我正在尝试在 Tikz 中构建下面的图像。除了手动计算每个笛卡尔坐标然后输入它们之外,我还想使用圆柱极坐标,因为这样会快得多。
\documentclass[tikz]{standalone}
\begin{document}
\begin{tikzpicture}
\node (1) at (0,0,0) {$\mathbb{Q}$};
\node (11) at (0,-1,0) {$\mathbb{Z}$};
\node (2) at (1.732,0,-1) {$\mathbb{Q}_{2}$};
\node (21) at (1.732,-1,-1) {$\mathbb{Z}_{2}$};
\node (22) at (3.464,0,-2) {$\mathbb{C}_{2}$};
\node (3) at (1.732,0,1) {$\mathbb{Q}_{3}$};
\node (31) at (1.732,-1,1) {$\mathbb{Z}_{3}$};
\node (32) at (3.464,0,2) {$\mathbb{C}_{3}$};
\node (5) at (0,0,2) {$\mathbb{Q}_{5}$};
\node (51) at (0,-1,2) {$\mathbb{Z}_{5}$};
\node (52) at (0,0,4) {$\mathbb{C}_{5}$};
\node (7) at (-1.732,0,1) {$\mathbb{Q}_{7}$};
\node (71) at (-1.732,-1,1) {$\mathbb{Z}_{7}$};
\node (72) at (-3.464,0,2) {$\mathbb{C}_{7}$};
\node (11) at (-1.732,0,-1) {$\mathbb{Q}_{11}$};
\node (111) at (-1.732,-1,-1) {$\mathbb{Z}_{11}$};
\node (112) at (-3.464,0,-2) {$\mathbb{C}_{11}$};
\node (10) at (0,0,-2) {$\mathbb{R}$};
\node (101) at (0,0,-4) {$\mathbb{C}$};
\draw (1) to (11);
\draw[->] (1) to (2);
\draw[->] (1) to (3);
\draw[->] (1) to (5);
\draw[->] (1) to (7);
\draw[->] (1) to (11);
\draw[->] (1) to (10);
\draw (2) to (21);
\draw[->] (2) to (22);
\draw (3) to (31);
\draw[->] (3) to (32);
\draw (5) to (51);
\draw[->] (5) to (52);
\draw (7) to (71);
\draw[->] (7) to (72);
\draw (11) to (111);
\draw[->] (11) to (112);
\draw[->] (10) to (101);
\end{tikzpicture}
\end{document}
这问应该在圆圈的中心R在它上面坐标-plane(尽管 tikz 似乎将其定位得很奇怪)第二季度,第三季度等等在里面转圈坐标-平面均匀分布。然后我想要是直接传播到其相应的问最后,我想要C距离中心 1 个单位问比其相应的问秒。
我很乐意减少“辐条”的数量以使图片更整洁,但希望至少有 5 个。
答案1
圆柱坐标系确实存在。
\documentclass[border=9,tikz]{standalone}
\usetikzlibrary{3d}
\begin{document}
\tikz{
\draw[->](0,0)--(1,0)node[right]{$x$};
\draw[->](0,0)--(0,1)node[above]{$y$};
\draw[->](0,0)--(0,0,1)node[below left]{$z$};
\fill foreach\i in{1,...,100}{
(xyz cylindrical cs:angle=\i0,radius=1,z=\i/10)circle(.1)
};
}
\end{document}
编辑
我希望这能给你一些关于使用\foreach
\tikz[x={(-.8cm,-.6cm)},y={(1cm,0cm)},z={(0cm,1cm)}]{
\draw[->](0,0)--(1,0)node[below left]{$x$};
\draw[->](0,0)--(0,1)node[right]{$y$};
\draw[->](0,0)--(0,0,1)node[above]{$z$};
\foreach\p[count=\i]in{2,3,5,7,9,\infty}{
\draw
(xyz cylindrical cs:angle=\i*60,radius=0,z=4)--
(xyz cylindrical cs:angle=\i*60,radius=2,z=4)node{$Q_\p$}--
(xyz cylindrical cs:angle=\i*60,radius=2,z=2)node{$Z_\p$}
(xyz cylindrical cs:angle=\i*60,radius=2,z=4)--
(xyz cylindrical cs:angle=\i*60,radius=4,z=4)node{$C_\p$};
};
}