ifthenelse{
|
||||
\i=5 \OR \i=10 \OR \i=15 \OR \i=20
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(\n + \i)}, #2) --
|
||||
({#1 + \sscalefn(\n + \i)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
% Submarks 60
|
||||
\foreach \i in {1,...,10}{
|
||||
\ifthenelse{
|
||||
\i=5 \OR \i=10
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(60 + \i * 2)}, #2) --
|
||||
({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.2);
|
||||
} {
|
||||
\draw[black]
|
||||
({#1 + \sscalefn(60 + \i * 2)}, #2) --
|
||||
({#1 + \sscalefn(60 + \i * 2)}, #2 + 0.1);
|
||||
}
|
||||
}
|
||||
}
|
||||
Loading…
x
Reference in New Issue
Block a user