python实现两个经纬度点之间的距离和方位角的方法


						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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