球面坐标的计算

问题描述

我不得不说，我很害怕，也很惊讶我对基础数学知之甚少。本质上，我得到的是一个原点(0，0，0)，我知道圆的半径(10)，我知道两个角度(theta和phi)。根据这个假设，我想计算球面上的投影点。我通过阅读https://stackoverflow.com/a/969880/1230358、https://stackoverflow.com/a/36369852/1230358、http://tutorial.math.lamar.edu/Classes/CalcII/SphericalCoords.aspx和https://en.wikipedia.org/wiki/Spherical_coordinate_system的答案得出了底层代码。

我当前的代码：

#!/usr/bin/env python3
import math


PI = math.pi
PI_2 = PI / 2

def calc_sphere_coordinates(radius, phi, theta):
    # see: https://stackoverflow.com/a/969880/1230358
    # see: https://stackoverflow.com/q/19673067/1230358
    # see: http://mathinsight.org/spherical_coordinates
    # see: https://en.wikipedia.org/wiki/Spherical_coordinate_system
    # see: http://tutorial.math.lamar.edu/Classes/CalcII/SphericalCoords.aspx

    # φ phi is the polar angle, rotated down from the positive z-axis (slope)
    # θ theta is azimuthal angle, the angle of the rotation around the z-axis (aspect)

    # z        
    # |  x
    # | /
    # |/
    # +-------- y

    # both angles need to be in radians, not degrees!
    theta = theta * PI / 180
    phi = phi * PI / 180

    x = radius * math.sin(phi) * math.cos(theta)
    y = radius * math.sin(phi) * math.sin(theta)
    z = radius * math.cos(phi)
    return (x, y, z)

if __name__ == "__main__":

    # calculate point position in hemisphere by rotating down from positive z-axis
    for i in (10, 20, 30, 40, 50, 60, 70 , 80, 90):
        print(calc_sphere_coordinates(10, i, 0))

    print("-"*10)

    # calculate point position in hemisphere by rotating around the z axis
    for i in (10, 20, 30, 40, 50, 60, 70 , 80, 90):
        print(calc_sphere_coordinates(10, 0, i))

    print("-"*10)

    # calculate point position by rotating in both directions
    for i in (10, 20, 30, 40, 50, 60, 70 , 80, 90):
        print(calc_sphere_coordinates(10, i, 90-i))    

代码输出如下：

(1.7364817766693033, 0.0, 9.84807753012208)
(3.420201433256687, 0.0, 9.396926207859085)
(4.999999999999999, 0.0, 8.660254037844387)
(6.4278760968653925, 0.0, 7.660444431189781)
(7.66044443118978, 0.0, 6.427876096865393)
(8.660254037844386, 0.0, 5.000000000000001)
(9.396926207859083, 0.0, 3.4202014332566884)
(9.84807753012208, 0.0, 1.7364817766693041)
(10.0, 0.0, 6.123233995736766e-16)
----------
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
(0.0, 0.0, 10.0)
----------
(0.30153689607045814, 1.7101007166283433, 9.84807753012208)
(1.16977778440511, 3.2139380484326963, 9.396926207859085)
(2.5, 4.330127018922192, 8.660254037844387)
(4.131759111665348, 4.92403876506104, 7.660444431189781)
(5.868240888334652, 4.92403876506104, 6.427876096865393)
(7.5, 4.330127018922192, 5.000000000000001)
(8.83022221559489, 3.2139380484326963, 3.4202014332566884)
(9.69846310392954, 1.7101007166283433, 1.7364817766693041)
(10.0, 0.0, 6.123233995736766e-16)

(10.0, 0.0, 6.123233995736766e-16)行的z坐标不应该是0而不是6.123233995736766e-16吗？无论使用什么角度(0.0, 0.0, 10.0)，绕z轴旋转都会得到相同的结果。

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解决方案

在我看来，您的代码运行良好。老实说，人们不得不承认6.123233995736766e-16对于所有实际应用来说几乎都是0，对吗？

您的问题可以归结为

为什么math.cos(math.pi / 2.0)不等于零

原因在于浮点数是如何在计算机中生成和存储的。尝试在python中计算0.1+0.2。惊讶？！如果您想了解更多信息，请访问Google浮点错误或任何相关内容。