from：http://blog.csdn.net/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