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我正在尝试计算两个大圆的交点（纬度和经度），每个大圆由圆上的两个点定义。我一直在尝试遵循此处概述的方法。但是我得到的答案是不正确的，我的代码在下面有人看到我哪里出错了吗？

import numpy as np
from numpy import cross
from math import cos, sin, atan2, asin, asinh

################################################
#### Intersection of two great circles.
# Points on great circle 1.
glat1 = 54.8639587
glon1 = -8.177818

glat2 = 52.65297082
glon2 = -10.78064876

# Points on great circle 2.
cglat1 = 51.5641564
cglon1 = -9.2754284

cglat2 = 53.35422063
cglon2 = -12.5767799

# 1. Put in polar coords.

x1 = cos(glat1) * sin(glon1)
y1 = cos(glat1) * cos(glon1)
z1 = sin(glat1)

x2 = cos(glat2) * sin(glon2)
y2 = cos(glat2) * cos(glon2)
z2 = sin(glat2)


cx1 = cos(cglat1) * sin(cglon1)
cy1 = cos(cglat1) * cos(cglon1)
cz1 = sin(cglat1)

cx2 = cos(cglat2) * sin(cglon2)
cy2 = cos(cglat2) * cos(cglon2)
cz2 = sin(cglat2)


# 2. Get normal to planes containing great circles.
#    It's the cross product of vector to each point from the origin.

N1 = cross([x1, y1, z1], [x2, y2, z2])
N2 = cross([cx1, cy1, cz1], [cx2, cy2, cz2])


# 3. Find line of intersection between two planes.
#    It is normal to the poles of each plane.

L = cross(N1, N2)


# 4. Find intersection points.

X1 = L / abs(L)
X2 = -X1


ilat = asin(X1[2]) * 180./np.pi
ilon = atan2(X1[1], X1[0]) * 180./np.pi

我还应该提到这是在地球表面（假设是一个球体）。

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2 回答 2

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DSM 在上述评论中的解决方案，您的角度以度为单位，而 sin 和 cos 期望弧度。还有线

X1 = L / abs(L)

应该，

X1 = L / np.sqrt(L[0]**2 + L[1]**2 + L[2]**2)
于 2013-09-20T16:02:04.707 回答
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需要进行的另一项更正是在 x 和 y 维度的“lon”之前更改 cos/sin：

x = cos(lat) * cos(lon)
y = cos(lat) * sin(lon)
z = sin(lat)

这是因为从角度到球面系统的原始转换是使用极坐标/方位角球面角完成的，它们与纬度/经度角不同（wiki it https://en.wikipedia.org/wiki/Spherical_coordinate_system）。

于 2016-07-04T14:20:51.737 回答