9

I hope the solution to my problem is not too complicated or tedious. I am drawing a diagram which has one nuisance in it:

enter image description here

Every time the sine curve touches the axis TiKz tries to draw a line of 0 length and therefore it produces the arrow tip at that location

How do I remove these arrow tips, while not listing out each individual line (i.e. using something like \foeach \x in ... or similar. So essentially remove every arrow that is a multiple of 180

enter image description here

Here is the code for drawing the diagram:

\documentclass[10pt]{article}
\usepackage{tikz}
\usetikzlibrary{calc,patterns}
\usepackage{tikz-3dplot}
\usepackage[left=0.00cm, right=0.00cm]{geometry}
\usepackage{physics}
\usepackage{bm}
\usepackage{rotating}

%Defining Diagonal Arrows:
\newcommand{\neswarrow}{%
    \begin{turn}{45}
        \raisebox{-1ex}{$\leftrightarrow$}
    \end{turn}
}
\newcommand{\nwsearrow}{%
    \begin{turn}{45}
        \raisebox{0ex}{$\updownarrow$}
    \end{turn}
}
\begin{document}
        \tdplotsetmaincoords{75}{135}
\begin{tikzpicture}[tdplot_main_coords,scale=0.5]
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=0:720,samples=360] (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});
\foreach \x in {45,90,...,720} { %LOOK HERE FOR FIRST SET OF ARROWS
    \draw[-latex,help lines,red] (0,{\x*(2*3/360},0) -- (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});}
\draw[very thick,red,latex-latex,densely dashed] (-2,12,2) -- (2,12,-2);
\begin{scope}[canvas is xz plane at y=12,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\node[anchor=south,transform shape,scale=1.5] at (0,3.5) {\large Polariser};
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (-90:1.5) arc (-90:-135:1.5) -- cycle;
\draw[blue] (-112.5:1.3) ..controls +(-112.5:0.7) and +(180:0.7).. (0.7,-1.7) node[right] {$\alpha$};
\end{scope}
\draw[very thick] (0,12,-3) -- (0,12,3);
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=720:1440,samples=360] (0,{\x*(2*3/360)},{2*sin(\x)});
\foreach \x in {720,765,...,1440} {%LOOK HERE FOR SECOND SET OF ARROWS
    \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- (0,{\x*(2*3/360)},{2*sin(\x)});}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,densely dashed,latex-latex,red] (-90:2) -- (90:2);
\end{scope}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (90:1) arc (90:30:1) -- cycle;
\node[blue] at (60:1.4) {$\theta$};
\draw[very thick] (-150:3) -- (30:3);
\node[anchor=south,transform shape,scale=1.5] at (90:3.5) {\large Analyser};
\end{scope}
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=1440:1800,samples=360] ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});
\foreach \x in {1440,1485,...,1800} { %LOOK HERE FOR THIRD SET OF ARROWS
    \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});}
\draw[-latex] (0,0,0) -- (0,32,0);
\draw[-latex] (0,0,-3) -- (0,0,3);
\draw[-latex] (3,0,0) -- (-3,0,0);
\begin{scope}[canvas is xz plane at y=1.5,xscale=-1]
\draw[red] (2,2) .. controls +(45:0.5) and +(-120:0.5).. (3,2.7);
\end{scope}
\node[scale=0.75,red] at (-3,4,3.7) {$\bm{E}=A_{\neswarrow}\cos\left(2\pi f t+\phi_{\neswarrow}\right)\ket{\neswarrow}+0\ket{\nwsearrow}$};
\draw (0,12,-2.7) .. controls +(0:1) and +(-135:2).. (4,12,-3) node[below,scale=0.75] {Axis of Polarisation};
\begin{scope}[canvas is xz plane at y=16.5,xscale=-1]
\draw[red] (0,-2) .. controls +(-90:1) and +(0:0.5).. (-2,-4.4) node[left,scale=0.75] {$\bm{E}=A_{\neswarrow}\cos(\alpha)\cos\left(2\pi ft+\phi_{\neswarrow}\right)\ket{\updownarrow} + 0\ket{\leftrightarrow}$};
\end{scope}
\end{tikzpicture}
\tdplotsetmaincoords{0}{0}
\begin{tikzpicture}[remember picture,overlay]
\begin{scope}[xshift=-5.8cm,yshift=4cm]
\draw (1,1) circle (1.1);
\draw[red,thick,latex-latex] (0.5,0.5) -- (1.5,1.5);
\draw[thick] (1,0.25) -- (1,1.75);
\draw[red,densely dashed,thick] (1.5,1.5) -- (1,1.5) (1,0.5) -- (0.5,0.5);
\node[left,scale=0.6,red,fill=white] at (1,1.5) {$A_{\neswarrow}\cos(\alpha)$};
\draw[blue,pattern=north west lines,pattern color=blue] (1,1) -- +(45:0.25) arc (45:90:0.25) -- cycle;
\draw[blue] (1,1) +(67.5:0.2) ..controls +(45:0.3) and +(180:0.1).. (1.4,1) node[right] {$\alpha$};
\node[below,scale=0.75] at (1,-0.1) {Polariser};
\end{scope}
\begin{scope}[xshift=-0.8cm,yshift=4.1cm,rotate=-60]
\draw (0,0) circle (1.1);
\draw[red,thick,latex-latex] (150:0.7071067812) -- (-30:0.7071067812);
\draw[thick] (0,-0.75) -- (0,0.75);
\draw[red,densely dashed,thick] (150:0.7071067812) -- (0,0.3535533906) (0,-0.3535533906) -- (-30:0.7071067812);
\draw[blue,pattern=north west lines,pattern color=blue] (0,0) -- (150:0.25) arc (150:90:0.25) -- cycle;
\draw[blue] (120:0.2) ..controls +(120:0.3) and +(-90:0.1).. (-0.3,0.6) node[above right=-0.07cm] {$\theta$};
\node[below,scale=0.75] at (-30:1.1) {Analyser};
\end{scope}
\end{tikzpicture}
\end{document}

Sorry about the code being very messy, I have very bad habits when it comes to writing in LaTeX. I have annotated the 3 lines of code which produce the arrows that define the wave.

Me just thinking about the problem

I think maybe the use of a double \foreach stack could work. The first one where some variable \a in a list {45,225,405} and then under that something like \foreach \x in {\A,\A+45,\A+90}{...} so it would look like this:

\foreach \a in {45,225,405} {
    \foreach \x in {{\a},{\a +45},{\a +90}}{
    \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- (0,{\x*(2*3/360)},{2*sin(\x)});
    }
}

For the first set of helplines.

Unfortunately this attempt does not work for me

EDIT: It does work (see answer below), but is there a more efficient or neat way of solving this problem, one that works without knowing the location of the 0 tip arrows (for example if they were randomly generated)?

  • In your double \foreach put \x in parentheses (0,{(\x)*(2*3)/360},0) -- (0,{(\x)*(2*3/360)},{2*sin(\x)}). And you don't need curly brackets in {\a,\a+45,\a+90}. – Kpym Jan 12 at 14:35
  • Yes I just found that out now instead you require parenthesis, I will post a solution very soon – sab hoque Jan 12 at 14:37
  • 2
    you can try \pgfmathparse{equal(mod(\x,180),0)} \ifnum\pgfmathresult=0 \draw[-latex,help lines,red] (0,{\x*(2*3/360},0) -- (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)}); \fi – touhami Jan 12 at 14:52
6

Here is an alternative. Whether or not it is more tidy, I don't know. In principle, tips=on proper draw from p. 187 of the pgfmanual, which Paul Gaborit pointed out here, should do the trick. But I couldn't make this work, so I implemented a length check by hand. This might be more useful in situations in which it is not so easy to seen when the zero-length paths occur analytically.

\documentclass[10pt]{article}
\usepackage{tikz}
\usetikzlibrary{calc,patterns}
\usepackage{tikz-3dplot}
\usepackage[left=0.00cm, right=0.00cm]{geometry}
\usepackage{physics}
\usepackage{bm}
\usepackage{rotating}

%Defining Diagonal Arrows:
\newcommand{\neswarrow}{%
    \begin{turn}{45}
        \raisebox{-1ex}{$\leftrightarrow$}
    \end{turn}
}
\newcommand{\nwsearrow}{%
    \begin{turn}{45}
        \raisebox{0ex}{$\updownarrow$}
    \end{turn}
}
\begin{document}
        \tdplotsetmaincoords{75}{135}
\begin{tikzpicture}[tdplot_main_coords,scale=0.5]
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=0:720,samples=360] (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});
\foreach \X in {45,90,...,720} { %LOOK HERE FOR FIRST SET OF ARROWS
    \path let \p1=($(0,{\X*(2*3/360},0) -
    (-{2*sin(\X)},{\X*(2*3/360)},{2*sin(\X)})$),\n1={veclen(\x1,\y1)} in 
    \pgfextra{\xdef\mylen{\n1}};
    \ifdim\mylen>1pt
    \draw[-latex,help lines,red] (0,{\X*(2*3/360},0) -- (-{2*sin(\X)},{\X*(2*3/360)},{2*sin(\X)});
    \fi
    }
\draw[very thick,red,latex-latex,densely dashed] (-2,12,2) -- (2,12,-2);
\begin{scope}[canvas is xz plane at y=12,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\node[anchor=south,transform shape,scale=1.5] at (0,3.5) {\large Polariser};
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (-90:1.5) arc (-90:-135:1.5) -- cycle;
\draw[blue] (-112.5:1.3) ..controls +(-112.5:0.7) and +(180:0.7).. (0.7,-1.7) node[right] {$\alpha$};
\end{scope}
\draw[very thick] (0,12,-3) -- (0,12,3);
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=720:1440,samples=360] (0,{\x*(2*3/360)},{2*sin(\x)});
\foreach \X in {720,765,...,1440} {%LOOK HERE FOR SECOND SET OF ARROWS
    \path let \p1=($(0,{\X*(2*3)/360},0) - (0,{\X*(2*3/360)},{2*sin(\X)})$),
      \n1={veclen(\x1,\y1)} in  \pgfextra{\xdef\mylen{\n1}};
    \ifdim\mylen>1pt    
    \draw[-latex,help lines,red,] (0,{\X*(2*3)/360},0) -- (0,{\X*(2*3/360)},{2*sin(\X)});
    \fi}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,densely dashed,latex-latex,red] (-90:2) -- (90:2);
\end{scope}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (90:1) arc (90:30:1) -- cycle;
\node[blue] at (60:1.4) {$\theta$};
\draw[very thick] (-150:3) -- (30:3);
\node[anchor=south,transform shape,scale=1.5] at (90:3.5) {\large Analyser};
\end{scope}
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=1440:1800,samples=360] ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});
\foreach \X in {1440,1485,...,1800} { %LOOK HERE FOR THIRD SET OF ARROWS
    \path let \p1=($(0,{\X*(2*3)/360},0) -  ({-0.7071067812*sin(\X)},{\X*(2*3/360)},{0.5*sin(\X)})$),   
    \n1={veclen(\x1,\y1)} in    
    \pgfextra{\xdef\mylen{\n1}};
    \ifdim\mylen>1pt
    \draw[-latex,help lines,red] (0,{\X*(2*3)/360},0) -- ({-0.7071067812*sin(\X)},{\X*(2*3/360)},{0.5*sin(\X)});
    \fi}
\draw[-latex] (0,0,0) -- (0,32,0);
\draw[-latex] (0,0,-3) -- (0,0,3);
\draw[-latex] (3,0,0) -- (-3,0,0);
\begin{scope}[canvas is xz plane at y=1.5,xscale=-1]
\draw[red] (2,2) .. controls +(45:0.5) and +(-120:0.5).. (3,2.7);
\end{scope}
\node[scale=0.75,red] at (-3,4,3.7) {$\bm{E}=A_{\neswarrow}\cos\left(2\pi f t+\phi_{\neswarrow}\right)\ket{\neswarrow}+0\ket{\nwsearrow}$};
\draw (0,12,-2.7) .. controls +(0:1) and +(-135:2).. (4,12,-3) node[below,scale=0.75] {Axis of Polarisation};
\begin{scope}[canvas is xz plane at y=16.5,xscale=-1]
\draw[red] (0,-2) .. controls +(-90:1) and +(0:0.5).. (-2,-4.4) node[left,scale=0.75] {$\bm{E}=A_{\neswarrow}\cos(\alpha)\cos\left(2\pi ft+\phi_{\neswarrow}\right)\ket{\updownarrow} + 0\ket{\leftrightarrow}$};
\end{scope}
\end{tikzpicture}
\tdplotsetmaincoords{0}{0}
\begin{tikzpicture}[remember picture,overlay]
\begin{scope}[xshift=-5.8cm,yshift=4cm]
\draw (1,1) circle (1.1);
\draw[red,thick,latex-latex] (0.5,0.5) -- (1.5,1.5);
\draw[thick] (1,0.25) -- (1,1.75);
\draw[red,densely dashed,thick] (1.5,1.5) -- (1,1.5) (1,0.5) -- (0.5,0.5);
\node[left,scale=0.6,red,fill=white] at (1,1.5) {$A_{\neswarrow}\cos(\alpha)$};
\draw[blue,pattern=north west lines,pattern color=blue] (1,1) -- +(45:0.25) arc (45:90:0.25) -- cycle;
\draw[blue] (1,1) +(67.5:0.2) ..controls +(45:0.3) and +(180:0.1).. (1.4,1) node[right] {$\alpha$};
\node[below,scale=0.75] at (1,-0.1) {Polariser};
\end{scope}
\begin{scope}[xshift=-0.8cm,yshift=4.1cm,rotate=-60]
\draw (0,0) circle (1.1);
\draw[red,thick,latex-latex] (150:0.7071067812) -- (-30:0.7071067812);
\draw[thick] (0,-0.75) -- (0,0.75);
\draw[red,densely dashed,thick] (150:0.7071067812) -- (0,0.3535533906) (0,-0.3535533906) -- (-30:0.7071067812);
\draw[blue,pattern=north west lines,pattern color=blue] (0,0) -- (150:0.25) arc (150:90:0.25) -- cycle;
\draw[blue] (120:0.2) ..controls +(120:0.3) and +(-90:0.1).. (-0.3,0.6) node[above right=-0.07cm] {$\theta$};
\node[below,scale=0.75] at (-30:1.1) {Analyser};
\end{scope}
\end{tikzpicture}
\end{document}

enter image description here

  • Just trying to understand the if statement. What does the line \n1={veclen(\x1,\y1)} mean. I know that you are defining a vector from the y axis to the wave, but where exactly is x1 and y1 because they're not referenced anywhere else in the code? – sab hoque Jan 12 at 22:52
  • @sabhoque This just measures the length of the line you are about to add an arrow tip to. If the length is below (the arbitrarily chosen scale of) 1pt, it won't draw the arrow. In more detail, \p1 is the vector from the start to the end of the line, and correspondingly \x1 and \y1 are its coordinates and therefore \n1={veclen(\x1,\y1)} its length. – marmot Jan 12 at 23:05
  • oh I see and is there a reason for using \n1, \x1 and \y1 as opposed to \n, \x, and \y – sab hoque Jan 12 at 23:12
  • 1
    @sabhoque Yes. The calc syntax requires you to name the points/vectors \p1, \p2 etc., and their coordinates will then be stored in \x1,\y1, \x2,\y2 and so on. (This is, BTW, the reason why I prefer to call the \foreach variable \X rather than \x. A \x interferes with this notation, as well as the standard variable in plots.) – marmot Jan 12 at 23:19
7

Using the \foreach in a double stack works. Essentially like this:

\foreach \a in {1485,1665} {
    \foreach \x in {{\a},(\a +45),(\a +90)}{
        \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});
    }
}

So the final code looks like this:

\documentclass[10pt]{article}
\usepackage{tikz}
\usetikzlibrary{calc,patterns}
\usepackage{tikz-3dplot}
\usepackage[left=0.00cm, right=0.00cm]{geometry}
\usepackage{physics}
\usepackage{bm}
\usepackage{rotating}

%Defining Diagonal Arrows:
\newcommand{\neswarrow}{%
    \begin{turn}{45}
        \raisebox{-1ex}{$\leftrightarrow$}
    \end{turn}
}
\newcommand{\nwsearrow}{%
    \begin{turn}{45}
        \raisebox{0ex}{$\updownarrow$}
    \end{turn}
}
\begin{document}
        \tdplotsetmaincoords{75}{135}
\begin{tikzpicture}[tdplot_main_coords,scale=0.5]
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=0:720,samples=360] (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});
\foreach \a in {45,225,405,585} {
    \foreach \x in {{\a},(\a +45),(\a +90)}{
        \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- ({-2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});
    }
}
\draw[very thick,red,latex-latex,densely dashed] (-2,12,2) -- (2,12,-2);
\begin{scope}[canvas is xz plane at y=12,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\node[anchor=south,transform shape,scale=1.5] at (0,3.5) {\large Polariser};
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (-90:1.5) arc (-90:-135:1.5) -- cycle;
\draw[blue] (-112.5:1.3) ..controls +(-112.5:0.7) and +(180:0.7).. (0.7,-1.7) node[right] {$\alpha$};
\end{scope}
\draw[very thick] (0,12,-3) -- (0,12,3);
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=720:1440,samples=360] (0,{\x*(2*3/360)},{2*sin(\x)});
\foreach \a in {765,945,1125,1305} {
    \foreach \x in {{\a},(\a +45),(\a +90)}{
        \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- (0,{\x*(2*3/360)},{2*sin(\x)});
    }
}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,densely dashed,latex-latex,red] (-90:2) -- (90:2);
\end{scope}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (90:1) arc (90:30:1) -- cycle;
\node[blue] at (60:1.4) {$\theta$};
\draw[very thick] (-150:3) -- (30:3);
\node[anchor=south,transform shape,scale=1.5] at (90:3.5) {\large Analyser};
\end{scope}
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=1440:1800,samples=360] ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});
\foreach \a in {1485,1665} {
    \foreach \x in {{\a},(\a +45),(\a +90)}{
        \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});
    }
}
\draw[-latex] (0,0,0) -- (0,32,0);
\draw[-latex] (0,0,-3) -- (0,0,3);
\draw[-latex] (3,0,0) -- (-3,0,0);
\begin{scope}[canvas is xz plane at y=1.5,xscale=-1]
\draw[red] (2,2) .. controls +(45:0.5) and +(-120:0.5).. (3,2.7);
\end{scope}
\node[scale=0.75,red] at (-3,4,3.7) {$\bm{E}=A_{\neswarrow}\cos\left(2\pi f t+\phi_{\neswarrow}\right)\ket{\neswarrow}+0\ket{\nwsearrow}$};
\draw (0,12,-2.7) .. controls +(0:1) and +(-135:2).. (4,12,-3) node[below,scale=0.75] {Axis of Polarisation};
\begin{scope}[canvas is xz plane at y=16.5,xscale=-1]
\draw[red] (0,-2) .. controls +(-90:1) and +(0:0.5).. (-2,-4.4) node[left,scale=0.75] {$\bm{E}=A_{\neswarrow}\cos(\alpha)\cos\left(2\pi ft+\phi_{\neswarrow}\right)\ket{\updownarrow} + 0\ket{\leftrightarrow}$};
\end{scope}
\end{tikzpicture}
\tdplotsetmaincoords{0}{0}
\begin{tikzpicture}[remember picture,overlay]
\begin{scope}[xshift=-5.8cm,yshift=4cm]
\draw (1,1) circle (1.1);
\draw[red,thick,latex-latex] (0.5,0.5) -- (1.5,1.5);
\draw[thick] (1,0.25) -- (1,1.75);
\draw[red,densely dashed,thick] (1.5,1.5) -- (1,1.5) (1,0.5) -- (0.5,0.5);
\node[left,scale=0.6,red,fill=white] at (1,1.5) {$A_{\neswarrow}\cos(\alpha)$};
\draw[blue,pattern=north west lines,pattern color=blue] (1,1) -- +(45:0.25) arc (45:90:0.25) -- cycle;
\draw[blue] (1,1) +(67.5:0.2) ..controls +(45:0.3) and +(180:0.1).. (1.4,1) node[right] {$\alpha$};
\node[below,scale=0.75] at (1,-0.1) {Polariser};
\end{scope}
\begin{scope}[xshift=-0.8cm,yshift=4.1cm,rotate=-60]
\draw (0,0) circle (1.1);
\draw[red,thick,latex-latex] (150:0.7071067812) -- (-30:0.7071067812);
\draw[thick] (0,-0.75) -- (0,0.75);
\draw[red,densely dashed,thick] (150:0.7071067812) -- (0,0.3535533906) (0,-0.3535533906) -- (-30:0.7071067812);
\draw[blue,pattern=north west lines,pattern color=blue] (0,0) -- (150:0.25) arc (150:90:0.25) -- cycle;
\draw[blue] (120:0.2) ..controls +(120:0.3) and +(-90:0.1).. (-0.3,0.6) node[above right=-0.07cm] {$\theta$};
\node[below,scale=0.75] at (-30:1.1) {Analyser};
\end{scope}
\end{tikzpicture}
\end{document}

And that fixes my problem:

enter image description here

Still wondering if there is something more tidy, neat or general than my solution?

5

I used the tip= on proper draw key indicated by @marmot in his answer.

  • It works fine for the first loop.
  • But for the other two, I had to rewrite the coordinates and divide by 60 instead of multiplying by 6 and then dividing by 360, which gives the right result.

Instead of:

 \draw[-latex,help lines,red] (0,{\x*(2*3)/360},0) -- (0,{\x*(2*3/360)},{2*sin(\x)});

I write:

\draw[-latex,help lines,red,tips=on proper draw] (0,{\x/60},0) -- (0,{\x/60},{2*sin(\x)});

I suppose this is a problem related to the way TeX is calculated.

screenshot

\documentclass[10pt]{article}
\usepackage{tikz}
\usetikzlibrary{calc,patterns}
\usepackage{tikz-3dplot}
\usepackage[left=0.00cm, right=0.00cm]{geometry}
\usepackage{physics}
\usepackage{bm}
\usepackage{rotating}

%Defining Diagonal Arrows:
\newcommand{\neswarrow}{%
    \begin{turn}{45}
        \raisebox{-1ex}{$\leftrightarrow$}
    \end{turn}
}
\newcommand{\nwsearrow}{%
    \begin{turn}{45}
        \raisebox{0ex}{$\updownarrow$}
    \end{turn}
}
\begin{document}
        \tdplotsetmaincoords{75}{135}
\begin{tikzpicture}[tdplot_main_coords,scale=0.5]
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=0:720,samples=360] (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});

\foreach \x in {45,90,...,720} { %LOOK HERE FOR FIRST SET OF ARROWS
    \draw[-latex,help lines,red,tips=on proper draw] (0,{\x*(2*3/360},0) -- (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});}

\draw[very thick,red,latex-latex,densely dashed] (-2,12,2) -- (2,12,-2);
\begin{scope}[canvas is xz plane at y=12,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\node[anchor=south,transform shape,scale=1.5] at (0,3.5) {\large Polariser};
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (-90:1.5) arc (-90:-135:1.5) -- cycle;
\draw[blue] (-112.5:1.3) ..controls +(-112.5:0.7) and +(180:0.7).. (0.7,-1.7) node[right] {$\alpha$};
\end{scope}
\draw[very thick] (0,12,-3) -- (0,12,3);
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=720:1440,samples=360] (0,{\x*(2*3/360)},{2*sin(\x)});

\foreach \x in {720,765,...,1440} {%LOOK HERE FOR SECOND SET OF ARROWS
    \draw[-latex,help lines,red,tips=on proper draw] (0,{\x/60},0) -- (0,{\x/60},{2*sin(\x)});}

\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,densely dashed,latex-latex,red] (-90:2) -- (90:2);
\end{scope}
\begin{scope}[canvas is xz plane at y=24,xscale=-1]
\draw[very thick,fill=white,fill opacity=0.6] (0,0) circle (3.5);
\draw[blue,pattern=north west lines, pattern color=blue] (0,0) --  (90:1) arc (90:30:1) -- cycle;
\node[blue] at (60:1.4) {$\theta$};
\draw[very thick] (-150:3) -- (30:3);
\node[anchor=south,transform shape,scale=1.5] at (90:3.5) {\large Analyser};
\end{scope}
\draw[red,very thick,-latex] plot[smooth,variable=\x,domain=1440:1800,samples=360] ({-0.7071067812*sin(\x)},{\x*(2*3/360)},{0.5*sin(\x)});

\foreach \x in {1440,1485,...,1800} { %LOOK HERE FOR THIRD SET OF ARROWS
    \draw[-latex,help lines,red,tips=on proper draw] (0,{\x*(2*3)/360},0) -- ({-0.7071067812*sin(\x)},{\x/60},{0.5*sin(\x)});}

\draw[-latex] (0,0,0) -- (0,32,0);
\draw[-latex] (0,0,-3) -- (0,0,3);
\draw[-latex] (3,0,0) -- (-3,0,0);
\begin{scope}[canvas is xz plane at y=1.5,xscale=-1]
\draw[red] (2,2) .. controls +(45:0.5) and +(-120:0.5).. (3,2.7);
\end{scope}
\node[scale=0.75,red] at (-3,4,3.7) {$\bm{E}=A_{\neswarrow}\cos\left(2\pi f t+\phi_{\neswarrow}\right)\ket{\neswarrow}+0\ket{\nwsearrow}$};
\draw (0,12,-2.7) .. controls +(0:1) and +(-135:2).. (4,12,-3) node[below,scale=0.75] {Axis of Polarisation};
\begin{scope}[canvas is xz plane at y=16.5,xscale=-1]
\draw[red] (0,-2) .. controls +(-90:1) and +(0:0.5).. (-2,-4.4) node[left,scale=0.75] {$\bm{E}=A_{\neswarrow}\cos(\alpha)\cos\left(2\pi ft+\phi_{\neswarrow}\right)\ket{\updownarrow} + 0\ket{\leftrightarrow}$};
\end{scope}
\end{tikzpicture}
\tdplotsetmaincoords{0}{0}
\begin{tikzpicture}[remember picture,overlay]
\begin{scope}[xshift=-5.8cm,yshift=4cm]
\draw (1,1) circle (1.1);
\draw[red,thick,latex-latex] (0.5,0.5) -- (1.5,1.5);
\draw[thick] (1,0.25) -- (1,1.75);
\draw[red,densely dashed,thick] (1.5,1.5) -- (1,1.5) (1,0.5) -- (0.5,0.5);
\node[left,scale=0.6,red,fill=white] at (1,1.5) {$A_{\neswarrow}\cos(\alpha)$};
\draw[blue,pattern=north west lines,pattern color=blue] (1,1) -- +(45:0.25) arc (45:90:0.25) -- cycle;
\draw[blue] (1,1) +(67.5:0.2) ..controls +(45:0.3) and +(180:0.1).. (1.4,1) node[right] {$\alpha$};
\node[below,scale=0.75] at (1,-0.1) {Polariser};
\end{scope}
\begin{scope}[xshift=-0.8cm,yshift=4.1cm,rotate=-60]
\draw (0,0) circle (1.1);
\draw[red,thick,latex-latex] (150:0.7071067812) -- (-30:0.7071067812);
\draw[thick] (0,-0.75) -- (0,0.75);
\draw[red,densely dashed,thick] (150:0.7071067812) -- (0,0.3535533906) (0,-0.3535533906) -- (-30:0.7071067812);
\draw[blue,pattern=north west lines,pattern color=blue] (0,0) -- (150:0.25) arc (150:90:0.25) -- cycle;
\draw[blue] (120:0.2) ..controls +(120:0.3) and +(-90:0.1).. (-0.3,0.6) node[above right=-0.07cm] {$\theta$};
\node[below,scale=0.75] at (-30:1.1) {Analyser};
\end{scope}
\end{tikzpicture}
\end{document}
  • That is quite strange. I had a 2*3 because I originally had something like \length=2 and period=6 at the beginning, which I guess explains the origin of (2*3)/360. – sab hoque Jan 12 at 22:58
  • 1
    @sabhoque No, it's not strange, you multiply \x by 1/60. So TeX first calculates 1/60 and gets an approximation. The error made in this approximation is then multiplied by \x and becomes non-zero for large values of \x. By dividing by 60, the error is always smaller. – AndréC Jan 13 at 5:27
  • So something like \length*\x*(1/360) would work – sab hoque Jan 16 at 4:05
5

You can use xintexpr for that, but surely your nested \foreach is more natural in your context.

\edef\mylist{\xinttheiiexpr seq((x/:180)?{x}{omit}, x = 45..[45]..720)\relax}
\foreach \x in \mylist { %LOOK HERE FOR FIRST SET OF ARROWS
    \draw[-latex,help lines,red] (0,{\x*(2*3/360},0) -- (-{2*sin(\x)},{\x*(2*3/360)},{2*sin(\x)});}

Explanation: x/:180 is the remainder, we want to eliminate when it is zero, this is what the ? operator and the omit keyword do. And of course 45..[45]..720 is generator of arithmetic sequence.

It is possible to use \xintFor syntax, but it expands only once its list argument, so like this is needed:

\xintFor #1 in {\romannumeral-`0\xintiieval{seq((x/:180)?{x}{omit}, x = 45..[45]..720)}}:
{ %LOOK HERE FOR FIRST SET OF ARROWS
    \draw[-latex,help lines,red] (0,{#1*(2*3/360},0) -- (-{2*sin(#1)},{#1*(2*3/360)},{2*sin(#1)});
}

The \xintiieval requires xint 1.3d. Else use \xinttheiiexpr, no braces, and \relax at the end.

As I said, this is only because of a rainy afternoon. Double \foreach avoids extra package...

enter image description here

  • I didn't really like my double foreach solution because what if I didn't know where the zero length arrows were? I would require something like the your, Marmot's or AndreC's answer. – sab hoque Jan 12 at 22:49
  • Also for future reference, is it possible to make it omit say every time the remainder is 90 when divided by 180? – sab hoque Jan 12 at 23:01
  • 1
    @sabhoque, yes, use for example seq((x/:180 != 90)?{x}{omit}, x =, or seq((x/:180 == 90)?{omit}{x}, x =, or seq(((x-90)/:180)?{x}{omit}, x =, the latter because /:180 is really periodical modulo 180 you don't have to worry about using it with a negative argument, it is really the mathematical remainder in the sense of the mathematicians. (actually that remark applies to x/:180 != 90 test too, which translates into "if the remainder is not 90 then choose first branch else second branch``.) – user4686 Jan 13 at 8:39

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