• 数学图形之球面,椭球面,胶囊体,刺球


    从这一节开始讲3D的数学图形,为此我准备了好久.这一节中将为你展示如何生成球面,椭球面,胶囊体,刺球,圆弧面.

    相关软件参见:数学图形可视化工具,使用我自定义语法的脚本代码生成数学图形.

    在我刚学计算机图形学的时候,就写过生成球面的程序,代码曾发布在球(Sphere)图形的生成算法.

    球与圆很相关,一个是三维,一个是二维,可以参考下:圆,椭圆

    (1)sphere的第一种写法

    vertices = D1:100 D2:100
    
    t = from 0 to (PI*2) D1
    r = from 0 to 1 D2
    
    x = 2*r*sin(t)*sqrt(1-r^2)
    y = 2*r*cos(t)*sqrt(1-r^2)
    z = 1-2*(r^2)

    球的网格线:


    (2)sphere的另两种写法

    vertices = dimension1:36 dimension2:72
    u = from 0 to (2*PI) dimension1
    v = from (-PI*0.5) to (PI*0.5) dimension2
    r = 10.0
    x = r*cos(v)*sin(u)
    y = r*sin(v)
    z = r*cos(v)*cos(u)
    vertices = dimension1:36 dimension2:72
    u = from 0 to (2*PI) dimension1
    v = from 0 to (PI) dimension2
    r = 10.0
    x = r*sin(v)*sin(u)
    y = r*cos(v)
    z = r*sin(v)*cos(u)

    两种写法生成的图形是一样的


    (3)彩色球

    在脚本中给rgb变量设值,就能设置顶点色.

    vertices = dimension1:72 dimension2:72
    
    u = from 0 to (2*PI) dimension1
    v = from (-PI*0.5) to (PI*0.5) dimension2
    
    x = cos(v)*sin(u)
    y = sin(v)
    z = cos(v)*cos(u)
    
    a = 10.0
    
    r = (x+1.0)/2
    g = (y+1.0)/2
    b = (z+1.0)/2
    
    x = a*x
    y = a*y
    z = a*z


    (4)圆弧面

    将球的第二维度范围减小,即得到圆弧面

    vertices = dimension1:36 dimension2:72
    u = from 0 to (2*PI) dimension1
    v = from (PI*0.1) to (PI*0.5) dimension2
    r = 10.0
    x = r*cos(v)*sin(u)
    y = r*sin(v)
    z = r*cos(v)*cos(u)

    (5)椭球面

    #http://www.mathcurve.com/surfaces/ellipsoid/ellipsoid.shtml
    
    vertices = D1:100 D2:100
    
    u = from 0 to (2*PI) D1
    v = from (-PI*0.5) to (PI*0.5) D2
    
    a = rand2(1, 10)
    b = rand2(1, 10)
    c = rand2(1, 10)
    
    x = a*cos(v)*sin(u)
    y = b*sin(v)
    z = c*cos(v)*cos(u)

    (6)胶囊体

    将球面向上下两头拉伸,即得到胶囊体,我也曾经写过胶囊体的生成算法胶囊体(Capsule)图形的生成算法

    (7)刺球

    将球面上顶点到球心的距离,有规律地变化,可以得到多变的球,如刺球

    vertices = dimension1:129 dimension2:65
    
    u = from 0 to (2*PI) dimension1
    v = from (-PI*0.5) to (PI*0.5) dimension2
    
    n =4
    
    a = from 0 to 128 D1
    b = from 0 to 64 D2
    
    t = (mod(a, n) + mod(b, n))/n*4
    
    r = 10.0 + t
    
    x = r*cos(v)*sin(u)
    y = r*sin(v)
    z = r*cos(v)*cos(u)

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  • 原文地址:https://www.cnblogs.com/WhyEngine/p/3864824.html
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