quadratic bezier curve

hbghlyj

从$P_{1}(x_{1},y_{1})$到$P_{2}(x_{2},y_{2})$画一条抛物线, 使其在$P_1,P_2$的切线的交点为$S(x,y)$
抛物线方程为$\begin{cases}x={\frac {m_{2}x_{2}-m_{1}x_{1}-(y_{2}-y_{1})}{m_{2}-m_{1}}}\\y=y_{i}+m_{i}(x-x_{i});\quad (i=1,2\,{\hbox{ gives same solution}})\end{cases}$
其中$m_1,m_2$分别为$SP_1,SP_2$的斜率.

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绘制一个带有波浪形顶部的矩形

Catenaries of Constant Length
悬链线具有以下奇怪的特性：两点之间的悬链线的长度等于这两点之间的悬链线下方的面积。
证明非常简单：\[\sqrt{1 +\left(\frac{d}{dx} \cosh(x) \right)^2} = \sqrt{1 + \sinh^2(x)} = \cosh(x)\]
现在对等式两端在区间 $[a,b]$ 上积分。注意前一个积分给出的是$a$和$b$之间$\cosh x$的弧长，后面的积分给出的是$a$和$b$之间$\cosh x$图象下的面积。画出长度相同的悬链线。
方程
$$\frac{y}{a}=\cosh \frac{x}{a}-1$$
使用$\ln$表示$\arccosh$得
\[
\left\{
\begin{aligned}
v & = \sqrt{L^2+a^2}-a, \\
u&=\frac{L^{2}-v^{2}}{2 v} \ln \frac{L+v}{L-v}\end{aligned} \right.
\]计算cubic Bezier curve控制点
\[\left\{
\begin{aligned}
\text{m1}&=0\\
\text{m2} & =\sinh\frac{u}{a}
\end{aligned} \right.
\]

将picture环境转换为TikZ

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Graphics in LaTeX2ε Appendix A
The Java program started with the line java qbezier x1 y1 m1 x2 y2 m2 writes a file qbezier.tex which can be pasted into a picture environment, whence it draws the desired curve.

1. import java.io.*;
2. class qbezier {
3. static void main(String[] args) {
4. if(args.length != 6) {
5. System.out.println("Gebrauch: java qbezier x1 y1 m1 x2 y2 m2");
6. return;
7. }
8. File outfile = new File("qbezier.tex");
9. try {
10. double x1 = Double.parseDouble(args[0]);
11. double y1 = Double.parseDouble(args[1]);
12. double m1 = Double.parseDouble(args[2]);
13. double x2 = Double.parseDouble(args[3]);
14. double y2 = Double.parseDouble(args[4]);
15. double m2 = Double.parseDouble(args[5]);
16. if(m1 == m2) {
17. System.out.println("Keine Auswertung mglich:");
18. System.out.println("m1 und m2 mssen verschieden sein");
19. return;
20. }
21. String outstring = "% qbezier P1=("+x1+"/"+y1+") m1="+m1+" P2=("+x2+"/"+y2+") m2="+m2;
22. outstring += "\n\\qbezier";
23. double x = (m2*x2-m1*x1+y1-y2) / (m2-m1);
24. double u = y1+m1*(x-x1);
25. double v = y2+m2*(x-x2);
26. double y = 0.5*(u+v);
27. outstring += "(" + runde(x1) + ", " + runde(y1) + ")";
28. outstring += "(" + runde(x) + ", " + runde(y) + ")";
29. outstring += "(" + runde(x2) + ", " + runde(y2) + ")";
30. FileWriter out = new FileWriter(outfile.getPath(), false);
31. out.write(outstring);
32. out.close();
33. }
34. catch(NumberFormatException e) {
35. System.out.println("Eingabe nicht korrekt");
36. System.out.println("Korrekte Eingabe: x1 y1 m1 x2 y2 m2");
37. }
38. catch(IOException e) {
39. }
40. }
41. static String runde(double t) {
42. String s = "";
43. int T = (int)(Math.round(10000 * t));
44. double d = T/10000.0;
45. s = String.valueOf(d);
46. return s;
47. }
48. }

Copy the Code

hbghlyj

hbghlyj 发表于 2023-3-9 17:48
将picture环境转换为TikZ

虽然看上去差不多, 检查SVG发现,

picture生成的SVG的曲线是由很多点拼成的	TikZ把1条cubic bezier转换为SVG的1条cubic bezier

说明TikZ比picture更好