您的位置：首页 > 编程语言 > Python开发

python实现两个经纬度点之间的距离和方位角的方法

2019-07-05 14:51 2036 查看

最近做有关GPS轨迹上有关的东西，花费心思较多，对两个常用的函数总结一下，求距离和求方位角，比较精确，欢迎交流！

1. 求两个经纬点的方位角，P0(latA, lonA), P1(latB, lonB)（很多博客写的不是很好，这里总结一下）

def getDegree(latA, lonA, latB, lonB):
"""
Args:
point p1(latA, lonA)
point p2(latB, lonB)
Returns:
bearing between the two GPS points,
default: the basis of heading direction is north
"""
radLatA = radians(latA)
radLonA = radians(lonA)
radLatB = radians(latB)
radLonB = radians(lonB)
dLon = radLonB - radLonA
y = sin(dLon) * cos(radLatB)
x = cos(radLatA) * sin(radLatB) - sin(radLatA) * cos(radLatB) * cos(dLon)
brng = degrees(atan2(y, x))
brng = (brng + 360) % 360
return brng

2. 求两个经纬点的距离函数：P0(latA, lonA), P1(latB, lonB)

def getDistance(latA, lonA, latB, lonB):
ra = 6378140 # radius of equator: meter
rb = 6356755 # radius of polar: meter
flatten = (ra - rb) / ra # Partial rate of the earth
# change angle to radians
radLatA = radians(latA)
radLonA = radians(lonA)
radLatB = radians(latB)
radLonB = radians(lonB)

pA = atan(rb / ra * tan(radLatA))
pB = atan(rb / ra * tan(radLatB))
x = acos(sin(pA) * sin(pB) + cos(pA) * cos(pB) * cos(radLonA - radLonB))
c1 = (sin(x) - x) * (sin(pA) + sin(pB))**2 / cos(x / 2)**2
c2 = (sin(x) + x) * (sin(pA) - sin(pB))**2 / sin(x / 2)**2
dr = flatten / 8 * (c1 - c2)
distance = ra * (x + dr)
return distance

以上这篇python实现两个经纬度点之间的距离和方位角的方法就是小编分享给大家的全部内容了，希望能给大家一个参考

您可能感兴趣的文章:

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签： python 经纬度距离方位角

相关文章推荐

新的分享

章节导航