如何检查经度/纬度点是否在坐标范围内？

Question

我有许多经度和纬度坐标组成一个多边形区域.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

这实质上是球体上的多边形点问题.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对相邻坐标,在它们之间绘制一个大的圆弧段.
2. 选择不在多边形区域内的参考点.
3. 绘制一个从参考点开始并在车辆终点结束的大圆段.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,则车辆在多边形之外.

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.如果您对非欧几里德几何体感到不舒服,这是更可取的,但是由于弧会扭曲,因此您的多边形边界周围会有一些扭曲.

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,法向量)

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

class GreatCircle{
    Vector normal;
}

任何两个非对映的弧度/经度坐标都只分享一个大圆.要找到这个大圆,请将坐标转换为穿过地球中心的线.这两条线的叉积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.该向量的对映体通过另一个点.

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整个大圆的交集b.
• 找到载体d,对映体c.
• 如果c位于a.start和之间a.end,或d位于a.start和之间a.end,则a与之相交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);
}

如果出现两个大圆段a并b相交:

• a与b整个大圆相交
• b与a整个大圆相交

 

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

现在,您可以构造多边形并计算参考线在其上经过的次数.

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;
}