数学图形之水滴

前几天看科幻小说<三体>,讲到有种武器叫水滴,是三体人用于对付地球人的.这一节将介绍几种水滴形的数学公式.

<三体>书中对水滴的描述如下:

当全世界第一次看到探测器的影像时，所有人都陶醉于它那绝美的外形。这东西真的是太美了，它的形状虽然简洁，但造型精妙绝伦，曲面上的每一个点都恰到好处，使这滴水银充满着飘逸的动感，仿佛每时每刻都在宇宙之夜中没有尽头地滴落着。它给人一种感觉：即使人类艺术家把一个封闭曲面的所有可能形态平滑地全部试完，也找不出这样一个造型。它在所有的可能之外，即使柏拉图的理想国中也没有这样完美的形状，它是比直线更直的线，是比正圆更圆的圆，是梦之海中跃出的一只镜面海豚，是宇宙间所有爱的结晶...美总是和善联在一起的，所以，如果宇宙中真有一条善恶分界线的话，它一定在善这一面。

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我之前写过有关水滴线的文章,数学图形(1.15) 水滴线,那是讲的二维曲线,而这一节给大家带来的是三维水滴形.当然水滴形就是由一个水滴线旋转180度,或半个水滴线旋转360度而生成的.

(1)水滴

vertices = D1:100 D2:100

u = from 0 to (PI) D1
v = from 0 to (2*PI) D2

a = 10
b = rand2(0.25, 1)

m = a/2*(1 + cos(u))
n = a*a/64*(sin(2*u) + 2*sin(u))*b

x = n*cos(v)
z = n*sin(v)
y = array_max(m)-m

(2)Ding-Dong

#http://mathworld.wolfram.com/Ding-DongSurface.html
# x^2+y^2=(1-z)z^2 

vertices = 360

v = from -1 to 1 D2

x = v*sqrt(0.5 - v/2)
y = v

a = 10

x = x*a
y = y*a

#http://mathworld.wolfram.com/Ding-DongSurface.html
# x^2+y^2=(1-z)z^2 

vertices = D1:100 D2:100

u = from 0 to (PI*2) D1
v = from -1 to 1 D2

x = v*sqrt(0.5 - v/2)*cos(u)
y = v
z = v*sqrt(0.5 - v/2)*sin(u)

a = 10

x = x*a
y = y*a
z = z*a

(3)kiss

#http://mathworld.wolfram.com/KissSurface.html
# x^2+y^2=(1-z)z^4 

vertices = 360

v = from -1 to 1

x = v*v*sqrt(0.5 - v/2)
y = v

a = 10

x = x*a
y = y*a

#http://mathworld.wolfram.com/KissSurface.html
# x^2+y^2=(1-z)z^4 

vertices = D1:100 D2:100

u = from 0 to (PI*2) D1
v = from -1 to 1 D2

x = v*v*sqrt(0.5 - v/2)*cos(u)
y = v
z = v*v*sqrt(0.5 - v/2)*sin(u)

a = 10

x = x*a
y = y*a
z = z*a

(4)Larme

#http://www.mathcurve.com/courbes2d/larme/larme.shtml

vertices = D1:100 D2:100

u = from 0 to (PI) D1
v = from 0 to (2*PI) D2

n = rand2(1, 10)

m = 10*cos(u)
n = 10*sin(u)*pow(sin(u/2), n)

x = n*cos(v)
z = n*sin(v)
y = m

(5)Pear

#http://www.2dcurves.com/quartic/quarticp.html#pearshapedcurve

vertices = D1:100 D2:100

u = from 0 to (2*PI) D1
v = from 0 to (PI) D2

m = -1 - sin(u)
n = 0.5*(1 + sin(u))*cos(u)

x = n*cos(v)
z = n*sin(v)
y = array_max(m)-m

(6)Teardrop

#http://mathworld.wolfram.com/TeardropCurve.html

vertices = D1:100 D2:100

u = from 0 to (PI) D1
v = from 0 to (2*PI) D2

m = rand2(0, 10)

n = 10*sin(u)*pow(sin(u/2), m)
y = 10*cos(u)

x = n*cos(v)
z = n*sin(v)