如何检查，如果经/纬度点是坐标范围之内？(How do I check if a longitude

我公司拥有一批经度和纬度坐标，使一个多边形区域。我也有一个经度和纬度坐标来定义车辆的位置。如何检查车辆位于多边形区域内？

这实质上是在多边形点上的球的问题。所以它使用大圆，而不是线段的弧可以修改光线投射算法。

对于每一对，使您的多边形相邻的坐标，画出它们之间的大圆段。
选择一个参照点，是不是多边形区内。
绘制开始于参考点和在车辆点结束的大圆线段。算上这部分有多少次跨越的多边形的部分。如果总次数为奇数，则车辆处于多边形内。如果连，车辆是多边形的外面。

另外，如果坐标和车辆的距离足够近，而不是极或国际日期变更线附近的任何地方，你可以假装地球是平的，并使用经度和lattitude简单的X和Y坐标。这样一来，你可以使用简单的线段的光线投射算法。这是最好，如果你是与非欧几里得几何不舒服，但你必须在你的多边形的边界有些失真，因为弧线将被扭曲。

编辑：更小几何球体上。

大圆可以由位于垂直于圆位于（AKA，所述平面中的矢量来识别正常矢量）

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

class GreatCircle{
    Vector normal;
}

任何两个lattitude /经度坐标不在对映份额只有一个大圆。为了找到这个大圆，转换的坐标，通过地球的中心线上。所述叉积这两行的是坐标的大圆的法线矢量。

//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;
}

两个伟大的圆相交于球体上的两个点。圆的横产物形成，通过这些点中的一个传递的载体。该矢量的对映体通过另一点。

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;
}

大圆段可通过穿过分段的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;
}

可以测量大圆段的弧形尺寸，或者任何两个矢量之间的角度，使用点积。

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()));
}

如果一个大圈段可以测试a相交的大圆b方式：

找到矢量c ，的交点a的整个大圆和b 。
找到矢量d ，的对映体c 。
如果c介于a.start和a.end ，或d介于a.start和a.end ，然后a与相交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);
}

两个大圆段a和b相交，如果：

a与相交b的整个大圆
b与相交a的整个大圆

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

现在，您可以构建您的多边形数一数你参考线多少次越过它。

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;
}