Question

我需要从lat / lng点到线的距离。当然需要遵循Great Circle。

我在http://www.movable-type.co.uk/scripts/latlong.html

找到了一篇很棒的文章

但代码无效。我做错了什么或者有什么东西丢失了。这是有问题的功能。如果需要，请参阅其他功能的链接。

    var R = 3961.3
    LatLon.crossTrack = function(lat1, lon1, lat2, lon2, lat3, lon3) {
     var d13 = LatLon.distHaversine(lat1, lon1, lat3, lon3);
     var brng12 = LatLon.bearing(lat1, lon1, lat2, lon2);
     var brng13 = LatLon.bearing(lat1, lon1, lat3, lon3);
     var dXt = Math.asin(Math.sin(d13/R)*Math.sin(brng13-brng12)) * R;
     return dXt;
    }

lat / lon1 = -94.127592,41.81762

lat / lon2 = -94.087257,41.848202

lat / lon3 = -94.046875,41.791057

报告0.865英里。实际距离是4.29905英里。

有关如何解决此问题的任何线索？我不是数学家，只是牙齿程序员中的一员。

Answer 1

大多数trig函数需要弧度。你的角度测量是度数吗？也许他们需要使用通常的公式进行转换：

2 *π弧度= 360度

如果你看一下Haversine公式的公式，你会看到：

（注意角度需要以弧度表示才能传递给trig函数）。

Answer 2

您的函数是否为这些坐标返回相同的值：

crossTrack(0,0,0,1,0.1,0.5);
crossTrack(0,0,0,1,0.1,0.6);
crossTrack(0,0,0,1,0.1,0.4);

我认为应该但我的不应该。第3点始终从赤道向北0.1。只有经度变化才不会影响结果。看起来确实如此。

Answer 3

我尝试了这个pointlinedistancetest发送aalatlon等

private static final double _eQuatorialEarthRadius = 6378.1370D;
private static final double _d2r = (Math.PI / 180D);
private static double PRECISION = 1;





// Haversine Algorithm
// source: http://stackoverflow.com/questions/365826/calculate-distance-between-2-gps-coordinates

private static double HaversineInM(double lat1, double long1, double lat2, double long2) {
    return  (1000D * HaversineInKM(lat1, long1, lat2, long2));



}

private static double HaversineInKM(double lat1, double long1, double lat2, double long2) {
    double dlong = (long2 - long1) * _d2r;
    double dlat = (lat2 - lat1) * _d2r;
    double a = Math.pow(Math.sin(dlat / 2D), 2D) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r)
            * Math.pow(Math.sin(dlong / 2D), 2D);
    double c = 2D * Math.atan2(Math.sqrt(a), Math.sqrt(1D - a));
    double d = _eQuatorialEarthRadius * c;





    return d;
}

// Distance between a point and a line

public static double pointLineDistanceTest(double[] aalatlng,double[] bblatlng,double[]cclatlng){



    double [] a = aalatlng;
    double [] b = bblatlng;
    double [] c = cclatlng;




    double[] nearestNode = nearestPointGreatCircle(a, b, c);
    //        System.out.println("nearest node: " + Double.toString(nearestNode[0])
    + ","+Double.toString(nearestNode[1]));
    double result =  HaversineInM(c[0], c[1], nearestNode[0], nearestNode[1]);

       //        System.out.println("result: " + Double.toString(result));



          return (result);






}

// source: http://stackoverflow.com/questions/1299567/how-to-calculate-distance-from-a-point-to-a-line-segment-on-a-sphere
private static double[] nearestPointGreatCircle(double[] a, double[] b, double c[])
{
    double[] a_ = toCartsian(a);
    double[] b_ = toCartsian(b);
    double[] c_ = toCartsian(c);

    double[] G = vectorProduct(a_, b_);
    double[] F = vectorProduct(c_, G);
    double[] t = vectorProduct(G, F);

    return fromCartsian(multiplyByScalar(normalize(t), _eQuatorialEarthRadius));
}

@SuppressWarnings("unused")
private static double[] nearestPointSegment (double[] a, double[] b, double[] c)
{
   double[] t= nearestPointGreatCircle(a,b,c);
   if (onSegment(a,b,t))
     return t;

   return (HaversineInKM(a[0], a[1], c[0], c[1]) < HaversineInKM(b[0], b[1], c[0], c[1])) ? a : b;
}

 private static boolean onSegment (double[] a, double[] b, double[] t)
   {
     // should be   return distance(a,t)+distance(b,t)==distance(a,b), 
     // but due to rounding errors, we use: 
     return Math.abs(HaversineInKM(a[0], a[1], b[0], b[1])-HaversineInKM(a[0], a[1], t[0], t[1])-HaversineInKM(b[0], b[1], t[0], t[1])) < PRECISION;
   }


// source: http://stackoverflow.com/questions/1185408/converting-from-longitude-latitude-to-cartesian-coordinates
private static double[] toCartsian(double[] coord) {
    double[] result = new double[3];
    result[0] = _eQuatorialEarthRadius * Math.cos(Math.toRadians(coord[0])) * Math.cos(Math.toRadians(coord[1]));
    result[1] = _eQuatorialEarthRadius * Math.cos(Math.toRadians(coord[0])) * Math.sin(Math.toRadians(coord[1]));
    result[2] = _eQuatorialEarthRadius * Math.sin(Math.toRadians(coord[0]));


    return result;
}

private static double[] fromCartsian(double[] coord){
    double[] result = new double[2];
    result[0] = Math.toDegrees(Math.asin(coord[2] / _eQuatorialEarthRadius));
    result[1] = Math.toDegrees(Math.atan2(coord[1], coord[0]));

    return result;
}


// Basic functions
private static double[] vectorProduct (double[] a, double[] b){
    double[] result = new double[3];
    result[0] = a[1] * b[2] - a[2] * b[1];
    result[1] = a[2] * b[0] - a[0] * b[2];
    result[2] = a[0] * b[1] - a[1] * b[0];

    return result;
}

private static double[] normalize(double[] t) {
    double length = Math.sqrt((t[0] * t[0]) + (t[1] * t[1]) + (t[2] * t[2]));
    double[] result = new double[3];
    result[0] = t[0]/length;
    result[1] = t[1]/length;
    result[2] = t[2]/length;
    return result;
}

private static double[] multiplyByScalar(double[] normalize, double k) {
    double[] result = new double[3];
    result[0] = normalize[0]*k;
    result[1] = normalize[1]*k;
    result[2] = normalize[2]*k;
    return result;
}

点到线的距离大圆功能无法正常工作。

3 个答案: