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我有许多构成多边形区域的经度和纬度坐标。我还有一个经度和纬度坐标来定义车辆的位置。如何检查车辆是否位于多边形区域内？

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1 回答 1

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这本质上是球体上的多边形中的点问题。您可以修改光线投射算法，使其使用大圆弧而不是线段。

  1. 对于构成多边形的每一对相邻坐标，在它们之间画一个大圆段。
  2. 选择不在多边形区域内的参考点。
  3. 画一个从参考点开始到车辆点结束的大圆。计算该线段跨越多边形线段的次数。如果总次数是奇数，则车辆在多边形内。如果是偶数，则车辆在多边形之外。

或者，如果坐标和车辆足够靠近，并且不在两极或国际日期变更线附近，您可以假装地球是平的，并使用经度和纬度作为简单的 x 和 y 坐标。这样，您可以将光线投射算法与简单的线段一起使用。如果您对非欧几里得几何体不满意，这是最好的选择，但是由于弧线会变形，因此您的多边形边界周围会有一些变形。

编辑：更多关于球体上的几何。

一个大圆可以通过垂直于圆所在平面的向量来识别（AKA，法线向量）

class Vector{
    double x;
    double y;
    double z;
};

class GreatCircle{
    Vector normal;
}

任何两个不是对映的纬度/经度坐标都共享一个大圆。要找到这个大圆，请将坐标转换为穿过地球中心的线。这两条线的叉积是坐标大圆的法向量。

//arbitrarily defining the north pole as (0,1,0) and (0'N, 0'E) as (1,0,0)
//lattidues should be in [-90, 90] and longitudes in [-180, 180]
//You'll have to convert South lattitudes and East longitudes into their negative North and West counterparts.
Vector lineFromCoordinate(Coordinate c){
    Vector ret = new Vector();
    //given:
    //tan(lat) == y/x
    //tan(long) == z/x
    //the Vector has magnitude 1, so sqrt(x^2 + y^2 + z^2) == 1
    //rearrange some symbols, solving for x first...
    ret.x = 1.0 / math.sqrt(tan(c.lattitude)^2 + tan(c.longitude)^2 + 1);
    //then for y and z
    ret.y = ret.x * tan(c.lattitude);
    ret.z = ret.x * tan(c.longitude);
    return ret;
}

Vector Vector::CrossProduct(Vector other){
    Vector ret = new Vector();
    ret.x = this.y * other.z - this.z * other.y;
    ret.y = this.z * other.x - this.x * other.z;
    ret.z = this.x * other.y - this.y * other.x;
    return ret;
}

GreatCircle circleFromCoordinates(Coordinate a, Coordinate b){
    Vector a = lineFromCoordinate(a);
    Vector b = lineFromCoordinate(b);
    GreatCircle ret = new GreatCircle();
    ret.normal = a.CrossProdct(b);
    return ret;
}

两个大圆在球面上的两点相交。圆圈的叉积形成一个通过这些点之一的向量。该向量的对映点通过另一点。

Vector intersection(GreatCircle a, GreatCircle b){
    return a.normal.CrossProduct(b.normal);
}

Vector antipode(Vector v){
    Vector ret = new Vector();
    ret.x = -v.x;
    ret.y = -v.y;
    ret.z = -v.z;
    return ret;
}

一个大圆段可以用穿过该段起点和终点的向量来表示。

class GreatCircleSegment{
    Vector start;
    Vector end;
    Vector getNormal(){return start.CrossProduct(end);}
    GreatCircle getWhole(){return new GreatCircle(this.getNormal());}
};

GreatCircleSegment segmentFromCoordinates(Coordinate a, Coordinate b){
    GreatCircleSegment ret = new GreatCircleSegment();
    ret.start = lineFromCoordinate(a);
    ret.end = lineFromCoordinate(b);
    return ret;
}

您可以使用点积来测量大圆弧段的圆弧大小或任意两个向量之间的角度。

double Vector::DotProduct(Vector other){
    return this.x*other.x + this.y*other.y + this.z*other.z;
}

double Vector::Magnitude(){
    return math.sqrt(pow(this.x, 2) + pow(this.y, 2) + pow(this.z, 2));
}

//for any two vectors `a` and `b`, 
//a.DotProduct(b) = a.magnitude() * b.magnitude() * cos(theta)
//where theta is the angle between them.
double angleBetween(Vector a, Vector b){
    return math.arccos(a.DotProduct(b) / (a.Magnitude() * b.Magnitude()));
}

您可以通过以下方式测试大圆段是否与大圆a相交b：

  • 找到向量c，a的整个大圆的交点 和b。
  • 找到向量d， 的对映点c。
  • 如果在andc之间，或在and之间，则与 相交。a.starta.endda.starta.endab

 

//returns true if Vector x lies between Vectors a and b.
//note that this function only gives sensical results if the three vectors are coplanar.
boolean liesBetween(Vector x, Vector a, Vector b){
    return angleBetween(a,x) + angleBetween(x,b) == angleBetween(a,b);
}

bool GreatCircleSegment::Intersects(GreatCircle b){
    Vector c = intersection(this.getWhole(), b);
    Vector d = antipode(c);
    return liesBetween(c, this.start, this.end) or liesBetween(d, this.start, this.end);
}

a如果满足以下条件，则两个大圆线段b相交：

  • ab与整个大圆相交
  • ba与整个大圆相交

 

bool GreatCircleSegment::Intersects(GreatCircleSegment b){
    return this.Intersects(b.getWhole()) and b.Intersects(this.getWhole());
}

现在您可以构建多边形并计算参考线经过它的次数。

bool liesWithin(Array<Coordinate> polygon, Coordinate pointNotLyingInsidePolygon, Coordinate vehiclePosition){
    GreatCircleSegment referenceLine = segmentFromCoordinates(pointNotLyingInsidePolygon, vehiclePosition);
    int intersections = 0;
    //iterate through all adjacent polygon vertex pairs
    //we iterate i one farther than the size of the array, because we need to test the segment formed by the first and last coordinates in the array
    for(int i = 0; i < polygon.size + 1; i++){
        int j = (i+1) % polygon.size;
        GreatCircleSegment polygonEdge = segmentFromCoordinates(polygon[i], polygon[j]);
        if (referenceLine.Intersects(polygonEdge)){
            intersections++;
        }
    }
    return intersections % 2 == 1;
}
于 2012-07-16T18:51:52.000 回答