如何检查经度/纬度点是否在坐标范围内

这本质上是球体上多边形中的点问题。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照点
3. 绘制一个从参考点开始到车辆点结束的大圆段。计算该线段与多边形的某个线段相交的次数。如果总次数为奇数，则表示车辆在多边形内。如果是偶数，则表示车辆在多边形之外

或者，如果坐标和车辆距离足够近，并且不在极点或国际日期线附近，你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，就可以将光线投射算法用于简单的线段。如果您对非欧几里德几何体不满意，但由于圆弧会失真，因此多边形边界周围会有一些失真，则此选项更可取。

编辑：关于球体上的几何体的更多信息。

大圆可以通过垂直于圆所在平面的矢量（AKA，法向量）来识别

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

任何两个纬度/经度坐标都不是对足的，它们正好共享一个大圆。要找到这个大圆，请将坐标转换为穿过地球中心的直线。这两条线的叉积就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点相交。圆的叉积形成一个通过其中一个点的矢量。该矢量的反极穿过另一点。

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;
}

大圆线段可以用通过线段起点和终点的矢量来表示。

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段的圆弧大小或任意两个矢量之间的角度。

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：

• 求出向量c，即a的整个大圆与b的交点
• 找到矢量d，即c的反极
• 如果c位于a.start和a.end之间，或者d位于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的整个大圆相交
• b与a的整个大圆相交

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

现在你可以构建你的多边形，并计算你的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;
}