1

How can I remove the decimal 0 from the angle?

\documentclass[11pt,a4paper,dutch]{article} \usepackage[per-mode=symbol,mode=text,per-mode=symbol,exponent-product=\cdot]{siunitx} \usepackage[T1]{fontenc} \usepackage[left=1.2cm,right=3.6cm,top=1.8cm,bottom=2.0cm,marginparwidth=2.4cm]{geometry} \usepackage{amsmath} \usepackage{contour} \usepackage{physics} \usepackage{xparse} \usepackage{amssymb} \usepackage{xcolor} \usepackage{rotating} \usepackage{float} \usepackage{subcaption} \usepackage{caption} \usepackage{pgfplots} \usepackage{tikz} \tikzset{every picture/.style={line width=0.75pt}} \usetikzlibrary{arrows.meta,calc,math,quotes,angles} \usepackage{fancyhdr} \tikzset{>=latex} % for LaTeX arrow head \usepackage{xcolor} \colorlet{veccol}{green!70!black} \colorlet{vcol}{green!70!black} \colorlet{xcol}{blue!85!black} \colorlet{projcol}{xcol!60} \colorlet{unitcol}{xcol!60!black!85} \colorlet{myblue}{blue!70!black} \colorlet{myred}{red!90!black} \colorlet{mypurple}{blue!50!red!80!black!80} \tikzstyle{vector}=[->,very thick,xcol] \sisetup { input-decimal-markers={.}, output-decimal-marker = {.}, exponent-to-prefix = true, round-mode = figures, round-precision = 3, scientific-notation = engineering } \begin{document} \begin{figure}[h] \centering \begin{minipage} [h]{0.4\textwidth} \centering \begin{tikzpicture} [x=0.75pt,y=0.75pt] \def\R{154.483} \def\ul{0.52} \def\az{15} \coordinate (O) at (0,0); \coordinate (A) at (0,100); \draw[dashed] (A) -- (0,-100); \path (O) node[left=4] {A}; \begin{scope}[xscale=1,yscale=1,rotate around={-\az:(O)}] ; \coordinate (B) at (0,100); \draw[gray, fill=green!30!white] (O)--(0,100)--(-5,100)--(-5,-100)--(0,-100)--cycle; \draw pic[<-,"$\textbf{\small{\ang{\az}}}$",draw=black,angle radius=60,angle eccentricity=1.1]{angle=B--O--A}; \coordinate (R) at (\az:\R); \coordinate (X) at ({\R*cos(\az)},0); \coordinate (Y) at (0,{\R*sin(\az)}); \draw[projcol,dashed] (X) -- (R); \draw[projcol,dashed] (Y) -- (R); \draw[vector,red] (O) -- (R) node[right] {$\SI{835.974}{\newton}$}; \draw[vector,<->,projcol](X) node[scale=0.9, below =-1,right] {$\SI{807.48}{\newton}$} -- (O) -- (Y) node[scale=0.9,below=4, left] {$\SI{216.365}{\newton}$}; \end{scope} \end{tikzpicture} \end{minipage} \end{figure} \end{document}
Improve this question

1 Answer 1

Reset to default
1

For the use case at hand, I can see no advantage in specifying the siunitx options round-mode = figures and round-precision = 3. It's better to write \SI{835}{\newton} instead of \SI{835.974}{\newton}, etc. Oh, and there's no need to encase the \SI directives in $ math mode shifters. As a side effect of making this change, you won't have to worry about typesetting "15.0" instead of just "15" as the angle of rotation.

Incidentally, \SI is a legacy macro that will eventually go away. Do get used to writing \qty instead of \SI.

enter image description here

Aside: You also may want to think about replacing

 node[scale=0.9, below = 4, left] {\qty{216}{\newton}\kern0.4em};

with

 node[scale=0.9, above=4, right] {\qty{216}{\newton}}; 
\documentclass[11pt,a4paper,dutch]{article} \usepackage[T1]{fontenc} \usepackage[left=1.2cm, right=3.6cm, top=1.8cm, bottom=2.0cm, marginparwidth=2.4cm] % are you sure you need this option? {geometry} %% The following packages aren't needed for this example %\usepackage{amsmath} %\usepackage{contour} %\usepackage{physics} %\usepackage{xparse} %\usepackage{amssymb} %\usepackage{xcolor} %\usepackage{rotating} %\usepackage{float} %\usepackage{subcaption} %\usepackage{caption} %\usepackage{fancyhdr} \usepackage{xcolor} \colorlet{veccol}{green!70!black} \colorlet{vcol}{green!70!black} \colorlet{xcol}{blue!85!black} \colorlet{projcol}{xcol!60} \colorlet{unitcol}{xcol!60!black!85} \colorlet{myblue}{blue!70!black} \colorlet{myred}{red!90!black} \colorlet{mypurple}{blue!50!red!80!black!80} \usepackage{pgfplots} \usepackage{tikz} \tikzset{every picture/.style={line width=0.75pt}} \usetikzlibrary{arrows.meta,calc,math,quotes,angles} \tikzset{>=latex} % for LaTeX arrow head \tikzstyle{vector}=[->,very thick,xcol] \usepackage{siunitx} \sisetup{per-mode = symbol, mode = text, exponent-product = \cdot, exponent-to-prefix = true, % the following options are unnecessary or counterproductive: %per-mode = symbol, % redundant %input-decimal-markers = {.}, %output-decimal-marker = {.}, %round-mode = figures, %round-precision = 3, %scientific-notation = engineering } \begin{document} %\begin{figure}[ht] \begin{center} %\begin{minipage}[h]{0.4\textwidth} %\centering \begin{tikzpicture}[x=0.75pt,y=0.75pt] \def\R{154.483} \def\ul{0.52} \def\az{15} \coordinate (O) at (0,0); \coordinate (A) at (0,100); \draw[dashed] (A) -- (0,-100); \path (O) node[left=4] {A}; \begin{scope}[xscale=1,yscale=1, rotate around={-\az:(O)}] ; \coordinate (B) at (0,100); \draw[gray, fill=green!30!white] (O)--(0,100)--(-5,100)--(-5,-100)--(0,-100)--cycle; \draw pic[<-,"\small\ang{\az}", draw=black, angle radius=70, % use 70, not 60 angle eccentricity=1.1]{angle=B--O--A}; \coordinate (R) at (\az:\R); \coordinate (X) at ({\R*cos(\az)},0); \coordinate (Y) at (0,{\R*sin(\az)}); \draw[projcol,dashed] (X) -- (R); \draw[projcol,dashed] (Y) -- (R); \draw[vector,red] (O) -- (R) node[right] {\qty{836}{\newton}}; \draw[vector,<->,projcol](X) %% Why 'below=-1" in the following line? Why not "below=1"? node[scale=0.9, below =-1, right] {\qty{807}{\newton}} -- (O) -- (Y) node[scale=0.9, below = 4, left] {\qty{216}{\newton}\kern0.4em}; \end{scope} \end{tikzpicture} %\end{minipage} \end{center} %\end{figure} \end{document}
Improve this answer
2
  • 1
    thnx. this latex is generated by C# code. So chopping off the decimals has to occur explicitly in either latex or C#. I found this: \def\Z{\SI[round-mode = places, round-precision = 0, round-integer-to-decimal]{15}{\degree}} \draw pic[<-,"\textbf{\tiny{\Z}}",draw=black,angle radius=60,angle eccentricity=1.1]{angle=B--O--A}; thnx for pointing out qty. Commented Feb 19, 2023 at 13:44
  • 1
    @Erik - round-mode = places, round-precision = 0 is a nice solution. Commented Feb 19, 2023 at 14:29

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.