Add Advanced/Fast Inverse Root #4
78
src/Warm-Ups/Slide Rules/main.tex
Executable file
78
src/Warm-Ups/Slide Rules/main.tex
Executable file
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% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions,
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shortwarning
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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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% 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}
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\uptitler{\smallurl{}}
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\title{Warm-Up: Slide Rules}
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\subtitle{Prepared by Mark on \today}
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\begin{document}
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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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\vspace{1ex}
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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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% Make sure the slide rule is on an odd page,
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% so that double-sided printing 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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\end{document}
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6
src/Warm-Ups/Slide Rules/meta.toml
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6
src/Warm-Ups/Slide Rules/meta.toml
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[metadata]
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title = "Slide Rules"
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[publish]
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handout = false
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solutions = true
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63
src/Warm-Ups/Slide Rules/parts/0 logarithms.tex
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63
src/Warm-Ups/Slide Rules/parts/0 logarithms.tex
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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 $\log_b{c} = p$. \\
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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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\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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\problem{}<logids>
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Prove the following:
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\begin{enumerate}[itemsep=2mm]
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\item $\log_b{(b^x)} = x$
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\vfill
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\item $b^{\log_b{x}} = x$
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\vfill
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\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
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\vfill
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\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
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\vfill
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\item $\log_b{(x^y)} = y \log_b{(x)}$
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\vfill
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\end{enumerate}
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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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\note{(note that these rules start at 0)}
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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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 rules, offset one, and you add.
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Do the same with a log scale, and you multiply! \\
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\linehack{}
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After assembling the paper slide rule, you can make a visor with some transparent tape.
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\end{instructornote}
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\pagebreak
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43
src/Warm-Ups/Slide Rules/parts/1 intro.tex
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43
src/Warm-Ups/Slide Rules/parts/1 intro.tex
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\section{Introduction}
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Mathematicians, physicists, and engineers needed to quickly compute products long before computers conquered the world.
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\medskip
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The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
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\medskip
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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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299
src/Warm-Ups/Slide Rules/parts/2 multiplication.tex
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299
src/Warm-Ups/Slide Rules/parts/2 multiplication.tex
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\section{Multiplication}
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We'll use the C and D scales of your slide rule to multiply. \\
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Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} 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 $1.15 \times 2.1$? \\
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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 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 $32 \times 210$? \\
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\problem{}
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Using your slide rule, calculate $32 \times 210$. \\
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%\hint{$32 = 3.2 \times 10^1$}
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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. \\
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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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%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
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%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
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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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\pagebreak
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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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%\hint{See \ref{logids}}
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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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\medskip
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||||
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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}. 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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||||
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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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||||
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||||
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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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||||
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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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||||
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Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$. We can do this by hand to verify our answer. \\
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\medskip
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||||
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||||
\problem{}
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Why does this work? \par
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\hint{Add a second $D$ scale.}
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\begin{solution}
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||||
Consider the following picture, where I've put two D scales next to each other:
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||||
|
||||
\begin{center}
|
||||
\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[
|
||||
draw=black,
|
||||
]
|
||||
(0, 0)
|
||||
--
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||||
(0, -0.3)
|
||||
node [below] {D};
|
||||
|
||||
\draw[
|
||||
draw=black,
|
||||
]
|
||||
(-10, 0)
|
||||
--
|
||||
(-10, -0.3)
|
||||
node [below] {D};
|
||||
|
||||
\slideruleind
|
||||
{-10 + \cdscalefn(7.2)}
|
||||
{1}
|
||||
{7.2}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(7.2)}
|
||||
{1}
|
||||
{7.2}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(3.96)}
|
||||
{1}
|
||||
{3.96}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\medskip
|
||||
|
||||
The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$ smaller than it should be.
|
||||
|
||||
\medskip
|
||||
|
||||
\medskip
|
||||
In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
|
||||
This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}
|
||||
Compute the following using your slide rule:
|
||||
\begin{enumerate}
|
||||
\item $9 \times 8$
|
||||
\item $15 \times 35$
|
||||
\item $42.1 \times 7.65$
|
||||
\item $6.5^2$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $9 \times 8 = 72$
|
||||
\item $15 \times 35 = 525$
|
||||
\item $42.1 \times 7.65 = 322.065$
|
||||
\item $6.5^2 = 42.25$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Compute the following using your slide rule. \\
|
||||
|
||||
\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
|
||||
BIN
src/Warm-Ups/Slide Rules/resources/rule.pdf
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src/Warm-Ups/Slide Rules/resources/rule.pdf
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src/Warm-Ups/Slide Rules/resources/rule.svg
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24144
src/Warm-Ups/Slide Rules/resources/rule.svg
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534
src/Warm-Ups/Slide Rules/sliderule.sty
Executable file
534
src/Warm-Ups/Slide Rules/sliderule.sty
Executable file
@ -0,0 +1,534 @@
|
||||
\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);
|
||||
}
|
||||
}
|
||||
}
|
||||
Reference in New Issue
Block a user