最近做有關GPS軌跡上有關的東西，花費心思較多，對兩個常用的函數總結一下，求距離和求方位角，比較精確，歡迎交流！

1. 求兩個經緯點的方位角，P0(latA, lonA), P1(latB, lonB)(很多博客寫的不是很好，這里總結一下)

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. 求兩個經緯點的距離函數：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

以上這篇python實現兩個經緯度點之間的距離和方位角的方法就是小編分享給大家的全部內容了，希望能給大家一個參考，也希望大家多多支持腳本之家。

總結

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