如何检查经度/纬度点是否在坐标范围内？

Question

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？

Answer 1

这本质上是球体上的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。

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. 选择一个不在多边形区域内的参考点。
3. 绘制一个从参考点开始到车辆点结束的大圆段。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接近，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感到不满意，这是最好的选择，但是由于圆弧会变形，因此您会在多边形的边界周围产生一些变形。

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，normal vector）

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

class GreatCircle{
    Vector normal;
}

任何不是antipodal 的两个纬度/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cross product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;
}

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

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段 a 是否与大圆 b 相交：

• 找到向量c，a的整个大圆与b的交点。
• 找到向量d，即c的对映点。
• 如果c介于a.start和a.end之间，或者d介于a.start和a.end之间，则a与b相交。

 

//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);
}

两个大圆段a 和b 相交，如果：

• 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;
}