Question

我有一些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？

Answer 1

这实际上是球体上的Point in polygon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。

对于构成多边形的每对相邻坐标，在它们之间绘制一个很大的圆弧段。
选择不在多边形区域内的参考点。
绘制一个大圆段，从参考点开始到车辆终点。计算此段跨越多边形段的次数。如果总次数是奇数，则车辆在多边形内。如果均匀，车辆在多边形之外。

或者，如果坐标和车辆足够靠近，而不是靠近极点或国际日期线，您可以假装地球是平坦的，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。如果你对非欧几里德几何不熟悉，这是更好的选择，但是你的多边形边界会有一些扭曲，因为弧会扭曲。

编辑：关于球体几何的更多信息。

可以通过垂直于圆所在平面的矢量来识别大圆（AKA，normal vector）

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

class GreatCircle{
    Vector normal;
}

任何两个不是antipodal的纬度/经度坐标只分享一个大圆。要找到这个大圆，请将坐标转换为穿过地球中心的线。这两行的cross product是坐标大圆的法线向量。

//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;
}

两个大圆在球体上的两个点相交。圆的叉积形成通过这些点之一的矢量。该向量的对映经过了另一个点。

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;
}

大圆段可以通过段的起点和终点来表示。

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dot product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圈a是否与大圆b相交：

找到向量c，a的整个大圆与b的交集。
找到向量d，对象c。
如果c位于a.start和a.end之间，或d位于a.start和a.end之间，则a相交与b。

//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);
}

如果出现以下两个大圆圈a和b相交：

a与b的整个大圈相交
b与a的整个大圈相交

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;
}

如何检查经度/纬度点是否在坐标范围内？

1 个答案: