I want to draw a cylinder in Tikz. There are a lot of easier ways to do it on tex.stackexchange.com, but I want to do it with the 3d library so that I can (hopefully) change the direction of the axes and have it automatically redrawn. I have set it up so that the y-axis is (in 2D) at 45° from the x-axis and at $\frac{1}{2}$ the length of the other two coordinates. The parameters look like this [y={(225:-0.5 cm)},x={(0:1cm)},z={(90:1cm)}, ...], which means the y-axis is actually at 225° but in the other (negative) direction and with a "scale" factor of $\frac{1}{2}$. Then I drew the circle in the x-y plane (called "canvas" in Tikz) with a given center and radius. How Tikz then transforms this into the ellipse shown, I don't really know, otherwise I could work with the angles and probably figure it out. The complete Tikz code is:
\documentclass[border=1mm]{standalone}
\usepackage{tikz}
\usepackage{amsmath}
\usetikzlibrary{3d}
\begin{document}
\begin{tikzpicture}[y={(225:-0.5 cm)},x={(0:1cm)},z={(90:1cm)}, thick, join=round, scale=2]
\def \r{ 2.50 } % Defines the radius of the cylinder
\def \h{ 7.50 } % Defines the heigth of the cylinder
\begin{scope}[canvas is xy plane at z=0]
\draw (0,0) circle (\r);
\end{scope}
\begin{scope}[canvas is xy plane at z=\h]
\draw (0,0) circle (\r);
\end{scope}
\draw[->] (-4,0,0) -- node[pos=1,above] {{\Large $x$}} (4,0,0);
\draw[->] (0,-4,0) -- node[pos=1,above] {{\Large $y$}} (0,4,0);
\draw[->] (0,0,-4) -- node[pos=1,above] {{\Large $z$}} (0,0,4);
\draw (\r,0,0) -- (\r,0,\h) (-\r,0,0) -- (-\r,0,\h);
\end{tikzpicture}
\end{document}
Which gives:
That obviously doesn't look nice. How do I find the points through which the vertical lines should go in order to make the cylinder look "closed"? Or a related question is: how do I find out the parameters of the ellipse (major and minor axes and angle with the horizontal line) given the definition of the axes above?
You can think of the ellipse as obtained from a circle (which I'll assume of radius $r=1$), first by squeezing the circle to an axis-aligned ellipse, then performing a shear transformation $(x,y)\to (x+ky,y)$ (sheared minor axis will form an angle $\theta=90°-\arctan k$ with $x$ axis, hence $k=1$ for $\theta=45°$).
To obtain a sheared semi-minor axis with length $l$ ($l=1/2$ in the question), one must start with a semi-minor axis with length $b=l/\sqrt{1+k^2}$ (this is then the squeezing factor). Hence the equation of the ellipse before shear is: $$ x^2+{y^2\over b^2}=1. $$ The tangent to the sheared ellipse is "vertical" at points where the slope before shear was $-k$. Differentiating the above equation with respect to $x$ and setting $dy/dx=-k$ one gets $$x={k\over b^2}y.$$ Inserting this into the equation of the ellipse one then finds $$ y=\pm{b^2\over\sqrt{k^2+b^2}}\quad\text{and}\quad x=\pm{k\over\sqrt{k^2+b^2}}. $$ Finally, as you need the coordinates before the circle was squeezed to an ellipse, just un-squeeze them dividing $y$ by $b$: $$ y=\pm{b\over\sqrt{k^2+b^2}}\quad\text{and}\quad x=\pm{k\over\sqrt{k^2+b^2}}. $$ These are the starting coordinates for the lateral sides of the cylinder. With the given data $k=1$, $b=\sqrt2/4$ this translates into: $$ y=\pm{1\over3}\quad\text{and}\quad x=\pm{2\sqrt2\over3}. $$