Python 距离（以英里为单位）到两个 gps 坐标之间的欧氏距离

Question

我正在尝试想出一个函数...
输入：以英里或公里为单位的测地距离
输出：输入距离相隔的任意两个 gps 点之间的欧氏距离

我觉得我有一些组件

import numpy as np
from numpy import linalg as LA
from geopy.distance import geodesic

loc1 = np.array([40.099993, -83.166000])
loc2 = np.array([40.148652, -82.903962])

这是这两点之间的欧氏距离

LA.norm(loc1-loc2)
#0.2665175636332336

这是这两点之间的测地距离（以英里为单位）

geodesic(loc1,loc2).miles
#14.27909749425243

我的脑子运行现在没电了，任何人都对我如何制作这样的函数有任何想法：

geodesic_to_euclidean(14.27909749425243)
#0.2665175636332336

Answer 1

如果您同意 great-circle 距离，如评论中所述，那么这应该可行。这是haversine距离：

def haversine(origin, destination, units='mi'):
    # Radian deltas
    origin_lat = radians(float(origin[0]))
    origin_lon = radians(float(origin[1]))
    destination_lat = radians(float(destination[0]))
    destination_lon = radians(float(destination[1]))
    lat_delta = destination_lat - origin_lat
    lon_delta = destination_lon - origin_lon

    # Radius of earth in meters
    r = 6378127

    # Haversine formula
    a = sin(lat_delta / 2) ** 2 + cos(origin_lat) * \
        cos(destination_lat) * sin(lon_delta / 2) ** 2
    c = 2 * asin(sqrt(a))
    meters_traveled = c * r

    scaling_factors = {
        "m:": 1,
        "km": 1 / 1000,
        "ft": 3.2808,  # meters to feet
        "mi:": 0.000621371  # meters to miles
    }

    return meters_traveled * scaling_factors[units]

如果您已经有了以米为单位的测地线（大圆）距离并且想要弦长，那么您可以执行以下操作

def chord(geodesic_distance):
    """
    Chord length
    C = 2 * r * sin(theta/2)

    Arc length; which is geodesic distance in this case
    AL = R * theta

    therefore
    C = 2 * R * sin(AL/(2*R))
    """
    r = 6378127  # Radius of earth in meters

    return 2 * r * sin(geodesic_distance / (2 * r))

Python 距离（以英里为单位）到两个 gps 坐标之间的欧氏距离

Python distance in miles to euclidean distance between two gps coordinates

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