diff --git a/src/Warm-Ups/Slide Rules/main.tex b/src/Warm-Ups/Slide Rules/main.tex new file mode 100755 index 0000000..f2f5471 --- /dev/null +++ b/src/Warm-Ups/Slide Rules/main.tex @@ -0,0 +1,78 @@ +% use [nosolutions] flag to hide solutions. +% use [solutions] flag to show solutions. +\documentclass[ + solutions, + shortwarning +]{../../../lib/tex/ormc_handout} +\usepackage{../../../lib/tex/macros} + + +\usepackage{pdfpages} +\usepackage{sliderule} +\usepackage{changepage} + +% Args: +% x, top scale y, label +\newcommand{\slideruleind}[3]{ + \draw[ + line width=1mm, + draw=black, + opacity=0.3, + text opacity=1 + ] + ({#1}, {#2 + 1}) + -- + ({#1}, {#2 - 1.1}) + node [below] {#3}; +} + + +\uptitlel{Advanced} +\uptitler{\smallurl{}} +\title{Warm-Up: Slide Rules} +\subtitle{Prepared by Mark on \today} + +\begin{document} + + \maketitle + + + \begin{center} + \begin{minipage}{6cm} + Dad says that anyone who can't use + a slide rule is a cultural illiterate + and should not be allowed to vote. + + \vspace{1ex} + + \textit{Have Space Suit --- Will Travel, 1958} + \end{minipage} + \end{center} + \hfill + + \input{parts/0 logarithms.tex} + \input{parts/1 intro.tex} + \input{parts/2 multiplication.tex} + + % Make sure the slide rule is on an odd page, + % so that double-sided printing won't require + % students to tear off problems. + \checkoddpage + \ifoddpage\else + \vspace*{\fill} + \begin{center} + { + \Large + \textbf{This page unintentionally left blank.} + } + \end{center} + \vspace{\fill} + \pagebreak + \fi + + \includepdf[ + pages=1, + fitpaper=true + ]{resources/rule.pdf} + +\end{document} \ No newline at end of file diff --git a/src/Warm-Ups/Slide Rules/meta.toml b/src/Warm-Ups/Slide Rules/meta.toml new file mode 100644 index 0000000..59669b6 --- /dev/null +++ b/src/Warm-Ups/Slide Rules/meta.toml @@ -0,0 +1,6 @@ +[metadata] +title = "Slide Rules" + +[publish] +handout = false +solutions = true diff --git a/src/Warm-Ups/Slide Rules/parts/0 logarithms.tex b/src/Warm-Ups/Slide Rules/parts/0 logarithms.tex new file mode 100644 index 0000000..172fade --- /dev/null +++ b/src/Warm-Ups/Slide Rules/parts/0 logarithms.tex @@ -0,0 +1,63 @@ +\section{Logarithms} + +\definition{} +The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\ +In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\ + +\problem{} +Evaluate the following by hand: + +\begin{enumerate} + \item $\log_{10}{(1000)}$ + \vfill + \item $\log_2{(64)}$ + \vfill + \item $\log_2{(\frac{1}{4})}$ + \vfill + \item $\log_x{(x)}$ for any $x$ + \vfill + \item $log_x{(1)}$ for any $x$ + \vfill +\end{enumerate} + +\problem{} +Prove the following: + +\begin{enumerate}[itemsep=2mm] + \item $\log_b{(b^x)} = x$ + \vfill + \item $b^{\log_b{x}} = x$ + \vfill + \item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$ + \vfill + \item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$ + \vfill + \item $\log_b{(x^y)} = y \log_b{(x)}$ + \vfill +\end{enumerate} + + +\begin{instructornote} + A good intro to the following sections is the linear slide rule: + \note{(note that these rules start at 0)} + \begin{center} + \begin{tikzpicture}[scale=0.5] + \linearscale{2}{1}{} + \linearscale{0}{0}{} + + \slideruleind + {5} + {1} + {2 + 3 = 5} + \end{tikzpicture} + \end{center} + + Take two linear rules, offset one, and you add. + Do the same with a log scale, and you multiply! \\ + + \linehack{} + + After assembling the paper slide rule, you can make a visor with some transparent tape. +\end{instructornote} + +\pagebreak \ No newline at end of file diff --git a/src/Warm-Ups/Slide Rules/parts/1 intro.tex b/src/Warm-Ups/Slide Rules/parts/1 intro.tex new file mode 100644 index 0000000..8d839c8 --- /dev/null +++ b/src/Warm-Ups/Slide Rules/parts/1 intro.tex @@ -0,0 +1,43 @@ +\section{Introduction} + +Mathematicians, physicists, and engineers needed to quickly compute products long before computers conquered the world. + +\medskip + +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. + +\medskip + +There are four scales on your slide rule, each labeled with a letter on the left side: + +\def\sliderulewidth{13} +\begin{center} +\begin{tikzpicture}[scale=1] + \tscale{0}{9}{T} + \kscale{0}{8}{K} + \abscale{0}{7}{A} + + \abscale{0}{5.5}{B} + \ciscale{0}{4.5}{CI} + \cdscale{0}{3.5}{C} + + \cdscale{0}{2}{D} + \lscale{0}{1}{L} + \sscale{0}{0}{S} +\end{tikzpicture} +\end{center} + +Each scale's ``generating function'' is on the right: +\begin{itemize} + \item T: $\tan$ + \item K: $x^3$ + \item A,B: $x^2$ + \item CI: $\frac{1}{x}$ + \item C, D: $x$ + \item L: $\log_{10}(x)$ + \item S: $\sin$ +\end{itemize} + +Once you understand the layout of your slide rule, move on to the next page. + +\pagebreak diff --git a/src/Warm-Ups/Slide Rules/parts/2 multiplication.tex b/src/Warm-Ups/Slide Rules/parts/2 multiplication.tex new file mode 100644 index 0000000..fee1493 --- /dev/null +++ b/src/Warm-Ups/Slide Rules/parts/2 multiplication.tex @@ -0,0 +1,299 @@ +\section{Multiplication} + +We'll use the C and D scales of your slide rule to multiply. \\ + +Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$: + +\def\sliderulewidth{10} +\begin{center} +\begin{tikzpicture}[scale=1] + \cdscale{\cdscalefn(2)}{1}{C} + \cdscale{0}{0}{D} +\end{tikzpicture} +\end{center} + +Then we'll find the second number, $3$ on the C scale, and read the D scale under it: + +\begin{center} +\begin{tikzpicture}[scale=1] + \cdscale{\cdscalefn(2)}{1}{C} + \cdscale{0}{0}{D} + + \slideruleind + {\cdscalefn(6)} + {1} + {6} + +\end{tikzpicture} +\end{center} + +Of course, our answer is 6. + +\problem{} +What is $1.15 \times 2.1$? \\ +Use your slide rule. + +\begin{solution} + \begin{center} + \begin{tikzpicture}[scale=1] + \cdscale{\cdscalefn(1.15)}{1}{C} + \cdscale{0}{0}{D} + + \slideruleind + {\cdscalefn(1.15)} + {1} + {1.15} + + \slideruleind + {\cdscalefn(1.15) + \cdscalefn(2.1)} + {1} + {2.415} + + \end{tikzpicture} + \end{center} +\end{solution} + +\vfill + +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. \\ + +\pagebreak + +Look at your C and D scales again. They contain every number between 1 and 10, but no more than that. +What should we do if we want to calculate $32 \times 210$? \\ + +\problem{} +Using your slide rule, calculate $32 \times 210$. \\ +%\hint{$32 = 3.2 \times 10^1$} + +\begin{solution} + \begin{center} + \begin{tikzpicture}[scale=1] + \cdscale{\cdscalefn(2.1)}{1}{C} + \cdscale{0}{0}{D} + + \slideruleind + {\cdscalefn(2.1)} + {1} + {2.1} + + \slideruleind + {\cdscalefn(2.1) + \cdscalefn(3.2)} + {1} + {6.72} + + \end{tikzpicture} + \end{center} + + Placing the decimal point correctly is your job. \\ + $10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$. +\end{solution} + +\vfill + +%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}. \\ + +%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact. + + +\problem{} +Compute the following: +\begin{enumerate} + \item $1.44 \times 52$ + \item $0.38 \times 1.24$ + \item $\pi \times 2.35$ +\end{enumerate} + +\begin{solution} + \begin{enumerate} + \item $1.44 \times 52 = 74.88$ + \item $0.38 \times 1.24 = 0.4712$ + \item $\pi \times 2.35 = 7.382$ + \end{enumerate} +\end{solution} + +\vfill +\pagebreak + +\problem{} +Note that the numbers on your C and D scales are logarithmically spaced. + +\def\sliderulewidth{13} +\begin{center} +\begin{tikzpicture}[scale=1] + \cdscale{0}{1}{C} + \cdscale{0}{0}{D} +\end{tikzpicture} +\end{center} + +Why does our multiplication procedure work? \\ +%\hint{See \ref{logids}} + +\vfill +\pagebreak + +Now we want to compute $7.2 \times 5.5$: + +\def\sliderulewidth{10} +\begin{center} +\begin{tikzpicture}[scale=0.8] + \cdscale{\cdscalefn(5.5)}{1}{C} + \cdscale{0}{0}{D} + + \slideruleind + {\cdscalefn(5.5)} + {1} + {5.5} + + \slideruleind + {\cdscalefn(5.5) + \cdscalefn(7.2)} + {1} + {???} + +\end{tikzpicture} +\end{center} + +No matter what order we go in, the answer ends up off the scale. There must be another way. \\ + +\medskip + +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, $7.2$: + +\begin{center} +\begin{tikzpicture}[scale=1] + \cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C} + \cdscale{0}{0}{D} + + \slideruleind + {\cdscalefn(7.2)} + {1} + {7.2} + +\end{tikzpicture} +\end{center} + +Now find the smaller number, $5.5$, on the C scale, and read the D scale under it: + +\begin{center} +\begin{tikzpicture}[scale=1] + \cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C} + \cdscale{0}{0}{D} + + + \slideruleind + {\cdscalefn(7.2)} + {1} + {7.2} + + \slideruleind + {\cdscalefn(3.96)} + {1} + {3.96} + +\end{tikzpicture} +\end{center} + +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. \\ + +\medskip + +\problem{} +Why does this work? \par +\hint{Add a second $D$ scale.} + +\begin{solution} + Consider the following picture, where I've put two D scales next to each other: + + \begin{center} + \begin{tikzpicture}[scale=0.7] + \cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C} + \cdscale{0}{0}{} + \cdscale{-10}{0}{} + + \draw[ + draw=black, + ] + (0, 0) + -- + (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 \ No newline at end of file diff --git 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+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/src/Warm-Ups/Slide Rules/sliderule.sty b/src/Warm-Ups/Slide Rules/sliderule.sty new file mode 100755 index 0000000..13e3995 --- /dev/null +++ b/src/Warm-Ups/Slide Rules/sliderule.sty @@ -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); + } + } +}