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点到线的距离大圆函数不工作。

  •  4
  • Brad Mathews  · 技术社区  · 16 年前

    我需要知道从一个lat/lng点到一条线的距离。当然需要跟随大圆。

    我在上找到了一篇关于这个的好文章 http://www.movable-type.co.uk/scripts/latlong.html

    但这一准则并不奏效。要么是我做错了什么,要么是我丢了什么。这里是有问题的函数。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;
        } 
    

    纬度/经度1=-94.127592,41.81762

    lat/lon2=-94.087257,41.848202

    纬度/经度3=-94.046875,41.791057

    报告显示0.865英里。实际距离为4.29905英里。

    有关于如何解决这个问题的线索吗?我不是数学家,只是个长牙程序员。

    3 回复  |  直到 16 年前
        1
  •  4
  •   Brian Webster Jason    14 年前

    大多数三角函数都需要弧度。你的角度度量单位是度吗?可能需要使用通常的公式转换它们:

    2*π弧度=360度

    如果你看哈弗斯林公式的公式,你会看到:

    (请注意,角度需要以弧度表示才能传递到trig函数)。

        2
  •  0
  •   Taryn Frank Pearson    13 年前

    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。第三个点总是从赤道向北0.1。只有经度的变化不应影响结果。似乎是这样。

        3
  •  0
  •   Tomsmith    10 年前

    我尝试了这个点线距离测试发送它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;
    }