7

I would like to put 1cm isometric dot paper into an exercise of the exam for my class, so my students can draw on. My apologies for no MWE. I haven't got a clue where to start. See image

enter image description here

3
  • When you say 1cm, is that the vertical or horizontal difference between two dots? Or is 1cm the vertical distance as well as the distance to the above left and above right dot?
    – TobiBS
    Aug 22, 2020 at 16:30
  • In isometry, the dot distances (let's call it the unit vectors) along the 30 deg lines and the vertical lines are the same. I guess it is this distance OP wants to be 1 cm.
    – AlexG
    Aug 22, 2020 at 16:35
  • Does this answer your question? creating a dotted paper pdf Related enough to get you started but not isometric dots.
    – Alan Munn
    Aug 22, 2020 at 16:36

4 Answers 4

8

Isometric dots can easily be created by modifying the x and y coordinates and then putting dots on a grid. However we would need to clip the result, because we basically rotated the coordinate system. The scale has to be adopted to your liking, as I asked how the 1cm shall be defined and now used the assumption of @AlexG.

\documentclass[tikz]{standalone}

\begin{document}
\begin{tikzpicture}[x={(0.86cm,0.5cm)},y={(-0.86cm,0.5cm)}]
\clip (0,12.5) rectangle (25,12.5);
\foreach \x in {0,...,25}
    \foreach \y in {0,...,25}
    {
    \fill (\x,\y) circle (2pt);
    }
\end{tikzpicture}
\end{document}

enter image description here

Edit after the comments of Jon

I took some measurements with Adobe Acrobat in the PDF created from the code above and this is the result: enter image description here

So which measurement needs to be 1cm, if you take this image as a reference? enter image description here

The isometric A4-paper

\documentclass[tikz,border={0.23cm 0.25cm}]{standalone}

\begin{document}
\begin{tikzpicture}[x={(0.86cm,0.5cm)},y={(-0.86cm,0.5cm)}]
\clip (0,25.5) rectangle (37.5,29);
\foreach \x in {0,...,50}
\foreach \y in {0,...,50}
{
    \fill (\x,\y) circle (2pt);
}
\end{tikzpicture}
\end{document}
10
  • I wanted 1 cm going across and down, also how did you know to use those particular coordinates (0.86cm. 05cm)?
    – Jon
    Aug 23, 2020 at 13:19
  • 1
    @Jon This is due to trigonometry: sin(60°) = 0.866 and cos(60°) = 0.5.
    – AJF
    Aug 23, 2020 at 14:07
  • @Jon AJ explained it already, it is based on the isometric projection. I have seen you tried to edit my answer instead of adding a comment, I will follow up on that later. Basically to not cut between dots, use a coordinate that doesn‘t cut through, so x.5 helps.
    – TobiBS
    Aug 23, 2020 at 14:50
  • TobiBS, I only understood your answer when I removed the clip I got a "rhombus" with the corners at (0,0), (0,25), (25,25), (25,0), then I \clip (0,16.5) rectangle (25,8.5); this gave me the top 10 cm for the top of my page, which I wanted, but when I print the whole page I do get isometric paper but 2 cm in length going across and down.I will wait for your follow up and then vote as solved for me
    – Jon
    Aug 24, 2020 at 14:16
  • @Jon Exactly, this is what you would do, basically the new definitions for x and y are defining a new coordinate system which is defined by new vectors. However I still don't get which measurement you want to be 1cm. Please see my edit of the answer, it shows an actual measurement from Adobe Acrobat. So which measurement do you want to change? Please indicate red, green or blue as in my second additional image.
    – TobiBS
    Aug 24, 2020 at 16:40
9

Update 2: new coordinate system defined with pgfkeys

Coordinates are given in the same style as the native implicit coordinates of tikz, i.e. 3 numbers separated by commas. They are prefixed by iso cs: as for example: (iso cs:0,1,7)

screenshot

\documentclass[tikz,border=5mm]{standalone}
\usetikzlibrary{arrows.meta}

\pgfkeys{/isometrique/.cd,
      coordonnee/.code args={#1,#2,#3}
        {
            \def\myx{#1}
            \def\myy{#2}
            \def\myz{#3}
        }
}

\tikzdeclarecoordinatesystem{isometric}
{
    \pgfkeys{/isometrique/.cd,
              coordonnee={#1}}
    \pgfpointadd{\pgfpointxyz{0}{\myz}{0}}{\pgfpointadd{\pgfpointpolarxy{-30}{\myx}}{\pgfpointpolarxy{30}{\myy}}}
}
\tikzaliascoordinatesystem{iso}{isometric}

\begin{document}
\begin{tikzpicture}[>={Triangle[angle=45:4pt 3]}]

\newcommand{\nbx}{11}%<--number of point on one row
\newcommand{\nby}{9}%<-- number of point on one column

\foreach \j in {0,...,\the\numexpr\nby-1} {
  \foreach \i in {0,...,\the\numexpr\nbx-1} 
  {\fill[black](90:\j)++(0:{2*\i*cos(30)})circle[radius=1pt]+(30:1)circle[radius=1pt];
}}

\draw[very thick,red,->](0,0)--node[sloped,below]{$y=6$}(iso cs:0,4,0);
\draw[very thick,blue,->](iso cs:0,4,0)-- node[sloped,above]{$x=2$}++(iso cs:2,0,0);
\draw[very thick,red,->](iso cs:2,4,0)-- node[sloped,below]{$z=3$}++(iso cs:0,0,3);

% Arrows showing the newest coordinate system "iso"
\draw [blue,thick,->](0,4)--node[below]{x}++(iso cs:1,0,0);
\draw [red,thick,->](0,4)--node[left]{y}++(iso cs:0,1,0);
\draw [violet,thick,->](0,4)--node[left]{z}++(iso cs:0,0,1);
\node[below,align=center,draw,fill=white] at (iso cs:0,1,2.7){New \textbf{iso} \\ coordinate system};

\begin{scope}[shift={(iso cs:2,4,3)}]
\draw[blue,thick] (iso cs:0,0,0)--++ (iso cs:3,0,0)
--++ (iso cs:0,3,0)
--++ (iso cs:0,0,3)
--++ (iso cs:-3,0,0)
--++ (iso cs:0,-3,0)
--++(iso cs:0,0,-3)
(0,3)--++(iso cs:3,0,0)--+(0,-3)
(iso cs:0,3,0)--+(iso cs:0,3,0);
\end{scope}
\end{tikzpicture}
\end{document}

Update Addition of another coordinate system with a vertical key z (at Tobi's request)

Its disadvantage is to be more verbose since you have to write 3 coordinates instead of 2.

With keyvals since here the keys are defined with keyval package, we can define default values and write for example (trio cs:x,y=2,z) instead of (trio cs:x=0,y=2,z=0). Here, the keys have default values, that is to say that if no value is given, they are worth the default value.

screenshot

\documentclass[tikz,border=5mm]{standalone}

%\usepackage{tikz}
\usetikzlibrary{arrows.meta}

\makeatletter
\define@key{triangularokeys}{x}[0]{\def\myx{#1}}
\define@key{triangularokeys}{y}[0]{\def\myy{#1}}
\define@key{triangularokeys}{z}[0]{\def\myz{#1}}
\tikzdeclarecoordinatesystem{triangularo}%
{%
\setkeys{triangularokeys}{#1}%
\pgfpointadd{\pgfpointxyz{0}{\myz}{0}}{\pgfpointadd{\pgfpointpolarxy{-30}{\myx}}{\pgfpointpolarxy{30}{\myy}}
}
}
\makeatother
\tikzaliascoordinatesystem{trio}{triangularo}
\begin{document}

\begin{tikzpicture}[>={Stealth[]}]

\newcommand{\nbx}{11}%<--number of point on one row
\newcommand{\nby}{9}%<-- number of point on one column


\foreach \j in {0,...,\the\numexpr\nby-1} {
  \foreach \i in {0,...,\the\numexpr\nbx-1} 
  {\fill[black](90:\j)++(0:{2*\i*cos(30)})circle[radius=1pt]+(30:1)circle[radius=1pt];
}}

\draw[very thick,red,->](0,0)--node[sloped,below]{$y=6$}(trio cs:x=0,y=4,z=0);
\draw[very thick,red,->](trio cs:x,y=4,z)-- node[sloped,above]{$x=2$}++(trio cs:x=2,y,z);
\draw[very thick,red,->](trio cs:x=2,y=4,z)-- node[sloped,below]{$z=3$}++(trio cs:x,y,z=3);


% Arrows showing the newest coordinate system "trio"
\draw [blue,thick,->](0,4)--node[below]{x}++(trio cs:x=1,y,z);
\draw [red,thick,->](0,4)--node[left]{y}++(trio cs:x,y=1,z);
\draw [violet,thick,->](0,4)--node[left]{z}++(trio cs:x,y,z=1);
\node[below,align=center] at (trio cs:x,y=1,z=3){New trio \\ coordinate system};

\begin{scope}[shift={(trio cs:x=2,y=4,z=3)}]
\draw[blue,thick] (trio cs:x,y,z)--++ (trio cs:x=3,y,z)
--++ (trio cs:x,y=3,z)
--++ (trio cs:x,y,z=3)
--++ (trio cs:x=-3,y,z)
--++ (trio cs:x,y=-3,z)
--++(trio cs:x,y,z=-3)
(0,3)--++(trio cs:x=3,y,z)--+(0,-3)
(trio cs:x,y=3,z)--+(trio cs:x,y=3,z);
\end{scope}
\end{tikzpicture}

\end{document}

First answer With a coordinate system called tri with the x and y keys.

screenshot

In addition to the Cartesian coordinates, I have defined a new coordinate system that makes it "simpler" to draw figures on this grid. It is called triangular and its alias is tri.

For example, the first red arrow is drawn like this:

\draw[very thick,red,->](0,0)--(tri cs:x=0,y=7);

The second arrow is defined as follows:

\draw[very thick,red,->](tri cs:x=0,y=7)--++(tri cs:x=2,y=0);

You'll notice that you can mix the two coordinate systems in the same path and use the relative coordinate.

Code

\documentclass[tikz,border=5mm]{standalone}

%\usepackage{tikz}
\usetikzlibrary{arrows.meta}

% new coordinate system called triangular 
\makeatletter
\define@key{triangularkeys}{x}{\def\myx{#1}}
\define@key{triangularkeys}{y}{\def\myy{#1}}
\tikzdeclarecoordinatesystem{triangular}%
{%
\setkeys{triangularkeys}{#1}%
\pgfpointadd{\pgfpointpolarxy{-30}{\myx}}{\pgfpointpolarxy{30}{\myy}}
}
\makeatother
% end of new coordinate system 

\tikzaliascoordinatesystem{tri}{triangular}%<-- define the alias tri for triangular
   

\begin{document}

\begin{tikzpicture}[>={Stealth[]}]

\newcommand{\nbx}{11}%<--number of dots in a single row
\newcommand{\nby}{9}%<-- number of dots in a single column

% Drawing of the isometric grid
\foreach \j in {0,...,\the\numexpr\nby-1} {
  \foreach \i in {0,...,\the\numexpr\nbx-1} 
  {\fill[black](90:\j)++(0:{2*\i*cos(30)})circle[radius=1pt]+(30:1)circle[radius=1pt];
}}

% The following code below shows how to draw on this grid

% Arrows showing the new coordinate system
\draw [blue,thick,->](0,4)--node[below]{x}++(tri cs:x=1,y=0);
\draw [red,thick,->](0,4)--node[left]{y}++(tri cs:x=0,y=1);

% Big red arrow going from the bottom left to the perspective cube
\draw[very thick,red,->](0,0)--node[sloped,below]{$y=7$}(tri cs:x=0,y=7);
\draw[very thick,red,->](tri cs:x=0,y=7)-- node[sloped,above]{$x=2$}++(tri cs:x=2,y=0);

% Cube perspective drawing
\begin{scope}[shift={(tri cs:x=2,y=7)}]
\draw (tri cs:x=0,y=0)circle(3pt)--++ (tri cs:x=3,y=0)
--++ (tri cs:x=0,y=3)
--++ (0,3)
--++ (tri cs:x=-3,y=0)
--++ (tri cs:x=0,y=-3)
--++(0,-3)
(0,3)--++(tri cs:x=3,y=0)--+(0,-3)
(tri cs:x=0,y=3)--+(tri cs:x=0,y=3);
\end{scope}


\end{tikzpicture}

\end{document}
15
  • That’s a nice solution, but shouldn’t there be also a z key?!
    – Tobi
    Aug 23, 2020 at 8:29
  • Why not, but I'm not sure what use would be made of it? What's your idea about that?
    – AndréC
    Aug 23, 2020 at 9:15
  • 1
    It could be used for al lines that go straight upwards. As the isometric view is a projection of 3D in 2D all three axis should be available. I’d say …
    – Tobi
    Aug 23, 2020 at 10:06
  • @Tobi I just finished it, but its disadvantage is that it is more verbose since you now have to write 3 coordinates instead of 2.
    – AndréC
    Aug 23, 2020 at 12:06
  • 1
    @Sebastiano I am sorry that you are grieving. I hope that you will get through this ordeal, whatever it is.
    – AndréC
    Apr 5, 2021 at 20:13
9

Just for fun, a pure PostScript solution for making 1-cm-scale isometric dotted paper in A4 (595 bp * 842 bp). Can be directly sent to a PostScript printer.

Use ps2pdf if you need a PDF; but it is much bigger [38 kB] than the PS [242 B]. (The PS code was somewhat optimized for size, though not too aggressively in order not to sacrifice legibility.)

isometricdottedA4.ps:

%! 
<</PageSize [595 842]>> setpagedevice 
/cm {28.346457 mul} def 
[.866 .5 -.866 .5 595 2 div 842 41 cm sub 2 div] concat 
0 1 41 { cm 
 0 1 41 { cm 1 index exch moveto 
  gsave initmatrix currentpoint 2 0 360 arc fill grestore 
 } for pop 
} for 

enter image description here

3
  • +1 It's interesting, but how do you actually print a postscript file from, say, a Windows 10 computer?
    – AndréC
    Aug 23, 2020 at 9:17
  • No idea, I can only speak for Linux where printing is centred around the PS format: lpr isometricdottedA4.ps (Of course, the default printer must be a PS printer. But most Laser printers in office environments, e. g. HP LaserJets, are.)
    – AlexG
    Aug 23, 2020 at 9:24
  • 1
    @AndréC : Found this on the web: COPY file.ps \\servername\printername. Seems to be the way to print a PS file on a PS network printer on Windows, at the DOS prompt.
    – AlexG
    Aug 23, 2020 at 9:43
2

A PSTricks solution only for either fun or comparison purposes.

\documentclass[pstricks,border=12pt]{standalone}
\begin{document}
\begin{pspicture}(10,10)
    \multips(0,0)(0,1){11}{%
        \multips(0,0)(1,0){11}{%
            \qdisk(0,0){2pt}\qdisk(.5,.5){2pt}}}
\end{pspicture}
\end{document}

enter image description here

4
  • Yes, fun! hahahaha (fun). :-(
    – user213378
    Aug 23, 2020 at 12:39
  • @user213378: Chishijimotoji. Aug 23, 2020 at 12:41
  • This grid is not isometric (triangles are not equilateral).
    – AndréC
    Aug 23, 2020 at 15:05
  • @AndréC: A good comment. Thank you! Aug 23, 2020 at 15:45

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