我想创建一个“不完美的圆圈”的创作者,圆圈有点扭曲,更随机,但仍然看起来有点像圆圈或云。
这就是我所说的不完美的圆圈:
我想创建一个能够获得“不完美圆圈”的最大和最小比例的函数,并得到它的所有点。我知道一个圆的公式: X ^ 2 + Y ^ 2 = R ^ 2但我想不出一种让它更随机的方法。有人有什么想法吗?
编辑:试图用点绘制一个完美的圆圈,但它不起作用:
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();
答案 0 :(得分:3)
圆形的另一个等式,它更容易绘制,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0
和2 * pi
之间的参数。所以,你可以在圆周上绘制一个像这样的“完美”圆圈:
for (int step = 0; step < NSTEPS; ++step) {
double t = step / (double) NSTEPS * 2 * pi;
drawPoint(R * cos(t), R * sin(t));
}
根据@nikis的建议,您可以通过在圆的半径上添加随机数量来使圆圈“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));
}
然而,这可能会给你非常尖锐的形状,因为没有任何东西可以使step
和step + 1
的半径相似。在不失一般性的情况下,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
生成半径的函数。现在,您可以选择t
的任何函数,但您可能希望选择在圆上连续的内容,即f(t)
突然改变值没有任何意义。
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]
应设为R
。这里的要点是余弦函数的周期为2 * pi
,因此f(alpha) = f(alpha + 2 * pi)
满足连续的要求。
然而,这远非唯一的选择。你可以选择像高斯内核之和那样的东西,它将“凸起”置于w[i]
的中心,并在圆周上放置sigma[i]
:
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;
}
(这不起作用,它不处理t
)
您需要四处游戏,看看哪些功能以及哪种随机选择的值可以为您提供所需的形状。