python实现两个经纬度点之间的距离和方位角的方法
作者:bboyzqh
今天小编就为大家分享一篇python实现两个经纬度点之间的距离和方位角的方法,具有很好的参考价值,希望对大家有所帮助。一起跟随小编过来看看吧
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
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