本文结合Vincenty公式计算椭圆形地球模型目标点的方法
/**
* 已知一点经纬度及与另一点距离和航向,求另一点经纬度
*
* @param lon 已知一点的经度
* @param lat 已知一点的纬度
* @param brng 已知一点与另一点的方向 (角度)
* @param dist 已知一点与另一点的距离 :米
* @return
*/
def computerOffsetPosition(lon: Double,
lat: Double,
brng: Double,
dist: Double): (Double, Double) = {
//1. Calculate the destination point useing Vincenty's formula.
// 角度转弧度
val lat1 = lat * math.Pi / 180
val lon1 = lon * math.Pi / 180
val brg = brng * math.Pi / 180
// 扁率
val flat = 298.257223563
// 地球 半长轴
val a = 6378137.0
//地球 半短轴
val b = 6356752.314245
val f = 1 / flat
val sb = math.sin(brg)
val cb = math.cos(brg)
val tu1 = (1 - f) * math.tan(lat1)
val cu1 = 1 / math.sqrt((1 + tu1 * tu1))
val su1 = tu1 * cu1
val s2 = math.atan2(tu1, cb)
val sa = cu1 * sb
val csa = 1 - sa * sa
val us = csa * (a * a - b * b) / (b * b)
val A = 1 + us / 16384 * (4096 + us * (-768 + us * (320 - 175 * us)))
val B = us / 1024 * (256 + us * (-128 + us * (74 - 47 * us)))
var s1 = dist / (b * A)
var s1p = 2 * math.Pi
var cs1m = 0.0
var ss1 = 0.0
var cs1 = 0.0
var ds1 = 0.0
//
while (math.abs(s1 - s1p) > 1e-12) {
cs1m = math.cos(2 * s2 + s1)
ss1 = math.sin(s1)
cs1 = math.cos(s1)
ds1 = B * ss1 * (cs1m + B / 4 * (cs1 * (-1 + 2 * cs1m * cs1m) - B / 6 * cs1m * (-3 + 4 * ss1 * ss1) * (-3 + 4 * cs1m * cs1m)))
s1p = s1
s1 = dist / (b * A) + ds1
}
val t = su1 * ss1 - cu1 * cs1 * cb
val lat2 = math.atan2(su1 * cs1 + cu1 * ss1 * cb, (1 - f) * math.sqrt(sa * sa + t * t))
val l2 = math.atan2(ss1 * sb, cu1 * cs1 - su1 * ss1 * cb)
val c = f / 16 * csa * (4 + f * (4 - 3 * csa))
val l = l2 - (1 - c) * f * sa * (s1 + c * ss1 * (cs1m + c * cs1 * (-1 + 2 * cs1m * cs1m)))
val lon2 = lon1 + l
(lon2 * 180 / math.Pi, lat2 * 180 / math.Pi)
}
参考资料
http://www.movable-type.co.uk/scripts/latlong-vincenty.html
http://www.geomidpoint.com/destination/calculation.html
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