[Tex/LaTex] xyz coordinates rotation

3dtikz-pgf

is it possible to set a different orthonormal base for the xyz coordinate space? for example, the default one is x=(1,0,0) y=(0,1,0) and z=(0,0,1).

I would want to set this base: x=(1,0,0) y=(0,0,1) and z=(0,1,0) so that when i draw a line from (0,0,0) to (0,1,0) it will be rendered as a line from (0,0,0) to (0,0,1) using default xyz coordinates.

i tried setting

    \begin{tikzpicture}[x={(1,0,0)},y={(0,0,1)},z={(0,1,0)}]

but it seems only to distort the drawing.
I would like to avoid using pgfplots, if possible.

thank you in advance for your attention

Best Answer

According to the manual, the default z unit vector is (-3.85mm, -3.85mm). If you set y={(-3.85mm, -3.85mm)} and z={(0cm,1cm)} the default y and z axis will be switched, if I'm not mistaken. TikZ draws everything in 2D space, so the unit vectors are 2D as well.

Related Solutions

[Tex/LaTex] TikZ: Cropping the Bounding Box

The code adds some completely useless invisible (or rather white) stuff. The lines

\clip(0pt,403pt) -- (389.957pt,403pt) -- (389.957pt,99.6166pt) -- (0pt,99.6166pt) -- (0pt,403pt);
\color[rgb]{1,1,1}
\fill(3.76406pt,399.236pt) -- (380.923pt,399.236pt) -- (380.923pt,253.19pt) -- (3.76406pt,253.19pt) -- (3.76406pt,399.236pt);
\fill(53.4497pt,394.719pt) -- (374.901pt,394.719pt) -- (374.901pt,289.325pt) -- (53.4497pt,289.325pt) -- (53.4497pt,394.719pt);

draw a white background that is larger than the actual picture. TikZ sees that and thinks it is part of the picture. Simply removing/uncommenting these lines removes most of the whitespace.

Near the end of the first scope,

\color[rgb]{1,1,1}
\fill(3.76406pt,249.426pt) -- (386.193pt,249.426pt) -- (386.193pt,103.381pt) -- (3.76406pt,103.381pt) -- (3.76406pt,249.426pt);

does the same. Additionally (near the end of the second scope),

\pgftext[center, base, at={\pgfpoint{220.95pt}{106.392pt}}]{\sffamily\fontsize{9}{0}\selectfont{\textbf{ }}}

adds a blank node below the picture, again enlarging the bounding box.

Removing all those lines gives a tight bounding box.

As far as I know, TikZ cannot do the cropping for you, as it can't know whether the white stuff is intentional or not (there might for example be a dark background behind the image so that white is visible).

[Tex/LaTex] Using TikZ coordinates in expressions

You can use the calc library for this. See section 13.5 Coordinate Calculations of the manual.

To get the halfway point between two coordinates, you can use the syntax ($(A)!0.5!(B)$) or ($0.5*(A)+0.5*(B)$). For the rotated vector, you can use ($(A)!<length>!<angle>:(B)$).

The midpoint between two coordinates can also be found using a \path command, as Altermundus points out in a comment.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc}


\begin{document}
\begin{tikzpicture}
\coordinate (C1) at (0,0);
\coordinate (C2) at (3,2);

%% Use one of the next two lines for calculating the midpoint between two coordinates using the calc library...
\coordinate (C3) at ($(C1)!0.5!(C2)$);
%\coordinate (C3) at ($0.5*(C1)+0.5*(C2)$);

%% Or create coordinates C1, C2 and C3 in one go using a path command (thanks, Altermundus!)
%\path (0,0) coordinate (C1) -- coordinate (C3) (3,2) coordinate (C2);

\coordinate (C4) at ($(C3)!2cm!90:(C2)$);

\draw (C3) -- (C2);
\draw [red] (C3) -- (C4);

\node at (C1) {C1};
\node at (C2) {C2};
\node at (C3) {C3};
\node at (C4) {C4};
\end{tikzpicture}


\end{document}

Best Answer

Related Solutions

[Tex/LaTex] TikZ: Cropping the Bounding Box

[Tex/LaTex] Using TikZ coordinates in expressions

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