数学图形(2.2)N叶结

上一节讲的三叶结,举一反三,由三可到无穷,这一节讲N叶结

再次看下三叶结的公式:

x = sin(t) + 2*sin(2*t)
y = cos(t) - 2*cos(2*t)

将其改为:

x = sin(t) + 2*sin((n-1)*t)
y = cos(t) - 2*cos((n-1)*t)

就变成了N叶结了,如此简单.

N叶结:

vertices = 12000

t = from 0 to (20*PI)

n = rand_int2(2, 24)

x = sin(t) + 2*sin(n*t - t)
y = cos(t) - 2*cos(n*t - t)
z = -sin(n*t)

r = 5;
x = x*r
y = y*r
z = z*r

另一种写法:

vertices = 12000

t = from 0 to (20*PI)

n = rand_int2(2, 24)

x = (2 + cos(n*t))*cos((n - 1)*t)
y = (2 + cos(n*t))*sin((n - 1)*t)
z = sin(n*t)

r = 5
x = x*r
y = y*r
z = z*r

四叶结

#http://www.mathcurve.com/courbes3d/noeuds/noeuddetrefle.shtml

vertices = 1000

t = from 0 to (2*PI)

r = 10;
x = r*(cos(t) + 2*cos(3*t))
z = r*(sin(t) - 2*sin(3*t))
y = r*sin(4*t)