tikz-3dplot accumulates the transformation every time let operator is used

Question

In follow up to my previous question, I now have the following problem with the tikz-3dplot library. Every time I use the "let" operator, the transformation matrix seems to act again on the coordinate frame, while computing \x1,\y1:

Edited code

\documentclass[convert,border=.3cm]{standalone}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{positioning,calc}

\begin{document}
\tdplotsetmaincoords{0}{10}
\begin{tikzpicture}[tdplot_main_coords]
\foreach\x in {0,1,2}
    \draw[gray,thin] (\x,-.3,0) -- ++(0,2.6,0);
\foreach\y in {0,1,2}
    \draw[gray,thin] (-.3,\y,0) -- ++(2.6,0,0);

\coordinate (A) at (1,1);
\draw (0,0,0)--(1,1);
\path let \p1=(A) in coordinate (B) at (\x1*1pt/1cm,\y1*1pt/1cm,0);
\draw[red] (0,0,0)--(B);
\path let \p1=(A) in coordinate (C) at (\x1,\y1);
\draw[green] (2,0,0)--(C);

\foreach\x in {0,1,2}
    \draw[gray,thin] (\x,-.3,0) -- ++(0,2.6,0);
\foreach\y in {0,1,2}
    \draw[gray,thin] (-.3,\y,0) -- ++(2.6,0,0);
\end{tikzpicture}
\end{document}

In the example above \tdplotsetmaincoords{0}{10} should (and does) just turn 10 degrees the coordinate frame around the z-axis (originally perpendicular to the screen)

This accumulation does not seem to affect any other element in the tikzpicture, except for the coordinates involved in "let". One can see this because the two "test grids" at the begining and the end are completely aligned.

Notice, that the same effect happens also when I just experiment with the first input of \tdplotsetmaincoords (rotation around x-axis), or with both.

Answer 1

In your code, you compute the projections of the 3d coordinates on the screen. That is, the calc syntax returns the x and y components 2-dimensional screen coordinates, which are not to be confused with the components of the 3d or main coordinates, i.e. those defined by calling \tdplotsetmaincoords{...}{...}. You take these projections of the 3d coordinates to form new x and y components of new 3d coordinates.

To illustrate this, consider

\documentclass[convert,border=.3cm]{standalone}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{positioning,calc}

\begin{document}
\tdplotsetmaincoords{0}{10}
\begin{tikzpicture}[tdplot_main_coords]
\coordinate (A) at (1,1);
\path let \p1=(A),\n1={\x1*1pt/1cm},\n2={\y1*1pt/1cm} in (\n1,\n2,0) coordinate
(B);
\coordinate (C) at (1,1,0);
\begin{scope}[tdplot_screen_coords]
\coordinate (D) at (1,1);
\coordinate (E) at (1.16,0.81);
\end{scope}
\draw let \p1=(A),\n1={\x1*1pt/1cm},\n2={\y1*1pt/1cm} in (0,0,0)--
node[pos=0.95,above left,sloped] {\pgfmathsetmacro{\myx}{\n1}
\pgfmathsetmacro{\myy}{\n2}$A=(\pgfmathprintnumber{\myx},\pgfmathprintnumber{\myy})$}(A);
\draw[red] let \p1=(B),\n1={\x1*1pt/1cm},\n2={\y1*1pt/1cm} in
(0,0,0)-- node[pos=0.95,below left,sloped] {\pgfmathsetmacro{\myx}{\n1}
\pgfmathsetmacro{\myy}{\n2}$B=(\pgfmathprintnumber{\myx},\pgfmathprintnumber{\myy})$}(B);
\draw[blue] let \p1=(C),\n1={\x1*1pt/1cm},\n2={\y1*1pt/1cm} in (2,0,0)
-- node[pos=0.95,above right,sloped] {\pgfmathsetmacro{\myx}{\n1}
\pgfmathsetmacro{\myy}{\n2}$C=(\pgfmathprintnumber{\myx},\pgfmathprintnumber{\myy})$}(C);
\draw[green] let \p1=(D),\n1={\x1*1pt/1cm},\n2={\y1*1pt/1cm} in (2,2,0)
-- node[midway,above,sloped] {\pgfmathsetmacro{\myx}{\n1}
\pgfmathsetmacro{\myy}{\n2}$D=(\pgfmathprintnumber{\myx},\pgfmathprintnumber{\myy})$}(D);
\draw[orange] let \p1=(E),\n1={\x1*1pt/1cm},\n2={\y1*1pt/1cm} in (-2,2,0)
-- node[midway,above,sloped] {\pgfmathsetmacro{\myx}{\n1}
\pgfmathsetmacro{\myy}{\n2}$E=(\pgfmathprintnumber{\myx},\pgfmathprintnumber{\myy})$}(E);
\foreach\x in {0,1,2}
    \draw[gray,thin] (\x,-.3,0) -- ++(0,2.6,0);
\foreach\y in {0,1,2}
    \draw[gray,thin] (-.3,\y,0) -- ++(2.6,0,0);
\end{tikzpicture}
\end{document}

Here,

1. A is defined to sit at (1,1) in the main coordinates. They correspond to (1.16,0.81) in the screen coordinates.
2. Now (1.16,0.81) is used to define B, where the coordinates have their meaning in the main coordinates. So you are telling TikZ to place B at (1.16,0.81,0) in the main coordinates.
3. C is defined at (1,1,0) in the main coordinates. It coincides with A.
4. D is defined to sit at (1,1) in the screen coordinates. It does not coincide with any of the points before, but the angles DOA and AOB coincide. This is the difference between active and passive transformations.
5. E is defined to sit at (1.16,0.81) in the screen coordinates, and thus coincides with A and C.

Overall, the point is that there are two coordinate systems. The main coordinates system which is used to define the points and which originates from an orthographic projection of 3d coordinates. And then there are the screen coordinates, which are those in which calc works.

Actually, there is one more thing to consider in your MWE.

\documentclass[convert,border=.3cm]{standalone}
\usepackage{tikz}
\usepackage{tikz-3dplot}
\usetikzlibrary{positioning,calc}

\begin{document}
\tdplotsetmaincoords{0}{10}
\begin{tikzpicture}[tdplot_main_coords]
\foreach\x in {0,1,2}
    \draw[gray,thin] (\x,-.3,0) -- ++(0,2.6,0);
\foreach\y in {0,1,2}
    \draw[gray,thin] (-.3,\y,0) -- ++(2.6,0,0);

\coordinate (A) at (1,1);
\draw (0,0,0)--(1,1);
\path let \p1=(A) in coordinate (B) at (\x1*1pt/1cm,\y1*1pt/1cm,0);
\draw[red] (0,0,0)--(B);
\path let \p1=(A) in coordinate (C) at (\x1,\y1);
\draw[green] (2,0,0)--(C);

\foreach\x in {0,1,2}
    \draw[gray,thin] (\x,-.3,0) -- ++(0,2.6,0);
\foreach\y in {0,1,2}
    \draw[gray,thin] (-.3,\y,0) -- ++(2.6,0,0);
\end{tikzpicture}
\end{document}

When you define C, you define it with two components both of which have units. TikZ takes these to be screen coordinates again. Why? Because these are just some dimensionful coordinates and this is how the parser treats it. On the other hand, and yes, this is really confusing, if TikZ sees a dimensionful 3d coordinate, it will strip of the dimensions and takes the resulting coordinates to be in the main coordinate system. Yes, this is truly confusing. This behavior is indeed a bit odd. Here is an attempt to summarize it.

\documentclass{article}
\usepackage{array}
\begin{document}
\begin{tabular}{|r|>{\centering\arraybackslash}p{5cm}|>{\centering\arraybackslash}p{5cm}|}
\hline
 & dimensionful & dimensionless \\
\hline
2D & screen & main \\
3D & convert dimension to points, strip units off, and interpret the result in
main coordinate system &  main\\
\hline
\end{tabular}
\end{document}

The maintainers of TikZ are actually aware of this, see this commit, which deals with the fact that a projection is a projection and implies loss of information.

See here for a way to store the 3d coordinates in pgfkeys.