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 = 12000t = 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*ry = y*rz = z*r

另一种写法:

vertices = 12000t = 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 = 5x = x*ry = y*rz = z*r

 

四叶结

#http://www.mathcurve.com/courbes3d/noeuds/noeuddetrefle.shtmlvertices = 1000t = 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)