java - 计算不完美的圆

Question

我想创建一个“非完美圆”的生成器，圆有点扭曲且更随机，但看起来仍然有点像圆或云。

这就是我所说的不完美的圆:

我想创建一个函数来获取“非完美圆”的最大和最小比例并获取其所有点。我知道圆的公式:X^2+Y^2=R^2 但我想不出一种方法让它更随机一些。有人有什么想法吗？

编辑:试图用点画一个完美的圆但它不会工作:

    for (int step = 0; step < 300; ++step) {
         double t = step / 300 * 2 * Math.PI;
         c.drawPoint(300+(float)(33 * Math.cos(t)), 300+(float)(33 * Math.sin(t)), p);
    }

编辑 2:

    for (int step = 0; step < 20; ++step) {
          double t = step / 20.0 * 2 * Math.PI; 
          double imperfectR = 50.0+randInt(10, 50);
          //I do it here?
          points[step]=new PointF();
          points[step].set((300+(float)(imperfectR  * Math.cos(t))), 300+(float)(imperfectR  * Math.sin(t)));
          if(step==0){
                pp.moveTo(points[step].x, points[step].y);
          }
          else
                pp.quadTo(points[step-1].x, points[step-1].y,points[step].x, points[step].y);

    } 

编辑 3:

double t=0;
for (int i = 0; i < points.length/4; i++) {
        if(t==1){
            t=0;
        }
        t+=0.10;
        double imperfectR=0.5*((2*points[i+1].y)+(-points[i].y+points[i+2].y)*t+(2*points[i].y-5*points[i+1].y+4*points[i+2].y-points[i+3].y)*(t*t)+((-points[i].y+3*points[i+1].y-3*points[i+2].y+points[i+3].y)*(t*t*t)));
        newPoints[i].set((300+(float)(imperfectR  * Math.cos(t))), 300+(float)(imperfectR  * Math.sin(t)));
        t+=0.10;
        imperfectR=0.5*((2*points[i+1].y)+(-points[i].y+points[i+2].y)*t+(2*points[i].y-5*points[i+1].y+4*points[i+2].y-points[i+3].y)*(t*t)+((-points[i].y+3*points[i+1].y-3*points[i+2].y+points[i+3].y)*(t*t*t)));
        newPoints[i+1].set((300+(float)(imperfectR  * Math.cos(t))), 300+(float)(imperfectR  * Math.sin(t)));
        t+=0.10;
        imperfectR=0.5*((2*points[i+1].y)+(-points[i].y+points[i+2].y)*t+(2*points[i].y-5*points[i+1].y+4*points[i+2].y-points[i+3].y)*(t*t)+((-points[i].y+3*points[i+1].y-3*points[i+2].y+points[i+3].y)*(t*t*t)));
        newPoints[i+2].set((300+(float)(imperfectR  * Math.cos(t))), 300+(float)(imperfectR  * Math.sin(t)));
        t+=0.10;
        imperfectR=0.5*((2*points[i+1].y)+(-points[i].y+points[i+2].y)*t+(2*points[i].y-5*points[i+1].y+4*points[i+2].y-points[i+3].y)*(t*t)+((-points[i].y+3*points[i+1].y-3*points[i+2].y+points[i+3].y)*(t*t*t)));
        newPoints[i+3].set((300+(float)(imperfectR  * Math.cos(t))), 300+(float)(imperfectR  * Math.sin(t)));
        if(i==0){
            pp.moveTo(newPoints[i].x, newPoints[i].y);
        }
        pp.lineTo(newPoints[i].x, newPoints[i].y);
        pp.lineTo(newPoints[i+1].x, newPoints[i+1].y);
        pp.lineTo(newPoints[i+2].x, newPoints[i+2].y);
        pp.lineTo(newPoints[i+3].x, newPoints[i+3].y);

}
pp.close();

Answer 1

另一种更容易绘制的圆方程是其参数形式之一:

x = R * cos(t);
y = R * sin(t);

其中 R 是标称半径，t 是介于 0 和 2 * pi 之间的参数。因此，您可以像这样在圆周上画点“完美”圆:

for (int step = 0; step < NSTEPS; ++step) {
  double t = step / (double) NSTEPS * 2 * pi;
  drawPoint(R * cos(t), R * sin(t));
}

按照@nikis 的建议，您可以通过向圆的半径添加随机量来使圆“不完美”:

for (int step = 0; step < NSTEPS; ++step) {
  double t = step / (double) NSTEPS * 2 * pi;
  double imperfectR = R + randn();  // Normally distributed random
  drawPoint(imperfectR * cos(t), imperfectR * sin(t));
}

但是，这可能会给您带来尖尖的形状，因为没有什么可以使 step 和 step + 1 处的半径相似。不失一般性，您可以将上面的代码重写为:

for (int step = 0; step < NSTEPS; ++step) {
  double t = step / (double) NSTEPS * 2 * pi;
  double imperfectR = f(t);
  drawPoint(imperfectR * cos(t), imperfectR * sin(t));
}

其中 f(t) 是为参数 t 生成半径的某个函数。现在，您可以为 t 选择任何函数，但您可能希望选择在圆上连续的函数，即 f(t) 没有变化的点值突然。

这里有很多选择。我上面建议的例子是建议使用余弦函数的总和:

double f(double t) {
  double f = 0;
  for (int i = 0; i < N; ++i) {
    f += A[i] * cos(i * t + w[i]);
  }
  return f;
}

其中A和w是随机选择的值； A[0] 应设置为 R。这里的重点是余弦函数的周期为2 * pi，所以f(alpha) = f(alpha + 2 * pi)，满足被连续的。

然而，这远非唯一的选择。您也许可以选择像高斯内核之和这样的东西，它将“凸起”放置在以 w[i] 为中心的位置，并在 sigma[i]圆圈:

double f(double t) {
  double f = 0;
  for (int i = 0; i < N; ++i) {
    f += A[i] * exp(-Math.pow(t-w[i], 2) / sigma[i]);
  }
  return f;
}

(这不太行，它不处理 t 的环绕)

您需要尝试一下，看看什么样的函数和什么样的随机选择值可以为您提供所需的形状。