问题
我想知道如何获得 2 个 GPS 点之间的距离和方位。我研究了haversine公式。有人告诉我,我也可以使用相同的数据找到方位。
编辑
一切正常,但轴承还不能正常工作。轴承输出负值,但应在 0 - 360 度之间。设置的数据应该使水平方位 96.02166666666666
并且是:
Start point: 53.32055555555556 , -1.7297222222222221
Bearing: 96.02166666666666
Distance: 2 km
Destination point: 53.31861111111111, -1.6997222222222223
Final bearing: 96.04555555555555
这是我的新代码:
from math import *
Aaltitude = 2000
Oppsite = 20000
lat1 = 53.32055555555556
lat2 = 53.31861111111111
lon1 = -1.7297222222222221
lon2 = -1.6997222222222223
lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * atan2(sqrt(a), sqrt(1-a))
Base = 6371 * c
Bearing =atan2(cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1), sin(lon2-lon1)*cos(lat2))
Bearing = degrees(Bearing)
print ""
print ""
print "--------------------"
print "Horizontal Distance:"
print Base
print "--------------------"
print "Bearing:"
print Bearing
print "--------------------"
Base2 = Base * 1000
distance = Base * 2 + Oppsite * 2 / 2
Caltitude = Oppsite - Aaltitude
a = Oppsite/Base
b = atan(a)
c = degrees(b)
distance = distance / 1000
print "The degree of vertical angle is:"
print c
print "--------------------"
print "The distance between the Balloon GPS and the Antenna GPS is:"
print distance
print "--------------------"
atan2(sqrt(a), sqrt(1-a))
与 asin(sqrt(a))
相同
这是一个 Python 版本:
from math import radians, cos, sin, asin, sqrt
def haversine(lon1, lat1, lon2, lat2):
"""
Calculate the great circle distance in kilometers between two points
on the earth (specified in decimal degrees)
"""
# convert decimal degrees to radians
lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
r = 6371 # Radius of earth in kilometers. Use 3956 for miles. Determines return value units.
return c * r
这些答案中的大多数都是“四舍五入”地球的半径。如果您将这些与其他距离计算器(例如 geopy)进行检查,这些功能将被关闭。
这很好用:
from math import radians, cos, sin, asin, sqrt
def haversine(lat1, lon1, lat2, lon2):
R = 3959.87433 # this is in miles. For Earth radius in kilometers use 6372.8 km
dLat = radians(lat2 - lat1)
dLon = radians(lon2 - lon1)
lat1 = radians(lat1)
lat2 = radians(lat2)
a = sin(dLat/2)**2 + cos(lat1)*cos(lat2)*sin(dLon/2)**2
c = 2*asin(sqrt(a))
return R * c
# Usage
lon1 = -103.548851
lat1 = 32.0004311
lon2 = -103.6041946
lat2 = 33.374939
print(haversine(lat1, lon1, lat2, lon2))
还有一个矢量化实现,它允许使用 4 个 numpy 数组而不是坐标的标量值:
def distance(s_lat, s_lng, e_lat, e_lng):
# approximate radius of earth in km
R = 6373.0
s_lat = s_lat*np.pi/180.0
s_lng = np.deg2rad(s_lng)
e_lat = np.deg2rad(e_lat)
e_lng = np.deg2rad(e_lng)
d = np.sin((e_lat - s_lat)/2)**2 + np.cos(s_lat)*np.cos(e_lat) * np.sin((e_lng - s_lng)/2)**2
return 2 * R * np.arcsin(np.sqrt(d))
您可以尝试使用 hasrsine 包:https://pypi.org/project/haversine/
示例代码:
from haversine import haversine
haversine((45.7597, 4.8422),(48.8567, 2.3508), unit='mi')
243.71209416020253
方位计算不正确,您需要将输入交换到 atan2。
bearing = atan2(sin(long2-long1)*cos(lat2), cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(long2-long1))
bearing = degrees(bearing)
bearing = (bearing + 360) % 360
这将为您提供正确的方位。
haversine formula
我第一次听到这个,谢谢。
这是@Michael Dunn 给出的Haversine 公式的一个numpy 向量化实现,比大向量提高了10-50 倍。
from numpy import radians, cos, sin, arcsin, sqrt
def haversine(lon1, lat1, lon2, lat2):
"""
Calculate the great circle distance between two points
on the earth (specified in decimal degrees)
"""
#Convert decimal degrees to Radians:
lon1 = np.radians(lon1.values)
lat1 = np.radians(lat1.values)
lon2 = np.radians(lon2.values)
lat2 = np.radians(lat2.values)
#Implementing Haversine Formula:
dlon = np.subtract(lon2, lon1)
dlat = np.subtract(lat2, lat1)
a = np.add(np.power(np.sin(np.divide(dlat, 2)), 2),
np.multiply(np.cos(lat1),
np.multiply(np.cos(lat2),
np.power(np.sin(np.divide(dlon, 2)), 2))))
c = np.multiply(2, np.arcsin(np.sqrt(a)))
r = 6371
return c*r
您可以通过添加 360° 来解决负方位问题。不幸的是,这可能导致正向轴承的轴承大于 360°。这是模运算符的一个很好的候选者,所以总而言之,您应该添加该行
Bearing = (Bearing + 360) % 360
在你的方法结束时。
默认情况下,atan2 中的 Y 是第一个参数。这是documentation。您需要切换输入以获得正确的方位角。
bearing = atan2(sin(lon2-lon1)*cos(lat2), cos(lat1)*sin(lat2)in(lat1)*cos(lat2)*cos(lon2-lon1))
bearing = degrees(bearing)
bearing = (bearing + 360) % 360
这实际上提供了两种获取距离的方法。他们是Haversine 和Vincentys。根据我的研究,我知道 Vincentys 是相对准确的。也可以使用 import 语句来实现。
这里有两个计算距离和方位的函数,它们基于之前消息中的代码和 https://gist.github.com/jeromer/2005586(为清楚起见,为两个函数添加了 lat、lon 格式的地理点的元组类型)。我测试了这两个功能,它们似乎工作正常。
#coding:UTF-8
from math import radians, cos, sin, asin, sqrt, atan2, degrees
def haversine(pointA, pointB):
if (type(pointA) != tuple) or (type(pointB) != tuple):
raise TypeError("Only tuples are supported as arguments")
lat1 = pointA[0]
lon1 = pointA[1]
lat2 = pointB[0]
lon2 = pointB[1]
# convert decimal degrees to radians
lat1, lon1, lat2, lon2 = map(radians, [lat1, lon1, lat2, lon2])
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
r = 6371 # Radius of earth in kilometers. Use 3956 for miles
return c * r
def initial_bearing(pointA, pointB):
if (type(pointA) != tuple) or (type(pointB) != tuple):
raise TypeError("Only tuples are supported as arguments")
lat1 = radians(pointA[0])
lat2 = radians(pointB[0])
diffLong = radians(pointB[1] - pointA[1])
x = sin(diffLong) * cos(lat2)
y = cos(lat1) * sin(lat2) - (sin(lat1)
* cos(lat2) * cos(diffLong))
initial_bearing = atan2(x, y)
# Now we have the initial bearing but math.atan2 return values
# from -180° to + 180° which is not what we want for a compass bearing
# The solution is to normalize the initial bearing as shown below
initial_bearing = degrees(initial_bearing)
compass_bearing = (initial_bearing + 360) % 360
return compass_bearing
pA = (46.2038,6.1530)
pB = (46.449, 30.690)
print haversine(pA, pB)
print initial_bearing(pA, pB)
考虑到您的目标是测量两点之间的距离(由地理坐标表示),将留下以下三个选项:
使用 GeoPy 测地线距离的 Haversine 公式 使用 GeoPy 大圆距离
选项1
Haversine 公式将完成这项工作,但重要的是要注意,这样做会将地球近似为球体,并且存在错误(see this answer) - 因为地球不是球体。
为了使用Haversine公式,首先需要定义地球的半径。这本身可能会引起一些争议。考虑以下三个来源
美国宇航局戈达德太空飞行中心:6371公里
维基百科:6371 公里(3958.8 英里)
谷歌 - 6371 公里
我将使用值 6371
km 作为地球半径的参考。
# Radius of the Earth
r = 6371.0
我们将利用 math
模块。
在半径之后,一个移动到坐标,一个开始将坐标转换为弧度,以便使用 math's trigonometric functions。为此导入 math.radians(x)
并按如下方式使用它们
#Import radians from math module
from math import radians
# Latitude and Longitude for the First Point (let's consider 40.000º and 21.000º)
lat1 = radians(40.000)
lon1 = radians(21.000)
# Latitude and Longitude for the Second Point (let's consider 30.000º and 25.000º)
lat2 = radians(30.000)
lon2 = radians(25.000)
现在准备应用Haversine Formula。第一个将点 1 的经度减去点 2 的经度
dlon = lon2 - lon1
dlat = lat2 - lat1
然后,这里有几个三角函数要使用,更具体地说,math.sin()
、math.cos()
和 math.atan2()
。我们还将使用 math.sqrt()
# Import sin, cos, atan2, and sqrt from math module
from math import sin, cos, atan2, sqrt
a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))
d = r * c
然后通过打印 d
获得距离。
可能会有所帮助,让我们将所有内容收集到一个函数中(受 @Michael Dunn's answer 启发)
from math import radians, cos, sin, atan2, sqrt
def haversine(lon1, lat1, lon2, lat2):
"""
Calculate the great-circle distance (in km) between two points
using their longitude and latitude (in degrees).
"""
# Radius of the Earth
r = 6371.0
# Convert degrees to radians
# First point
lat1 = radians(lat1)
lon1 = radians(lon1)
# Second Point
lat2 = radians(lat2)
lon2 = radians(lon2)
# Haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
c = 2 * atan2(sqrt(a), sqrt(1 - a))
return r * c
选项 2
我们将使用 GeoPy's distance,更具体地说,是 geodesic
。
我们可以在公里或英里 (Source) 上获得结果
# Import Geopy's distance
from geopy import distance
wellington = (-41.32, 174.81)
salamanca = (40.96, -5.50)
print(distance.distance(wellington, salamanca).km) # If one wants in miles, change `km` to `miles`
[Out]: 19959.6792674
选项 3
我们将使用 GeoPy's distance,更具体地说,是 great-circle
。
我们可以在公里或英里 (Source) 上获得结果
# Import Geopy's distance
from geopy import distance
newport_ri = (41.49008, -71.312796)
cleveland_oh = (41.499498, -81.695391)
print(distance.great_circle(newport_ri, cleveland_oh).miles) # If one wants in km, change `miles` to `km`
[Out]: 536.997990696
import math
,则必须指定math.pi
、math.sin
等。使用from math import *
,您可以直接访问所有模块内容。查看 Python 教程中的“命名空间”(例如 docs.python.org/tutorial/modules.html)