计算地球上两个坐标之间的距离 [英] Calculate distance between two coordinates on a globe

问题描述

我得到两个坐标对，它们的形式为 90°0′0″ N 0°0′0″ E 作为字符串，并希望计算这些点之间的距离一个半径为R = 6371km的球体。

I get two coordinate pairs in the form 90°0′0″N 0°0′0″E as string and want to calculate the distance between those points on a sphere with radius R=6371km.

我在互联网上发现了两个公式此处， haversine和余弦的球形定律，但它们似乎无效。对于应返回 2 * pi * R / 4 的90°角，haversine操作正确，但余弦失败并返回0。具有更多随机坐标的其他点返回false

I found two formulas on the internet here, the "haversine" and the "spherical law of cosines", but they don't seem to work. For a 90° angle which should return 2*pi*R / 4, the haversine operates correct but the cosines fail and return 0. A different point with more random coordinates returns false values with both algorithms: the haversine is too high and the cosines are too low.

我的实现是错误的还是选择了错误的算法？

Is my implementation wrong or did I chose an incorrect algorithm?

我应该如何进行这些计算（坐标对与地球表面上的距离）？

How should I make these calculations (coordinate pairs to distance on globe surface) instead?

（是的，我知道我我还没有检查N / S和E / W，但测试的坐标都在东北半球。）

(And yes, I know that I'm not checking for N/S and E/W yet, but the tested coordinates are all in the north-eastern hemisphere.)

这是我的Python 3代码：

Here's my Python 3 code:

import math, re
R = 6371
PAT = r'(\d+)°(\d+)′(\d+)″([NSEW])'

def distance(first, second):
    def coords_to_rads(s):  
        return [math.radians(int(d) +int(m)/60 +int(s)/3600) \
                for d, m, s, nswe in re.findall(PAT, s)]

    y1, x1 = coords_to_rads(first)
    y2, x2 = coords_to_rads(second)  
    dx = x1 - x2  
    dy = y1 - y2  

    print("coord string:", first, "|", second)
    print("coord radians:", y1, x1, "|", y2, x2)
    print("x/y-distances:", dy, dx)

    a = math.sin(dx/2)**2 + math.cos(x1) * math.cos(x2) * math.sin(dy/2)**2  
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a))  
    haversine =  R * c  

    law_of_cosines = math.acos( math.sin(x1) * math.sin(x2) + \
                                math.cos(x1) * math.cos(x2) ) * R

    print("HS:", round(haversine, 2), "LOC:", round(law_of_cosines, 2))

    return haversine
    #return law_of_cosines

if __name__ == '__main__':
    def test(result, correct):
        print("result: ", result)
        print("correct:", correct)

    test(distance("90°0′0″N 0°0′0″E", "0°0′0″N, 0°0′0″E"), 10007.5)
    test(distance("51°28′48″N 0°0′0″E", "46°12′0″N, 6°9′0″E"), 739.2)
    test(distance("90°0′0″N 0°0′0″E", "90°0′0″S, 0°0′0″W"), 20015.1)
    test(distance("33°51′31″S, 151°12′51″E", "40°46′22″N 73°59′3″W"), 15990.2)

这是一些输出：

coord string: 90°0′0″N 0°0′0″E | 0°0′0″N, 0°0′0″E
coord radians: 1.5707963267948966 0.0 | 0.0 0.0
x/y-distances: 1.5707963267948966 0.0
HS: 10007.54 LOC: 0.0
result: 10007.543398010286
correct: 10007.5

coord string: 51°28′48″N 0°0′0″E | 46°12′0″N, 6°9′0″E
coord radians: 0.8984954989266809 0.0 | 0.8063421144213803 0.10733774899765128
x/y-distances: 0.09215338450530064 -0.10733774899765128
HS: 900.57 LOC: 683.85
result: 900.5669567853056
correct: 739.2

推荐答案

您似乎混淆了 x 和<$在计算 a 时得出c $ c> y 。您应该采用纬度（ y ）的余弦，而不是经度（ x ）的余弦。

It looks like you mixed up x and y in your calculation of a. You're supposed to take the cosine of latitude (y), not longitude (x).

我是通过将您的距离更改为 angular_distance 来发现这一点的（即， t乘以 R ）并添加一些其他测试：

I discovered this by changing your distance to angular_distance (i.e. don't multiply by R) and adding some additional tests:

test(angular_distance("90°0′0″N 0°0′0″E", "89°0′0″N, 0°0′0″E"), math.radians(1))
test(angular_distance("90°0′0″N 0°0′0″E", "80°0′0″N, 0°0′0″E"), math.radians(10))
test(angular_distance("90°0′0″N 0°0′0″E", "50°0′0″N, 0°0′0″E"), math.radians(40))
test(angular_distance("90°0′0″N 0°0′0″E", "50°0′0″N, 20°0′0″E"), math.radians(40))

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