python基于经纬度距离汇聚点_python实现两个经纬度点之间的距离和方位角

from：/zhuqiuhui/article/details/53180395

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